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0};7 observe that li,n ^-- (Bn;
n=(2),p>2, IIp:=(ZED.I(pxp,C):Z'=-Z, IIp-ZZ*>0}; n=(PZl),p2 1,IIIp:=(ZE DvI(pxp,C):Z' = Z, IIp-ZZ*>0); IV" := 1Ln.
The domains of types 1-117 are called classical. They are balanced - cf. Definition 1.4.14. n = 16, an exceptional domain V16; n = 27, an exceptional domain V127. 'For A E M(m x m,C), "A > 0" means that A is positive definite. i.e. X'AX = m E'P t ai.k X; Xk > 0 for all X 6 (C"')*.
Chapter 2. Biholomorphisms of Reinhardt domains
178
The above Cartan domains are not biholomorphically equivalent except for the following cases: bih
(a) 12,2 '=' La,
(b) 112
ID, 113 Lh (B3, 114
(c) III1 ^_- I), II12
bib
bib
L6; thus type 11 is essential only for p ? 5,
L3; thus type III is essential only for p > 2,
bih
(d) L t = 9), L2 '_- ID2; thus type IV is essential only for n > 5.
Let 1/i(n) denote the number of biholomorphically non-equivalent canonical bounded symmetric domains G C C" (from the above list) and let W(n) be the number of non-equivalent bounded symmetric domains G C C" (which are biholomorphic to Cartesian products of canonical symmetric domains). The following table describes the situation for I < n < 30.
n
I
*(n) 1 1 1 11)
1
1
2
1
3
2
4
2
5
2
6
4
136,12.3
7
2
8
9
II
(112D) (1111=D)
182 183
N
111
(113.183)
1112
1 104,12.2
Y/VI
SY(n)
(LI=D)
1
(L2-D2)
2
(L3
1112)
4
(L4
12.2)
7
L5
11
L6
21
187
L7
31
3
188,12,4
L8
51
3
189, 13.3
L9
80
10
5
181o, 12,5
Llo
126
11
2
B11
L11
187
12
4
1812, 12.6, 13.4
L12
292
13
2
1813
L 13
427
14
3
1 1 1814,12.7
L14
638
15
5
615,13.5
16
5
816,12.8,14.4
L16
17
2
18j7
L 17
1960
18
4
818,12.9, 13.6
LI8
2843
191 1
2
B19
L19
4024
20
4
1820,12.10,14.5
L20
5724
21
5
1821,13,7
12,
8046
L22
11303
L23
15687
22
3
23
2
185
1
822, 12.11 1
1823
L6)
(114
115
116
117
1113
1114
1115
1116
935
L15 V16
1371
2.2*. Cartan theory
I
n 1 1*(n)J I
IV :'l (n)
Ill
11
24
5
1 1IB24,12.12,13.8,14,6
C24
21840
25
3
1 1821, 11,5
L25
30058
26
3
1 1 ®26,12.13
126
41366
27
4
827, 13.9
28
6
829,
29
2
30
5
56525
V127
127 128
77126
829
L29
1 1104490
830,
_30
1
1117
118
179
11415261
Remark 2.2.3. The biholomorphisms in (a)-(c) are given by the following formulas: (a)
iZ2+Z31 E 12,2
+iZ4
L4 B (Z 1, Z2, Z3, Z4) H IZ1 iZ2-Z3
Z1-iZ4J
(EXERCISE).
(b)
0
63 9 (Z1,Z2,Z3) H -Z1 L-z2
ZI
Z2
0
Z3
-Z3
0
E 113
(EXERCISE),
and (cf. [Mor 1956]) 0
+iZ2) L6 9 (ZI....,Z6) H -(Z1 -(z3 +iZ4) -(Z5 + i Z6)
Z1+Iz2
Z3+iZ4
0
Z5-IZ6
-(Z5 -IZ6) -(-Z3 + iZ4)
-(Z1 - iZ2)
0
Z5+iz6 -Z3+iZ4 EII4. Z 1 -122 0
(c)
13 9 (ZI. Z2, Z3) -*
LZ1
2iZ3
Z1
1.Z3] E 1112.
Exercise 2.2.4. Find a formula for fl n). Remark 2.2.5. Let us mention the following two results related to various characterizations of a bounded domain D C C" by its automorphism group Aut(D). (a) Assume that b E aD is a point such that aD is strongly pseudoconvex at
b (such a point always exists if aD E e2). Moreover, assume that there exist a compact K C D and sequences (ak)k° I C K, ('k)°k° 1 C Aut(D) such that bih
Ok(ak) -+ b. Then D ^- IB" ([Ros 1979]). (b) We say that a bounded domain D C C" has piecewise ek-boundary if aD is a topological (2n -1)-dimensional manifold, and there exist an open neighborhood pm = 0 on aD and for U of aD and p E ek (U, Qt"') (with some m) such that p1 every
allzEUwith p11(z)_
have .
=ple(z)=0.
180
Chapter 2. Biholomorphisms of Reinhardt domains
Let D C C" be a bounded homogeneous domain with piecewise e2-boundary. bih
B", x ... x (Bno ([Pin 1982]). In particular, every bounded homogeneous Then D domain D C C" with smooth boundary is biholomorphically equivalent to B". (c) In virtue of Theorem 2.1.17 (and Remark 2.1.18), the above result implies that the boundary of 1" (n > 3) is not piecewise e2.
Exercise* 2.2.6. Prove directly (without using Remark 2.2.5 (c)) that the boundary of 1,, (n > 3) is not piecewise e2.
2.3 Biholomorphisms of bounded complete Reinhardt domains in C2 Let us look more thoroughly into the problem of biholomorphic classification of
bounded Reinhardt domains D C C2 with Vj n D # 0, j = 1.2. The case where DI, D2 C C2 are bounded convex complete Reinhardt domains was first considered by K. Reinhardt in [Rei 19211. The general case was completely solved by P. Thullen in [Thu 1931 ] (see also [Car 1931 ]). Observe that, by Proposition 1.12.8, we may always assume that D 1, D2 are bounded complete Reinhardt domains of holomorphy. By rescaling variables, we may reduce the situation to the case where D j is no)7alized, i.e.
{Z E C : (ZI,O) E Dj} = {z2 E C : (0,Z2) E Dj} = D.
j = 1.2.
(2.3.1)
In particular, D C D2.
Lemma 2.3.1. Let D C C2 be a normalized complete Reinhardt domain of holomorphy. Then 1D2 D = {(z1. z2) E : Izi l < R(Iz21)}. where R = RD: [0. 1) -+ (0, 1] is a continuous function with R(0) = 1. Prof. Since log D is convex, we conclude (EXERCISE) that for every u E (0, 1) there exists exactly one t =: R(u) E (0, 1) such that (t. U) E 8R(D) (recall that R(D) := {(Iz11,1z21) : (zI, z2) E D}). It remains to observe that the function R: [0, 1) -+ (0, 1] is continuous (EXERCISE).
Example 2.3.2. Let IEP = {(Z1 2) E C2 ; 1ZI 12P1 + IZ212P2 < l },
P = (PI P2) E
be a complex ellipsoid; cf. (1.18.5). If (PI = 1, P2 # 1) or (pi
lR>0,
1, P2 = 1),
then EP is traditionally called a Thullen domain. The domain IEP is a normalized complete Reinhardt domain of holomorphy (cf. Exercise 1.18.7). Notice that
RE,,(t) := (I -t2P2)I/(2P1),
t E [0, 1).
2.3. Biholomorphisms of bounded complete Reinhardt domains in C2
181
Exercise 2.3.3. Determine the function RD for the domain D :_ {(z1, z2) E ID 2 : Izr dal Iz2Ia2 < 9},
where 011, a2 > 0, 0 < 9 < 1.
Recall that for i; = i;2) E T2, we have put TT(z) _ z = (z1, z2) E C2. Let S(zr, z2) := (z2. z1), TT := TT O S.
z = (6Z I, 6z2),
The following three results due to P. Thullen [Thu 1931) give the full characterization of biholomorphic equivalence of bounded normalized complete Reinhardt domains of holomorphy in C2.
Theorem 2.3.4. Let a > 0, a # 1. Then the group Aut(l£(a.1)) coincides with the set of all mappings of the form Ic12
iE(a,l) 9 Z H
(.Izl((_)2)
p .S2hc(z2))
E I(a.l)
(2.3.2)
where c E ID, (l; I, 2) E T2 , and the branch of the power () 11(2a) may be arbitrarily chosen.
In particular, IE(a,l), a $ 1, is not homogeneous. Observe that Aut(i£(a,t)) (a 34 1) depends on four real parameters (cf. Example 2.1.12 (d)).
Theorem 2.3.5. Let D C C2 be a normalized bounded complete Reinhardt domain of holomorphy. (a) The following conditions are equivalent:
(i) Aut(D) acts transitively on D;
(ii) either D = ID2 or D = 82 (cf Example 2.1.12 (a), (b), (d), Theorem 2.1.17).
(b) Assume that D
{ID'. 82}. Then the following conditions are equivalent:
(i) there exist b E V and Ob E Aut(D) such that Ob(z) _ (zl fb(22), mb(z2)),
z = (z1, z2) E D,
wherefb E 0* (D) 8 and mb E Aut((D), mb (b) = 0 (note that fib (0, b) _ (0, 0));
(ii) for every c r= lD there exists a 1C E Aut(D) such that
0c(z) _ (z1fc(z2), mc(z2)).
z = (zl , z2) E D,
where ff E 0*(ID) and m, E Aut(D), mc(c) = 0 (45c(0, c) = (0, 0));
80*(G):=If e0(G): f(z)
0,
zeG).
182
Chapter 2. Biholomorphisms of Reinhardt domains
(iii) D = IE(a,t) for some a # 1. (c) The following conditions are equivalent:
(i) there exists an a E D such that Aut(D) = Auta(D); (ii) Aut(D) = Auto(D); (iii) any automorphism P E Aut(D) is of the form TT or T, with second case is possible only if S (D) = D.
E
T2 (the
Obviously, (b) may be formulated also with respect to the second variable.
Theorem 2.3.6. Let Dt, D2 C C2 be normalized bounded complete Reinhardt domains of holomorphy and let F E Bih(D1, D2). Then we have the following possibilities:
Dc =D2=1D2;
Dt=D2=1B2; DI = D2 = IE(a,1) with a 34 1 and F = tyc,C for some c E ID and i E T2, where tfic, is as in (2.3.2);
DI = IE(a,t), D2 = lE(t,a) with a # 1 and F = S o tPc,t, for some c E ID and E T2, where +Lc, is as in (2.3.2);
DI = D2 = IE(t,a) with a # 1 and F = S o tPc,t o S for some c E ID and E T2, where 'J' , is as in (2.3.2); in all remaining cases, either D2 = DI and F = TT with i; E T2, or D2 = S(D1) and F = T* with E T2; in particular, F E Biho,o(DI, D2). Our method of proof of the Thullen theorems needs the notion of the so-called Wu ellipsoid introduced by H. Wu in [Wu 19931, see also [Joh 1948] and [Jar-Pfl 2005], § 1.2.6.
Exercise 2.3.7. Let L, Lt, L2 E GL (n, C). Then:
(a) L-t (IBn) = {z E C"
:
zt M2 < 1), where M := Lt L, and therefore,
L-t ((n) is given by the Hermitian scalar product (z. w) H z' Mw.
(b) A2n(L(IBn)) = d (c) L-1(0,) is a Reinhardt domain iff L-t((Bn) = {z E C" : r z E B.) for some r E IR;o.
(d) Li t(lBn) = L2 t(8,) if L2 o L, t E OJ(n). Lemma 2.3.8. For every bounded domain 0 96 D C C", the family
{L-'(ln) : L E GL(n,C), D C L-' (B,,))9 contains exactly one domain (the Wu ellipsoid) IE(D) with minimal volume. 9Observe that the family is not empty because D is bounded.
2.3. Biholomorphisms of bounded complete Reinhardt domains in C2
183
Proof. Let .F(D) :_ {L E GL(n,C) : D C L (ln)). By Exercise 2.3.7(b), we want to maximize the function .F(D) 9 L H I det LI. Let B be the smallest balanced domain containing D, B = int n GDD G. Then B C L-' (S.) for G is balanced
every L E .P(D). Hence IIL(z)II < hB(z), z E Cl, where hB stands for the Minkowski function of B. In particular, the set Y(D) is bounded in D,I(n x n, Q. Consequently, there exists an Lo E F (D) such that I det Lol = C := sup{ I det L I : L E F(D)} (EXERCISE).
Let F0 (D) := J L E F (D) : I det L I = C). We have to show that M LZ o L I E OJ(n) for any Lt, L2 E .Fo(D) (Exercise 2.3.7(d)). We may assume that det L I = det _L2 = C. Suppose that the Hermitian matrix M1 M ,54 IIn. Then we can write MI M = PAP-I , where P E OJ(n) and A is a diagonal matrix with elements dl,..., d" > 0. Since det M = 1, we have d, . d" = 1. Moreover, since M' R # IIn, we conclude that dJ 54 1 for at least one j. Observe that
17'(L;LI + L2' T,2)2 = 2(IILI(Z)II2 + IIL2(=)II2) 2(hB(z) +hB(z)) = hB(z).
Z E C".
Write 1(L i L1 + LZL2) = L' L with L E GL(n, C). The above inequality shows that L E F (D ). Thus I det L I < C. On the other hand we have
IdetLI2 = 2" det(L'LI +L2L2)
= 2" det(Li)det(1, + (Li)-IL2L2(LI)-I)det(LI)
= ZC2 n det(II" + MM) = I C2det(P)det(IIn + P-I M'MP)det(P-I ) = ZnC2det(II"+A)=C21
2dI
...
1
2d"
>CZ dI...dn =C2:
a contradiction.
Remark 2.3.9. (a) If A E GL(n, C), then for every bounded domain D C C" we have A(L(D)) = L(A(D)). In particular, if A(D) = D, then A(I=(D)) = E(D). (b) If D is a bounded Reinhardt domain, then lE(D) = (z E C" : r - z E (Bn} for some r E UZ;o.
Lemma 2.3.10. Let D 1, D2 C C" be bounded Reinhardt domains with 0 E D I fl D2. Let F E Biho,o(DI, D2). Then
F(z) = p-t U(r z). where U E OJ(n), r. p E UZ;o, and p I := (1/pl,..., 1/pn).
184
Chapter 2. Biholomorphisms of Reinhardt domains
Proof By Remark 2.3.9 (b),
IE(D1)={z E
GL(n.C). By Remark 2.3.9(a), F(IE(D1)) _
E
IE(D2). Consequently, the linear isomorphism
C"
Z
p F(r-I Z) E C"
0
maps Mn onto 1Bn, and therefore, it belongs to U(n).
Proposition 2.3.11. Let DI, D2 C C2 be bounded complete Reinhardt domains of holomorphy and let F E Biho,o(D1, D2). Then: either Dj = E(DD ), j = 1, 2, and F has the form from Lemma 2.3.10, or
IE(Dj), j = 1, 2, and F(zi, z2) = (rI I za(1), E T2, a E 62 (r1, r2) E R2 >O' Dj .
where
Proof. Using Remark 2.3.9 (b) and Lemma 2.3.10, we may assume that, after rescaling variables, we have DJ IE (DJ) = 82, j = 1, 2, and F E (IJ (n). Next, by permuting and rotating variables, we may also assume that
F(zI,z2) = (z1cosa+z2sina,-zIsina+z2cosa) with a E [0,;r/2) (EXERCISE). We have to prove that a = 0. Suppose that a E (0, 7r/2) and consider the following construction.
T a k e a point x° = (r cos 9, r sin 9) E Q Q. n aDI (r >0,0< 9 <7r/2). Fix an arbitrary tp = (tpI, (ft) E 1R2 and consider the point (e"°' FI (x°), e`2 F2 (x°)) E aD2. Put
(YI(W),YAM := F-I(e"PI
Fi(x°),e"°2F2(x°))
E aD1
and, finally, let
NO = (6(04 2(0) := 01 (01- IY2(SP)I) E IR . n aD1. Direct calculations give
I (Ip) = r c ost 9 + 2 sin 2(9 - a) sin 2a (1- cos((p1 - 92)),
V
1;2((p) = r
sin2o-
2
sin 2(9 - a) sin 2a (1- cos((pi - (P2)).
2.3. Biholomorphisms of bounded complete Reinhardt domains in C2
185
Put d(8) := sin 2(8 - a) sin 2a E (-1, 1). We have proved that for any point x° = (r cos 8, r sin 8) E Qt+ n aD 1, the points
x(t) := (r cost 8 + d(8)1, r
sine 8
- d(8)t ),
0 < t < 1.
belong to Qt+ n aD1. Observe that x(0) = x° and 11x(t)II = r = IIx°II. Consequently, if 8 96 a, then the boundary of I+ n D 1 contains the are
I(x°) := {x(t) : 0 < t < 1}. Thus, there exist r_, r+ > 0 such that {(p cos *, p sin
(p = r_, 0 < * < a) or (p = r+, a < >/i < it/2)}
ctR+naD1. Finally, using the completeness of D 1, we get D, = 82 - a contradiction.
0
Corollary 2.3.12. Let D1, D2 C C2 be normalized bounded complete Reinhardt domains of holomorphy and let F E Biho,o(D1. D2). Then either D1 = D2 = 182 or F = TT (and D2 = D 1) or F = T,* (and D2 = S (D 1)) for some E T'2.
Proof. The result follows from Proposition 2.3.11 and the fact that Dj = F(Dj) if Dj = 132 (because Dj is normalized), j = 1.2. 0
Proposition 23.13. Let D 1, D2 C C2 be normalized bounded complete Reinhardt domains of holomorphy, D2 0 132, and let F E Bih(D1, D2) be such that F(0, b) = (0, 0) with b 0. Then either
F(z) _ (ztf(z2),m(z2)), z = (zt,z2) E D1,
(2.3.3)
F(z) _ (m(z2),z1 f(z2)), z = (z1,z2) E D1.
(2.3.4)
or
where m E Aut(D), m(b) = 0, and f E O"(D).
Proof. For an arbitrary A E T, let Fx := F o T1x,11 o F-1 E Auto(D2). By Corollary 2.3.12, there exists a (),) E T2 such that either Fx = Tf(x) or Fx = T(x) . The case where Fx = TE(A) for uncountably many A E T. We have
F(Azi,z2) = (1(A)F1(z), 2(A)F2(z)),
z E D1, A E I c T.
where I is uncountable. Observe that if 1; (A) = (1, 1). then F(Azt, z2) = F(z), z E D 1, which implies that A = 1. Thus at least one of the sets
12:_{AEI:tt(A)=1, t2(A)01} is uncountable.
Chapter 2. Biholomorphisms of Reinhardt domains
186
Case 1. 11 is uncountable. We have (Ft (O, z2). F2(O, z2)) = (r (A)F1(O, z2), 2(A)F2(O, z2)) Hence F t (0,
z2 E ID, A E I1.
0, i.e. Ft (z) = zt G1(z) with G E 0(D1). Since Ft (0, ) = 0
)
and F is a biholomorphism, we conclude that F2(0, - ) = m E Aut(ID), m(b) = 0. Consequently, F2(z) -m(z2) = z1G2(z) with G2 E 0(D1). Thus (Az1 G1(Az1, z2), Az1G2(Az1, z2) + m(z2))
= (t1(A)z1G1(z). 2(A)z1G2(Z)+42(A)m(z2)) Taking zt = 0, we get m
z = (z1,Z2) E D1, A E It.
2(A)m, so i2(A) = 1 for A E J. Hence
AG1(Az1,z2) _ t1(A)G1(z),
AG2(Azl,z2) = G2(z).
First consider the second equation. In terms of the power series expansion of G2 we get 00 G2,i,k(Aj+t
- 1)z1 Z2 = 0,
Z = (Z1, Z2) E D1, A E I1.
j,k=0
Since 11 is uncountable, there exists a A E It with Aj+r 56 1, j = 0, 1, 2, ... . Hence G2 0. Now we come back to the first equation. We have aF
aF
0 0 det
FF
(0, z2) = det
rG1(0, -2) L
m,0 2), = G1(0. z2)m'(z2).
Hence Gt(0,z2) 0 0, z2 E ID. We have AG1(0,z2) = 1(A)G1(0,z2). Hence i t(A) = )L, ,k E 11. Therefore, G1(Azt,z2) = G1(z), z = (z1,z2) E D1, A E I. By a power series argument we see that G 1 depends only on Z2. Finally,
F(z) = (zrf(z2).m(z2))
z = (z1,z2) E D1.
where f E 0"(ID) and m E Aut(ID), m(b) = 0. Case 2. 12 is uncountable. We have F1(Azl,z2) = F1(z), z = (zr,z2) E D1, A E 12, which implies that F1 depends only on z2. Hence F1(z) = m(z2) with m(b) = 0. Furthermore, F2(0, Z2) = 2(A)F2(0, Z2), which implies that
F2(0, ) - 0. Consequently, M E Aut(D) and F2(z) = zt G2(z) with G2 E (9(D1). Moreover, 0 34 det
OO
(0, z2) = det I G2(0, L
z2)
m/002)]
-
G2(0,
z2)m'(z2),
187
2.3. Biholomorphisms of bounded complete Reinhardt domains in C2
which says that G2(0, z2) # 0, z2 E D. Since AG2(0, z2) = 2(,X)G2(0, z2), we
get 42(A) = ;L, A E 12. Therefore, G2(Az1,z2) = G2 W, z = (zl,z2) E D1, A E 12. Hence G2 depends only on z2. Finally,
F(z) _ (m(z2), z1f(z2)),
z = (z1, z2) E D1,
where f E O`(ID) and M E Aut(D), m(b) = 0. The case where FA = TE(A) for uncountably many A E T. We have
F(Azt,z2) = (1(A)F2(z), 2(A)F1 (z)),
z E D1. A E I C T.
where I is uncountable. In particular, F1(0,z2) _ 1(A)F2(O,z2),
F2(O,z2) _ 2(Z)F1(0,z2).
Observe that F1 (0, - ) 0- 0 (even more, if F1 (0, c) = 0 for some c E ID, then F2(0, c) = 0 and hence F(0, c) = (0.0), which implies that c = b). Consequently, 1. We get F1(Azt,z2) = S1(A)F2(z1,z2) = t1(A)52(A)F1((1/A)z1,z2) = F1((1/A)zl,z2),
(Z1,Z2)ED1, AEI. Thus Ft and F2 depend only on z2 - a contradiction.
Proposition 2.3.14. Let D C C2 be a normalized bounded complete Reinhardt domain of holomorphy. (a) Assume that for a b E ID, there exists a 4'b E Aut(D) of the form
1b(Z) _ (Z1 fb(Z2), mb(z2)),
z = (zt. z2) E D,
(2.3.5)
where mb E Aut(D), m(b) = 0, and fb E (9*(O)). Then for any c E ID there exists a 4Sr E Aut(D) of the form
z = (21,22) E D,
'PC (Z) =
where m, E Aut(DD), m(c) = 0, and ff E O`(ID). Moreover, either D = ID2 or D = B2 or D = IE(a,l) with a # 1 and 2
Ob(Z1,Z2) = (1z1((I -bz2)2) a. 2hb(Z2)).
Z = (Z1,Z2) E E(a,t),
where (6 , 6) E V. (b) Assume that for a b E ID. there exists a Ob E Aut(D) of the form Ob(Z) = (mb(z2), z1 fb(z2)),
z = (Z1, Z2) E D,
(2.3.6)
188
Chapter 2. Biholomorphisms of Reinhardt domains
where Mb E Aut(D), Mb (b) = 0, and fb E O' (D). Then there exist a point a E D. and an automorphism &a E Aut(D) of the form
(1 (z)=(m;(zl),Z2fa(zl)) z=(z1,z2)ED. where mo E Aut(D), ma (a) = 0, f,, E O`(D). Consequently, after permutation of variables, we are in the situation as in (a).
Proof (a) Step 1. Let 1' := d5b'. Observe that 111(z) = (zlg(z2),k(z2)), z = (ZI, Z2) E D, where k = mb' E Aut(D), k(0) = b, and g = I /fb o k E 0*(D). Define W ,x := T(1,,,) o cpb o T(t,,k) o'// E Aut(D), n, A E T. Observe that the required 0, exists for every c E D such that there exist r), % E T with (0, 0) _ (0, c), and then `f'n,x(zt, z2) = Ob o T(I,l/x) o l o T(I,l/n)(zl, z2)
Pc(zl, z2)
='b o T(I,1/x) o W(zt, (1/)l)z2) = Ob o (1/)L)k((1/rl)z2)) = (zlg((1/rl)z2)fb((I/A)k((1/71)z2)),mb((I/A)k((I/r1)z2))) =: (zl fc(z2), mc(z2)), (Z1, °2) E D.
Thus, every point c E D such that there exists a A E T with Icl = Im(Ab)I is "accessible". Direct calculations show (EXERCISE) that
{Imb(Ab)I : A E T} _ (0. I
Ibi2) =: (0. rl).
+ Observe that rt > Ibi.
Repeating the above procedure with Ob substituted by 0, with 0 < Ic I < rl
leads to a new set of accessible points of the form {d E D : 0 < Id I < r2} 2r
I
> rt. Let
r
-!2
(EXERCISE).
Step 2. Write D = {(Z1,z1) E D2 : Iz1I < R(Iz21)},
where R: [0, 1) -+ (0, 1] is a continuous function with R(0) = 1 (Lemma 2.3.1). Let z° = (z°. z2) E 8D n (D x D), Izo I = R(Izi I). Let 0, be as in Step 1. Then Oc(z°) E 8D n (D x D) and therefore Iz° fc(z2)I = R(Ihc(zi)I), which gives the equation R(1z21)Ifc(:2)I = R(Ihc(z2)I),
z2,c E D.
(2.3.7)
189
2.3. Biholomorphisms of bounded complete Reinhardt domains in C2
Take c = reie, Z2 =
re`(`+B). Observe that the function I
I
= R(r)I.fc(re'tr+B))I - r2e'r )
is of class C°O. Consequently, R is e0O on (0, 1). Indeed, write (p,. (t) = R o *r(COSt), where
t E IR,
1/2
2 - 2u
Vir(u):= r(1 +r4-2r2u)
U E (-1.1).
A short calculation shows that *r '(u) < 0, u E (-1. 1). Moreover,
+
r2) .- Ir.
Consequently, R is C°O(10. Letting r / 1, we conclude that R E C°°(0, 1). Step 3. Put
U(t) := log R(t).
Q(t) := U"(t) + (1/t)U'(t).
t E (0, 1).
We have
Q(IzI) = AU(IzI).
Z E m.,,
82
where ,d := e + ay (EXERCISE). Moreover, since log 1 f 1 is a harmonic function ([Con 1973, Chapter X), we get
Q(1z1) = A log R(Izl) + A log Ifc(z)I = A log R(Ih,(z)I)
=
1 - IC12
I(1 -cz)2I
Z - C I11\,
1 -cz
Z E D., C E 1D \ {z}
(EXERCISE).
(2.3.8)
We are going to determine the function R. First consider the special case Q = 0. Then the equation
U"(t)+(1/t)U'(t)=0 gives U(t) = Co logt + logCt, and hence R(t) = Cltc0, 0 < t < 1. The continuity of R and condition R(0) = 1 imply that Co = 0, CI = 1, i.e. R = 1. Consequently, in this case we get D = D2. Now, consider the case where Q 0- 0. Observe that if Q(to) 34 0, then for every t E (0. 1) there exists a c E ID such that Ihc(to)I = t. Hence, using (2.3.8), we conclude that Q(t) 0 0.
190
Chapter 2. Biholomorphisms of Reinhardt domains
Observe that if QI, Q2 are two functions of this type, then, using (2.3.8), we get
_
QI(t) Q2(1)
QI(IiCCZI) Q2(1 1 cz
t=IZIE(0,1). CE(D\{z}.
I)
Fix a to E (0, 1). We have already observed that for any t E (0, 1) there exists a C E
( such that I he (to) I = t. Then Q I (to)/ Q20o) = Q I W/ Q2(t). Consequently,
QIIQ2 = const. Step 4. The domain D := E(,,,, 1) satisfies the assumption of (a). Indeed, for any b E ID and (S I, 2) E T2, the mapping
F(zl.z2)
1
6Z I
12
-bz2)2)
2hb(Z2)),
is an automorphism of E(1 ,I) with F(0, b) = (0, 0) (EXERCISE). Direct calculations show that, for R(1) = (1 - 12)I/(2a), the corresponding
function Q has the form
all
Q(t)
212)2
(EXERCISE).
Step 5. By Steps 3 and 4, for any domain with Q 0- 0 we have
Q(t) =
2
C(1 -12)2'
where C E Qt, is a constant. Hence U(t) = ilog(1 - t2) + logC1, and so R(t) = CI(1 - 12)I/(2C), 0 < t < 1. The condition R(0) = I implies that CI = 1. Since D is bounded, we have C > 0. Thus D = IE(c,I). Step 6. Observe that if D = E(,. 1) with a 96 1, then Ob(ZI,Z2)
_
(IzI(()2) 11 b y
12
.
2hb(z2)).
z = (zl,z2) E IE (a,I),
where (6, %2) E T2. Indeed, since (1 - 1b12)(1 - 1z212)
1 - 1hb(z2)12 =
z2 E (D.
(2.3.9)
11 - bz212
we get (cf. (2.3.7))
Ilb(Z2)I -
R(Ihb(z2)1) R(IZ2I)
(I - Ihb(z2)12 I - 12212
21.
)
I - IbI2 2L. = (11 -bZ212)
Z2 E D.
2.3. Biholomorphisms of bounded complete Reinhardt domains in C2
191
(b) Let +P := Ob I. Observe that W(z) = (z2g(zt),k(zt)), z = (zt,z2) E D, where k = Mb I E Aut(ID), k(0) = b, and g E O `(ID). Fix A E T \ (1 } and put a := mb(,kb) E D.. Then (Pb o T(t,I/,A) o (P(a,0) _ Ob o T(I,/A)(0,k(a))
= 'Pb o T(I.I/x)(0.116) = 45b(0, b) = (0.0). Moreover,
45Q (zt, z2) : = 'Pb o T(I,I/x) 0 111(21, 22) = 45b o T([,1/x)(Z2g(z0 k(zt))
_ Ob(z2g(z1), (l/A)k(zI)) = (mb((1/A)k(z,)),z2g(z,).f((1/A)k(z,))) (mQ(za), z2fa (z1)). (z1, z2) E D.
0
Proof of Theorem 2.3.4. We already know (cf. the proof of Proposition 2.3.14 (a), Steps 4 and 6) that 2
Aut(E(a,I)) J I (E(a,1) 3 Z H
(,zl((l1 - I cZ2)2)
-
.
J
2hc(z2)) E E(a,t)
CE(D.
(ry
111.
2)ET2}
yy
_ (45 E Aut(IE(a,t)) : $ has form (2.3.5)} Moreover, (S' is a subgroup of Aut(E(a,I)) (EXERCISE).
Fix a 45 E Aut(IE(a,t)). If 45(0, 0) = (0, 0), then, by Corollary 2.3.12, either 45 = TT or 45 = T . The second case is impossible because E(., 1) is not symmetric. Hence 45 E 6r.
Now, assume that 0(0,0) = (a, b) 0 (0, 0). Then F := +vb,l 0 45 E Aut(IE(a,I)) and F(0, 0) = (c, 0) for some c E D. If c = 0, then F E (Sf and hence 45 E (SS.
Suppose that c 54 0. Put G := S o
F-I
o S E Aut(I_(t,a)). By Proposition 2.3.13, G is either of the form (2.3.5) or (2.3.6). In the first case Proposition 2.3.14 (a) implies that E(1,«) = E(p,t), which is impossible. In the second case, since G(8E(, )) = 8E(,,,) we get Ihc(z2)I2 + (1-
IZ212a)aIfc(z2)12a
1,
z2 E D.
Hence
1- Ihc(z)I2 (2.3.9) 1- IC12 Ifc(z)I2a = ( I - IZI2a)a I
Consequently, a = 1; a contradiction.
1-1=12 a
ZEID.
192
Chapter 2. Biholomorphisms of Reinhardt domains
Indeed, suppose that 0 < a < 1. Then for every
E T we get
I
lim
II-
(1 - ta)a
= 0.
Hence, by the maximum principle, fc - 0; a contradiction. If a > 1, then lim
1
- 11 -
(1 - ta)a
1- Ic12 lim r,l- 1 -t
=mot Iff(z)120
-0
and we have again a contradiction.
Proof of Theorem 2.3.5. (a) Assume that D is homogeneous and D # 182. In
particular, for any b E D there exists a (bb E Aut(D) such that Pb(0, b) = (0, 0). By Proposition 2.3.13, 'b is either of the form (2.3.5) or (2.3.6). Now, by Proposition 2.3.14, either D = ID2 or D = lE(«,t) or D = IE(t,a) with a 96 1. By Theorem 2.3.4, the only homogeneous case is D = ID2. (b) follows from Proposition 2.3.14 (a) and Theorem 2.3.4. (c) follows from Corollary 2.3.12 with DI = D2 = D.
Proposition 2,3.15. Let D C C2 be a normalized bounded complete Reinhardt domain of holomorphy and let ' E Aut(D) be such that 0(0.0) = (a. b) with ab $ 0. Then there exists a W E Aut(D) such that l'(0, c) = (0.0) or P(c, 0) _ (0, 0) for some c E D..
Proof. Put Vo := ((zl,z2) E D : ztz2 = 0) = (D x {0)) U ((0) x ID), V. Vo \ {(0, 0)) = (D. x (0)) U ((0) x DD.). Suppose that the result is not true, i.e. (*) F(0, 0) ll V, for every F E Aut(D) (equivalently, (0, 0) 0 W(V.) for every E Aut(D)). Define Wt
M
0-t o Tt 0 0 E Aut(D), {T,r(P(t)) : q, E T2},
lpt(0,0),
E T2, 1' t (Vo),
E T2.
Vo for all E T2 \ ((1,1)). Indeed, in view of (*), P(1;) E Vo if Note that (1, 1). W (0, 0) = (0, 0), which means that TT o 0(0, 0) = cP(0, 0) and hence
Moreover, m fl Indeed, it is clear that
E M fl Fix q. l; E T2 and suppose that i.e. W I(T0(P())) = p t o T. o i' (0.0) E Vo. By (*) we E get W t (T,(P())) = (0,0). Thus T,7(P()) = P(i;). We are going to show that
(**) there exists a point
NCM.
which lies on a smooth 3-dimensional surface
2.3. Biholomorphisms of bounded complete Reinhardt domains in C2
193
Assume for a moment that (**) is already proved. Then the intersection N fl cannot be a point, which leads to a contradiction. Indeed, N' :_ PP(N) is also a smooth 3-dimensional surface. It suffices to
prove that N' fl Vo 0 ((0, 0)). Assume that
N' = {(xl,YI,x2,q(xt,Yt,x2)): (XI,Yt,x2) E U3), where U C IR is an open neighborhood of 0 and (p is a smooth function in U. Then
N' fl Vo contains the curve {(0, 0, x2, 00, 0,x2)) : x2 E U). All other cases are similar (EXERCISE).
We come back to (**). Consider the mapping
i
(0, 2,r) x (0, 2n) D a i
(I
(P(e;a))1I
I(P(eia))21)
E (>o.
Put W := f((0, 2ir) x (0, 27r)). Observe that if W contains a smooth curve, then M contains a smooth 3-dimensional 2-circled surface. Indeed, suppose that W contains a graph u = ap(t), t E U, where U C IR>o is open, .p is smooth in U and V(t) > 0, t E U. Then M contains the set
M'
(e'# t, e'Y(p(t)) : t E U,
y E IR}.
Consider the mapping Ux!R2
(t,#,y) H (e'Ot,e'YV(t))_(tcosP,tsin$,cp(t)cosy,(p(t)siny) E 1R4
and calculate g'(t, P, y):
g,(t
y) _
cos$
-tsinP
sin fl
t cos 6
Ip'(t) cosy _V (t) sin y
0 0
0 0
-V(I) sin y VQ) Cosy
Then rank g'(t, f, y) = 3 (EXERCISE), which implies that M' locally contains a smooth 3-dimensional surface. Now we prove that W contains a smooth curve. The mapping f is real analytic (EXERCISE). If there exists an a with rank f'(a) = 2, then W contains an open set and, therefore, a curve. Thus we may assume that rank f'(a) < 1, of E (0, 27r) x (0, 2n). Obviously, if rank f'(a) = I on a non-empty open set, then W contains a curve. It remains to exclude the case where rank f' = 0 on (0, 2n) x (0, 27r). Then
f is constant. Thus I P(%) I I = cI > 0, 1 P(02I = C2 > 0. ' E (T \ 11))2. By continuity, (0.0) = (I P((1, 1))1I,IP((l, l))21) = (cl.c2); a contradiction. 0 Proof of Theorem 2.3.6. Since F E Bih(DI, D2), we see that DI is homogeneous if D2 is homogeneous. Thus, by Theorem 2.3.5 (a), DI E {ID2, 'B2) if D2 E
Chapter 2. Biholomorphisms of Reinhardt domains
194
{m2, 82}. By the Poincare Theorem 2.1.17, the only possible cases are D1 = D2 = ID2 and D, = D2 = lB2. Assume that Dj is not homogeneous, j = 1, 2. If there exists an a E Dt such that Aut(D1) = Auta(Di), then Aut(D2) = AutF(a)(D2). Consequently,
by Theorem 2.3.5 (c), a = F(a) = 0 and by Corollary 2.3.12, F = TT (and D2 = DI) or F = TT (and D2 = S(D1)) for some E 72. It remains to consider the case where Dj is not homogeneous and Aut(Dj)
Auta (Dj) for any a E Dj, j = 1, 2. Then, by Theorem 2.3.5 and Proposition 2.3.15, Dt = ln, D2 = E. for some
p,q E (11) x (IR>o \ {1))) U ((IR>o \ {1)) x {1)). In view of Theorem 2.3.4, we only need to prove that p = q or p = S (q).
The case F(0, 0) = (0, 0) follows from Corollary 2.3.12, so assume that F(0, 0) # (0, 0). In the case where F(0, b) = (0, 0) for some b E ID we use Proposition 2.3.13 and we conclude that F is either of the form (2.3.3) or (2.3.4). In fact, substituting D2 by S (D2), if necessary, we may assume that
F(z) = (Z1fb(z2),mb(Z2)),
Z = (Z1,Z2) E DI,
where fb E 0*(D) and mb E Aut(tD). Mb (h) = 0. Recall that
Dj _ 1(Zt.z2) E
[D2
. 1211 < Rj(I221)),
j = 1,2,
where
RI(t) := (I - t2P2)1/(2P1)
R2(t) := (1 -f2g2)t/(2qI)
I E [0, 1).
Since F(8Dt fl (D x ID)) = dD2 fl (D x ID), we get
Rl(I:21)Ifb(z2)I = R2(Ihb(z2)I), (1 _ IZI2v2)1/(2vl)Ifb(z)I
Z2 E D,
= (1 -
ZEID,
and, consequently, IA(z)1 =
(I - Ihb(Z)I2g2)t/(2qi) (I - 1z12P2)h/(2Pt)
ZEID.
We have to consider the following three cases: F E Bih(IE(a,,). E(p,t)). Then we have 2
Ifb(2)I = (Il -162112)
(1- IZ12)
-
,
ZEID.
2.3. Biholomorphisms of bounded complete Reinhardt domains in C2
195
Since fb E 0`(D), letting I z I -+ 1, we conclude that a = . F E Bih(IE(a,l), (E(1,p)). Then we have
Ifb(Z)I =
1/2 (
( I - I hb(z)12'0 I - Ihb(z)12
1 -1b 12
1/2
E D.
11 -bzI2
Consequently, a = 1; a contradiction. F E Bih(IE(l,a), IE(l,p)). Then we have
(
I fb(z)1=
1 - Ihb(z)I2 112 I - IzI2a ) I - Ihb(Z)I2p
1 - lhb(z)120 I - Ihb(z)12 I - IZ12 1/2 IzI2a ) 1-1=12
- (1 - Ihb(z)12
I - Ib12
I - IZ12
1/2
t-Ihb(z)12 I1-6:121-1z12a) Letting z -*
E lp.
,
E T, we conclude that I -1b 12
Zm I fb(z)I=(811 -
-
1 a)1/2_
I1const
l
-hi;12
ET.
Hence, by the identity principle, we get
fb(z)=
I
nh`,
ZED,
with rl E C.. We have - Z12* Ig12I11-Ibz12
= 1 - Ihb(:)I20,
Z E D.
Since both sides of the above equality are real analytic functions on C \ { 1/b}, we get 1?1121-1Z12a
1l -hz12
= 1-Ihb(z)I20.
2EC\{I/b}.
Letting z -+ 1/b, we conclude that $ = I (EXERCISE); a contradiction. The case where F(a, 0) = (0, 0) for some a E D. is analogous.
In the case where F(a, b) = (0, 0) for ab # 0 we may assume that D1 = IE(a,1). Then +Irb,I(a,b) = (a*,0) (Wb,t is as in Theorem 2.3.4). Consequently, F o J/ (a*, 0) = (0, 0). Thus the problem is reduced to the previous situation, for some ' E T2. Finally, which implies that D2 = E(a,l) and F o Wbl = F = Wb,1 0 't'a* t E AUt(IE(a,I)).
Chapter 2. Biholomorphisms of Reinhardt domains
196
2.4 Biholomorphisms of complete elementary Reinhardt domains in C2 Recall that a Reinhardt domain of the form Da,c = {z E C2 : IZI la' IZ2Ia2 < ec},
a = (a1,a2) E (IR+)., C E IR,
is a so-called elementary Reinhardt domain. Because of the restriction on the exponent a it is complete; moreover, it is a domain of holomorphy. Remark 2.4.1. Observe that Da,c is algebraically equivalent (cf. Definition 1.5.12) to Da = Da,o (EXERCISE). Therefore, we will only study domains of type Da, a E (IR+).. In fact, we will only consider the following three types of normalized elementary Reinhardt domains, namely:
a l a2 = 0: then either D. = D(l,o) = ID x C or D. = D(o,1) = C X ID (obviously, both domains are biholomorphically equivalent);
Of l a2 # 0 and a 1102 = p/q with p, q E 04. p, q relatively prime: then D. = {z E C2: IZII°IZ2Iq < 1}; ala2 36 0 and a1/a2 0 0: then Da = D(r,ll with t := at/a2 E IR,o \ Definition 2.4.2 (Cf. Definition 1.4.8). Let a = (a1,a2) E (IR+).. (a) If a 102 = 0 or a r , a2 E W, a 1, a2 relatively prime, then the domain Da :_ (z E C2 : Izl Ia' Iz2Ia2 < 1) is called an elementary Reinhardt domain of rational type. (b) If ala2 0 and at 0 Q, a2 = 1, then D. = {z E C2 : Izl Ia' Iz2I < 1) is called an elementary Reinhardt domain of irrational type. Remark 2.4.3. Let D. be an elementary Reinhardt domain. Then its logarithmic image contains the straight line L := ((t1, t2) E IR2 : all + a2S2 = 1), 1 < 0, if a 1 a2 0 0, or { (S l . ?; 2) : 52 E R), t j < 0. if a = (1, 0). Conversely, any unbounded
C2 whose logarithmic image contains a straight line is of the form D = Da,c (EXERCISE) and so biholomorphic
complete Reinhardt domain of holomorphy D
to D. Exercise 2.4.4. Let Da, a E W2 (a1, a2 relatively prime). For an f E 0*(ID) Put SI>
Of (z) := (ZI (f (za))-a2. z2(f (Zf))a' ).
z E Da.
Prove that g := {g f : f E 0*(ID)} is a subgroup of Aut(DD). In the following theorem all automorphisms of an elementary Reinhardt domain, normalized as before, are described.
Theorem 2.4.5 ([Shi 1991], [Shi 1992]). Let D. be a complete elementary Reinhardt domain.
2.4. Biholomorphisms of complete elementary Reinhardt domains in C2
197
(a) If Da = D(t,o), then
Aut(DD) = {Da -3 z H (m(zt) f(zt)z2 + g(zl)) m E Aut(D). f E 0*(®), g r= O(D)). (a') If Da = D(o, I), then
Aut(D) = S o Aut(D(t,o)) o S. to (b) If Da is of rational type with at , a2 E W, relatively prime, then
Aut(Da) = {T o gf o a f E O*(D), i; E T2, a E Fs(Da)). where V (Da)
IS. id) if at = a2 = I. {id}
if ala2 54 1.
(c) If D. is of irrational type (i.e. a = (at,1), at f 0), then
Aut(Da)={Da -3 zH
8>0}.
Exercise 2.4.6. Let Da be as in (b). Prove that F E Aut(Da) iff there exist E T, f E O*(D), and A E GL(2, z) n MI(2 x 2, Z+) with aA = a such that F = T(t,t) o of o OA (for the definition of OA := PI,A see Definition 1.5.12). Moreover, the following equivalence result will be discussed.
Theorem 2.4.7 ([Shi 1991], [Shi 1992]). Let D. and Dp be normalized complete elementary Reinhardt domains (in the sense of Remark 2.4.1). (a) If Da is of rational and DD of irrational type, then D. is not biholomorphically equivalent to Ds. (b) D(t,o) and Da, a = (at, a2) E N2, at, a2 relatively prime, are not biholomorphically equivalent.
(c) If D. and Dp are biholomorphically equivalent, then either Da = Dp or D. = S(DO). The proof will be based on the following notion of a Liouville foliation.
Definition 2A.8. Let D C Q;" be a domain. A system (F,t),,EA (A a suitable index set) of sets F,j C D is called a holomorphic (resp. psh) Liouville foliation of D if the following conditions are fulfilled:
F,,1nFee=fifnt0n2, D = UjEA Fn,
10Recall that S (z, w) = (w, 4
198
Chapter 2. Biholomorphisms of Reinhardt domains
if u E 3t°°O(D) (resp. U E 98J{(D), bounded from above), then uI F is identically constant, q E A, for q, . R2 E A, q, # q2, there exists a u E MOO (D) (resp. an upper bounded
U E PSX(D)) such that uIF,, 0 uIFn2'
Example 2.4.9 (A holomorphic Liouville foliation). Let D = Da C C2 be an elementary Reinhardt domain of rational type.
If a = (1,0), put FF := {i;} x C,
E D. Then (Ft)tEu is a holomorphic
Liouville foliation of D. In fact, if u E 3e°O(D), then, ford E lD, u(t;, ) E 3e°°(C). Hence, in virtue of the Liouville Theorem, it follows that u IF, is constant. Finally,
observe that the function D a z H z, is a bounded holomorphic function on D which separates the fibers FF.
If a1a2 # 0, put Ft := 1z E D : za = }, E D. Then, again, (FF)tEp is a holomorphic Liouville foliation of D. In fact, we mention that for E Q) \ (0) the map Wt : C. -> D, (pt (A) := (.1-'12, All ), where f 12 = , is holomorphic. Therefore, if u E .lf°O(D), then u o (Pt E 3e°D(C.). Note that q (C.) = Ft. Applying the Riemann removable singularity theorem and then the Liouville theorem, we conclude that u IF. is identically constant. In case of % = 0 the fiber F0
equals (10} x C) U (C x 10}). By the same reasoning as above it is easily seen that if u r= .7e°O (D ), then u I F0 is a constant function. Moreover, the bounded holomorphic function g: D C, g(z) := z", separates the different fibers. In order to be able to present an elementary Reinhardt domain of irrational type as an example of a psh Liouville foliation we will need the following result due to Kronecker (cf. (Har-Wri 1979], see also p. 97).
Lemma 2.4.10. Let C E IR \ 0, b E C. Moreover, put
Lc,b := {z E C2 : Cz, + z2 = b} and P : C2 -+ C;, 4S(z) := (e2"zl, Then 4'(Lc,b) is a dense subset of F := 1z E 2 : izt I`Iz2I = e2'rReb} e2'rz2).
Proof. Take a point z E Lc,b. Then c Re z, + Re Z2 = Re b, hence dy(L',b) C F. On the other hand fix a point z° E F. Then choose an w E C with e2"w = z2. Setting (b - w)/c we have that (l;, w) E L,,b and so ({; + i I. W - i ct) E Lc,b, t E IR. Then ds((, (o) = (z1 e's, z2) for a suitable s E R. Moreover, it is well known (recall that the number c is irrational) that the set
1'P(t; + i t, w - ict) : t E R) =
{(z°ei(s+t),
zee-ict) :
t E IR)
is dense in F.
Example 2.4.11 (A psh Liouville foliation). Let D = D. C 2 (Cl = (a,, 1)) be a normalized complete elementary Reinhardt domain of irrational type. Put
2.4. Biholomorphisms of complete elementary Reinhardt domains in C2
199
Ft := {z E D : Izt Ial IZ21 = t }, t E (0, 1). Then (Fr),Elo,1) is a psh Liouville foliation.
In fact, if t = 0, then Fo is as in Example 2.4.9. Therefore, if u E PSJ{(D) is bounded from above, then u(0, - ) E S3f(C) and 0) E SJ{(C) are upper bounded and so, in virtue of the Liouville theorem for psh functions (see Remark 1. 14.3 (g)), identically constant. Hence, uIF0 is a constant function.
Now let t E (0, 1). Fix a U E PSJ((D) bounded from above. Using the holomorphic mapping 0 : C2 __+ C2, O(z) := (e2rr2, , e2' 2), we see that 0 maps the domain S2 := {z E C2 : at Re zt + Re z2 < 0} holomorphically onto D fl C. Thus u o 45 is a psh function on 92 which is bounded from above. Fixing a point b = (e2irf1, e2i $2) E Ft we define La1 2L105r
E C2 : ate+(0 = 2a log1) C 12.
Then C D A H u o O(A, -L 2jr log t - a t A) is an upper bounded subharmonic function and therefore identically constant. Hence, u is constant on the 0-image of Lat log, that is dense in Ft. Applying that u is upper semicontinuous we conclude that u(b) < u(p) for any p E Ft. Changing the role of b and p we see that UI IF, is identically constant.
Finally, it remains to mention that u : D - IR, u(z) := Iz1 Ia' IIz21, is bounded psh and separates different fibers. In the sequel the following observation will serve as a basic argument.
Lemma 2.4.12. Let W : D -* D' be a biholomorphic mapping between domains
D. D' C V. Assume that (Fa)aEA (resp. (F)pes) is a holomorphic Liouville foliation of D (resp. D'). Then there exists a bijective map r : A -+ B such that W (F,) = F'(a), a E A. The same result is true for psh Liouville foliations. Proof. We restrict ourselves to proving this lemma for holomorphic foliations. The analogous argument in the case of psh foliations is left as an EXERCISE. In a first step we assume D = D' and = ido. Observe that if F. fl F' 34 0, then Fa = F'. Indeed, suppose that both fibers are different. Then we may assume
that F. \ F' ¢ 0. Fix points p E Fa fl F, and q E F. \ F'. In view of the last condition in Definition 2.4.8 there is a bounded holomorphic function h on
D with h(p) # h(q). On the other hand, p.q E Fa, therefore, h(p) = h(q); a contradiction. The remaining properties of Definition 2.4.8 then prove the lemma.
Now let D and D' be arbitrary. We have only to observe that (P(F,)),EA defines a holomorphic Liouville foliation of D' (EXERCISE). Then the first step 0 completes the proof. Proof of Theorem 2.4.5 (a) and (a'). The proof will be based on the holomorphic x C (see Example 2.4.9). Let rp E Aut(D,). foliation (FF)tEg>, where FF :=
200
Chapter 2. Biholomorphisms of Reinhardt domains
In virtue of Lemma 2.4.12, there is a bijective mapping r : ID (p(FF) = FFq), E ID. Therefore, (p may be written in the form
D such that
(P(z) = ((Pt(z),(P2(z)) = (r(z0.(P2(z))
Therefore, r - (pt ( , 0) is a biholomorphic map from D to D. What remains is to describe the second component function. Let us fix a Ao E ID;
then (p2(Ao, ) is a biholomorphic map from C to C, i.e. (p2(Ao, w) = y(Ao)w + S(A0), w E C, where y(Ao) E C., S(zo) E C. Now, we write 502 as its Hartogs E O(D)." Then yi(Ao) = 0, j > 2. series (p2(z) = Since A0 was arbitrarily chosen, we get 62(z) = yo(zt) + y1 (zt)z2. Observe that
yt E O`(U) Obviously, any mapping given in the lemma is an automorphism of Da. It remains to mention that S gives a biholomorphic mapping between Da and
0
CxlD.
To be able to continue the proof of Theorem 2.4.5, it is necessary to study another automorphism group.
The automorphism group of D. Let a = (at, a2) E Z2, a
(0, 0), at, a2
relatively prime in the case where a t a2 0 0. We set
D.:_(zEC*:(zlI"Iz2112 <1)=D.\Vo. An automorphism of D,* is called an algebraic one if it is the restriction of an algebraic mapping (Pa,A (cf. Definition 1.5.12). Then we obtain the following automorphism groups.
Lemma 2.4.13.
Aut(D(i o)) = Aut(D* x C.)
={D(i,0)9Z
(gzt,f(zt)z2):gET,f EO*(ID.), e=f1}.
x C., E ID.. Then is a holomorphic Liouville Proof. Put FF foliation of D(,.0) (EXERCISE). Let (p E Aut(D(i o)). In virtue of Lemma 2.4.12 there exists a bijective map r : ID. -* D. such that (p(FF) = F q), E ID.. Observe that (pt (z1,.) is a bounded holomorphic function on the whole plane and, therefore, identically r(zt), i.e. V(z) = (r(zi),62(z)). Since (pt(zt,0) = r(zt), zt E ID., the function r is holomorphic and hence a biholomorphic map from ID. onto U).. Therefore, r is a rotation of LD., i.e. r(zt) = qzt, where jql = 1. where Using Laurent expansion we may write 602(z) = C E )((D.). In virtue of Lemma 2.4.12, (p2(zt, ) is a biholomorphic mapping
"Recall that yi(z))= z f8K(r) x`+ )dr;,zi ED.
2.4. Biholomorphisms of complete elementary Reinhardt domains in C2
201
from C * onto itself. Hence, rp2 (z) = f (zi )z:L' , f(z1) # 0. Then the uniqueness
of the Laurent expansion leads to cj = 0, j # ± 1, and c_ t = f, c I = 0 or c_1 = 0, c1 = f , zI E D. So we are led to the following shape of tp2, namely 1, pp2(z) = .f(z1)z2 z E D(i o), where f E O*(D*). In order to continue we observe that any domain D.*, a E Z;, al , a2 relatively prime, is biholomorphically equivalent to D(i,a). In fact, choose integers c, d such
that a1d - a2c = 1 and define the following mapping rp: C; -+ C' , V(z) (za, zi z2 ). Then V gives a biholomorphic mapping from D.* onto D(i,o).
Corollary 2.4.14. Let Da*, D;, a, f E Z2, where a I, a2, respectively 0 1, )62, are assumed to be relatively prime. Then Da* and DS are biholomorphically equivalent and a biholomorphic mapping is given by 4~C, where C E GL(2. Z) and flC = a.
Proof. Choose c, d, c, d E Z with at d - a2c = fl I d - 021E = 1. Recall that 'PA and "B induces biholomorphic mappings from D.* to D(i,o) and from D; to D(i,o), where A
cu
d
] and B := [fit d2], respectively. Then set C :=
B-1 A.
In the case where a E 71;, ala2 36 0, al, a2 relatively prime, we have the following description of the automorphism group of D.*.
Lemma 2.4.15. Let a _ (al, a2) E Z2* such that a 1, a2 are relatively prime. Then
Aut(D.) _ {TT o gf o OpI D. : t; E T2, f E 49 *(1D*), t P E GL(2, Z) with aP = a).12 Proof. Let X : D.* -+ D(*, o) be the biholomorphic mapping from above, i.e. X(z) =
'A (z) = W, zi z2 ), where c, d E Z with a I d -a2c = 1. Observe that X-1 (z) = (zd z2 °`2 ,
ziczZ at ). Then any automorphism (p of D.* can be written in the form
(P =X-1 o
o X, where i E Aut(D(*j o)). What remains is to apply Lemma 2.4.13
(EXERCISE).
Conversely, any map given in Lemma 2.4.15 belongs to Aut(D,*) (EXERCISE). 11
Now we turn to the irrational case, i.e. a = (a 1, 1), a 1 E IR \ (. Here the method of proof has to be changed; one has to use a covering argument.
Lemma 2.4.16. Let a = (a 1, 1), a I E R \ Q. Then
Aut(D:) = {0 A :
E C2 A E GL(2, Z) such that 'P ,A E Aut(D.)},
za2 i.2 yp2z'2 ' za2.2)In particular, any automorphism where 4s;,A(z) := {fit z°1.1 1 2 of DQ is an algebraic one. 12Recall that Op (z) _ (z zZ ,Z j z2). Moreover, note that g f is defined on Da*, a E Z2.
202
Chapter 2. Biholomorphisms of Reinhardt domains
In order to prove Lemma 2.4.16 we need the following auxiliary considerations. Put S2a := It (-= 1R2 : attt + t2 < 0} and T« := S2a + i IR2. Then we study the following holomorphic mapping
0: Ta -+
0('):=
Observe that (EXERCISE) 0 is a covering map, i.e. for any point z E D. there is a suitable open neighborhood UZ C Da such that 45-I (UZ) is a union of pairwise disjoint open sets Vj, j E Z, such that 0 1Vj is a biholomorphic mapping from Vj onto Ua. Moreover, it is easily seen that S2a and hence Ta is convex; in particular, it is simply connected. Therefore, Ta is the universal covering of Da (cf. [For 19811).
Recall a few properties of the universal covering 0: T. D.': For any simply connected domain D C C2, any holomorphic mapping
f : D -> Da, any point z° E D, and any point w° E T. with t(w°) _ f(z°) there exists a uniquely determined holomorphic function f : D -> T. with f (z°) = w° such that c o f = f . f is called the lifting off . We advise the reader to look into general books on topology for this result. In particular, for any pair of points w', w" E Ta with O(w) = O(w") there
is an f E Aut(Ta) such that 0 o f= 0 and f (w') = w"; f is uniquely defined. In fact, f is uniquely defined since it is the lifting of 0: T. -> Da. In virtue of the former property of the universal covering we find a holomorphic map f : T. -->
T. with f (w') = w" such that 0 o f = (P. We have to show that f E Aut(Ta). Changing the role of w' and w" we also have a holomorphic g : T. -+ T. with
(w") = w' such that 'P = 0 o g. Then 0 o (f o g) _ 0, 4i o (g o f) = Qi, o f (w') = w', and o g(w") = w". Using again the first property of the
f
universal covering we conclude that f o g = id ITa and g o f = id ITa. Therefore,
f E Aut(Ta). Moreover, it is easily seen that
Aut"(Ta):={* where a,,(z) := z + iq, z E Ta. Now we are going to apply the above lifting properties for a given q. E Aut(Da ). Then there is a lifting rp E Aut(Ta) such that 0 o Sp = (p o 0 (EXERCISE). Moreover, for a fixed q E Z2, 0 o Cr'? 0 0-I E Aut"(Ta). Therefore, we find an q' E 72 such that 0 o an = or,,, o gyp. It is easily seen that the mapping q --> q' leads
to a group isomorphism of Z2. Therefore, there exists a matrix P E GL(2, Z)
such that oa,,=alp00,gE7L2. So we are led to study the group of automorphisms of the domain Ta. We get the following lemma.
Lemma 2.4.17. Let (p E Aut(D.'). Assume that its lifting gyp: T. + Ta is a complex affine transformation, i.e. 0 (i;) = A + f, where A E GL(2, C) and
2.4. Biholomorphisms of complete elementary Reinhardt domains in C2
203
al.1'al.2,a2Za2.1Za2.2
E C 2. Then P is of the form rp(z) = = 0a,n(z). where A E 4L(2, Z), a = (a,, a2) a C2., i.e. rp is an algebraic automorphism ofDa'. Proof. From the discussion before we get
i; ETa,rlEZ2.
0
Using the form of cp it follows that A = P. Finally, the equality gP o 0 = ds o leads to the form of (p claimed in the lemma. After all these preparations we proceed with the proof of Lemma 2.4.16.
Proof of Lemma 2.4.16. Let V E Aut(D,x) and 0 E Aut(Ta) be its lifting. In virtue of Lemma 2.4.17 we have to show that 0 is a complex affine transformation. In
a first step we will show that there are a r E Aut(H-), an f E O*(H-), and an h E O(H) (lw- := {z E C : Re z < 0}) such that
(f (g*)gt +
where _ (6, 6) E T. and (to simplify notation) i * := at Sl + In fact, put tib : C2 - C2, *(%) := (i;1, i;*). Obviously, * is a biholomorphic mapping and 1r IT,, maps T. biholomorphically onto C x H -. Its inverse mapping is given by *-1(z, w) = (z, w - at z). Thus, Aut(Ta) = 1/r-1 o Aut(C x Oil-) o *.13 Moreover, let g: Of- -+ O) be any biholomorphic map. Then
g: C x 0h- -+ C x 0),
8(z, w) := (z, g(w)),
is also biholomorphic. So Aut(Ta) = 0: 1 o k-l o Aut(C x O)) o k o 1/r. Then, in virtue of Theorem 2.4.5 (a), there are f E O*(O)), h E 0(O)), and m E Aut(D) such that for i; E Ta we get
OW = 1/r-1 ° k-1 (f ° W*)6 + h ° and therefore, for
m ° W*)),
E Ta,
0(0 = (f°g(t*)St+hog(t*) g-lomogW)-at (f°g(t*)St+hoga*))), which proves the above claim with f := fog, h := h o g, and z := g-1 o m 0g. In virtue of Lemma 2.4.17 it remains to verify that cp is a complex affine mapping.
Indeed, in virtue of the properties of the covering mappings we have a matrix P = [ n s ] E GL(2, Z) such that for all pairs (k. e) E Z2 the following identities are true:
i k, y2 + it) = 13To be precise {b should be understood as {Gl rQ.
i (k, e) P,
i; E Ta.
204
Chapter 2. Biholomorphisms of Reinhardt domains
In particular,
+ik)+h(i;*+i(atk+l)) i(kp + er), r(r* + i (atk + e))
=
+ik)+h(i;*+i(a2k+t)) =
h(i;*)) + i(kq + fs).
From these identities we deduce that f is a constant function. In fact, fix a point wo E H-. Then, applying the first of the above identities for points (fit, wo - a t S t) E Ta, gives
f(wo + i (atk +
=
ik) + h(wo + i (atk + e)) h(wo) + i(kp + tr), t E C.
Hence, f (wo + i (at k + e)) = f(wo) =:Ao 0 0, k, e E Z. Recall that the number at is an irrational one. So the set (wo + i(atk + 8)) : k, j E Z} has an accumulation point in the plane. Then, in virtue of the identity theorem, it follows
that f = Ao. Applying the first of the above identities, we claim that h is a complex linear function. Indeed, the above identity implies that
i.lok+h(Z+i(atk+t)) = h(Z) + i(kp + fr) for all Z = * E H-. Differentiation in direction of Z leads to
h'(Z + i (atk + 8)) = h'(Z). Fixing some Z = Zo E H- we have h'(Zo + i (at k + e)) = h'(Zo) := µt . So h' is constant, i.e. h(Z) = µt Z + µo for a suitable µo. It remains to show that r is a complex affine mapping. In fact, using the second identity, we arrive at the following equality:
r(Z) - at
µt Z + µo) + i (kq + ts),
Z E H-.
Again differentiation gives r'(Z + i (atk + e)) = r'(Z), Z E H-. As above, fixing Z = Zo and using the identity theorem, we arrive at r' - r'(Zo) =: Ti 36 0 (recall that r is a biholomorphic mapping). As a consequence we conclude that r(Z) = rt Z + to for a suitable TO. Finally, rewriting gyp, we see that
OM =
+A I W) + µo, rt W) +ro -of 1 (Ao)t +111 W) + µo)).
2.4. Biholomorphisms of complete elementary Reinhardt domains in C2
205
or
µatk t EA°
at (riTI
t
A0 _ I
LIal)] + (120. To -
Obviously, the matrix is a non-singular one, and therefore, fP is complex affine. Hence, Lemma 2.4.17 gives the end of the proof. Now we return to the proof of Theorem 2.4.5.
Proof of Theorem 2.4.5 (b). Obviously, any of the given mappings belongs to Aut(DD). To prove the converse we will use the holomorphic Liouville foliation
where FF := {z E C2 : za = }. Fix a yP E Aut(DD). Then, applying Lemma 2.4.12, there exists a bijective mapping T : (D -+ ID such that rp(FF) = Fm),
E 0). Observe that the fiber F0 is the only one with a "singularity". So one concludes that r(0) = 0 (EXERCISE) which means that rplD defines an automorphism of D.*. Using Lemma 2.4.15 shows that
z E D.
w(z) = where
(yI, S2) E 72, f E O*(D ), and P = [; s ] E GL(2, Z) with
aP = a. Since aj E Z+, one concludes that P = II2 if ata2 # 1, and (P = II2 or p = s = 0, q = r = 1) if aI = a2 = 1, which gives the description of a. We will only discuss the case when a = id (the case when a = S may be taken as an EXERCISE). Observe that K := 1 L* x (1) C Da. Therefore,
(p2(z1, 1) = i;2(f(zz1 E
zL*, is bounded. Applying the Riernann theorem of removable singularities we see that f extends holomorphically to D. Taking into account that rp is bijective, it even follows that f E O*(U). Finally, a continuity argument leads to the description of rp on the whole of Da.
Finally, we discuss the case of normalized elementary Reinhardt domains of irrational type. Proof of Theorem 2.4.5 (c). Here weusethepshLiouvillefoliation (FF)rElo,I)from Example 2.4.11. Let rp E Aut(Da). Then there is a bijection r: [0. 1) -> [0, 1)
such that (p(FF) = Fy), t E [0, 1). In particular, F0 is homeomorphic to F,(o), which implies that r(0) = 0. So (pIDa E Aut(D,*,). Applying Lemma 2.4.16, cIDa is of the form
IP(z) _ (fit
z°,., Za,.2
t
2
z 2.2), 2 za2., t 2
zE
Da*,
where A = [ a2:; a2 .2 ] E GL(2, 7L). Observing that the coordinate axes belong
to Da, it follows that a'j E Z+ and, therefore A = II2, which implies Theorem 2.4.5 (c) (EXERCISE).
206
Chapter 2. Biholomorphisms of Reinhardt domains
Summarizing, Theorem 2.4.5 has been completely proved.
Now we turn to the proof of Theorem 2.4.7. We start with the discussion of bounded holomorphic functions on elementary Reinhardt domains. First, recall that an elementary Reinhardt domain D" of rational type carries at least one bounded holomorphic function that is not identically constant. On the other hand, we have
Lemma 2.4.18. Any bounded holomorphic function on an elementary Reinhardt domain of irrational type is identically constant.
Proof. Take an f E 3e°o(D"). Then If I E PSJ{(D"). Therefore, in virtue of Example 2.4.11, f IF, is identically equal to a constant st, where (Ft)tE[o.t) denotes the psh Liouville foliation from that example. Recall that
Ft= {zED":Iztl"1Iz21=t}. In particular, f(zt. 1) = st whenever Iztl"1 = t. Applying the identity theorem, it follows that st = s, t E [0, 1). Hence, f = s on D. Proof of Theorem 2.4.7. (a). In virtue of Lemma 2.4.18 and the remark before, it is clear that D" and DS are not biholomorphically equivalent. (b) Suppose that there is a biholomorphic map tp : D(o, t) -+ D. Then, using the holomorphic Liouville foliation (FF)tED of D. and (F )tE[) of D(t,o), respectively (see Example 2.4.9), there is a bijection r : ID -+ ID such that tp(FF) = F=(t), E D. In particular, tp I Ft, = F. Using that Fo has a singularity at (0.0) we get, as before, a contradiction. (c) Assume that D" and Dp are biholomorphically equivalent. In virtue of (a) there are two cases. Case 1: Assume that D. and Dp are of rational type. Suppose that a = (1, 0), then f = (1, 0) or P = (0, 1) and a biholomorphic map is given either by the identity
or by S. Therefore we only have to discuss the case where ala2 54 0 0 0102, a t , a2, resp. f t , i62, relatively prime. Take a biholomorphic map tp : D" -+ D'8. As in the proof before using holomorphic Liouville foliations we conclude that svl Da E Aut(D°*). Following Corollary 2.4.14 there is a biholomorphic mapping tJr : D; - D°* of the form t/r ='A, A E GL(2, 71) with det A = 1 and aA = P. Hence, , := tP o >(i E Aut(DS ). In virtue of Lemma 2.4.15, >1i may be written as
Vf(z)=Tt-gf-Op(z), zED*. Therefore,
tpIDa =TtoSforhpoOA-i = TToSfo4PA-1 and PA -I = a. Thus, Tt
' 0 (P(z) = ((f(z$))-112zr zZ, (f(zP))"1 zrz2),
z E D°*,
2.5*. Miscellanea
207
PA-1 where =: [ p s ]. Observe that T o (P defines a biholomorphic mapping from Da onto Do. Recall that the left-hand side is holomorphic on D. In particular, the functions
D. a l i-*
(f(A$1+82))-Q2Ap+q
and D. 3 X H
(f(AP1+52))S1Ar+s extend
holomorphically to D. Therefore, f has a pole at 0, i.e. f(A) = Ak f (A), A E lD*, where f E O*(ID) and k E Z. Hence, TTI
o
(p(Z) = ((f
(-10))-Q2Zp-kQ1fi2Z4-kfi
(f (-S) )S'
-1
+k, 1 `2+kfl I 02 ) .
Z E Da .
Finally, we define an automorphism of DS, namely,
X(z) := g1(z),
: E D.
Then, for : E D,,*,. we get
X o Ti o (P(Z) = (:Ip-kflr#2Z29-k#
r+kp _s+k0192 ).
Z1
..2
Taking into account that the mapping on the left-hand side is holomorphic on Da, it is easily seen that X : = X o TT o (p = id I Da or X = S I Da . Hence, Case I is verified.
Case 2: Assume that D. and DS are of irrational type, i.e. a = (&1, 1), P = (#1, 1), where a,, fl, r= IR+ \ Q. Applying psh Liouville foliations, one gets that cOI D : D." -> D6 is a biholomorphic map. Following the proof of Lemma 2.4.16, one may show (EXERCISE) that (PI Da = Applying now that the left mapping is holomorphic on Da, it follows that either (P(:) = TT(z) or co(:) _ TT o S(z) whenever z E Da. In the first case observe that ICI IS' 1R2I02 = 1. Then of-12 IDq E Aut(DS), 1, X21). Hence 0 where 0 (P = id on Da. A similar argument for the ><2
second case is left to the reader (EXERCISE).
0
Remark 2.4.19. An independent proof of Theorem 2.4.7 may be found in [Edi-Zwo 1999].
2.5* Miscellanea Besides the problem of biholomorphic equivalence of Reinhardt domains D1 , D2 C
C", one can try, for instance, to characterize all proper holomorphic mappings F: DI -+ D2.14 In the remaining part of this chapter we collect several results related to this area of problems. More precisely: "Recall that a mapping f : X -+ Y is proper if f -1(K) is compact for every compact K C Y. Every homeomorphism is proper.
208
Chapter 2. Biholomorphisms of Reinhardt domains
§ 2.5.1 Biholomorphic equivalence of Reinhardt domains. § 2.5.2 Automorphisms of Reinhardt domains. § 2.5.3 Proper mappings. § 2.5.4 Non-compact automorphism groups. In general, the methods of proofs of the presented results (based, for example, on the Lie theory or rescaling methods) are beyond the scope of the book. Nevertheless, we decided to put them here as illustrations of various streams of research. The results in this section may be also a starting point for further studies of the reader.
2.5.1 Biholomorphic equivalence of Reinhardt domains It seems that in the category of Reinhardt domains one has bih
alg
DI 'vD2#==> DI a-D2 (cf. Definition 1.5.12, Theorems 2.3.6, 2.4.7). Several particular cases are known (they were proved by methods based on the Lie theory). Q We like to point out that, unfortunately, we do not know any alternative methods of proof (without the Lie theory).
0
Theorem* 2.5.1 ([Sun 19781). Let D j C Cm be a bounded Reinhardt domain with blh D2 iff there exist rl , ... , rn > 0 and a permutation 0 E D j, j = 1, 2. Then D 1
a E Ch such that D2 = {(rl zv(1).... , rnza(n)) : (z1, ... , Zn) E D1 }. In particular, D1
bih
aig
D2 q D1 ^_- D2.
Observe that the case n = 2 was already discussed in Theorem 2.3.6.
Definition 2.5.2. For k = (kl, ... , ks) E DJs and p = (p l , ... , ps) E Quo, let s
IEk,P := t (z1 , .... zs) E Ck, X ... X Cks :
Ilzl ll2Pi
1}
j=1
be the generalized complex ellipsoid. In the case where k 1 = . . . = ks = I the generalized complex ellipsoid reduces to the standard complex ellipsoid Ip; cf. (1.18.5). Theorem 2.5.1 implies the following classification theorem for generalized complex ellipsoids (the case where p E W, q E W' was solved in [Naru 19681).
2.5*. Miscellanea
209
Theorem 2.5.3. Let Ek,p, Et,q C C" be two generalized complex ellipsoids with:
k=(k1....,ks)E Ns, I=(el....,et)E II`,
n=kt+...+kset+...+ft,
pl <...
n=
Eq if
p = q up to a permutation (cf also [Jar-Pfl 1993], Theorem 8.5.1).
0
Proof. Use Theorem 2.5.1 - EXERCISE.
Let D C C" be a Reinhardt domain satisfying the Fu condition. Recall (cf. Remark 1.5.11(a)) that, after a permutation of variables, we may always assume that: (*) there exists k = ap(D) E (0.....n) with D n V # 0, j = I..... k,
DnV =0,j =k+1,...,n.
Observe that if 0 E D. then a-(D) = n.
Exercise 2.5.4. Let T := { (z l , 22) E 02 : I z i I < 1z21 } be the Hartogs triangle
and let T* := T \ (10} x i)) = ((zl, z2) E Ifl2 : 0 < Izl l < Iz21). Observe that a-(T) = 1 and a(T *) = 0. Prove that T and T * are not biholomorphically equivalent.
Hint: Observe that T
n_n
ID x D* and T' 2b
0)
x m*.
Theorem* 2.5.5 ([Bar 1984]). Let D1. D2 C C" be bounded Reinhardt domains buy
satisfying the Fu condition with N. Then D1 D2 iff a(D1) = K(D2) =: k and D1, D2 are algebraically equivalent via a biholomorphism (P,,A such that 1
if
i
a;,i = 0 i f i
if i >k,
where a E Gk-
Observe that if ap(D1) = a-(D2) = n, then A(z) _ (zQtlt, ... , zat"1) for a a E G. Thus the above result generalizes Theorem 2.5.1. Theorem* 2.5.6 ([Shi 1988]). 7%vo bounded Reinhardt domains D 1, D2 C C" are biholomorphically equivalent iff they are algebraically equivalent.
Theorem* 2.5.7 ([Kru 1988]). Two hyperbolic (cf. § 4.7) Reinhardt domains D1, D2 C C" are biholomorphically equivalent if they are algebraically equivalent.
Notice that any hyperbolic Reinhardt domain of holomorphy is algebraically equivalent to a bounded domain (see Theorem 4.7.2), so, in fact, Theorem 2.5.7 follows from Theorem 2.5.6.
210
Chapter 2. Biholomorphisms of Reinhardt domains
Theorems 2.5.8 ([Sol 20021). Two Reinhardt domains D1, D2 C C2 are biholomorphically equivalent i¶ they are algebraically equivalent.
0 We do not know whether Theorem 2.5.8 remains true for n > 3. 2.5.2 Automorphisms of Reinhardt domains Theorem* 2.5.9 ([Naru 1968]). If P E N. then Aut(lk. ) 2 T". Theorem* 2.5.10 ([Lan 1984]). Assume that 0 < k < n > 2, p E { 1 }k x
W2-k
Then
Aut(lp) = {FH,t : H E Aut(IBk), t; E Tn-k),
(2.5.1)
where
FH.t (z)
n ,n 1 1- Ila'112 ) (1 - (zl. a'))21
yp-kl 1- pa'p2 +I (H(Z'), 4+t Zk+1f (1- (`,, a,))2 J
I
= (z', zk+1..... an) E IEP C Ck X
C"-k, and a'
2
:= H(0').
In particular, the group Aut(IEp) depends on k2 +k + n real parameters (cf. Ex-
ample 2.1.12(b)); ifk = 0, then Aut(IEp) = T" (cf. Theorem 2.5.9).
Remark 2.5.11. Notice that in general, for arbitrary pk+l, ... , pn E IR>o \ 11), the set {FH, : H E Aut(ISk). E Tn-k) is a subgroup of Aut(IEp) (ExERctsE); Cf. [Jar-Pfl 1993]. Lemma 8.5.2. Q We do not know whether (2.5.1) remains true.
0
Theorems 2.5.10, 2.1.20 and Lemma 2.1.21 imply the following
depends on d = n + Ek=1 mi (mj + 1) real parameters. For instance, for arbitrary P1, P2. P3, P4, Pj, P2' E W2, we have:
n = 2: d 2 4
(E(
I.p )
E(1.
6 8
k=2
k=1 )
ID2
82
2.5*. Miscellanea
211
n = 3:
k=2
k=1
d 3
E(
5
E(1,
)
ID x E( ,,,p,)
IDxE(1
7
9
k=3
E (, 1
) ID 3
)
IDx82
11
B3
15
n = 4:
d
k=1
4
E(PI.P2.P3.Pa)
'E(P,.
E(P
x E(1.P2.P3), E(I.P2) x IE(1.P') 10
IE(I.I.P3.P4)
I(P..P2) X B2
14 16
ID2 X E(I.P2) ID4
ID2 X 1112
IE(I.1.1.P4)
B2 X 1132
mxB3
18 24
ID2 X
IDXE(1.1.P3),E(1.P2)XB2
12
k=4
2) x IE(p.P')
x
6
k=3
k=2
84
Remark 2.5.13. Let D C C" be a hyperbolic Reinhardt domain such that Aut(D) depends on d real parameters. Recall that the group Aut(IB") depends on n2 +2n real parameters. There are the following general results (cf. [GIK 2000], [Isa 2007]):
If d > n2 + 2, then D = S. up to rescaling of variables. If D IB", then d E [n, n2 + 21 and d is of the same parity as n. d = n2 + 2 if D = ID x IBn_1 up to permutation and rescaling of variables. If d = n2, then D is algebraically equivalent to one of the following domains:
(i) {z E C" : r < IIzil < R}, 0 < r < R < +oo (cf. Exercise 2.1.13); (ii) ID3 (n = 3); (iii) (82 x(62 (n = 4); (iv) E(t ,,. p" 96 1 (cf. Theorem 2.5.10);
212
Chapter 2. Biholomorphisms of Reinhardt domains
(v) {(z',zn) E (n-1 x C : r(1 - IIz'112)a < Iznl < R(I - IIz'II2)*), 0 < r <
R<+00,aECR; (vi) {(z'. zn) E C" x C : re°fIZ'12 < Iz,, I < Re°COZ'02}, 0 < r < R < +oo,
aEIR.and(R=+oo= a>0). Some intermediate cases where n2 < d < n2 + 2 are also discussed in [GIK 2000] and [Isa 20071.
For p = (p l , triangle
.... p,,) E Rn and l < s < n -1, define the generalized Hartog s s
n
II2p' <1}.
Fp,s:={(zl..... zn)EC":1:Izjl2p' < j=s+I
j=1
If n = 2, then 0(I,I),I is the standard Hartogs triangle (cf. Remark 1.5.11 (c)).
Theorem* 2.5.14 ([Lan 1989]). Let 0 < k < n - 1, p E { 1)k x N"-k. Then
Aut(Fp,n-1) = {FH, : H E Aut(Ik).
E T"-k},
(2.5.2)
where, for z = (z', zk+1.... zn) E [Fp,n-1 C Ck x C"-k and a' := H-1((Y), we put FH,t(z)
Zn" H(z'/zn") yk+lZk+l }}
G-
1k+1 Z'/Zn".a'))2) i /
}y
2n p=
112
(1 - (z'/zn
a'))2 /
.
.
nZn).
In particular, the group Aut(Ip,n_1) depends on k2 + k + n real parameters. Remark 2.5.15. (a) To prove that FH, t (FF,n _ 1) C F,,,,,_1, note that n-l
I(FH,t)n(Z)I-2p" >2 I(FH,C)j(z)I2Pj j=1 -(z'/l oll
=
a') 12
Il
> IZj12p' j=k+1
n-1 -1-(t-IIU'II )(1-Ilz'/-n"112)+IZ"I-2pn 1-IIa'II2 I1-(z'/:, .a')12 I1-(z'/.nPM ,a')12j= I Pn
-
1
I1 _ (`,/
a')" I2 \1
1:
- IznI-2p"(IIz'II2 + j=k+1
I
I2p')).
Izj l2p'
213
2.5*. Miscellanea
(b) Observe that in general, for arbitrary pk+1,... , pn-l E IR,o \ (11, the set
{FH, : H E Aut(IBk),
E T"-k}
is a subgroup of Aut(lFF,n-i) (EXERCISE). 2 We do not know whether (2.5.2) remains true.
0
Theorem* 2.5.16 ([Che-Xu 2002]). Let 2 < s < n - 2, 0 < k < s, 0 < f < n - s,
pE(1)kxN2
Then
Aut(IFp.s) = {FH',H", : H' E QJ(k). H" E QJ(f).
E
T"-k-t}.
where, for : = (z', zk+l ..... Zs. Z" Zs+t+l .... , Zn) E Fp,s C Ck x
(2.5.3) Cs-k X CI X
C'-'-t, we put FH',H".l; (Z) := (H'(z'), 4+1Zk+1.... , szs, H"(Z" ), Ss+t+I s+t+1, ... , nzn). In particular, the group Aut(IFp,s) depends on k2 + f2 + n - k - f real parameters.
Remark 2.5.17. Observe that in general, for arbitrary pk+1 pn E R,o \ { 1 }, the set { FH,C : H E AUt(IBk ).
is a subgroup of true.
(EXERCISE).
0
. Ps. Ps+t+1.
.
E Tn-k }
0 We do not know whether (2.5.3) remains
2.5.3 Proper mappings Theorem* 2.5.18 ([Bar 1984]). Let D1. D2 C C" be bounded Reinhardt domains satisfying the Fu condition. Then any proper holomorphic mapping F : D I - D2 extends holomorphically to a neighborhood of D1.
Theorem* 2.5.19 ([Lan 1984]). Assume that n > 2. For arbitrary p. q E N" the following conditions are equivalent: (i) there exists a proper holomorphic mapping F :
(ii) (PI/ql.
IEp
Eq;
Pn/gn) E N".
Moreover, any proper holomorphic mapping F : lE p -+ IEq is, up to an automorphism of IEq, of the form
F(z) =
(zr"
.
Znnl4n ).
Z = (ZI ..... Z") E Ep.
In particular, any proper holomorphic mapping F : IEp -+ IEp is an automorphism (see [Ale 1977] for the case E p = IBM ).
214
Chapter 2. Biholomorphisms of Reinhardt domains
Remark 2.5.20. The implication (ii) (i) is obvious and remains true for arbi p, q E Ut;o. Q We do not know whether the implication (i) (ii) remains true.
Theorem* 2.5.21 ([Lan 1989]). (a) If n following conditions are equivalent:
-
3, then for arbitrary p. q E N" the
(i) there exists a proper holomorphic mapping F : (ii) A It E N : s j := (eqn - Pn)/qj E Z rj Pjlgj EN, j =1.....n-1.
(FP,,,-1
+ Fq,n-1;
j = 1,.. ,n - 1 } # 0 and is, up to an
Moreover, any proper holomorphic mapping F: IFP,n_1 automorphism of E9,,,_ I, of the form
F(z) F( _ (ZI z n . , ... z n-1 '
zl). e E A.
(b) If n = 2, then for arbitrary p, q E W2 the following conditions are equivalent:
(i) F : Fp,1 -> Fq,l is a proper holomorphic mapping;
(ii) F(zl,z2) if P21PI
=
W.
eq2/ql - kP2/PI E Z, (6 zZ9z/4jB(zl z2 D2/Pl ). 2ZD if P2/PI E W. eq2/ql E W.
where S I S2 E T, k, e E N. and B is a finite Blaschke product.
Remark 2.5.22. (a) In the case n > 3 the implication (ii) = (i) is elementary and remains true for arbitrary p. q E IRao. Indeed n-1
n-1
IFn(Z)I-2q,, E IFj(Z)I2qj
j=1 n-1
1:IZn
=IZnI-2tq.
EIZnl2sjqjIZlI2r'qj
j=11
n-1 12(fgn/qj -Pn/9j )qj -2tgn
1Zj 12Pj = IZn I-2P" E IZj I2P' .
j=1
j=1
Observe that F is biholomorphic if e = r1 =
= rn_ I = I iff p j = q j .
j = L.... n - 1, and (pn - qn )/ p j E Z. j = 1, ... , n -
1.
In particular,
there are p,q E N" such that IFP,n_1 0 Fq,n_I but tP,n-1 (Fq,n_I. Take for instance p i = qj, j = 1, ... , n - 1, and pn 0 qn such that (pn - qn )l Pi E Z. n - 1.
0 We do not know whether the implication (i) = (ii) remains true.
(b) In the case n = 2 the implication (ii) for arbitrary p, q E R2>01 Indeed
(i) is elementary and remains true
IF2(z)I-292IF1(Z)I291 = (Iz1IIz2I-P2/PI)2kg1 IB(ZIZ2P21P1)I2gl
if p2/PI 0 N, ifp2/PI E W.
2.5*. Miscellanea
215
Observe that F is biholomorphic if
£ = k = 1, q2/q - P2/P1 E 7L f= 1, B E Aut(D),Q2/q1 E N
if p2/P1 0 W. if p2/p1 E W.
0 We do not know whether the implication (i) = (ii) remains true. 0 Theorem* 2.5.23. Assume that p E W", 2 < s < n - 2. (a) ([Che-Xu 2001 ]) The following conditions are equivalent:
(i) there exists a proper holomorphic mapping F : [Fp,s -* Fp,s:
(ii) there exist permutations or E G5 and 8 E G"_S such that pa(j)/pj E N, j = 1, .. ..s, Ps+8(k)/Ps+k E N, k = n - s. (b) ([Che-Xu 2002]) Any proper holomorphic mapping F : 1Fp,s -+ Op,s is an automorphism (cf Theorem 2.5.16). Let V E e°0([0, 1], 1R+) be such that there exists an h E (0, 1) for which (PI[a,h] = 0, co(t) = 1,
(p'>0and cp">0on [0,1], gyp'>0andlp">0on(h,1). Define D,e,h :_ ((z1, Z2) E D2 : IZ1 I2 + Oz2I2) < 1).
Exercise 2.5.24. (a) Dp,h is a normalized (cf. (2.3.1)) bounded pseudoconvex complete Reinhardt domain with (8D) x (h FD) C dD,e,h.
(b) D,y,h 0 {D2, IE(l,a), E(a,,), a > 0). Consequently, by the Thullen Theorem 2.3.6, every biholomorphic mapping F : D9)1 ,h D,e2,h2 is of the form
,-
bih
F = Ti for some ' E T2. Hence, D(p1.h1 Dm2,h2 iff hl = h2 and cO1 = c02 (c) D9,,h is strongly pseudoconvex at a boundary point a = (a,. a2) E BD,P,h if Ia2l > h (cf. § 1.18*). In particular, the set of weakly pseudoconvex boundary points is not contained in Yo.
Theorem* 2.5.25 ([Lan-Pat 1993]). The following conditions are equivalent: (1) there exists a proper holomorphic mapping F : D,,, ,h, -> D02,h2; (ii) there exist m E N and ti 1. 6 E T such that: t E [0. 1], h2 = hm, (p,(t) = F(c) =1 (6z1,6z2 ), Z = (zI,:2) E Dwl,hi. p2(tm),
Theorem* 2.5.26 ([Lan 1994]). Let D C C2 be a bounded smooth pseudoconvex complete Reinhardt domain whose weakly pseudoconvex boundary points are contained in Yo. Then any proper holomorphic mapping F : D -> D is an automorphism.
216
Chapter 2. Biholomorphisms of Reinhardt domains
Theorem* 2.5.27 ([Lan-Pin 1995]). Let D1, D2 C C2 be bounded pseudoconvex complete Reinhardt domains such that there exist a complex analytic variety W and an open neighborhood U of a point a c- 8D 1 such that W fl U c 8D 1. Then any
proper holomorphic mapping F = (Fl, F2): D1 -- D2 is such that Fl and F2 depend only on one variable. Moreover, if D1 = D2 is not a bidisc, then F has the form
F(zl,z2) = (6Za(1),S2za(2)), where Sl, 6 E T, Cr E 62 Theorem* 25.28 ([Ber-Pin 1995], [Lan-Spi 19961, [Spi 1998]). Let D1, D2 C C2 be bounded complete Reinhardt domains such that at least one of them is neither a bidisc nor a complex ellipsoid. Then any proper holomorphic mapping F : D1 -> D2 has the form F(z1, z2) = (cl a(i), C2 0(2)
where c3 E C, mj E W, j = 1, 2, a E 62. Moreover, if D1 = D2, then F has the form F(z1, z2) = (6Za(1), S2Za(2)),
where 1,1;2ET,QE82. Remark 2.5.29. The full description of proper holomorphic mappings F : D,
D2, where D1, D2 C C2 are bounded Reinhardt domains, may be found in [Isa-Kru 2006].
For the case of proper holomorphic mappings between unbounded Reinhardt domains we mention the following result.
Theorem* 2.5.30 ([Edi-Zwo 1999]). Let a, f E Z and let
C2JDaF ) DpCC2 be a proper holomorphic mapping. Then
F(z) _ (H 11,01 (za)z21, zi2 H-1/fl2(za)). where H E O(D,C.), E T, and kI,k2,E1,E2 E Z+ are such that a2181k1 = aif2k2, compare with Theorems 2.4.5, 2.4.7.
alrllel = a2182e2:
2.5*. Miscellanea
217
Theorem* 2.5.31 ([Din-Pri 1987]). Let D C C" be a Reinhardt domain and let
F: D -+ IEp be a polynomial proper mapping with p E N". Then F(z) _ (zit , ... , zn") with d l , ... , d" E W, up to action of T" on D and an automorphism bih of Ep. 'S Moreover, D IEq with qj := dj pj, j = I.... , n. Theorem* 2.5.32 ([Din-Pri 1988]). Let D C C" be a Reinhardt domain with 0 E D and let p r= N". Then the following conditions are equivalent:
(i) there exists a proper holomorphic mapping F : D -+ lip; (ii) there exists a proper polynomial mapping F : D -+ CEp.
Theorem* 2.5.33 ([Din-Pri 1989]). Let D1 C C" be a Reinhardt domain and let D2 C C" be a bounded simply connected strictly pseudoconvex domain with
C' boundary. Then any proper holomorphic map F : D 1 -+ D2 is, up to an automorphism of D2, of the form F(z) = (zi' , ... , zn") with dl .... , do E N. Remark 2.5.34. For a = (al, a2) E (IR2)* and 0 < r- < r+ < +oo let Da,r-,r+ := {(Z1,z2) E C2(a) : r- < Izal < r+}. Recently L. Kosinski [Kos 2007] gave the full characterization of all proper holo-
morphic mappings F : Da,r-,r+ -+
Pr :='4(1/r, r),
D$,R-,R+. More precisely, let
Dy,r := {(Z1, z2) E C2 : 1/r < IZIIIZ2Iy < r},
yEIR\Q,r>1. One may prove (EXERCISE) that Da,r-,r+ is algebraically equivalent to a domain of one of the following three types:
PrxC,
ifala2=0,
Pr X C,.,
if a2/al E Q,,
Dy,r,
if Y := a2/a1 $ Q.
(2.5.4)
If D 1, D2 are of type (2.5.4), then there are no proper holomorphic mapping F : D 1 -+ D2 except for the following four cases: (1) D, = Pr xC, D2 = Prm XC (m E N), F(z) = (l;zf"'. P(z)), where l; E T.
s E {-I, I}, P(z) = E50P(z1)z, N E N, Pp,... , PN E O(Pr), P(z1, ) const, z1 E P. (2) D1 = Pr x C., D2 = Pr- X C (m E N), F(z) = z2 k P(z)), where E Ir, s E (-1, 1), P(z) _ FN O Pi (zl)z2, k, N E N, 0 < k < N, Po,..., PN EO(Pr),Ek=OIPi(Zl)I >0,FNk+lIPi (Z1)I>0,Z1 E Pr. '5That is, there exist
ZED.
E T" and 0 E Aut(IEP) such that df o F o Tt(z) = (z d' , ... , zd,"),
218
Chapter 2. Biholomorphisms of Reinhardt domains
(3) D1 = Pr X C., D2 = Prm X C. (m E W), F(z) =
zig(zl)), where
ET,eE{-1.1),kEZ.,gE(9*(Pr).
(4) D1 = Dr,r, D2 = Da,R with tog R
=
kl + 18.
l + .e28 } tog r = k2
for some k = (k1, k2), t = (e 1, e2) E Z2, F(z) = (azsk, bzse), E E {-1,1}, a,b E C, lallbl8 = 1.
2.5.4 Non-compact automorphism groups Theorem* 2.5.35 ([Bed-Pin 1998] (see also [Bed-Pin 1988])). Let D C C2 be a bounded domain with real analytic boundary such that Aut(D) is non-compact.t6 Then
D 2h (l,m) = {(z1, Z2) E C2 : 12112 +
11,
where m E W. In particular, D admits a proper holomorphic mapping onto 132.
We say that a bounded domain D C C" with smooth boundary is of finite type if there exists an m E N such that for every point a E aD and for every complex one-dimensional manifold V passing through a, the order of contact of aD and V does not exceed m, i.e. for any a E 8D and (P E O(1D, C") with (p(0) = a we have ordo(u o (p) < m for any local defining function u: U -+ Q( defined in a neighborhood U of a with gp(m) C U. Notice that Theorem 2.5.35 remains true if D is a pseudoconvex domain with smooth boundary of finite type.
Theorem* 2.5.36 ([Bed-Pin 1991]). Let D C C"+1 be a bounded pseudoconvex domain with smooth boundary of finite type such that Aut(D) is non-compact. Assume that the Levi form of a defining function of D has rank at least n - 1 at each boundary point. Then X (C
bih
{(Z1..... Zn, W) E C"
D
n
: 1w12m + E iZj 12 < 1 }. J=1
where m E N.
Theorem* 2.5.37 ([Bed-Pin 1994]). Let D C C"+1 be a convex bounded domain with smooth boundary of finite type such that Aut(D) is non-compact. Then there exist m 1, ... , m" E N and aa,p = apq,a E C such that bih
D = {(z. W) E C" X C : 1w12 + I: aa,pza $ < 11. 'That is, there exist a point a E D and a sequence
C Aut(D) such that
8D.
2.5*. Miscellanea
219
Qwhere the sum is taken over all a. # E Z with a 1 / m 1 + ...+a,, /m = 1 and
N1/ml +-+Pn/mn = 1. Theorem* 2.5.38 ([FIK 1996a]). Let D C Cn+1 be a bounded Reinhardt domain with C°O-smooth boundary such that Aut(D) is non-compact. Then D
Nh
{(z, w) E C" X C : Iwi2 + P(IZI I..... IZnI) < 1}.
up to permutation and rescaling of variables, where P is a non-negative polynomial with real coefficients.
The case where 3D is only of class Ck was solved in [Isa-Kra 19971 - in this case P is a non-negative Ck-function.
Theorem* 2.5.39 ([Isa-Kra 1998])). Let D C C2 be a hyperbolic Reinhardt domain with Ck-smooth boundary (k > 1) such that D fl Yo # 0 and Aut(D) is non-compact. Then D is algebraically equivalent to one of the following three types of domains:
{(z1,z2) E C2 : IZ1I2 +
Iz2I2,
< 1), wherem < 0orm > k or In E M;
{(zl,z2) E ID x C : t1-I 17J.- < IZ2I < (I_Ia }. where I < R < +oo, >
a0;
Q Q {(z1.z2) E C2 : exp(NIz1I2) < IZ21 < Rexp(iBIZ1I2)}. where I < R < +00, ,8 E IR. and (R = +oo =" > 0).
Remark 2.5.40. (a) Some of the above results may give the impression that every domain with non-compact automorphism group is biholomorphic to a Reinhardt domain. This is not true - the following bounded pseudoconvex circular domain with real analytic boundary and non-compact automorphism group is not biholomorphic to any Reinhardt domain ([FIK 1996b]): {(zl. Z2, Z3) E C3 : IZ1 I2 + IZ2I4 + IZ3I4 + (52Z3 + Z3Z2)2 < I }.
(b) In [KKS 20051 the reader may find a characterization of those "analytic polyhedra" in C2 whose automorphism groups are not compact. (c) For general domains with non-compact automorphism groups the reader may contact the survey article [Isa-Kra 1999].
Chapter 3
Reinhardt domains of existence of special classes of holomorphic functions 3.1 General theory Let D be a Reinhardt domain of holomorphy and let 8 C 0 (D) be a natural Frrchet space (cf. § 1.10). e.g. 8 = ,7e-A(D), Ak(D), Lh.k(D), O(N) (D), O(o+)(D);
cf. Example 1.10.7. Our aim is to find geometric characterizations of those Reinhardt domains D which are 8-domains of holomorphy. We like to point out that such geometric characterizations are not known for more general classes of domains (e.g. balanced domains of holomorphy). Except for § 3.1, all results presented in this chapter are more elaborated and detailed versions of some results from [Jar-Pfl 2000], § 4.1.
Remark 3.1.1. Consider the case where 8 = O(N)(D) (Example 1.10.7 (f)). (a) First recall some known general results. Let G C C" be a domain of holomorphy (Reinhardt or not). Then: ([Jar-Pfl 2000], Corollary 4.3.9.) G is an Ot2n+6)(G)_domain of holomorphy for any e > 0. ([Jar-Pft 2000], Corollary 4.3.9.) If G is a bounded domain, then G is an 00+e)(G)-domain of holomorphy for any e > 0. ([Jar-Pft 2000], Theorem 4.2.7.) If G is bounded and fat, then G is an L2 (G)-domain of holomorphy; in particular, in this case G is an 0(")(G)-domain of holomorphy (cf. Example 1. 10.7 (c), (f), (g)). We do not know whether the above results are optimal, e.g. whether there exists a µ < n such that every bounded fat domain of holomorphy is an 0 (A) -domain of holomorphy. (b) In contrast to the above general situation, in the case where D is a Reinhardt domain of holomorphy, we are able to show that: D is an 0(') -domain of holomorphy (Theorem 3.4.4).
0
If D is fat, then D is an 04)-domain of holomorphy for any e > 0 (Theorem 3.4.3).
The following notion will be useful in the sequel. Definition 3.1.2. Let D C C" be a Reinhardt domain. We say that a natural Frdchet space 8 C 0(D) is regular if for every function f E 8 with the Laurent expansion
f(z) _
afza, z E D. arez"
3.1. General theory
221
we have:
Z" E $,a E E (f) _ (a E Z" : a f # 0), the set {at za : a E E (f) } is bounded in $ (cf. Definition 1.10.3). Remark 3.1.3. Observe that there are natural Frdchet spaces which are not regular.
For example, 8 := C f , where f E 0(D) is not a "monomial" of the form cz". Example 3.1.4 (Examples of regular natural Frdchet spaces). (a) 8 = O(D). Indeed, by the Cauchy inequalities, for any Reinhardt compact set K C D we
have Ilafz"IIK < IlfIIK,a E E(f). (b) $ = 3e1°o°(D).
Indeed, by the Cauchy inequalities, we have Ilafzal1B(r)nD < III IIB(r)f1D
aEE(f),r>0.
(c) $ _ MI(D). Indeed, by the Cauchy inequalities we have II a f za lID < II f IID, a E E (f ).
(d) $ = O(N)(D) (N > 0). Indeed, the function 8D is invariant under n-rotations. Hence, using once again the Cauchy inequalities, for r E D fl R'O, a E E (f ), we get
sD(r)Iafral <8D(r)IIfllaor(r) = IISDIIIaor(r) < II8DfIIDThus Ilspaaz`IID II8D.fIID,a E E(f). (e) $ = LP (D) (1 < p < +oo) (cf. Example 1. 10.7 (c)). (f) 8 = .4(D) in the case where D satisfies the Fu condition. _ Indeed, in virtue of (b), we only need to observe that z" E C(D), a E E(f) (EXERCISE).
Remark 3.1.5. (a) Let $j be a natural Frdchet space in 0(D) with the topology T(Qj) generated by a countable family Qj of seminorms, j E W. Consider the space 8 := njE. 8 with the topology generated by the family Ujt I Qi 1, 8. We know that 8 is also a natural Frdchet space in 0(D) (Remark 1. 10.7 (h)). Observe that if each space $j is regular, then 8 is regular. In particular, the spaces L*(D), 0(O+)(D) are regular. (b) Let A C Z11 , 0 E A, and suppose that $ is a natural Frdchet space in O(D) with the topology given by a countable family of seminorms Q.
vEA. Define WA:={f E0(D):D°f E$,,, vEA). We know that,8A is a natural Frdchet space with the topology generated by the family of seminorms $A E) f H q(D°f ). V E A, q E Q (Remark 1. 10.7 (i)). Observe that if each space $ is regular, then $A is regular. as z", z E D, then Indeed, if f (z)
D°f(z) _
D%(af za) _
aEE(f)
(-V)960 v!(,,)aa z"aEE(f),
z E D.
222
Chapter 3. Reinhardt domains of existence
Hence, for every v E A, the set { D °(af za) : a E E (f)) is bounded in 8, which implies that the set {a, za : a E E(f )} is bounded in $A. In particular, the natural Frechet spaces J(oo.k(D), je.Z'k(D), Lh'k(D) and L," (D) (k E N U {oo}) are regular. Moreover, the space Ak (D) is regular provided that D satisfies the Fu condition.
Proposition 3.1.6. Let 0 0 D
C" be a fat Reinhardt domain and let $ C O(D) be a regular Frechet space. Then the following conditions are equivalent-
(i) D is an 8-domain of holomorphy; (ii) there exists an f E 8, f (Z) = F-aEZ" as za, z E D, such that the set E (f ) is unbounded and
D = {Z E C"(E(f )) : v`(Z) < 1), where v(z) := lim sup IaazaItilal, Z E G "(E(f)); (ii') there exists an f E 8, f (Z) = F-aeZ" of za, z E D. such that the set E (f ) is unbounded and 00
D = U (int
n
U=1
aEE(f):lal>v
{z E C" (E (f )) : Ia f za I < 1));
(iii) there exist an unbounded set E C (Z"). (E C (Z n), if 0 E D) and (ca)aEE c IR>o with (1.15.1)1 such that:
- D = {z E C"(E) : v`(z) < 1), where v(z) := limsup ,cazaItllal, Z E C"(E), za E 8, a E E, and the set {caza : a E E} is bounded in $;
(iii') there exist an unbounded set E C (7L"). (E C (71+), if 0 E D) and (ca)aeE C P>o with (1.15.1) such that. - D = U0,°=1 (lnt naEE: Ial>v{Z E (C"(E) : CalZal < 1}),
- za E 8, a E E, and the set {caza : a E E} is bounded in $;
(iv) for every point a E C \ D there exist sequences (a(k))k 1 C (Z").
((a(k))0 C (7L+). if 0 E D) and (d(k))k 1
, C IR>o such that:
- a(k)I -> +oo,
- DCC"(E),where E:={a(k):kE N},
- Za(k) E 8, k E DJ, and the set {d(k)za(k) : k E DJ} is bounded in $, - d(k)Iaa(k)I -> +oo. 'That is, sup(c,,,"«' : a e E) < +oo.
3.1. General theory
223
Observe that condition (iii') gives an effective geometric characterization of $-domains of holomorphy. Notice that the result need not be true for non-regular natural Fr6chet spaces (cf. Remark 3.1.3). Proof. The equivalences (ii)
' (ii') and (iii) q (iii') follow from Lemma 1.15.13. (ii): By Proposition 1.11.11, there exists an f E 8 such that D is the domain of existence of f. Let f (z) = LaEZn of za, z E D, be the Laurent (i)
expansion off . By Proposition 1.11.6 the domain of convergence V y of the above series is a domain of holomorphy. Thus D = Df. In the case where E(f) is finite we have D f = C" (E (f) ), which contradicts our assumption that D C" is fat. Now, it remains to use Proposition 1.15.15.
(ii) = (iii): Put E := E (f)*, ca := laa 1. The regularity of 8 implies that the set {af za : a E E} is bounded in 8. _ (iii) (iv): Fix a point a E C; \ D. By Lemma 1. 15.13 (h), v(a) = v* (a) > n > 1. Thus, there exists a sequence (a(k)) t C E such that la(k)l -> +oo and la(k)I -> +oo, k -, +00. (iv) (i): Suppose that D is not an 8-domain of holomorphy and let Do, D be as in Proposition 1.11.2(*). Since D is fat, we may assume that D C C; and that there exists a point a E D \ D. Let Q = (q; : i E N) be a countable family of seminorms generating the topology of 8 with q; _< qi+t, i E W. Since 8 is a natural Frechet space, the extension operator 8 g H g E O(D) is Ca(k)Iaa(k)I
continuous (Remark 1. 11.3 (n)). In particular, there exist C > 0 and io E W such that Ig(a)I < Cq;o(g), g E 8. Since the set {d(k)za(k) : k E W) is bounded in 8, there exists a constant M > 0 such that q;o (d (k)za(k)) < M, k E W. In particular, d(k)Iaa(k)I < CM, k E W; a contradiction.
Proposition 3.1.7. Let 0 0 D C C" be a Reinhardt domain and let 8 C MOO (D) be a natural Banach algebra which is moreover regular (e.g. 8 = with the norm III 118 := 2k max{ 1 1 D f 11D : 1 v 1 < k); cf. Example 1.10.70)). Then the
following conditions are equivalent:
(i) D is an 8-domain of holomorphy; (ii) there exists an f E $, II f 11,8 < 1 II f 11D < 1, f (z) _ such that the set E (f) is unbounded and
aEzn as za, ZED,
D = {ZEC"(E(f)):u*(Z)<1), where u(z) := sup{laf zal'/IaI : (Y E E(f)*}, Z E C"(E(f)); 01
(ii') there existsan f E 8, 11f 118 _ 1, 11f II D < 1, f(z) = EaEZn of Za, ZED, such that the set E (f) is unbounded and
D = int n {z E C"(E(f)) : laf zal < 1); aEE(f )
224
Chapter 3. Reinhardt domains of existence
(iii) there exist E C (Z")* (E C (Zi.)* if 0 E D) and (ca)aEE C lZ>o with (1.15.1) such that:
D = {z r= C"(E) : u*(z) < 1}, where u(z) := sup{Icazalt/Ial : a E
i }, Z E C"(E),
z" E 8 and IIcaZ" II8
1, a E E;
(iii') there exist E C (Z")* (E C (Z"). if 0 E D) and (ca)aEE C IR>o with (1.15.1) such that.
D = intnaEE{Z E C"(E) : calz,I < 1},
ZaES and llcazall8<1,aEE. Proof. The equivalences (ii) * (ii') and (iii) 4 (iii') follow from Lemma 1.15.13.
(i) = (ii): There exists an f E 8 such that D is the domain of existence of f (Proposition 1.11.11). We may assume that 11f 118 < 1, 11f 11D < 1. Since f is not holomorphically continuable beyond D, D coincides with the domain of convergence of the Laurent series EaEZn at Z" of f . Observe that E (f) is unbounded (because D is fat (cf. Corollary 1. 11.4 (a))). It remains to apply the second part of Proposition 1.15.15. The implication (ii) = (iii) follows from the regularity of 8 (cf. the proof of Proposition 3.1.6). (iii) (i): Notice that, by (iii'), D must be fat. Suppose that D is not an $-domain of holomorphy and let Do and D be as usual. We may assume that
D C C" (E ). Since 8 is a natural Banach algebra, we have IIS II p s II S II S, g E'8 Ilcazal1Dlal < 1, a E E. Hence u* < 1 on D. (Remark 1. 11.3 (n)). In particular,
Since 0 # Do C D fl D, the maximum principle implies that u* < 1 on D. Thus D C D; a contradiction.
Corollary 3.1.8. Let D
C" be a Reinhardt domain and let k E Z+. Then the
following conditions are equivalent: (i) D is an 3e°O.k-domain of holomorphy; (ii)
D = int n Da,c(a), aEE
where E C (Z")*, c: E -* R, and 2kf!I(0)le-c(a)IIZa-$IID
In particular, Da,c (a 0 0) is an 3f
1,
aEE. I#I
,k -domain
of holomorphy i,ff k = 0 and
aEIR.Z". Exercise 3.1.9. Prove that 3f oo,k (Da,c)
C fork > 1.
3.2. Elementary Reinhardt domains
225
3.2 Elementary Reinhardt domains This section is devoted to the most elementary case where D = Da,c is an elementary Reinhardt domain. The reader should consider the results below as an illustration of problems we will meet in the sequel.
Theorem 3.2.1 ([Jar-Pfl 1987]). Let D := Da,c be an elementary Reinhardt do(R")*.2 Then: main with a E (a) For any N > 0 the domain D is an O1Nl -domain of holomorphy. (b) For every k E 71+ the domain D is an .k -domain of holomorphy. (c) The following conditions are equivalent:
(i) D is an N'-domain of holomorphy; (ii) D is an 0(0+)-domain of holomorphy; (iii) a E LR Z".
(d) If a f I Z", then 3f°O(D) ^_- C.
(e) If a = (a1 .... , a") E 71" and al , ... , an are relatively prime, then the operator X°O(L)) 9 g H g E MOO(D),
where g(z) := g(e`za), z E D, defines a Banach algebra isomorphism
(f) If a E IR Z", then 3P°O(D) 010+1(D) and, consequently, J°O(D) is of the first Baire category in 0(0+)(D). (g) LP (D) = (0), so D is never an LP -domain of holomorphy, 1 < p < +oo.
Observe that the only problem in (e) is to prove that ! is surjective. Indeed, since m* C (e-`za : z E D), we see that U 11D = IIgII,). Proof. We may assume that a E Rn (EXERCISE). Moreover, we may assume that a,, . . . . . . . . . . . . . . . . aan, , < 0 for some 0 < s < n. (a) Fix an N > 0 and suppose that D is not an O(N)-domain of holomorphy. Let D0, D be as in Proposition 1. 11.2 (*) with 8 = O(N)(D). Since D is fat, we
may assume that D C C . Put a := N/(3n). By the Kronecker theorem (cf. p. 97), there exist sequences
j
(gv)0O_I C QJ
-q,..Iaill < s,
j = 1,...,n, q -> +oo.
(pi") 001 C QJ, such that IPi,v 'That is, D
C". Recall that D is a fat log-convex domain with log D = Ha.c.
226
Chapter 3. Reinhardt domains of existence
W e may assume that y > s(l la l + - - + 1 /as ), v = 1, 2, .... Put gv(Z) := a-q, cZI I.v ... ZPs.v
.
... Zn Pn.v
Zs+1-+l.v
Z E C"(a).
Observe that Igv(Z)l1/4,' _ a-'IZa1 =: 9(z),
ZE
C" (a) = C' X
Suppose f o r a moment that we already know that 8^D' Ig,, I < 1 ,
*-s
v = 1. 2, ... .
Then, by Remark 1.11.3(n) (with 8 = O(N)(D)), for every compact k C D there CK, v = 1.2,.... Hence 9(z) < 1 exists a constant Cg > 0 such that Ilg, II K for z E K and, consequently, 9(z) < 1 for z E D. The maximum principle for the plurisubharmonic function 0 gives 0 < 1 on D. which implies that D C D; a contradiction. We move to the proof of the estimate 8D (z) Ig (z) I _< 1, z E D, v = 1.2.... .
Fix an a E D. We may assume that a $ V0 (EXERCISE). Let q := 9(a) E (0, 1).
For j E {1,...,n}, put j := (a1,...,aj-1,q 1101iaj,aj+1,....a.) E 8D. Consequently, pD (a)
I1-7-t/aiIlaji
Ila - YjII =
Put I := { j E { 1, .... n) : I a j I > 1 }. For V E N we have
Igv(a)I =
9i,
Igv(a)I q4v
=
4v
17
l 1j=1
IajlPj.v-gvai
nj=+1
IajIpi.v+9vai
< j7gv([1jE1 IajI r1j f, laj I
Finally, 8o2ne(a)PD 8D (a)lgv(a)I
(a)Igv(a)I
(30 (a)(jII_,j I/ajllajl)nj,11ajI)s /
j=I 3
< lev (I7-(1/ai+...+I/as)8on(a) fl laj I2)s jE1
< nqv-(I/ai+...+1/a5)s <
1,
1
v = 1, 2, ....
which finishes the proof of the estimate.
(b) Fix a k E Z+ and suppose that D is not an gk-domain of holomorphy. Let Do, D be as in Proposition 1.11.2 (*) with 8 = Ak(D). We may assume that
D C C. Take an e > 0 so small that D St Da,c+s =: G. By (a) (applied to Da,c+s) and Proposition 1.11.11, there exists an f E 0(1)(G) such that G is the domain of existence of f. We are going to show that z(3k+3)t f E Ak (D), which obviously will contradict the fact that f is not continuable beyond G.
'Wehav'eII-q-"'I:,j
forj = 1,...,s,and
II-q-flay<1forj =s+l,...,n.
227
3.2. Elementary Reinhardt domains
It suffices to prove that if I < s < n - 1, then z(3k+3)1-aDrf(z)
lim DnC"(a)9z-+a
= 0,
a E (aD) \ C"(a), ° Ia I + Ir1 < k. (3.2.1)
First observe that there exists a neighborhood U of the set (aD) \ C"(a) such that
Z E U n,5 n C" (a).
dG(z) > Iz21,
(3.2.2)
Indeed, fix an a = (a1,...,an) E (aD) \ C"(a). Note that a1 as = . an = 0. We only need to prove that there exists a neighborhood U of a such that U'(z,1z21) C G for any z E U n D \ Yo (EXERCISE). Let U be a neighborhood of a such that Iz2-ej I < 1, z E U, j = I--, n, and as+1
s
n
fl(1+Iz2-ej1)aj
j=I
(I-1Z2-ejI)aj <e8, fl j=s+1
- EU.
Then n
s
fl (Izj I + IZ21)aj
J1 (IZj I -
IZ21)aj
j=s+1
j=1
S
n
11 (1 = 1=1Ia1 ...IZnlan 11 (1 + 1Z2-ej I)ai j=I j=s+1
Iz2-ej
I)aj < ec+e,
zEUnDu,which shows that We need the following lemma.
Lemma 3.2.2. Let S2 C Cn be open and let S : S2 -+ (0, 1) be a function such that S _< pn,
f
IS(=') - 3(Z" )l < IIZ' - Z"II, Z' E S2, Z" E IB(:', As1(z')) (c. Example 1.10.7 (g)). Then
IISN+ITIDrgll.n < r!(,/n)Ir12N+1rIll8"glls2 N> 0. T E Z", g E (9(Q). Proof of Lemma 3.2.2. Fix N, r, g, and a E S2. Observe that S(z) > 8N+IrI(a)I
Let r := Z(f <
2dn(a).
28(a), z E IP(a, r). By the Cauchy inequalities we get
Drg(a)1 < sN+lTl(a)i=1 Ilgllp(a.r) <_ r!(2,`)IrIS"(a)IISIIP(a,r) < r!(2,/n)Irl2N118Ng1111(a,r) <-
r!(,/n)lrl2N+ITIIISNgIIn
which finishes the proof of the lemma. °Ifa e (8D)nC"(a),thena E Gand,conseyuently,thefunction continuous at a. Observe that (8D) \ C" (a) C G \ C" (a) C 8G.
z(3k+3)1-(Drf
is obviously
228
Chapter 3. Reinhardt domains of existence
We come back to (3.2.1).
Fix an a E (dD) \ C"(a). By Lemma 3.2.2, we get
IrI
Sc kIDrfI
where co is a constant. Consequently, for Z E U fl D fl C" (a), z near a (z should be so near a that SG(z) = pG(z)), using (3.2.2) we get Iz(3k+3)1-aDr f(z)I
CoIZ(3k+3)1-aISG(1+k)(Z)
< <
C1Iz(3k+3)1-o-2(1+k)1I
C1IZ(3k+3)1-aldG(1+k)(Z)
<
=
C11Z(k+1)1-a1
0.
where c1 is independent of z, which proves (3.2.1). (c) The equivalence (i) * (iii) follows from Corollary 3.1.8. The implication (i) = (ii) is obvious. It remains to show that (ii) (iii). Suppose that a 0 IR Z". Take an f E 0(0+)(D),
ZED.
ofZv,
.f(z) vEE(f )
To get a contradiction we will show that f - const.
Suppose that there exists a v E E (f ),. Let W E IR" be such that w 1 a,
IIwII = 1, ands := (w, v) > 0. Fix 0 < N < s and x° E I. Note that x° + tw E III,,, t E IR. Put r(t) := ex°+tw, t E IR. Since r(t) E D, the Cauchy inequalities imply (cf. Example 3.1.4 (d)): l
a
fI<
II.f IIaoF(r(t)) < 1 1 8 V
r(t)v
= e-(x°,v)
- r(t)vSD (r(t)) IISN
f II D
< e-(xo y) Il"DDf II D
(etsINSD(r(t)))N
MN
where M := sup{SD(r(t))ets/N : t E IR). It suffices to show that M = +oo. Suppose that M < +oo. Put T It E IR>o : PD(r(t)) > So(r(t))}. If t E T, then
"
e-2ts/N + `e2(xc+twj-ts/N) > M-2 Consequently, T is bounded. Therefore, 8D(r(t)) = PD(r(t)) fort > to. Now, we estimate PD.
Let d := dist(x° + IRw, 2II.,,) = dist(x°, (8D) \ Vo such that pD(r(t))
Fix a t E Qt and Z E 2IIz - r(t)II. Write Iz11 = rj(t)e"i (u1 E R),
j = 1--n. Note that I I u I I > d. Fix a j = j(1) such that I ujI ? d / /
.
Then
we obtain
pD(r(t)) ? ilzz - rj(t)I ?
2IIzzI -r3(t)I = Zrj(t)Ie"; - lI > dorj(t),
3.2. Elementary Reinhardt domains
229
where d° := i(1 - e-d/1). Finally, choose jo such that there is a sequence (tk)k C [t0, +oo) with j(tk) = jo for all k and limk,,, tk = oo. Then 1
doe"'/p+tkwjp+tk{w.v}/N
M > PD(r(tk))etkslN >
) 00; k-P. oo
a contradiction.
(d) Let f E 3eO0(D), f(z) =
z E D. In view of Proposi-
av z
tion 1.6.5 (b), for v E E (f ), we have
c(v) := log IIIIID
Ia,c C Hv,c(v),
Consequently, E (f) C (1R+ a) n z,. In particular, if a V IR . Z", then f = const. (e) For f = FpEZ" of zfl E 3e°O(D) define 00
ofkaekckk
g(A)
AEID.
k =O
Since D. C (e-cza : z E D) C ID, for every A E D. there exists a z E D such that k = e-cza. We know that E(f) C (IR+ a) f1 Z". Observe that in fact
E (f) C Z+ a (because 00
are relatively prime). Thus 00
laf ekc(e-cza)k I =
la aekcAk I = k=0
k=0
laf -01 < +oo OEZ"
and, therefore, g is well defined, g E 0(D), and g(z) = g(e-`za) = f (z), z E D. Hence, g E 3e°O(D) and fi(g) = f . (f) Recall (Example 1.10.7 (j)) that the inclusion 3e°O(D) -+ O10+1(D) is continuous. Consequently, by the Banach theorem (Theorem 1.10.4), either 3e°° (D) _ O10+i(D) or 3e°°(D) is of first Baire category in the Frdchet space 0(0+)(D). Define
f(z) = Log 1- e-cza I
ZED.
where Log stands for the principal branch of the logarithm. Obviously, f is holomorphic and unbounded. We are going to prove that f E O (°+) (D). Fix an N > 0. Then
If W1 <
+ log
1
1 - 6(z)'
z E D.
where 6(z) := e-`lzal, z E D. In particular,
3N(z)If(z)1
230
Chapter 3. Reinhardt domains of existence
Recall (cf. the proof of (a)) that
ZED.
PD(z) < I1 - 0-1/a"(Z)IIZ,I, Suppose that 1/2 < 0(z) < 1. Then
PD(Z) < 2Y(1 - 0Y(z))Iz"j, where y := I/ 10f" 1. Finally, 2/N
1
8 (z)11(z)12/N < 8o(z)21'(l
-
0Y(z))Iz"I(ir + log 1
< 2'(l - O)'(z))7r + log
1 - 0(z) YIN.
)
1 - 0(z)
2/N
Bl m (1 - OY) (n + log
1
=0
1 0)
(EXERCISE),
we conclude that 8p f is bounded.
(g) Suppose that LP (D) # (0) for some 1 < p < +oo. Then, by Example 3.1.4 (e), there exists a v E (Z"). such that z" E LP (D). On the other hand h
we have
J jzvjP dA2n(z) = (2n)" D
= (2n)"
f
rpv+t dA"(r) (D)
r
e(x.Pv+2)
dA"(x).
Nn.r
We may assume that a" 0 0. Changing the variables H«.c 30 x = (x', x") H (x', (a, x)) E
IV-I x (_oo, c)
and, next, applying the Fubini theorem shows that the latter integral is infinite
0
(EXERCISE); a contradiction.
3.3 Maximal affine subspace of a convex set II The present section is a continuation of § 1.4.
Definition 3.3.1. A log-convex Reinhardt domain D C C" is of rational (resp. irrational) type if E (log D) is of rational (resp. irrational) type. Exercise 3.3.2 (Cf. Exercise 1.4.9). Let
D :_ ((z1-z2) E C2 : CIZ1II` < IZ2I < dIztIlL) Decide when D is of rational type.
(c,d,µ > 0).
3.3. Maximal affine subspace of a convex set 11
231
Remark 3.3.3. Let D C C" be a fat log-convex Reinhardt domain. Then D is of rational type if D = int I(«,c)EA Da,c, where A C Z" x IR (cf. Lemma 1.4.11 (iv), I
Remarks 1.5.7 (e), 1.5.8 (b)).
Definition 3.3.4. Let X C IR" be a convex domain.
F(X):=(E(X)1n(Q")1. Ko(X) := X.
Define
Z(X):=X+F(X),
KK(X) :=
j = 1,2....,
Z(Ki-t(X)).
00
K.(X) := U Kj(X), M(X) := E(K.(X)). i=o
Remark 3.3.5. (a) Recall that E (X) C F (X) and E (X) = F (X) if X is of rational type (Remark 1.4.4 (c) (v)).
(b) Z(X) is a convex domain, X C Z(X ), and F(X) C E(Z(X)).In particu-
E(Ko(X)) C F(Ko(X)) C E(KI(X)) C F(KI(X)) C E(K2(X)) C
(c) Ki(X) = X + F(Kj_I(X)), j = 1.2..... (d) Kp(X) Kp+t(X) a F(Kp_t(X)) E(Kp(X))
.
F(KK(X)). In
particular, if Kp(X) = Kp+t(X), then: Kp(X) is of rational type,
Kp(X) = K(X) for every j > p + 1, Koo (X) = Kp(X). (e) If Ko(X) {0}
KI(X)
E(Ko(X))
Kp(X), then
F(Ko(X))
...
E(Kp_i(X))
F(Kp_t(X)).
" _< LZJ (f) If X C Y, then Z(X) C Z(Y) and, consequently, K0,, (X) C Koo (Y).
-
Consequently, dim E (X) > l and p <
L"-dim E (X )+ I 2
J
(g) K .(X) = K (X) = the smallest convex domain of rational type containing X (cf. Remark 1.4.10).
Indeed, since K,(X) is of rational type, the definition of K (X) implies that
K(X) C KO(X). On the other hand, since X C K(X), we get Ko,,(X) C Koo(K(X)) = K(X). Example 3.3.6 (W. Jarnicki). There exists a convex domain X C IR" such that Ko(X) !c KI (X) ... K,(X) with p = LZ J.
232
Chapter 3. Reinhardt domains of existence
Indeed, let r := N/r2-, s := ,/3-. Consider the following basis (b t.... , b") of Qt":
bl n-1
k=I,2... 2k-1
I
I
L2J'
n-2k+1
b2k+1
Ln 2 1 J.
n-2k-1
2k-1
Put
Xt1bj1+1I1 jI= < 12i
112j +ll
1'J12kI<1.
1, 2,. . .. k,
whenn
= 2k + 1
J=1
Observe that X is convex. The equality p = L J follows directly from the
i
identities below.
(1)t: Kt(X)=X+>J t lcRbi,I=0,1.2....,L21], (2)t: E(Kt(X)) = Ej +l IRb1, f = 0,1,2,..., (3)C
F(K1(X)) =
2f j2
Il"21 ],
Rb1, t = 0. 1,2...., "22].
Since IRb I C X, we get (1)o. We will use induction: (1)t
(3)t
(2)t
(1)t+1 (2)1: The inclusion "J" in (2)1 follows trivially from (I)t. Fix an
(1)t
a = Fj=1 tjbi E E(Kt(X)). Suppose that there exists a j > 2f + I with 76 0. Let jo be the smallest of such j's. We may assume that tj Put x := b2 + b4 + b6 + + b2" = Ej=t xjbi, where q := ["21 J. Then x E X C Kt(X). Consider the vector x + a E Kt(X). Using (l)t, write x + a = z + c with z = En=, zz bi E X and c = Ej ti 1 ca b' . Observe that t1
z j = z, + c. = x, + aj, j = 2e + 2, ... , jo. In each of the following three cases we get a contradiction with the definition of X:
jo is odd: Then zjo = ado = -1, zj0_1 = xjo-t = 1; jo < n and jo is even: Then zio = I - 1 = 0; jo = n and jo is even: Then z1o = an = -I. (2)1
(3)t: It suffices to observe that (2)1 implies
E(Kt(X))1non ={a=(at.....a")EOn :(a,bl)=...=(a,b2t+1)=0)
aEQ"
=0
ifkC+1/2
a2kr + a2k+l s + a2k+2 + ... + an = 0 ifk < f _ (a E On : a t = ... = a2t+1 = 0, a2t+2 + a2t+3 + ... + an = 0).
I
3.3. Maximal affine subspace of a convex set II
233
(1)t+l: The inclusion "C" in (I)j+1 follows trivially from (3)t. To (3)t prove the opposite inclusion, take x x j bf E X and t = 21+3 t j bf . Define t+l
z := x1b' + E((Ix2jl + It2j+lI)b2' j=1 n
xjbf E X,
+ (x2f+l + 121+1)b2l+l) + j =2f+4 t+l
i:= x1b' + 1:(x2j - Ix2 j I + 12j - It2j+l Db2j E F (K1(X )) 1=1
Since x + t = i + 1, we conclude that x + t E X + F (Kt (X )) C Kt (X) +
F(Kt(X)) = Kt+l(X) Exercise 3.3.7. Let X be the domain from the above example. Write X in the form
X=hic,n...nHU2p,C2p, p=
n 2
so that
Kt(X)=Halt+I,c2t+I n...nlj2,c2p, a=0, 1,2,...,L Hint. It suffices to choose of, c j , j we have
n-1 2
= 1, ... , 2p, so that for every x =
(x, a2j-1) < C21_1 q t21+1 < t2j,
(x, a2j) < C2j q t2j+1 > -12j, (x , a"-') < Cn -
l q to < 1, (x, a") < Cn to > -1,
F_'=1 ij b1
n j=1,2,....LI-1 2
j
Ln 2 1 J,
n = 2p, n = 2p.
Proposition 3.3.8. Let X = I1, , n
n I.N,CN, where a1... . aN E (Ot"), and let r := rank [a' , .. . , a"]. Further, let
J:={1 =(i1,...,Ir):1
I E J,
MI:={vEIR": (v, a)=0, aEAI)=Af, I EJ, M:=nMt=(UAI)1.
IEJ
IEJ
234
Chapter 3. Reinhardt domains of existence
Then M (X) = M and, consequently, K (X) = X + M. In particular, K (X) = IR"
i f`A, = {0}, I E J. Proof. Observe that for every I = (i 1, ... , ir) E J there exists an xl E R" such that
Haikci ={xEIR":(x-xl,aik)<0}=HXk, k=1,.. ,r. k Put
YI := int n Ha', XI := n Ha'k,cik, I = (i1... .ir) E J. k=1
aEA1
Obviously, YI is of rational type. Moreover, X C X1 C YI and E (YI) = M! . Thus M(X) C M(YI) = MI, I E J, which implies that M(X) C M. Now assume that M (X)
R" and let
K(X) = int n HP, (fl,c)E B
where B C (Z'). x IR. Then M(X) = {U E IR" : (v,,8) = 0, P E prRn(B)}. To continue we need the following lemma.
Lemma 3.3.9. Let X j := I>lj cj C IR", aj E ((R")., cj E E, j = 0, ... , N, r := rank[a1,...,aN], and 0 0 XI fl ... fl XN C Xo. Then there exist 1 < it < < it < N such that r = rank[a't, ... , alr] and Xi, fl fl Xir C Xo. Proof of Lemma 3.3.9. We use induction on n. The case n = I is trivial. Assume that the result is true in IR, ... , IRn-I (n > 2). We may assume that N > 2 and
n
Xj0Xo, k=1,...,N.5
(3.3.1)
It suffices to prove that r = N. Consider the following two cases:
Case 1. r
smaller. Nevertheless, if we find 1 < 11 < . . . < is < N, s < r. with s = rank[a'" , ... , a's J and Xi, n . n Xi, C Xo. then it suffices to take arbitrary air so that rank [a'" , ... , ai,1 = r.
3.3. Maximal affine subspace of a convex set II
235
Xj = Yj x IR"-r, where Yj is an open halfspace in Rr with pi 1 8Yj
Clearly, r = rank [f1, ... , fN]. Moreover,
n
Yj¢Yo.
k=I,...,N.
j El t,...,k-I.k+i,...,N)
Hence, by the inductive assumption, r = N. Case 2. r = n: We may assume that Xo = {x" < 0} and (a N, ... , an 1) Let aj = W, ag ), j = 0, .... N. Observe that f N # 0. Let
0.
Yj:={yEl2"-1:(y,0)EX1}={yEIR"-1:(y,pi)
yo
Observe that
N-i
OonYjCYo.6 j=1
... , flN-1]. By the inductive assumption, there exist l < i I < Let s := rank ,--< is < N -I such that Y;, n...n Y;s C Yo. Consequently, Xi, n.. n xi, n xN c X0.7 Hence N = s + 1. On the other hand s < n - 1 = r - 1. Thus r = N. 1
Fix a (fi, c) E B. Since X C Hp,,, Lemma 3.3.9 implies that there is an I = (it....,ir) E J such that X1 C Hp,,. Hence ii E E(Hp,e)1 C E(Xr)1 = + Ara'r, k := (AI. . Ar) E + IRair. Consequently, fi = Mail + (Rail + (Rr ).. It remains to show that A E R+ (then prlR" (B) C U, e j A 1, which implies
that M C M(X)). Let L : IR" -> Rr be given by
L(x) :_ ((x.ait ),.... (x, air)),
x E IR".
Then
L(XI) = { =
E IRr
: tk < (X1,a'k), k = 1... ,r}
and
L(Hp,c) = {t E Rr : (, A) < 0). Now, the inclusion L(X) C L(Hp,e) implies that A E Rr (EXERCISE). I'If y, n . . . n YN-1 = 0. then (X I n . . . n XN_I) n (x" = 0) = 0.
X, n .. nxN-I C Xo; a contradiction. If y e
Y,
XN_,) n {x = 0). Consequently, (y,0) 0 XN. Thus Y, n
'(X;,
nXN)n(x" =o}=Y;,
which implies that
n ... n YN_,. then (y, 0) E (X, n ... n n YN_, C Yo, and finally.
n(QZ"-I \YN)=0.
236
Chapter 3. Reinhardt domains of existence
The following examples illustrate Proposition 3.3.8.
+ 1, 0),a2 := at + 1 = ( + 1, p + 2,1),
Example 3.3.10. Let at where , E IR>° \ 0. Put
X := Halo n Hat 0 = {(x1,X2,X3) E IR3 : &1 + (f + 1)X2 < 0, (fl + 1)x1 + (fl + 2)x2 + X3 < 0).
Then K(X) = IR3. Indeed, rank [a1, a2] = 2 and A(1,2) = 713 fl (IR+ a1 + IR+ a2) = (0). Exercise 3.3.11. Let a1
:= (at,a2,a3),
a2 := 1-at = (1 -at, l -a2, I -a3),
where 0 < aj < 1, j = 1, 2.3, a, # 012, (al - a3)/(a4 - CO 0 Q. Put
X
H. 1,0 fl !2,0 = ((x,,x2,x3) E IR3 : a1x1 +a2x2 +a3x3 <0, (1 - al)x, + (1 - a2)x2 + (1 - a3)x3 < 0).
Then
K (X) _ ( (x,, x2, x3) E IR3 : x1 + x2 + x3 < 0)-
3.4 XO0-domains of holomorphy In this section we characterize the most important class of special Reinhardt domains of holomorphy, namely MI-domains of holomorphy (cf. [Jar-Pfl 20001, § 4.1, for the general theory). Recall that we already presented a general characterization in Proposition 3.1.7.
Theorem 3.4.1 ([Jar-Pfl 1987)). Let D following conditions are equivalent:
C" be a Reinhardt domain. Then the
(i) D is an MI-domain of holomorphy; (ii) D is an 0(0+)-domain of holomorphy; (iii) D is a fat domain of holomorphy of rational type; (iv) D = int n(a,c)EA Da,c for some A C (71" ). X R.
Proof. The implication (i) = (ii) is trivial. The equivalence (iii) q (iv) follows from Lemma 1.4.11. The implication (iv) (i) follows from Theorem 3.2.1(c) and Remark 1. 11.3 (e). So, we only need to prove that (ii) = (iii). It is clear that D is fat (cf. Proposition 1.9.12). Put X := log D and F := F (X) (cf. Definition 3.3.4). Assume that E (X) is not of rational type. Then dim F >
dim E(X). Let f E 0(0+)(D),
f(z) _ E vEE(f)
ZED.
3.4. MOO-domains of holomorphy
237
We will show that
E(f) C E(X)l.
(3.4.1)
Suppose for the moment that (3.4.1) is true. Then for X E X, V E F we get Rye(x+v,v) = ae(x.V). E E VEE(f) VEE(f)
Consequently, the series is summable in the domain exp(X + F). Since D is an
i9(0+)-domain of holomorphy, exp(X + F) C D, and hence, X + F C X; a contradiction.
Now we are going to prove (3.4.1). Take a v $ E(X)-L. Choose w E E(X),
IIwOI = 1, such that s := (w, v) > 0. Fix 0 < N < s and x0 E X. Put r(t) := ex°+rw, t E R. Now, we continue exactly as in the proof of Theorem 3.2.1 (c).
0 Proposition 3.4.2. Let D C" be a Reinhardt domain that is an .°O-domain of holomorphy. Then 3e°°(D) 0('+)(D). Proof. In view of the proof of Theorem 3.2.1 (f), we only need to prove that there
exist an a E (8D) fl C; and a a E (Z")* such that D C Da,c with c := log Iaal. Indeed, by Theorem 3.4.1, D = int n(a,c)EA Da,c with A C (7L")* X R. It remains to use Remark 1.5.9.
Theorem 3.4.3. Let D
C" be a fat Reinhardt domain of holomorphy and let N > 0. Then D is an O (N) -domain of holomorphy.
Proof. Since D = int n(a,c)EA Da,c, where A C (R")* x IR, and
U O(N)(Da,c)ID C O(N)(D), (a,c)EA
we only need to use Theorem 3.2.1(a) and Remark 1. 11.3 (e).
Theorem 3.4.4 ([HDT 2003]). Every Reinhardt domain of holomorphy D C C" is an O(1)-domain of holomorphy (cf. Theorem 3.4.3). Proof By Theorem 1.11.13, D = D * \ M, where D * is a fat domain of holomorphy and M = Uj: VjnD=0 V . Suppose that Do, D are as in Proposition 1.11.2(*) with ,8 =_0(1)(D). We may assume that Do is a connected component of D fl D.
Letb E DfldDo. Ifb E 3D*, then, by Theorem 3.4.3, D C D*. Thusb E D*f1M,
say b E Vj. Then the function f (z) := 1 /z j belongs to 00)(D) (recall that V fl D = 0). Since f does not extend through b, we get a contradiction.
238
Chapter 3. Reinhardt domains of existence
Remark 3.4.5. Observe that if D is a non-fat Reinhardt domain of holomorphy, then by Proposition 1.9.12, D is not an O(tr)-domain of holomorphy forO < N < 1. Theorem 3.4.6 ([Jar-Pfl 1987]). Let D C C" be a Reinhardt log-convex domain. Then
E (D. 3e °O (D)) = int exp K (log D) ) (cf. Definition 1.12.1).
Proof. Put Gt := E(D. 3C°O(D)), G, := in( exp(K(log D)). Recall that
K(log D) = K(log G,) is the smallest convex domain of rational type that contains X := log D. Note that both domains Gt and G, are fat and of rational type. Obviously, G, C Gt. It remains to show that .7C°°(D) C (9(G,). Fix an f E Je°°(D),
E
A z) =
of za,
z E D.
aEE(X)1f1Z"
Looking at the definition of Z (X) and the form of the series it is clear that f extends to a bounded holomorphic function on exp(Z (X)).8 Repeating this argument leads
to a bounded extension of f on exp(K(X)). Finally, the Riemann theorem on removable singularities gives the extension to G,.
Example 3.4.7 (Cf. Example 3.3.10). For f E IR>o \
let al
+ 1, 0)
anda2:=(fl+1.f +2,1). Put D:=Da, nDa2 _ {(Z1. Z2, Z3) E
C3
: IZ1 Ifl IZ2IB+1 <
1
IZ1 I'0+l IZ2I0+2IZ3I < 0-
Then 3e°O(D) a-- C.
Exercise 3.4.8 (Cf. Exercise 3.3.11). Let
al := (al,a2,a3), a2 := (1 -a,,1 -a2,1-a3), where O < a J < l, j = 1 , 2, 3, a, 0 a2, (al - 0 1 3 ) / ( a l - a2)
Put
D : = Da, n Da2 1(Z1 Z2, Z3) E
C3
: IZ, Ian IZ2112IZ3Ia3 < 1, IZ1
I1-"
IZ2I
I-a2IZ3I1-a3 < 11.
Then
E(D.3e'(D)) = {(ZI,Z2,Z3) E C3: IZ1Z2Z3I < 0. 8For x° E X. V E F(X), and a E E(f ), we have Ia f I(e-t°+L)a = Ia f Iaa le(x°.a)
3.5. Ak-domains of holomorphy
239
3.5 ,k-domains of holomorphy The next important space after the space Je°O(D) is the space Ak(D), 0 < k < +oo. We begin with the case where k E Z+. Theorem 3.5.1 ([Jar-Pfl 1997]). Let D C C" be a Reinhardt domain of holomorphy. Then the following conditions are equivalent:
(i) D is an d.'-domain of holomorphy for every k E Z+; (ii) D is an .7fIo°'k-domain of holomorphy for every k E Z+; (iii) D is an X ,a -domain of holomorphy; (iv) D is fat. In particular, if D is an JC°O-domain of holomorphy, then D is an Ak -domain of holomorphy, k E Z+.
Proof The implications (i) = (ii) = (iii) are obvious. The implication (iii) (iv) follows from Corollary 1. 11.4 (a). The implication (iv) Remark 1. 11.3 (e) and Theorem 3.2.1(b).
(i) follows from
0
We move to the case where k = +oo. Theorem 3.5.2 ([Jar-Pfl 1997]). Let D C C" be a Reinhardt domain of holomorphy. Then the following conditions are equivalent: (i) D is fat and satisfies the Fu condition; (ii) D is an Je1a'0O-domain of holomorphy; (iii) D is an .4°O-domain of holomorphy; (iv) D is an (9(D)-domain of holomorphy.
Moreover, if D is an X'-domain of holomorphy, then each of the above conditions is equivalent to the following one:
(v) D is an JeO°(D) fl 0(D)-domain of holomorphy. We need the following two lemmas.
Lemma 3.5.3. Let D
C" be a Reinhardt domain, 8 E 0(D), and k E W. Then
the following conditions are equivalent:
(i) D is an 8-domain of holomorphy;
(ii) D is an 7-domain of holomorphy, where F :_ {D0f : f E 8. fl E Sk}, where Sk :_ {13 E 71+ : 1 1 = k}.
Proof The implication (ii) (i) is obvious (Remark 1. 11.3 (a)). (i) = (ii): Suppose that D is not an F-domain of holomorphy. Let a E D and r > dD (a) be such that each derivative DOf extends holomorphically to IP(a. r), 13ESk.
240
Chapter 3. Reinhardt domains of existence
Observe that if g E O(D) is such that each derivative a extends to a function g j E 0(UP(a, r)), j = 1 ,n , then the function g itself extends holomorphically to IP(a. r). Indeed, the extension may be given by the formula n
g(z) = g(a)+J(zj -aj)
r
I
gj (a+t(z-a)) dt, z E IP(a,r)
(EXERCISE).
o
j=I
Consequently, using the above remark inductively, we easily conclude that every
function f E 8 extends holomorphically to P(a, r); a contradiction.
O
Lemma 3.5.4. Let D C" be a Reinhardt domain. Assume that D is an domain of holomorphy,9 where S is such that there exists a ko E Z+ such that Sko C S. Then
.7(',S-
D = int
n {z E V(a) : Iaf f fEXoo.s(D)
!(a)za-fi
< IIDfif IID}
aEE(f), flES
a0fl, (s)#o
Moreover, if S = (Z")., then D satisfies the Fu condition.
Proof. By Lemma 3.5.3, D is an .F :_ {Dpf : f E ,7(°O,S(D), 6 E S}-domain of holomorphy. Observe that Y C J°O(D) is invariant under n-rotations (in the sense of (1.12.1)). Thus we may apply Proposition 1. 12.6 (b) (EXERCISE). Now, assume that S = (Z"). and let (SD) fl V j o 0 0 f o r some jo E 11, ... , n). By Remark 1.5.7 (e), to prove that D (jo) = D we only need to show that a jo -fjo
Oforanya E E(f), fi E S witha # )4, (s) 96 0. Suppose that there exists f E J'°O'S(D) and a E E(f) with ajo < 0. Put max{0, av }, v = 1, ... , n. Then iB a, and 0. # = (ft, .... /+n), Therefore
za-p =
T7
1!
z.0v
vE(I..,n) av <0
is bounded on D. Consequently, ajo > 0 because of (SD) fl Vo # 0; a contradic-
0
tion.
Now we are able to verify Theorem 3.5.2.
Proof of Theorem 3.5.2. We may assume that D V. The implications (v) (iii) (iv) (ii) are evident. (ii) = (i): Recall that D is fat (Corollary 1.11.4). Suppose that aD fl V0 0 0 for some jo E (1, ... , n ). Observe that for any r > 0 the open set Dr := D fl F(r) 'Recall that
J(oo.s(D)
:_ (f E O(D) : Vaes : DO1f E Je°"(?)}, 0 96 S C Z.
3.6. Lh -domains of holomorphy
241
is fat and log-convex. We know that D, is an JOo.Z+-domain of holomorphy (cf. Remark 1. 11.3 (e)). Hence, by Lemma 3.5.4, if D, fl Vo 0 0, then Drj°) _ Dr. Consequently, D(i0) = D. (i) (iv) (resp. (i) (v) if D is an X'-domain of holomorphy)): Suppose that D is not an O(D)-domain of holomorphy (resp. Jt°(D) fl 0(b)-domain
of holomorphy). Let Do, D be as in Proposition 1.11.2(*) with 8 = O(D) (resp. $ = J'°O(D) fl O(D)). Recall (cf. Theorem 1.1 (b) (resp. 3.4.1 (iv))) that D can be written as
D = int n Da,c, (a,c)EA
where A C (tR"). x R (rest. A C (7L"). x R). Fix (a, c) E A and a> 0 such that D C Da,c C Da,c+s and D 0 Da,c+,. Since Da,c+e is a domain of holomorphy (resp. an J°O(Da,c+e)-domain of holomorphy), we only need to observe that, by Remark 1.5.11 (b), we have D C Da,c C Da,c+e; a contradiction. 0
Example 3.5.5 ([Sib 1975]). Let T = {(z1, z2) E D X D.: Izt I < Iz21) be the Hartogs triangle. Then: (a) T is an Ak-domain of holomorphy for arbitrary k E Z+ (cf. Theorem 3.5.1). (b) T is not an A'-domain of holomorphy (cf. Theorem 3.5.2).
Exercise 35.6. Complete the following direct proof showing that T is an Ak (T)domain of holomorphy for every k E Z +. Fix a k E 71+ and let Do, D be as always with D =_T and 8 = .Ak(T). We may assume that D ¢ T. Let a = (a 1, a2) E (D \ Yo) \ T . We have the following two cases:
tat I > 1: Then the function f(zt, z2) := 1/(z1 - a1) belongs to O(T) C dk (T) and is not extendible to D; a contradiction. 1a21 < lai I < 1: Then the function f(zt, z2) := zi+2/(a1z2 - a2z1) belongs to .,6k (T) and is not extendible to D ; a contradiction.
Exercise 3.5.7. Find a power series f(z) _ Fk oakzk, z E m, such that f E db°O(D) \ 0(i)).
3.6 L h -domains of holomorphy Our aim in this section is to discuss the problem of geometric characterization of Reinhardt $-domains of holomorphy D C C" with 8 C Lh (D). Recall the general Proposition 3.1.6 and Theorem 3.2.1(g) for the elementary Reinhardt domains. The following lemma, which will be used in the proof of Theorem 3.6.4, is of independent interest.
242
Chapter 3. Reinhardt domains of existence
Lemma 3.6.1. Let D C C" be a Reinhardt domain. Then Moo,st (D) C db(D). Consequently, 3oo.k(D) C ,bk-I(D) fork E N. In particular, MOO-°°(D) C Exercise 3.6.2. Find a fat domain D C C \ (-oo, O] such that Log E .3e°D'S1(D) \ .4(D), where Log stands for the principal branch of logarithm. Before presenting the proof we recall a few well known facts from the theory of metric spaces. Let (X, p) be an arcwise connected metric space. For a continuous curve y : [a, b] -+ X define its p-length L,p(y) E [0, +oo] by the formula N
Lp(Y) := sup { E i=I
: a = to < ... < tN = b. N E N}.
Obviously, p(y(a), y(b)) < Lp(y). One can easily prove (EXERCISE) that
a < c < b.
Lp(Y) = Lp(YI [Q,c]) + Lo(YI [c,bl),
We define the inner metric p' associated to p by the formula
p'(x, y) := inf{Lp(y) : y E C([a, b], X), y(a) = x. y(b) = y},
x, y r= X.
Clearly, p < p'. One can easily prove (EXERCISE) that p' is a metric. In the case where X is a subdomain of E, where (E, 11 11) is a normed space, with P(x,Y) = IIx -YII, x, y E X, we have 11x - YII = P(x,Y) provided
that the segment [x, y] is contained in X. In particular, if B(x°, r) _ {x E E
IIx-x°II
(EXERCISE).
Nevertheless, one can easily find (EXERCISE) a bounded domain X C E such that the metric p' is unbounded.
One can prove (EXERCISE) that if X is a subdomain of IR', f : X - C is Frdchet differentiable, and II f'(x)II < C. X E X, then
I.f(x) -.f(Y)I 5 CP'(x,Y),
x. y E X.
Proof of Lemma 3.6.1. Since D has a univalent envelope of holomorphy (cf. The-
orem 1.12.4), we may assume that D is a domain of holomorphy.10 Since the 10Let D be the envelope of holomorphy of D. Then, by Lemma 3.5.3. ,7(oo.s' (D)ID = 3L°°.s, (D). Moreover, ..4(D)ID C .4(D).
3.6. L '-domains -domains of holomorphy
case D = C" is trivial, we may assume that D
243
C". Fix an f E M00,S1 (D).
Obviously,
(maxIIDafIID)-D(z',z")
If(z')_
E D,
aES,
where ED denotes the arc-length distance on D with respect to the e 1-norm 11111. To show that f extends continuously to D it suffices (EXERCISE) to prove that for any a E aD there are a constant c > 0 and a neighborhood U. of a such that
Z'.Z"E DnUa.
ED (Z',--")
where
J:= (jI. ,js), I <j1 <... <js _
PJ(zl,.. .zn)
(z11,..., Zj,)
Po:=0.
Fix a point a E aD. We may assume that J = 0 or J = (1.... , s) (1 < s < n). Take z', Z" E D.
IfJ = 0, then put w' := z',w":=z". If J # 0, then put W, := (IzI1...
IZsl,zs+1
,zn),
z"
w
E D.
Obviously,
uD (Z' w') < 27r(Iz I + ... + 1
Iz
s I)
-D(z W"):<
s
It remains to show that there is a constant c' > 0 and a neighborhood U. such that
'ED W, w") < c'IIZ' - z"II
z' Z" E D n U
Using continuity it suffices to consider only the case where 0 0 Iz j 154 Iz! 10 0,
j = 1--n. n. Let L I =
= Ls = Log be the principal branch of the loga-
rithm and let L,+1..... Ln be arbitrary branches of the logarithm defined in small neighborhoods of a,,+1, ... , an, respectively. Define
y = (yl.... yn): [0, 11 --- Cn
yj(t) := e(1-r)L;(wj)+tL;(w ), j = 1.....11.
Since D is logarithmically convex, y([0, 1]) C D. What is left to be shown is that for each j there is a cj' > 0 such that the length ej of y, is estimated by
244
Chapter 3. Reinhardt domains of existence
tj
provided that z', z" are near a. We have
=
f
f
t
IY;(t)I dt =
t
IwwI-tIwj" I'ILj(wj)- Lj(wj')I dt
ILj(wl)-Lj(wi)I
1,
- Iloglwjl -loglwj'llllwll - Iwfll
ifj <s,
1zi -z7I Ilz' I-Iz"II I log I zj, II-log
II
1L(z) - Lj(zJ )I if j > s + 1.
The case j < s is trivial. If j > s + 1, then we only need to observe that
ILi(zj)-Lj(z7)1
Iu - vi - Ia,1' u,ot a;I Ilogu - logvi
-
O
I-Izr
so the term I log Iz; [-log lz; remains bounded near aj. The proof is completed. 11
Remark 3.6.3. (a) Recall that if D is a bounded (Reinhardt or not) domain, then Ak(D) C 3eO0,k(D). Consequently, if D is a bounded Reinhardt domain, then, by Lemma 3.6.1, 3CO0"k(D) C Ak-t (D) C ieoo,k-1(D), k E N. In particular, if D is a bounded Reinhardt domain, then 3e°°' (D) = AO°(D). (b) Let
D=T={(z1.z2)EIDxID:IziI
jk(Z) := Zk ,
Z = (Z1,Z2) E D. k E H.
2
Then fkE
,°o.k(D) \ Ak(D) C Ak-1 (D) \ Ak(D).
Indeed, k
az; aZ2
-k Z2) = P!(P)q!(q)
2k-P
2k
(ZI,
_k+q
(k.P q)
`2
zt -P k+q
Z2
Consequently, if p + q < k, then
ap+qf aZ aZ2
(zt.22)
Ic(k. p,q)IIz,-(P+q)I < Ic(k,p.q)I
(zt.Z2) E D.
3.6. Lh -domains of holomorphy
245
and so fk E 3£°O'k(D). Moreover, if p + q = k, then aP+q fk aZi aZZ
(Z1,Z2) _ c(k
r Z, 2k-P
42
which shows that a°z,1,4 o C(D). 2 (c) In the case where D is an unbounded Reinhardt domain, Lemma 3.6.1 implies that 3ez,S' (D) C A(D) if D satisfies the following condition:
(*) for every point a E aD there exists a bounded Reinhardt neighborhood Ua of a such that D n Ua has a finite number of components.
In fact, take f E X a'S' (D) and a E OD. Let Ua be as in (*). If S is a connected component of D n Ua, then Lemma 3.6.1 shows that f E A(S). Suppose that u Sk. D n Ua has a finite number of connected components, D n Ua = Sr u Fix ao E D and ai E Sj, j = 1, ... , k. Let y : [0, 1] --* D be a curve connecting
ao, a',..., ak and let r > 0 be so big that y([0, 1]) U (D n Ua) C IP(r). Then D n Ua is contained in one connected component S of D n P(r). Hence, by the first part of the proof, f E A(S) C A(D n Ua). In particular, if D satisfies (*), then .7e 'k(D) C Ak-t (D), k E W. Observe that (*) is satisfied if D is log-convex - we take Ua = F(ra) with sufficiently large ra > 0 and observe that D n IP(ra) is log-convex and therefore connected (cf. Remark 1.5.6 (d)).
We do not know whether the inclusion eia°'S' (D) C A(D) is true for arbitrary (unbounded) Reinhardt domains.
t
Theorem 3.6.4 ([Jar-Pfl 1997]). Let D C" be a Reinhardt domain of holomorphy. Then the following conditions are equivalent: (i) for each k E Z+ the domain D is an Lh'k -domain of holomorphy; " (ii) D is fat and there exists a p E [1, oo) such that LP(D) # {0}; (iii) D is fat and E (log D) = {0);'2 (iv) for each k E 71+ the domain D is an Lh,k n Ak-domain of holomorphy; (v) D is fat and for every a 0 D U Vo there exist sequences (cj)jtt C IR>o, (f J) j __ l C Z" ((fl! 1 C 7L+ provided that D is complete) such that
D C C"(PJ).
IIcizfli. 1IL2(D)
Proof The implications (v) = (ii) and (iv) cation (i)
1. j E W,
ilm cila0' I = +oo.
(i) = (ii) are obvious. The impli-
(iv) follows directly from Lemma 3.6.1.
(cf. Example 1.10.7(h)). "Recall that Loh .k(D) = L,'' (D) ( 12Recall that E (log D) = (0) for any bounded Reinhardt domain.
246
Chapter 3. Reinhardt domains of existence
(iii)
(v): Fix a 0 D U Y° and j E W. Put X := log D and let x° := (log jai 1, ... , log Ia"1)
Note that x°
X. Since E (X) = {0}, there exist linearly independent vectors a 1,
an E Z" (a 1..... a" E Z+ provided that D is complete) such that n
x C n Ha° _: Xo; i=1
cf. Lemma 1.4.11 (iii). Let A := [ak]i,k=1,...,n E GL(n, Q. We may assume that
IdetAI > j2n"1aI ...an12. Put pj =: a1 +---+a" -1,
X1 ={WEIR":ti<(x°.ai),i=1,.. ,n}. Then we get Il-fl' 11L2(D) _ (2;T )n ZRI(D)
r2fij+1 dAn(r) = (27r )n r e(X'2(fl'+1)) dAn(x)
e(x,2(9'+1)) dAn(x) _ (2,r)"
< (27r )n fXO
(2,r)n
e2(b,l>dAn(
IdetAI Jx,
xo
e(x,2(a'+...+a",)) dA
(x)
" e2(x°,a1+...+a") IdetAl
n"
12.
l2
1 det Al
f
Let cj := j la-,O' 1. Then cjZ81 ll
IIL2(D) < I
and
cjla6' I = j.
Since D is fat, we have D C f,"=1 Da; .(ai.xo) C fi"=t C. (ai) C C" (pi) (EXERCISE), which finishes the proof of (v). (ii) (iii): We argue as in the proof of Proposition 1. 13.13 (b) (using Remark 3.1.4 (e) instead of Example 1. 10.7 (c)) - EXERCISE. (iii) = (i): Fix a k E 71+ and suppose that D is not an LO'k _domain of
holomorphy. Then we find domains Do, D such that 0 0 Do C D n D, D ¢ D. and for any f E L*,k (D) there exists an f E (9(D) with f = f on Do. Since D is fat, D ¢ D. Moreover, we may assume that D n Yo = 0. Since E(log D) = (0) and D is fat, there are linearly independent vectors a1, an E 71" , C 1, ... , C" E IR, ands > 0 such that n
n
DCGo:= nDa;ci c nDaf,cj+6=: G1, j=1
j=1
3.6. Lh -domains of holomorphy
247
and D G1. Using the fact that the al 's are linearly independent, we may assume that c 1 = = cn = 0.13 Now, we fix an a E D \ G 1 and then a jo such that I a"0 I > es. Moreover, we seta :=al For N E N we define ZNa
za' -aa&o'
fN(Z)
Z E G1.
Obviously, fN E (9(GI). Observe that if IN E L/'k(D), then
fN(z)(za'o -aaJo) = zNa on D and we have a contradiction.
Consequently, it remains to prove that fN E L/'k(D). Observe that D#fN with fi E Z+, < k, is a finite sum of terms of the form (EXERCISE) Na+tajo-fi d
(Za'0 - aabo)t+1 '
where d E Z, f E {0, .... k). Thus it suffices to find an N such that
I#I
1,
IIZNa-0IILP(Go)
Let A := [ae ]f,t=l,...,n E GL(n. C), B := A-1. Put n
Tj(x)
)Bt,ixt,
x = (x,-..x.) E 1R", j = 1.....n.
t=1
If p = 1 and v E 71", then we have e(x,v+2)
J Iz"I dA2n(z) = (27r)" f o
dAn(x)
og Go
= (27r)" J
e(Bt,v+2)I
, <0,...,1"<0) (27r)n
det BI dAn(l;)
provided that T j (v + 2) > 0, j = 1, ... , n. In particular, if
Tj(v) >
27r
det All/n
-Tj(2),
j
13Use the biholomorphic mapping C" a (z],..-,z") F- (rtzt,...,r"z") E C", where
rt,...,r" > 0are such that rj' rn" =e-`'. j = 1,...,n.
248
Chapter 3. Reinhardt domains of existence
then IIZ' IIL1(Do) < 1. Hence, if v = Na - fl and if
2n
N ? Tj(fl)+max!0,
I det A I t/"
1
IIzN"-fi
then
-Tj(2)
: j = 1,....n, IfI
-k
}
IIL' (Go) < 1 for arbitrary If I < k.
It remains to prove that IIzN"-' 11x-(Go) < 1 for all If I < k. Fix such a f .
Write Na - B = Alai +
+.lna",
A,...... ln E R. Then A j = Tj (Not - fl) = N - Tj (fl) > 0, j = 1, ... , n. Consequently, IIxoe(co) < 1. IIzN«-s
Remark 3.6.5. Notice that the following general result holds. If D C C" is a bounded domain of holomorphy, then D is an Lh-domain of holomorphy if U \ D
is not pluripolar (Definition 1.14.18) for any open set U such that U fl D # 0 (cf. [Pfl-Zwo 2002]).
Exercise 3.6.6. Using the above remark prove that every bounded fat Reinhardt domain of holomorphy is an Lh-domain of holomorphy.
Proposition 3.6.7 ([Jar-Pfl 1997]). Let D C C" be a Reinhardt domain of holomorphy. Then the following conditions are equivalent:
(i) D is fat and there exist 0 < m < n and a permutation of coordinates such
that D = D' x C"-` with E (log D') = {0}; (ii) D is an M', '-domain of holomorphy;
(iii) D is an J Proof. (i) (ii)
,k-domain of holomorphy for any k E Z
(iii) follows from Theorem 3.6.4. The implication (iii) = (ii) is trivial.
(i): Let F := E(log D) and m := dim F1. We may assume that
et..... es E F J-, es+t , ... , en 0 F -L for some 0:5 s < m. The cases m = 0 and m = n are trivial. Assume 1 < m < n - 1. Since D is an Jt'°°'t-domain of holomorphy, Proposition 1. 12.6 (b) implies that
n
D = int
{z E C"(a) : Iair z?I < 1}.
f Ex°O.1(D), MILD=t
aEE(f)»
Thus
Fl = span{a : 3fExoo.i (D) : Of E E(f):}. By Proposition 1.6.5 (b) we get
D C int
IEx
n
D ,O, aeE(f), jE{t,...,n}
a#ej, aj 960
ZE 1
"(a) : IQ fa j za_ej I
f
< II a2! IID I
249
3.6. Lh -domains of holomorphy
Hence
F1 D (a -ej :
3jc-(1,...,n) : a E E(f). a # ej, aj # 0}.
7'°°,1(D). Take an f E If a E E(f) is such that a 96 ej and aj 0 0, then a, a - ej E F1 and, consequently, ej E F1. Thus,
E(f) C {es+1.....e"} U (Zs x 10}"-s) and, therefore, the Laurent expansion of f has the form a (0,0)z'$) + ae +, zs+1 + ... + ae z,
.f(z) $EZS
z=(Z'.Zs+1,...,z")EDCCsxC"-s. Since D is the domain of existence of Moo,1(D ), we conclude that D = D' x C"-s
Clearly, F = E (log D') x i2"-s. Hence s = m and therefore E (log D) = {0}.
0 Proposition 3.6.8 ([Jar-Pfl 1997]). Let D following conditions are equivalent:
C" be a Reinhardt domain. Then the
(i) D is an MIA -domain of holomorphy; (ii) there exist A C (Z"), and functions c1, ... , c" : A -+ Qt such that
n
D = int
{Z E C" (a) : IZa-ej I < ecj(a)}.
(3.6.1)
aEA, jE(1,...,n}
aOej, aj#0
Proof The implication (i) = (ii) follows from Lemma 3.5.4. To prove that any domain D of the form (3.6.1) is an -domain of holomorphy, observe that
D = int n int n aEA
{z E C" (a) : Iza-e, I < ecj («) } _: int n Ga.
je(1,...,n)
aEA
a#ej,aj#0
Thus, it suffices to consider only the case where
G. = n jE(1,.,n)
{z r= C"(a) : Iza-ej I < ecj(a)}.
a,-4j, aj 960
We may assume that a j 0 0, j = 1, ... , n (otherwise G.
G' X Ck and we
consider a lower dimensional case). In particular, a 96 ej, j = 1, ... , n. Since G. is fat, it is enough to prove that for any point a 0 Ga U V there exists a function
Chapter 3. Reinhardt domains of existence
250
f' E MIA (D) such that f cannot be continued across a. Fix such an a and let jo E { 1, ... , n) be such that 3 := ecjo (a) - l aa-ejo I > 0. Then the function ,a f (Z)
Za-ej0 - aa-ejo
ZEGa,
belongs to ,ie°O,S' (Ga) and evidently cannot be continued across a. Indeed,
a za-ej
a
()z = dzj
Za-ej0
-
(a - e j0 )j
za-ei za-eJo
(Za-ejo -as-ej0)2
as-ejo
'
z E Ga,
and so
If II
aZ j
(a - e10)jecj(a)+cj0(a)
a1ecj(a) I
ca -
+
8
82
Remark 3.6.9. There exists an a E (71"). such that ,Je°O-SI (Ga)
3f°O(Ga)
For example a := (1, -1), cI (a) = c2(a) = 0. Then G. = {(z1, z2) E
(C2
: IZ21 > 1, Iz1I < Iz212},
and the unbounded function f(z) := z1/z2, z = (Z1,z2) E Ga, belongs to MOO-S, (Ga) (EXERCISE).
The problem of characterizing circular (in particular, balanced) F -domains of holomorphy is still open for many of the natural Frdchet spaces .' we discussed so far. Q (Cf. [Sib 1975], [Sic 1982], [Sic 1984], [Sic 1985], [Jar-Pfl 1996] for positive results.)
Chapter 4
Holomorphically contractible families on Reinhardt domains
4.1 Introduction Recall from Chapter 2 that Bih(O". U3,,) = Bih(l8", lL") = 0 for n > 2 and Bih(f". L") = 0 for n > 3 (see Theorem 2.1.17). In this chapter we will study other methods which may be useful to decide whether two given domains in C" are not biholomorphically equivalent. The idea here is that two biholomorphically equivalent domains should have the same amount of functions of a special class (e.g. bounded holomorphic functions or psh functions with specific singularities) or geometric data (e.g. analytic discs through corresponding pairs of points). To give a rough idea of what we are going to deal with let us discuss again whether ID" and B" are biholomorphically equivalent domains. We introduce the following function:
inD : D -+ [0, oo).
AD (z) := SuP{I f(z)I
:f
E 0(D,1)). f(O) = 0),
where D is a domain in C" with 0 E D. Let D C C" and G C C"' be domains, both containing the origin. Note that if F E O(G, D), F(0) = 0, then (EXERCISE)
rD(F(z)) < mr,(z),
z E G.
(4.1.1)
In particular, if F is biholomorphic, then inD o F = mG In the case where D E (D', lB") we get
AD(z) = qD(z),
z E D,
(4.1.2)
where
qD(z) :=
Ilzll00
if D = lD",
Ilzll
ifD=(B",
ZEC".
Indeed, let f E 0(D, ID), f (0) = 0. Then, in virtue of Proposition 2.1.9, it follows that I f(z)I < qD(z), z E D. Hence, AD < qD, D E (ID", lB"). On the other hand, in the case of D = lD" and z E lD" \ {0) put
gz:
gz(w):=wj,
252
Chapter 4. Holomorphically contractible families on Reinhardt domains
when qpn (z) = Iz j I. Therefore, mpy (z) > Ig.(z)I = qyn (z). Now let D = I" and Z E ie" \ (0). Choose a rotation AZ such that Azz = (21, 0); in particular, lid = IIA=zII = I4iII. Put gz: 1B. -+ D,
gz(w) :_ (Azw)i.
Obviously, gz E O(C3,,, ID), g,(0) = 0; hence men(z) > Igz(z)I = 1IzII = gen(z) So the above equations are verified. Exercise 4.1.1. Prove formula (4.1.2) for an arbitrary norm ball in C".
Now let F : to -+ ID" be a biholomorphic mapping, n > 2. Using a Mobius
transform (p: Dn -+ IDn, (G(z)
Fn(z) - F.(0)
F1 (z) - F1(0)
I-
71(0)F1(z)'...
1- F.(0)F.(z)
"
Z E O,
we may even assume that F(0) = 0. Then, by (4.1.1),
mmn(F(z)) =
z E 1$n.
Therefore, we get the following equation:
IIF-'(t, 1/2,..., 1/2) 11 = max{t, 1/2),
t E (0, 1).
Note that the left function is differentiable on (0, 1), but, obviously, the right one is not; a contradiction. Observe that in the above argument the number of bounded holomorphic functions on ID" and I" is compared and this strategy finally has led to the result that both domains cannot be biholomorphically equivalent. Instead of dealing with the function AD. i.e. with the family of bounded holomorphic functions, we may take all analytic discs p E 0 (1), D) in D through two given points, where D is a domain in C" with 0 E D. We define
kD(z) := inf{r E [0. 1) : 3q,Eo(n,D) : p(0) = 0, W(r) = z},
z E D.'
Remark 4.1.2. Let G C C'", D C C" be domains both containing the origin. If F E O(G, D), F(0) = 0, then (EXERCISE)
kD o F < kG; see (4.1.1). In particular, if F is biholomorphic, then kD o F = kG. Note that ko (0) = 0.
(4.1.3)
253
4.2. Holomorphically contractible families of functions
We claim that kD = qD, where D E {m", In }. Indeed, fix a Z E D \ {0}. Then V(A) := A9v(s) gives a function rp E O(U). D)
with (p(0) = 0 and rp(gD(z)) = z. Therefore, kD(z) < qD(z). On the other hand, let tp E O(ID, D) with rp(0) = 0 and p(r) = z, r > 0. If D = D", then using the Schwarz lemma for (pj we get that IzjI _ Iq. j(r)I < r, j = 1, ... , n. Therefore, q 1D. (z) < k®n (z). In the case when q E 0(i), is") with
p(0) = 0 and V (r) = z we see that qe" (z) = qe ((p(r)) = rqs" (ip(r)), where p(A) = AO(A), A E ID, and W E 0(U),C"). Observe that qe"(ip(A)) < l/1A1, A E iJ \ {0}. Applying the maximum principle for the subharmonic function qe" o SP we obtain that 0 E 0 (m, [B" ). Hence, qen (z) < r and since r was arbitrarily chosen,
we have qen (z) < ken (z). Observe that we may also use this geometric function to disprove the biholomorphic equivalence of U)" and B,, (EXERCISE).
Let us summarize what we have done so far. We have introduced a family (dD)oEDCCn. nEN
of functions dD : D [0, 00) (dD E (AD, kD)) satisfying the following property: (*) for any domains G C C'", 0 E G, D C C", 0 E D. and for any
F E O(G, D), F(0) = 0, we have dD(F(z)) < dG(z), z E G. In particular, dD(F(z)) = dG(z), z E G, if F E Biho,o(G, D). Moreover, these functions were explicitly described in terms of the geometry of D.
4.2 Holomorphically contractible families of functions Let us begin with the following definition of a holomorphically contractible family which puts the functions of the introduction in a general context. The interested reader is referred to [Jar-Pfl 19931 and [Jar-Pfl 2005] for more information than is given in this chapter.
Definition 4.2.1. A family (dD)D of functions dD : D x D --> R+, where D runs over all domains D C C" (with arbitrary n E N), is said to be holomorphically contractible if the following two conditions are satisfied: (A) dm (a, z) = m (a, z) _ I l aza2 1, a, z E D (m is the Mobius distance),
(B) for arbitrary domains G C C"', D C C", any F E O(G, D) works as a contraction with respect to dG and dD, i.e.
dD(F(a), F(z)) < dG(a.z),
a,z E G.
(4.2.1)
(Compare condition (B) and (*) from Section 4.1.) Notice that there is another version of the definition of a holomorphically contractible family in which the normalization condition (A) is replaced by the condition
254
Chapter 4. Holomorphicaily contractible families on Reinhardt domains
(A') dm = p, where p =
2
log i±' is the Poincare distance on D.
Both definitions are obviously equivalent in the sense that (dD)D fulfills (A) and (B) iff the family (tanh-I dD)D satisfies (A') and (B). In our opinion the normalization condition (A) is more handy in calculations.
Remark 4.2.2. (a) Recall that m and p are distances on m. (b) If F E Bih(G, D), then F-I E O(D, G). Therefore,
dD(F(a), F(z)) = dc(a,z),
a, z E G.
In particular, if F E Aut(D), then dD(a, z) = dD(F(a), F(z)), a, z E D. (c) Moreover, if Di C C"J, j = 1.2, are domains, then
dD1XD2((a,,a2),(bi,b2)) ? max{dD1(a,,bl),dD2(a2,b2)},
(4.2.2)
whenever (a 1,a2), (b1,b2) E DI x D2 (EXERCISE, use (4.2.1) for the projection maps). If in (4.2.2) always equality holds, then we say that (dD )D satisfies the product property. The following holomorphically contractible families of functions seem to be the most interesting ones.
Example 4.2.3 (Mobius pseudodistance). MD (a, z) :
= sup{m(f (a), f(z)) : f e 0(D, m)} = sup{I f(z)I : f E 0 (D,1D), f(a) = 0},
(a, z) E D x D;
the function CD := tanh-I mD is called the Carathesodory pseudodistance. Indeed, to prove (B) it suffices to note that for F E O(G. D) and f E O(D, ID) one has f o F E O (G, ID). Moreover, the fact that
m(a. z) = m(f (a). f (z)),
f E Aut(D),
a, z E m,
gives the second equality in the definition of mD. To obtain condition (A) it suffices to observe that m m (0. ) = m (0. ). And this equation follows immediately by using the Schwarz lemma. Observe that mD (resp. CD) is positive semidefinite, symmetric and it satisfies the triangle inequality (EXERCISE). So mD and CD are, in fact, pseudodistances
on D. For D = C, Liouville's theorem immediately gives that me = 0; thus, in general, mD (resp. CD) is not a distance.
Example 4.2.4 (Mobius function of higher order).
mDk)(a. z) := suP{If(z)II1k : f E O(D, ID), orda f > k},
(a,z)EDxD, kEW,
4.2. Holomorphically contractible families of functions
255
where orda f denotes the order of zero of f at a. Indeed, in order to see (B) it suffices to observe that for a, z E D, F E O (G, D) and f E O(D, iD), ordF(a) f > k, one has f o F E O(G, ID) and orda f o F > k.
In particular, mD)(a, z) = mD)(F(a), F(z)) if F E Aut(D). Therefore, to see (A) it suffices to show m(k)(0, z) = m(0, z) which is a simple consequence of the Schwarz lemma.
Remark 4.2.5. Note that, in virtue of Montel's theorem (see Theorem 1.7.24), there exist extremal functions for mD), i.e. for any domain D C C", any pair
(a, z) E D x D, and any k E N there exists an f E O(D, D) with orda f> k such that mD)(a, z) = I f(z)I'lk Example 4.2.6 (Pluricomplex Green function).
gD(a,z) := sup(u(z) : u: D -+ [0. 1), logu E TSX(D),2 3C=C(u,a)>0 WED : u(w) < CIIw - all),
(a,z) E D x D.3
The point a is called the pole of the pluricomplex Green function.' Indeed, to see (B) let F E O(G. D), where D C C" and G C C'" are arbitrary domains. Fix an a E G. If u : D --> [0,1) is log-psh satisfying u < C II -F(a) II on D, then log u o F E PSX(G) and
(u o F)(z) = u(F(z)) < CIIF(z) - F(a)II < CIlz -a1I,
z E IB(a,r) Cs G,
where e and r are suitably chosen. Therefore, u o F < gC, (a, z). Since u was arbitrarily chosen, it follows that gD (F(a), F( - )) < gG (a, ). In the case where D = ID we fix a u : U) -* [0,1) such that log u is psh and u(A) < C IAI. Observe that then u/I ido I E SJ{(!D,) and that this function is locally bounded in D. Therefore, it extends to a sh function on the whole of D. By the maximum principle it follows that u < I id[) I. Hence we have that u = I idm I, i.e. gp(0, - ) = m(0, ). The situation for a general pole follows immediately using (B) and a MObius transformation.
While (dD)D, dD E {MD, mD), gD}, are based on families of functions we turn now to families defined by geometric conditions, namely by a set of analytic discs. 2Recall that 9S9-C(D) denotes the family of all functions plurisubharmonic on D. ;Note that it suffices to have u(z) < C (I z - a i( for all z E I(a, r) \ {a } when r > 0 is sufficiently small. Moreover. u(a) = 0 (Ext=atcISE). 4For relations between the pluricomplex and the classical Green functions in the unit ball see (Car 1997]. For a different pluricomplex Green function see [Ceg 1995]. [Edi-Zwo 19981.
256
Chapter 4. Holomorphically contractible families on Reinhardt domains
Example 4.2.7 (Lempert function).
ko(a, z) := inf{m(1l, µ) : A, /I E m, 30EO(D.D) : w(),) = a, (p(/L) = z) =inf{µ E [0,1) : 3yvEo(D,D) : V (O) = a. (p(µ) = z) = inf{µ E (0, 1) : W(0) = a, V(µ) = z),
where (a, z) E D x D. Put k'D := tanh-t ko. Indeed, first we have to show that the above definition makes sense. So let us fix
points a, z E D. Connect them by a continuous curve, i.e. take a y E e([0,11, D) with y(0) = a and y(1) = z. In virtue of the Weierstrass approximation theorem, we may approximate y uniformly by a sequence of polynomial mappings (p1 )j. Taking a sufficiently large j we may assume that 411 Pi
- YII[o,l] < dist(y([0,1]), 8D).
Put p := pj and then
P(A) := P(A) + (a - P(0))(1 -A) +A(z - p(l)),
A E C.
Note that p([0,1]) C D, p(0) = a, and p(1) = z. Then p maps even a simply connected domain U, [0, 1] C U, into D (EXERCISE). Applying the Riemann E O (ID, U) with *(0) = 0 and W(p) = 1 for a mapping theorem leads to a suitable A E (D. Hence, 0 := p o tlr gives an analytic disc through the points a and z.5
Observe that if F E O(G, D), a, z E G, and Q E O(D, G), (p(0) = a and V(µ) = z for some it E [0, 1), then F o rp E O(0). D) with F o cp(0) = F(a) and F o V (It) = F(z). Hence (B) is fulfilled. To prove (A) use the Schwarz lemma in order to see that k®(0, z) > Izl = m(0, z), z E m \ {0}. To get the inverse inequality take simply qp c (9(U, ID), p(A) := x 1 1.
Exercise 4.2.8. Prove that
k p(a, z) = inf{lul : µ E m, 3,Eo(m,D) : qp(0) = a, (p(µ) = z),
a. Z E D.
For many purposes it is important to know whether the infimum in the definition of the Lempert function is taken by some analytic disc.
Definition 4.2.9. Let D C C" and a, b r= D. A mapping (p E O(U, D) is called a k'D-geodesic for the pair (a, b) 6 if there are 1t, A2 E 1) such that cp(.AI) = a, co(A2) = b, and kp(a, b) = m(AI, A2). 'Note that the same remains true in the case when D is a connected complex manifold (see [Win 2005)). 6We also say that rp is an extremal disc through a and b.
4.2. Holomorphically contractible families of functions
257
Remark 4.2.10. In general such a geodesic need not exist. For example, take D := iB2 \ ((1/2,0)). Observe that D is a non-taut domain. Fix now the points (1/4, 0) from D. For R E (0, 1) put (PR E 0(U), D), a (0, 0) and b (PR(A) :_ (RA, s(R)A(A -1/(4R))), where s(R) << 1. Then kv (a, b) < 11(4R). Since R was arbitrarily chosen, we have ko(a, b) < 1/4. _ (*1, *2) E 0(U, D) for Suppose now that there exists a kD-geodesic
(a,b) with *(A1) = a, *(-k2) = b, and 4(a, b) = m(At,)L2). Using Aut(D) (it) = b, and it = k p (a, b). Note that t t c-
we may assume that >/r(0) = a,
0 (U, U)) with Ort (0) = 0. The Schwarz lemma gives 1 /4 < kD (a, b) and therefore, *1 = idp. Then, taking into account that * maps ID into 1B2 leads to the fact that
0 (use the maximum principle). And therefore, *(1/2) = (1/2,0); a *2 contradiction. In the case of taut domains we always know that such geodesics exist.
Proposition 4.2.11. Let a, b be two points of a taut domain D C V. Then there exists a kD-geodesic for (a, b). Proof. By definition we have a sequence ((p) j C 0(ID, D) such that rpj (0) = a,
kn(a,b). By assumption, D is taut and V(Qj) = b with aj E (0, 1), and aj (p j (0) = a, j E N. Therefore, we may choose a subsequence (V jk) such that rpjk -+ tp E 0(ID, D) locally uniformly. Then (p(0) = a and rp(k,(a, b)) = b (EXERCISE), i.e. (p is a geodesic we were looking for.
0
Exercise 4.2.12. (a) Let D C C" be a domain and let a, b E D. A map rp E O(U, D) is an MD-geodesic for the pair (a, b) if there exist At, A2 E lD such that (p(At) = a, 9P(A2) = b, and mD(a, b) = m(A1, A2) Prove that any MD -geodesic for (a, b) is a k; -geodesic for (a, b). (b) Let cP E (9(1), D) be an MD-geodesic for ((p(A1), O A2)), whereAI 96 A2. Prove that mD(9 (A'), (P(A")) = m(A', A"), A', A" E i7, i.e. (p is an MD-geodesic for all pairs (cp(A'), Sometimes such a q is simply called an MD -geodesic. and use properties of Hint. Study the function m \ (At } 9 A H m subharmonic functions. (c) Prove that any complex mD-geodesic op E O(0), D) is a proper injective mtv(p
mapping.
The next example relies on the normalization (A'). Example 4.2.13 (Kobayashi pseudodistance).
kD(a, z) := sup{d(a. z) : dD a pseudodistance on D, d < kD} _: tanh-t kD(a, z), a. z E D.
258
Chapter 4. Holomorphically contractible families on Reinhardt domains
Indeed, let F E O(G, D) be as in (4.2.1). To any pseudodistance dD < kD on D x D we associate a new pseudodistance dG on G x G by dG (w', w") dD(F(w'), F(w")). Then
dG(w', w") < kD(F(w'), F(w")) < kG(w', w"). Therefore, dD(F(w'), F(w")) < kG(w', w"), w', w" E G. Since dD was arbitrarily chosen we end up with (4.2.1). For the normalization (A') it suffices to mention that p is a distance.
_ (a) kD is the largest pseudodistance on D below of kD; (b) CD
Remark 4.2.14. Note that (EXERCISE):
(d)kp(a.z)=inf{FNtkD(zj-l,Zj):NEM,a=20....,ZN=ZED), a,' E D. To any pseudodistance dD E (CD, kD), D C C", one associates the dD -length of a curve a : [0, 1] -+ D as N
LdD (a) := sup {
dD (a(t j-1). a(lj )) : N E M, 0 = t0 < ... < tN = l t 7
j=l Exercise 4.2.15. Calculate Lp([0, s])," where s E (0, 1). Here [0, s] is just an abbreviation for the curve a : [0. 1] -+ ID, a(t) := ts. It is clear (use (4.2.1)) that LdD (F oa) < Ldc (a) whenever F E O(G, D) and
a: [0, 1] -+ G. Moreover, by the triangle inequality, dD(a(0),a(l)) < LdD(a). A more precise result is true in case of the Kobayashi pseudodistance.
Proposition 4.2.16. Let D C C" and a, b E D. Then kD(a,b) = inf{LkD(a) : a: [0, 1] -> D continuous and II . II-rectifiable,
a(0) = a. a(l) = b}. Proof. By Remark 4.2.14(d) it is clear that kD(a, b) is less than or equal to the right-hand side. Now fix an s > 0. By definition we find points sj E [0, 1) and analytic discs (pj E O(ID, D), j = 1.... , k, such that Wj (0) = a.
cpj (sj) = (pj+t (0), 1 < j < k,
cPk
b.
k
p(0,sj)
j=l 7By Corollary 4.2.25, the length is finite if the curve a is assumed to be II . 11-rectifiable, i.e. there
existsanM>0suchthatFN 1IIa(tj)-a(tj-t)8<Mwhenever NEW.0=to
259
4.2. Holomorphically contractible families of functions
Obviously, we may assume that all the sj's are positive. Put
a(t) := rpj((t - )ksj), ift E [if, J] and j = 1,...,k. Then a is a piecewise real analytic curve in D connecting the points a, b. Therefore, we have k
Lko(a) < E Lko(alll j1) j=I k
k
Lkm([O,sj]) < FP(O,sj) < kD(a,b) +e. j=1
j=1
0
Since the choice of e was arbitrary, the proof is finished.
Looking at the proof of Proposition 4.2.16, one easily concludes the following corollary (EXERCISE).
Corollary 4.2.17. Let D C C" and a. b E D. Then kD(a. b) = infILkp (a) : a: [0, 1] -> D piecewise real analytic,
a(0) = a, a(1) = b}. Remark 4.2.18. We have to point out that for the Carathdodory pseudodistance Proposition 4.2.16 is no longer true. Already for the very simple domain D = A(1/R, R) a counterexample can be given. For more details see [Jar-Pfl 1993], Example 2.5.7.
Lemma 4.2.19. For any domain D C C" the following inequalities are true:
mD = mod <
mD)
< gG < kD,
CD < kD < kD,
and for any lwlomorphically contractible family (dD )D we have
mD < dD < kD.
(4.2.3)
i.e. the Mbbius fancily is minimal and the Lempert family is maximal.
Proof. Fix an a E D. Let f E O(D, ID) with f (a) = 0. Then f k E O(D. m) with orda f k > k. Therefore, I f(z)I = I fk(z)I'ik < mD,(a, z), z E D. Since f is arbitrarily chosen we end up with MD (a, ) <
Now let f E O(D, U) with orda f > k. put U := Illtjk Then logu is psh and u(z) < C IIz -all and so l f(z)lI/k = u(z) < 9D (a, z), z E D. Hence m(k)(a, ) < gG(a, . ).
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Chapter 4. Holomorphically contractible families on Reinhardt domains
Fix a z° E D. Let u: D -+ [0,1) be such that logu E PSJ{(D) and u(z) < CIIz - all, z E D, and let O E 0(m, D) with rp(0) = a and W(s) = z° for a certain µ E [0, 1). Then u o V E PS3{(D), u o (p(0) = 0. Therefore, applying the Schwarz lemma for psh functions, we get u o V(A) IA , A E m. In particular, u (z°) = u -(p(µ) < A. Since u and rp were arbitrarily chosen, we have gD (a, z°) k'D(a, z°). Now let (dD)D be an arbitrary holomorphically contractible family. Fix a domain D C C" and points a. Z E D. Let now f E O(D, ID) with f (a) = 0. Then
dD(a, z) > d®(0, f(z)) = m(0, f(z)) = I.f(z)I. Hence, dD(a, z) > MD (a, z). Finally, let (p E O(U), D) with V(0) = a and V(µ) = z for a certain µ E [0. 1). Then dD (a, z) = dD (V(0), (p(µ)) < d® (0, µ) = m(0,µ) = µ. And so dD(a, z) < k *D (a. z).
Remark 4.2.20. (a) Observe that k c = 0 on C x C (EXERCISE). Then also k c, = 0 via (4.2.1). Therefore, kD is, in general, not a distance. (b) Moreover, the following result is true (see [Jar-Nik 2002], [Nik 2002]): Let
F j C C be a closed subset, j = 1, ... , n, n > 2, such that Ft # C 0 F2. Put D := C" \ (Ft x . . . x F"). Then kD = 0 on D x D. In fact, for any two points a, b E D there exists a q E O(C, D) such that cp(0) = a and Cp(l) = b. For balanced domains we have the following formulas.
Proposition 4.2.21. Let D C C" be a balanced domain given as
D=(zEC":h(z)
gD (0, ) = k v (0,-)=h on D. (c) Even more, if D is a convex domain, then hID.9
MD(0,
Proof. (a) Fix a z° E D. If h(z°) = 0, then Cz° C D. Therefore, using the holomorphic contractibility, k; (0, z°) < kZ(0,1) = 0 = h(z°). So we may A E U), gives a (p E O(ID, D) with assume that h(z°) # 0. Then V(A) := rp(0) = 0 and (p(h(z°)) = z°. Hence, £D(0, z°) < h(z°). h(sz°,
9Recall that a convex (pseudoconvex) complete Reinhardt domain is in particular a convex (pseudoconvex) balanced domain.
4.2. Holomorphically contractible families of functions
261
(b) Recall that by assumption log h E PS7{(C") (see Proposition 1.15.11) and
h(z) < Ilzllh( TZT ) < Cllzll, Z E (C"). (use that h is upper semicontinuous and
therefore bounded on dIB). Since 0 < h < 1 on D it follows that h < 9D(0, - ). (c) In the last case just apply the Hahn-Banach theorem to get h ID < mD (0, ).
0 In particular, (b) implies the following result for biholomorphic mappings.
Corollary 4.2.22. Let Df = {z E C" : hj (z) < 1) be a pseudoconvex balanced domain, j = 1, 2. If F E Bih(D1, D2) with F(0) = 0, then h2 o F = hl on D1.
Remark 4.2.23. In fact, much more is true, namely if Bih(D1, D2) # 0, then Biho,o(Di, D2) # 0 (see [Kau-Upm 1976], [Kau-Vig 1990]), where Dj are pseudoconvex balanced bounded domains in V. Later we will even see that if Bih(DI, D2) 0 0, then DI is linearly equivalent to D2 (see Proposition 2.1.9 in the case of norm balls). Moreover, we have the following explicit formulas.
Corollary 4.2.24. Let a, z E IB". Then
k' (a, z) = k' (a, z) _ B
(1-
(1- IlaII2)(1- Ilzll2)11/2,
I1-(z,a)I2
1
where ke = tanh-t kB". In particular, all functions introduced so far coincide on IBn.
Proof For a E IB \ {0}, apply ha E Aut(IB" ),
ha(z) =
1 - llail2(z - Bali a) - a + Ilall a
I - (z,a)
ZE8"
(cf. Example 2.1.12 (b)), and the above proposition.
Corollary 4.2.25. Let D C C" be a domain and 1B (a, r) C D. Then
z) < 11z -aall , z E IB(a, r). r In particular, kD is locally bounded from above by the Euclidean norm.
Proof. Fix azElB(a,r)CD.Then k,(a, z) < k®(a,r)(a. z) = ke(r)(0, z - a) = liz - all /r, since helrlW =
IIII/r
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Chapter 4. Holomorphically contractible families on Reinhardt domains
Remark 4.2.26. (a) Put D := C2 U (({0} x m) U (ID x {0})). Then D is a balanced domain which is not pseudoconvex (EXERCISE). For R > 1 put ccR(A) (A(RA-1), 2A),A E ID. ThentpR E O(ID.D)withcpR(0) = (0,0)andcpR(1/R) _
(0, 1/2). Therefore kkD((0.0), (0, 1/2)) = 0 < hD(0, 1/2).
(b) Let D := {z E C2 : Iztl < 1, Iz21 < 1, Iztz21 < r}, where 0 < r < 1/4. Obviously, D is a pseudoconvex complete Reinhardt domain (in particular, a balanced domain) which is not convex. Its Minkowski function is given by h(z) := max{Izt I, IZ21
r-t Iztz2l} Therefore, ko(0, (t, t)) = L-, r < t < f. Then
kD(0, 0,0) < kD(0, 0,0)) + kD((0, t), (t, t)) <
tanh_t
(m (0, t)) + tanh-t (mK(r/r) (0, 0)
= tank- t (t) + tanh- t (t 2/ r) = < 2log
+t
t2
1
2 log
Wt
r+t2I
= kD(0, (t, 0),
when t is sufficiently near r. In the second inequality we have used that the above functions are holomorphically contractible. Note that this example shows: kD is not identically equal to tanh-t oh ID. m_ D is not equal h I D
kD does not satisfy the triangle inequality. Hence the introduction of the Kobayashi pseudodistance is justified.
Corollary 4.2.27. Let D C C" be a domain and a E D. Then
gD (a, z) < C 11z -all, z E D.
Proof. Set u : = 9D (a, ). Then u*, its upper semicontinuous regularization, is log-psh. Moreover, if IB(a, r) C D, then u(z) = gD(a, z) < gB(a,r)(a. z)
1Z;a
,
z E Ia(a, r).
In virtue of the maximum principle, we get u* < 1 on D. Therefore, u* = 9D (a, ), which obviously implies the corollary.
Remark 4.2.28. There exists a pseudoconvex bounded balanced domain D C C", n > 2. such that its Minkowskifunction h is not continuous (see Proposition 1.15.12). Therefore, 9 D( 0 ,-) = kD (0, - ) is not continuous.
Applying Corollary 4.2.24 we get
4.2. Holomorphically contractible families of functions
263
Proposition 4.2.29. Let D C C". Then: (a) The functions MD, kD are continuous. (b) The functions mD and kD are upper semicontinuous. (c) If, in addition, D is assumed to be taut, then kD is continuous.
(d) For a E D, mD (a, ) is continuous. Proof. (a) Fix points a, a', b, b' E D and let dD E {kD, MD). Then I dD (a, b) - dD (a', b') I
dD (a, a') + dD (b, b').
Therefore it suffices to apply Corollary 4.2.24 and the fact that MD < k *D'
(b) The case mDl: Let D 3 aj -+ a and D -3 bj -+ b. According to the remark concerning extremal functions there are fj E 0(D, ID) with ord01 fj > k
and Ifj (b j) I t/ k = m
l (aj, by), j E N. Using a Montel argument gives an D f > k and f E O(D, D) with orda
mD)(ajv,bj») =
Ifi"(bjv)I1/k
If(b)I'l"`
s+Dl(a.b)
for a suitable subsequence, i.e. mD) is upper semicontinuous. The case in : For an arbitrary s > 0 choose an analytic disc cp E O (lD, D)
with cp(0) = a, (p(s) = b, and (0,1)Dµ < k,(a, b) + E. Then V(FD) is compact and therefore, dist(V(D), 8D) _: r > 0. Fix a' E IB(a, pr/6) C D and b' E IB(b,.tr/2) C D. Now we define a new analytic disc * E O(0), D) by
a ((µ - )(a' - a) + l(b' - b)).
A E ID.
Therefore, kL(a', b') < µ < kD(a, b) + e.
(c) Assume that kD is not lower semicontinuous at (a, b) E D x D. Then kD (a, b) > 0 and there are sequences (aj) j , (b) j C D with a j -> a and b j -> b such that for all j,
kD(aj,bj)
All the functions introduced so far in this section behave well under union of domains; to be precise we have the following result.
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Chapter 4. Holomorphically contractible families on Reinhardt domains
Lemma 4.230. Let D = Ujro°_ I Dj C C", Dj C Dj +t , be the increasing union of the domains Dj, j E N. Then dD1 - dD if j -* oo, where d E {m(k), g. k*, k).
Proof. We restrict ourselves to prove this lemma only for d = k*; the remaining cases are left as an EXERCISE for the reader. Obviously, we have
j E N.
kDi > kDi+, > k*D,
In particular, we havej -+oo lim kD. > kD. Now fix points a, z E D and choose a jo
'
Jim k )., (a. z) > k,(a. z). Then
such that a, z E Dj, j > jo. Suppose that p
there exist an analytic disc tp E O (U), D) and a number 00 r E (0. p) such that cp(0) = a
and rp(r) = z. Select an e > 0 such that (1 + s)r < p. Put O(A) := VOA/(1 + s)), A E [D. Then 0 E O(ID, D), 0 (0) = a, and 0 (r(1 + s)) = z. Note that 0 (lD) C= D. Therefore, 0 E O(UD, Dj), j >> 1, which implies that kDj (a, z) < (I + s)r < p; a contradiction.
Recall that the pluricomplex Green function gD (a, - ) = h, a E D, need not be continuous (see Remark 4.2.28), where D = Dh denotes a pseudoconvex balanced domain with Minkowski function h. With the help of Lemma 4.2.30 we get the following continuity result for the pluricomplex Green function.
Proposition 4.2.31. Let D C C" be a domain. Then gD is upper semicontinuous
onDxD. Proof. In view of Lemma 4.2.30 we may restrict ourselves to study only a bounded domain D. To be able to continue we need the following lemma.
Lemma 4.2.32. Let D C C" be bounded, assume that IB(a, r) C D, and lets > 0. Then there exists a 8 E (0, r) such that (gD(z, W))' +e < 9D (a. w).
z E B(a. 8), w e D \ 8(a, r).
Proof. Puts := r/3 and R := diam D. Then, by (4.2.1) and Proposition 4.2.21,
gD(z. w) < ge(z,2s)(z, w) <
2sw11
Ilz
,
z. w E Q3(a, s).
Now fix an s > 0 and choose a positive 8 E (0, s/3) such that
33)1+6
S
2s
u(z) _
z-° m ax I(gD (b, z))t+8, =-Q }
if z E l8(a, 28),
if z r= D \ !(a, 28).
4.2. Holomorphically contractible families of functions
265
z_Q
Note that (gD(b, z))1+E < for z E al8(a, 28). Therefore, u is log-psh on D. Moreover, it fulfills all other conditions to be a competitor in the definition of the pluricomplex Green function with pole at a. Thus,
(gD(b, w))t+E < u(w) < gD(a, w),
w E D \ IB(a, 28).
It remains to mention that D \ IB(a, r) C D \ IB(a, 28). Obviously, gD is continuous at points (a, a) E D x D (EXERCISE). So without
loss of generality, let (a, b) E D x D with a 0 b. Choose an r > 0 such that b 0 IB(a, r) C D. Assume now that 9D (a, b) < a < f < 1. Then fix an s < 0 such that a < 9D (Z, w)
01+E
Taking the corresponding 8 from Lemma 4.2.32 we see that (9D (a,
w))1/(1+E)
z E IB(a, 8). W E D \ IB(a, r).
Recall that gD (a, -) E T85C(D); in particular, gD (a, -) is upper semicontinuous in b. Therefore, gD (a, w) < a, when w E IB(b, >?) C D \ IB(a, r) for a sufficiently small >) > 0. Hence, gD(z, w) < a'/(1+E) < P, z E IB(a, 8), w E IB(b, >)), which proves the upper semicontinuity. Exercise 4.2.33. Prove the following slight generalization of Lemma 4.2.32. (a) Let D, a, r, and s be as in Lemma 4.2.32. Then there is a 8 E (0, r) such that (9D (Z, w))1+E < 9D W, w),
Z. Z' E @(a, 8), WED \ R(a. r).
(b) Show (using (a)) that for a bounded domain D C C" the function gD (-, w) is continuous if w E D is fixed. Moreover, we have the following deep result due to Demailly (cf. [Dent 1987]; see also [Kli 1991]) which we will not prove in this book.
Theorem* 4.2.34. Let D C C" be a bounded hyperconvex domain. Then the function gD is continuous on D x D, where 91DxaD
I-
To get explicit formulas for the "invariant" objects on Cartesian products we discuss the following result which is extremely useful.
Proposition 4.2.35. The family (k p)D satisfies the product property; i.e. for all pairs of domains Dj C C"J, j = 1, 2, and points (a1, a2), (b1, b2) E D1 x D2 one has
kD,xD2((al,a2),(b1,b2)) = max{kp1(a1,b1),ko2(a2,b2)}.
266
Chapter 4. Holomorphically contractible families on Reinhardt domains
Proof. Because of (4.2.2) only the remaining inequality has to be verified. Suppose that
p := kD,XD2((a1,a2) (b1.b2)) - r
> max{i 1(a,,b1),kD2(a2,b2)} = kD1(a,,b1) for some r > 0. We find analytic discs (pj E O(ID, Dj) with (pf (0) = aj and
(pj(p,) =bb,where kD,(a1,b1) <µI <pandµ2 E(0,A1). Put (P(A) := ((P1(),), (p2(u?A)).
)L E ID.
Then (p E (O(D, DI x D2), 00) = (a,,a2), and (p(s) = (b1,b2). Hence, p > kD,,,D2((a,, a2), (b1, b2)); a contradiction.
0
Exercise 4.2.36. Use Proposition 4.2.35 and Corollary 4.2.24 to prove that ID" x IB," is not biholomorphically equivalent to EBm+".
Remark 4.2.37. Note that also the family of Mobius pseudodistances (resp. of pluricomplex Green functions) fulfills the product property; for details see [Jar-Pfl 2005] (resp. [Jar-Pfl 1995], [Edi 1999], and [Edi 2001]). Obviously, if the product property holds one can get new formulas for the invariant functions for Cartesian products.
Proposition 4.2.38. Let Di C C" be a domain, j = 1, 2, and let F E 49 (D1. D2) be such that daED2 dbEDl, F(b)=a d9)EO(D,D2),Qp(°)=a 3*EO(®,D1) : *(0) = b, F o * = (p. (4.2.4)
Then, for points a1, a2 E D2 and bI E DI with F(b1) = a1, one has
kD2(a1, a2) = inf{kD, (b1, b2) : b2 E D1, F(b2) = a2}:
(4.2.5)
kD2(a1,a2) = inf{kDI (bl, b2) : b2 E D1, F(b2) = a2}.
(4.2.6)
Remark 4.2.39. (a) Recall the following definition: an F E O(DI, D2), DI, D2 domains in C", is said to be a holomorphic covering if for any z E D2 there exists a neighborhood V = V(z) C D2 such that F-I (V) = UJEJ U1, where Uj is an open subset of D 1, such that F I u, : Uj - V is biholomorphic, j E J. Any holomorphic covering F : D I -+ D2 satisfies (4.2.4). The reader is referred
to [Con 1995]. Even more is true: for any b E DI, A E ID, and 1p E O(U, D2) with (p(2.) = F(b) there exists a unique tfi E O(1), DI) such that *(A) = b and F o * = (p; * is called the lifting of (p with respect to F. (b) Recall that for any plane domain D C C there is a simply connected domain
D° E {C, ID) and a mapping F E O(D°, D) with the property (4.2.4). D° is the
4.2. Holomorphically contractible families of functions
267
universal covering domain of D (cf. the uniformization theorem in the classical complex analysis of one complex variable). (c) In Example 4.4.16 (see [Zwo 1998]), an example will be given where the infimum in equation (4.2.5) is not attained. This gives a negative answer to a long standing question asked by S. Kobayashi. Proof of Proposition 4.2.38. In view of (4.2.1) we obviously have
kD2(a,,a2) < inf{kD, (bl, b2) : b2 E D1. F(b2) = a2}. Assume that the above inequality is a strict one. Then there exists an analytic disc cP E O(0). D2) with (p(0) = al and (p(µ) = a2, where
inf{k,,(bl,b2) : b2 E D1, F(b2) = a2} > µ > kD2(a1.a2). Applying property (4.2.4), we find an analytic disc * E 19(0), D1) with Or(0) = b1
and F o /r = V. Therefore, A
kD1(bt, *(A)) > inf{kDl (bl, b2) : b2 E DI, F(b2) = a2} > /A:
a contradiction. Hence (4.2.5) has been verified. The equality (4.2.6) could be proved in a similar way. Its proof is left as an exercise for the reader.
Exercise 4.2.40. Find the formula for
D
where O+ :_ {A E C : Red, > 0).
One of the most important results in the theory of invariant functions is the following one due to Lempert. Its proof is beyond the scope of this book. Therefore, the reader is referred to [Jar-Pft 1993].
Theorem* 4.2.41 (Lempert theorem). (a) Let D C C" be abounded convex domain
and let a, b E D. Then there exists acomplex MD-geodesic W E O(0), D) such that a, b E (p(ID). In particular, CD = kD on D x D. (b) Assume that D C C" is a domain which can be exhausted by an increasing sequence (Dj) j of domains Dj C C", where each of them is biholomorphically equivalent to a convex domain. Then CD = kD on D x D. Note that (b) is a simple consequence of (a) and general properties of the MObius and the Lempert functions (EXERCISE).
As an application for pseudoconvex Reinhardt domains we have
Theorem 4.2.42. Let D C C be a pseudoconvex Reinhardt domain. Then
kD=kDon DxD. In particular, kD is continuous.
268
Chapter 4. Holomorphically contractible families on Reinhardt domains
Proof. Recall that the logarithmic image log D is convex. Hence the tube domain
TD := log D + i IR" is convex. Therefore, by the Lempert theorem, we have kTD = kTo on TD X TD. D, F(z) := eZ, is a On the other hand, observe that the mapping F: TD holomorphic covering. Therefore, Proposition 4.2.38 immediately gives the proof of the equality in the theorem. Recall from Remark 4.2.26 that for a general pseudoconvex Reinhardt domain D the Kobayashi pseudodistance kD and the Lempert function kD are, in general, different. Example 4.2.43. Another application of Theorem 4.2.41 gives the following result (see Exercise 2.1.13). Let N be a complex norm on C", n > 2. Recall that !B = (z E C" : N(z) < 1),
B(r)={zEC":N(z)
w :_ p(f). Then, in virtue of Exercises 1.1.1(c) and 4.2.12 (b),
p(0, ll) = cB(0, O(w)) = cB(a, w) = p(a, fl) = p(0, P) - p(0 a). Therefore, 0 = p(0,a) = cB(0,a), i.e. a = 0. Hence, 0 E Auto (0). Remark 4.2.44. (a) We mention that recently a domain G2 in C2 has been found for which mq2 -k G2 , but which does not fulfill the assumption of Theorem 4.2.41(b). Here we only give the definition of G2,
G2 := {z E C2 : IzI - 2IZ21 + IZ212 < 1}.
Let 7r: C2 -+ C2,,r(z1,z2) := (ZI +z2,ztz2). Then G2 = 7r(D2) and 7r: D2 G2 is proper (EXERCISE).
For more details and other sources the reader may contact [Jar-Pfl 2005]. This is, at least at the moment, the only known example (up to simple modifications) with these properties. (b) The notion of a holomorphically contractible family (dD )D (Definition 4.2.1) can be extended to the case where D runs through all connected complex manifolds, complex analytic sets, or even complex spaces. In particular, one can define the Mobius pseudodistance mm, the Lempert function k;, (defined as 1 for pairs of points for which there is no analytic disc passing through them), and the Kobayashi
4.3'. Hahn function
269
pseudodistance km for an arbitrary connected complex analytic set M. For recent results in case of the Neil parabola M := {z E C2 : zi = z2 } see, for example, [Kne 20071, [Nik-Pfl 2007], [Zap 2007).
4.3' Hahn function Note that in Definition 4.2.1 one can also consider conditions that are weaker than (B), for instance: (B') Condition (4.2.1) holds for every injective holomorphic mapping F : G --> D.
Example 4.3.1 (Hahn function). HL (a, z) := inf{m(,1, µ) : 3,Eo(1D,D) : tp is injective, 1p(A) = a, (P(µ) = z) = inf{IuI : 34 eO(D.D) : sp is injective, V(0) = a, (p(µ) = z),
where (a, z) E D x D, satisfies (A) and (B').10
Remark 4.3.2. Observe that the infimum in the above definition is taken over a
non-empty set. Indeed, fix points a, b E D. a 0 b. Then there is an injective C1-curve a : [0, 1) -+ D connecting a and b such that a'(t) 0 0, t E [0, 11. By the Weierstrass approximation theorem, we find a sequence (p j) j e 14 of polynomial
mappings p j : C -> C" such that
Pj (0) = a.
P1(l) = b,
11
k)
- a(k) II
Io.11
0,
k=0,1.
and
pj([0, 1]) C D,
j E N.
If j >> 1, then pj I(o,11 is injective. Indeed, suppose the contrary, i.e. there exist 1 ', with ,o1 tj', tj" E [0, 1],1 j j E N. By the compactness of [0, 1] P1 we may assume that ti -. t' and t". Then the uniform convergence of (pj)j
implies that a(t') = a(t"). Applying the fact that a is injective gives t' = t". Therefore, n
0
IIPi(ti)-Pi(tj")II2
= IPj.k(ti.k)I2Itf - tjlz k=1
n
Ilak(rj,k)I - I PJ,k(tj,k) -
a'(ij,k)II2Itjl - tfil2
k=1
2IIa'II[0,1 It'
_,j"12, if j >> 1,
where tj,k is between tj' and r,". Hence tj' = tj for j >> 1; a contradiction. 1°Observe that the definition of HD is similar to the one of k,.
270
Chapter 4. Holomorphically contractible families on Reinhardt domains
Fix a j such that pj is injective on [0, 11. Then there exists a simply connected
domain G C C with [0, 1] C G, pj (G) C D, and pj lG injective (EXERCISE). Arguing as in the case of the Lempert function we end up with an injective analytic disc in D passing through a and b.
0 0 H.
Remark 4.3.3. Obviously, kD < H. But
Indeed, fix two different points a, b E C,. Let V E O(U, C.) be injective with cp(0) = a, rp(µ) = b for a suitable µ E D. Applying the Koebe distortion theorem (see [Pom 1992], Theorem 1.3 and Corollary 1.4) we have lb - a I = I q (µ) - V(0)I < Iw'(0)I
1111
(I - (µl)2
< 4dist(a. af(tD))
1µl
(I -
1µl)2.
Taking into account that f (D) is simply connected we get
lb - al < 4lal
1µ1 -1µl)2.
(I
Hence Hey (a, b) > 0. On the other hand, the following result for n > 3 is due to M. Overholt.
Theorem 4.3.4 ([Ove 1995]). If D C C", n > 3, is a domain, then ko = HD on D x D.
Proof. Fix a, b e D, a # b. Without loss of generality, we may assume that a = 0 E D (EXERCISE). Let e > 0. Then there is an analytic disc (p E 49(0), D) with cp(0) = 0, cp(µ) = b for a suitable µ E m such that 0 < 1µI < kD (0, b) +e/2.
We choose an R E (0. 1) such that µ/R E U) and lµ/RI < k,(0, b) + e. Put cPR(A) := w(RA), IAI < 1/R. Obviously, VRlp E O(m, D) with WR(0) = 0 and cpR(µ./R) = b. Since 'pR is continuous on ID, we have dist('R(lD), aD) =: 2s > 0. Now we take a polynomial mapping p : C -+ C" coming from the power series expansion of'pR such that dist(p(m), aD) < s/2 and (PRCR)
PCR)II<2R 111,
Finally, put
P(A):=p(A)+
AA
((PR`R/-P`RII' AEC.
Hence PID E O(m, D) with p(0) = 0 and p(µ/R) = b. Observe that p = (P1 , ... , p") is a polynomial mapping with m
pj(A) = 1: aj,kAk, k=1
A E C. j = 1....,n,
4.3*. Hahn function
271
where m > n is sufficiently large. Put A:= [a j,kII<j
1: aj,k.lk)I<j
P(A) :_
I E C.
k=1
gives an injective mapping from ID to D with p(0) = 0 and p(µ) = b. µ := µ/R, i.e. a j,1 has to satisfy the equation m aj,kµklµ,
aj,I = bj - E
j = I..... n.
k=2
Assume that p is not injective. Then p(,X l) = p(A2) for certain A 1, A2 E C, A 1 36 A2. Therefore,
m
m
k=1
k=l
E aj,k'I =
j = 1.....n.
Or, after dividing by .l I - A2, m
-aj,1 =
E
k-1
aj,k
As Ak-1-s
12
).
j = I....,n.
s=0
k=2
Now, for an arbitrary n x (m - I) matrix A = [a j,k ] 1 < j
M(A) :=
{(z2.....
zm) E Cm-1 : jajkzk = -a1,1. j = I... .n), k=2
1
where
m
aj,I _ (bj
- 1: aj,kAk)/µ.
j = 1,...,n.
k=2
Observe that M(A) is an (m - I - rank A)-dimensional affine subspace of Cm -1. Therefore, there is a dense subset of matrices A in IM(n x (m - 1); C) such that the corresponding affine subspace M(A) has dimension m - 1 - n < m - I - 3. Moreover, define t
S
Lrr w1 I\ S=0 (
2 -s
E 1 <1 <m-1
cm-1
: w = (wI , w2) E C2, wl # W2
272
Chapter 4. Holomorphically contractible families on Reinhardt domains
Note that the map t
'P(wl,w2)
w1w2 s)1
Cm-1,
w1 1 # w2,
s=0
is a regular holomorphic mapping onto S with 45(wl, w2) = P(wl, W2) if and only if (wl, w2) = (iul, 02) or (w), W2) = (w2, w1). Hence, S is a 2-dimensional complex submanifold in Cm_i. It suffices to find a sequence of matrices (A(f))t C M(n x (m - 1);C) such that A(t) -- A when t --> oo such that s fl M(A(f)) = 0. 0 So the dimension 2 is left in the general comparison of the Lempert function and the Hahn function. Here we present an answer to what happens with the Hahn function for the product of two plane domains. It shows that both functions can be different also in the 2-dimensional case. Before stating this result we ask the reader to solve the following exercise which will be important in the proof of the following proposition.
Exercise 4.3.5. Let D C C" be a domain. Then the following properties are equivalent:
(a) ko = Hp (b) for any rp E O(U), D), 0 < a < 8 < 1 with cp(0)
ye(a), there exists an injective * E O(U), D) with t('(0) = cp(O) and *(8) _ sp(a).
Theorem 4.3.6 ([JarW 20011). Let Dj C C be a domain, j = 1, 2. (a) If at least one of the Dj 's is simply connected, then kDt xD2 = HDt xD2-
(b) If at least one of the D1 's is biholomorphically equivalent to C., then kD1 xD2 = HD1 xD2
(c) Otherwise, kvl xD2 $ HD, xD2
The proof of (c) will be based on the following nice lemma from classical complex analysis and the uniformization theorem.
Lemma 4.3.7. Let D j C C be a non-simply connected domain that is not biholo-
morphically equivalent to C., j = 1, 2. Denote by p j : m -> D j the universal covering mapping." Then there are two different points q1, q2 E lD and automor-
phisms fj E Aut(ID), j = 1, 2, such that pj (.f (q 1)) = pj (fj (q2)), j = 1, 2, and det
r(pi ° fi)'(gi) (P1 ° fi)'(g2)1 (P2 ° f2)'(g1)
(P2 ° f2)'(g2)J # 0.
Note that, in virtue of the uniformization result, the universal covering of D1 is given by the unit disc.
4.3*. Hahn function
273
Proof. By assumption the map p j is not injective, j = 1, 2. Therefore, there exists tlrj E Aut(ID) \ {idm} such that pj o 1lrj = pj, j = 1, 2; in particular, 1/rj is a lifting of p j . Note that *j has no fixed points in lD (otherwise, applying the uniqueness of the lifting, it would be equal to ide). Therefore, it has one or two fixed points on am (see Exercise 2.1.4 (b)). Fix A' E am with t/rj (A) # A' for j = 1.2. Then m(tA', * ( : A ' ) ) -+ 1 when t / 1, j = 1, 2. Hence we find zI, z2 E m with m(Z1, *I (Z 0) = m(z2, 0'2 (Z2)) E (0,1).
Let d E (0, 1) with m (-d , d) = m (zl , *I (z1)). Then, by Exercise 2.1.4(b), there exist h j E Aut(ID) with
hj(-d) = zj, hj(d) = IGj(zj) j = 1,2. Assume that (p j o h j)'(-d) 0 ±(p j o h j)'(d) for at least one of the j's, say for j = 1. Then one of the following determinants does not vanish: det
det
(PI °hI)'(-d) (pI ohi)'(d) (P2 o h2)'(-d)
(P2 o h2)'(d)
L(PI ohI o (- idr))'(-d) (Pt ohI o (- idr))'(d) JJ (P2 ° h2)'(-d) (P2 ° h2)'(d)
L
(EXERCISE, use that (P2 o h2)'(d)
0.)
So we may put fi = h 1, f2 = h2 (resp. f, = h I o (- id[), f2 = h2) and
qt=-d,q2=d.
Now, for the remaining part of the proof we may assume that
((pj ohj)'(d))2 = ((pj ohj)'(-d))2,
j = 1,2.
(4.3.1)
Put t%rj := hi I o t/ij o hj and Pj := pj o hj, j = 1,2. Then t%rj(-d) = d
and Pj'(-d) = (Pj o j)'(-d) = Pj'(jj (-d ))t/rj'(-d ).
Taking squares on
both sides we get ((-d))2 = I (see (4.3.1)). Therefore, either i/rj (-d) = d, jj(-d) = -1 or I%rj(-d) = d, t%rj(-d) = 1. Applying Exercise 2.1.5 (b) to 'p = - idm (in case of -) or rp = he and the fact that 3/j has no fixed points in ID, it follows that
-2d I =Vr2=hc with c:= 1+d 2' Now fix an a E D and choose cp E Aut(D) such that W(a) = I/r (a) and qp(* (a)) = a (see Exercise 2.1.4 (b)). Note that such a 'p exists.
Suppose that Sp'(a) = *'(a). By Exercise 2.1.4(b), SP = t/' and therefore (r o t/r(a) = a. So t/r o >/i has a fixed point in m and, therefore, none on ao). On the
274
Chapter 4. Holomorphically contractible families on Reinhardt domains
other hand, Or is without fixed points on D. So it has at least one fixed point on 8D, say b E 8U). Then i/i o tf(b) = b; a contradiction. Fix an ao E ID n IR. Let rp E Aut(D) with tp(ao) = }G(ao) and tp(t/r(ao)) = ao. Then gp = h_a0 o (- ide) o hhao(*(ao)) o hao. By a direct calculation it follows that
g'(ao) -Vr'(ao) Summarizing, we know that if tP E Aut(D) is such that g(ao) = t/r(ao) and g(tf (ao)) = ao, then g'(ao) # ±t/r'(ao). Then, by Exercise 2.1.4(b), tp o (p = id® (note that gp o go has two fixed points in U)) and so tp'(t/r(ao)) _5. Finally, we put ql := ao, q2 := *(ao), .ft := ht, and f2 := h2 o ip. Then Pt (fi (g2)) = (P1 o h l)(f(ao)) = (Pt o t/rl)(h1(ao)) = (P1 o h1)(g1) = (PI o ft)(g1), P2(f2(g2)) _ (P2 o h2)((P(tG(ao))) = (P2 o h2)(ao) = (P2 0 *2)(h2(ao))
= (P2 o h2)(tr(ao)) = (P2 o (h2 o (p))(ao) = (P2 0 f2)(gl)Moreover, we have Idet
(P1 0 .fi)'(gt) 0 .f2)'(g0
r
(P1 0 fi)'(g2)(P2
(P2 0
f2)'(q2)]
(Pt oh1)'(ao) o
h2)'(go(ao))go'(ao)
r(P1 ohl)'(tr(ao))tr'(ao)
det lL(P2
0
(P1
(p2° (pt oh1)'(tf(ao)) 1 (P2° h2)'(Vr(ao))/go'(ao)]
_ (P1 ohl)'(*(ao))(P2oh2)'(*(ao))det
*'(ao) Sv (ao)
1
Vr (ao)/go
(ao)]
0.
0
Hence this lemma is proved.
Proof of Theorem 4.3.6. (a) Without loss of generality, we may assume that D1 is simply connected (EXERCISE). Our task is to apply Exercise 4.3.5. So let (p =
(g1, g2) E 0(ID, D1 x D2) and 0 < a < 8 < 1 with tp(0) 0 to(a). Assume that tpl (0) 54 VI (a). Recall that koi = HD1. Hence there exists an injective t%rl E 0 (U), D 1) with t (0) = rpt (0) and it (8) = (Pt (a). Put
* (A) :_ ( l (A), p2('A)). Then
A E ID.
E 0(ID, D1 x D2), t1' is injective, and one obtains *(0) = (p(0) and
tr(8) _ 0a) Now let (pl (0) = VI (a) and tp2(0)
(p2(a). Take a d E (0, dist((pl (0), i)D1))
275
4.3*. Hahn function
and put h(A)
V2(04 -'i(0)
(4.3.2)
W2 40 - 'P2(0)
*I (A) := VIP + M8 + 1(h (A)
A E ID,
(4.3.3)
S
where M := IIhII fi. Observe that I/it E O(L, DI). Finally, define *(A) :=
(i/ri(A). (M'10), A E D. Then Ir E O(D, D1 x D2) with *(0) = V(O) and cp(a). Moreover, one easily sees that I/r is an injective analytic disc. Hence, (a) is proved. C (EXERCISE). Let, as in (a), (b) We may assume that DI = C. and D2 54 '(a). Moreover, ' P = ('PI , V2) E 0 (1D, DI x D2), 0 < a < 8 < 1 , a n d
applying a suitable automorphism of C. we may even assume that 'PI (0) = I (EXERCISE). _ rp2(0) + dist('p2(0), aD2)I). In the case where w2(0) = 'p2(a) define D2 Obviously, D2 is a simply connected domain, 0 _ (VI, 02) E 0(D, DI X D2), where 02(l) := V2(0), A E D. In virtue of (a), there exists an injective analytic disc /r E O(D, D1 X D2) with I/r(0) = (p(0), *(8) = sp(a) = cp(a). Next, we discuss the situation when Ip2 (0) # cp2 (a). For the moment we assume,
in addition, that (pt (a) = 1 + 8. Put
I/i(A) :_ (1 + A, (2(gA)),
A E D.
Then ly E O(D, C* x D2) is injective and satisfies VV (0) = p(0) and I/r(8) = Ip(a). Now we turn to the remaining case IpI (a) # 1 + S. Then, fork E N, we choose and Arg(dk) 0 when k -> oo. numbers dk E C \ (1) such that dk = Note that dk -+ 1. Put (P2(00 - 'P2(0) Ck
Since Ick I
1 - dk
k E W.
oo we choose a ko such that Ickol > M := sup{IIp1(A)I : IAI <
u}.
Define
fr(A) := ((1 + A)hk°(A),'2(gA)),
A E ID,
where h(A)
(P2(PA)-Cko,
A E D.
'P2(0) - cko
Then h E 0(U), C*) and so >/r E 0(iD, DI x D2) with IJr(0) = (1, W2 (0)) = V(0). Moreover, a short calculation leads to'r(8) = rp(a). If *(A') = *(A"), then h(A') = h(A"), and therefore, A' = A", i.e. 1/r is also injective. Hence, the proof of (b) is complete.
Chapter 4. Holomorphically contractible families on Reinhardt domains
276
(c) Recall that the universal covering of D j is ID and that the covering mapping pj : ID -+ Dj is locally biholomorphic and surjective, but both are not injective, j = 1, 2. Applying Lemma 4.3.7 we find a point q = (q1, q2) E 0)2, qI q2, and
automorphisms fj E Aut(ID), j = 1, 2, such that with fj := pj o fj, j = 1.2, the following is true:
Pj(gt) = Pj(g2), j = 1,2,
and
det p; (qt)
P2(qt)
P;(42) # O. P2(g2)
Moreover, choose an r E (0, 1) such that both mappings fj are injective on K(r) and
put a := (a,, a2) = (P1(0), h(0)), b := (bt. b2) = (fit (r) P2(r)) E DI x D2. Note thataj 36 bj, j = 1,2. Then, in virtue of Proposition 4.2.35, Proposition 4.2.38, and the choice of r, we have
kptxD2(a,b) = max{k,1(a,,bl),kD2(a2,b2)} = r. Assume now that k,1 xD2 = Hot xD2 in particular, r = kpl D (a, b) _ HDIXD2(a.b). Then there exist a sequence of analytic discs ((pj)j C ( (ID, DI x D2) and a sequence of numbers (aj )j C (1, 1 / 4) with a j \,, I such that (p j (0) _ a and (p j (a j r) = b for all j. Then, applying Exercise 4.3.5, we find /,, = (j(lj,I, *j,2) E O(ID. DI x D2) injective such that *j (0) = a and t/rj (a4 r) = b, j E W. Recall that P j are covering mappings. Therefore, we can lift the functions *j,k, k = 1, 2, i.e. there are holomorphic mappings 1Jij,k E 19(0), ID) such that Pk ° %j,k = *j,k and tyj,k(0) = 0. Note that (fk ° t%ij.k)(c r) = fk(r). Recall that fk is injective on K(r) and therefore injective on K(r +s), where e E (0,1-r) is sufficiently small (EXERCISE). Then, for large j, we have that 1/ij,k (a?r) = r,
k=1,2. By the Montel theorem we may assume that tifrj,k - t%rk E O(ID, ID) locally
uniformly, k = 1, 2. Since >%i(0) = 0 it follows that, in fact, j E 0(ID, ID2). Moreover, because of the previous remark, ik (r) = r, k = 1, 2. Then, by the Schwarz lemma, we have 1/rk = idp, k = 1, 2.
Put g = (gI.g2): 0)2 C2, gk(A./.t) := Pk(A) - Pk(µ). Note that g(q) = 0 with q := (q1, q2) and det g'(q) 9& 0. Hence we find neighborhoods U = K(ql, s) x K(q2, s) C D2 of q and V = V(O) C C2 such that g maps U biholomorphically to V and K(qt, s) fl K(q2, s) = 0.12 Letnowgj: 0)2 C2, j E W, g j (A, A) := (*j',1(A) - *j,1(l4 Vuj,2(A) - *j,2(IL))
(A. A) E D2.
By the result before we conclude that gj -). g uniformly on U. Then, in virtue of Theorem 1.7.28, there exists a large index jo such that g1° vanishes in at least one 12Recall that q1 96 q2.
4.4. Examples I - elementary Reinhardt domains
277
point (ti, t2) E U, i.e. >lijo(tt) = t/'jo(t2), which contradicts the injectivity of 'f
.0
4.4 Examples I - elementary Reinhardt domains In this section we will establish effective formulas for dq,,, a E R", where k
dDa E and
Da={ZEC"(a):IZIa<1}, is an elementary Reinhardt domain. Note that Da,c : = {z E C" (a) : IZ Ia < e` } and D. are biholomorphically equivalent; so it suffices to study only Da. Generalizing
Definition 3.3.1 we say that D. is of rational type if a E IR Z" and of irrational type if it is not of rational type.
Note that if n = I and a > 0, then D. = ID and all the invariant functions coincide with m on ID x D. If n = I and a < 0, then D. = C \ ID. Thus D. is biholomorphically equivalent to D.. Then it is easy to prove that m = mo. = gp. on D. X lD* (EXERCISE). Moreover, applying Proposition 4.2.38, we are able, at least in principle, to calculate k*.. Firstly, we give the formula for the Lempert function even for an arbitrary annulus.
Theorem 4.4.1. For R > 1 put A = A(1 /R, R). If a E (11R. R), then
x2 + 1- 2x cos(n(s - t))11/2 l X2 + 1- 2x cos(rr(s + t))1 where a =
Rt-2s
IzI =
RI-2r, and x
= exp (
z = IzIe'B(Z) E A. -7r < 9(z) < s,
Ilo' gR).
Proof. Note that
A3A-*A/aEQ:={wEC:rl
k,(a,z) =kQ(l,z/a),z E A. Put S {z E C : log rt < Re z < log r2}. Note that 1 E Q and that exp Is: S
Q is a holomorphic covering. Moreover, observe that (w-Ior )f
e' 109r2 rt
S3wH
.(w-Io8r n .
e b9rZ rl where
a
in log(r2/rt)
-t +i
and
eaw- e1'
= eaw + eip
_ H(w) E ID,
= !t logrt 1 it 2 + log(r2/ri))
278
Chapter 4. Holomorphically contractible families on Reinhardt domains
D. Note that FI (0) = I+e;,6. Then, after gives a biholomorphic mapping H : S a suitable MObius transformation, we get the following biholomorphic mapping
H:S - m, H(w)
_
e°fw - I earw
-
W E S,
A'
with H(0) = 0, where A0 := e` bgZTT r2 l Hence, h := exp is o H-1: tD -+ Q is a holomorphic covering with h(0) = I (EXERCISE). Consequently, by Proposition 4.2.38, we get
kQ(1,0 = inf{IAI : A E h-I(t;)} = inf{IH(w)I : w E S, exp(w) = i;) = inf{IH(log ICI + i (0 + 2irk))I : k E Z). where = Mere with -ir < 0 < it. Calculating the last term leads to the function
f(t):
-I erei"p-1
2
etei0-ei*
,
tEUt,
I
where log M
tP - n log(r2/rI)
and
2n
log rl
log(r2/rI ) Note that f(t) = f(-t). A simple calculation shows that f'I(o,,,) > 0, i.e. f IR+ is strictly monotonically increasing (EXERCISE).
Note that t in f corresponds to -iog(r'
2kn), k E Z. Therefore, we
have the following possibilities:
(a)if0 =0, then kQ(l,i;) = (b) if 0 E (0, ir], then
I (1, ') = min{IH(log ICI + i0)I, IH(log ICI + i(0 - 2ir))I}; (c) if 0 E (-ir, 0), then
kQ(1,t;) = min{IH(logI I +i0)I,
+i(0+2ir))I}.
Observe that f(t) > f(t + x) if (I - ex)(e2t+x - 1) > 0. In (b) (resp. 109(r2 r 0 < 0 , x = io 2 r 2 7r, an d so 2 t + x > 0 0 > 0, x = iog 2 r, 2 ir, and so 2 t + x < 0) We get
in (c)) w e h ave t = ( res p. t = - iog 2 r.
.
k*(l,%) = IH(loglr;l +i0)I. Hence, z l (a,z)= H{log al +iArg (z (Q))'
where the argument is chosen in (-n, 7r]. What remains is to evaluate the right-hand side which is left as an EXERCISE for the reader.
4.4. Examples I - elementary Reinhardt domains
279
Corollary 4.4.2. For any a E (0, 1) we have k*D. (a. z) =
(02(z)+(loglzl-loga)2lt/2 l02(z) + (log IzI + log a)2 J
whenever z = jzje'e(Z) E D. and -R < 0(z) < jr. Proof. Use either a covering argument as in the proof of the former theorem or Lemma 4.2.30 (EXERCISE).
As an immediate consequence of Corollary 4.4.2 we get the following identities.
Corollary 4.4.3. (a) If a E (0, 1) and k E M, then
km.(ak,zk) = min{k;.(a,ze T') ') : 0 < f < k - 1),
z E D,.
(b) If a, b E (0, 1) and t E QZ, then
kp.(a',b') =
(a. b).
Proof. The proof is left as an EXERCISE.
If D = D" X Ck, then (EXERCISE)
dD((a',b'), (a". b")) = dDa(a',a"). a', a" E D",
b'.b" E Ck.
Hence for the remaining part of this section we will always assume that:
n>2; a1....,as <0anda,+1,...,an > 0forans =s(a) E {0.1.....n); if s < n, then t = t (a) := min{as+t , ... , an );
a = (at..... an) E D",aI...ak i4 0,ak+t = ... = an = 0fora k = k(a) E {s,
,n);
if k < n, then r = r(a) = r. (a) := ak+I +
+ an; if k = n (in
particular, if s = n), then r = r(a) = r"(a) := 1; observe that if a E Z", then
r(a) = orda(za _a"); if D" is of rational type, then of E Z" and al,. .. , an relatively prime; if D" is of irrational type ands < n, then t(a) = 1. We are able to describe effectively some holomorphically contractible functions of D. - the following formulas are known and will be discussed in the sequel.
Theorem 4.4.4. Under the above assumptions we have: a
mD.,(a,z)
gDa(a,z)
Rational type
(m(a", z"))f1 %1
(m(a", z"))t/r
Irrational type, k < n
0
Izalt/r
Irrational type, k = n
0
0
280
Chapter 4. Holomorphically contractible families on Reinhardt domains
k
a min
k = n, z 0 Vo
Iz"I'k S2)},
aa_tr Rational, s < n
k
(a, z)
min x {m(T;I, 2)}
z"=
km (aa, z")
kp. (a", za)
Rational, s = n
m(laal, IzaI),
Irrational, s < n
IZaillr+
Irrational, s = n
m(la"I, Iz"I)
k
ko.(Ia"I,
k®'(la"I, (z" I) I
Note that, in fact, the above formulas cover all possible cases. Before we start to prove this theorem we present some applications.
Remark 4.45. (a) If Da C Cn (resp. Dp C C") is an elementary Reinhardt domain of rational (resp. irrational) type, then these domains are not biholomorphically equivalent (EXERCISE).
(b) If Da C C", n > 2, is an elementary Reinhardt domain of rational type with
0 E D" (i.e. s(a) = 0) and if e > 2, then
= (m(0,za))7
Ta7
= IZaI'r1 1.
Z E D. \ Vp.
mt(z,0) = m(za,0) = Iz"1 < m(O,z),
z E D" \ Vo.
mlpa(0,Z)
Observe that Y (1 1 < 1. Therefore,
Hence we conclude that, in general, the function mD) is not symmetric.
(c) Let Da be as in (b). Fix a b e Da \ Vo and a sequence (a) j C Da \ Vo which converges to 0. Then
mtom,(aj,b) =m(aj,b") -+ m(0, b") = Ibal <mt")(0,b). Hence, the function m ( ( , b) is not continuous, e > 2. (d) Let Da C C" be as in (b). Then
gDa,(0,b) = (m(0,b"))I/1a1 > IbaI = m(b,0) = g,%(b,0).
b E Da \ V0.
This shows that, in general, the pluricomplex Green function is not symmetric. (e) Let Da be as in (b). Fix a b E Da \ Vo and a sequence (a1)1 E Da \ Vo with
a j - 0. Then gDa(aj,b) = m(aj,b") -> m(0, b") < (m(0,ba))UI"1 = gDa(0,b)
4.4. Examples I - elementary Reinhardt domains
281
Therefore, in general, gD ( , b) need not be continuous. Recall that also gD (a, is not necessarily continuous.
)
Using the formulas for the pluricomplex Green function for elementary Reinhardt domains, we get the following result (see Lemma 2.4.12).
Lemma 4.4.6. Let Da, Do be elementary Reinhardt domains in C", n >_ 2, of rational type with s(a) = s(fl) = 0, i.e. 0 E Da n Dp, and let F E O(Da, Dp). Then there exists a V E 9 (D, D) with F(V (Da, A)) C V (DD, (o(A)),
A E D.
where V(Da,A):={zEDa:za=A). Proof. Fix a A E D. Put
a = a(X) :=
1,A11a"),
where At/an is a certain root of A. Obviously, a E V (Da, A) and k = k(a) = n. Set
(p(A) := Fj"'(a)...Fhn(a) = F" (a) E D. Then F(a) E V (Do. rp(A)). Now take another point z in V (Da, A). Then
0
I
as - za - aaza
11/rcr(a)
= gDa (a, z)
gD8(F(a), F(z)) =
_F# (a) - Ffi(z)
I1/rs(F(a))
1 - F$(a)FP(z)
Therefore, F(z) E V (DO, V (X)). Note that by taking locally a holomorphic root A1/a^ in D. it follows directly from its definition that V is holomorphic in D.. Hence it extends holomorphically
to the whole of ID and this extension coincides with (p(0) = lim FO(a(t)) = lim to(t). So rp E O(D, ID).
r-o+
Corollary 4.4.7. Let Da and DD be as in the previous lemma. Assume now that
F E 9(Do , DO) is even biholomorphic. Then there is a V E Aut(ID) with F(V (Da, A)) = V (DS, (p(A)),
A E ID.
To show how to use invariant functions like the pluricomplex Green function we are going to prove the following result [Edi-Zwo 1999] (see also Section 2.3).
Theorem 4.4.8. Two elementary Reinhardt domains Da, Do C C", s(a) = s(p) = 0, are biholomorphically equivalent if and only if there exist a permutation a of { l .... , n) and a t > 0 such that aj = t f, (J ), j = 1, ... , n.
Chapter 4. Holomorphically contractible families on Reinhardt domains
282
Proof. From the very beginning we may assume that n > 2 and that both domains are either of rational or of irrational type. In the first case we know by Lemma 4.4.6 that F(Vo) = V (Do, A) " for a certain it E D. Observe that if µ E D*, then the latter set is an analytic set with only regular points, but Vo has 0 as an irregular point. Therefore, µ = 0, or F(Vo) = Vo. Now let both domains be of irrational type. Suppose that F(V0) ¢ Vo, i.e. there is an a E Vo with F(a) 0 Vo. By Theorem 4.4.4 we conclude that
0 # gDa (a, -) = gDe (F(a), -) = 0: a contradiction. So, F(Vo) = Vo also in this case. 1,
Hence, there is a permutation a of { 1.... , n) such that F(Vj) = VQ(j), j = ... , n. In particular, F(0) = 0. Applying the formulas for the pluricomplex
Green function it follows that
(IZllat
...IZ.1,,)1/(a1+...+.an)
= gDa(0.z) = gDS(0, F(z))
1
IFn(z)1'in)11(B1+...+Bn)
= (IFI(z)Ist ...
ZEDa.
Moreover, for the points dd = (1, ... ,1, 0, 1, ... , 1) E V1 we have that F(dj) is contained in Va(t) \ Ucn=1, t &a(j) V . Therefore,
Izalt/aj = gDD(dj, z) = gDS(F(dj). F(z)) = I FB(z)11 /Ba(i).
Z E Da.
Combining the last equalities gives
Oil + ... + en
O!,
fit +-'-+On
A, (j)
j=
n,
0
which finishes the proof
Now we come back to prove almost all of the formulas stated in Theorem 4.4.4.
Proof of Theorem 4.4.4. The proof will be given in several steps.
Proof for m Da - the rational case. Define
f(W)
/ wa - a°`\ f r l :
1\
1 - aallJa
wEDa.
Then f E O(Da, ID) with orda f = r [ l > e. Hence m() (a, z) > If(z)I 1l I fr . which implies that mDa (a. z) > (m(aa, za))7 1-'Note that Vo C D. fl Dg.
283
4.4. Examples I - elementary Reinhardt domains
Now let f E O(Da, U)) with ordr f > P. Put 45(w) := wa, w E Da. Applying Theorem 3.2.1 (e), we get f = tp o 4>, where (P E O (ID, D) and order f = r ord,a V.
Hence, ordaa V > r i and therefore, If f(z)I" t < (m (aa za)) l f;1, which proves the remaining inequality. Proof for m( .?f°O(Da)
- the irrational case. According to Theorem 3.2.1(d) we know that C, which immediately proves the claimed formula.
ProofforgDa -therationalcase. Let # :_ (la1I
Ob-
serve that Da and DO := Dp fl C; are biholomorphically equivalent via the following mapping
F: Cn(a) -* Cn(a),
z H (zi I,...,zy.
zs+l,...,zn)
Hence gga(a, z) = gDUo(F(a), F(z)), Z E D, In virtue of Proposition 1.14.25, we even know that gD0o(F(a), ) = gDS(F(a), - ) (EXERCISE).14 Therefore, from now on we assume that s = 0. Using the equation for m D. from above, we know that gDa (a, z) > m D. (a, z) = (m (a', za)) tl'. To get the converse inequality let u : D.
[0, 1) be a log-psh function satisfying
u(z) < C 11z -all, Z E Da; in particular, u(a) = 0. Fix a A E ID. with µan = A. Then C n-1 -r Da,
1/r(wl, ... , wn-1)
-nn
(w1 .... , wn-1' WIat
It an_t wn-t
is a holomorphic mapping onto V (Da, A) (EXERCISE15). So u o t/r E 3'8J{(C;-1) is
bounded. By Proposition 1.14.25 we conclude that this function is, in fact, psh on the whole of Cn-1. Moreover, it is bounded from above. Therefore, the Liouville type theorem for psh functions gives that u o v(A) E D. So we have constructed a function v : ID. -+ ID. such that uIV(Da,x)
vW,
A E D*-
v is sh on iD,. Indeed, fix Ao E D, and p > 0 with K(Ao, p) C D.. Choose a holomorphic an-th root of idx(xo,p), i.e. a g E 0(K(Ao. p), ID) with gan (A) = A. A E K(Ao, p). Then
v(A) = u(1, ... , 1, g(A)),
A E K(Ao, p).
14Note that Do \ Vo = D. " Use the assumption that or t , ... , an are relatively prime and therefore Z = a 1 Z +
+ an Z.
284
Chapter 4. Holomorphically contractible families on Reinhardt domains
Hence, v is locally log-sh on D. and therefore log-sh on DD,.. As above, v extends to a log-sh function v on D.
First we discuss the case k < n, i.e. a01 = 0. Recall that r = ak+I + For A E D. we have
+ an.
v(A) = v(A) = u(al....,ak,Allr....,A'/r,Al/r(a11 ...akk)-'Ian) (a71...akk)-'/an)II
AEID.,then (gD(O,Za))I1r
u(Z) = v(Za) 5
= IZaI'/r
Therefore, gDD(a.z) < Izall/r = (m(aa,za))'/r on Da \ Yo. If z E Da n Vo, then the mean value inequality for psh functions leads to gq. (a, z) = 0 = (m (a', za))1Ir. Now let k = n, i.e. as 0 0. Then we get for A E D. near Ao := a"
v(A) = u(al.....an-1,)L'/an(al' ..aa Alan < Cla
an-1- I
1
an
an
-anl < CIA
1)-l/an) 1/art
- l"an Ao I < CIA -AoI,
which implies v(A) < gp(aa, A) = m(Ao, A), A E D. Hence, u(z) = v(za) < m(a", za), Z E Da n C. By the same reasoning as above it follows that the above inequality holds also on Da (EXERCISE).
For further purpose we add the following observation.
Lemma 4.4.9. Ifs = 0 and k < n, then gDa (a, z) ? I za I1"r, Z E Da.
Proof. Note that the function u: Da -* [0, 1), u(z) := 1z' I'/", is log-psh. For Z E Da, z near a, one has u(z) < (IZII"I ... IZklak)'lr(IZk+1 -Olak+l ...lZn
-0lan)Ilr
Recall thatTa =
E T, j = 1,...,n).where a E C.
Then Ta is a group with the following multiplication:
(1a1.....nan) (>)ta1....,lnan) := (trltal.....n1]nan)
4.4. Examples I - elementary Reinhardt domains
285
Let a E IR;. Define Ta(a) to be the subgroup of Ta that is generated by the set
t(ei a 2knai... ,
a 2knna,) ei an
I < Jt..... jn < n, kt...
,
kn E 7L
Note that if a is of rational type, then
Ta(a) _
1,
1 < j < n).
i.e. T0(a) is finite. If a is of irrational type, then Ta(a) = Ta (cf. p. 97). To get some information on the Lempert function we need the following result on analytic discs.
Lemma 4.4.10. Let a, z E Da, Z E IR;, and 2 E T (a). Then for any rp E O((D, Da) with rp(A1) = a, 0A2) = z, 11 # A2, Ai E ID, j = 1, 2, there is a a and O(A2) _ In particular, k)a (a, z) = kDa (a, i).
r p E 0(m, Da) such that rp(A1)
Proof. First recall that the strip H := {A E C : -I < Re A < 11 is biholomorphically equivalent to ID (use the Riemann mapping theorem). Therefore, it suffices to prove the lemma when we substitute ID by H. So we may assume that rp E 0(H, Da) with rp(0) = a and (p(i r) = z for a certain r > 0. For kn E Z and j E ( 1, ... , n ), let 0: H --> D. be defined as
pA:_ {p1
2knaA (A),....wn-2(A),a-:pn-r(A),a
aj 2knirA anT
1
E O(H,D.),hp(0) =a,andhp(iv) _ (zr,....z.-I.e an 2knn i
T h e n
(pn(A))
Now we continue modifying the other coordinates in the same way as above 0 which finishes the proof. In the same spirit is the following lemma.
Lemma 4.4.11. Let L 1, L2 C= I), L C= C., and a E IR;. Assume that
m(L1, L2) = inf{m(Ai,A2) : )Li E Li, j = 1,2} > 8 > 0. Then there exists a set L C C., L C L, such that for any 71- z2 E L and Ai E Li,
j = 1, 2, thereisa V1 EO(ID,4.)suchthat*(A3)=zi, j = 1.2,andi/'(D)C L. Moreover, there is a set K C= C. such that for any z r , ... , :n E L with I za l =
1
and any wr.... , wk E L, k < n, there are fimetions h/ri E 0(m. C.) such that V/' (ID) C K, V*1(A1) = zi, j = I..... n, and V' (A2) = wi, j = I..... k, and ji r (A) ... *an (A) = e!6 A E 9).
286
Chapter 4. Holomorphically contractible families on Reinhardt domains
Prof. Without loss of generality we may assume that Lt = (A1 = 0) and L2 =
0-2 = 5).16 Put L := exp 1(L) n (IR + i [0.2n)). Then L C (log El, log E2) + i [0.27r), where 0 < e1 < e2 and the e1's depend only on L. Set
L :_ (eaz+h : A E D, a, b E C such that aA1 + b E L, j = 1, 2). Then L C= C.. What remains is to choose 11i := eh, where h is an appropriate function of the form as it appears in the definition of L. wn_1 E L To prove the last part of the lemma we take, in addition, in an arbitrary way. For the pairs z1, w1 we fix functions 1Jrj E O(D, C.) with
f1(l)CL,
Y'17.1)=z1,
1Gi(A2)=wf.
Put
eiB (,I,i 1 (A) ...
*I,
1Yn
jnar^
t
I
AEID,
where the branches of the pow ers are chosen arbitrarily and 0 is taken such that *n (An) = zn. Exercise 4.4.12. Prove the following statement using the ideas from the proof of Lemma 4.4.11. (a) Let a E C., X E C, and A E ID be given. Then there exists a t/' E O(D, C.) such that V'(A) = a and *'(A) = X.
(b) Moreover, if A E ID, a E C', X E Ck, where k < n, and a E IR; with Iaa I = 1, then there is a 1/r =
t//n) E O(ID, C;) such that
4/(A) = a, (t/r1(A).... ,1//k(A)) = X, V,01' ... *n- = eiB idm
.
Applying the previous lemmas in the case of elementary Reinhardt domains leads to the following results.
Lemma 4.4.13. Let Da be of irrational type and a, z E Da n CZ. Then
k
(a,z)=k,(a',z'). a'ETa, z'ET.
Proof. Note that it is enough to prove that k, (a. z) = k% (a, z') (use the symmetry), whenever z' E T,. Suppose the contrary, i.e. there are points z', z" E T with
k , (a z') < k, (a, z") =: e. '6To get the general case. recall that if m(A 1, )2) > 8 = m(0, 8), then there is a 4p c- 490D, ID) with (p(.A 1) = 0 and rp(.l2) = 8.
287
4.4. Examples I - elementary Reinhardt domains
In virtue of Lemma 4.4.10, we have k ,(a, z) = C, z E TZ""(a). Observe that z' E T = Ti" = Therefore, z' is an accumulation point of Ti""(a), D
which contradicts the upper semicontinuity of k a (a, ).
Corollary 4.4.14. Let a E Da fl C. where D. is of irrational type. Then k
(a,z) = 0,
z E Ta.
Proof for k Da - the case k < n. In virtue of Lemma 4.4.9, we know that
kDa(a,z) > gD (a.z) > 1z.111'. In order to prove the converse inequality for a z E Da \ Vo, put r Then all the points z1, ... , zk, z* ...... Aa belong to C. with n
k
ITIj)=i.
(1I,Ia1)( j=1
IZ XI
j=k+1
Moreover, a 1, ... , ak E C.. Then, in virtue of Lemma 4.4.11, we find functions
*j E O(D, C.) such that
ft *7'(A) = e'6,
AEID,
j=1
*j (r) = zj, 1 , , ( r ) =
- .
*j (0) = aj,
j = 1.....k, j = k + 1,...,n,
j = 1....,k.
Put
(p(A) := (*1(A)..... IVk(A), A1//k+1(A)......X*n(A)), A E D. Then (P E 0(Q), Da) with tp(0) = a and (p(r) = z. Hence, k%(a, z) < r, i.e. the proof is complete. Now we discuss the remaining case when z E D. f1 V0. Note that necessarily
we have -j # 0, j < s. First suppose that there is a coordinate zj = 0 with j > k + 1. Then the holomorphic mapping
CsX C n-s-1 3w = (wt .... , wj-l, wj+l.... , wn) (Wl,...,wj-1.0,wj+l,..., Wn) E Da leads to the following inequality
k%(a, z) < kCs xC^-S-t (a. i) = 0, "Recall that one can map C onto C via the exponential map.
17
288
Chapter 4. Holomorphically contractible families on Reinhardt domains
where a :_ (at,...,aj-l,aj+t,...,an)and1 :_ (zt,...,zj-t,zj+t,...,zn) What remains is the case where I z j I + 1a11 0 for all j 's. For fi E (0, 1) we are going to define a SP = (apt .... , 0.) E O(m, D.), where
i = *j (A)
if a j = 0,
*j (A)
if Zj = 0,
Vj
.X E D.
ifajzj 54 0,
*j (A)
The *j E O (D, C,) have to be chosen in a correct way.
We need that the *j's satisfy rj;=t e'9 on m and that tp(f) = a, and (p(- fi) = z. Note that then we would get k, (a. z) < m(fl, -P) -+ 0 if P -+ 0. Fix some jt such that a j, = 0. Then we would like the functions tlrj to attain the following values in C.:
'Vi(i) =
ifajzj # 0,
aj a(
ifzj =0,0aj,
20
if aj = 0 0 zj, j 0 ji,
1
*h(8)
f I*j(0)lai rj ItVj(0)lai zi =0
aj zj #0
Moreover, at -P we only need to have
tVi (-f) =
zj
ifajzj 76 0,
z, (t-0 -20 2)
if aj = 0
zj
Note that there are fewer than n values we want to specify at -P. Therefore, Lemma 4.4.1 1 works and gives such a mapping t/r E O (D, Cl.) which completes the proof.
The remaining case, i.e. s = n or k = n, for elementary Reinhardt domains will be discussed later in this section. First we establish the formulas for the pluricomplex Green function in the irrational case.
Proof for gam, - the irrational case. As in the proof of Theorem 4.4.4 we may as-
sume that s = s(a) = 0. In the case where k = n we have gD (a. z) = 0, z E Ta (see Corollary 4.4.14). Then the maximum principle for psh functions implies that g D,,, (a, - ) = 0 on (P(0, (Ia i E, .... lan p). Recall that log g j (a, ) E P83{(Da) and that either log g
(a, ) _ -oo or the level set {z E D.
pluripolar set. Thus, gI (a, ) = 0 on D..
:
log g,Q (a, z) = -oo) is a
4.4. Examples I - elementary Reinhardt domains
If k < n, then, by the formula for k
289
in that case, we know that
gj (a, z) < kva(a, z) =
Iza1l/r.
To conclude the proof apply Lemma 4.4.9.
The last part in this proof is devoted to prove some of the remaining formulas for the Lempert function.
Lemma4.4.15.Ifa,zED=DtnYo (1=(1....,1)),then
(a,z)=0. If aEDtnC;,zEDi,then k,, (a,z) _
(m(aI...an,zr...zn))r/`
where r := max{#{ j : zj = 0}, 1). Proof. The first formula is a direct consequence of the one for k* which has been proved before. Now we will discuss the case where a, z E C n.. In a first step suppose that µ := a r an = zr zn. Then the holomorphic mapping "-I
BO (w1....,wn-t) H (e'I....,eWn-I,µewI-..._wn_I) E DI
is onto V (Dt, µ), i.e. F(w) = a and F(w") = z for certain w', w" E C"-r. Hence,
kvt (a. z) < kin-I (w , w") = 0. zn. Put So we may assume that, from now on, a t . an zr A l :=a, ... an E I).
A2 := Zr ... Zn E D.
Applying Lemma 4.4.11 we find a i/rj E O(I), C) such that
fj(A1)=a1,'fj(A2)=zj, j =1,...,n-1, and *1
*n(Ai)=(at...a"-t)-'
;fin - ere on D1. Put
(A)
(A), ... , n-t (A)
(A)),
A E ID.
Then V E O(U), DI) such that (p(A1) = a, t/r(A2) = z. Hence,
m(ar...an,zr...Zn)?kD,(a,z)?m(ar...an,zr...zn), where the last inequality is a consequence of the property of holomorphic contractibility.
It remains the case that a E C and Z E Vo. Then again Lemma 4.4.11 gives the desired formula.
290
Chapter 4. Holomorphically contractible families on Reinhardt domains
What we have just discussed is the simplest case of an elementary Reinhardt domain of rational type. The proof of the formula in case s < n needs deep results on geodesics which are beyond the scope of this book. Therefore we skip its proof. Details may be found in [Pfl-Zwo 19981 and [Zwo 2000]. Note that the case k < n is contained in Theorem 4.4.4. So the difficult case is the one with k = n. Now we turn to the irrational case for s < n.
Proof for k* - the irrational case with s < n. The case k < n was already verified. So we assume that k = n and z 0 V0. Recall that
kpa(a,z) = kpa,((IaiI...., lanl),(Iz1I,....Iznl)) (see Lemma 4.4.13). Now we approximate the ay's by rational vectors. We choose
a sequence (a('))i C 0' such that
a(i)>a, t(a(i))=1,
a1
j E N.
Applying Theorem 4.4.4 for points x, y E Da(i) fl IR+ we get a(i) , y1 U)
*
ynn
kD (.) (x. y) = m(xl
a(i) ... yn
jEN.18
(4.4.1)
By employing a biholomorphic reordering of the coordinates we may assume that
t(a)=a,,. Now suppose that
k,(a,z) < m(Iaal, Izal). Then there exist an analytic disc (p E O(1, Du) and A 1. A2 E ID such that cp(A 1) _
(Jail,..., IanI),W(A2) =
IznI),and
m(A1, A2) < m(laaI, Izal). Since tp(ID) is a compact subset of D. we can choose a large jo such that VIm E O(ID, Da(io)). Therefore, (io)
m(A1, A2) < m(I ai Ia(
(io
(io)
... Ian la"
,
Izi Ia,
(io)
... Izn Ian
):
a contradiction to (4.4.1).
On the other hand put Al := Iaal,
A2 := IzaI. If At # A2, then we find
analytic discs 1/!j = ehi E O (D, C.) with h j (,11) = log l a j l and hi (A2) = log Izi I,
j = 1, ... , n - 1 (use Lemma 4.4.11). Moreover, define eXP(-alh1
A E ID.
"Note that m(x,e'By) > m(x, y), t9 E IR, when x, y E [0, 1) (ExERcisE).
4.4. Examples I - elementary Reinhardt domains
291
Then A 1 i/r" (A 1) = Ian 1. Put
(P(A) := (* (A), ....1//n-1(A), A*n(A))
A E 0).
Then (pn (A 1) = Ian l and r,n (A2) = Izn 1. Hence,
k, (a, z) <m(laa1.1za1).
IfA1 =A2,putAE:=A2+EEQ),e>0small. Put hs(A)=
A-A1 - As I -A1A I -AsA'
AE[.
Then h, E O (D. D) with hE (A 1) = hs (As) = 0. Then, using an appropriate Mobius transformation, we find an hs E O(D, I)) such that hr(A1) = hv(AE) = Al.
As in the case before there are 1/ri E O (ID, C.) with Or, (A 1) = l ai I and *j (AE) = Izi 1 , j = 1.... , n - 1. Put now W(A) := (Or1(A).... , *.-I (,X), hE(A)(
1
A E D.
()L) ...1/in"1
Then V E O(D, Da), (P(A 1) = (la l l.... , Ian l), and (p(AE) = (lz1 I, ... , Izn l). Taking into account that s may be taken arbitrarily small, we end up with k % (a, z) = 0, which finishes the proof. 0
In a last step we discuss the case when s = n, i.e. ai < 0 for all j.
Proof for k, - the case s = n. First observe that the map F : C"-1 x D. FA
Da,
A
is a holomorphic covering. Note that F(A) = w iff
Ai =
an
(log lw1I+i(Argwi+21in)), n-1
1
An
where 11,
j = 1....,n-1.
n Ej=isi(Argaj +21i rr) .
= 1=1
... , In_1 E Z. Applying Propositions 4.2.38 and 4.2.35 we are led to
k% (a. z) = inf ji*.
i (la'l-Vane_«n
Za -1/aye-an
Ejnn=,
aj Area,
Ej-;aj(Argz;+2rjn)
.1
In the rational case we apply Corollary 4.4.3 and get k*,(a, z) = k*. (aa. za).
292
Chapter 4. Holomorphically contractible families on Reinhardt domains
In the irrational case we conclude via the Kronecker theorem that
k
(a z) = ko+(laal-an, IzaI-a").
It just remains to mention that kpy(x, y) = W;. (x'. y') fort > 0. Applying this remark with t = -a gives the desired formula. In a final step we discuss the Kobayashi pseudodistance in the case of elementary Reinhardt domains with s < n, which so far we have not mentioned.
Proof for k
- the case s < n. In the rational case it is easy to see that
Da x Da 9 (z, w) H
y2) : yt, y2 E D, as
za = 2})
=: d(z. w) satisfies the triangle inequality and is majorized by i,%. Hence it follows that kDo, > d and both are equal outside of the axes. Then the continuity of the Kobayashi pseudodistance gives the desired result. In the irrational case the reasoning is analogous to the one before and therefore left to the reader as an EXERCISE.
Proof for k I - the case s = n. This step is left as an EXERCISE. In particular, the effective formulas from above make it possible to give a negative answer to the following old question asked by S. Kobayashi (see [Kob 19701, p. 48), namely: is the infimum in Proposition 4.2.38 taken by a certain point in the holomorphic covering.
Example 4.4.16 ([Zwo 1998]). Let D = DI_,i _tl and >(r : C x D. -* D,
(e1,_'_e'11). 112
be a holomorphic covering. Take a := (r, r) E D with r > 0 and z := (r, ir). Then kD (a, z) = 0. Fix the following preimage (- log r, r-t -") of the point a. Then
0 = inf kE,o, ((- log r, keZ
(-logr + 27rik, ri exp(/(-logr + 2nik)))) = inf k p, (r-t kEz
r
exp(f(-log r+21rik))).
293
4.5. Holomorphically contractible families of pseudometrics
Suppose that there is a k' E Z such that
kD.
exp (v(- log r + 2nik'))) = 0.
Then 1/2 + k'J E 71;19 a contradiction.
4.5 Holomorphically contractible families of pseudometrics Recall from classical analysis that the Euclidean distance between two points x. Y E
G = lR" can be also given by the "minimal" length of all piecewise el-curves in G connecting these points. Let y : [0,1 ] --)- G be such a curve. Then the length of y is given by L(y) = fo II y'(t)Ildt, i.e. along the curve the lengths of its tangent vectors II y'(t) II, t E [0,1 ], are summed up. Hence we have an assignment
G x R" 9 (x, X) H ac(x;X) :=
him
IIx-(x+tX)II
hR.3t-+o
ItI
with the following property: a(x; sX) = IsIa(x; X), X E G, S E QR, and X E IR". This procedure will be transformed into the context of families of holomorphically contractible families of functions (dD)D. Let us start with a general definition.
Definition 4.5.1. A family (8D)D of pseudometrics SD : D x C" domain in C", i.e.
Z E D, A E C, X E C",
SD(Z; AX) = IAISD(Z; X),
9t+, D a (4.5.1)
is said to be holomorphically contractible if the following two conditions are satisfied:
(A) 81D(a; X) = y(a; X) := 1
I
IQI2,
a E D, X E C.
(4.5.2)
(B) for arbitrary domains G C C, D C C", any F E O(G, D) works as a contraction with respect to SG and SD, i.e.
SD(F(a); F'(a)X) < & (a; X), a E G, X E C"'.
(4.5.3)
Note that the Hermitian pseudometrics discussed in Section 1.19 are pseudometrics in the sense of Definition 4.5.1. In the following we will discuss the most important holomorphically contractible families of pseudometrics. 19Note that m(A, µ) = kp(A, µ.) <_ k;. (A, µ) = 0 implies that A = µ, when A, p, E D..
294
Chapter 4. Holomorphically contractible families on Reinhardt domains
Example 4.5.2 (Carathdodory-Reiffen pseudometric).
YD(a; X) := sup{If'(a)XI : f E O(D, ID), f(a) = 0},
a E D. X E C". (4.5.4)
where D C C" is a domain. Indeed, the properties (4.5. 1) and (B) are obvious. To prove (A) let F E O (UD, U))
with F(a) = 0 and X E C. Put
g(A):=F\1+j. AE(D. Then g E O((D, lD) with g(0) = 0. By the Schwarz-Pick lemma it follows that 1 > Ig'(0)i = IF'(a)I(1 - Ia12); hence, y(a; X) > yr (a; X). To get the converse A -a inequality just take the function F(A) A E D. We begin by stating some simple properties of the Carathdodory-Reiffen pseudometric (details are left as an EXERCISE): YD (a; - ) is a seminorm on C'.
By the Montel theorem, there exists an f E (9(D. D) with f(a) = 0 and I f'(a)X I = YD (a; X) (such an f is called an extremal function for YD (a, X)).
Put MD(a) :_ (If l :f E O(D, ID), f(a) = 0}. Then
YD(a;X)=sup
lim
u(a+AX):uEMD(a)I,
aED. XEC".
IAI
(4.5.5)
Note that any function u E MD (a) satisfies: u : D -> [0. 1), log u is psh, and u(z) < C IIz - all, when z E IB(a, r) C D for certain C, r > 0.
If D = Ujt1 Dj C C", where Dj C Dj +1, j E N, are domains, then (use Montel) YD (a; X) _ alt me yD; (a; X).
For a balanced domain Dh we have yD (0; ) < h. Indeed, suppose first that h(X) = 0. Then C D A H AX E Dh is a well-
defined holomorphic mapping. By (4.5.3), yDh (0; X) < yC (0: 1) = 0 = h (X).10
Now assume that h(X) > 0. Then D 9 A t-+ 4X E Dh is holomorphic and, therefore, yDh(0;X) < y(0;h(X)) = h(X). In particular, YD (a. X) :5 YIB(a,r)(a; X) = YB(r)(0, X)
X
r, a E D C C". X E V.
It turns out that the Carathdodory-Reiffen pseudometric is given as a derivative of the Mobius pseudodistance, even in a strong sense (see Lemma 4.5.3 (b)). mUse the Liouville theorem.
295
4.5. Holomorphically contractible families of pseudometrics
Lemma 4.5.3. Let D C C" be a domain. Then: (a) For any compact K C D and for any e > 0 there is a S > 0 such that
I mD(Z' z") - YD (a; z'-z") I < ellZ' -
Z"II,
a E K, Z'. Z" E IB(a, S) C D.
(b) For (a; X) E D x C", II X II = 1, one has
m W. Iiz' - Z"II
-- YD (a; X), when z', z" - a z' # z"
z
-z
X.
IIz' - Z"II
(c) In particular,
limo
YD(a; X) =
mD(a.a+AX)
a E D, X E C".
Proof. (a) Fix an r E IR>o such that IB(b, 4r) C D, b E K. Now, take a E K, z', z" E IB(a. r), and X E C". We may assume that YD (Z': X) > YD (z"; X). Choose an extremal function f E O (D, ID) for YD (z'; X), i.e. f (z') = 0 and I f'(Z')X I = YD (z'; X). Then
IYD (z':X)-YD
(z,';X)I
-YD(z":X) < If'(Z')XI - I f'(Z")XI
= If'(z')XI
< if I(zl)x -.f I(Z")XI < Ilf'(=') - f'(Z")IIIIXII < max{Ilf"(z)11: z E [Z'.z")}IIz' - z"11IIX11
(*)
<2r211Z'-Z"1111X11
Using (*) it follows that
ImD (=',z")
- YD(z';z -z") I+ IYD (z';z -z") - YD (a; z -z
< ImD(Z',z")-YD(z';Z'-Z") I + 2I
llz'- allllz'"II.
It remains to estimate the first term on the right-hand side. We may assume that MD (z', z") > YD (Z'; Z" - Z') (the other case follows in a similar way (EXERCISE)).
So let f E 0(D. ID), f(z') = 0, and f(z") = MD W, z"), i.e. f is an extremal function for MD. Then, by the Cauchy inequalities, we get with Y := z" - z' (IIYII <- 2r) ,
ImD(z,=")-
,
,
YD (z: Y) 1
r 1,
E k=2`
3r
-< 3r2 II Y II2
lll
k)
,
(z )II IIYII
k
296
Chapter 4. Holomorphically contractible families on Reinhardt domains
Hence,
2I llZ'-all)Ilz'-z"II.
ImD(z'.z")-YD(a:z'-z")1
which proves (a). (b) and (c) are easy consequences of (a) and therefore, the proof is left as an EXERCISE to the reader.
Corollary 4.5.4. The function YD is locally Lipschitz on D x C". Proof. Use (*) from the proof of (a) in the previous lemma. Remark 4.5.5. Lemma 4.5.3 is also true when we substitute MD by CD (EXERCISE).
Example 4.5.6 (k-th Reiffen pseudometric).
..(k)i... v.._ - _ II [-` Ial=k
-
I D
a!
f(a) X a Ilk f E I
O( D ,
ord Q
f - k },
aED, XEC". where k E M and D C C" is a domain. Note that yD = yo ) The proof of the fact that (y(k))o is a holomorphically contractible family of pseudometrics is left to the reader (EXERCISE).
First, let us state some simple properties of the k-th Reiffen pseudometric (ExERCISE):
There exists an f E O(D, ID), ordQ f > k, such that Y
)(a;X) =
rD"f(a)X"
I
Ilk
Ial=k
(use the Montel theorem); such an f is called an extremal function for yDk)(a: X). Put 1v[D)(a) {IIII/k : f E O(D, (D), ordQ f > k}. Then
yp)(a: X) = sup {
lim c.3x-+o
u(a +.XX)
: u E M(k)(a)y,
JAI
a E D, X E C".
Note that any function u E Mpk)(a) satisfies: u: D -> [0, 1), log u is psh, and u (z) < C 11z - all, when z E IB(a, r) C D for certain C, r > 0.
If D =
I
Dj C C", DJ C Dj +t then y..) (a; X)
Montel's theorem). Moreover, we have the following results.
yD)(a; X) (use
297
4.5. Holomorphically contractible families of pseudometrics
Lemma 4.5.7. Let a E D C C", D a domain. Then:
EC".
(a)yDk)(a;X)=C lira
(b) yD) (a; ) is continuous and yo) is upper semicontinuous. (c) If we additionally assume that D is bounded and II X II = 1, then yo) is continuous on D X C" and (k) (z I 9z "
IID'-z"i
, _z" IF,-ell
Y( )(a;X), when z',
X.
Proof. It is obvious that the left-hand side is majorized by the right-hand side. Now let C. 9 A, -+ 0. Choose extremal functions f , E O (D, m), orda f,, > k, such that
MD)(a,a+At,X) =
Ilk I
IaI=k
a.
f locally uniformly on D. Note that
By the Montel argument we get fv,U
orda f > k and f E C9 (D, D). Then 1
Yp)(a:X)> I EatDa.f(a)Xal
1/k
lal=k
= µm IAv
=µmIA
lfvu(a+AvuX)I1/k vu
IMD)(a,a +A,, X).
The proof of the remaining points are left to the reader as an EXERCISE.
What we saw up to now is that the Mdbius functions have led via (4.5.5) or (4.5.6) to holomorphically contractible families of pseudometrics. In the case of the pluricomplex Green function we put
XD (a) := {u: D -> [0,1) : logu E PSK(D), 3M,r>o : u(z) < M II z - all, z E IB(a, r) C D},
where D is a domain in C" and a E D.21 Recall from Corollary 4.2.27 that gD (a, - ) E BCD (a). Moreover, for any X E C" the limit I
lim sup u (a + AX) C.aA-0 IAI
always exists. Thus we can define 21 Notethat M '(a) C Xo(a).
298
Chapter 4. Holomorphically contractible families on Reinhardt domains
Example 4.5.8 (Azukawa pseudometric).
AD(a;X ) : =sup {limsup -u(a +AX) : u E 9Cp(u)} 1
(4.5.7)
IAI
= lim sup
1
I
gD(a, a+ AX),
a E D, X E V.
(4.5.8)
C.3A-+o IAI
Indeed, for D = ID we have lim sup
1
go)(a, a + AX) = lim sup
C.3A- 0 JAI
I
m(a, a + AX) = y(a; X).
c.91-0-o JAI
Note that u o F E XG(a) whenever F E O(G, D) and U E XD(F(a)). Hence, property (4.5.3) follows.
Lemma 4.5.9. Let D be as before. Then AD is upper semicontinuous.
Proof Fix a E 8(a, 2r) C D and Xo E C". Suppose that AD(a; X0) < M' < M. Then 19D(a.a + AX0) < M' when 0 < IAI < 2e for a certain positive E < 2(xo +1 In particular, fgD (a, a + AX0) < eM', IAI = E. Then, applying the upper semicontinuity of gD, there is a positive 8 < r such that
gD(b, b + AX) < eM'.
b E IB(a, 8). IIX
- XoII <8, IAI = e.
Observe that for such an X the function
AH
lfjgD(b,
b + AX)
AD (b; X)
if 0 < IAI < 2s, if A = 0,
is psh on K(2E). Hence the maximum principle leads to AD(b; X) < M for all b, X as above, which proves that AD is upper semicontinuous at (a, X0).
Example 4.5.10. Let D = Dh
0
(z E C" : h(z) < 1) be a pseudoconvex
balanced domain. Then AD(0, ) = h on C". In particular, AD(0: ) need not be continuous and is not necessarily a seminorm. Indeed, we only have to recall that gD (0, -) = h.
The following example ([Zwo 2000a]) shows that, in general, the "lim sup" in the definition of the Azukawa pseudometric cannot be substituted by "lim".
Example 4.5.11. Let h : C2
Qt+ be the function from Proposition 1.15.12. Recall that log h E P&H(C2), h(Az) = IAIh(z), h-t (0) is dense in C2, and h 0- 0.
Choose a E C2 with h(a) 34 0. Then h(z) = 1 h(ztat. z2a2) satisfies the same
4.5. Holomorphically contractible families of pseudometrics
299
properties ash but now h(1, 1) = 1. Finally, define h(z) := max{h(z), JZI) Note that h is not continuous at the point (1, 1). Then
D={zEC2:h(z)<1) is a bounded pseudoconvex balanced domain. Sinceh (0) is dense wechooseasequence (J ) withzi -+ (1, 1)andh(zj) _
1/4 for j >> 1. Then 0. Therefore, h(z) < 1 /5 for large j and so h (l. z2 /z i) there is a sequence (a) j C C, a j -> 0, such that e"i = zZ /z J . Using a similar argument we find another zero sequence (fl, ) j C C satisfying h (1, A) I (EXERCISE22).
Let F E Aut(C2), F(z) := (zI,z2exp(-zl)). By Corollary 1.15.6, D' F' (D) is a bounded pseudoconvex domain in C2. In virtue of Proposition 4.2.21 it follows that
-9D'(0, (ak, ak)) =
k
k gD (0, (ak. ak exp(ak))) h(ak,ak exp(ak)) = h(1,exp(ak)) < 1/4, 1 ak
when k oo. In a similar way, we get gD'(0, (bk, bk)) -> 1, when k -+ oo. Hence, this different behavior of (0, A0, 1)), when A --* 0, verifies that we cannot take the limit in (4.5.8). Exercise 4.5.12. Try to construct a simpler example of a bounded pseudoconvex balanced domain in C2 with the same phenomenon as above. Nevertheless, in the case when D is bounded and hyperconvex the "limsup" in the definition of the Azukawa pseudometric can be substituted by just taking the limit ([Zwo 2000a)).
Proposition 4.5.13. If D is a bounded hyperconvex domain in C", then: (a) AD is continuous,
(b) AD(a;X) =
lim Cs A- o
I gD(a,a+AX),
(a, X) E D X C".
We should mention that there are similar results even under weaker hypotheses; for more details see [Zwo 2000a]. The proof of the above proposition needs some preparation which will be discussed first. Let D be as in the proposition and a E D. Put
DE := D,(a) :_ (z E D : gD(a. z) < e-E), where e E IR,o. Obviously, D. is open, a E DE, and, by Theorem 4.2.34, DE C D. Even more is true. 22Use the mean value inequality for psh functions.
300
Chapter 4. Holomorphically contractible families on Reinhardt domains
Lemma 4.5.14. Under the above conditions, Ds is a domain. Proof. Suppose the contrary, i.e. that D. has a connected component U with a 0 U. e_8 We know that gD < on U and gD(a, z) > e-8, z E aU fl D. Consequently, the function if z E U, e-8 v(z)
gD(a,z) ifz E D \ U.
is psh on D (EXERCISE). Noting that a > U we have v < 9D (a. - ) on D. In particular, gD (a, - ) I U> e-8; a contradiction. 11
Lemma 4.5.15. Let a, D, D8 be as before. Then gDf(a, z) = 9D (a, z) ee,
z E D8;
ZED8, X EC".
(4.5.9) (4.5.10)
Proof. Note that gD (a, )e8 < 1 on D8. Thus, gD (a, z)ee < gDE (a, z), z E D8. On the other hand, gD (a, z) > e'e when z E aD8. Therefore, the function gDE(a,
v(Z)
z) e-8 if z E D8,
9D (a, z)
if z E D \ D8,
is psh on D. Hence, v < gD (a, - ) on D, which finally gives (4.5.9). The remaining equation is a simple consequence of the definition of the Azukawa pseudometric.
0 Proof of Proposition 4.5.13. (a) Fix (a, X) E D x C" with AD (a; X) # 0;23 in particular, X 54 0. Suppose that there is a number M > AD (a; X) and a sequence ((aj. Xj))j E D x (C" \ {0}) converging to (a, X) such that AD(aj; Xj) > M, j E W. Fix then e E R>0 such that M > AD(a; X) el. Put e' := 2E. Then D8'(a) C= D8(a) (note that gD is continuous on D x D, where gD = 0 on D x aD). Now we choose affine isomorphisms Fj E Aut(C") (j >> 1) such that
Fj(aj) = a,
Fj(aj)Xj = X,
D6'(a) C Fj(D6(aj)),
j E N large enough.
Then, by (4.5.10),
AD(aj;Xj)-e8 = AD,(aj)(aj;Xj) = AF,(De(aj))(Fj(aj):FJ(aj)Xj) = AFj(D.(a,))(a: X) < AD, (a) (a; X) = AD(a; X) e8', i.e. AD (a j ; X j) < AD (a; X) e8 < M ; a contradiction. Hence, A D is continuous. 23Note that AD is upper semicontinuous.
4.5. Holomorphically contractible families of pseudometrics
301
(b) Without loss of generality we may assume that a = 0 and AD (0; X) > 0 (EXERCISE); in particular, X # 0. Suppose now that there is a sequence (A j) j C C., A ,j -+ 0, such that 1
gD(O,O+AjX) < AD(0:X)e- 28,
I
j E N.
Since DE = D8(0) C= D, we find a Oo E (0, rr) such that e'0De 9 D, 101 < Oo. Moreover, we may assume that A j X E D8, j E N. Then l gD(0,e'BAjX) < 1 IAjI IAj1
gei6Dr(0,ei8),jX)
=
141
a; X e-8.
e8 < A
O. A X
9DE(O+AjX)
0< 00,
E N.
Fix r > 0 such that 8(r11 X 11) C D. Put TffgD(0, M
AD (0; X)
if E K.(r), if l; = 0.
Then log u E S}((K(r)). By the upper semicontinuity of u there is a jo such that A j E K(r), j > jo, and
u(e'B)j) < u(0)e
2! e35
u(0)es,
0 E [-7r, 7r].
On the other hand we already know that
u(ef°Aj) <
u(0)e_8.
j E N. 101 < 0o.
Applying the mean value inequality for the subharmonic function u yields for large j,
jlogu(e9Aj)dO n
2rru(0) < <
f
(log u(0) - e)d0 + f _
log u(e'BAj)d0
201>Bp
91<60
< 200(u(0) - e) + (2n - 200)(u(O) + e) = 2nru(0);
0
a contradiction.
Remark 4.5.16. Using the former argument one can even prove (EXERCISE) the following sharper version of Proposition 4.5.13 when D is a bounded hyperconvex domain in C" (cf. [Zwo 2000a]).
AD(a: X) =
lim
i-x r
n
9D W, Z") 11Z' - z"11
(a, X) E D x V. 11XI1 = 1.
Chapter 4. Holomorphically contractible families on Reinhardt domains
302
Example 4.5.17 (Kobayashi-Royden pseudometric).
xD(a; X) := inf{t > 0 : 3toEO(D,D) : (p(0) = a, ttp (0) = X), 24
(4.5.11)
where D C C" is a domain and (a, X) E D x C". Indeed, it is easily seen that XD is a pseudometric. Now, let D = D 31 a. If W E O(D, ID) is such that X0(0) = a and trp'(0) = X E C,, t > 0, then iu
A
*
SP(A) -a 1 - iiW(A)
belongs to O(m, ID). t/i(0) = 0. Therefore, by the Schwarz lemma, we have I > ,ial
and sot = I X I/ko'(O)I ? IXI/(1- IaI2). Hence, xm(a; X) > y (a; X). To get the converse inequality for X = 1 it suffices to take V(AA) _,
I*'(0)I =
t
,
AED. The proof of (4.5.3) is simple and therefore left as an EXERCISE.
First we collect a few simple properties of the Kobayashi-Royden pseudometric.
Exercise 4.5.18. Prove the following statements:
(a) ND(a; X) := inf{t > 0 : 3veO(U,D) : p(O) = a, ttp'(0) = X}. (b) If (SD)D is a holomorphically contractible family of pseudometrics, then YD $ 8D 5 MD, D C C" a domain, i.e. (YD)D (resp. (XD)D) is the minimal (resp. maximal) holomorphically contractible family of pseudometrics.
(c) If Dj / D and (a, X) E D x C", then xD, (a; X) \ xD (a; X) when
j - oo.
(d) If D is taut and (a, X) E D X C", then there exists an extremal analytic disc tP E O(D, D), i.e. rp(0) = a and XD (a: X)cp'(0) = X. Such a rp is called a ND-geodesic for (a, X). (e) Let D. G C C" be domains and let F : G -+ D be a holomorphic covering.
Assume that (i, X) E G X C". Then xG(i: X) = xD(F(z): F'(i)X). (See Proposition 4.2.38.) Moreover, we have:
Lemma 4.5.19. (a) If D = Dh C C" is a pseudoconvex balanced domain with Minkowski function h, then
ND(0; X) = h(X),
X E C".
In particular, Ni, (a; - ) need not be continuous. (b) xD : D X C" --> IR+ is upper semicontinuous. (c) If D is taut, then ND is continuous on D x C. 24Note that the infimum is taken over a non-empty set of analytic discs.
4.5. Holomorphically contractible families of pseudometrics
303
Proof. (a) By Example 4.5.10, XD (0; X) > AD (0; X) = h (X), X E C".
To discuss the converse inequality assume first that h(X) # 0. Then ID a A . AX/h(X) gives an analytic disc in D with (p(0) = 0 and h(X)p'(0) = X, i.e. xD(0; X) < h(X). If h(X) = 0, then ID 9 A A. AkX, k E N, gives an analytic disc in D with }< , (0) = X, i.e. xD (0; X) = 0(b) Fix a point (a, X) E D x C" and assume that xD(a; X) < A for a certain real number A. By Exercise 4.5.18 (d), there is an analytic disc (p E O(0). D) such that (p(0) = a and trp'(0) = X, where IR>0 9 t < A is suitably chosen. Then (p(lD) is a compact subset of D. In particular, a full s-neighborhood of V(FD) is contained in D. Now take z E IB(a. E14) and Y E C" with (1/t)11Y - X 11 < s/4. Then
*(A) :_ (p(A) + (z - a) + (Alt)(Y - X)
leads to a t/i E O(ID, D), ;fr'(0) = z and tt/r'(0) = tgo'(0) + Y - X = Y, i.e. MD (Z; Y) < t < A. (c) is left as an EXERCISE.
As a direct application we get
Corollary 4.5.20. Let Dj = Dhj C C" be balanced pseudoconvex domains, j = 1,2. If F E Biho,o(D1, D2), then F'(0): DI --* D2 is a linear isomorphism. Proof. Let X E D1. Then
h2(F'(0)X) = xD2(0; F'(O)X) = xDt (0; X) = ht(X),
0
which immediately gives the proof.
Remark 4.5.21. Let Dj be bounded balanced pseudoconvex domains in C". By a deep result of Kaup-Upmeier (see [Kau-Upm 1976], [Kau-Vig 1990]) it is known
that if Bih(D1, D2) 36 0, then Biho,o(D1, D2) j4 0. In particular, if Dt is biholomorphically equivalent to D2, then DI is linearly equivalent to D2. Exercise 4.5.22. Applying Lemma 4.5.19. prove:
x
2
2
1/2
42)2) . 25 (a) Yen (a, X) = xe, a: X) = 2 + (b) IB" is not biholomorphically equivalent to ID", n > 2. (c) Decide whether 033 and thedomainD := {z E C3 : IziI4+Iz214+Jz3I4 < 1) are biholomorphically equivalent (use Remark 4.5.21). t_1t
There is also another way to define the Kobayashi-Royden pseudometric which will be important in the proof of Proposition 4.5.25. 25To get the equation for y use the fact that l6 is convex.
304
Chapter 4. Holomorphically contractible families on Reinhardt domains
Proposition 4.5.23. Let (a, X) E D x C", X # 0, where D is a domain in C". Then
xD (a; X) = inf {t E 1R,0 : 3FEo(Bn,D) : F(0) = a, t
- ( 0 ) = X. det F'(0) # 0}. 1
Proof. Only during this proof we will denote the right-hand side by WD (a; X). Obviously, any mapping F in the formula for ID (a; X) induces an analytic disc !p E O(D, D) by (p(A) = F(A, 0, ... , 0). Hence, xD(a; X) < iD(a; X).
Now suppose that xD (a; X) < m < x`D (a; X). Then there exist a t E (xD(a; X). m) and a V E 0(D, D) such that qp(0) = a and t(p'(0) = X. Put zE(DxCn-1,
FE(Z):=(SPI(Zt).
s>0.
Obviously, det F;(0) # 0, F&( , 0..... 0) = (p on D, t (0) = X. Now fix r < 1, near 1. Then F1(rID. 0,.... 0) is a compact subset in D. Thus we can choose a 8 small enough such that F1(rD x 80) x ... x SD) C= D. Hence, if a is sufficiently small, Fs E O(In(r), D). Finally, define F(z) := FE(z/r), Z E Bn, which gives the desired contradiction for r very near to 1.
Lemma 4.5.24. Let a. Z E 8n and let (P E O(O, B,), (p(A') = a, (p(Ag) = z, and Ao # Ao such that k® (a, z) = k1Bn (cp(A0), (p(A0)) = m(A0, Aa). Then A E D. In particular, (p is a xen -geodesic for (rp(A). V'(A)), A E D.
xBn (co(A); (p'(A))
Proof Observe that kBn = men and xen = Yen (EXERCISE). Put
u(A) .- mm'A' A ))
A E ID \ (A0').
Recall that men (a, - ) is continuous; so, by its definition, it is log-psh. Therefore, u is sh, u < 1, and u(A") = 1. Then, by the maximum principle, it follows that
u=Ion ID \{Ao}. So men (a, rp(A)) = m(Ao, A),
A E ID.
Now we can repeat the same argument w.r.t. the first variable to get finally m(A', A"),
men
A'. A" E ID.
Fix Ao E D. Then, by Lemma 4.5.3, we have
Y(Ao; l) =
lim
lim-
m()Lo' A)
IAo - Aj men ((POLO), V(A)) PLO -'XI
= Yen
i
co (Ao)).
4.5. Holomorphically contractible families of pseudometrics
305
Note that the definition of the Kobayashi-Royden pseudometric is of different type than the ones of the previous pseudometrics. Nevertheless, if the domain D is taut, then we have the following result (see [Pan 1994]).
Proposition 4.5.25. Let D C C" be a taut domain. Then
xD(a;X)_C.liim lim
oI
IkD(a,a+AX) 1
RI
k(a,a + AX), (a, X) E D x V.
0 We do not know any example of a non-taut domain for which this result becomes false although such an example seems very probable. 0 Proof. Note that it suffices to prove only the second formula. Suppose it is not true.
Then there are a point (a, X) E D x C", X 96 0, and a sequence C. a Aj - 0 such that
' kp(a,a+AjX)-xD(a:X)I ?so>0 1
(4.5.12)
IAiI
for some so. Applying Proposition 4.2.11, we may choose k;-geodesics (pj for
(a,a + AjX), i.e. cpj E (9(D, D), cpj(0) = a, cpj(tj) = a + AjX, and tj = kv (a, a + A j X) > 0 (recall that D is bounded). Moreover, D is taut, so we may, without loss of generality, assume that rpj -> .p E O(U), D), op(0) = a, locally unifonnly. Fix IB(a, r) C D. Then, if j is sufficiently large, 1 _ k* (a, a + Ai X) < "IB(a,r)((Pi (0), Vj (tj))
tj
tj < 111 Vj (o) - cci (ti) II r tj
.- 1 Ilty to)II r
Hence, wp'(0) 96 0.
Fix an s E IR,o. Then, by Proposition 4.5.23, we find an F E 0(8, D) and a t > 0 such that F(0) = a, det F'(0) , 0, t (0) = V(0), and 0 < ND (a: c'(0)) < t < XD (a: V(0)) + s.
Now we choose open neighborhoods U = U(0) C IB" and V = V(a) C D such that Flu: U -* V is a biholomorphic mapping. Moreover, fix Jo such that
a+AjX E Vand define gj :=(Flu)-t(a+AjX), j > jo. Note that l3" is taut. Hence we have &
-geodesics for all pairs (0, qj), i.e. there exist >lij E (9 (ID, (B"),
f(rj(0)=0,*j(rj)=gj,andkBn(0,gj)=rj,J ?Jo.
306
Chapter 4. Holomorphically contractible families on Reinhardt domains
In virtue of Lemma 4.5.24, we conclude that men (0: tlr (0)) = 1, j ? jo. Applying again that @n is taut we may assume (without loss of generality) that *j -> * E O(DD, llln) locally uniformly. Obviously, *(0) = 0. Then, because of Lemma 4.5.19, we obtain
I = xBn (0: V'j (0)) - men (0 Vr'(0)) = I, i.e. >/i is a xBn-geodesic for the pair (0. Vr'(0)). Note that
9j -0
(Flu)-'(a+,XjX)-(FIu)-'(a)
tj
tj _ (Flu)-'((pj(tj)) - (Flu)-'((pj(0)) -+ tj
(F-' o(p)'(0)
In particular, this limit exists and it is different from zero. Observe that F o *j (O) = F(0) = a = SDj (0),
Fot/rj(Tj) = F(9j) =a+A1X =cpj(tj). Therefore,
tj
for j > jo. In particular, 1 < Recall that Fo*j(ij)-Fo*j(0)
lim
j*oo
=(Fov/r)'(0)#0
Tj
and Jim'pi(ts)-Vj (0)
0 54 (R' 0
= lim
a+AjX - a
= Jim F kj(Tt)-Fo%rf(0)Ti j-+00
Ti
tj
Hence, lim 1 =: A > 1 exists. Moreover, we have
i
V'(0) = A(F o *)'(0) = AF'(0)*'(0)
Taking into account that t-(0) = cp'(0) finally leads to Ai/r'(0) = tet = t (1, 0, ... , 0). Then I = xmn (0; Vr'(0)) = AxBn (0: et) =
A.
4.5. Holomorphically contractible families of pseudometrics
307
i.e. A = t and, therefore, I < A = t < XD (a; cp'(0)) + e. Then, letting E \ 0, gives 1 < XD (a; q'(0)) < 1. Hence, cp is a XD-geodesic for the pair (a, (p'(0)). Note that
fpj(tj!pj(0) _
V (O) = lim
tj
1-400
a+AjX -a = X lim .lj j- 00 j-.oo tj lim
tj
So we conclude
k,(a,a+AjX)
II
ju +00 rn
t
= XD (a; (p'(0)) jl m
I = xD (0; X):
a contradiction to (4.5.12).
11
Working with the Kobayashi pseudometric we always have
Proposition 4.5.26. Let D C C" be a domain and (a, X) E D x C". Then 1
kD(a,a+AX) lim sup C.3z-.o IAI
Proof. If e > 0 is given, then we find an analytic disc Sp E O(ID, D) with the following properties:
tcp'(0) = X,
(p(0) = a,
0 < t < XD (a; X) + E.
Then rp can be written as
rp(A) = a+
i X+ A20(A),
A E ID,
where 0 E 0(1D, C"). Fix IB(a, r) C D and then take only such A E C. with IAIt < 1 and a + AX E 1B (a. r). Hence,
IkD(a,a +.1X) < I,XIkD(a,v(tA)) + I IkD(w(tA),a + AX) I
I1 kD(co(0).rp(ll)) + rill IIt2A2w(1A)II ICI p(0,tA.) +
121AI II0(1A)II
r
Letting A -+ 0 leads to lim
I IAI
kD(a,a+AX)
Since s was arbitrarily small, the proposition is verified.
Chapter 4. Holomorphically contractible families on Reinhardt domains
308
But even if one sharpens the notion of the "derivative" of kD there is, in general, no equality with the Kobayashi-Royden pseudometric as the following example shows.
Example 4.5.27. Put D := {z E ID2 : IzIX21 < r2}, where 0 < r < 1/2.
Fix points z' z" E (D2, Iz I < r2, Iz71 < r2, j = 1, 2. Then
kD(z', z") :5 kD(z', (zi, zi)) + kD((z' z2'), z") < P(zi z') + P(4-4)(EXERCISE). If we discuss the following general differential quotient at 0 in direction
of (r, r), then by the former inequality we obtain
limsup
IIIkD(a,a +AX)
X -(r,r)
C.
lim sup
a-O
X-+(r,r)
1 p(al. at + AX1) + lima-Osup 1 p(a2, a2 + AX2) = 2r. I1kI
X-+(r,r)
1A1
On the other hand, MD (0, (r, r)) = I (use Lemma 4.5.19), i.e. the "differential quotient" of the Kobayashi distance is different from the Kobayashi-Royden pseudometric.
The defect shown in the example has led S. Kobayashi to introduce a new pseudometric (see [Kob 1990]).
Example 4.5.28 (Kobayashi-Busemann pseudometric).
xD(a: ) := sup{q : q a C-seminorm, q < XD(a: )},
a E D,
(4.5.13)
where D C C" is a domain. Indeed, in the case where D = ID we know that XD(a; ) = yc (a; - ) is a norm. Hence, x® = y. To see (4.5.3) let F E 9(G, D) and a E G. Take a seminorm q with q < xD (F(a); ). Then
q(F'(a)X) < xD(F(a); F'(a)X) < xG(a: X). In particular, 4 := q(F'(a) - ) is a seminorm below of MG (a: ). Therefore, q < xG (a: ). Since q was an arbitrary seminorm, we have (4.5.3).
Note that, by definition, £D (a. ) is a seminorm.
4.5. Holomorphically contractible families of pseudometrics
309
Exercise 4.5.29. Let D = Dh C C" be a pseudoconvex balanced domain. Give a formula for xD (0; X), X E C'. Finally, we mention the following result by M. Kobayashi (see [Kob 2000]) that makes it possible to think of the Kobayashi-Busemann pseudometric as a derivative of the Kobayashi pseudodistance.
Proposition 4.5.30. Let D Cc C" be a taut domain. Then
XD(a;X) =
1
lim
(b.Y)-+(a.X) I)'
kD(b,b+AY),
(a,X) E D xC".
What we have seen is that sometimes the pseudometrics introduced here are in a strong relation with certain pseudodistances, i.e. they might be thought of as a derivative of the corresponding pseudodistances. Conversely, one may associate to a pseudometric its so-called integrated form to get either a new pseudodistance or even the one we start from. Here we will restrict ourselves only to the case of the Kobayashi-Royden pseudometric. For further information the reader is referred to [Jar-Pfl 1993] or [Jar-Pfl 2005].
Let D C C". Note that XD is upper semicontinuous. Hence, if a: [0, 1) -* D is a piecewise e'-curve, then the integral LID (a) := fo xD(a(t): a'(t))dt exists. We call the number LID (a) the XD-length of the curve a. Using this notion we define.
Definition 4.531. Let D be as above. Put f xD : D x D -+ U2+ as
(f xD)(a,b) := inf(LID (a) : a E C'([0.1]. D), a(0) = a, a(l) = b),
a,bED. f XD is called the integrated form of ND. Note that (f XD)D is a holomorphically contractible family of pseudodistances (EXERCISE). Therefore, f ND < kD. Even more is true, namely the integrated form leads back to the Kobayashi pseudodistance.
Proposition 4.5.32. If D C C", then f xD = k D.
Proof. It remains to show that kD < f MD. Suppose that the opposite is true, i.e. there are points a, b E D and a piecewise C' -curve a : [0.1 ]
D connecting
a and b such that LID (a) < kD (a. b). Observe that there are numbers I = so < S I < < SN = I such that all a1 := are C' -curves. Then Lxo(a) = ENI Lxo(aj) < :NI kD(a(sJ_I).aJ). Therefore, without loss of generality, we may assume that a is a e I -curve.
Put f(t) := kD(a,a(t)), I E [0, 1].
310
Chapter 4. Holomorphically contractible families on Reinhardt domains
Fix a to E [0, 1] and IB(a(to).2r) C D. Then, if S is sufficiently small, we conclude that a(t) E IB(a(to), r) whenever It - toI < S and t E [0. I]. If now t, s E [0, 1] n (to - S, to + S), then
If(s) - f(t)I = IkD(a,a(s)) -kD(a,a(t))I kD(a(s),a(t)) <- Ctolla(s) -a(t)II :< el, IS -11, i.e. the function f is locally Lipschitz. Therefore, f is almost everywhere differentiable and 1
f'(t)dt.
kD(a, b) = J0
What remains is to estimate f':
I f(t + hh - f(t)I
If'(t)1 < lim to+
< lim sup kD(a(t hh),a(t)) h--+O+
< lim sup
kD(a(t).a(t) + ha'(t))
h-'O+
h
+ lim sup
kD (a (t + h), a(t) + ha'(t)) h
h-.o+
< xD(a(t);a'(t)) + lira sup
CtIIa(t) + ha'(t) -a(t + h)II
h-+o+
h
= XD (a (t); a' (0)
for almost all t E [0, 1). Here we have used Proposition 4.5.26 in the last line. Hence, kD(a, b) < LID (a); a contradiction. Exercise 4.5.33. Formulate what is meant by f RD and prove that kD = f RD.
4.6 Examples II - elementary Reinhardt domains In this section we briefly discuss the formulas for some families of holomorphically contractible pseudometrics with respect to the elementary Reinhardt domains. In the one-dimensional case we have
Proposition 4.6.1. (a) Let A = A (I / R, R), where R > 1, and a E A n IR+. Put
a = R t _2s. Then
xA(a; l) =
n 4a log R sin(gs)
(b) Let a E lD. n IR+, then xD. (a: 1) = 2a loga'
Proof. Note that A and D. are taut domains (EXERCISE). Hence the formulas follow directly from Theorem 4.4.1, Corollary 4.4.2, and Proposition 4.5.25.
4.6. Examples 11 - elementary Reinhardt domains
311
Observe that yp, = yo, A[). = AID on ID, X C. Now we turn to the discussion of elementary Reinhardt domains Da in higher
dimensions. Let D. C C" be given and fix a pair (a, X) E D« x C". For the remaining part of this section we will always assume that (see Section 4.4):
n>2; al,...,a., <0andas+i.....an>0foran s=s(a)E{0,1.....n); ifs < n, then t = t(a) := min{a5+1.... , a.); a = (al,....a") E D«, a, ...ak 54 0, ak+1 = ... = a" = 0 for a
k = k(a) E {s,...,n);
if k < n, then r = r(a) = r«(a) := ak+1 + . + a"; if k = n (in particular, if s = n), then r = r(a) = r«(a) := 1; observe that if a E Z", then r(a) = orda(z« - a«); if D« is of rational type, then a E Z" and a,,. - -a" relatively prime; if D. is of irrational type and s < n, then t (a) = 1. The following effective formulas for holomorphically contractible pseudometrics for D. are known ([Jar-Pfl 1993], [Pfl-Zwo 1998], and [Zwo 2000]).
Theorem 4.6.2. Under the above assumptions we have: a
11
Rational type
Irrational type, k < n Irrational type, k = n
y(aa;aa E"=1 a)
Rational type, s < n
Rational type. s = n
Irrational type, s < n Irrational type, s = n
(y(aa; F(r)(a)(X)))11r
0
([k=1 I ai I «' nJ=k+1 ix ;«; )t/r
0
0
where F(z) := z« and F(r)(a)(X) :_ a
ADa(a;X)
yam,(a;X)
DIOF(a)X#
xi (a; X) y((aa)11t; (aa)1/r Fj_1 «
),
k=n
(Tat I« ... lak lak l Xk+l I«k+1 ... IX"Ian)1/r. k < n
xm, (aa; a« F"1=1 «,ax; ) y(Ia« I' Ia« I E j-I a (Tall «1 .. Iakl «k IXk+11 «k+1 ... IX"I a ) 1/r ,
k=n k
xm,(la«J; la«l j=1 «;x, )
Proof. In the case of the Carathdodory-Reiffen (resp. the Azukawa) pseudometric use Theorem 4.4.4 and Lemma 4.5.3 (c) (resp. (4.5.8)).
312
Chapter 4. Holomorphically contractible families on Reinhardt domains
To prove the corresponding formulas for x we will need several steps.
Proof for xDa - the case k < n. The estimate from below follows immediately by applying xDa (a; X) ? AD,, (a; X) and the formula for A Da, . So the upper estimate remains. In a first step assume that Xk+1 . . . X, 0 0. Put r := (Iai IQ" ... Iak Iak IXk+1 lak+, ... IXn Ia" )1"' > 0.
In virtue of Exercise 4.4.12, we find functions t/r = (1fr1, satisfying
*j(0) = aj, *j(0) = Xj/r, j = 1 ,
.
.
..
k,
.... *n) E
O (1D. C * )
fij(0) = Xj/r, j =k+
n,
1/i,°,`" = e'0 idm for some 8. Then the holomorphic mapping
and 0,0"
ID 3 .Z H (P(A) :_ (W 1(A).... , Vk
E Da
fulfills the following properties: V (O) = a and r(p'(0) = X. Hence, xDa, (a; X) < r. If X jo = 0 for some jo E {k + 1, ... , n }, then we have the holomorphic mapping k X
n-k-i 3(Z1.....Zjp_1,Zjo+l,...,Zn H (z1,...,Zjo_1,O,Zjo+1.....Zn) E Da.
Therefore,
0=
(X1,... ,Xjo_1.Xjo+1.....Xn)) ? xD,,(a;X), which proves the remaining case.
Lemma 4.6.3. Let a E D. n C and X E C" such that Ej=1 !LL = 0. Then l
xi (a;X)=0.
Proof. Observe that for U E iD* the mapping Fµ : CI-1 --> C", F, (z1,...,zn-1) := (eanz`,
..,ea"zn-1,µe-a,z,-..._an-izn-I
belongs to O(Cn-1, Da). Then there are a µo E I* and a 1 = (z1, ... , Zn_1) E
C"-1 such that FN,o(z) = a and F 0(2)Y = X, where Y = (X1, ... , Xn_1). Hence,
0 = xCn-1 (2; Y) ? xDa (a; X), i.e. the proof is finished.
4.7. Hyperbolic Reinhardt domains
313
Proof for xq, - the case s < n = k. First we recall that (Proposition 4.5.26)
xDa,(a;X)> lim sup
k°r°'(a.a+AX) ICI
Evaluating the right-hand side leads (by a trivial calculation) to the claimed formula for xD, (a; X). Hence the estimate from below is verified. We continue with the estimate from above. In virtue of Lemma 4.6.3, we may assume 54 0 and t = an (recall that t = 1 in the irrational case). Ejn= t °`
By the symmetry of Da we also assume that aj > 0, j = 1, ... , n. Now put AO := (aa)t/«n E ID n (0, 1) and r := k0 Ej_t ° Exercise 4.4.12, we find a V E 0(ID, D") such that (p(Ao) = a,
.
In virtue of
rfp'(ko) = X.
Hence, r > xD,, which gives the desired estimate from above.
Proof for xD. - the case s = n. We leave this last case as an EXERCISE for the reader. Use the ideas of the corresponding case for the Lempert function and Exercise 4.5.18 (e).
4.7 Hyperbolic Reinhardt domains We know that, in general, CD (resp. kD) need not be distances. We define
Definition 4.7.1. Let D C C" be a domain and let dD E {CD, kD, k;). D is said to bed -hyperbolic if dD (a, b) = 0, a, b E D, implies that a = b.
In particular, D is c- (resp. k-) hyperbolic if and only if CD (resp. kD) is a distance on D (in the sense of metric spaces). Note that (EXERCISE)
any bounded domain is c-hyperbolic; any c-hyperbolic domain is k-hyperbolic; any k-hyperbolic domain is k-hyperbolic. In the class of pseudoconvex Reinhardt domains we have a complete description of those domains which are hyperbolic.
Theorem 4.7.2 ([Zwo 19991). Let D C C" be a pseudoconvex Reinhardt domain. Then the following properties are equivalent:
(i) D is c-hyperbolic; (ii) D is k-hyperbolic; (iii) D is k-hyperbolic; (iv) D is Brody hyperbolic;
314
Chapter 4. Holomorphically contractible families on Reinhardt domains
(v) there exists A = [a j,k ] t < f,k
'PA(Z) := (Z' . ... Z'"),
Z E D,
maps D biholomorphically onto its image 'PA (D) which is a bounded Reinhardt domain of holomorphy. (iii), and (v) Proof. Obviously, (i) = (ii), (ii) (i). Now assume (iii). Suppose D is not Brody hyperbolic. Then we find a tp E O(C, D), gp 0 c. Therefore, kD(0A), 00)) < kE(A.0) = 0, A E C. Since D is k-hyperbolic, it follows that (p is the constant function (p(0); a contradiction. Recall that (iv) (v) follows directly from Theorem 1.17.1 1.
Consequently, this result allows us to speak only of hyperbolic pseudoconvex Reinhardt domains. In the general situation there is a reformulation of k-hyperbolicity in terms of the Kobayashi-Royden pseudometric. More precisely, the following statement is true.
Lemma 4.7.3. For a domain D C C" the following properties are equivalent:
(i) D is k-hyperbolic:
(ii) for any point a E D and any neighborhood U = U(a) C D there exist neighborhoods U = U(a) C U and V = V(0) C ID such that if (P E O(m, D), V(O) E U, then V(V) C U. (iii) D is x-hyperbolic, i.e. VaED 3U=U(a)CD 3c>o : ZD(z: X) > clIX il, z E
U.XEC". Proof (i)
(ii): Fix a and U as in (ii) and choose an r > 0 such that S(a, 2r) C U.
Then kD(a, z) > 0 whenever z E al6(a, r). By assumption, kD(a,.) > c > O on 81B(a, r) (recall that kD(a, - ) is continuous).
Now take a point z E D \ IB(a, r) and a piecewise et-curve a: [0, 1] - D. which connects a and z. Then
f
I.
LXD (a) >
XD (a(t ); a'(t ))dt > kD (a. co(to)) > c,
wherea(t) E 8(a, r), t E [0, to), and a(to) E 31B(a, r). Therefore, kD(a. z) > c. Hence we have shown that the kD-ball BkD(a,c) C IB(a.r) C U.6 Put U _ Bk0 (a, c/2) and V := Bkm(0. c/2) C D. Now let (p E O(0). D) with tp(0) E U. If A E V, then kD(a.co(A))
kD(a,to(0))+kD(V(0),V(A))
26Recall that Bd (a, r) := (x E X : d (x, a) < r).
c/2+ko(0,A) $c.
4.7. Hyperbolic Reinhardt domains
315
Hence, rp(V) C U. (ii) (iii): Fix an a E D and put U := B(a, r) C= D. Then choose V = V(O) C m and Cl = CI (a) C U according to the assumption in (ii). If now
rp E O(lD, D) with V(0) E U and tcp'(0) = X, then rp(V) C U. Fix an s > 0 such that K(s) C V. Then we conclude that VI K(,) E O(K(s), U). Put O(A) := rp(sA). Then cp E O(1D, U) satisfying 0(0) = rp(0) and sip'(0) = X. Hence
xD(Z;X)>SxU(z:X)>CIIXII,
zEU, X EC",
where c is a certain positive number. (iii) = (i): Fix two different points a, b E D. By assumption, we may choose a neighborhood U = U(a) C D such that XD (z; X) > c II X II for a certain c > 0,
z E U. Fix now an r > 0 such that b 0 B(a, 2r) C U. Let a : [0, 1] -+ D be a piecewise e t -curve in D which connects a and b. Let to be that time such that a(t) E (B (a, r) for all t E [0, to) and a(to) E 8B (a, r). Then to
LXD(a) > Lxo(aI[o401) > cJ
IIa'(t)Ildt > cr > 0. 0
Hence, kD (a, b) > Cr; in particular, D is k-hyperbolic.
0
Exercise 4.7.4. (a) Prove that a domain D is k-hyperbolic if top D = topkD. Here top D means the standard topology on D, where top kD is the topology on D that is induced by the Kobayashi pseudodistance. (b) Prove the following generalization of Cartan's theorem (see Theorem 2.1.7):
Let D C C" be a k-hyperbolic domain, let a E D. and let 0: D -+ D be a holomorphic mapping such that 45(a) = a and df'(a) = id. Then ds = id. Use (a) to get the bounded situation. Moreover, we have
Proposition 4.7.5. Any taut domain D C C" is k-hyperbolic. Proof. Suppose that D is not k-hyperbolic. Then, in virtue of Lemma 4.7.3 (ii), we
find a z' E D, a neighborhood U = U(z') C D, a sequence Aj -+ 0 in D, and a sequence (rpj)j C O (m, D) such that tpj (0) -+ z', but rpj (,kj) 0 U, j E W. Since tp, (0) -+ z', there is no subsequence which is locally uniformly divergent. And because of rpj (Aj) 0 U, j E W, there is no subsequence tending locally uniformly to a rp E 0(lD, D); a contradiction. To get another large class of k -hyperbolic domains we prove the following result which is due to [DDT Tho 1998].
316
Chapter 4. Holomorphically contractible families on Reinhardt domains
Proposition 4.7.6. Let u : lD --> [-oo, oo) be upper semicontinuous and locally bounded from below. Then
D := {zEIDxC:lz21e"(z1) <1) is a k-hyperbolic domain.
Proof. In virtue of Lemma 4.7.3 it suffices to show that XD is locally positive definite, i.e. for any point z' E D there exist U = U(z') C D and c > 0 such that
xD(Z;X)>ClIX11,ZEU,XEC2. By assumption we have
g(r) := inf{u(;L) : RI < r} > -oo,
r E (0, 1).
Fix S E (0, 1), z' E (sD x C) fl D, and X E C2, X # 0. Now choose an analytic disc (p = ((pt, V2) E O(Q), D) such that V(O) = z' and tV'(0) = X for a certain t E (0, 1). In virtue of the Schwarz lemma we have ico (0)1 < I - Izi 12 < 1. Put so := 1+2s s0 for a A0 E D. Then )I _ 2+s Note that so < 1. Suppose hvt (Ao> the Schwarz lemma implies that
p1(Ao) - z'1 > Iwt (Ao)I -Izi 1 > so -Izi 1 > so - s IAoI > ll-z,V,(,Xo)l- 1-Iwt(Ao)IIz'l - I-so1z'l - 1-sos
Put S2 := (A E ID
2
IVI(A)l < so). Then I(p2(A)I < e-g(so), A E S2, and
:
K(1/2) C S2. Thus, I(pl (0)l < 2e-g4O) (Schwarz lemma). Hence
t > max I IX1I,
se-94lo)
>
I
I
min
{
1,
2e-s(so) IIXII =: t(s)II X II
Since c was arbitrarily chosen we have
XD(z; X) 2 t(s)IIX II,
Hence, D is k-hyperbolic.
(z, X) E ((sm x C) fl D) x C2.
0
The result allows us to present a k-hyperbolic pseudoconvex domain which is not c-hyperbolic. Thus the general situation is more complicated than the one inside the class of pseudoconvex Reinhardt domains.
Example 4.7.7 ([Sib 1981]). Choose a sequence (a1)j C ID of pairwise different points aj such that any boundary point E 8D is the nontangential limit of a subsequence of (aj)j. The reader is asked (EXERCISE) to construct such a sequence. Moreover, we choose natural numbers m j and nj, j E W, such that
nj <mj, j E N, 2 < -00, E1=, nt log -lug
4.8. Caratheodory (resp. Kobayashi) complete Reinhardt domains
K(a j . 3e-!'j)
317
fl K(ak, a-kmk) = 0, j # k,
K(aj, 3e-jmi) C D. Finally, we define
u(,.):=En maxjmj,logJA j=I i
)LE(.
2 ajI
Then U E e(ID) fl 83{(ID) (EXERCISE). In particular, u is locally bounded from below. Therefore, by Proposition 4.7.6, we see that
D:={Z E IDxC:I:2Ieu(=t) < 1} is a k-hyperbolic domain. Observe that D is even pseudoconvex (EXERCISE).
On the other hand, let f E X1 (D). Then f(z) =
r
hj(zI)z2, where the
h j's are holomorphic on D. Then, using the Cauchy inequalities and the maximum principle for holomorphic functions leads to h j = 0, j ? 1. So f depends only on
the variable z1. In particular, CD((zt,z2),(zt,0)) =0whenever z = (zi,z2) E D. Hence, D is not c-hyperbolic.
4.8 Caratheodory (resp. Kobayashi) complete Reinhardt domains Let D C C" be a bounded domain. Then we know that (D, CD) (resp. (D, kD)) are metric spaces in the usual sense. In general, let dD E {CD, kD }. We define
Definition 4.8.1. Let D C C" be a given domain.
(a) D is said to be d -complete if it is d -hyperbolic and any dD-Cauchy sequence (zj)JEN C D converges in the standard topology to a point z* E D, i.e. Ilzj - z*II -> 0. (b) D is said to be d -finitely compact27 if it is d -hyperbolic and if for any a E D
and any r > 0 the dD-ball Bdo (a, r) is relatively compact in D in sense of the standard topology of D.
Remark 4.8.2. (a) Observe that if D is d-finitely compact, then D is d-complete (use that dD is continuous). (b) Any c-complete domain is d-complete. (c) Any c-complete domain is a domain of holomorphy (EXERCISE).
(d) Recall that if D is k-hyperbolic, then top D = topkD. Therefore one may formulate k-complete (resp. k-finitely compact) by using top kD instead of top D. Note that in case of c there are examples showing that the topologies top CD and top D are different (see [Jar-Pfl 19931). 27This notion is taken from standard differential geometry. see the theorem of Hopf-Rinow.
318
Chapter 4. Holomorphically contractible families on Reinhardt domains
(e) For a c-hyperbolic plane domain D we have: D is c-complete if D is c-finitely compact. This result is due to [Sel 1974] and [Sib 1975] (see also [Jar-Pfl 1993], Theorem 7.4.7). (f) Q In the case of a domain D in C', n > 2, it is still an open question, whether c-completeness implies c-finitely compactness. Q On the other hand, in the class of general complex spaces there are counterexamples; see [Jar-Pfl-Vig 1993]. (g) If F E Bih(D1, D2) and if DI is d-complete, then D2 is d-complete, where
d E {c, k). Dealing with the Kobayashi distance we have that both notions are the same.
Proposition* 4.8.3. Let D be a k-hyperbolic domain in C". Then the following properties are equivalent: (i) D is k-complete; (ii) D is k-finitely compact. Here we will not present a proof of this result. But we mention that the former result is a particular case of a result where one deals with continuous inner distances
(note that kD satisfies these properties). The main idea is taken from differential geometry. Details may be found in [Jar-Pfl 1993], Theorem 7.3.2. Next we mention the following necessary conditions for a domain to be k-complete.
Lemma 4.8.4. Any k-complete domain is taut. In particular, it is a domain of holomorphy (use Exercise 1.17.21 and the solution of the Levi problem).
Proof Take a sequence (Vj)jew C (9(ID, D). Suppose that this sequence is not locally uniformly divergent. So we find compact sets K C ID and L C D such that, without loss of generality, qp j (A. j) E L, where k j E K. Fix a point a E L and an r E (0, 1) such that K C K(r). Then, for any A E K(r), we have
kD((oj(A),a) < kD(cof(A),coJ(Af)) +kD(j(AJ),a) < p(,1Jj) + sup{kD(z,a) :: E L} $ C(r). Hence,
U cpj(K(r)) C BkD(a,C(r)) C= D. jeN What remains is to apply Montel's theorem.
0
In case of Reinhardt domains even the following converse statement is true.
Theorem 4.8.5. Any hyperbolic pseudoconvex Reinhardt domains D C C" is k -complete.
4.8. Carathdodory (resp. Kobayashi) complete Reinhardt domains
319
In order to be able to prove Theorem 4.8.5 we need the following localization result due to Eastwood (see [Jar-Pfl 1993], Theorem 7.7.5).
Lemma 4.8.6. Let D C C" be a bounded domain. Assume that for any b E aD there exists a bounded neighborhood U = U(b) of b such that any connected component U' of D fl U satisfies the following condition: if a E U' and U' 3 bk -+ b, then kU-(a, bk) _+ oo. The,, D is k-complete. Proof Suppose the contrary. Then we find a b E aD and a kD-Cauchy sequence
(zJ)jEN C D such that z1 -+ b E aD. Let U = U(b) be the neighborhood whose existence is known from the assumption. Choose R > 0 such that U U D C 1B" (R) =: V. By Exercise 4.7.4, U is open in topkv, i.e. Bkt, (b, 2s) C U for a certain positive s. Then we find a jo E D 1 such that
kD(zj, zt) < s/3
Z1 E Bk1, (b, s/3).
and
j, e > jo.
Fix a j > jo. Then there exist k E N, V,, E O(D, D), and A E 1D, v = 1.... , k, such that (P1(0) = Z
Wv+1 (0),
v = 1.... , k - 1.
k
p(0. Av) < s/3.
cPk(Ak) = Z', v=1
LetAEBp(0,s)CIDandI
k
<s+1: p(O.A)+s/3<2s, u=1
C U, v = 1,.. ,k.
i.e.
Note that B. (0, s) is a disc with center 0. We choose a biholomorphic dilatation
y: D - Bp(0,s) and put 01(0) = zio
(p, o y E 9(D, D fl U). Observe that 0g+1 (0),
p = 1, ... , k - 1,
0k(Y-1(Ak)) = z'.
and Eµ
p(0, y-t(.lµ)) < c Ek=1 p(0,.1µ) < cs for some c > 0 which is
independent of j. Note that, by construction, the points z-, j jo. belong to one connected component U' of D fl U. Hence, kU,(zJ0, z3) < cs, j > jo; a contradiction.
320
Chapter 4. Holomorphically contractible families on Reinhardt domains
Proof of Theorem 4.8.5. By the hyperbolicity condition we may assume that D is bounded and D C D". The proof is done by induction. The one-dimensional case is obvious (use the explicit formulas from Section 4.4). Now let D C C", n > 2, and assume that the theorem is true for all smaller dimensions. We will show that D fulfills the condition in Lemma 4.8.6. So let us fix a b E 8D. There are different cases to discuss. Case b E C;: Let us choose a polycylinder I'(b, r) C= C. Put
V := {zEC":Ilzkl-IbkII
of D n V. Then V' is a Reinhardt domain of holomorphy and fulfills the Fu condition. Fix an a E V' and a sequence V' bk -> b (if it exists). In virtue of Theorem 1.13.19, there is a function f E O(V', D) such that f (a) = 0 and I f (bk) I -> 1. Therefore,
kv,(a.bk) > tanh-t mv,(a,bk) > tanh-t If(bk)I -> oo, i.e. b fulfills the condition in Lemma 4.8.6.
Case b E Yo \ {0}: We may assume that b = (bt , ... , bk, 0,.... 0) = (b', 0) E
C; XC"-k,where I
Assume that there is a jo E {k + 1.....n} with D n Vjo = 0; without loss of generality, let jo = n. Then D 3o z -t z,, defines a holomorphic map
F E O(D,C,) and F(D) is bounded. Therefore, F(D) C K, (r). Note that K,(r) is k-complete. Hence, if a E D and D E) bk -r b, then
kD(a,bk) > kx.(r)(F(a), F(bk)) -> oo, i.e. b fulfills the condition in Lemma 4.8.6 with U = U(b) = C". Case b = 0: If D n Vo = 0 for a jo, then one argues as just before. On the other hand, D n Vj # 0 for all j = 1, .... n is impossible, since D is a Reinhardt domain of holomorphy. In case of Carath6odory finitely compactness we have the following reformulation in terms of bounded holomorphic functions.
Lemma 4.8.7. Let D be a c-hyperbolic domain in C". Then the following properties are equivalent:
4.8. Carathdodory (resp. Kobayashi) complete Reinhardt domains
321
(i) D is c-finitely compact; (ii) for any a E D and any sequence (zj)j EN C D without accumulation points in D there exists an f E 0 (D, D) with f (a) = 0 and sup{ I f (zz) I : j E
N} = I. In particular, any c-finitely compact domain is X'-convex. Proof. Obviously, (ii) implies (i). For (i)
(ii) just apply Proposition 1.13.18.
Remark 4.8.8. According to this result one can conclude that a lot of smooth pseudoconvex domains whose boundary points are general peak points are c-finitely compact. For example, any strongly pseudoconvex domain is c-finitely compact. Recall that a boundary point a of a domain D is said to be a general peak point if
for any sequence (zj)jEN C D, zj -- a, there exists an f E 0(D, ID) such that sup{I.f(zi)I : j E N} = 1. In case of Reinhardt domains the following geometric characterization is true (see [Fu 1994] and [Zwo 2000b]).
Theorem 4.8.9. Let D be a pseudoconvex Reinhardt domain in C. (a) If D is hyperbolic, then D is Kobayashi complete. (b) The following properties are equivalent:
(i) D is c-finitely compact; (ii) D is c-complete;
(iii) there is no sequence (zj)j C D having no accumulation points in D such that E,'?°_t gD(zJ,z1+1) < oo; (iv) D is bounded and satisfies the Fu condition. Proof. (a) Note that D is biholomorphically equivalent to a bounded pseudoconvex
Reinhardt domain. Hence, from the very beginning we may assume that D is bounded. Suppose that D is not k-finitely compact, i.e. there exist a point a E D fl C;, an R > 0, and a sequence (zj)jE N C D such that
kD(a,zj)
and
Zj -+Z* EM
Assume for a moment that z* E C'. Without loss of generality, we may assume
that z* = 1 E 8D. Then, because of Remark 1.4.1, we find an a E 6t" \ {0} such that D C Da and 1 E BD.. Again, without loss of generality, let a be of the following form: at ak 0 0 and ak+l = = a,, = 0, where 1 < k < n. Thus we have D. = D(.,,...,a11) x C"-k. Applying the product property for the Kobayashi distance we get
kD(a.z!) > kDa(a,z1) =kD(a,....ak1(a,zj) =: rj,
322
Chapter 4. Holomorphically contractible families on Reinhardt domains
where a :_ (at,...,ak) and 'i :_ (zi,t,...,zi,k), j E N. Because of the formulas of the Kobayashi pseudodistance for elementary Reinhardt domains this yields that rj -* oo; a contradiction. Therefore we conclude that z* is lying on some of the axes. Assume, without
loss of generality, that z, zk 54 0 and zk+t = = z,*, = 0 for a certain k E {0, .... n - 1 }. There are two cases to discuss, namely: (i) 3te(k+I,...,n) : V n D = 0; (ii) dt=k+t,...,n :
V n D # 0.
Assume first (i). Let D' be defined as the projection of D in t-th direction. Hence D' is a bounded domain in C and zt E dD. In particular,
D' C K(ze , R) \ {zt)
for a certain R > 0.
Therefore, using Theorem 4.2.42 and Corollary 4.4.2, we get
kD(a,zi) ? kD'(at,zi,t) -> oo; a contradiction.
In the case (ii), if k = 0, then z* = 0 E D; a contradiction. So we only have to discuss the case when k > 0. Then kD(a, zi) > kprck (D)(prck (a), prck ("i ))
Note that prck (zi) -+ prCk (z*) E 8(prck (D)) and the limit has no vanishing coordinates. This means we are back in the situation we started with; a contradiction. (b) The implication (i) = (ii) is trivial, (ii) (iii) is a consequence Of MD < gD,
the triangle inequality for CD, and the growth of tanh-t at 0. Finally (iv) =(i) follows directly from Theorem 1.13.19 and Lemma 4.8.7. So it remains to prove (iii) = (iv). Suppose (iv) is not true. In case that D is not bounded we know that D is an unbounded hyperbolic pseudoconvex Reinhardt domain. Then, in virtue of Proposition 1.1 7.12, we conclude that D is algebraically equivalent to a bounded pseudoconvex Reinhardt domain that does not satisfy the Fu condition. Hence for the rest of the proof we may assume that D is bounded and does not fulfill the Fu condition. So, without loss of generality, it suffices to deal with the following situation:
DnV 34 0, DnV. =0, 1 < j
DnV=O,
k+l :s j
DnVV 36 0,
e+1 <j <-n,
where I < k_< f < n. In case when e < n we can even simplify the situation. Namely, put D := {z E Ct : (z, 0) E D}. Obviously, D is bounded and does not fulfill the Fu condition. Moreover, note that 5 has property (iii).
4.8. Carathdodory (resp. Kobayashi) complete Reinhardt domains
323
Summarizing we may assume, without loss of generality, that
DnI; 0, DnVj=0, 1 <j
by Lemma 1.4.17, we find V = (-r, r)" C log D and V E lR"- \ {0} such that V + (R+v E log D. Note that vj = 0, j = k + L..., n. Without loss of generality, we may assume that vj < 0, j = 1, .... e < k,
<j
v t = - l , and v j (ex' a-f eX2et V2,
ext of of ,
ex(+1 -e Xn ) E D.
I > 0. X E V.
Put a := -v. Then there exists an e > 0 such that
(ez, µ2eAaz..... µleAaf. 1.-, 1) E D. e-f < Iµj I < es, j = 2, .... P. A, tt j E C. Re A < 0. Put
A := (A = (µ2.
.
. I2e) E
Ce-t : e-e < 'I < e£. j = 2..... a),
HR:={AEC:ReA
F(A, µ)
µ2exa,
(eA.
.... p. ezaf ).
Note that F is a locally biholomorphic mapping and DR := F(HR x A) is a pseudoconvex Reinhardt domain (EXERcisE). Moreover, we have DR / D :_ F(C x A) C C;, when R -+ oo, and D,,. is a pseudoconvex Reinhardt domain. Hence, we get
kD,,.(F(-1. 1...., 1). F(A. 1,.... 1)) < kC(-1.A) = 0.
A E C.
In virtue of Theorem 4.2.42 we know that kD,o is continuous. Therefore,
1).z)=0, zED,, nF(Cx{(1..... 1)})=:M. Note that Do x (1) C D but (0, ... , 0, 1) 0 D, where 1 E C"-e t rj < oo. To get a contradiction to Choose positive numbers rj such that (iii) it remains to find points Z j E Do, j E IN, such that -j -> 0 and
j
9D((zj,1), (zj+t, 1)) <
rj.
j E N.
324
Chapter 4. Holomorphically contractible families on Reinhardt domains
Recall that kDR is continuous on DR x DR and that
kDR(F(-1, 1...., 1), z) \ kD.(F(-l, 1,..., 1).z) = 0, z E M. Then, by Dini's theorem, we conclude that this convergence is locally uniform. Hence we find a sequence (R), 0 < R j / oo, such that
kDRj(F(-1.1,...,1),F(A,I....,1))
-2
Now note that the mapping *R: Do --> DR,
IkR(Z) := (zieR,z2ea2R
...zte°fR
is biholomorphic. Therefore,
kDo(F(-1 - R. 1,.... 1), F(A, 1...., I))
= kDRJ (F(-I, L...,1).F(A+ Rj, I,..., 1)) < rj,
-2-Ri
uj(A) := loggDo(F(-1 - Rj, 1,..., 1),(F(A, 1..... 1)),
A E Ho.
Then U j E SX(H0). By Exercise 1.14.10, we conclude that
uj(A)
ReA<-1-Rj.
Therefore, we may take zj := F(-l - Rj, 1-- 1) as the points we were looking
0
for. Hence the proof is complete.
Corollary 4.8.10. Let D j C C", j = 1, 2, be biholomorphically equivalent Reinhardt domains. If D 1 is bounded and satisfies the Fu condition, then D2 is bounded and it satisfies the Fu condition.
Remark 4.8.11. Moreover, for a pseudoconvex Reinhardt domain D the following
properties are equivalent: (i) D is c-complete, (ii) D is c' -complete, where cD denotes the so-called associated inner distance, i.e.
cD(a,z) := inf(LCD(a) : a: [0, 1] -- D continuous and 11 11-rectifiable, a(0) = a, a(1) = z},
a. Z E D,
where N
LCD(a) := SUP I EcD(a(ti-1),a(tj)) : N E N. 0 = to < tl < ... < tN = l i=1 See [Zwo 2001 ].
.
4.9*. The Bergman completeness of Reinhardt domains
325
We conclude with a few remarks on completeness for pseudoconvex balanced domains.
Remark 4.8.12. Let D = Dh be a bounded pseudoconvex balanced domain in C". Then: If D is k-complete, then h is continuous. For any n > 3 there exists a bounded pseudoconvex balanced domain with continuous h which is not k-complete. Q Whether such an example does exist in dimension 2 is open. Q It is also an open problem whether there exists a characterization of a c- (resp.k-) complete pseudoconvex balanced domains D = Dh in terms of properties of the Minkowski function h. For more details see [Jar-Pfl 19931.
0
0
4.90 The Bergman completeness of Reinhardt domains In the last section of this book we briefly introduce the Bergman metric for bounded
domains in C" and present (without proofs) a full characterization of Bergmancomplete bounded Reinhardt domains due to Zwonek ([Zwo 1999a], [Zwo 2000]). For more details on the Bergman metric we refer the reader also to [Jar-Pfl 1993], [Jar-Pfl 20051.
Let D C C" be a domain. Then L2 (D) is a Hilbert space with the scalar ID f(z)g(z)dA2n(z) (cf. Example 1.10.7(c)). Recall product (f.g)L2(D) that the mapping L2 (D) B f H f(a) E C is a continuous linear functional. Therefore, by the Riesz representation theorem, there exists a uniquely defined E L2 (D) such that
(f KD(..a))L2(D) = fD .f(z)KD(z.a)dA2n(z) = f(a)
f E L2(D)
11
The function KD : D x D -* C is the Bergman function for D. Recall from the Hilbert space theory that there is another description of the Bergman function via a complete orthonormal system ((pj) jEN C L2h (D), where N C W.28 Namely, we have KD(Z,W) _ E'pj(z)Wj(w). (Z. W) E D x D, jEN
where the convergence here is meant as the one in the Hilbert space Lh(D). In the case of a Reinhardt domain D C C" there is a complete description of those monomials z" which belong to Lh(D). To be able to formulate this result we 28Note that Li (D) is a separable Hilbert space: therefore, it has a countable complete orthonormal system.
326
Chapter 4. Holomorphically contractible families on Reinhardt domains
have to introduce some terminology. Let a E D fl C. Put
E(D.a) := {v E IR" : loga + IR+v E log D}. Observe that 1. (D, a) = E(D. b) whenever b E D fl C . Hence we are allowed to set
S(D) :_ 'I(D. a), when a E D fl C;. Lemma* 4.9.1. Let D C C" be a pseudoconvex Reinhardt domain. Then for an a E Z" the following conditions are equivalent:
(i) z°` E L2(D),
(11) (a+1.v) <0, v E.(D)\{0}. Therefore, for a pseudoconvex Reinhardt domain its Bergman function can be written as
KD(Z, W) = F, aazawa, where the sum is taken over all a E Z" such that (a + 1, v) < 0, v E 3(D) \ {0}, and the aa's have to be determined. Remark 4.9.2. Note that there exist domains Dk C C2, not pseudoconvex, such
that dim L2(Dk) = k, k E W (see [Wie 1984]). 0 It is not known whether dim L2 (D) = oo for any pseudoconvex domain D in C2 with L2 (D)
{0}.
0
The above equation immediately leads to KD(z. w) = KD(w, z), (z, w) E D x D. Moreover, it can be understood in the sense that Ej=1 (pj (z)Oj (w) converges to KD(Z, w) locally uniformly (in the case when N = W) (EXERCISE). Note that this partial sum is a function which is holomorphic on D x D. Therefore, by the
Osgood theorem, D x b a (z, w) H KD(z, w) is holomorphic. In other words, we may say that KD is holomorphic in the first variable and antiholomorphic in the second variable on D.
Remark 4.9.3. There are effective formulas for the Bergman kernel for standard domains like the Euclidean ball, the polydisc, or the minimal ball; for more details see [Jar-Pfl 1993] and [Jar-Pfl 20051:
KIB,, (z, w) = n (1 - (z, w))-1, KD+(z. w) _
fl(1 -zjwj)!1.
_" j=1
Z. W E l6":
z,w E lD".
The following result describes the behavior of the Bergman kernel under biholomorphic mappings.
4.9*. The Bergman completeness of Reinhardt domains
327
Proposition* 4.9.4. Let F : D -+ G be a biholomorphic mapping. Then
KG(F(z), F(w)) det F'(z)det Fl(w) = KD(Z. w),
Z. WE D.
Remark 4.9.5. It should be mentioned that there is a similar formula in case when the mapping F is not biholomorphic but proper holomorphic (see [Bel 1982] or [Jar-Pfl 1993]).
Exercise 4.9.6. Calculate KT, where T is the Hartogs triangle, i.e. T = {z E (D2 Izil < IZ21}.
In the following we put kD(z) := KD(Z. z); note that logkD E PSJ(D). kD is called the Bergman kernel of D. In case that Lh (D) description of the Bergman kernel, namely lf(Z)12
kD(z) = Sup {
112
II
(0) one has an alternative
:f EL 2 (D) \ {0}} .
Z E D.
h(D
From this equality it follows that kG ID < kD if D C G C C". In the study of the Bergman kernel it is important to know its boundary behavior.
Definition 4.9.7. Let D be a domain in C" and let zo E aD. D is said to be Bergman exhaustive at z° if lim kD(z) = oo. DDz-+z0
There is a vast literature on conditions which are sufficient for Bergman exhaustiveness. Here we only mention the following one which combines properties of the Green function and Bergman exhaustiveness (compare [Che 1999], [Her 1999]).
Theorem* 4.9.8. Let D be a bounded pseudoconvex domain in C" and let zo E aD.
Put Az(D):={
If lim
A2n(Az(D)) = 0.
D is Bergman exhaustive at zo. In particular, any bounded hyperconvex domain is Bergman exhaustive.
In what follows we will always assume that kD is positive on D, i.e. for any point a E D we can find an f E L2 (D) with f (a) 0 0. Note that this condition is always true if D is bounded. Put n
flD (Z; X) j,k = r
a2 log kD az j aZk
1/2
(Z) Xj xk )
,
ZED. XE
n.
Then fD gives a Hermitian metric on D. OD is called the Bergman pseudometric on D.
328
Chapter 4. Holomorphically contractible families on Reinhardt domains
Example 4.9.9. Here we present effective formulas of OD for the standard domains
D=lBnand D=ID":
fa"(z;X)= n+1
(z,X)EIBnxC"; n
(y(Zj:Xj))
R fD"(z:X)=VL
2
ZE(D",
X =(X1....,Xn)EC".
j=1
Observe that fir = may. Moreover, we have
Lemma* 4.9.10. Let F E Bih(D, G). Then ia(F(z); F'(z)X)) = OD (Z; X),
ZED,XEC". Hence, in the class of domains we are discussing, we have a family (8D)D of pseudometrics bD which are invariant under biholomorphic mappings such that fD = y. Observe that this family is not holomorphically contractible as the following example shows.
Example 4.9.11. Put F : ID2
ID2, F(z) := (zl. zl) and X := (1, 0). Then
fiD2 (F(0); F'(0) X) = 2 > N L = YD2 (0; X)
(EXERCISE).
We mention that there is also a different way to describe PD which is often very useful.
Lemma* 4.9.12. Let D C C" with kD > 0 on D. Then PD (Z; X) =
kD(Z) suP{I
f'(z)X I : f E L2 (D), 'If IIL2(D) = 11 .f(Z) = 0),
ZED,XEC". With fD we can measure the length of all tangential vector X E C" at any point z E D. Therefore, we introduce a new pseudodistance on D, the Bergman pseudodistance bD, defining
bD(z, w) := inf $ fo f D(a(t):a'(t))dt : a E el([0,11, D), 1
a(0)=z,a(l)=w
Z,WED.
It is easy to check that bD is a pseudodistance on D and that bG(F(z). F(w)) _ bD(z. W), z, w E D, whenever F E Bih(D, G), i.e. the Bergman pseudodistance is invariant under biholomorphic mappings.
4.9*. The Bergman completeness of Reinhardt domains
329
Exercise 4.9.13. Prove that the sequence ((0, ))JEt.J is a bT-Cauchy sequence, where T is the Hartogs triangle. In particular, T is not Bergman complete (see Definition 4.9.15). In comparison to the objects we have discussed before we have the following result (see [Jar-Pfl 1993]).
Lemma* 4.9.14. Let D C C" be such that kD(z) > 0, z E D. Then (a) YD (b) CD
8D:
CD < bD
We should mention that there is no comparison between the Bergman pseudometric and the Kobayashi pseudometric. Now we start to discuss the notion of Bergman complete domains. To be more precise we set
Definition 4.9.15. Let D C C" such that kD > 0. D is said to be Bergman complete if bD is a distance, any bD-Cauchy sequence (z j) j C D (i.e. bD (z f , zk)
0 if j, k
oo)
converges to a point a in D (i.e. lim zJ = a in the topology of D). J-+00
There is a long history of studying Bergman complete domains. An old result by Bremermann shows that any Bergman complete bounded domain is necessarily pseudoconvex (see [Jar-Pfl 1993]).
Proposition 4.9.16. If D C C" is a bounded Bergman complete domain, then it is a domain of holomorphy.
Proof. Suppose the contrary. Then there are a point a E D and positive numbers r < R such that
i'"(a,r) C D, i'"(a, R) ¢ D, df EO(D) IP"(a,r) _ .f I P (a,.). In particular, by the Hartogs theorem, the Bergman kernel function KD extends to IP" (a, R) x IP" (a, R), or more precisely, there exists an f : IP"(a, R) x Fn (a. R) -+ C such that
.f [R. (a, R) x Fn (a, R) 9 (z, w) H f (z, w) is holomorphic. By construction we find a point b r= IP" (a, R) f13D such that [a, b) C D. Apply-
ing that log kD (z) = log f (z, z), z near [a, b), leads to the fact that PD (a + t(b - a); b - a) is bounded on (0, 1). Hence, (a + (1 - 1/j)(b - a))J is a bD-Cauchy sequence tending to the boundary point b; a contradiction. Therefore, in order to characterize the Bergman complete domains it suffices to restrict on domains of holomorphy.
330
Chapter 4. Holomorphically contractible families on Reinhardt domains
The most useful sufficient criteria for being Bergman complete is that of Kobayashi (see [Jar-Pfl 1993], [Blo 2003), and [Blo 2005]).
Proposition* 4.9.17. Let D C= C" be a bounded domain. (a) If D3zmaD If(z)1
< Ilf IILh(D)
f E Lh(D) \ {0},
then D is Bergman complete. (b) Assume for a dense subspace H C Lh (D) that for any sequence (z j ) C D, zj -+ zo E aD, and any f E H there exists a subsequence (zjk )k such that lim k-'oo
If(zik)I
= 0.
kD (zik )
Then D is Bergman complete.
After a long development the following result was found (see [Blo-Pfl 19981, [Her 1999]). Theorem* 4.9.18. Any bounded hyperconvex domain is Bergman complete.
But conversely, there are a lot of Bergman complete domains which are not hyperconvex; examples will be given later. In the class of bounded pseudoconvex Reinhardt domains there is a complete characterization of Bergman complete domains in geometric terms due to Zwonek (see [Zwo 1999a], [Zwo 2000]). To do so we have to introduce the set
E(D) := (v E (K(D) : exp(a + R+v) C D}, where a E D fl C'. Then we have the following result. Theorem* 4.9.19. Let D be a bounded pseudoconvex Reinhardt domain. Then the following conditions are equivalent:
(i) D is Bergman complete;
(ii) ti:'(D) fl on = 0, where V(D) :='(D) \ 5(D). We conclude this discussion presenting two examples.
Example 4.9.20. (a) Put Dt := {Z E C2 : IZ112/2 < Iz2I < 21zt I2. Izt I < 1). Then D is a bounded pseudoconvex Reinhardt domain. Note that
IS'(D1) = IR >o - (-l, -2).
4.9*. The Bergman completeness of Reinhardt domains
331
In particular, W(D) contains the rational vector (-1. -2). Hence, DI is not Bergman complete. We add also a direct proof which may give some idea of how to prove (i) = (ii)
in Theorem 4.9.19. Recallthatza E L2(Dt)iff(a+1,(-1,-1)) <0. Therefore,
aazaiva =
KD1(z, w)
a., zawa,
z, w E D.
aEZ"
aEZ"
-3
vE(E(D)\{0} (a+1,v)
Put (p(A) :_ (,1, A2), A E (D.. Then V E 49 (0)., D 1) and aaIZI2ai+4a2 = r bjlAI-'.
kD, °(P(A) = E -3
A E (D.
1>10
where jo > -3 and bb0 0 0. Therefore, a2
( log
bi i>-io
IAI2(i-io)1
I
A E ID.,
meaning that PD1 (co(t); rp'(t)) is bounded on (0, 1/2). Note that limt\o ap(t) E 8Dt. Hence, bD1((p(1/2), ap(t)) is bounded on (0, 1/2) which implies that Dt is not Bergman complete. (b) Put
D2 := {Z E C2 : Izt If /2 < IZ2I < 2IZt I'r. Izt I < 1). Obviously, D2 is a bounded pseudoconvex Reinhardt domain. Calculation then leads to (f;'(D) = R>o (-1, -/), i.e. V(D) does not contain any rational vector. Hence, D2 is Bergman complete. It is easy to see that D2 is not hyperconvex. Iz2I2e-t1IZi 12 < 1). Calculate log D3, Exercise 4.9.21. Put D3 := {z E m. x C : ((D3), and decide whether D3 is Bergman complete. Is D3 hyperconvex?
Note that D1, j = 1, 2, 3, does not fulfil the Fu condition, hence it is not c-complete.
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Symbols
General symbols N := the set of natural numbers, 0 0 N; No Z := the ring of integer numbers; 0 := the field of rational numbers; I2 := the field of real numbers; It_,p := {-oo} U IR, R+".:= It U {+oo}; C := the field of complex numbers;
In E N n > k};
W U {0}; Wk
LxJ := max(k E 7 : k < x) = the integer part of x E It;
(xl:=min(kEZ:k>x),xEIR; Re z := the real partof z E C, lm z :=the imaginary partof z E C; 2:=x - t the conjugate of z = x + iy; IzI := x2 + y2 = the modulus of a complex number z = x + iy; the Cartesian product of n copies of the set A, e.g. C"; M(m x n; A) = the set of all (m x n)-dimensional matrices with entries from a set
A"
ACC; II" := the (n x n)-dimensional unit matrix; (L E IM(n x n; I") : det L 0 0), I' E {0, R, C}; GL(n, F) (L E Dal (n x n; Z) : I det LI = 1); GIL(n, Z)
x_y:sxj
A_:={aEA:a
(a E A : a > 0), e.g. It>o;
A,o(aEAa<0);
A>o :_ (A>o)", e.g. It>o;
A+B:={a+b:aeA, bEB},a+B:={a}+B, A,BCX,aEX,Xis a vector space;
8, k :=
,
bEB), ACC,BCC"; if j 96 =k
j
= the Kronecker symbol;
=k e = (el,...,en) := the canonical basis inC",ee := 1 = In = (1, ... ,1) E W"; 2:= 2.1 = (2, ... , 2) E W"; I.
if
81,n), j = 1,...,n;
zj wj = the Hermitian scalar product in C"; (z, w) :_ w := (01, ... ,1,On), w = (W1,..., wn) E C";
z=(z1,...,zn),w=(w1....,w")EC";
e=:=(es',...,ez^), z=(z1 .,zn) C"; IiZll
(z z)1"2 = (Ej=1 IZi I2)1/2 =the Euclidean norm in C";
Ilzllao := max{Iz1 I
Iz"1} = the maximum norm in C";
346
Symbols
Ilz1I1 :=
=theft-norm in C";
idA,x : A -> X, idA,X(x) := x, x E A; idA := idA,x if A = X or it is clear what the outer space X is, #A := the number of elements of A; diam A := the diameter of the set A C C" with respect to the Euclidean distance; cony A := the convex hull of the set A; A C= X :4==> A is relatively compact in X ;
prx : X x Y X, prx(x, v) := x, or prx : X ®Y -)-X, prx(x + y) := x; Bd (a, r) := {x E X : d (a, x) < r), a E X, r > 0 ((X, d) is a pseudometric space; d:XxX IR+, d(x, x) = 0, d(x, y) = d(y. x), d(x. y) < d(x, z) + d(z, y)); Bq(a, r) :_ {x E E : q(x - a) < r}, a E E, r > 0 ((E, q) is seminormed space; q : E --> IR+ q(0) = 0, q(Ax) = IAIq(x), q(x + y) 5 q(x) + q(y)); Euclidean balls (B(a, r) = M,, (a, r) := {z E C" : 11z - a11 < r} = the open Euclidean ball in C" with center_a E C" and radius r > 0; IBn(a.0) := O; IB(a, +oo) := C"; l3(a, r) = lin (a, r) := LB (a. r) = {z E C" : 11z -aII < r} = the closed Euclidean (a); ball in C" with center a E C" and radius r > 0; lBn(a.0)
B(r) _ tn(r) :_ tn(0,r); IB(r) _ IB (r) :_ n(O,r); : = ,, (1) = the unit Euclidean ball in C'; K(a, r) := IB, (a, r); K(r) := K(0. r); K(a, r) K(r) := K(0, r); 1Bl (a, r); IB = iB
K. (a. r) := K(a, r) \ (a); K. (r) := K.(0, r);
lD:=K(1)=(AEC:I.lI<1)=theunitdisc; T := a0); T. C" ;
y ={(Slat,
} y p ET),a=(a1 . .Snan):S.... .yn
.a")E
Polydiscs
IP(a, r) = IPn (a, r) := {z E Cn : IIz - all < r} = the polydisc with center a E C" and radius r > 0; IPn (a, +oo) := C"; IP(a,r) = lPn(a.r) := lPn(a,r); in(a,0) := {a); IP(r) = lP (r) := (P,. (0. r); Fn := lPn (1) = ID" = the unit polydisc in Cn ;
IP(a,r) = Fn(a.r) := K(a1,rl) x x K(an,rn) = the polydisc with center a E Cn and multiradius (polyradius) r = (rl, ... , rn) E IRn 0; notice that IP(a, r) = IP(a, r 1); to simplify notation we will frequently write Fn (a. r) instead of (Pn(a. r) (in particular, in all the cases where it clearly follows from the context that r is a multiradius);
IP(r) = (n(r) := IPn(0,r); aolP(a, r) := aK(al, r1) x . . . x dK(an, rn) = the distinguished boundary of IP(a, r);
Symbols
347
Annuli
A(a,r-,r+):_{zEC:r-
r+ > 0; if r- < 0, then A(a, r-, r+) = K(a, r+); A(a, 0, r+) = K(a, r+) \ {a};
A"(a.r-,r+) := A(a1,r-, r+) x ... x A(a",rn r,+), a = (a,,...,an) E C",
r+ = (r, ..... rn ), -oo < rj < r < +oo, rf > 0,
r- = (ri .....
j=1,
A" (r-, r+) := A" (0. r-. r+); Laurent series
za:=ZI
...zn". Z=(Z1,. ,Zn)EC",a=(a,.....a")E7" (00:= 1);
a! ali...a"!, a = (a1,....00 E Z+; IaI := Iai I + ... + Ian I , a = (a1..... an) E IR"; () := a s-1 a- +1), a E 7, N E 7+;
(0) '-(#I)...(fin), a=(al,...,an)EZ" Functions
11fIIA :=sup( If(a)I :a E A}, f: A -+ C; supp f (x : f (x) 0 0) = the support off ; .P (C") := the space of all polynomials f : C" -* C;
Rd(C"):_ {F E.P(C"):degF
f
k(X, y) := the space of all Ck-mappings f : X -* Y, k E 7L+ U {oo} U {w} ((o stands for the real analytic case); Ck(S2) ('k(52,C);
Co(Q):=if ECk(S2):suppf C_ S2);
C 1([a. b], Y) := the space of all piecewise (°,' -mappings a : [a, b] -+ Y; AN := Lebesgue measure in (RN; LP(Q) := the space of all p-integrable functions on S2; the norm II II
in LP(Q); LP(S2, loc) := the space of all locally p-integrable functions on S2;
O(X, Y) := the space of all holomorphic mappings f : X -t Y; 0(Q) := O(S2, C) = the space of all holomorphic functions f : S2 -+ C; f (k)(a) := the k-th complex Frechet differential off : S2 --> C" at a; a (a) . __ lim J(a+he )-J' (a) = the j -th complex partial derivative of f at a; C
o
---
Jf(a) := det [(a)]fk_.,.....,, , f = (f1, ... fn): S2 -> CO;
348
Symbols
DO :_ (a )a' o Taf(z) d(T0 f)
o(
-)°f" = a-th partial complex derivative;
gz.
Eaez+. I Daf (a)(z - a)a = the Taylor series off at a; sup(r ? 0 : Ta f (z) is uniformly summable for z E IP(a, r)} = the
radius of convergence of the Taylor series TO f ;
LP(S2) := O(n) fl LP(S2) = the space of all p-integrable holomorphic functions on S2;
Oak>(S1.8) :_ {f E 0(S2) : I18kf IIn < +oo};
6)(A):= U UJA
U open in C"
O(U)IA;
3l°O(92) := the space of all bounded holomorphic functions on S2; Ak(S2) (f E O(Q) : VaEZ+, lal_k : Daf E e(b)} Aut (Q) := the group of all holomorphic automorphisms of S2;
Auta(S2) := {h E Aut(S2) : h(a) = a}; SJ{(Q) := the set of all subharmonic functions on S2, S2 C C; SH(S2) := the set of all plurisubhannonic functions on S2; Xu(a; X) :_ Fl,k=1 i (a)XjXk = the Levi form of u at a.
List of symbols
Chapter 1
a-(I):= (ACI:A#0, #A<+oo) ..
.
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.. .. ... ... .....
fax:=0 ....................
fA:=E;EAf,AE a-(I),
f/ = EJE f 8(1, C2) := the space of all uniformly summable families 8(1, T Z) := the space of all uniformly summable families .
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6 6 7
Z -> C, i E 1
7
Z -+ T, i El
7
f =EiEJfs, #JCI . . .... ... .. .. ... . . .... .. .
DS := the domain of convergence of a power series S = E.EZ, aaz°`
8
.
13
[a,b]:=((1-t)a+tb :t E[0.11) .....................
........................
21
. . . . . .
21
14,,{xE(R"(x,a)
{X E (R" : (x - a, a) < 0)
.
[a.b):={(1-t)a+tb:tE[0,1)) Al :_ {X E IR" : VaEA : (x, a) = 01
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prF ...................................... .
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................................. K(F) ..................................... E(X) ..................................... [A] = span A
EH(X) .................................... K(X) .....................................
.. .. ... .... . .... ... .... . R: C" -+ 1R+, R(zt,...,z") := (IzIII,.. .Iz"I) .............. hX = the Minkowski function
TA(z) := A z = n-rotation, A. z E C" ................... xmx{1}"-j, zEA} A
(x E I.R
exp B = {(z1...
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.............
ex e A) = the logarithmic image of A . Z.) E C". : (log Izt I... , log IZ.1) E
30 30
B) .........
30
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31
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.. .. .... . NO =Vo{(zt....z")EC":ZI...Z.=0} ............... C(x):=Cifx>0; C(x):=C.ifx <0 . . ... . . ... ... .
....................... .
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C"(E):=n.,EC"(a),ECIR" ...................... IzaI := Izt1", ...Iz"Ian ............................ D.,
22 22 22 22 24 24 24 25 28 29 30
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. .
Vf=Vj":=((zt,...,z")EC":zj =0),j =1... .n C"(a):=C(at)x...xC(a")
21
:_ {z E C"(a) : IzaI < ec} = the elementary Reinhardt domain
. .
.
31
32 32 32 32 33
350
List of symbols
. ................................ D .................. int D = int a(D) ..................................... T :={(zI.Z2)(=- ID xID :Iz,I°
Da.o
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D \ Yo = int exp log
D*
Z E C"(A) .
Oa,A(a) .= (a1Za1,..
.. .......... . ...
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E(S):={a EZ" :aa #0} ......................... DS := the domain of convergence of a Laurent series S = EaEzn aaza UaEA IP(a, r) ............................. A(r) dn(a) := sup{r > 0 : IP(a, r) C Sl} ..................... . . as = a'! = aa(f) = aa(f, r) :_ (2rri)" faou(r) *' d . . .
. .
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. ... . ............ ... .. ... .. hD = the Minkowski function .. .... ..... ..... ... .. .. Bih(S2, S2') .... .
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.............. Reg(M) .................................... .
dist(z, A) := inf{Ilz - III :SEA}, Z E Cn D A .
Sing(M) .................................... dim.M .................................... dim M ..................................... maxi.=max{q:gE1} . ... ..
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.. .... .. .. .... . .....
Bq(fo.r):={f E.F:q(f-fo)
PQ ....................................... E(f):={aEZ":aa 96 0}
.........................
(f g)L2(n) := ff2 f g dA2n ......................... Je,
(S2) := If E O(S2) : VKCi2: IIf IIKna < +oo} ...........
8o(z) :=
35 37
37 38 38
aig
D = G ..... .
33
39
41 41
42 48 51
54 56 62 63 63 63 63 64 64 65 67 68 68 69
t+11=112
ps2(a):=sup{r>0:IB(a.r)CS2} .....................
Sn := max{pn, 80} .............................. 0(k)(S2) := O(k)(S2,Ss2) ...........................
. ... ... .. ... . .... . ....
nt
s) := nk>o
..
.. ... .. ...... ..... ..
0(0+)(S2) := p(o+)(S2, Ss2) . . . jeoo.k(S2) If E O(S2) : VaEz'+:jal
. ... ..
69 69 69 70 70 70 71
L(SZ) := If E 19(S2): V«ez"+:lal
71
Ak(S2) := If E O(Q) : Vaez+:laI k : D" f E A(S2)}
71
3'a0o,k(S2)
:_ If E 0(Q) :
........... Da f E X°°(12)} ........
71
List of symbols
351
Lh'k(S2) :_ (f E 0(S2) : daEZ+*,l
71
E(D, 8) := the $-envelope of holomorphy of D E(D) := F, (D, 0(D)) = the envelope of holomorphy of D
84
dD(A):=inf{dD(z):zEA},ACD
90
orda f = the order of zero off at a
91
.. ... ... .... ..
......... .................... . ..... ...............
71
84
. .... .. ... .......... 96 M (U) = the space of all harmonic functions .. ........ .. ... orda f = the order of zero of f at a ..................... 103 ,z_z_a.l H+:={AEC:Re),>0}
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..
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101
. .
P(u:a,r:z)
2n,,f[o.2n]n(I1j=1In::I J(u: a, r) := P(u: a, r; a) _ (2n )n I[0,2n]" u(a + r eie) dAn(0) ..... 104 A(u;a. r)
u dA2n = a" fp" u(a + r w) dA2n(w)
(nr`)
?X(S2) = the set of all pluriharmonic functions on S2 z
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the real Hessian of v at x
..
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104 .
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.......
8D,x(a) := sup(r > 0: a + K(r) X C D), a E D, X E C" dD,q(a) := sup{r > 0: Bq(a, r) C D}, a E D u(z)<maxKu}
109
114 118
.. .. ........... 119 ....... ... ........ 119
hN. ...................................... 131 hND
...................................... 131
Tb (8D) := {X E C" : F"=,
(b)XX = 0) = the complex tangent space
to 8D at b .............................. 143
EP := {(z1.... , zn) E C" : Ej=1 I zj 12Pj < 1) = the complex ellipsoid
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145
g = (9v,µ)1
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146
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. .... ....... .... 146 .. .. .. .. ... .. ....... .. .. 146 dg(zi, z2) ... . .. f-'(g) .. ... .. . ... . .. .. ... ........ ..... 147 Lg (y) ..... .
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Chapter 2
.. .. ... .... .... ....... ..... 160 .. ..... ...... .. ....... ..... 160 .. ... ...... .......... .. .. 160 .. .. .. ... ...... . ... .. .. 160 Fix(O):={zEG:'(z)=z) .. ............ ... .. 160 C5n := the group of all permutations of n-elements ............. 166 U(n) := the group of all unitary automorphisms of C" ........... 167 Biha,b (G, D)
... .
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Aut(G) := Bih(G, G) . Auta,b(G) := Biha,b(G, G) Auta(G) := Auta,a(G)
L,, = the Lie ball
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............................... 168
352
List of symbols
L. = the Lie norm ... .. ... .. .. ... ....... .. ... ... 168
.... .... 168
O(n) = the group of all orthogonal isomorphisms IR" -+ Ut"
F. := {q : q: C" -* IR+ is a C-norm. dxER" : q(x) = IIxil} ...... 170 170
II
Ymin := (q E Finax : dzec- : q(z) < Izl) IIZllmin := inf{q : q E .m;.}
M" =(ZEC":llzllmin<1) S(z1,z2) := (Z2,Z1) Tt := TT o S .
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172
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181
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181
O`(G):={f EO(G): f(z)&0. zEG} IE(D) Da
..................................... 182
....................................... H-:={zEC:Rez <0} .. .. . ... .. ..... ........ . .................. .
.
IFk.p = the generalized complex ellipsoid ffp,s
200
203 208
...................................... 212
Chapter 3
F(X) := (E(X)1 n (Q")1 . Z(X) := X + F(X) . . Ko(X) := X, Kj(X) := .
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Koo(X) := Ujto KJ(X).
Sk:={PEZ+:Irll=k}
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. ............. 231 .
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Z(Kl-1(X)), j = 1,2 ....
M(X) := E(Koo(X))
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231
..... .. ... 231
............
231
.......................... 240
Chapter 4 mD := SUP( Ilf II
: f E 0(D, IQ), f(0) = 0} .... .. .. ........ 251
kD(z) := inf{r E [0, 1) : 3yoeo(o,D) : (p(0) m(a, z) := II i z U = the MObius distance
= 0. (p(r) = z) ........ 252
... ....... ..... ... 253 p := i logI-M1+' = the Poincare distance .... ........ . ...... 254 MD = the Mdbius pseudodistance . ........... .. ....... 254 CD = the Carath6odory pseudodistance . ..... .... ......... 254 .
mD)
= k-th Mobius function .... . .... ..
. .... .. .. . .... 255
9D = the pluricomplex Green function .................... 255
.. .. ... .......... ........ 256 .. .. ...... .. . ..... ....... 256 k D = the Kobayashi pseudodistance .... .... ... .. ....... . 257 LdD(a) = the dD-length .. . ........ . ......... .. .... 258 Ho = the Hahn function .. ..... ....... ..... .. ...... 269 ko = the Lempert function kD .= tanh-1 kD . . . .
.
.
.
V(Da,)) := {z E Da : za =A)
....................... 281
353
List of symbols
y(a; X) :=
t
lal2
.. . ..... ... .... 293
= the Mobius pseudometric
yD = the Caratheodory-Reiffen pseudometric .... .. ... ... ... . MD(a) {I f I : f E O(D, m), f(a) = 0} . yD = the k-th Reiffen pseudometric ... .
.. .. ... .. . ... .. .
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.. ..... .. ....
.
MDl (a) := (If It1k : f E O(D, ID). order f > k) ............. XD(a) := {u: D -+ [0, 1 ) : logu E TSx(D), 3M,r,o u(z) < MIIz --
294 294 296
296
.. . .... .. ... .. ...................... XD = the Kobayashi-Royden pseudometric ................. 302 ................ ...................... 309 aII, z E B(a,r) C D}
... .
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297
.
AD =the Azukawa pseudometric
298
xD = the Kobayashi-Busemann pseudometric f ND = the integrated form of xD
308
top ....................................... 315 KD = the Bergman function
...... .. .. . .... . ... .. .. .
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(£(D)_E(D,a):_{vEIR":log a+IR+vElog D},aEDnC,
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.. ... ... . .. .. . ... . . ... .. .. .. .. . ... .. ... ... . . ... .. . ... ... . {v E E(D) : exp(a + I+v) C D) . .... . . ... ... .
kD = the Bergman kernel OD = the Bergman pseudometric bD = the Bergman pseudodistance
a (D) W(D)
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325 326 327
327 328 330
.(D) \ .(D) ........................... 330
Subject index
Abel's lemma, 1.3 absolute convergence, Ll
image, 30 absolutely
summable family, 9 series, 9 uniformly summable family, 9 series, 9 algebraic automorphism, 200 equivalence, 34 mapping, 38 analytic set, 63 automorphism, 2 group
of I,,, 162 of ID", L66 of Ln , L6.8
Azukawa pseudometric, 298 balanced domain, 56 Banach theorem, 65 Bedford-Taylor theorem, 1118 Bergman exhaustive, 32.2 function, 325 kernel, 32.2 pseudodistance, 328.
pseudometric, 322 biholomorphic mapping, 54 boundary of class Ck, 143 bounded holomorphic function, 62 set, 65 Brody hyperbolicity, L33
Carathdodory-Reiffen pseudometric, 294 Carath6odory pseudodistance, 254 Cartan domains, 177 theorem, 164, 165, 177, 315
Cauchy condition, 8 criterion, 2 inequalities, 43, 52 integral formula, 47 48 product, 16 c-complete, 317 c-finitely compact, 317 c-hyperbolic, 313 circular domain, 85 Ck-smooth boundary, L43 domain, 143 CW-Kahler metric, 141 complete circular domain, 55 Hermitian metric, L42 Kahler metric, L42 a-circled domain, 3 15 set, 15 Reinhardt
domain, 3 15 set, 15 complex derivative, 1 ellipsoid, 145 Frdchet differential, 16 manifold, 53 partial derivative, 16 tangent space, 143 contraction, 253, 293
356
Subject index
convex function, L14 set, 21
C-seminorm, 63 curve rectifiable, 258
function for YD, 294 function for m(k), 255 fat
hull, 35 set, 14
finite type, 218
dD -length, 25$ defining function, 143, L44 3-tempered holomorphic function, 69 dimension of an analytic set, 63 domain, 3 n-circled, 30 balanced, 56 circular, 85 complete n-circled, 3 circular, 56 Reinhardt, 3 homogeneous, L60 of convergence of a Laurent series, 4 41 of a power series, 1, 3, 13
of existence of f, 22 of holomorphy, 5 22 Reinhardt, 30 relative complete, 5, 8Q starlike, 85 symmetric, 161 weakly relative complete, 84 D-point, 15$ elementary Reinhardt domain, 5, 33 normalized form, 196 of irrational type, 196. 222 of rational type, 126, 222 entire holomorphic function, 42 envelope of holomorphy, 84 equivalent families of seminorms, 64 exhaustion function, 119 extension operator, 73 extremal disc, 256
first Baire category, 65 formal partial derivatives, 17 Frdchet differentiability, 16 differential, 16 space, 65 Fu condition, 32 function of slow growth. 62
generalized complex ellipsoid, 208 Hartogs triangle, 212 geometric series, 12 g-length, L46 global defining function, L44 g-pseudodistance, L46 g-pullback, 142 Green function, 255 pole, 255.
group of automorphisms of IB", 162
ofd",166 of IL" , L68
Hahn function, 269 halfspace, 21 harmonic function, 1D1 Hartogs extension theorem, 5a lemma for psh functions, LLl theorem on separate holomorphy, 51 triangle, 32 Hermitian metric, L46 pseudometric, 146
Subject index
Hessian, L14 higher order complex Frdchet differentials, 12 partial derivatives, 12 holomorphic covering, 266 automorphism, 2 convexity, 89 function, 42 with polynomial growth, 69 Liouville foliation, 197 mapping, 42 holomorphically contractible family of functions, 253 pseudometrics, 293 homogeneous domain, 160 Hurwitz-type theorem, 55 hyperbolicity, 313 hyperconvexity, 130
identity principle, 50 inner metric, 242 integrated form of XD, 309 inverse mapping theorem, 54 irrational type, 22 25, 230 Josefson theorem, L07
Kahler metric, 147 KAhler pseudometric. 147 XD-geodesic, 302 k-complete, 317 k-finitely compact, 317 k-hyperbolic, 313 k_-hyperbolic, 313 kD-geodesic, 256 Kobayashi pseudodistance, 252 Kobayashi-Busemann pseudometric, 308 Kobayashi-Royden pseudometric, 302 Kronecker theorem, 97 k-th M6bius function, 255 k-th Reiffen pseudometric, 296
Laurent series, 2 441 Lempert function, 256 Levi condition, 144 form, 102 problem, 122 Lie ball, 168 norm, 168 lifting, 202 266 linearly generated, 152 Liouville theorem, 52 for psh functions, L01 local defining function, L43 potential, L48 pseudoconvexity, 119 locally complete pluripolar set, 106 normal convergence, 43 normally summable family, 9 series, 9 uniformly summable family, 9 series, 2 logarithmic convexity, 4 31 image, 352
plurisubharmonicity, 191 lower semicontinuity, 28 matrix orthogonal, 168 positive definite, 177 unitary, 162 maximal norm. 110 maximum principle, 53 for psh functions, 102 mD-geodesic, 252 minimal ball, 123
357
358
Subject index
norm, 172 representation of a convex set, 21 of a Reinhardt domain. 36 Minkowski function, 28 5L MSbius distance, 2 253 function of higher order, 255 pseudodistance, 254 Montel theorem, 54 natural Banach space, 66 Frdchet space, 66 Hilbert space, 65 a-circled
domain, 4 30 set, 30 n-fold Laurent series, 4 power series, 3 13 normalized elementary Reinhardt domains, 196 Reinhardt domain, 180 normally summable family, 9 series, 9 a-rotation, 29
open halfspace, 21 order of zero, 91, 103, 255 orthogonal complement, 22 matrix, 1.68 operator, L68 Osgood theorem, 511
overshears, 163 partial derivative, 116
Pfaffian form, 151 piecewise ek-boundary, 179 p-integrable holomorphic function, 57 pluricomplex Green function, 255
pluriharmonic function, 149 pluripolar set, L06 plurisubharmonic function, 1514
measure, 131 Poincard
distance, 2 254 theorem, L24 positive definite, 177
power series, 1.3 13 product property, 254 proper mapping, 207 pseudoconvexity, 118 pseudodistance, 254 pseudometric, 293 psh Liouville foliation, L92 radius of convergence of a power series, I of a Taylor series, 1$ rational type, 22. 25, 234 real Frdchet differential, 16 partial derivative, 16 tangent space, L44 regular Frdchet space, 220 point- 63 regularization of a function, 112 regularized relative extremal function, 131 Reinhardt
domain, 4 30 normalized, 180 set, 30 relative completeness, 5 80 extremal function, 13.1 removable singularities of psh functions, 108 p-length, 242
Subject index
Riemann removable singularities theorem, 60
8-convexity, 89 .8-domain of holomorphy, 72 segment, 21 seminorm, 29 63 $-envelope of holomorphy. 84 separately holomorphic function, 42 mapping, 42 shears, 153 Sierpinski theorem, 10 singular point, b3 smooth boundary of class ek, 143 starlike domain, 28, 85 Stein neighborhood basis, 13.1 strict hyperconvexity, 135 R-hyperconvexity, L3.6 strictly plurisubharmonic function, 103 strong convexity, L44 pseudoconvexity, L44 sum of a summable family, 2 series, Z summable family, 6 series, 6 symmetric domain, 150
tautness, 141 Taylor series, 18 theorem Banach theorem, 65 Bedford-Taylor theorem, 108 Cartan theorem, 164, 165, 177, 315 Hartogs extension theorem, 58 Hartogs' theorem on separate holomorphy, 51
359
Hurwitz-type theorem, 55 inverse mapping theorem, 54 Josefson theorem, 102 Kronecker theorem, 92 Liouville theorem, 52 for psh functions, 191 Montel theorem, 54 Osgood's theorem, 50 Poincard theorem, 174 Riemann removable singularities theorem, 60 Sierpitiski theorem, 10 Thullen theorem, 181. 182 Vitali theorem, 54 Weierstrass theorem, 53 thin set, 59 Thullen domain, 180 theorem, 181. 182 topology generated by seminorms, 64 transitivity, 160 triangle inequality, 29 unconditional convergence, 11 uniformly summable family, 6 series, 6 unit disc, 2 unitary matrix, 162 operator, 152 upper semicontinuity, 28 semicontinuous regularization, 105 Vitali theorem, 54 weak Fu condition, 92 psh barrier function, 132 relative completeness, 80 Weierstrass theorem, 53 Wu ellipsoid, 182
Marek Jarnicki Peter Pflug
First Steps in Several Complex Variables: Reinhardt Domains This book provides a comprehensive introduction to the field of several complex variables in the setting of a very special but basic class of domains, the so-called Reinhardt domains. In this way the readers may learn much about this area without encountering too many technical difficulties. Chapter 1 describes the fundamental notions and the phenomenon of simultaneous holomorphic extension. Chapter 2 presents a fairly complete discussion of biholomorphisms of bounded (complete) Reinhardt domains in the two-dimensional case. The third chapter gives a classification of Reinhardt domains of existence for the most important classes of holomorphic functions. The last chapter deals with invariant functions and gives explicit calculations of many of them on certain Reinhardt domains. Numerous exercises are included to help the readers with their understanding of the material. Further results and open problems are added which may be useful as seminar topics. The primary aim of this book is to introduce students or non-experts to some of the main research areas in several complex variables. The book provides a friendly invitation to this field as the only prerequisite is a basic knowledge of analysis.
ISBN 978-3-03719-049-4
www.ems-ph
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