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)(X). Proof: It is clear that CM(< P > ) ( X ) _C C M ( P ) ( X ) . Conversely, it is easy to see that for any G-set X, I have
(Y)=
~
]:U~Z
9:Z--Y
so that any element c~ of < P >
(Y) can be written as c~ = E A.(gi)A*(fi)(p{) i
for suitable elements g{, fi and Pi E P. But if m 6 C M ( P ) ( X ) , then o~ • m = ~ d . ( g i ) g * ( f i ) ( p i ) • m = ~ M.(gi • I d ) M * ( f i • Id)(pi • m) . . . . i
i • i
tlx
And for f : U ---+ Z and g : Z --+ Y, I have
KJUX
KX~L/
p{)
CHAPTER 6. CONSTRUCTION OF GREEN FUNCTORS
144
. . . . M*(gxld)M* (ZX) M*(Idx f)
M*(gxId)M* (XZ) M*(Id•
. . . . . 'l'(XY) M*(Id•
....
f)
Thus
i
• gi)M*(Id
yx
x f/)(m •
X
yx
Pi) . . . .
M * (g)M*(fi)(p i i ))
=M,
zy) (m x c~) yx
which proves the lemma.
P r o p o s i t i o n 6.5.2: T h e p r e v i o u s d e f i n i t i o n s t u r n C M ( L ) i n t o a s u b - M a c k e y f u n c t o r of M, c a l l e d t h e c o m m u t a n t of L in M. Proof: If f : X ~ Y is morphism of G-sets, and if m r I have
~ • V.(f)(-O = M.(ld • f)(a •
CM(X),
M.(IU • f ) M . ( x ~
then for a r
L(U),
(~•
\ ~x J
. . . . M. uy and this shows that
uy
M~(f)(CM(L)(X)) C C . ( L ) ( Y ) .
a x M*(f)(m) = M*(Id x f)(a • m) = M*(Id x f)M.
....
M.
Similarly, if m E
M(Y)
uy
M*._. ~
•
X?2
which shows that M*(f)(CM(L)(Y)) Mackey functor of M. k
/
c_ CM(L)(X),
hence that
CM(L)
is a sub,,
In the special case when M is the functor A, viewed as an A-module-A, there is a little more:
P r o p o s i t i o n 6.5.3: Let L be a s u b - M a e k e y f u n c t o r of t h e G r e e n f u n c t o r A. T h e n CA(L) is a s u b - G r e e n f u n c t o r of A. Proof: I must check that if X, Y and (J" are G-sets, if a E A(X) and /? E A(Y) commute with l E L(U), then ~ • also commutes with I. But
\uxy/
\uxy/
6.5. COMMUTANTS =
\uxy/
145
~ • A.
(1 • ~)
yu
\uxy/ \uxy/
\xyu/
)
Ix<
= A.
\uxy/
A.
(~•215
\xyu/
. . . .
\xyu] \xuy/
\uxy/
which proves that CM(L) is closed under x. Since is it clear that ~ C CM(L)(*), the proposition follows. 9 Notation:
If M and N are modules for the Green functor A, I denote by
~A(M, N) the commutant of A in the bimodule H( M, N). By definition, if I set 7-I = H(M, N), then for any G-set X
H A ( M , N ) ( X ) = { f e H ( X ) I c~ x f = H. (xu~ ( m x
~)
\ ttX /
VU, Va C A(U)}
Let r~ (resp. l~) be the element of H(M, M)(U) (resp. of H(N, N)(U)) associated to a. Then c~ • f is the morphism from M to Nux defined for m E M ( Z ) by
(~ x f ) z ( m ) = (I~ 5 f ) z ( m ) = N. ( z x u ~ o l~,zx o f z ( ~ )
. . . .
\ZUX/
Similarly, the morphism f x a from M to Nxu is given by
=
o fzg o r~,z(m) = N. , z u x \ZXU/
\ZX?s
o fzu o M.
Z'U
(a x m)
Since f is a morphism of Mackey fnnctors, I have
and then \ZX'a/
o iV.
\Z'tlX/
o fuz(o~ • m) = N.
\ZXtt/
o fuz(oe • m)
Thus f is in t t A ( M , N ) ( X ) if and only if for any U and any a E A(U), I have
\zux/
\zux/
\zxu/
\zux/
which reduces to • f~(m) = f~(~
•
.~)
Thus HA(M, N ) ( X ) is just the set of A-module homomorphisms from M to Nx. P r o p o s i t i o n 6.5.4: Let A b e a G r e e n f u n c t o r for t h e g r o u p G. If M a n d N are A - m o d u l e s , a n d X is a G-set, t h e n
HA(M, N ) ( X ) = Homa-Mod(M, Nx) If moreover M = N, then the product functor.
5
turns
~A(M,M)
into a Green
146
6.6
CHAPTER 6. CONSTRUCTION OF GREEN FUNCTORS
T h e functors M |
Let A and B be Green functors for the group G. If M is an A-module-B, and N an Amodule, I have built 7~A(M, N), which is a priori a Mackey functor, Actually, in that case, it is a B-module: i f X and Y are G-sets, ifb E B ( X ) and f E HomA(M, Nz), I denote by fib the element of HomMack(a)(M, Mz) deduced from the action of B on the right on M. Then f 5 Pb C ~ ( M , N ) ( Y X ) . It is the morphism of Mackey functors from M to N z x defined on the G-set U by
( f S Pb)u(m) = N. k u y x / o fux o Pb,u(rn) = N. kUyX/ o f u x ( m x b) It follows easily that if I set
\uxy/ I obtain a morphism from M to N x y . Moreover, it is clear that this turns St(M, N) into a B-module. Finally, if f is a morphism of A-modules, so is b • f, since if a r A(Z), then
....
a x f u x ( m x b) = a x (b • f ) u ( m )
Now I have defined a B-module structure on ~A(M, N). The correspondence N "HA(M, N) is moreover functorial in N: if 0 is a morphism of A-modules from N to N', determined by morphisms 0z from N ( Y ) to N'(Y) for any G-set Y, then I define the morphism ~ A ( M , O) from ~HA(M, N) to ~A(M, N') in the following way: if Y is a G-set, if f is an element of HA(M, N ) ( Y ) = I-IomA(M, Nz), and if U is a G-set, then
It is then clear that 7-~A(M, 0 ) z ( f ) is a morphism of Mackey functors from M to N{., and since f and 0 are morphisms of A-modules, it is actually a morphism of A-modules. So there is a map 7-LA(M, O)y from 7-~A(M, N ) ( Y ) to (HA(M, N')(Y), which defines a morphism of Mackey functors from 7{A(M , N) to 7-~A(M, N'). In other words, the Mackey functor 7-LA(M, N) is a sub-functor of ~ ( M , N), which is moreover invariant under 7-t(M, 0) if 0 is a morphism of A-modules. So the bimodnle M defines a functor from A - M o d to B - M o d . A natural question is then to look for a left adjoint, as in proposition proposition 1.10.1. This question leads to the following definition: D e f i n i t i o n : Let A and B be Green [unctors for the group G, and M be an A-moduleB. If N is a B-module, and X is a G-set, I set
6.6. THE FUNCTORS M |
N
147
where J is the R-submodule generated by the dements
M.(f)(rn) @ n' - m | N*(f)(n') for f : (Y, ~5) ~ (Y', qS'), m E M(Y), n' E N(Y') M*(f)(m') | n' - rn | N.(f)(n) for f : (I/, 4') --+ (Y', 4/), n / E M(Y'), n E N(Y) r
-
|
M(Y), b
B(Y),
C N(Y)
L e m m a 6.6.1: T h e p r o j e c t i o n M~)N(X) --+ M@BN(X) turns M@BN into a M a c k e y functor, and t h e m o r p h i s m A -+ A~)B turns it into an A - m o d u l e , q u o t i e n t of M@N. M o r e o v e r , t h e c o r r e s p o n d e n c e N ~-~ M@sN is a f u n c t o r from B-Mod to A-Mod.
Proof: To say" that the structure of Mackey functor of M@N is compatible with the projection is equivalent to say that the elements
generate a sub-Mackey functor of M@N. But if f : X ~ X ' is a morphism of G-sets, then
( m ~ N ) . ( f ) ( [ m . b | hi(Y#) - [m | b.n](y,~)) = [rn.b| n](y,f~) - [m @ b.n](y,/~) Similarly, if now f is a morphism fl'om X ' to X, and if the square a
y'
,
X'
Y
I X f
is cartesian, then
....
[M*(a)(ra.b) | N*(a)(n)](v,,+,) -[M*(a)(m)|
But the proof of lemma 5.2.2 shows more generally that
m*(a)(m.b) = m*(a)(,n).B*(a)(b) which proves that To prove that that the elements element a E A(Z)
Since
X*(a)(b.,~) = B*(a)(b).N*(a)(n)
M Q s N is a quotient Mackey functor of M ~ N . MQBN is a quotient A-module of M ~ N , I must check similarly (E) generate an A-submodule of MQN. But the product of the by the element [ m | hi(y,,) of M ~ N ( X ) is given by
, , ] CHAPTER 6. CONSTR[ CTION OF GREEN FUNCTORS
148 I have
c~ x [m | n](Y,r = [(c~ x rr~)@ N* ( zyY) (n)](ZxY, ldxr In particular
a x [m | b'n](v'r = [(a x rn) (3 N* ( zY) Now the proof of lemma 5.2.2 shows that
Moreover =
zyzy
o, x . - ~
x
(b
. . . .
It follows that
• ([~.b e ~l(Y,~)- In | b.~](y,~0
. . . .
= [((OlXnZ).B. (~)(b,)@?Z](ZY,id;3)--[(o~• , @B* (?)(b,.N* (~)(~)](ZY, Id4)) which proves that the A-module structure of
M@eN passes down to its quotient
M@BN. It remains to observe that the construction N ~ M~)BN is functorial in N: if f : N --+ N ' is a morphism of B-modules, I can set for a G-set (Y, r over X
The m a p
Mc~sf is well defined, because
(M@Bf) ([m.b|162174
= [m.b| fv(n)l(y,~, )- [m| fv(b.n)l(v,r . . . . ....
The l e m m a follows.
[re.b| fy(n)](y,r - [m |
b.,fv(n)](y,~) ,,
6.6. THE FUNCTORS M |
N
149
P r o p o s i t i o n 6.6.2: Let A and 13 G r e e n f u n c t o r s for t h e g r o u p G. If M is a n A-module-/3, if N is a /3-module and P an A - m o d u l e , t h e n t h e r e are i s o m o r p h i s m s of M a c k e y f u n c t o r s
~A(M+,,N, P) ~_ n,, (N, ~A(M, P)) M,N and P. In p a r t i c u l a r , the f u n c t o r N ~-+ M@BN is left adjoint to the f u n c t o r P ~-+ n (M, P). which are m o r e o v e r n a t u r a l in
P r o o f : I have the following diagram of Mackey functors:
UA(M6.N,P)
1 (M6N, P)
1
-(7,
where the bottom isomorphism ~r comes from proposition 1.10.1. I need to show that this isomorphism maps the left column to the right one. Let X be a G-set, and f be a morphism of Mackey functors from MQN to Px. It follows from proposition 1.8.3 that f is determined by bilinear maps
]y : M(Y) • N(Y) + P,:(Y) = P(YX) More precisely, the image under f of the element [m | n](u,~) of
To say that f passes down to the quotient
for any G-set (U, r over Y and
(U, r = (Is, Id),
MQN(Y) is given by
Mc~BN(Y) is equivalent to say that
b E/3(U). This is also equivo,lent to the special case
i.e.
= j (m, b.n)
(6.5)
On the other hand, the element corresponding to f under (7 in
~(N, 7-t(M,P))(X) = HomM=&(a)(N,7-t(M,P)x) = HomM~&(a)(N,'H(M, Px)) can also be defined using f: if Y is a G-set, and if n E N(Y), then (r(f)(n) is the morphism from M to (Px)z = Prx defined for a G-set Z and m E M(Z) by
150
CHAPTER 6. CONSTRUCTION OF GREEN F U N C T O R S
Now cr(f)(n) is a morphism of B-modules if and only if for any G-set V and any b r B(V), I have ~(f)(b x n) = b • ~r(f)(n) But
whereas
(6.7) Moreover, the expression of the product • from the product ".', and lemma 5.2.2 show that
Now equality of (6.6) and (6.7) becomes
f~v~(M. (7)(~,, ~ (7)~,.~-(7)(~,)
. . . .
It is now clear that (6.5) implies (6.8). Conversely, if I set f z , g ( m , n ) = cr(f)(n)z(m) 9 P ( Z Y X ) I know that f is a bilinear morphism from M, N to Px, and that I can recover f by the formula
Equality of (6.6) and (6.7) can also be written
Then i f Z = V = Y ,
Ihave
Since f is bifunctoriat, this is also
6.6. THE FUNCTORS M | By
a
N
151
similar computation =
=
N*
YY
YY
Y
(bxn)
. . .
YY
which proves the equivalence of (6.5) and (6.8), and the isomorphism
7~(M6BN,P) -~ ~ . (X, ~(M, P)) It remains to prove that this isomorphism maps 7-gA(M~BN, P) into ~ e (N, 7-/a(M, P)). Let X be a G-set, and f be a morphism from M Q B N to Px, determined by
Then f is a morphism of A-modules if and only if for any G-set Z and any a C A(Z) I have
s y(o x
| <(u,+,)=a• sy(i |
which can also be written
( The right hand side is equal to
P* \ zr
]
and f is a morphism of A-modules if and only if
fzu(a • m, CB,Z • n) = a • ]u(m, n)
(6.9)
It corresponds to f under c~ an element of HomB(N,~(M, Px)), which maps n E N ( Y ) on the morphism from M to (Px)Y = PYx defined by
This morphism is a morphism of A-modules if and only if for any G-set U and any a E A(U) I have
....
fuzY ( M * ( u z Y ~ ( a • ....
fuzy(axM*(?)(m),N*(ugY)(n))
.... (6.10,
152
CHAPTER 6. CONSTRUCTION OF GREEN FUNCTORS
It is then clear that (6.9) implies (6.10), since
Conversely, the image of the left hand side of equation (6.10) in the case Y = Z under the map P* (~;~) gives
"" = a > < P * ( Y ) f Y 2 ( M * ( Y l Y 2\ ) YI
\ Y2 f
which, thanks to condition iii) of proposition 1.8.3, can also be written
The image of the right hand side is
. . . .
•
•
and equation (6.10) is equivalent to equation (6.9), which proves the isomorphism
nA(M+sN, P) ~- 7-@(N, hA(M, P)) Those isomorphisms are deduced from those of proposition 1.10.1. So they are natural in M, N and P. Now evaluation at * gives the claimed adjunction property, and completes the proof of the proposition. 9
Chapter 7 A Morita theory 7.1
C o n s t r u c t i o n of b i m o d u l e s
Let A, B and C be Green flmctors for the group G. If M is an A-module-B and N is a B - m o d u l e - C , then M is in particular an A-module, and N is a C~ so M Q N is an A~C~ that is an A-module-C. It is easy to check that this structure passes down to the quotient and turns M ~ B N into an A-module-C. Moreover Proposition
7.1.1: L e t A, B, C b e G r e e n f u n c t o r s for t h e g r o u p G.
1. I f M is a n A - m o d u l e - B , if N is a B - m o d u l e - C , a n d if P is a C - m o d u l e D, t h e n t h e r e a r e i s o m o r p h i s m s o f A - m o d u l e s - D
( M ~ B N ) ~ c P ~- M ~ B ( N ~ c P ) w h i c h a r e m o r e o v e r natural in M, N a n d P . 2. If M is an A - m o d u l e - B , t h e n t h e r e a r e i s o m o r p h i s m s o f A - m o d u l e s - B
A d A M ~-- M
MQBB ~ M
w h i c h a r e n a t u r a l in M . P r o o f : The first assertion follows from proposition 5.3.2 and from the fact that the isomorphisms
(M6N)~P
~_ M ~ ( N ~ P )
are compatible with taking quotient, and are natural in M, N, and P. The second assertion follows easily by adjunction fl'om the fact that HomA(A, M x ) = M x ( . ) = M ( X ) which clearly implies 7-~A(A , M) ~_ M.
9
Similarly, if P is an A-module-C, then ~ A ( M , P ) is a B-module-C: if X , Y and Z are G-sets, if b E B ( X ) , i f f E ~ A ( M , Pz), and c ~ C(Z), I can define b • f • c on the G-set U by
(b • f • c)u(m) = f u x ( m • b) • c E P ( U X Y Z ) = P x y z ( U )
154
CHAPTER 7. A MORITA T H E O R Y
7.2
Morita
contexts
Those constructions of bimodules over Green functors lead to try to generalize the notion of a Morita context(see Curtis-Reiner [4] 3.53), in the following way: Definition: Let A and B be Green functors for the group G. A Morita context (M, N, O, @) for A and B consists of an A-module-B M and a B-module-A N, and morphisms of bimodules ~ : M @ s N ~ A and r : N@AM ~ B, which are balanced in the sense that the bilinear morphisms ~P and o2 associated to them are such that for any G-sets X , Y and Z
m • (gv,z(n,m') = ~x,y(m,n) x m' Vm C M ( X ) , n E N ( Y ) , m ' E M ( Z ) n • ~y,z(m,n') = ~ x , y ( n , m ) x n' Vn E N ( X ) , m c M ( Y ) , n' E N ( Z ) I will say that (M, N, ~, q2) is a surjective Morita context if ~ and 9 are surjective. L e m m a 7.2.1: Let (M, N, O, ~) be a M o r i t a c o n t e x t for A and B. Let (U, r be a G-set over X, and (V, ~b) be a G-set over Y. T h e n if m C M(U), n C N(U), p e M ( V ) and q e N ( V )
[m@n](v,r
~Y([p@q](v,r
= r162
x [p|162
in
(M@BN)(X x Y)
Proof: The equation relating ~ and ~ is
Setting P = [m|162
x Oy([P|162
I have P : [(m• CB,Y)@ ( n • A.(~)A* (v;)(~g'v(P'q))](UxY,r On the other hand nxA.(~)A*
vvV ~v'v(P'q)=N*
ur
....
N,
As (r x Id) o (Id • r
~(v)
nxr
y(p,q) . . . .
~w
= r • r I have also
~(~) Moreover
uvv
u~v
(v•
7.2. MORITA CONTEXTS
155
whereas expressing the product x using the product ".", I have
As the map N* (u;;) is compatible with the product ".", it is also
so
Similarly, if I set O : ~x ([m | n](v,~/) x [p | q](v,~,) I have SO
Moreover
uuU ~v,u(m,n)•
A.(r
A. r
uuv
. . . . M. r As
M*(UV) .... M*(Z)
' uuv
m x (gv,v(n,p)
.... (M'(U':12V)(m).B'(UluU~vV)~u,v(n,p)) . . . .
I have also
A.(r
(uu ) Spu,U(m,n) x p : M. r
(M* ( u v ) ( m ) . ~ u , v ( n , p ) )
Finally, I have
Q=
M*
(m).~g,y(n,p)) |
r
(SA,X x q) (vxy,r215162
and as gA.X xq=N*(~:,~,)(q),Ihaveaiso
so P = Q in MQBN(X • Y), and the Iemma follows.
9
C H A P T E R 7. A M O R I T A T H E O R Y
156
P r o p o s i t i o n 7.2.2: L e t A a n d B b e G r e e n f u n c t o r s for t h e g r o u p G, a n d let (M, N, ~, qJ) b e a M o r i t a c o n t e x t for A a n d B. 9 I f ~o is s u r j e c t i v e , t h e n ~ is an i s o m o r p h i s m . 9 I f ~P. a n d ~o are s u r j e c t i v e , t h e n 9 a n d q2 are i s o m o r p h i s m s . M o r e o v e r in t h a t case 1. T h e m o d u l e Ma is a p r o g e n e r a t o r for A - M o d a n d for M o d - B , a n d t h e m o d u l e Na is a p r o g e n e r a t o r for B - M o d a n d for M o d - A . 2. T h e r e are i s o m o r p h i s m s of b i m o d u l e s
N ~" 7-(A(M,A) ~_ ~Bop(M,B)
M ~_'HB(N,B) ~_ "HAop(N,A)
3. T h e r e are i s o m o r p h i s m s of G r e e n f u n c t o r s
A ~- ~ B o , ( M , M ) ~- (']-{B(N,N)) ~
B ~-- 7-{Aop(N,N) "~ (7-~A(M,M)) ~
4. T h e f u n c t o r s P ~ N | P a n d Q ~ M ~ B Q are m u t u a l i n v e r s e e q u i v a l e n c e s of c a t e g o r i e s b e t w e e n A - M o d a n d B - M o d . P r o o f : As q5 is a morphism of A-modules-A, and as A is generated as a bimodule by cA, if ~ , is surjective, then eA is in its image, and then ap is surjective. To prove the first assertion, it suffices then to prove that q~ is injective. Let X be a G-set. Let moreover [rei @ ni](u,,r for 1 < i < re be elements of M Q s N ( X ) such that
~X
rei @ ni](U,r
)
= 0
By hypothesis, the map ~. is surjective: let [pj | qJ](5,r of M @ N ( e ) such that
(the map r
for 1 < j <_ p be elements
is then the unique map from Vj to *). Setting m
v = F_.[.~ | ~d(v,,~,) i=1
it follows from the previous lemma that m
v=v
p
X eA = E ~-~[rniOni](ui,r
x r
([pjO qj](v,r
....
i=l j=l
i=1 j=l
j=l
7.2. MORITA CONTEXTS
157
which proves that r is injective, hence that it is an isomorphism. Now if ~o and q*o are surjective, then ~ and 9 are isomorphisms (indeed the 4-tuple (N, M, ~, (I)) is a Morita context for B and A, and ~ and 9 play symmetric roles). Assertion 4) follows then fi'om proposition 7.1.1, since
N@A(M@BP) ~-- (N@aM)~BP ~- B@BP ~- P L e m m a 7.2.3: Let M and N be Mackey functors. t h e n there are i s o m o r p h i s m s of M a c k e y functors
If X and Y are G-sets,
(M+N)xy ~- M.v+Ny which are natural in M, N, X and Y. P r o o f : This follows from the fact that
7~(Mx, N) ~_ 7-t(M, Nx ) naturally in M, N, and X. Then
(Mx Ny,
P))
Px)y)
Moreover, for any G-set Z
~(M, Px )y(Z) = ~(M, Px )(ZY) = HomM~k(o)(M, (Px)zz) . . . . ....
HomM~k(a)(M, PzYx) = HomM~ck(a)(M, (PYx)z) = "H(M, Pyx)(Z)
This gives the isomorphism 7-{(M, Px)Y ~- ~(M, PYX), and then ^/r
.
.
.
.
U(MGN, Pxy) and the lemlna follows.
"
R e m a r k s : 1) The isomorphisms of the lemma can be stated precisely. If Z is a o ) from U to Z X Y . G-set, then a G-set over Z X Y is determined by a map ( ~(~)~(~)~(~) I define then a map from (M@N)xr(Z) = M@N(ZXY) to (Mx@Ny)(Z) by
(u,-~) which makes sense because M,
~(,~) (m) r M(UX) = Mx(U), and N,
N(UY) = Xv(U). The inverse isomorphism is defined by
~(~)
(n) E
CHAPTER 7. A MORITA THEORY
158
2) The case X = 9 gives in particular the isomorphism
(M(~N)x ~- M+(Xx) ~- (Mx)+N and the naturality of this isomorphism shows that if M is an A-module-B, then for any B-module N
(M+BN)x ~- M+B(Nx) ~- (Mx)+~X P r o o f of p r o p o s i t i o n 7".2.2 ( p a r t 2): Assertion 1)follows from the previous lemma, because an equivalence of categories preserves modules of finite type, and because Ma is the image of Be under the functor P ~ M@sP, since m~ ~- (M@BB)a ~- M@B(Ba) Concerning assertion 2), I observe first that evaluating at the set 9 the isomorphism
~ , (N, 7~A(M, A)) ~- 7%(M@uN, A) gives HomB (N, 7-ta(M, A)) ~- HomA(M@BN, A) Then the morphism (I) gives by adjunction a morphism 0 from N to 7-{a(m, A), very easy to describe: if n E N(X), then the image of n in ttomA(M, Ax) is the morphism defined by
m E M(Y) ~ 3Pv,x(m,n) E A ( Y X ) = Ax(Y) The morphism O is injective: the element n E N(X) is in the kernel of Ox if and only if for any G-set Y and any element m E M(Y), I have ~v,x('~,~) = 0 But as 9 is surjective, there exists G-sets ~ and elements p~ E N(Y~) and q~ E M(YQ, for 1 < i < n, such that
i
YlYl
Then
The expression inside hooks is also
Pi • ~.y,,x(qi, r~) So it is zero for all i, which proves that n = 0 and that O is injective. To prove that O is also surjective, I must prove that if f E HomA(M, Ax), then there exists an element n E N(X) such that for any G-set Y and any m E M(Y), I have
fy(m) -= ~y,x(rn, n) If such an element exists, keeping the previous notations, I have
=
i
\ yiYiX ]
pi x ~y,,x (qi, n) . . . .
7.2. MORITA CONTEXTS
159
i
kyiyix/
Let then
i
\yiyix/
Then for any Y and any rn E M(Y), I have
'
i
\yiYiX/
'
As (~ is a bilinear morphism, it is also
{
\ yx I
\yy;y,:z
eY'h~x(rn'pi x fY,(qi))
As ~5 is a morphism of modules-A, I have
(~<<~x(rn,pi • fy'~(qi)) = ~y,y,(rn,pi) x fY,(qi) and as f is a morphism of A-modules, it is also
or
m;2
•
Finally
As f is a morphism of Mackey functors, it is also
Expression inside hooks is equal to
rnxB*(~)B*(Yi
)
so I have which proves that O is an isomorphism. It is easy to see that it is an isomorphism of bimodules. The other isomorphisms of assertion 2) now follow, by switching the roles of A and B, or replacing them by their opposite. Assertion 3) is proved by observing that
"]'{A(A, A) ~_ A ~
CHAPTER 7. A MORITA THEORY
160
Moreover, if I denote by F the fimctor N~A-- from A - M o d to B - M o d , the isomorphisms
(N+aP)x show that F(Px) ~- F(P)x. As F is an equivalence of categories, then HomB ( F ( P ) , F(Q) = HomA(P, Q) for any P and Q. It follows that
....
HomA(P, Qx) = HA(P, Q)(X)
and it is easy to see that those isomorphisms induce isomorphisms of Mackey functors
"HB(F(P), F(Q)) "Ha(P,Q) which are moreover compatible with the product 5 . Thus for any A-module, the Green functors "HB(F(P), iv(P)) and "HA(P,P) are isomorphic. Now for P : A, this gives
"HB(N, N) ~_ A ~ and assertion 3) follows. This completes the proof of the proposition.
7.3
9
Converse
The previous proposition has a converse: P r o p o s i t i o n 7.3.1: L e t A b e a G r e e n f u n e t o r for t h e g r o u p G, and M b e an A - m o d u l e s u c h t h a t Mu is a p r o g e n e r a t o r o f A - M o d . L e t N = "HA(M, A) and B = ('HA(M,M)) ~ Then there exists a surjective Morita context (M, N, ~, ~ ) for A and B. P r o o f : Let X and Y be G-sets. If rn r M(X) and f E B(Y) = H o m a ( M , M r ) , let
m • f = fx(m) E My(X) = M(XY) It is easy to see that this definition turns M into an A-module-B. "HA(M, A) is a B-module-A. As moreover
Then N =
"HA(M+sN, A) ~_ ~ s (N, "HA(M, A)) = "HB(N, N) the identity map of N gives a morphism of A-modules from M ~ B N to A, which can be described as follows: if X and Y are G-sets, let ~x,Y be the bilinear map from M(X) • N(Y) to A(XY) defined by
m E M(X), r E N(Y) = HomA(M, Av) ~ e x ( m ) E A t ( X ) = A(XY) It is easy to check that those maps define a bilinear morphism from M, N to A, associated to a morphism ~ from M+BN to A, which is a morphism of A-modules-A.
7.3. CONVERSE
161
Conversely, there is a bilinear morphism t) from N, M to B defined as follows: if X and Y are G-sets, if f E N ( X ) = HomA(M, Ax) and m r M ( Y ) , then ~ x , y ( f , m ) is the morphism from M to M x v defined for a G-set Z and m E M(Z) by
~ x . v ( f , m ) : 7Y~l ~ M ( Z ) ~ f z ( m , ) • ro E M ( Z X Y ) = M x v ( Z ) The morphism from N@M to B associated to it passes down to the quotient, and defines a morphism from N@AM to B, which is a morphism of B-modules-B. Before proving the proposition, I will give an equivalent formulation of the hypothesis on M, independent of fh L e m m a 7.3.2: L e t A b e a G r e e n f u n c t o r for G, and M b e an A - m o d u l e . T h e f o l l o w i n g c o n d i t i o n s are e q u i v a l e n t : 1. T h e m o d u l e Mn is a p r o g e n e r a t o r . 2. T h e m o d u l e M is a f i n i t e l y g e n e r a t e d p r o j e c t i v e m o d u l e , a n d t h e r e e x i s t s a G - s e t X s u c h t h a t A is a d i r e c t s u m m a n d of Mx. 3. T h e r e e x i s t s G - s e t s X and Y such t h a t A is a d i r e c t s u m m a n d a n d M is a d i r e c t s u m m a n d of A z .
of Mix
P r o o f i First I observe that if X and Y are G-sets such that X divides Y in CA, then M x is a direct s u m m a n d of M r : indeed, say that X divides Y in dA is equivalent to say that A x is a direct s u m m a n d of Av. But Mx is isomorphic to "HA(Ax, M), which is a direct s u m m a n d of "HA(Ay, M) ~- Mz. So if Ma is a progenerator, as 9 divides f~ in CA, it follows that M is a direct s u m m a n d of Ma, so M is a finitely generated projective module. Moreover, there exists a set I and a surjective morphism p: @ M# ) ~ A iE1
from a sum of copies of Ma to A. In particular, there is a finite subset ,1 C_ I and elements rnj E -/I/t(J) - a '( o" ) , for j E J, such that. ~A = P.(~'~ mj) j~_J
Then the image B of the restriction of p to | j) is an A-submodule of A, and B(*) contains CA. This proves that there exists an integer n such that A is a quotient, hence a direct summand, of nMfl ~- M,~fl. Thus 1) implies 2). If now hypothesis 2) holds, then as An is a progenerator, and as M is a finitely generated projective module, there exists an integer n such that M is a direct s u m m a n d of nAG ~- A,~a. So 2) implies 3). Finally if 3) holds, then M is a finitely generated projective module, because it is a direct s u m m a n d of the projective module Az. Then Mn is a direct s u m m a n d of ( A z ) a ~- Aay, and Ma is a finitely generated projective module. Moreover, the module An is a direct s u m m a n d of ( M x ) a ~- Max. As f / X divides a multiple of f~ in CA, it follows that Aa is a direct summand of a direct sum of copies of Ma. As Aa is a progenerator, so is Ma, and this proves the lemma. 9
CHAPTER 7. A MORITA THEORY
162
P r o o f o f p r o p o s i t i o n 7.3.1: I must prove that if Mn is a progenerator, then 02 and q~ are surjective. It is equivalent to say that CA lies in the image of 4p,, and eB in the image of ~o. I will show that if there exists X such that A is a direct summand of Mx, then 02 is surjective, and that if there exists Y such that M is a direct s u m m a n d of A r (which is equivalent to say that M is projective and of finite type), then 9 is surjective. So let X be such that A is a direct summand of Mx. Then there exists an element ct E HomA(A, Mx) and an element fl E HomA(Mx,A) such that
flo a = IdA But I have seen that HomA(A, M x ) ~- M(X). Let m E M ( X ) be the image of c~ under that isomorphism. Then c~ is defined on the G set Z by
a E A(Z) ~ a x m E M ( Z X ) = Mx(Z) On the other hand
HomA(Mx, A) ~_ HomA(M, Ax) Under this isomorphism, the element fl maps to 5 E Homm(M, Ax) = N ( X ) , and I know that for any G-set Z and any m' E M x ( Z ) = M ( Z X ) , I have
Now say that fl o c~ = IdA is equivalent to say that for any Z and any a E A(Z)
which can be written as
a=a•
X)
This is equivalent to
Then let = [m | ? ] ( x , ( : ) ) E M~cBN(.) The image of ~- under 02~ is precisely =
= XX
7x(m)
= cA
X32
and this proves that 02~ is surjeetive. Similarly, if M is a direct s u m m a n d of M y , then there exists a E HomA(Ay, M) and fl E HomA(M, Ay) such that c~ o/~ : IdM. But HomA(Ay, M) "~ M ( Y )
7.4. A R E M A R K ON BIMODULES
163
and c~ is determined by an element m E M ( Y ) , such that for a ff A y ( Z ) = A ( Z Y )
Now say that c~ o ~ = IdM means that for any G-set Z and any m' E M ( Z ) , I have
Let r ' = [3 | m](v,(~)) E N + A M ( . ) The image of r ' under q2 is the element of HA(M, M ) ( . ) = EndA(M) defined by
which can be evaluated at a G-set Z by
It follows that ~(r')(m') = m', so q2(r') is the identity of M, that is the unit e~a(M,M). This proves that ffJ is surjective, and completes the proof of the proposition. 9
7.4
A remark on bimodules
If A is a Green functor for G, theorem 4.3.1 states that evaluation at fl is an equivalence of categories from A - M o d to A(g/2)-Mod. If B is another Green functor, and M is an A-module-B, then M ( a ) is an A(a2)-module, and a B~ As B~ s) ~_ B(~s) ~ the module M(~ s) is a module-B(fF). However, the module M ( a ) is not in general an A(aS)-module-Btas): indeed, if a E A(aS), if m E A(f/), then
Then if b E
\ CO1 /
\COlt~20.JS
\ cos /
\colco:cos/
B(a 2)
The product inside hooks is equal to A. (colco2co3co4~ A* ( c o i c o 2 t M 3 c d 4 ) ( ~ x . l . x b ) \ colco3co4 / \t~lco2co20.)3t.04
As the square ~1&2~3 CO1~2~1~3/
) \COlcolco2/
CHAPTER 7. A MORITA THEORY
164 is cartesian, I have
and it follows that (a oa ~ ) oa b . . . .
\
/.,33
\ OJ1r 2/.,U2 (.O1 t.O3 /
A similar computation gives
The product inside hooks is equal to
As the square f~3
C2&)
, fp
\r
/
is ca.rtesian, I have \CUI(.U2CU2
\
0.)10J2CJ4
/
\
CU1CU2 /
\O-~ICO2CU3~2 /
SO a Of~ ( m 0 a b) . . . .
\
~,g1
\
Cgl0g2
/
\~,Ul~g20.)3Cd2 /
kt.U1CU20J3t~)3~.U 4/
There is no obvious reason for which this expression should be equal to (E), so M ( 9 ) is not a bimodule in general. This is because (AQB~ e) is the tensor product of A(fF) and B(fF) ~ over the non-commutative algebra b(92) (i.e. Mackey algebra) as shown in the next proposition: P r o p o s i t i o n 7.4.1: L e t A and B b e G r e e n f u n c t o r s for t h e g r o u p G. I f fl is a G - s e t , t h e n t h e m o r p h i s m f r o m b to A (resp. f r o m b to B) t u r n s A(f~ 2)
( r e s p . B ( f ~ ) ) into a module-b(Pt e) (resp. a b(f~Z)-module), a n d (A+B)(f~ 2) _~ A(f~2)(~b(a2)B(f~z) as R - m o d u l e s .
7.4. A R E M A R K ON BIMODULES
165
Proof: Lemma 7.2.3 shows that (A@B)a2 -~ Aa@Ba as Mackey fnnctors. Evaluation at the trivial G-set gives
(m~B)a2(,) = (m~B)(a 2) = (da6Ba)(~ and by definition of tensor product of Mackey functors
(do@Sn)(*) = da(f~) |
B~(f~) = d(f~ 2) |
B(f~ 2)
so proposition follows.
9
The algebra structure of A(f~2) | phism
r
B(f~2) |
B(~ 2) can be recovered using the isomor-
A(f~2) ~- (B@A)(f~2) ~- (A@B)(~22) -~ A(f~2) @b(n~)B(f12)
If a and a' are in A(f~), and b, b' in B(f~2), and if
r
@ a') = ~ ai" @ bi" i
then (a | b)(a' @ b') = y~(aai") | (bi"b') i
Chapter 8 Composition 8.1
Bisets
In [2], I have studied the following problem: if G and H are groups, what kind of functors F from G - s e t to H - s e t induce by composition a functor from Mack(H) to Mack(G)? It seems natural to ask that the functor F transforms a disjoint union into a disjoint union, and a cartesian square into a cartesian square. The functors from G - s e t to H - s e t having those two properties can be completely classified up to isomorphism by the (isomorphism classes of) H-sets-G: D e f i n i t i o n s : Let G and H be groups. An H-set-G is a set X with a left H-action
and a right G-action, which commute, i.e. such that if g E G, h C H and x E X
h (x g) -- (h.x) g f i G , H, and K are groups, if X is an H-set-G, and Y a K-set-H, I denote by Y OH X the K-set-G de/~ned by YOHX={(y,x)
EYxXIVhEH,
yh=y~
3gE G, h . x = x . g } / H
where the action of H is given by (y, x).h = (y.h, h-ix). The action of K and G on Y OH X is given
k.(y, x)g = (k.y,x4) If X is an H-set-G, and if G and H are clear from context, I will also say that X is a set with a double action, or biset for short. W i t h those notations, if U is an H-set-G, and X is a G-set, or G-set-(l), then U o a X is an H - s e t - ( i ) , that is an H-set. This construction gives a functor from G - s e t to H - s e t , that I denote by U oG - . The precise statement proved in [2] is then Theorem
8.1.1: L e t G a n d H b e f i n i t e g r o u p s .
* I f F is a f u n c t o r f r o m G - s e t t o H - s e t w h i c h t r a n s f o r m s d i s j o i n t u n i o n s into disjoint unions and cartesian squares into cartesian squares, then t h e r e e x i s t s a n H - s e t - G U, u n i q u e u p t o i s o m o r p h i s m o f H - s e t s - G , s u c h t h a t F is i s o m o r p h i c t o t h e f u n c t o r U oa - .
CHAPTER 8. COMPOSITION
168
9 C o n v e r s e l y , if U is a n H - s e t - G , t h e n t h e f u n c t o r U o a - t r a n s f o r m s dis-
joint unions into disjoint unions and cartesian squares into cartesian squares. I also proved that in these conditions, the set U induces a functor from Mack(H) to Mack(G), defined by composition, and denoted by
M~MoU If M is a Mackey functor for H, the Mackey functor M o U is defined over the G-set X by (M o U)(X) = M(U o~ x ) If f : X --+ Y is a morphism of H-sets, then U oG f : U oa X --+ U oa Y is defined by
x):
s(x))
and then
( m o U),(f) = M.(U oa f)
(M o U)*(f) = M*(U oa f)
E x a m p l e s : 1) If H is a subgroup of G, and if U is the set G, viewed as an H - s e t - G by multiplication, then the functor U oa - is tile restriction functor [rom G - s e t to H - s e t . If V is the set G viewed as a G-set-H, then V OH -- is the induction functor from H - s e t to G - s e t . The functor M H M o U is then the induction functor for Mackey functors, and the functor N ~ N o V is the restriction functor for Mackey functors. 2) If N is a normal subgroup of the group G, if H = G/N, let U be the set H, viewed as an H-set-G, the group G acting by the projection G ~ G/N, and let V be the same set viewed as a G-set-H. Then the functor U oa - is the "fixed points by N" functor. It is easy to identify the functor M H M o U as the inflation functor from Mack(H) to Mack(G), defined by Th6venaz and Webb (see [14], [15]). The functor V oa - is the inflation functor from H - s e t to G-set. The functor N ~ N o V is the "coinflation" functor for Mackey functors (that Th6venaz and Webb denote b y / 3 ! in [15] L e m m a 5.4, and I denote by Pg/N in [2]). 3) If U is an H-set-G, then U oa * ~- U/G. 4) If U is a G-set-G, and X a G-set, then
UoaX=
8.2
{(u,x) E U •
u.g=u~
g.x=x}/G
Composition and tensor product
Definition: Let G and H be groups, and U be an H-set-G. If M and N are Mackey functors for H, and if X and Y are G-sets, I denote by r X,Y U the map
YxY : M(U o~ X) • W(U oG Y) -+ (M+N)(U o~ (X • Y)) defined by
8.2. COMPOSITION AND TENSOR PRODUCT Lemma
169
8.2.1: T h e m a p s TU, y f o r m a b i l i n e a r m o r p h i s m f r o m M o U, N o U o U.
to (M@N)
P r o o f : The maps TUg being clearly bilinear, all I have to check is their bifunctoriality. So let f : X --+ X' and g : Y --+ Y' be morphisms of H-sets. If m E M(U oa X) and n E N(U oa X~), setting M' = M o U and N' = N o U, I have
t, x' ) M~*(f)(m) | m'*
\ yt )
m~(g)(n)
(U~
As the square
X x Y'
fxld
X
) X ~ x Y'
)
l,x(x'G
X ~
f is cartesian, its image under U oc - is also cartesian, and then
M* (U oa (x'Y'~ It follows that
~x,,~, (M:(f)(~), N:(g)(~)) ....
=
. . . .
M.(Uoa(fxId))M'*
(m)|
Y' ]N'.(g)(n)
N'(g)(,,)
\ Y' } \ y' ]
t, y' ]
(Uo~(X'xZ'),H)
....
. . . .
(UodX•215 J (UodXxY,),Uo~(S•
But for the same reason, I have
t,y') This gives TX, y, ( M : ( f ) ( m ) , N ' (g)(n)) . . . .
[
(? \ Y' ) )
.... J(Uoa(XxZ),Uo~(Sxg))
CHAPTER 8. COMPOSITION
170
proving finally that
so T is covariant. Now if rn' r M'(X') and n' E N'(Y'), then
r Xv, Y {M'*tr'(rn '']~ N " ( g ) ( n ' ) ) = k ~JJk
9 ""
On the other hand, as the square xY)
U oa ( f • g) ,
u oo ( x • Y)
,
Uoa(X
Uoc(X'•
,d I
lid u oG (f • 9)
u oa ( x ' • Y')
is trivially cartesian, I have
....
M'*(f x g)M'* \ x' J (rn') | N'*(f • g)N'* \ Y' ,] (n') (Uoa(Xxz),/d)
As moreover
f o
=
f(x)
\ x' ] o ( f x g)
go
=
g(Y)
\ Y' J o ( f x g)
I have
"rUz(M'*(f)(m'),N*(g)(n')) = ( ( M ~ N ) o U)*'rx, z,(rn',n') and r is also contravariant. The l e m m a follows.
8.3
"
Composition and Green functors
W h e n A is a Green functor for the group H, the product on A gives a morphism from A ~ A to A, hence a morphism from (A+A) o U to A o U. Composing this morphism with the morphism
(A o U)~(A o U) ~ (AQA) o U deduced from T gives a morphism
(A o U)6(A o U) -+ A o U
8.8.
171
COMPOSITION AND GREEN FUNCTORS
I view it as a product on A o U, that I denote by • It is natural to ask if A o U is a Green functor. To see this, I will first describe precisely the product: if X and Y are a-sets, if a C (A o U ) ( X ) and b C (A o U)(Y), the product a x v b is equal to
where the product "." on the right hand side is the product of A. Then
(~,x,y)(u,~,y)
\ (~,x) ] (a) •
\ (u,y) ]
(~)
....
=A*((u,x,y))A.((ul,xl,yl)(u~,xx,y2)~(axb)
(~, x, ~)(~, z, y)
"
~
(~l,x,)(~,y~)
]
whence finally
a•
(~,x,y) ~(a•
This leads to the following definitions: D e f i n i t i o n s : Let G and H be finite groups, and U be an H-set-G. If X1, . . . X~ are Gsets, I denote by f u 1,...,x. the map from U oa (Xl,., Xn ) to ( U oGX1) X . . . • ( U Og X n ) , defined by f f A is a Green functor for the group H, I set r
= A.(pv/a)(eA) e (A o U)(.) = A ( U / G )
where pv/a is the unique map U o 9 ~_ U/G to .. With those notations, I have a •
b = A * ( $ u z ) ( a x b)
where the product x on the right hand side is the product of A. The following remark will be useful: L e m m a 8.3.1: T h e m a p (~u X,Y is injective. P r o o f : Indeed, as the square
(?) X •
(x:) j
-~ X
l l:l
Y
~ 9
(:)
172
C H A P T E R 8.
COMPOSITION
is cartesian, so is its image under U oa -
Uoc(X
x Y)
, UoaX
UoaY
~
Uoa.
which proves that U oa (X • Y) maps into (U oG X) • (U oa Y), precisely by the map 5XU, Y " So 5XU, Y is injective. .. P r o p o s i t i o n 8.3.2: L e t G a n d H be finite g r o u p s , a n d U b e an H - s e t - G . I f A is a G r e e n f u n c t o r for t h e g r o u p H, t h e n A o U is a G r e e n f u n c t o r for G, for t h e p r o d u c t x U a n d t h e u n i t CAoU. P r o o f : The product x u is bifunctorial by construction. So I must check that it is associative and unitary. Let X, Y, and Z be G-sets. If a E ( A o U ) ( X ) , if b E ( A o U ) ( Y ) , and if c E (A o U ) ( Z ) , then (ax%)•
= A*(4•215 ....
= A . (~xx,.-z)(A u 9 (Sx.,,-)(a u • b) x c )
v z ) A 9(~SX, uY A 9(~Sxxy,
g because ( 5ux,Y x Iduooz) o 5x•
•
IdvoGz)(a • b x c)
=
=
...
A*(aUr, z)(a x b x c)
g since for ( u , x , y , z ) E U o G ( X x Y • = 5x,Y,Z,
( %u r • Zd~o~Z) o ~xe x~,Z(~, x, ~, ~) = ( %u ~ • I d v o c z ) ( ( u , x, y), (~, ~)) = ...
....
((~, x), (~, v), (~, ~)) = 4,~,~(~, x, v, z)
On the other hand
....
9
U
A (Sx,z•
since (Iduo~x x 6~z ) o 5x,yx a •
*
(Iduocx • 5u, z)(a • b • c) = A*(Su,z,z)(a • b • c)
= 5Ux,y,z by a similar computation. Finally
(b x U c) = A*(Sgv,y,z)(U,x,y,z ) = (a •
b) x U c
and the product • is associative. Moreover, if a 9 (A o U ) ( X ) , then eAoU •
v * (pu/~)(e) • a = A 9 (5o,x)(A
a)
v * (Pu/a • I d x ) ( a ) = A * (5.,x)A
But i f ( u , x ) 9 (Pu/G X I d x ) o , 5 [ x ( U , X ) = pu/a • I d x ( ( u , . ) , ( u , x ) )
= (., (u,x))
and with the usual identifications, this gives ~AoU • a =
a
A similar computation shows that eAoU is a right unit for A o U, and the proposition follows. ..
8.4. COMPOSITION AND ASSOCIATED CATEGORIES
8.4
Composition
and associated
173
categories
Let G and H be finite groups. If U is an H-set-G, and A is a Green functor for H, then A o U is a Green functor for G. Each of these Green functors has an associated category, and the functor U oc, - defines actually a functor from Caou to CA:
P r o p o s i t i o n 8.4.1: L e t G and H b e finite A b e a G r e e n f u n c t o r for H. T h e n t h e G-set X to t h e H - s e t U[X] = U oa X , and t o U[a] = A.(5~,x)(C~ ) + A((U oc X ) x (U oa CAoU t o C.
g r o u p s , let U b e an H - s e t - G , and correspondence which maps the t h e m o r p h i s m c~ E (A o U ) ( Y • X ) Y)) is an R - a d d i t i v e f u n c t o r f r o m
P r o o f : I must check that if X, Y, and Z are G-sets, if a C A ( Y X ) and t3 E A ( Z Y ) , then where o};7is the product o for tile category CAoU, Tills equality can also be written as
A.( ~,~. )(~) O~ooy A.(~,~)(~) = A.( e~,x )(~ o~ <~)
(s.~)
Let X ' = U[X], Y' = U[Y], and Z' = U[Z]. The left hand side of (8.1) is equal to
(\z'y'y'x')
A. \ z'x' )
A*(5~'r • fu':")(/3 x a)
The square
(c) Z'Y'X'
( z'y'y'x' X' ~
is c a r t e s i a n : i n d e e d , if ( ~ l , Z , y l )
~ U oo ( Zr}
Z'y'2x'
~)~ ( ~ 2 , y 2 , x )
Uoc,(YX) and(z',y,x')
Z ' Y ' X ' are such that
(=', v', v', .') = ((,,,, z), (~,, >), (,,2, y2), (~, x)) then there exists a E G such that U2 ~ ~18
8Y2 = Yl
Thus The element (ul,z, yl,sx) is in U oG ( Z Y X ) : indeed, if t E G is such that uit = ul, then as (ul,z,yl) E Y oa (ZY), I have tz = z and ty 1 = Yl" As m o r e o v e r u2.s-lt.s = ults = UlS = u> and as (u2, y~, s) E U oa (YX), I have also 8-1tsx = x, or t.sx = 8x.
174
CHAPTER
8.
COMPOSITION
Moreover
(~,z,y)(~,y,x) and (721, yl, .$3:) x ~, and then
=
(~/18, 8-1yl, X)
:
(~2, Y2, 5C). Similarly (u,, sx) = (uls, x) = (u2, x) =
--( ( ~ , , ~ ) , ( ~ , ~ l ) , ( ~ , ~ x ) )_- (~', ~,~) ' '
~zVy,x ( ~ ~ , ~ , , ~ ) ,
The map 5 Z,Y,X U is the product of two injective maps. So it. is injective, and (C) is cartesian. It follows that the left hand side of (8.1) is equal to
A.\ z'~' )
\(~,~,~)(~,~,x)
....
A.
(u,z)(u,:c)
\(u,z,y)(u,y,x)
On the other hand
zx /
\ zyyx /
....
c
(
,, z x
\ zyyx / /
which gives
(~,z,x) ]
\(~,z,y)(~,y,x)
(9 • 4)
The right hand side of (8.1) is then equal to
\(~, z)(~, x))
t(~,z,y)(~,y,x)
(~ • ~)
which is the right hand side of (8.2). Moreover for any G-set X, I have
9 .-
,
~,~,x~j
....
A.
~,.~j
r
A*
'o
. . . .
(cA) = 1A(l~o~X)~
So I have defined a functor from CAoU to CA. It is clear that this functor is R-additive, and this completes the proof of the proposition. 9
8.5. COMPOSITION AND MODULES
8.5
175
C o m p o s i t i o n and m o d u l e s
Let G and H be finite groups, let U be an H-set-G, and A be a Green fnnctor for H. The above functor from CAoU to CA induces by composition a functor between the associated categories of representations: if F is an R-additive functor on CA, then the functor F o U is an R-additive fimctor on CAoU: if X is a G-set, then (F o U)(X) = F(U oa X). As the category of R-additive functors on CA is equivalent to the category of A-modules, the functor t" corresponds to an A-module N, and then the functor F o U corresponds of course to the module N o U: P r o p o s i t i o n 8.5.1: Let G a n d H b e f i n i t e g r o u p s , let U b e an H - s e t - G , a n d A b e a G r e e n f u n e t o r for H. If N is an A - m o d u l e , t h e n N o U is an A o Um o d u l e , for t h e p r o d u c t x u d e f i n e d b y
( a , n ) E (A e U ) ( X ) • (N o U)(Y) ~ c~ x U n = N*(5~,z)(a • n) E (X o U)(X x Y) T h e c o r r e s p o n d e n c e N ~ N o U is a f u n c t o r f r o m A - M o d t o A o U - M o d . P r o o f : W i t h the notations of theorem 3.3.5, I have to find the A o U-module N ~ = MFNoU, and to prove that it coincides with N o U as Mackey functor. If X is a G-set, I have
X'(X)
= (FN o U ) ( X ) = r N ( ( : oa X ) =
N(U o~ X)
If f : X --+ Y is a morphism of G-sets, then by definition
N'.(f)
= ( FN o
U)( f~.r) = FN( U[ff.'])
where f.u is the element of (A o U ) ( Y X ) associated to f by lemma 3.2.3, i.e.
f~. = (Ao U).
(x) f(x)x
(Ao U)*
(:/
(CAoU)
In other words
((,,,x))
f a = A. \ ( u , f ( x ) , x )
A*
(~, o)] ,4" ( ....
A. (
(cA) . . . . (u,x)
then
=
*( Y,x)(f. ) = A.
(~, f ( x ) ) ( ~ , x)
'o
(E) = (U oc f ) .
SO
N ' ( f ) = FN((U oG f ) . ) = N.(U oa f) = (N o U).(f) A similar argument shows that if fu,. is the element of (A o U ) ( X Y ) corresponding to f by lemma 3.2.3, then fu,. = (U oG f)*, and then
N'*(f) = FN(f u'*) = FN((U oa f)*) = N*(U oa f) = (N o U)*(f)
CHAPTER 8. COMPOSITION
176
which proves t h a t N ' coincides as Mackey functor with N o U. T h e s t r u c t u r e of A o U - m o d u l e of N o U is then defined as follows: if X and Y are G-sets, if o~ E (A o g)(x) and n E ( N o U ) ( Y ) , then the p r o d u c t c~ x U n is the e l e m e n t of (N o U)(X x Y) given by
o~•
(AoU). xyy
Let
\ xyT/ /
A.
....
(~,,~,,y)(.~,y)]
\
(~,x)
]
(,)
then x ~ ~ = FN(fl)(~)
= fl Ouo~y ,, . . . . .
Moreover
/~ X T/,= N. ( Let f =
(ttltX'yl)(~2'~/2)
~ N* ((t~l'*t Ya)(?~2'y2)~ (dgx I%)
(~,=,y~)(~2,y2) '1 and g = (\(~l,x,~l)(~2,y2)(~>y2)]" (*~,=,y,)(~2,y2) ~ As the square (~l,=,y,)(~l,y=)(~2,y2)/~
(,,,x, ~)(~, y)
u o~(xY)
,
(u o~ (xY)) x (u o~ r )
g is cartesian, I have
N* (
(ul'x'yl)(U2'Y2)
I N, (
\(tz1,X,~ll)(tz2,Y2)(ll,2,Y2)/
(*q'x'Yl)(u2'Y2) l >)(u2, Y:)/
~(tzi , x, >)(u,,
....
N.
(~,.T, y)(~, y)
... --
(~,x,y)(~,y)
and finally c~ x U n is equal to the image of c~ x n under the m a p (~,x,y,)
X.
\(~,x,y)(~,y)
o...
8.6. FUNCTORIALITY
177
oN'*( (~,x,y) )N.((Ul,X, yl)('U,2,Y2))_~ "'"
(72, X, y)(tt, y)
k
(721, X)(2s Y2) )
''"
which proves the claimed formula
8.6
Functoriality
If G and H are finite groups, and U is an H-set-G, then I have a functorial construction X ~ U oa X from G - s e t to H - s e t . This construction is not quite functorial in U: if f : U --+ V is a morphism of H-sets-G, there is in general no associated morphism U oa X --+ V oa X: this is because if (u, x) E U oa X, i.e. if the right stabilizer of u is contained in the left stabilizer of x, generally, the right stabilizer of f(u) is not. I have studied this question in [2], and showed that it is natural to ask moreover that f is injective when restricted to each orbit of G (or equivalently to ask that the right stabilizer of f(u) is equal to the right stabilizer of u, for any u C U). Then, there is a morphism of functors f oa - from U oa - to V oa - defined on the G-set X by In those conditions (see [2] prop. 10 and 11), if M is a Mackey functor for H, then M o U and M o V are Mackey functors for G, and the morphism f induces two morphisms of Mackey functors (denoted by jr. and f* in [2], but differently here to avoid confusion): a morphism Mf from M o U to M o V, and a morphism M / from M o V to M o U. Those morphisms are defined for a G-set X by
Mf,x = M . ( f oa X) : (M o U)(X) ~ (M o V)(X) M~ = M*(f oc X ) : (M o V)(X) ~ (M o U)(X) W i t h those notations: P r o p o s i t i o n 8.6.1: L e t G a n d H b e f i n i t e g r o u p s , a n d l e t A b e a G r e e n f u n c t o r for H. L e t m o r e o v e r U a n d V b e H - s e t s - G , a n d f : U --* V b e a m o r p h i s m o f H - s e t s - G , w h i c h is i n j e c t i v e on e a c h r i g h t o r b i t . 9 I f A is a G r e e n f u n c t o r for H , t h e n A f is a u n i t a r y m o r p h i s m o f G r e e n f u n c t o r s f r o m A o V t o A o U. 9 I f M is a n A - m o d u l e , t h e n r e s t r i c t i o n a l o n g A f g i v e s M o V a n d M o
U structures of A o V-modules, and the morphisms morphisms of A o V-modules. Proof:
Mi and M f are
For the first assertion, I must show that if X and Y are G-sets, if a E
(A o V)(X) and b E (A o V)(Y), then A]x(a) •
A~(b) = A]x•
•
b)
CHAPTER 8. COMPOSITION
178 The left hand side is equal to
A ( 6Ux y ) ( A * ( f oG X)(a) • A*Cf oc Y)(b)) = A'(6~#y)A*((f oo X) x (fom Y))(a • b) and the
right hand side to A * ( f oc~ (X x Y))A*(6~y)(a x b)
Now equality follows from
since for ( u , x , y ) 6 U oG ( X x Y)
((lo~x) x (,focy))o4,y(~,x,~)= ((,-ocx) xisoc,,-)) ((~,=), (~,~))
. . . .
.... ((s(=), =), (s(~), ~)) : ~.L.-(,f(,,),.~,,~) = 4,~o(so~(x • Y))(.,,,,,~) Moreover
A{@AoV) = A*(f oc ")A*(py/a)(Cm) = A*(pu/c)@A) = CAoU since PU/G = Pv/c; o ( f oG .). So the morphism A I is a unitary morphism of Green funetors. For the second assertion, I must show that if a e (A o V)(X) and m 6 (M o U)(Y), then a x v M ! y ( m ) = My,xy(Af(a) x U m) (8.3) and that if m' C (M o V)(Y), then
MY(a x y m') : AY(a) x u MY(m ')
(8.4)
But
axVMym (= )M6('~Y':a()xM*X (~
'
(v,x)(f(u),y)](v'x)(u'Y) ](axrn.)
The square
U oG (XY) O~,x,y) V oa ( X Y )
(u,x,y) (f(~,), z)(,,, v)) )
((~,x)(~,y))
(V oo X) x (U oc Y)
\(~, x)(f(u), y)] (v oa x) • (v oc Y)
(c)
("'~'Y) /
is cartesian: if ((v,x), (u, y)) E (V oG X) x (U oo Y)and (v', x', y') 9 I/oG ( X Y )
such that (v',x') = (v,x)
(v',y') = (f(u),y)
are
8.6.
FUNCTORIALITY
179
then there exists s and t in G such that ?2t 7--- ~)S
SX t z X
v'= f(u)t
ty'= y
In those conditions, the element ( u , t x ' , y ) is in U oa (XY): indeed, if r E G is such that ur = u, then as (u,y) E U oa Y, I have ry = y. Moreover
v ' t - l r t = f ( u ) r t = f ( u r t ) -- f ( u t ) = f ( u ) t - v' and as (v', x/) E V oo X, I have t - % t x ' = ix', or r.tx' = tx'. Moreover
( f ( u ) , tx', y) = (v't -1, tx', y) = (v', x', t - l y ) = (v', x', y') and ((f(u),~Xt),('U,,,y))
~- ( ( V t ~ - i , t x t ) , ( ? j , ~ ] ) )
: ((Vt, x t ) , ( u , y ) )
....
Conversely, if (Ul, Xl, Yl) E U oa ( X Y ) is such that :
:
then the last equality proves that replacing (Ul, Xl, Yl) by (uls -1 , SXl, syl) for a suitable s E G, I can suppose Ul = u and y~ = y. Then (f(u),Xm) = ( v , x ) = ( f ( u ) , t x l ) . So there exists r E G such that
f(u)r = f(u)
r-ix1 = tx'
As f is injective on the right orbits, the first equality shows that ur = u, and since (u, Xl, y) E U oa ( X Y ) , I have rxl = Xl = tx', and
(u,, x,, y, ) : (u, tx', y) which proves that (C) is cartesian. It follows that
\(f(u),x,y)]
\(f(u),x)(u,y)]
But the right hand side of (8.3) is equal to
M f . x y ( A f ( a ) xU m) : M . ( f oa ( X Y ) ) M ' ( 5 ~ ; , y ) ( A * ( f o a X ) ( a ) • m) . . . .
"'"
(f(~l), X), ('U2, y)
9 which proves equality (8.3).
..=M,((u,~-,y)~ M" ( (f(u),x,y)]
(u,x,y)
\(f(u),x)(u,y)
) (a • ~)
CHAPTER 8. COMPOSITION
180 Similarly, the left hand side of (8.4) is
M/(axVm')=M*(f~
( ~z)(f(~), ';'Y)
y) )
(a •
whereas the right hand side is
A/(a) xVM/(m ') = M *(hx,y)(A ~ "(foaX)(a) " xUM*(foa r)(~')) . . . . .. : i*((hU,y)M * ( ( ~ l ' X ) ( ~ 2 ' ~ / ) ~ \ (f(ul), z)(f(u2), y)]
((/x;'7"/): ]~/./"(
(7.l,x,y) ) (a)<mt) \ (f(u), x)(f(u), y)
9
This proves equality (8.4), and the proposition.
8.7
Example:
induction
and
"
restriction
Let O be a finite group, and H be a subgroup of G. Let U be the set G, viewed as a G-set-H, and V be the set G, viewed as ai1 H-set-G. If X is an H-set, then U oa X identifies with I n d e X , functorially in X. It follows that if M is a Mackey functor for G, then M o U is isomorphic to Res~M. If Y is a G-set, then V oG Y identifies with Res~Y, functorially in Y, by the map (g, Y) ~ gY. Thus if N is a Mackey functor for H, then N o V identifies with Ind~N. So if A is a Green functor for G, then Res~A is a Green functor for H, and this is not surprising. The case of induction is less clear, but corresponds to what Th6venaz calls coinduction (see [13]):
P r o p o s i t i o n 8.7.1: L e t H b e a s u b g r o u p of t h e g r o u p G, a n d B b e a G r e e n f u n c t o r for H. then: 9 T h e f u n c t o r I n d ~ B is a G r e e n f u n c t o r for G. I f K is a s u b g r o u p of
G, t h e n there is an isomorphism of rings (with unit)
(In4B)(I,') =
[I
xEH\GII(
B(H n ~I~)
9 T h e f u n c t o r B H IndCHB, f r o m t h e c a t e g o r y Green(H) of G r e e n funet o r s for H, t o Green(G), is r i g h t a d j o i n t to the f u n c t o r A ~ Res~A f r o m Green(G) to Green(H). P r o o f : The formula giving ( I n d ~ B ) ( K ) is well known for Mackey functors. point is that is is still true for Green functors. By definition of induction (Ind~B)(h')
=
B
(Res~(G/K))
But
Res~(G/K) ~_
I_I H/(H ~eHW/I<
n
~tr
The
8.7. EXAMPLE: INDUCTION AND RESTRICTION
181
the isomorphism (from right to left) mapping h(H N ~:I'() E H / ( H N ~K) to hxK E Thus if a and a' are in A(K) = A ( G / K ) = B ( V oa (G/K)), setting F = G / K , and denoting by ,v,, the product "." on A, I have
~
\(~,~)(~,~)/
,
But I know that the ring ( B ( V oa F), .) is isomorphic to the direct product of rings (B(a3), .) for the various orbits w of H on V oG P. The first assertion follows. Similarly, the second assertion is clear for Mackey functors. I must show that the adjunction passes down to Green functors. If A is a Green functor for G, if B is a Green functor for H, and (I) is a morphism from A to Ind/~B, then for any subgroup K of G, I have a morphism
~,~-: A(K)
,
@
B(HNXK)
xEH\G/K
Say that r is a morphism of Green functors means exactly by the previous remarks that qSK,x is a morphism of rings (with unit) fi'om (A(K), .v) to ( B ( H n xK), .), for any x. But if L is a subgroup of H, the morphism q~L from A(L) = R e s t ( L ) to B(L) associated to (I) by adjunction is (I)L,1. So q? is a unitary morphism of Green functors. Conversely, if 9 is a unitary morphism of Green functors from ResgA to B, then for any subgroup L of G, I have a morphism q'L of rings (with unit) from A(L) to B(L). The morphism ~ associated to 9 by adjunction is then defined on the subgroup K of G by x
K
a
~K,::(a) = q~Hn~h'( rH,~nK( )) It is the product of three morphisms of rings (with unit). So it is a morphism of rings with unit, and gP is a unitary morphism of Green functors. This proves the proposition. 9 The adjunction property shows in particular that if A is a Green functor for G, then there is a unitary morphism ~a of Green functors, adjoint to the identity of Res~A r/A :A --+ IndGHReSaHA But a G = (A o U) o V = A o (U OH V ) IndHReSHA
It is easy to see that U OH V ~_ (G XH G), which is the quotient of G x G by the right action of H given by (gx,g2)h = (gxh, h-~g2). On the other hand, if I denote by I the set G, viewed as a "identity " G-set-G, the functor A is equal to A o I, and the morphism rlA comes from the morphism
f : U OH V ~ I
(gl,g2)H ~-~ 9t-92 E I
This morphism is injective on the right orbits: indeed, if f((gl,g2)g.g),
,, .~ f ( ( g l , g 2 g )
~ = g l g 2 g ~- f ( ( g l , g 2 ) g )
= gig2
182
C H A P T E R 8.
COMPOSfTION
then 91.92.9 = 91 .g2
which proves that g = 1. Now if M is an A-module, I know that Res~M is a Res~A module. Conversely, if N is a Res~A-module, then Ind~N is a Ind~Res~A-module, so an A-module by the morphism flA. The proposition 8.6.1 shows that those correspondences are functorial. Moreover P r o p o s i t i o n 8.7.2: Let A be a G r e e n f u n c t o r for the group G, and H be a s u b g r o u p of G. T h e f u n c t o r M H ReSaHM f r o m A-Mod to ResaHA-Mod is left and right adjoint to the f u n c t o r N H Ind~N from Res~A-Mod to A-Mod. Proof: Here again, this property is well known for Mackey functors. All I have to do is to keep track of the adjunction procedure: if M is an A-module, if N is a Res~Amodule, and f is a morphism of nes A modulesfrom R e s ~ M to N, then I n d ~ f is a morphism of Ind~Res/~A-modules from Ind~Res~M to Ind~N. Thus it is also a morphism of A modules from Ind~Res~M to Ind~N. The adjoint of f is obtained by composing this morphism with the morphism from M to IndGHResaM, which is a morphism of A-modules. So the adjunction at the level of Mackey functors maps morphisms of Res~A modules to morphisms of A-modules. Conversely, if g is a morphism of A-modules from M to Ind~N, then ResaHg is a morphism of Res~A-modules fi'om Res~M to Res~Ind~N. The adjoint of g is obtained by composing this morphism with the morphism p from Res/~IndaHN to N, which is a morphism of ResaHA-modules: it follows indeed from the morphism of H-sets-H @:hEHH(h,1) EVoGU which is injective: indeed, if hi and h2 in H are such that (hi, 1) = (h2, 1) in V oa U, then there exists g r G such that h2 = h~g and 1 = g.1. Sog = 1 and hi = h~. If B = Res~A, the morphism p is then a morphism of Res~Ind~B-modules from Res~Ind~N to N. As Res~Ind~B = Res~Ind~/Res~A I have the morphism ResaT1/from Res~A to Resalnd~Res~A, and the composite aes,~A aes~qA Res~Ind~Res~A
Be)Res~A
is the identity of B, because it expresses the adjunction of Res~ and Ind~ at the level of Mackey functors. So through Res~rlA I recover the same Res~A-module structure on N, and this proves that p is a morphism of Res~A-modules. Now the adjunction at the level of Mackey functors maps morphisms of A-modules to morphisms of Res~A-modules, and this proves that the functor Rest/is left adjoint to the functor Ind,. A similar argument. shows that it is also right adjoint. This completes the proof of the proposition. 9
Chapter 9 Adjoint constructions Notations: If G and H are groups, and L is a subgroup of H • G, I denote by p~(L) the projection from L to H, and p2(L) its projection onto G. I denote by kl(L) (resp. k2( L ) ) the normal subgroup of p~( L ) (resp. of p2( L )) formed of elements h E H (11esp. elements9 E G) such that (h, 1) E L (resp, such that (1,9) C L). The groups pl(L)/kl (L) and p2(L)/I~2(L) are canonically isomorphic, and 1 denote by q(L) this quotient. I denote by (H x G)/L the set of left classes of L in H • G~ viewed as an H-set-G
by h.(~,v)L.g : (h~,g-~v)L With those notations, I have shown ([2] Lemme 2) that if U : (H x G)/L, and if X is a G-set, then u 9 ~(C/ ,~, a , ~ ,)~ ( L ) U oa X _~ l ~. d p~(L)mIpdL)/k~(L)rL[lteSp2(5)A
denoting by TL the transport by the canonical isomorphismp~(L)/kl(L) ~_ p~(L)/k2(L)
of a p~(L)/k~(L)-set. It follows (see [2] 4.I.2) that if M is a Mackey functor for H, then M o U = Ind~(L)Inf~:(L{ M ( . / ka(L) OLj: ; ( (L) ) / k.. rL, . ResH,L' . . . where OL denotes the transport by isomorphism of Mackey %nctors for p~(L)/kl(L) to Mackey functors for p2(L)/k2(L). As any H-set-G U is a disjoint union of transitive H-sets-G, i.e. of the form (H • G)/L, it follows that the functor M o U is a direct sum of fnnctors composed of restriction, inflation, coinflation, and induction. I already know that the functors of restriction and inflation are mutual left and right adjoints. Thhvenaz and Webb (see [14] 5.1, [15] 2.) have built left and right adjoints for inflation, and they also mention ([15] 5.4 and 5.6) that coinflation has a left adjoint (and a right one if the ground ring is a field). It follows that the functors M ~-~ M o U always have a left adjoint. I will show that they always have a left and right adjoint, and describe their structure in terms of G-sets.
9.1
A l e f t a d j o i n t t o t h e f u n c t o r Z ~-~ U OH Z
Let U be a fixed G-set-H. [ denote by u ~ uH the projection from U to U/ H. If (Y, f) is a G-set over U/ H, I denote by U.Y the pull-back product of U and
Notation:
C H A P T E R 9. A D J O I N T C O N S T R U C T I O N S
184
Y over U / H , defined by U.Y = {(u,y) E U • Y I f ( Y ) = uH} I view U.Y as a G-set-H: i f g 6 G and h 6 H, then g.(u,y).h = (guh,gy) If (u, y) E U.Y, I denote by G(u, y) its orbit under G. I denote by G \ U . Y the set of orbits of G on U.Y, viewed as an H-set by h.a(u, y) = G ( u h - ' , y) This construction is functorial in Y: if a : (Y, f ) ---, (Y', f ' ) is a morphism of G-sets over U / H , then the map
a \ U . ~ : G i u . Y -~ G \ U . Y ' defined by is a morphism of H-sets. Conversely, if Z is an H-set, and Pz is the unique morphism from Z to *, then U o H p z is a morphism from UOHZ to U O H . = U/H. Thus Z ~ U o H Z is a functor from H - s e t to G-set~u/H. Actually: Proposition
9.1.1: T h e f u n c t o r ( Y , f ) ~ G \ U . Y f r o m G - s e t l u / H t o H - s e t is
l e f t a d j o i n t t o t h e f u n e t o r Z ~-~ U OH Z.
P r o o f : Let (Y, f ) be a G-set over U / H , and Z be an H-set. If r is a morphism of Gsets over U / H from (Y, f ) to (U OH Z, U OH PZ), and if (u, y) E U.Y, then f ( y ) = u Y . Moreover, as
(U o g Pz)r
= f(Y) = u g
there exists h E H and z E Z such that r = (uh, z). If h' E H and z' C Z are such that r = (uh', z'), then there exists h" E H such that
uh = uh'h"
z' = h" z
In those conditions uh = 4h.h-lh'h '', and as (4h, z) E U OH Z, it follows t h a t h - l h ' h ' z = z, hence that hz = h'h"z = h'z'. In particular the element hz C Z is well defined by the condition r = (uh, z) = (u, hz). I can then define O(u, y) C Z by the condition r
= (u,0(u,y))
Then as g(u, y) = (gu, gy) for g E G, if z = O(u, y), I have
r
= gr
= g(4, z) = (g4, z)
which proves that O(gu, gy) = O(u,y). So 0 is a map from G \ U . Y to Z. As moreover
h.G(~, y) = (~h-~, y), and as r
= (~,z) = (4h-l,~z)
9.1. A L E F T A D J O I N T TO THE F U N C T O R Z ~ U OH Z
185
I have O(h.G(u,y)) = hz = hO(u,y). So 0 is a morphism of H-sets from G \ U . Y to Z. Conversely, if 0 is a morphism of H-sets from G \ U . Y to Z, and if y C Y, let u E U such that f ( y ) = uH. Then (u,y) E U.Y. In those conditions, if z = O(G(u,y)), the element (u, z) is in U OH Z: indeed if h C H is such that uh = u, then h -1.u(u, y) = G(m~, y) = G(u, y) thus z = O(G(u,y)) = hO(G(u,y)) = hz. The element (u,z)does not depend on the choice of u such that f ( y ) = uH: if I change u to uh, for h E H, then
O(a(uh, y)) --
h-10(G(~,y))
and (u, z) = (uh, h-lz). I can then set r as O(G(gu,gy)) : OiG(u,y)) : z, I have
r
= (r
:
h-lz
= (4, z). As f(gy) = gI(Y) = gull, and
z) = .q(u, z) = gr
Thus r is a morphism of G-sets from Y to U OH Z. The correspondences r ~ 0 and 0 ~ r are inverse to each other: indeed, the equality defines 0 from r and r from 0. The proposition will then follow, if I know that the unit and co-unit of this adjunction are functorial:
N o t a t i o n : If (Y, f) is a G-set over U/H, I will denote by ~(y,S) (or ~,y if f is clear from context) the unit of the previous adjunction: it is the morphism
t,(yd) : Y ~ U OH (G\U.Y)
defined by ~(~,S)(Y)= (~, U(~, y)), if /(y)= ~H. If Z is an H-set, I will denote by 71z the co-unit of this adjunction: it is the morphism rlz : G\U.(U oH Z) ~ Z defined by 7jz(G(u',(u,z))) = h - l z i f h 9 H is such that u ' = uh. T h e n , is functorial: if c~: (Y, f) ~ (Z, g) is a morphism of G-sets over U/H, and if (u, y) 9 U.Y, then "(yd)(Y) = (u, G(u, y)). Thus
But gc~(y) = f(y) = uH, so t,(zm)c~(y) = (u, G(u, c~(y))), which shows that
.(~,.)~
: (u o. (GXU..))~(~,j)
S i m i l a r l y , is functorial: if f : Z ~ T is a morphism of H-sets, and if (u', (u, z)) E
U.(UoHZ),thenthereexistsh 9 H s u c h t h a t u ' = u h . As o,,
=
Then,z(G(u',(u,z)))
=h-lz.
CHAPTER 9. A D J O I N T CONSTRUCTIONS
186 I have
[G\u (u o. i ) ( G ( <
:
i
SO
,T (G\U.(U OH f ) ) = f~lz
9.2
The categories ~Du(X)
Definition: If X is a G-set, let 39u(X) be the following category: 9 The objects of D u ( X ) are the G-sets over X • (U/H). If ( Y , I ) is an object of 2)u(X), I denote by fu the map from Y to U/H obtained by composing f with the projection from X • (U/H) on U/H, and f x the composition product of f with the projection on X. Then (Y, fu) is a G-set over U/H, to which I associate the pull-back product U.Y. 9 A morphism ce from (Y,f) to (Z,g) is a mol'phism of G-sets from Y to Z such that g o c~ = f , and such that the morphism U.o from U.Y to U.Z associated to a is injective on each left orbit under G, i.e. such that (u, y) ~ u.Y, g ~ G, v~ = ~, ~(gy) = ~(y) ~
gy = y
R e m a r k : In particular, if a is injective, then U.c~ is injective on each left orbit. It is clear that ~Du(X) is a category: the product of two morphisms which are injective on the left orbits is injective on the left orbits, and the identity morphism is injective on the left orbits. If r : X --+ X ' is a morphism of G-sets, let r be the morphism r x Idu/H fi'om X • (U/H) to X ' • (U/H). There is an obvious functor from 7)u(X) to D u(X'), which maps (Y, f) to (Y, e l ) , and the morphism o~: (II, f ) + (Z,g) to a : (Y, Cf) -+ (Z, q~g). I will denote this functor by 7)u,,(r Similarly, there is a functor from Z)u(X') to :Du(X), defined by inverse image along r if (Y', f ' ) is a G-set over X ' • (U/H), let Y, f and a be such that the square a
Y
,
I,"
x • (U/H)
,
x ' • (U/H)
1
is cartesian. If a ' : ( Y ' , f ' ) ---+(Z',g') is a morphism of G-sets over X ' • (U/H), let Z, g, and b such that the square
Z
X x (U/H)
b
~
Z'
, X' x (U/H)
9.2.
THE CATEGORIES
Du(X)
187
is cartesian. Then as
g%'a = f ' a = C f there exists a unique morphism ~ : Y --, Z such that the diagram y
a
y,
Z
X x(U/H)
b
~)
/
9 Z'
, X'•
is commutative. Moreover, in those conditions, the square a
Y O~
, y'
1
l
Z
Ol
~ Z' b
is cartesian. Indeed, if z E Z and y' E Y' are such that b(z) = ~'(y'), then
g ' b ( z ) = Cg(z) = g ' a ' ( y ' ) = f ' ( y ' ) Then by definition of Y, there exists a unique y E Y such that
g(z) = f ( y )
y ' = a(y)
Moreover
ha(y) = c/a(y) = ce'(y') = b(z)
gc~(y) = f ( y ) = g(z)
Then z and a ( y ) have the same image under b and g, so z = c~(y). And if another element yl E Y is such that c~(yl) = a ( y ) and a(yl) = a(y), then
goe(yl) = gc~(y) = f ( y l ) = f ( y ) and unicity of y implies yl Y. Moreover, the morphism U.a is injective on the left orbits: =
L e m m a 9.2.1: L e t a
Y
, y'
Z
~ Z'
ol
1
b
b e a c a r t e s i a n s q u a r e o f G-sets over U/H. If U.a' is i n j e c t i v e o n each left o r b i t o f G on U.Y', t h e n U.a is i n j e c t i v e on e a c h left o r b i t o f G on U.Y
C H A P T E R 9. A D J O I N T C O N S T R U C T I O N S
188
P r o o f i Indeed, if ( u , y ) r U.Y and (gu,gy) have the same image under U.a, then gu = u and a(gy) = a(y). Taking the image under b of this relation gives
ba(gy) = a'a(gy) = ba(y) = a'a(y) so a ' ( g a ( y ) ) = a ' ( a ( y ) ) . As U.a' is injective on the left orbits, it follows that ga(y) = a(gy) = a(y). As moreover a(gy) = a(y), and as the square Y, Y', Z, Z' is cartesian, I have gy = y. .. It follows that if I set D~(r from 7 ) u ( X ' ) to D u ( X ) .
9.3
f ' ) = (Y, f ) and D ~ ( r
= a, I get a functor
Qu(M)
The functors
D e f i n i t i o n s : I will say that an H-set ( Z, a) over G \ U.Y is v(vj)-disjoint (or v-disjoint for short) if the square 0
,
Y
1
~ UoH(G\U.Y)
1
UOHZ UOHa is cartesian, or equivalently if
(U oH a)(U OH Z) n uv(Y) = If M is a Mackey functor for H, and (Y, f ) is a G-set over U / H , I set Qu(M)(Y, f) = M(G\U.Y)/
~ M.(a)M(Z) (z,~)
where the sum runs on the H-sets ( Z, a) over G k U . Y which are v-disjoint. R e m a r k : Say that (Z, a) is not v-disjoint means that there exists z C Z and y' ff Y, such that if u ' H = f u ( y ' ) and a(z) = G ( u , y ) , then there exists v C U such that (v,z) G U o H Z a n d
(v,
This means that there exists h E H such that v = u~h and
G(u, y) = h-lG(~ ', v') = G(u'h, y') So there exists g E G such that
uth = gu
y' = gy
Finally, there exists g E G and h E H such that
v = gu
u' = guh -1
y~ = gy
9.3. THE FUNCTORS Qu(M)
189
Now (v, z) : (g u, z) belongs to U OH Z if and only if (u, z) E U OH Z does. Conversely, if a(z) = G(u, y), and if (u, z) E U OH Z, then (u, y) E U.Y, and
vy(y) = (u, G(u,y)) = (U OH a)(u,z) Thus (Z, a) is v-disjoint if and only if g e e Z, a(z) = G(u,y) ~ (u,z) ~ U oH Z If X is a G-set, and Y is an object of Du(X), I set
Qu(M)(k] f) = Qu(M)(fu) Lemma
9.3.1:
L e t X b e a G-set. T h e c o r r e s p o n d e n c e
(Y, f) H Qu(M)(Y, f) a: ( Y , f ) + (Z,g)
~
M.(G\U.a): M(G\U.Y) ~ M(G\U.Z)
induces a functor from ~u(X) to R-Mod. I must prove that if a is a morphism in ?)u(X) from (Y,f) to (g,g), if (T,a) is a v-disjoint H-set over G\U.Y, and if m E M(T), then the image of M.(G\g.cOM.(a)(m ) in Qu(M)(Z, gu) is zero. This will be the case in particular if, denoting by /3 the morphism G\U.e, the H-set (T,/3a) is v-disjoint over G\U.Z.
Proof:
This is equivalent to say that the square 0
,
1
u oH T
Z
1
, U oH (3~)
(c)
u ou ( o \ u . z )
is cartesian. But this square is composed of the two squares O~
0
Y
~ Z
1
U OH (G\U.Y)
1
~ U on (G\U.Z)
U OH T
U OH a
1
U OH
The left square is cartesian if (T,c~) is v-disjoint. Thus (C) is cartesian, since the right square is cartesian by the following lemma: 9.3.2: L e t a : (Y, fu) --+ (Z, gu) b e a m o r p h i s m of G-sets o v e r U/H, such t h a t U.a is i n j e e t i v e on each left o r b i t of G on U.Y. T h e n t h e s q u a r e
Lemma
C~
y v"l
U OH (G\U.~") is c a r t e s i a n .
,
Z I vz
, U OH (G\U.Z) U oH (G\U.~)
CHAPTER 9. ADJOINT CONSTRUCTIONS
190
P r o o f : Let ('u,, G(u2, y)) E U OH (G\U.Y) and z E Z such that
If v 9 U is such that vH = gu(z), then uz(z) = (v,G(v,z)l. means that there exists h 9 H such that /
k
ulh = v
The above equality
h-lG(u2, e(y)) = c(,, ~) = c ( ~ h , ~ ( y ) )
So there exists x 9 G such that v = xu~h
z = xa(y) = a(xy)
The image of the second equality under gu gives ge(Z)
= vH
= guo~(xg)
=
fu(xy)
Then
Conversely, let yl and Y2 be elements of Y having the same image under a and uy. T h e n if u C U is such that fu(Yx) = u H , I have also
fu(Y2) = guc~(y2) = gua(y,) = fu(Y,) = uH Then uy(yl)= (u,G(U, yl)) and ur(y2)= (u,G(u, ye)). These elements are equal if there exists h E H such that tt ~-- tth
h - l G ( t t , yl) = G(tt, y2) = G(tth, yl) = G(tt, yl)
Then there exists x E G such that u = my and Y2 = my1. As m o r e o v e r c~(y2) = c~(yl) , and as U.c~ is injective on the left orbits of G on U.Y, it follows that Y2 = Yl, which proves the lemma. 9 Let r : X --+ X ' be a morphism of G-sets. If ( K f ) is an object of 19u(X), I denote by (Y', f ' ) = 2)u,.(r f ) = (Y, Cf) its image in Du(X'). As (r = fu, the sets U.Y and U.Y' are equal, and M(G\U.Y) identifies with M(G\U.Y'). Similarly, the u-disjoint H-sets over G\U.Y identify with u-disjoint H-sets over G\U.Y'. It follows that Qu(M)(Y,I) is naturally isomorphic to Qu(M)(Y',I'), i.e.
Qu(M) ~ Qu(M) o Du,.(r Conversely, if (Y', f ' ) is an object of Z)u(X'), I set (Y, f ) = :D~j(Y' , f ' ) . cartesian square a
Y
,
y'
x • (U/H)
,
X' • ( U / H )
I have a
9.3. THE FUNCTORS Qu(M)
191
If m E M(G\U.Y'), then M*(G\U.a)(m) E M(G\U.Y). If ( Z ' , a ' ) is a u-disjoint H-set over G\U.Y', I can fill with Z, ~, a the cartesian square Ct
Z
, G\U.Y
Z'
, G\U.Y'
I
Then (Z, (~) is a u-disjoint H-set over the square
G\U.Y: indeed, let T, 7, and 6 be such that
T
Y
U OH Z
U OH (G\U.Y)
1
U oH ol
is cartesian. I have then a commutative diagram 7
T
U oH Z
~}"
U oH ( yt
UoH3~ Uy,
U OH Z'
9
u o. ( a \ u Y ' )
U OH a'
As the bottom square is cartesian, I can fill this diagram by a morphism from T to 0, so T is empty, and (Z, c~) is u-disjoint. As M*(G\U.a)M.(c~') = M.(cjM*(fl), the image of M*(G\U.a)M.(~') in the quotient Qu(M)(Y, f) is zero. So I have built a morphism T:y, S') from Qu(M)(Y', f') to Qu(M)(Y, f). This construction is moreover functoraal in (Y, f'): if c~': (Y', f ' ) --* (Z',g') is a morphism in 7?u(X'), then the square
Qu(M)(Y',f') Qu(M)(a') 1 Qv(M)(Z',g')
Ti~,,y, ) )
)
T (Z,,g,) r
Qtr(M) o D~(r f') 1 Qv(M) o ~b(r Qu(M) 73b(C)(Z',~') o
(c)
C H A P T E R 9. A D J O I N T C O N S T R U C T I O N S
192
is commutative: indeed, I have a diagram y
a
X • (U/H)
},.-/
~
, X'•
and to show that (C) is commutative, it suffices to show that
M.(G\U.o~)M*(G\U.a) = M*(G\U.b)M.(G\U.c~') which will follow from the fact that, the square g.a
U.Y
i U.Y'
1
U.Z
,
U.Z'
U.b is cartesian, as the pull-back of a cartesian square. I will apply to this square the following lemma: L e m m a 9.3.3: Let
a
Y C~
~ y'
1
1 C~
Z
) Z' b
b e a c a r t e s i a n s q u a r e o f G-sets. If ~' is i n j e c t i v e on e a c h o r b i t o f G o n Y', then the square
c\Y G\,~ 1 G\z
, G\Y' r
)
I G\,~' G\Z'
is c a r t e s i a n . Proof: Let Gz r G \ Z and Gy' E G \ Y ' be such that (G\b)(Cz) : (C\~')(Oy')
: Cb(z) : C ~ ' ( y ' )
Then there exists an element g E G such that 9b(z) = b(gz) = c?(y'). So there exists a unique element y E Y such that
9.4.
THE FUNCTORS Lu(M)
193
In those conditions (a\a)(Gy)
: G~(~) : @ '
and moreover (G\~)(@)
= O~(~) = Ggz = Gz
Now if Gyl and Gy2 are elements of G \ Y having the same image under G \ a and G \ a , I have Ga(yl) -- Ga(y2) Ga(yl) = Ga(y2) So there exists elements g and g' of G such that ga(yl) = a(y2)
g'a(yl) = a(y2)
The image under b of this relation gives
a(yl)) Then as ~(y~) = a-~(y2), I have
As a ' is injective on the left orbits, I have g'g-la(y2) = a(y2), or
g-la(y2) = a(g-ly2) = a(yl) = 9'-la(y2) As c~(y~) = a(g'-ly2), it follows that YI = g'-iY2, and then Gy~ = Gy2, completing the proof of the lemma. " I have finally built a natural transformation T e from Q u ( M ) to Q u ( M ) o D~(r so I have proved the following lemma: L e m m a 9.3.4: I f ~p : X ~ i s o m o r p h i s m of f u n e t o r s
X'
is a m o r p h i s m
of G - s e t s , t h e n q~ i n d u c e s an
T, : Q u ( M ) --* Q u ( M ) o Z)u,,(r and a n a t u r a l t r a n s f o r m a t i o n
T~ : Q u ( M ) ~ Q u ( M ) o ~gb(~))
9.4
The
functors
s
The previous lemma leads to the following definition: D e f i n i t i o n : If X is a G-set, and M is a Mackey functor for H, I set
s
=
lira Qu(M)(M, f ) (v,f)Ov(x)
If ~ : X ~ X ' is a morphism of G-sets, then the isomorphism T~ induces a morphism lira Q ~ d M ) ( Y , f ) ~ lim Q u ( M ) o ~ g u , . ( ~ ) ( Y , f ) --* lim Qu(M)(Y',f') -------+ (v,l )e~,( x ) (v J )~u( x ) (Y',f')~u( x')
194
C H A P T E R 9. A D J O I N T C O N S T R U C T I O N S
i.e. a morphism s from s to E u ( M ) ( X ' ) . Conversely, the trans[ormat~ion T ~ induces a morphism lim Qu( M )( Y' , f )' - ~ ~lira cc-2u(M)oDu(r * (v',Y')e~)u(x ') (Y',f')evu( x')
" , f ' ) ---+ ~lim Q u ( M ) ( Y , f ) (v,De~v( x )
i.e. a morphism t;u(M)*(~) r,-om &~(M)(X') to & ; ( M ) ( X )
Proposition 9.4.1: The previous definitions turn s tot.
into a Maekey func-
Proof: It is clear that if 4) : X --+ X' and 0' : X' --+ X" are morphisms of G-sets, then s162
= s163
s162162
= 12u(M)*(r163
It is also clear that s and s are the identity morphisms. It suffices then to check axioms (M1) and (M2) for s I will start with (M2): let a
Z
---, d
T
be a cartesian square of G-sets.
Notations: Let ( E , e ) be a G-set over Y • (U/H), and m E M ( G \ U . E ) . I denote by m.(E,~) the image of rn under the composite morphism
M(G\U.~) --+ Qv(M)(E, ~) --+ &~(M)(Y) Furthermore, I denote by X ~ X the fimctor X ~-~ X • ( U / H ) from G-set to G-set.
To compute Eu(M)*(a)(rn(z,~)), I fill the cartesian square C~
The image of m(E,~) by s
F
,
E
X
------+
Y
is then
C~(M)*(a)(.~(~,~))
The image of this element under s
: M*(a\U.~)(,.)(Fj)
is
s163
= M*(G\U.a)irn)(FSf)
On the other hand, the image of m(E,e) under
]~u(M).(c) is
Z;v(M)~(c)(-~(u,~)) = m(u,~)
9.4. THE FUNCTORS Lu(M)
195
As the square o~
F
Z
,
E
,
T
J is cartesian, because it is composed of the previous one and of the cartesian square a )
l Z the image under s
,
g
T
of m(E,Te) is then
s163
=
M*(G\U.~)(~)(F,;s)
which proves that
s163 So s
=s
satisfies (M2).
To check axiom ( M 1 ) , I consider two G-sets X and Y, and their disjoint union Z = X LI Y. As 2 = X H,Y, it is clear that a G-set (E, e) over 2 is the disjoint union of a G-set (El, el) over X and a G-set (E2, ~2) over r . Then G\U.E is the disjoint union of G\U.E1 and G\U.E2. Let il and i2 be the respective injections from E1 and E2 into E. If rrz E M(G\U.E), let m~ = M*(G\g.il)(rn) and m 2 = M*(O\U.i2)(m). Then
m = M.(G\U.i,)(ml) + M.(G\U.i2)(m2) I can view this equality in s as m(u,r : M.(G\U.it)(rnj(u,~) + M.(a\u.ij(m~)(u,~) But as i1 is injective, it is a morphism in 2)u(Z) from (E,, eh) to (E, e), and in I have
s
a,(G\U, il)(rrq)(E,e ) : (TI%I)(EI,eil) Now if ix is the injection from X to Z, I have (rftl)(El,eh)
:
s
The same argument for is shows then that, denoting by iv the injection from Y to Z, I have /
%
s
Cu(M).(,r))
M).(iy )((mj(E~,~))
In particular, the map (s from Cu(M)(X) * C~(M)(Y) to s is surjective. The following lemma now shows that axiom (M1) holds for s and this completes the proof of the proposition. 9
CHAPTER 9. ADJOINT CONSTRUCTIONS
196
L e m m a 9.4.2: L e t L b e a b i f l m c t o r on G - s e t , s a t i s f y i n g a x i o m ( M 2 ) , a n d s u c h t h a t L({3) = {0}. If for a n y G - s e t s X and Y, t h e m a p
(L.(ix), L.(iy)) : L(X) 9 L(Y) ---+L(X I_I Y) is s u r j e c t i v e , or t h e m a p
L*(ix) | L*(iy) : L(X H Y) ~ L(X) | L(Y) i n j e c t i v e , t h e n L is a M a c k e y f u n c t o r for G. P r o o f : Let 0 be the map as the square
(L.(ix),n.(iy)).
1
Y
As L satisfies (M2), as L(0) = {0}, and
1
, XHY
iy
is cartesian, the products L*(ix)L.(Q) and L*(ir)L.(ix) are zero. As ix is injective, the square
X
Id
,
X
x
, XLIY ix is cartesian. It follows that L*(ix)L.(ix) is the identity of L(X), and that L*(iz)L.(iv) is the identity of L(Y). Now the map r = L*(ix) | L*(iy) is such that r = IdL(x)eL(Y). Then 0r = 0, and r162 = 0- If 0 is surjective (resp. if r is injective), the first (resp. the second) of these equalities implies 0r = IdL(xHz), so (M1) holds. This proves the lemma.
9.5
9
Left adjunction
If 0 : M ~ M' is a morphism of Mackey functors for the group H, and if X is a G-set, I de~ne a map s from s to s by setting', for an object, (Y, f) of:Du(X), and a~ Cement m E M(G\U.Y)
Notation:
s
= Oc\uY(~)(rj)
L e m m a 9.5.1: T h i s d e f i n i t i o n t u r n s tors from s to s P r o o f : First, the map s
s
s
i n t o a m o r p h i s m o f M a c k e y func-
is well defined: if a : Z ~ G\U.Y is u-disjoint, then
(M.(G\U.a)(m)(z,f)) = OG,\uyM.(a)(m)(zd) = (M.(a)Oz(m))(v,/) = 0
Moreover, if a : (Y,f) ~ (Z,g) is a morphism in Du(X), then
c . ( o ) ~ M.(~)(.~(~,j~) = ( O ~ w . Y . ( G \ U . ~ ) ( . ~ ) ) ( a ~ )
....
9.5. LEFT ADJUNCTION
197
Now let r : X ~ X ~ be a morphism of G-sets. Then
Z_u(M).(r
= s162 ....
....
OG\U.y(m)(y,~f)
= .~u(O)x,(m(y'~.f))
....
z:u (0)~,~:~ (M).(r
. . . .
If moreover (Y', f ' ) is an object of :Du(X'), and if m' ~ M(G\U.Y'), then let Y, f, and a filling the cartesian square a
Y
,
y'
X
~ XI
With those notations, I have
s
9
'
(r163
' Oa\u.v,(m )(y,,f,) ) . . . .
=s
....
s
(M*(G\V.a)(m')(yj)) = s
which proves the lemma.
.,
T h e o r e m 9 . 5 . 2 : L e t G a n d H he finite g r o u p s , a n d U b e a G - s e t - H . correspondence
The
V ~ Cu(M) 0 E Homu~,ck(u)(M, M') ~ s
C Homuack(a)(s
s
is a f u n c t o r f r o m Mack(H) t o Mack(G), w h i c h is left a d j o i n t t o t h e f u n c t o r
NHNoU. P r o o f i It is clear that the correspondence M ~-~ 12u(M) is functorial in M. The main point is the adjunction property. So let M be a Mackeyfunctor for H, and N be a Mackeyfunctor for G. I f 0 i s a morphism of Mackey functors from M to N o U, then for any H-set Z, I have a morphism
Oz : M(Z) --, (W o U)(Z) = N(U OH Z) In particular, if (Y, f) is a G-set over x • (U/H), I have a morphism
o~\~.y : M(a\U.V) ~ N ( u o,~ (G\U.V)) Composing this morphism with N*(ug) gives
N*(uy)Oa\v.v : M(G\U.Y) ~ N(U cu (G\U.Y)) ~ N(Y)
CHAPTER 9. ADJOINT CONSTRUCTIONS
198
If (Z, a) is a u-disjoint H-set over G\U.Y, I have the diagram OZ
M(Z)
M(a\U.Y)
,
N(U OH Z)
,
N(O)
N(U oH(G\U.Y))
)
N(Y)
This diagram is commutative: the left square is because 0 is a morphism of Mackey functors, and the right square is because (Z, a) is r-disjoint. It follows a morphism from Qu(M)(Y,f) to N(Y). Now I can compose this morphism with the morphism N.(fx) : N(Y) ---, N(X) deduced from f. If c~: (Y,f) ~ (Z,g) is a morphism in 7?u(X), I have the following diagram
M(G\U.Y)
M(G\U.Z)
oG\v.y
N*(.y)
N.(fx)
Oa\u.z' N(UoH(G\U.Z))
N*(uz)) N(Z)
X.(gx)'
, N(UoH(G\U.Y))
, N(Y)
N(X)
N(X)
The left square is commutative because 0 is a morphism of Mackey functors. middle square is commutative by lemma 9.3.2 because N is a Mackey functor. right square is commutative because gc~ = f. It follows a morphism ~bx from s to N(X). Now if r : X ~ X ' morphism of G-sets, if (I/, f) is a G-set over X • (U/H), and if m E M(G\U.Y), by definition of ~bx
The The is a then
Cx(m(y,j)) = N.(fx )N*(~'r )Oa\v.y(m) On the other hand
s
M).( r )(m(y,])) = m(g,-~])
SO
Cx,s162
= N.(r
= N.(r
and then r 1 6 3 1 6 2 = N.(C)r Conversely, if (Y', f ' ) is a G-set over X ' x (U/H), and if m' C M(G\U.Y'), then the image of mlg,,], ) under L:u(M)*(r is obtained by filling the cartesian square a
Y
, y'
X
, X'
si
?
More precisely, I have then
s162
= M*(G\U.a)(m')(y,])
9.5. LEFT ADJUNCTION
199
The image under ~bx of this element is by definition
Cx Cu( M)*( r )(m{y,,f,)) = N.( f x )N*( uy )Oa\u.z M*( G\ U.a )(rn')
(9.1)
As 0 is a morphism of Mackey functors, I have
Oa\u.zV*(G\U.a) = X* (U OH (G\U.a))Oa\u.z, Moreover, as the square
a
Y
y! l l/y1 + U oh, (G\U.Y')
u o, (a\u.z)
U oi~ (G\U.a) is commutative, I have
N*(uy)N*(U OH (G\U.a)) = N*(a)N*(uz,) Furthermore, it is easy to see that the square a
Y
~ y'
X
~ X'
is cartesian, and then N.(fx)N*(a) = N*(r (9.1) as
). Finally, I can write equation
= N "(d~)N.(f'x,)N ' *(uy,)Oa\u.z,(m)'
~xs162 On the other hand
Cx,(m(y,j,)) = W.(,f'~,)N'(~y,)Oawy,(,~') It follows that Oxf-.u(M)*(r = N*(r and this shows that the morphisms Cx define a morphism r of Mackey functors from Eu(M) to N. Conversely, let ~b be a morphism of Mackey functors from s to N. Then for any G-set X, I have a morphism ~bx from f-.u(M)(X) to N(X). In particular, if Z is an H-set, I have a morphism ~uoHz from f-.u(M)(U OH Z) to N(U OH Z ) = (N o U)(Z). B u t if m c M(Z), t h e n M*(vz)(m) e M(G\U.(U on Z)). Moreover, setting rrz(u,z) = ((u,z),uH), I define a morphism of G-sets fez from U OH Z to (U OH Z) • (U/H), and then U OH Z is an object of Du(U OH Z). In particular, I can consider the element
Az(m) = M*(~z)(m)(vo.z,,~z) E f-.u(M)(U OH Z) Finally, I have the composite morphism
Oz : M(Z)
)~z , s
on Z) CUo.Z N(U OH Z)
CHAPTER 9. ADJOINT CONSTRUCTIONS
200
If r is a morphism of H-sets from Z to Z', and if m' C M(Z'), then
OzM*(r
= CUo.zAzM*(r
Moreover
= M*(rlz)M*(r
AzM'(r
It is clear moreover that the square
G\u.(u o. z)
-
~z
~ z
[r
G\U.(U ~ r 1 G\U.(U
o.
z')
~Z'
, z'
is commutative. So I have
~M*(r
: M" (O\U.(U o. r
The square
UoHZ
UOHr , UoHZ'
UOHZ
UOH;
UoHZ'
is trivially cartesian, so the square
UoHZ
Uour
, UOHZ'
UOHZ
~ UoHZ' UoHr
is also cartesian. It follows that
: ~u(M)*(Uo.~)(M*(,~,)(-~')(~o.~, ~,)) :.
M*(C\U.(Vo.~))M*(,7~,)(.;)(~o.Z,.~)
....
s
on r
Finally, I have
OzM* ( r
= r163
M)* ( U oH r )A z,(,~')
and as r is a morphism of Mackey functors, it is also
OzM*( r )(m') = (N o U)*( r
AZ,(m') = (N o U)*( r )Oz,(m')
Thus OzM*(r = (N o U)*(r The previous proof shows that AzM*(8) = s
OH r
9.5.
201
LEFT ADJUNCTION
I will now prove that Az, M , ( r
oH r
= s
so that A will be a morphism of Mackey functors from M to s e U. Then 0 will be composed of A and of the morphism r o U from s o U to N o U. It will also be a morphism of Mackey functors. First I fill the cartesian square
(2 YI
a \ u . ( u oH z')
~
~lZ'
Z
, z'
and I denote by i the morphism from G \ U . ( U OH Z ) to II filling the commutative diagram
c \ u . ( u o . z)
o
Z
a \ u . ( u o. z')
rlz,
Z'
L e m m a 9.5.3: I n t h e p r e v i o u s d i a g r a m : 9 The morphism
i is i n j e c t i v e .
9 I f m o r e o v e r II = Im(i)]_I H', t h e n IY is a u - d i s j o i n t H - s e t o v e r t h e s e t
G\U.(U oH z').
Proo~ Let G(~', (~,~))and ~(~, (~,,z,))be two elements having the same image under i. If u' = uh and u~ = u l h l , this means that
G(ut,(?~,r
: G(~i, (ul, r
)
h-lz : hllzl
Then Zl = h l h - l z , and there exists g ~ G such that ( g u , r gu' = u't. So there exists h' C H such that gu = u , h ' and r
conditions
a(~'l, (~l,Zl)) = a(g,,', (~,,,,h-l~/) But
....
u l h ' h h l I = g U h h l 1 = gU'hl ~ = u lt h 1--1 = ul
= (ul,r = h'-lr
and In those
CHAPTER
202
9. A D J O I N T
CONSTRUCTIONS
As ('gl, Zl) ~ U OH Z, it follows that h'hh~lZl = z1, i.e. h'z = h l h - l z .
G('u/1, (Ul,Zl))
----
~'(lt", (~]),"-1, htz))
----
Then
G(/s (/s
and this proves that i is injective. To prove the second assertion of the lemma, I must prove that the square 0
U O H Z'
~
1
[VUo~Z, \
/
U o . n'
,
u o. (a\u.(v \
--
-
o. z'))
UoHb is cartesian. This means that if (u", 7r) 6 U OH II and (ul, z~) 6 U OH Z' are such that
(U OH b)(u', 71-) = //UoHZ,('U.1, Z'I)
(9.2)
then ~r is in the image of i. The element ~r is of the form
with (u,z') e U O H Z ' , and holz ' = r and then
G(u',(u,z')),
if ho e H and who = u'.
Now b0r ) =
(u o. On the other hand
'))]
= [~I,G ( "t/1,(Ul,Z 1
1]UoHZ,('ts
Now equality (9.2) means that there exists h 6 H such that
~" = ?.tlh
h-IG(ul,(tll,Z'I))
= G(uih,(Ul,Z;)):
G(?-t", (l/,,zt))
Then there exists g 6 G such that (gu~,z'l) = (u,z') and gulh = u'. Finally, there exists g 6 G, and h, h' 6 H such that
'a" : 'U1. ]I
gulh = u'
gulh' = u
h'-lz'l = z'
In those conditions
71"= [G(ut,(?~,Z")),Z] =
[G(gulh,(gll, lh",hor
] ....
.... [G('u.lh,(ulht, ho~(z))),z] : [G(u',(u'h-lht, ho~(z))),z] .... .... Furthermore, as (u",Tr) E U OH II, it follows that if hi 6 H is such that u"hl = u", then hl~r = % which implies in particular that hlz = z. So (u", z) E U o H Z . Moreover
u " h - l h ' h o = ulh'ho = g-luho = g - l u ' = ulh = u"
9.5. LEFT ADJUNCTION It follows that h-~h'hor
203
=r
= r
and then
which proves the lemma. To prove the equality Az, M.(~b) = s
o . ~)Az
I choose rn E M(Z). Then
Then U oH r is a morphism of G-sets over U oH Z' from (U OH Z , ( U oH r to (U OH Z', ~rz,). Moreover U.(U OH 5) is injective on the left orbits of G on U.(U OH Z): indeed, if (u,(u',z)) has the same image than g.(u,(u',z)) = (gu,(gu',z)), then gu = u and there exists h E H such that gu' = u'h and hS(z) = r But there exists h0 such that u = u'ho. As gu = u, I have also gu' = u', and then (gu, (gu', z)) = As Uo/4~ is a morphism in Z)u(UoHZ'), I have the following equality in f~u(M)(Uos
Z') M*(~z)(m)(uo.z,(uT-s
= M.(U oH r
Moreover
M.(U oH r
= M.(b)M.(i)M*(i)M*(a) . . . .
. . . . M.(b)M*(a) + M.(b) (1 - M.(i)M*(i))M*(a) If j denotes the injection from II' into l-I, I have 1 - M.(i)i*(i) = M.(j)M*(j), and M.(b) (1 - M.(i)M*(i))M'(a) = M.(bj)i*(aj) But ( I I ' , b j ) is a v-disjoint G-set over G\U.(U OH Z'). Qu(M)(U OH Z') is then zero. It follows that
The image of M.(bj) in
M.(U OH ~)M*(~z)(m)(uo.z,,~z,) = M.(b)M*(a)(m)(VosZ',~z,) . . . . ....
M*(~z,)M.(d))(m)(UonZ,,.z,) = Az, M . ( r
proving that Az,/.(r
= s
oH ~b)Az
so A is a morphism of Mackey functors from M to f~u(M) o U. Now I have a correspondence A : 0 ~
HomM~ck(H)(M, N o U) ~ HOmM~ck(a)(s
N)
CHAPTER 9. ADJOINT CONSTRUCTIONS
204
and a correspondence B in the other direction. It is easy to see that those constructions are functorial in M and N. The theorem will now follow, if I prove that they are inverse to each other. It is equivalent to check the relations on unit and co-unit (see Mac-Lane [10] chp. IV): if 0 is the identity of N o U, I must check that B o A(O) = O. Similarly, if ~b is the identity of s I must check that A o B(~b) = ~b. But A(O) is the morphism from s o U) to N defined on a G-set X by
m(yd) E s
o U)(X) ~ N.(fx)N*(uv)Oa\u.v(m) E N ( X )
where (!/, f ) is a G - s e t over x- , and m is an element of M ( G \ U . Y ). A(O)x is defined by m(v,S) C s
In other words
o U)(X) ~-+ X , ( f x ) N * ( u z ) ( m ) 9 N ( X )
Then B o A(O) is the endomorphism of N o U defined on an H-set Z by
,~ 9 (N o U)(Z) H A(O)uo,,Z~z(,j Moreover I have here Az(n) = (N o U)*(~lz)(n)(uo,z.,z) = N*(U OH •z)(n)(Uo,Z.,z) So
A(O)UogAz(n) : N.((~Z)Uonz) N*(uUo,z)N*(U OH ,z)(n) But (lrZ)UonZ is the identity of U OH Z. And as u and ~ are the unit and co-unit of the adjunction of Z ~-~ U OH Z and Y H G\U.Y, I have
(U OH TIz)IIUoHZ= IduoHz So B o A(O) is the identity of N o U. Conversely, if ~b is the identity of s s o U defined on the H-set Z by
then B(~b) is the morphism from M to
m 9 M(Z) ~
Iz(m)
The endomorphism A o B(~b) of s is then defined for a G set X , a set (Y, f ) over X , and an element ra 9 M(G\U.Y) by
re(v,]) C s
~ s
But
Aa\v.v(ra) = M*(Tlc\v.y)(m)(uon(a\u.z).,aw.r) Let e be the map from Y to Y defined by e(y) -- \(y, tidy)]. It is clear that the square
Y
l/y ,
U
oH t ~ \ u . Y )
e!
~TrG\U.Y
z
, Yy
u o,
(a\U.V)
9.6.
THE F U N C T O R S Su(M)
205
is cartesian, and then
s
= M*(G\U.uv)M*(~a\u.v)(m)(z,e)
Thus
s
)s
M)*(uz )Aa\v.y(m) : M*(G\U.uz )M*(rla\u.z)(m)(z, Txc)
But so f x e = f. Moreover rlatu.z(G\U.uy ) is the identity map, because u and r/ are the unit and co-unit of the adjunction of Z H U OH Z and Y ~ G\U.Y. Finally
s
M ) , ( f x )s
M )*( uy )AakU.Z(m ) = m(z,S)
and AoB('~b) is the identity morphism of s 9.5.2.
This completes the proof of theorem 9
Remark: Let X be a G-set. The expression of the colimit over D u ( X ) shows that if M is a Mackey functor for G, then
s
: (
M(G\U.Y))/fl
|
YLXx(UIH) where J is the submodule generated by the submodule
M.(<,)(M(z)) whenever (Z, a) is a u-disjoint H-set over G\U.Y, and by the elements
m - M,(G\U.a)(m) whenever .~ ~ M ( G \ U . Y ) and ~ : (Y, f) ~ (Y', f') is a morphism of G-sets over X x (U/H) which is injective on each G-orbit. 9.6
The
functors
Su(M)
To build right adjoints, I need the following dual definition: D e f i n i t i o n : If M is a Mackey functor for H, and (Y, f) is a G-set over Y / H , I set
S u ( M ) ( Y , f ) = ~ KerM*(a) (Z,a)
where the intersection runs over u-disjoint H-sets ( Z, a) over G\ U.Y . L e m m a 9.6.1: The correspondence
(Y, f) ~ S u ( M ) ( Y , fu) c~ : ( Y , f ) ~ (Z,g) H M * ( G \ U . a ) : M ( G \ U . Z ) ~ M ( G \ U . Y ) induces a contravariant functor from Du(X) to R-Mod.
CHAPTER 9, A D J O I N T CONSTRUCTIONS
206
[ have seen that if a is a morphism in /Pu(X) from (K.f) to (Z,g), and if (T, a) is a u-disjoint //-set over C\U.Y, then setting/3 = G\U.c~, the H-set (T,/3a) is a u-disjoint set over G\UZ. Thus if m. E $u(M)(Z,g), then M'(fla)(rn.) = 0 = M*(a)M*(/3)(m), and this proves that M*(~3)(n~)is in 8u(M)(Y, f). m Proofi
L e m m a 9.6.2: If 4) : X ~ isomorphism
X ~ is a m o r p h i s m o f G-sets, t h e n b i n d u c e s an
S~ : 8u( M ) o ~u,.( iJ) --+ 8u( M) and a natural transformation
s ~ : & , ( M ) o z~.(+) -+ $~,(~.v~) Proof: The first assertion is clear: let (}, f) be an object of ID0-(X), and let (Y', f ' ) =
Du,.(4)(Y,f) = (Y, 4f).
As (~f)u = fu, the sets U.Y and U.Y' coincide, and the u-disjoint sets over G\U.Y and G\U.Y' are the same. Thus Su(M)(Y, f) =
S~,(M)(Y',y). For the second assertion, let (Y', ]") be an object of lPu(X'). I fill the cartesian square g y ----, y'
X
, X'
Then if (Z',c~') is a u-disjoint H-set over G\UY', and if I fill the cartesian square cr
Z
, G\U.Y
,2 1
1 6 ' \ U.a
Z'
--, G\U.Y' I
I have seen that (Z, a) is u-disjoint over G\U.Y. Then if rn 6 $u(M)(Y, f), I have
M'(~')M.(C;\U.~)(,~)
: M.(#)M'(~)(~.) = o
and M.( G\ U.a ) induces a morphism S~,d, ) from 8u( M)(}s f) =$u( lVl)oD~r(~ )( Y', f') to Su(M)(Y', f'). It remains to see that this construction is functorial in (Y', J"): but if a ' : (Y',f') ---+(Z',g') is a morphism in Z)u(X'), then the diagram
b(y,.f,) Su(M) o Db(~)(Y',f' ) ------, Su(M)(Y',f')
S~(M) o ~(r
1
I $v(M)(~')
$u(M) o lPg.(c~)(Z',g')
-~ Su(M)(Z',9') ,5'[~x,,9,)
is commutative: indeed, this is equivalent to say that if m 6 8u(M)(Y,f), then
M*(G\U.a')M.(GiU.a)
=
M.(G\U.b)M*(G\U.a)
and this equality follows from lemma 9.3.3. This proves the second assertion.
,.
9.7. THE FUNCTORS Ru(M)
9.7
The
functors
207
gu(M)
The previous section leads to the following definition, which is dual to the definition of L;u(M): D e f i n i t i o n : If X is a G-set, and M a Machey functor for H, I set
Ru(M)(X) = If r
: X --+ X ~ is a
f)
morphism of G-sets, then the isomorphism Sr induces a morphism
Su(M)(Y',f') -+
lim
<--------
lim s + ______ (Y,I)~vu(x) ~
(Y-',f')eDu(X') ~
Su(M)oDu.(~)(Y,f) --+ '
lim
(
(Y,f)EDu(X) ~
8u(M)(Y,I)
lira
<.----_
(Y,f)eDu(X) ~
i.e. a morphism Tgu(M)*(~) from "R.u(M)(X') to Tgu(M)(X). Conversely, the natural transformation S r induces a morphism lira
(
Su(M)(Y,f)--+
(Y,f ) e D u ( X )~
lim
+-----
Su(M)oD~(r
lim
(
(Y',J')eDu( X') ~
Su(M)(Y',I')
( Y ' , f ' ) e D o ( X') ~
i.e. a morphism ~u(M).(c~) from 7r
to 7r
P r o p o s i t i o n 9.7.1: T h e above definitions turn T @ ( M ) tor.
into a M a c k e y func-
Proof: It is clear that T~u(M) is a bifunctor on G-set. axioms (M1) and (M2) of Mackey funetors. For (M2), let
It suffices then to check
a
X
Z
,
Y
,
T
1
d be a cartesian square of G-sets. The module T@(M)(Z) identifies with the set of sequences rn(s,~/, indexed by the objects of Du(Z), such that m(E,~l E M(G\U.E), and such that :0
whenever (T, a) is a u-disjoint H-set over G\U.E, and such that *
?
M (G\C.~)('~(F,~/) = m(E,~) if a is a morphism from (E, e) to (F, f) in the category D r ( X ) . The image under gv(M)*(b) of the sequence m(u,~) is the sequence m}F,i) indexed by the objects of D e ( X ) , and defined by !
~(F,f) = m(F,~i)
CHAPTER 9. ADJOINT CONSTRUCTIONS
208
The image under "R.u(M).(a) of this sequence is then the sequence ra{~,,y,) indexed by the objects of 2)u(Y), and defined by filling the cartesian square O 84
F
,
X
F'
~ Y E
and setting I! ~(~,,:,) = M.(a\U.~)(.<.,:)) = M.(G\U.~)(m(F,~:))
On the other hand, the image under T@(M).(d) of the sequence m(E,~) is the sequence nlE,,r ) indexed by the objects of Z)u(T), defined by filling the cartesian square
E C
, E'
1
l' e
Z
,
T
and setting
"{~',~'/= M.(O\U.5)(-~(E,~)) The image under T@(M)*(c) of this sequence is the sequence hi'r, y,) indexed by the objects of ~Du(Y), defined by l] n(F,,f,
! ) =
r~(F,,cf,
)
As the squares F
Z
C%
, F'
l
g
,
T
are cartesian, because they are composed of two cartesian squares, I have / '~(F,,~s,) = M*( O\f U ~ )(~(F,~:))
so I have
7r
= T@(M).(a)TCu(M)*(b), and (M2) holds.
To check (M1), I observe as before that an object (E,e) in 7)u(XIjY) is the disjoint union of an object (El, el) of Igu(X) and an object (E2, e2) o f / ) u ( Y ) . Let i~ and i2 be the respective injections from E1 and E2 into E. If m E g u ( X L[ Y), and if (E, e) is a.n object of Du(X II Y), let r~ = m(Ed), and Tt 1 =
M*(G\U.il)(n)
7~2= M*(G\U.i2)(n)
so that n = M.(G\U.il)(n~) + M.(G\U.i2)(n2).
9.8. R I G t t T A D J U N C T I O N
209
As il is injective, it is a m o r p h i s m in ~)u(XI_[Y) from ( E ~ , f i l ) to ( E , f ) . m(Ex,fil). B u t I have a cartesian square
Thus
rt 1 =
E1
X
il
7v
~
E
~ XIIY
ZX
which proves t h a t f i l = ix fl, and that
R v ( M)*( ix )(m )(El,f,) = m(e,,i~ fl ) = m(E~,:i~) = n~ Similarly, for a suitable m a p f2, I have
TCu(M)*(iy)(m)(E2,:~) = n2 T h e n if 7r = 0 and 7r n = 0 for any ( E , e ) , so m = 0. T h e n the m a p
n u ( M)*( ix ) | 7r
= O, I have n l = n2 = O, a n d t h e n
M)*( iz )
is injective, and l e m m a 9.4.2 shows that ~ u ( M ) is a Mackey functor.
9.8
Right adjunction
N o t a t i o n : If 0 : M ~ M ~ is a morphism of Mackey functors for the group H, a n d if X is a G-set, [ de/ine a m a p T~u(O)x from ~ u ( M ) ( X ) to "Fgu(M')(X) by setting, for m C T @ ( M ) ( X ) , and fox" an object ( Y , f ) o f ~ ? u ( X )
~v(O)x (m)(r,s) = Oa\v.r (m(r,s)) L e m m a 9.8.1: This definition turns tors f r o m ~v(M) to 7~u(M').
T~v(O)into
a m o r p h i s m of M a c k e y func-
Proof: First, the m a p Tgu(O) is well defined: if a : Z ~ G \ U . Y is u-disjoint, t h e n M*(a) T@(O)x(rrz)(rd) Moreover, if c~ : ( Y , I ) ~ then
= M*(a)Oa\uy(m(rd)) = Oa\u.vM (a)(
(rd)) = 0
(Z,g) is a m o r p h i s m in D u ( X ) , and if m E M ( G \ U . Z ) ,
M*(~) (~(0)~(.~)(~,~)) : M*(~)O~\~.~(.~(~,~))
. . . .
....
Oa\v.r M*( ~ )( m(z,~) ) = Oa\v.r(m(Y,S)) = ~r O)x(m)(yj)
Now let ~ : X ~ X ' be a m o r p h i s m of G-sets. T h e n , if m' E Tgu(M)(X')
~u(M)*(g))~u(O)x,(~')(r,f)
= 7r
)(r,~?) = Oa\v,r(
(r,;'s)) . . . .
CHAPTER 9. ADJOINT CONSTRUCTIONS
210
If moreover (Y', f ' ) is an object of Du(X'), and if m q M(G\U.Y'), then let Y, f, and a fill the cartesian square y
X
a
,
}i,
, X'
With those notations, I have
7 ~ ( M ) . ( ~ ) Z e ~ ( O ) x ( , . ) ( ~ , ~,) -....
M.(G\U.~)(r~u(O).~(~,~)(y,s)) ....
M.(G\U.a)Oc\u.y(m(y,f)) = Oa\v,M.(G\U.a)(m(y,f)) . . . . ....
r~u(o)x,r~.(M)(.~)o:,,:,)
o~\~, (7~:(M)(.~)(~, :,)) =
which proves the lemma.
"
T h e o r e m 9.8.2: Let G a n d H b e f i n i t e g r o u p s , and U b e a G - s e t - H . f u n e t o r M ~ Tgu(M) is right a d j o i n t to t h e f u n e t o r N ~ N o U.
The
The proof is dual of the proof of theorem 9.5.2. Let 0 be a morphism of Mackey functors from M o U to N. Then for any H-set Z, I have a morphism
Proof:
Oz : (M o U)(Z) = M(U o . Z) ~ X ( Z )
In particular, if (Y, f) is a G-set over X • (U/H), I have a morphism
Oc\u.y : M(U oH G\U.Y) --+ N ( G \ U . Y ) Composing this morphism with M.(vy), I have a morphism
Oa\u.yM.(vv) : M ( Y ) + M(U oH G\U.Y) + N ( G \ U . Y ) If (Z, a) is a u-disjoint H-set over G\U.Y, I have the diagram M.(uy)
M(~')
1 M(O)
0cW.r
, M ( U o , G\u.}")
, N(C,\U.Y)
M*(U~ ,
[ N'(~)
M(U OH Z)
,
N(Z)
Oz This diagram is commutative: the left square is because (Z, a) is u-disjoint, and the right square is because 0 is a morphism of Mackey functors. So this gives a morphism from M ( Y ) to S u ( N ) ( Y , f ) , that can be composed with the morphism M*(fx) from
re(x) to M(Y).
9.8. RIGHT ADJUNCTION
211
If oe: (Y, f) -+ (Z,g) is a morphism in ~9~(X), I have the diagram
M(X)
M(X)
M*(gx)
M*(fx)
, M(Z)
> M(Y)
M.(uz)
M.(uy)
, M(UoHG\U.Z)
Oa\g.z , N(G\U.Z)
, M(UoHG\U.Y)
, N(G\U.Y) Oa\u.y
The left square is commutative because gc~ = f. The middle square is commutative by lemma 9.3.2. The right square is commutative because 0 is a morphism of Mackey fimctors. So I have a morphism e x from M(X) to Tgu(N)(X). If 4) : X + X ' is a morphism of G-sets, and if rn C M(X), then ~x(rn) is the sequence indexed by the objects (Y, f) o f / ? u ( X ) defined by ~bx(m)(y,f) = Oa\urM.(ur)M*(fx)(m) Then 7~v(a).(4))Ox(m) is the sequence rn{r,,y, ) indexed by the objects of D u ( X ' ) , and defined by filling the cartesian square g
Y
,
y'
X
,
X'
More precisely
m}y,,f,) = M.(G\U.a)(~bx (re)if, i)) = N.(G\U.a)OG\uyM.(py)M*(fx)(m) On the other hand, the element r
is the sequence n(y,],) defined by *
!
n(y,,f,) = Oa\u.y,M.(uy,)M (fx,)M.(4)) As the square a
Y
X
~
r
y'
, X'
is cartesian, I have M*(f~,)M.(r
= M.(a)M*(fx)
Moreover as ~ , o a ; (U o~ ( G W . ~ ) ) o ~Y, Ihave M . ( ~ , ) M . ( a ) = M.(U o . ( C \ U . a ) ) M . ( ~ ) Finally as 0 is a morphism of Mackey functors, I have
Oa\v.y,M. (U OH (G\U.a)) = N.(G\U.a)Oa\u.y
CHAPTER 9. ADJOINT CONSTRUCTIONS
212 and finally
n(v,,l,) = N.(G\U.a)Oa\u.vM.(uv)M*(fx)(m) = m}v,4, ) This equality proves that 7~v(M).(r
= ~,x,M.(r
Now if rn' E M(X'), then ~'x,(m') is the sequence indexed by the objects (Y', f ' ) of Du(X'), defined by
~x,(rn')(v, f,) = Oa\v.y,M.(uy,)M*(f~x,)(rrz ') is the sequence indexed by the objects (Y, f) of Du(X),
Its image under 7r162 defined by
, =
But (~f)x, = efx, and then
(T@( M)*( r
(y,f) = Oaxu.yM.(uy )M*(fx )M'( r
= ~x (V*(r
which proves that
7r162
= $xM*(r
Thus ~b is a morphism of Mackey functors from M to ~u(M). Conversely, if ~b is a morphism of Mackey functors from M to 7~u(N), I have for any G-set X a morphism %bx from M(X) to "fCu(N)(X). In particular, if Z is an H-set, I hame a morphism ~ v o . z : M(U o . Z) -+ Tau(N)(U o . Z)
An element of TQs(N)(U OH Z) is a sequence n(r,f) indexed by the objects of 7)u(U OH Z), with n(v,f) E N(G\U.Y). I can then consider the element
and its image under N.OTz), which is an element of N(Z). I get a morphism Oz from M(U oH Z) = (M o U)(Z) to N(Z), defined by
To prove that 0 is a morphism of Mackey functors, it suffices to observe that 0 is composed of the morphism ~/Jo U from M o U to ~ u ( M ) o U deduced from ~b, and of the morphism O from 7r o U to N, defined on the set Z by
,~ ~ (T~u(X) o V)(Z) = 7r
o . Z) ~ Oz(~) = X.(~z)(n(uo.a.~) )
It suffices then to prove that this is a morphism of Mackey functors. So let r : Z --* Z ~ be a morphism of H-sets. The element
( ~ ( N) o u).( r )( ,, )(Vo,
9.8. RIGHT ADJUNCTION
213
is obtained by filling (trivially) the cartesian square
UOHr
U OH Z
) U OH Zt
UOHZ
) UOH Z! UOHr
Then: so that
o ~ , ( ~ ( N ) o v).(r As rlz, (G\U.(U OH r
= x.(,~,)N.(G\U (U o. ~))(~(~o.~,.z>)
: CzIz, this is also = N.(r
thus Oz, (7~u(N) o U),(r = N,(r Conversely, if n' E Tgu(N)(U OH Z'), then
N*(r
(,n(uouZ,,~,z))
= N (r
is the sequence indexed by the objects (Y, f)
On the other hand (TZu(N) o U)*(r of Du(U OH Z), defined by *
t
(Y,(Uonr It follows that *
!
!
But I have already observed in the proof of theorem 9.5.2 that the morphism U OH r is a morphism in 7?u(U OH Z') from (U OH Z, ((U OH r in Tgu(N)(U OH zr), I have
n'(Uo.Z,(UoHr ~
to (U OH Z',Tcz,). Then
: N.(U oH r
SO that *
l
r
I have again the commutative diagram a\u.(u o. z)
o G\U.(U oH Z')
Vz,
' Z'
CHAPTER 9. ADJOINT CONSTRUCTIONS
214
and lemma 9.5.3 shows that i is injective, and that if II = Ira(i) [I 11', denoting by j the injection from 11' into II, the set (II', bj) is v-disjoint. I have then
N*(r
Sz,) = N.(a)N*(b)(nluouz,,r
....
) ....
....
As n' E Tgu(N)(U oH Z'), and as (II',bj) is v-disjoint, the second term is zero, and this gives
N*(r
= N.(ai)N*(bi)(nluo,Z,fz)) . . . . ....
N.(..)N'(GW
(Vo.
Finally, I have proved that
and O is a morphism of Mackey functors. To complete the proof of the theorem, it remains to state that the correspondences A : 0 F-+ ~b and B : ~b ~-+ 0, which are clearly functorial in M and N, are inverse to each other. It suffices to check that if 0 is the identity, then so is (B o A)(O), and that if ~b is the identity, so is (A o B)(r Let 0 be the identity endomorphism of M o U, and ~b = A(O). If X is a G-set, the morphism ~bx from M(X) to 7r o U)(X) maps the element m C M(X) to the sequence '~bx(rn)(rj) defined by
~x(rn)(r,]) = M.(vy)M*(fx)(m) e M(U OH (G\U.Y)) = (M o U)(GkU.Y) \ I have for any H-set Z and any m E M(U oH Z)
Then if O' = B ( r
O'z(m ) = ( M o U).(7?z ) (ZbUo.z(m )(yo.z,,~z) ) which gives
e'z(m) : M.(U oH 7lz)M.(vUouz)M* ((~rz)x)(m) But I have seen that (U OH rlz)VUo.Z is the identity map, as well as (Trz)x. Thus O} is the identity, and so is 0'. Now if ~b is the identity endomorphism of TCu(M), and if 0 = B(~b), then for any H-set Z and any
m' C (Tiu(M) o U)(Z) = ~u(M)(U OH Z) I have If r = A(O), if m E R u ( M ) ( X ) , then r (I/, f ) of Du(X) defined by r
is the sequence indexed by the objects
= Oa\u.yT~u(M).(vy)TCu(M)*(fx)(m)
9.9. EXAMPLES Setting m ' =
But by
215
T~u(M).(uz)Tiu(M)*(fx)(m),
nu(M)*(fx)(m)
Moreover rn' is the
I have then
n(E,e)indexedby
is the sequence
n(E,e) = m(E,~xe) image of n under Tiu(M).(uy). The
the objects of
component
T)u(Y),
defined
m}uo~(a\u.z),,~a\uy )
is then obtained by observing that if e is the morphism from Y to Y defined by e(y) = ((y, fu(Y)), then the square fly
Y
,
U o.
e!
(u\u.~)
~rCaiu.Y
Y
, U OH (a\U.Y) Py
is cartesian. Then
,
m(uoH(G\U.y),rCc\uy ) =
As moreover
fxe
M.(G\U.uy)( m(y,~xe ) )
= f , I have then r
=
M.(rla\u.y )M.( G\ U.uy )(m(zj) )
As finally rla\u.z(G\U.uz ) is the identity map, I have r is the identity, which completes the proof of the theorem.
= re(y j ) . Thus ~ ' 9
R e m a r k : Let X be a G-set. The expression of the limit over :Du(X) shows that if M is a Mackey functor for G, then Tiu(M)(X) is the set of sequences m(gj), indexed by G-sets (Y, f ) over X x (U/H), such that rn(y,f) E M(G\U.Y), and M*(a)(-~(y,s)) = 0
whenever (Z, a) is a r-disjoint H-set over m(zj) =
G\U.Y,
and moreover
M*(G\U.a)(m(y,j,))
whenever a : (Y, f ) --+ (Y', f ' ) is a rnorphism of G-sets over X • injective on each G-orbit.
9.9 9.9,1
(U/H)
which is
Examples Induction and restriction
Let G be a group, and H be a subgroup of G. If U is the set G, viewed as a G-set-H, then the fimctor N ~ N o U is the restriction functor for Mackey functors from G to H.
CHAPTER 9. ADJOINT CONSTRUCTIONS
216
As U/H = G/H, an object ( Y , f ) over U/H is of the form I n d ~ Z , with Z = f - l ( H ) (see lemma 2.4.1). An object of 7?u(X) is then an H-set Z, with a morphism from Ind/~Z to X , i.e. a morphism from Z to Res/~X. Moreover, the group G acts freely on U, so on U.Y. Thus if c~ is a morphism of G-sets over X x (U/H), then U.c~ is injective on each left orbit. In other words, the category 7?u(X) identifies with
H-set~aes~X. Moreover, the set G\U.Y identifies with Z, by the map
G(u,y) E G\U.Y H u-ly E g This map is indeed surjective, because if z E Z, then z is the image of G(1,z). Conversely, if u-ly = u'-ly ', then G(u, y) = G ( u , u u ' - l y ') = G(1, u'-~y ') = G(u', y') It follows that
U OH (G\U.Y) ~- Ind~(G\U.Y) ~_ Ind/~Z ~_ Y and that vg is an isomorphism. Then if the H-set (T, a) over G\U.Y is v-disjoint, the image of U OH a is disjoint of the image of yr. As vy is surjective, I have U OH T ----0. But U OH T = Ind~rT, and then T = 0. In particular, I see that
Qu(M)(Y, f) = M(G\U.Y) = M ( Z ) = Sv(M)(Y, f) As ReSaHX is a final object of 7?u(X), I see that
Cu(M)(X) =
lim
M(Z) = M(Res~X)
ze~u(x) As ReSaHX is an initial object of :Du(X) ~ I have also
T@(M)(X) = M(ReSUHX ) and the following isomorphisms follow easily:
s
~- I n d , ( M ) _~ Tiu(M)
I recover that way the adjunction properties of induction and restriction. Now switching the roles of H and G, I consider V = G as an H-set-G. The functor N o V is then the induction functor for Mackey functors from H to G. As V/G = . , an H-set (]I, f ) over V / G is just an H-set, and V.Y = V • Y. The group H acts freely on V, so if a is a morphism of H-sets from Y to Z, then V.c~ is injective on each left orbit of H on U.Y (because h.u = u implies h = 1). The category :Dv(X) identifies then with H-setJ, x, and has a final object X. If Y is an H-set, then H \ V . Y identifies with Ind~Y. Let ( T , a ) be a v-disjoint G-set over H \ V . Y . If t E T and a(t) = H(v, y), then as G acts freely on V, I have (v, t) C V oG T, which contradicts the hypothesis on (T, a). So T = 13. In those conditions, it is clear that
Lv(M) ~ Res~,(M) _~ 7~v(M) and I recover once again the adjunction properties of induction and restriction.
9.9. EXAMPLES
9.9.2
217
Inflation
Let N be a normal subgroup of the group G, and H be the quotient G/N. If U is the set H, viewed as an H-set-G, then the functor M ~ M o U is the inflation functor for Mackey functors: indeed, if X is a G-set, then U oa X = X N. As U/G = *, an object of 7?u(X) is just an H-set over X. If (!/, f ) is such a set, then U.Y = U • Y, and H\U.Y identifies with I n k Y . As H acts regularly on U, if ~: (Y, f) ~ (Z,g) is a morphism of sets over X, then U.a is injective on each orbit of H on U.Y. So X is a final object in 7Pu(X), and then for any X, I have
s
= Qu(M)(X)
~ u ( M ) ( X ) = Su(M)(X)
Furthermore as U oc (G\U.Y) = ( I n ~ Y ) N _~ Y, the morphism uy is an isomorphism for any Y. So an object (T,a) over H\U.Y = In~HY is ,-disjoint if and only if U oa T = T N = O. Finally: P r o p o s i t i o n 9.9.1: L e t N b e a n o r m a l s u b g r o u p o f t h e g r o u p G, a n d H = GIN. I f M is a M a c k e y f u n c t o r for G, a n d X is a n H - s e t , I s e t
M N ( x ) = M(In~HX)/ ~ M.(a)M(T) (T,a)
Mar(X) = ~ KerM*(a) (T.~)
w h e r e t h e s u m a n d i n t e r s e c t i o n r u n o v e r t h e G - s e t s ( T , a ) o v e r InfaHX s u c h t h a t T N = O. I f ~ : X --~ X ' is a m o r p h i s m o f H - s e t s , t h e n t h e m a p s M . ( I n ~ r and M*(In~r i n d u c e m o r p h i s m s b e t w e e n MN(X) a n d MN(x'), a n d b e t w e e n Mar(X) a n d Mar(X'), w h i c h t u r n M N a n d MN i n t o M a c k e y f u n c t o r s for H. T h e f u n c t o r M ~-* M N is left a d j o i n t t o t h e f u n c t o r L ~-~ I n , L, a n d t h e f u n c t o r M ~ MN is r i g h t a d j o i n t t o it. R e m a r k : W i t h the notations of Th~venaz and Webb (see [14], [15]), it is easy to identify M x with M +, and MN with M - : any set T such that Tar = 0 is indeed isomorphic to a disjoint union of sets of the form G/K, for N g K , and it follows easily that if L/N is a subgroup of H = G/N
MN(L/N) = M ( L ) / ~ tL-M(I()
MN(L/N) = A Ker ri K
where the sum and intersection run on the subgroups K of L (which give morphisms from G/K to G/L) not containing N.
9.9.3
Coinflation
Let N be a normal subgroup of G, and H = G/N. Let V be the set H, viewed as a G-set-H. Then if Z is an H-set, the set V on Z identifies with I n , Z, and the functor M H M o V is the functor that I have denoted by p~ (and denoted by fl! by Th6venaz and Webb see[15].5). Here again, the set V/H is trivial, so if X is a G-set, an object (Y,f) of Dr(X) is just a G-set over X. The set V.Y is the product 1/ x Y. Let c~ : (Y,f) ---, (Z,g) be a
CHAPTER 9. ADJOINT CONSTRUCTIONS
218
morphism of sets over X. Then V.a is injective on the left orbits of G on V.Y if and only if the hypothesis imply gy = y. But the stabilizer in G of any point v of V is equal to N. Thus V.a is injective on the left orbits of G on V.Y if and only if c~ is injective on the orbits of N on Y. a morphism in l ) v ( X ) from ( Y , f ) to (Z,g) is then a morphism a of sets over X , which is moreover injective on each orbit of N. Now the set G \ V . Y identifies with N \ Y . Then V OH (G\V.Y) identifies with InfaH(N\Y). The morphism uy maps y to its orbit Ny by N. In particular, it is surjective. So an object ( T , a ) over G\V.Y is ~,-disjoint if and only if V OH T = In~HT = ~, i.e. if T = ~. It follows that
Qv(M)(Y, f) = M ( N \ Y ) = Sv(M)(Y, f) Finally: P r o p o s i t i o n 9.9.2: L e t N b e a n o r m a l s u b g r o u p of G, and H = G/N. I f V is t h e s e t H , v i e w e d as a G - s e t - H , a n d if X is a G - s e t , t h e n :Dv(X) is i s o m o r p h i c t o t h e c a t e g o r y w h i c h o b j e c t s are t h e G - s e t s o v e r X , a n d t h e m o r p h i s m s are t h e m o r p h i s m s of s e t s o v e r X w h i c h are m o r e o v e r i n j e c t i v e on e a c h o r b i t of N . I f M is a M a c k e y f u n c t o r for H, t h e n
s
=
M(N\Y)
lira
Tiv(M)(X) :
(vJ )~Vv(X )
lim
M(N\Y)
(v,f )~Vv(X )~
Let K be a subgroup of G. I denote by K, ordered by the following relation
Notations:
wN(K) the
set of subgroups of
LCL' L A N = L'NN
LZL'~{
If M is a Mackey functor for the group H = G/N, I denote by lim M ( L N / N ) L6wN(K) the quotient of OLc_KM(LN/N) by the submodule generated by the elements of the .L'N/N form ~ L N / N m - - m , for L ~_ L' and m 6 M ( L N / N ) . The group K acts on lim M ( L N / N ) , and I denote by LEwN(K) lira M ( L N / N ) ) ~ . L6wN(K)
<~(M)(K)=(
the biggest quotient on which K acts an element of M ( L N / N ) , I denote by ff K C _ K', I denote by t K' (resp. (resp. from L~4(M)(K' ) to f i ( M ) ( K ) ) "
K'
~K'z
trivially. If L is a subgroup of K, and if m is m f the image of ,~ in ~ ( M ) ( A ' ) . K' ) the m a p from L~(M)(K) to f i ( M ) ( K ' ) rK defined by
IK',
=
xEK\K'/L'
{x L'N/N ,~K ~ 7"(K~nLON/Nrn )KnxL,
9.9. EXAMPLES
219
[ f x ~ G, I s e t ~ ( m f ) = ~x v K
Dually I denote by
lira
(_._._ L6wN(K) ~
M(LN/N)
the set of sequences m[ indexed by the subgroups of K, such that m~ E M ( L N / N ) , L'N/NI--n~ and such that I'LN/N k,,~L,] = mhn" whenever L -< L'. The group K acts on this set, and I denote by fi(M)(K) = ( ~im M(LN/N)) K LEwN(K)~ the set of its fixed points. K' (resp. r~/) the map from fiH(M)(K) to ~ ( M ) ( K ' ) If I< C_ K', I denote by t K (resp. from ~ ( M ) ( K ' ) to Z~(M)(K)) defined by K'
K'
(tK m)u =
,x K ~ t(L,AXK)N/N( m,L,~nK )
z K' ,\K' ( r h. m )L :
-L'N/N
E
tK m L
xEK\K'/L'
Finally, if x E G, and if m is an element of fiH(M)(K), I denote by ~m the dement of ~a(M)(~K) defined by (~m)LK = ~(mn~ Proposition 9.9.3: Those definitions turn t~(M) and i~4(M) into Mackey f u n c t o r s f o r G.
T h e f u n c t o r M ~-~ t ~ ( M ) is left a d j o i n t t o t h e f u n c t o r
L ~-~ p(L), and the functor M ~-~ i~(M) is right adjoint to it.
Proof: I have to identify s and T i v ( M ) ( X ) i n the case X = G/K. If (Y, f ) is a G-set over G/K, I can choose a system of representatives S of G \ Y contained in f - l ( K ) . If s E S, then its stabilizer G, is contained in K , and I have a natural map #~ : gG, ~-~ g.s from G/G~ to Y. The union [ I , e s #~ is an isomorphism : 11
a/a
--, Y
sES
If m E M ( N \ Y ) , then M * ( N \ # ~ ) ( m ) E
and
N\(G/G~) ~_ (G/N)/(G~N/N) The m a p Av defined by
Ay(m) = ~ M*(N\#~)(m)~ E ~a(M)(K) sES
does not depend on the choice of the system S inside f - l ( K ) : indeed, changing s in g~s, so that f(g~s) = g~f(s) E K forces g~ E K. Then M*(N\#~)(m) is replaced by its conjugate under g~, which has the same image in f i ( M ) ( K ) . This m a p passes down to the quotient s if a : ( Y , f ) ~ (Z,g)is a morphism of G-sets over G / K which is injective on each N-orbit, I choose a system of representatives T of G \ Z contained in g-l(K). Then for any s E S, there exists a unique t~ 6 T such that a(s) E Gt~, i.e. a(s) = x~t~, for x~ E G. As ga = f , I have f(s) = x~g(t~), and then x, E If. Moreover, I have G~ C_~'Gt,, and if n E ~ G t , ( / N , then =
=
=
=
=
CHAPTER 9. ADJOINT CONSTRUCTIONS
220
If a is injective on each N-orbit, I have ns = s, which proves that N n ~:'Gt, = N MG~, so
Then let 7r~ be the map gG~ ~ gx~Gt, from G/G~ to G/Gt,, and 7r be the m a p
II
~s: II
c/c~
~
sES
sES
II
c/c,
tET
As The square #Y
YLes G/G~
, Y
1o -
I~teTG/Gt
#z
, Z
is commutative. As #y and #z are isomorphisms, it is cartesian. As moreover a is injective on N-orbits, by lemma 9.3.3 the square
N\#y
N\I_I,~C/C~
-
N\Y
N\LIt~TG/G~
N\#z, N\Z
,
is also cartesian. Then I have
M*(N\#z)M.(N\a) = M.(N\~r)M*(N\#z) But if m E M ( N \ Y ) , I have
M*(N\fy)(m) = (~ M*(f=)(m) sES SO r/
xs G
*
\xsni(
M.(N\~r)M*(N\fy) = @ [(tc, *"M (N\#v)(m)) Jc,, sES
As xs E I f , this is also
~ t ~ " M*( N\#y )(m )~:a,.. sES
and as G~ ~ ='Or,, this is
( ~ M*( N \ f y )(m)~ sES
i.e.
Ay(m). Furthermore M * ( N \ # z ) i . ( N \ a ) ( m ) = ( ~ M*(ft)) M.(N\a)(m) = AzM.(Nka)(m) tET
which proves t h a t A passes down to the quotient.
9.9. EXAMPLES
221
Conversely, to the element ,rz~ E ~C~(M)(K) I associate the image A'(rn) of rn in s corresponding to the G-set G/L --+ G'/K over G/K. This map is welt defined, because if L ~ L' then the projection G/L --+ G/L' is injective on N-orbits: indeed, if gL' = ngL' for n E N, then n g E L' N N = L Cl N, so gL = ngL. Similarly~ if k E K, then the conjugation G/L ~ G/kL is bijective, so it is injective on the N-orbits. If L C K, I can take {L} as a system of representatives of G\(G/L) for the computation of Ac/c, and then it is clear that AA' is the identity. Conversely, the isomorphism #y : I]l~es G/G~ --+ Y is an isomorphism of sets over G/K, and it follows that A'A is the identity. This states the isomorphism
s
~- ,~(M)(K)
Tile formulae giving the transfers and restrictions from c~(M)(K) to ~a(M)(K') follow easily from isomorphisms k and A'. The isomorphism
and the corresponding formulae can be proved by an argument dual to the previous one.
9
R e m a r k s : 1) The set aaN(K) is seldom connected: indeed if L and L' are in the same connected component, then L N N = L' N N. For a subgroup J of K FI N, if I denote by co~,(J, K) the set of subgroups L of K such that L f3 N = J, then coN(J, K) is the connected component of coN(K) containing J, and I have the isomorphism
JC_KNN J rnod. K
LEwN(J,K)
and ~ similar isomorphism for ;(M)~(K). 2) If K N N = {1}, then wN(K) has a biggest element K, and then
4(M)(K) = M(KN/N) = It follows in particular that ~ ( M ) ( 1 ) = M(N/N). 3) It is easy to see that for any Mackey functor L for H G G puInitt(L) _~ L
This follows from the fact that, a.t the level of H-sets (Inf~Z) 'v~" Z By adjunction, it follows that ~C(L) ~'ft
~- L
That can be checked directly: let K be a subgroup of G containing N. If L is a subgroup of If not containing N, and if rn r M(LN/N), the element m~" of z.~(M)(K) is equal to t~(m~). And if H _D N, then H < Ix', and then m.~ = tH/NUn)I"'',~ ~/N" So if rr is the projection from ~ ( M ) ( K ) onto its quotient ~v(M)~V(K/N), tile morphism
c M(K/N)
c 4(M)N(K/N)
222
C H A P T E R 9. A D J O I N T C O N S T R U C T I O N S
is surjective. Conversely, it is easy to see that the morphism mr[ E ~ ( M ) ( K )
~ { 0K/Nif N ~ L tn/N(m ) otherwise
passes down to the quotient t~v(M)N(K/N), and induces an isomorphism inverse of the previous one. 4) The associativity of the product OH and the adjunction show that if G, H, and K are groups, if U is a G-set-H, and V is an H-set-K, then s
os
~- s
7{u o Tgv ~- T~uouv
5) The Burnside functor b for the group H is such that HomM~ck(u)(b, Nz) ~- N(Z)
if N is a Mackey functor for H, and Z is an H-set. Then if U is a G-set-H, and M a Mackey functor for G HomM~k(,) (b, ( M o U)z) ~- M ( U OH Z) ~_ HomM~ck(a)(b, Muo.~Z) ~-- . . . ... ~-- HomM~k(G)(buo,Z, M ) ~ HomM~k(H)(bz, M o U) ~- ... "~HomM:ck(a)(f-u(bz),M) As this isomorphism is natural in M, it follows that s
~- bUo.Z
where b in the left hand side is the Burnside functor for H, and in the right hand side the Burnside functor for G.
Chapter 10 A d j u n c t i o n and Green functors 10.1
Frobenius morphisms
Let G and H be groups, and U be a G-set-H. If X and Y are H-sets, I have defined the maps 6Xv, Y : O o H (X x Y ) --+ ( g OH X ) x (U OH Y ) by
4,~(~,x,y) ((~,,.), (~,y)) =
L e m m a 10.1.1: I f f : X --* X ' a n d g : Y --~ Y' a r e m o r p h i s m s of H - s e t s , t h e n the square
U oH (X x Y)
~U X,Y )
(U oH X) x ( g oH Y)
/
I
U 0H ( f X g) / U oH (X' x Y')
)
(U OH f ) • (U OH g)
(c)
(u o . x') • (u o . y')
~Tu,y,
is cartesian. Proof: Indeed, if
(~X,,y,(~,(xl, y')) = (([.70 H f)X
(Uo H g))
((UI,X),(u2, y))
then in U o H X ' , I h a v e ( u , z ' ) = ( u l , f ( x ) ) , and in U o H Y ' I Then there exists elements s and t of H such that
u.s=ul
x ' = s . f ( )x
u.t=u2
have (u,y') = (u2,g(y)).
y ' = t . g ( y)
In those conditions, the element (u, (sx, ty)) is in U OH (X x Y): indeed, if r C H is such that u.r = r, then U l s - l r s = u> As (u,~,x) E U oH X, I have s - i r s . x = x, so r.sx = sx. Similarly, since t t 2 ~ - l v t = U2, and since (u2,t) C U o H Y , I have t - l r t . y = Y, or r.ty = ty, which proves that r.(sx, ty) = (sx, ty) if u.r = u. Moreover
uo,, (s •
= (,,,
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
224 and
The injectivity of 6X,Y u (see lemma 8.3.1) now shows that the square (C) is cartesian. The lemma follows. 9 Observing that if P is a Mackey functor for G, then
P ( U Olt ( x • YI) = (P o u ) I x • Y) = (P o u ) y t x ) P ( U oH x x U oH Y ) = Puo,,r(U o . x ) : (PuoHZ o U ) ( X )
lemma 10.1.1 shows that the maps P . ( ~ y ) : (P o U ) r ( X ) ~ (P,o,~. o U)(X) and
p, (~x,Y) u : (Puo~,r o u)(x) ---, ( P o U ) ~ ( x ) induce for any Y morphisms from (P o U)y to PUoHr o U. Those morphisms are moreover functorial in Y and P, in an obvious sense. Moreover, the injectivity of f xU y shows that so (P o U)r is a direct summand of PUo,V o U. Then if M is a Mackey functor for H, there are morphisms HOmM~k(H) (M, (P o U),~) ~ HOmM~k(H)(M, P u o j o U)
(10.1)
The left hand side is also HomM~k(H)(Mr, P o U)
~- UOmMack(G)( ~-.u(My ), P)
and the right hand side is
HomM~k(a)(s
~- HomM~k(a)(s
P)
Thus the functor HomM~fG) ( s is a direct summand of the functor
Now Yoneda's lemma shows that s is a direct summand of s The left hand side of (10.1) is also equal to ?-f(M, P o U)(Y), and the right hand side is equal to HomM~k(H)(M, PUoHY o U) ~_ H o m f ~ k ( c ) ( s
PUo.Y) . . . .
10.1. FROBENIUS MORPHISMS
225
The functoriality in Y shows that the above morphisms induce morphisms of Mackey functors
a
,p:
(M,P o u)
P) o U
and such that VM,pAM,p = [d. Those morphisms are moreover functorial in M and P. If N is a Mackey functor for H, then by composition I have morphisms
By proposition 1.10.1, the left hand side is
HomM~r
P o U) ~- HomM~k(a)(s
P)
and the right hand side is HomM~ok(a)(s
7-/(s
P)) ~- HomM~c~(a)(s163
P)
Thus the functor
HomM~k(~)(s is
a
direct summand of the functor HomM~k(c) (s
M)+s
Now Yoneda's lemma shows that the functor functor s M)+s Furthermore I have morphisms HomM~k(.) ((P
s
N),-) is a direct summand of the
o U)v, M) ~ HOmM~k(H)(PUo.r o U, M)
The left hand side is also HomM~k(H)(P o U, My) -~ HomM~k(c)(P, Teu(My)) and the right hand side is HomM~(G)(Pcro/~Z,7~u(M)) -~ HOmM~,a)(P,~u(M)uoj) It follows that Teu(My) is a direct summand of 7~u(M)UoHY. The left hand side of (10.2) can be written as Homi~k(u)(P
o U, Mz ) = 7-t(P o U, M)(Y)
and the right hand side as
HOmMadc(G')(Puo,V,T'gu(M)) "" HOmMa~k(G)(P,~g(M)uong) . . . .
(10.2)
CHAPTER JO. ADJUNCTION AND GREEN FUNCTORS
226 So
7Y(P oU, M)is a direct summand of [TY(P,T@(M))] oU. Then
is a direct summand of
HOmM~k(H)(N, 7-liP, T~u(M))oU) ~_ HomM~(a)(L;u(N), N(P, T~v(M)) ) ~- . . .
It follows that (P o U)@N is a direct summand of [P@s o g. Then denoting by M I N the relation "M is a direct summand of N", I have also
N@( P o U) I [s
N)+P] o U
So for any M
which can be written as
HomM~k(H)(P o U, ~( N, M)) I HomM~k(a,(/2u(N)@P, TQr(M)) or
it foHow that In the case
M))
a direct s mma d of
U/H = 9 (i.e. when H is transitive on U), as the square
U o H ( X X Y)
'
UOHY
, U~
V OH X
UoH,~_,
(Y.)
is cartesian, it follows that
U oH (X • Y) ~_ (U oH x ) • (U oH Y) so 5 X,Y U is an isomorphism. Then all the previous split monomorphisms are isomorphisms. Finally, I have proved the P r o p o s i t i o n 10.1.2: L e t G a n d H b e g r o u p s , and U b e a G - s e t - H . I f P is a M a e k e y f u n e t o r for G, if M and N are M a e k e y f u n e t o r s for H, a n d if Y is an H - s e t , t h e n
(P o U)y t Puo.r o U
s
Is
~u(Mr) l~v(M)uoHr
10.2. LEFT ADJOINTS AND TENSOR PRODUCT
227
~(P o U, M) I "H(P,Tiu(M)) o U
n(M, P o U) I n(C~(M), P) o V
Zu(M)+s
s
M+(P o U) (C~(M)+P) oV "H(s M o r e o v e r , if
U/H =
Tiu(M))
*, all t h e s e split m o n o m o r p h i s m s are i s o m o r p h i s m s .
E x a m p l e : In the case when H is a subgroup of G, if U is the set G, viewed as an Hset-G, I have U/G = 9 (the roles of H and G have to be switched in the proposition). The functor - o U is the induction functor for Mackey functors, and the functors Z;u and 7~u are equal to the restriction fnnctor. The previous isomorphisms give
(Ind~.P)v ~ Ind~(Ma~gy)
G G ~(M, Ind~P) ~_ IndH~(ResHM , P)
7-{(Ind~P, M) -~ I n d ~ ( P , Res~M)
a e s ~ ( M ~ N ) ~_ ReSaHM~)aes~H M ~ I n d ~ P -~ Ind~(Res~(M)~N) R e s ~ ( N , M) ~ H(Res~N, ResiN) The last but one of these relations explains the name of "Frobenius morphisms" for this section.
10.2
Left adjoints and tensor product
Let G and H be groups, and U be a G-set-H. If M and N are Mackey functors for H, proposition 10.1.2 shows that there are natural morphisms
I will describe those morphisms explicit]y. Let P be a Mackey functor for G. I have the following diagram
HomM~c~(H)(N,7-t(M,P oU))
HomM=~k(H)[N,7-I(s
Hom.ock(,
HomMa~k(H)(M @H, P o U) -___+
0
HomM~k(a)
oU]
P)] (s
i
M)@s N), P)
where ~ = HomM~ck(H)(N,--/-/-~M,P),and O is the map obtained by composition of (I) with the isomorphisms of the diagram. Then the morphism I am looking for is
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
228
obtained by taking for P the functor s identity map of s By adjunction, I have the morphism
it is the image under @ of the
~M6N : M[~N ~ s
oU
that I already used in the proof of theorem theorem 9.5.2. It is defined for an H-set V by (M| (rlv)(t)(vouV,~v) The associated bilinear morphism is then defined for H-sets V and W by
m E M(V), n E N ( W ) ~ (M+N)*(,lv•
w(m,n))(Uo.(V•215
)
where I set
\ W /
Now this gives through q) a morphism fl'om N to to the element
J ( V x W , Id)
~(s
P)oU: it maps n E N(W)
p E [7-{(f~u(M),P) o U] ( W ) = HOmM~k(c)(s obtained by adjunction from the element
which maps the element m E V x U OH W) defined by
Now the definition of
M(V) to the element rv, w(m, n) of s
OH
U P*(@,w) for P = f~u(M@N) shows that
Then if X is a G-set, and (S, f) is an object of element m E M(G\U.S) to
7?u(X), the morphism p maps the
P.(fx x Iduo.w)P*(v(s,]) • Iduo.w)ra\u.s,w(m,n) E PUo.w(X) 7-g(f-.u(M),P) oU gives by adjunction a morphism from s to ~(s P ) , o r a bilinear morphism fl'om s s to P, defined as follows: if Y is a G-set, if (T,g) is an object of Du(Y), and if n E N(G\U.T), then the image of m(s,f), n(T,g) is the element This morphism fl'om N to
P. (Idx•
)P* (Idx•
P. (fxxlduo~(C\u.r))P* (U(S,s)Xlduohia\U.T))ra\u.s,a\u.r(m, n )
I0.2. LEFT ADJOINTS AND TENSOR PRODUCT
229
of P ( X x Y). As the square
Ids x P(T,g) SxT
)
i
S • U o~ (G\U.T)
f x x IdT )
l fx x Iduo~(a\U.T)
XxT
X • U o , (G\U.T) Idx • u(T,g)
is cartesian, it is also
P,(Idx • gz )P,(.fx x IdT)P*( Ids x U(T,9))P*(u(Sj) X IduoNG\U.T))rG\U.S,G\U.T(m, n) or // P.(fx • g~,)P ((s,s) • .(~,~))~a\~,~w~( m, ~) *
To compute this expression, I must fill the cartesian square a
r
,
u o , [(a\U.S) •
s • T
(G\U.T)]
~ U o , [(G\U.S) ~ (a\U.T)] I/(S,I)
•
P(T,9)
where I denote by d the map (Sa\U.S,a\U.T)~ra\u.s,G\U.T. u Let (s, t, u ,, H) E S ~ - T , and u, u' E U such that fu(s) = uH and gu(t) = u'H. Then
(.(s,s)-~(~,.))(~, t, u"H) = ((~, a ( ~ , . ) ) , (~', a(~', t)) Furthermore, let (u,, G(u2, so), a(u3,to)) e g oH [(G\U.S) x (G\U.T)]. Its image under d is equal to It is equal to (u(Sj)~'U(r,g))(s, t, u"H) if and only if
This is equivalent to say that there exists elements h and h' in H such that 4, = uh
a ( u 2 , So) = a ( ~ h , ~)
~, = ~'h'
a(~3,to) = a(~'h,t)
u"H = ~H
In those conditions, I have
uH = ulH = u'H = u"H and then fu(s) = gu(t). Thus (s, t) is in the pull-back of S and T over U/H. If (S, f) and (T,g) are G-sets over U/H, I denote by S.T their pullback over U/H. If s E S and t E T are such that (s,t) ~ S.T, I denote by s.t the couple (s,t), and G.s.t its orbit by G. Is X and Y are G-sets, if ( S, f) is a G-set over
Notations:
C H A P T E R 10. A D J U N C T I O N AND G R E E N F U N C T O R S
230
Z = X x (U/H), and (T, g) is a G-set over ? = Y x (U/H), I denote by f.g the map from S.T to X x Y x (U/H) defined by
With these notations, I can define a map ~_ : S . T --* U OH [(GkU.S) • ( G \ U . T ) ]
by setting
a(~.t) = (u, Cm.s,G.u.t)
if fu(s) = gu(t) = u H
Indeed, if h 6 H is such that uh = u, then
h( a.~,.s, a.u. 0 = (a.~,h -~.s, 6'.~,h -~.t) = (a.u.s, G.u. 0 Thus a(s.t) 6 U OH [(G\U.S) x (G\U.T)]. And if I replace u by uh, then
(uh, G.uh.s, C.uh.t) = (uh, h-lC.u.s, h - l G . u . t ) = (tt, O.tt.8, C.tl.t) I have also a map/3 from S.T to S ' ~ T , defined by
It is clear that the square (3:
S.T
U oH [(G\U.S) • (C\U.T)]
s • T
,
U o , [ ( a \ U . S ) • (G\U.T)]
I/(S,]) • 1/(T,9)
is commutative. The previous argument shows that the associated map i from S.T to F is surjective. As/3 is injective, and factors through i, it follows that i is injective. Thus F is isomorphic to S.T, and the above square is cartesian. In those conditions, I have P * ( I/(S,] ) • I](T,g))FG\U.S,G\U.T( KO., • ) . . . . ^
....
.
(MEN) (G\U.a)(M|
^
.
07a\U.S•
Furthermore, if G.u.s.t 6 G\U.S.T, then
Notation: If ( S , f ) and (T,g) are G-sets over U/H, I denote by x S,T u the map US,T : GkU.(S.T) --* (G\U.S) x (G\U.T)
defi, ed br ,~,~( a.~.~.t ) = ( C.~,.~, a.~,.t ). So I have
P*(V(S,I) X V(T,g))rG\u.s,G\U.T(m, n) = (M@N)*(~UT)WGkU.S,G\U.T(m, n)
10.3. THE GREEN FUNCTORS Lu(A)
231
As
[LM* k
O)G\U.S,G\U.T(~7~,t
G.u.s
]
\
G.u.t
]
(aXU.Sxa\u.r,Ia)
I have finally (M|
^
'
*
:
U
( ~s,r )a~aXU.S,CXU.r(m,n) . . . .
[M.( UT)M. (G.u.s G.u.t'~ L
~,
\
G.~.~
/ ('~) | N * ( ~ g , ~ ) N *
(G.u.s G.u.t'~ \ G.u.t /
(?%)]J (a\u.s.T,1d)
It is also the element
o(m, ?%)
[
t M* k G.u.s /
\ G.u.t /
J (GXU.S.T,Id)
Finally, I see that
P*(u(sj) • U(T,g))raxu.s,aXU.r(rn,?%) -= O(m, 7?,)(S.Tfl ) and the image of m(s,l), n(T,g) is equal to
P.(fx x gy)(O(m, n)(s.:
I have proved the following lemma: L e m m a 10.2.1: Let X and Y be G-sets. Let (S,f) be an o b j e c t of 19u(X) and (T,g) be an o b j e c t of 7)u(Y). T h e n the m a p which to m E M(G\U.S) and ?%6 N(G\U.T) associates
G.u.s.t'~
N* G.u.s.t~ (n)]
induces a bilinear m o r p h i s m from s
s
E (M~)N)(G\U.S.T) to s
defined by
(m(s,l), n(r,g)) ~ O(m, n)(S.T,l.g)
10.a
The Green funetors s
When A is a Green functor for H, and M is an A-module, I can compose the above bilinear morphism O
s
s
with the morphism from s to s I obtain a product that I denote by Ux
s163
~ s induced by the product A@M ~ M.
-~ s
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
232
L e m m a 10.3.1: L e t X a n d Y b e G - s e t s . I f ( Z , f ) is a G - s e t o v e r X , if (T,g) is a G-set over ? , if a E A(G\U.Z) a n d m C M(G\U.T), t h e n
a(z,f) U•
m(T,g ) =
M.(~cUr)(a • m)(z.TJ.g)
P r o o f : Indeed, by definition
O(a(z,f), m(T,g)) = O(a, m)(Z.T,y.g ) and
[A* (a.~.z.t) (a) o M* (a.~ z.t) (.~)] ~ (A6M)(G\U.Z.T) \ G.u.z ] \ G.u.t ] (a\U.Z.rdd)
O(a, m)
By composition with the product, I obtain the element
(G.~.z.t~ (a).M* (a.~.z.~)
M.(Id) [A* \ G.u.z ]
\ G.u.t ]
(m)]
But
A* (G'u'z't'l (a).M* (G'u'z't'l \ G.u.z ] \ G.u.t ] (m) . . . .
"'"
: M* ( G. .z.t • M* (m)) : \G.u.z.t G.u.z.t/ (A* \ G.u.z ] (a) \ G.u.t ] "'"
..=M*( 9
a.~.z.t ~M.(a.~l.~l.tla.u2.z2.t~(axm) . . . .
\G.u.z.t G.u.z.t]
\
G.Ul.Z 1 G.tt2.t 2
M * ( . G.u.z.t ~(a• . . \ G.u.z G.u.t /
.
.
,] u M 9(~;Z,T)(a • m)
and the l e m m a follows.
9
N o t a t i o n : Let Pa\u be the unique morphism from G\U to .. Then (U/H, Id) is a G-set over 9 x (U/H), that is an object o f / : ) v ( . ) . Moreover U.(U/H) ~_ U. I denote
by Cs
) = A*(pG\u)(CA)(U/H,Id )
the image of the element A*(pa\u)(eA) of A(G\U) in s Proposition
10.3.2: L e t G a n d H b e g r o u p s , a n d U b e a G - s e t - H .
9 I f A is a G r e e n f u n c t o r f o r H , t h e n s is a G r e e n f u n c t o r f o r G for the product u• with unit r T h e c o r r e s p o n d e n c e A ~-+ s is a f u n c t o r f r o m Green(H) t o Green(G). is a n A - m o d u l e , t h e n s is a n s r e s p o n d e n c e M ~-+s is a f u n c t o r f r o m A - M o d t o s
9 If M
and the cor-
10.3. THE GREEN FUNCTORS Lu(A)
233
Proof." The product e• is bifunctorial by construction. I must check that it is associative and unitary. To check associativity, I consider a-sets X, X' and X ' . Let (Y, f), (Y', f') and (Y",f") be G-sets over X • (U/H), X'• (U/H) and X" • (U/H) respectively. Let a E A(G\U.Y), a'C A(G\U.Y'), and m" E M(G\U.Y"). Then
a' "• ,~"
:
:~I*(,~,~,,)(~'
~")
•
So a
Ux
(a' Ux r n " ) = M(~y,z,.y,,)(a• * g M.(~v,z,,)(a,x m")) . . . .
. . . . M*(tC~,y,.y,,)M*(Id
x
t~u,,v,,)(a
x
a'
x
rn")
•
m")
On the other hand
a
u•
a'= A*(nuy,)(a
•
a')
•
Wt')
which gives
(a
~•
a')
~x
U U ,~" : M 9 (~yy, y,,)(A * (~y,~.,)(a
•
a')
:
u * (ny, u y, • . . . . M 9 (l.Cy.y,,y,,)M
. . .
Id)(a •
But after identification of (Y.Y').Y" with Y.(Y'Y"), I have U
U
which proves that a ux (a' ux r n " ) = (a u• a') u• m". Moreover, if (Y,f) is an object of Du(X), and if m E M(G\U.Y), then
u 9(pa\u)(e) • m) = M*(nU/H,y)M*(p)(m) eCu(A) v• m = M 9(aU/H,y)(A where p is the projection from (G\U) x (G\U.Y) onto G\U.Y. But identifying (U/H).Y with Y, and G\U.((U/H).Y) with G\U.Y, and U.(U/H) with U, the map U ~U/H,Z is the map
G\U.Y-~ (G\U)
x
(G\U.Y)
which maps G(u, y) to (Gu, G(u, y)). Then p o tcU/H,y is the identity, and it follows that 5s
) UX m = rZt
The previous arguments, in the case M = A, show that Z;u(A) is a Green functor (it remains to check by a similar argument that eCv(A) is also a right unit). This proves the first part of the first assertion. The case of an arbitrary A-module M proves the first part of the second. If 0 : A ~ B is a unitary morphism of Green functors from A to B, then 0 induces a morphism of Mackey functors s from s to Z:u(B). If X and Y are G-sets, if (Z,f) is an object of ~u(X) and (T,g) an object of s if a E A(G\U.Z) and a' E A(G\U.T), then
c~(o)x•
Ux ~,~)) : O~\~.~.~(A*(~,~)(~
x
~)(~.~,s.~))
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
234
As 0 is a morphism of Mackey functors, this is also
B*(tcU, T)OG\u.ZxG\U.T(a • at)(Z.T,.f.g) and as 0 is a morphism of Green functors, it is
so that
~u(O)x•
u• a~T,g))= s
U• s
Moreover
s
=s
(A*(PG\U)(r ....
= OG\uA*(pc\u)(eA)(U/H,rd) . . . .
B*(PG\u )O,(c A )(U/H,Id) ---- B*(PG\U )(r )(U/H,Id) = gs
Thus s is a unitary morphism of Green functors, if 0 is. This proves the first assertion. And if 0 : M ~ N is a morphism of A-modules, then 0 induces in particular a morphism of Mackey functors s from s to s defined for an object (Y,f) of 7?u(X) and a element m of M(G\U.Y) by
il) = Oaw.y(m)(vj)
s
Then if (Y', f ' ) is an object of Du(X'), and if a' E A(G\U.Y'), I have ' v~ s a(y,,],)
= N*(nu,,y )( a' • Oa\u.y(m /) (y,.y,f,.y) . . . . ....
....
U N * (xz,,y)O(a\u.y,)•215 *
U
I
t
• m)(y,.y,y,.y) . . . .
= OG\u.(y,.y)(at U• m)(y,.y,f,.y) . . . .
OG\U.(y,.y)M (t~y,,v)(a •
. . . . z~u(O)x'• which proves that the correspondence M H s
10.4
s
~• "~(Y,s)) is functorial in M.
9
and adjunction
Let G and H be groups, and U be a G-set-H. Let moreover A be a Green functor for the group H. In the proof of theorem 9.5.2, I have considered the morphism deduced by adjunction from the identity morphism of s It is the morphism
AA : A ~ s
oU
defined as follows: if Z is an H-set, then (U OH Z, ~rz) is an object of 7)u(U OH Z). If a 6 A(Z), then
A*(qz)(a) e A(G\U.(U OH Z)) and I set
,~A,Z(a) = A*(rlz)(a)(vo~z,~z) Similarly, if M is an A-module, I denote by /~M the morphism of Mackey funct-or-sM ~ s o V. The module s is an s so the module s oU is an s o U-module. In those conditions:
10.4. L v ( A ) - M O D U L E S AND A D J U N C T I O N
235
L e m m a 10.4.1: Let Z and Z' be H-sets. If a C A(Z) and m E M(Z'), t h e n ~M,ZxZ,(a
• m ) = )~A,Z(a) •
/~M,Z,(rrt )
w h e r e t h e p r o d u c t in the right h a n d side is the p r o d u c t of s A) o U-module.
o U as
s
Proofi With the notations of the lemma, I have AM,Zxz,(a
• rrt) =
M*(rlZxZ,)(a x m)(uo~(zxz'),,~z•
On the other hand, I have AA,z(a) E ( s
o U)(Z), and AM,z,(m) E ( s o U is equal to
U)(Z'). Their product for the functor s AA,z(a) x u AM,Z,(m)= s
z,)(AA,z(a) ux AM,Z,(m))
where the product ux in the right hand side is computed inside s
i.e.
A*@z)(a)(voj,~z) vx M*(~z,)(m)(Vo.Z,,,~z, ) . . . .
)~A,Z(a) UX )~M,Z,(ra) = 9 .. z
o
v*( I,ZUoHZ,UoHZ u ( ) ( )a X M *(r/z, )()) ,) ( A *~]Z m ((Uo.Z).(uo.z'),.z.,~z,) ~ ....
9
g
...
*
M (auouz,uouz,)M (71z x ~Tz,)(a • m)((uo~z).(uouz'),,~z.~z,)
Finally /~A,Z(a) •
....
/~M,Z,(m) . . . .
s
x ~?z,)(a x m)((UonZ).(UoHZ'),,~z.,~z,)) (10.3)
To compute this expression, I note that the square
U oH (Z x Z') (zz"~
,
UoHZ'
U oH Z
~ Uou'~-U/H
is cartesian, as the image under U oH - of a cartesian square. Thus 6~,z, induces an isomorphism
O~,z, : U OH (Z • Z') ~_ (U oH Z).(U o . Z') Then the square OU, z ,
u o. (z • z')
, ( u o~ z ) . ( u o . z ' )
7rZ• ~
I 7rZ'TrZP
U oH ( Z x Z')
, ~SU,z ,
U oH Z x U oH Z'
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
236
is cartesian, and expression (10.3) can be written
AA,z(a) • AM,Z,(m) . . . . *
U
*
U
*
. . . . M (G\U.Oz,z,)M (~Uo,Z,Uo,z,)M (~z x ~lz,)(a • m)(zxz',,,z• Let G(u',(u,(z,z'))) be an element of such that u ~ = uh. Moreover
G\U.(U oH(Z • Z')). Then there exists h E H
[G(u',(u, (z,z')))] : Hence =
and finally
....
(h-lz, h-lz t) z h - l ( z , z ,) : ,ZxZt[G(ttt,(U,(Z, Zt)))]
Now it follows that
)~A,Z(a) •
)~M,Z,(m) = M*(rlzxz,)(a x m)(z•215
= AM,Zxz,(a x m)
which proves the lemma.
9
It follows in particular that the morphism AA is a morphism of Green functors from A to s o A, non unitary in general: indeed, by definition AA,.(CA) = A*(rl.)(cA)(uo, .... ) Identifying U OH 9 with U/H, the map ~r. is the diagonal injection from U/H into U/H • U/H, and r/. is the (unique) morphism from G\U.(U/H) "~ G\U t o . , that I denote by pa\u. Thus AA,.(eA) = A*(pa\u)(U/H,~.) On the other hand, the unit of
s
o U is by definition
f--u(A)*(PU/H)(eCv(A) = f~u(A)*(PU/H)(A*(pa\u)(CA)(U/H,H)) I compute this element with the cartesian square
U/H x U/H
ulH u2H'~ u2H ) , U/H
Id~ U/H
lid , PU/H
i
So ~s
~-
(ulH u2H'~,A . . . . . A*(G\U. ~ u2H )) (PG\U)(eA)(U/HxU/H,Id)
10.4. Lu(A)-MODULES AND A D J U N C T I O N
237
or
eCv(A)oV = A*(pa\u.(V/H)2)(e A)(V/H• which is not equal to
)~A,.(CA) in
general.
N o t a t i o n : If N is an s the module N o U is an s o U module, which gives an A-module by restriction along AA, that I will denote by x o g l A , defined on the H-set Z by
(N o UtA)(Z ) = ~..(eA) •
(N o U)(Z) C_ (N o U)(Z)
where the product in the right hand side is the product of N o U as an s module.
o U-
P r o p o s i t i o n 10.4.2: Let G and H be groups, and U be a G-set-H. Let A be a Green f u n c t o r for H. T h e n the c o r r e s p o n d e n c e
N ~ NOUIA is a f u n c t o r f r o m s M ~ Co-(M).
t o A - M o d , which is right adjoint to the funetor
Proof: It is clear that the correspondence N ~ N o UIA is a functor, because it is composed of the functor N ~ N o U and of the functor of restriction to A along tAI must then prove the adjunction property. Let ~ be a morphism of s modules from s to N. Then ~ o U is a morphism of s o U-modules from s o U to N o U. Restricting along I , I obtain a morphism ~b o UIA of A-modules from s o UiA to N o UIA. But lemma 10.4.1 shows that the morphism from M to s o UIA is a morphism of A-modules. Composing with ga o UIA, I obtain a morphism 0 of A-modules from M to N o UIA. The square c___+
Homcv(A)(s
HomM~&(a)( s
N)
1
HomA(M, N o U]A)
c_+
M ), N)
HomMack(H)(M, N o U)
is commutative: indeed, any homomorphism 0 from M to N o U which is compatible with the product of A is a morphism of A-modules from M to N o UIA, since if Z is an H-set, and if m 6 M ( Z ) , then OZ(~Tt ) = OZ( s A X TrZ) = /~A,.(gA) x U OZ(fO') C7_( N o UIA)( Z )
It follows that the correspondence ~ H 0 defined above is injective. Conversely, if 6 is a morphism of A-modules from M to NOUIA , then I can compose 0 with the inclusion N o UIA ~ N o U, and obtain by duality a morphism of Mackey functors ~b from s to N. If I know that ~b is a morphism of s the proposition will follow. Let X be a G-set. If (I/, f ) is an object of 2)u(X), and if m 6 M ( G \ U . Y ) , then by construction of ~b, I have
~x(rn(y,f)) = N , ( f x )N*(y(gj))Oa\u.y(m)
(10.4)
23s
CHAPTER I0
ADJVNCT
ON AND GREEN FVNCTORS
10.4.3: L e t dy : Y --+ Y • (U/H) be the m a p (v]~(v)). T h e n if a C
Lemma
A(G\U.Y) and m E M(G\U.Y), and if 0 E HomA(M, N o UIA), I have N*(u(y,f))OG\u.y(a.m) = ay, dy.N*(u(yj))OGku.y(m) w h e r e t h e p r o d u c t in the right hand side is the p r o d u c t "." for t h e s module N.
Proof: I denote by Z the set G\U.Y. If 0 E HomA(M,N o UjA), then
Oz(a..~) : aYOz(.~) where the product ,.u,, is the product of the A-module N o U. By definition, I have
Oz(m))
zz /
9 = N* U OH
N (hz,z)(AA,Z(a) X
zz
where the product in the right hand side is the product of the s
N. As
AA,z(a) = A*(,z)(a)(~o.Z,~z) I have finally
N*(u(y,f))Oz(a.m) =X'(u(y,f))X* U OH
zz
(
N*(hu, z) A'(qz)(a)(go,Z,,~z)xOz(m)
As moreover, if y 9 Y and fu(Y) = uH, I have 5Uz,z
U OH
...
ZZ
U(yj)(y)
Z,Z
ZZ
5g
U,
=
=
=
this gives
N*(u(Y"o)Oz(a'm) = N* (
Yuu(y) ) (A*(rlz)(a)(U~
... = N" ( ) Y
YY it is also
and as N is an s
x
....
N*(uy x uy)(A *(Tlz)( a )(Uo.Z,~,z) • Oz(m))
N*(u(yj))Oz(a.m) : N * ( yyy)[s
....
. . . . s (A)*(uy)(A" (Vz)(a)(uo.z,~,z)).S'(uY)Oz(m) Now the lemma follows from the next lemma: Lemma
10.4.4: L e t Z = G \ U . Y .
s
Then
A)*(uy ) (A*(qz)(a)(vozZ,.z)) = a(y,dv)
)
10.4. Lu(A)-MODULES AND ADJUNCTION P r o o f : Indeed the square
239
//y
Y
~ UoHZ
Y
~ UOHZ
1]y
is cartesian. Then
= (A'(G\U
dy)
and the lemma follows, since rIz o (G\U.uy) is the identity.
9
Let then X ' be a G-set. If (Y', f ' ) is an object of Du(X'), and if a E A(G\U.Y'), then the product of equality (10.4) by a(y,j,) gives
a(z,j,) x ~bx(rn(z,f)) = a(z,,s,) x N.(fx)N*(utz,]))Oa\u.z(m) Moreover, setting dz, =
(10.5)
y' Sy,(y') ' I have
a(g,j,) = s As N is an s
equality (10.5) becomes
ao,., f, ) x ~x(rn(yj)) = N . ( S X, • fx) (a(y,,dy,) x X*(u(z,f))Oa\u.y(rn)) Let e be the unit of the ring (A(G\U.Y),.),
equal to
e ----A*(pG\u.y)(eA) denoting by PG\u.Y the unique morphism from G\U.Y to .. Then by lemma 10.4.3
N*(,(Y,s))Oaxv.y(,~)
= N*(~'(y,i))Oaxv.y(~.m)
= ~(y,e~).N*(~,(y,j))Oaxv.Y(m)
and then
a(y,j,) • g,x(rn(gj)) = N.(f}, x fx) [a(z,,dy,) x e(z, dv).N*(u(zj))OGxu.z(m)] Lemma
(10.6)
10.4.5: L e t A b e a G r e e n f u n c t o r for G. If X a n d Y are G - s e t s , if
a E A(X) a n d b,c E A(Y), t h e n
Proof." It suffices to compute
a x (b'c) = a x A* ( Y ) (b x c) = A* ( xyy xY ) (a x b x
....
A*(xY)(
\ xyl ]
....
\
Y2 ]
....
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
240
Thus in equality (10.6), the expression inside hooks is equal to
a(Y'~dY~ • e~Y~1Y).N~(~Y~f~)0~.Y(m) ~ (a~Y'~1~.~• e~d~).N* ( y;y) N ~ ( ~ Y d ) ) 0 ~ . Y ( m ) I set
Now equality (10.6) becomes
~(Y,j,) • ~ ( ~ ( , . , j ) )
= x.tf.i., • .f~.)((~(Y,,~,) • ~(Ys,-)).T)
By definition, I have
a(y',dw) • e(z, dy) = A*(t~:,,y)(a • e)(Z'.Z.dy,.d,) I set P = A*(~,u,z)(a • e). I denote by i the injection from Y'.Y into Y' • Y, and J the complement of Y ' . Y in Y' • Y, If j is the injection from J to Y' • Y, and dy,.y is the map from Y'.Y to (Y'.Y) • (U/lt) defined by dy, y ( ~,, ~ ) =
(,y , y , f~(y )) = ('Y,Y, fb(Y ,,))
then As
and as
s
dy,) x e(y,dy)) = s
y, dy, y) ) = 0
since the images of i and j are disjoint, I have
But the identity of N(Y' • Y) is equal to N.(i)N*(i) + N.(j)N*(j). It follows that
a(y,,f,) •
= N . ( f ) , x fx)(N.(i)N*(i)+N.(j)N*(j))((a(y,,dy,)•
....
N.(f' X, • Jx)N.(i)X*(i)((a(y,,dy,) • e(z, dr)).T)
As moreover by lemma 10.4.5
(a(y',dr,) • eO1,dy)).T = a(y,,dr, ) • e(y, dy).N*(~(y,y))OG\u.y(m) . . . . ....
a(y,,dr, ) • N*(~(y,f))Oa\u.y(m)
=..
10.4.
Lu(A)-MODULES AND ADJUNCTION
241
I get finally a(g,j,) •
= N.(f~x , x f x ) N . ( i ) N * ( i ) ( a ( y , , d y , ) x N * ( v ( z j ) ) O c \ u . z ( m ) )
(10.7)
On the other hand
a(y,j, / x m(v,s/= M'(~:~, y)(a • rn)cy,.y,s,.s/ Thus ~ x , x x ( a ( z , d , ) x m(yd) ) = N . ( ( f ' . f ) x ' •
N *(vy,.y)OG\u.(y,.y)M . ( ~ yu, , y ) ( a • m )
As 0 is a morphism of Mackey functors, I have U OG\U.(y,y)M 9 (gy,,y) = ( N o U) *(Ny,,y)O(G\U.Y,)x(G\U.y)(a u • Wt)
As 0 is a morphism of A-modules, I have O(a\v.z,)•
x m) = a x u Oa\u.z(m)
where the product x u is the product of the A-module N o U. Then
,
•
0G\u.y(
)
u = N . (~G\v.Y,,cxv.Y)
•
Oaxu.Y(~))
It follows that ~bx,•
:,) • re(r,:)) . . . .
But the composite map
U y )~,y,.y 8~\u.y,,c\u.~,(U oH ~y, has the following effect on (y',y) E Y ' . Y , if fu(Y) : f~'(Y') = u H
. . . .
So it is the restriction to Y ' . Y of ~,y, • ~:),, or (~:z, • z:z) o i. Then
@X'xX (a(y,,y,) • rn(y,f)) . . . . ....
N.((f'.f)x,•215215
....
It remains to observe that setting Z' = G \ U . Y ' , lemma 10.4.4 shows that s163
: s
: a(z,,dy,)
Finally
As moreover ( f ' . f ) x ' • = (fix, • f x ) o i, this expression is equal to the right hand side of (10.7), and then
which proves that ~ is a morphism of s proposition.
and completes the proof of the 9
242
10.5
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
Right adjoints and tensor product
Let G and H be groups, and U be an H-set G. If M is a Mackey functor for H, and P is a Mackey functor for G, I have built morphisms
(s
M)&2P) o U ~ M ~ ( P o U)
In the case P = Tiu(N), for a Mackey functor N for the group H, I have in particular a morphism Composing this morphism with the morphism
M + ( R u ( N ) o U) ---+M ~ N deduced from the co-unit g u ( N ) o U --* N, I obtain a morphism
ou
M~N
By adjunction, I have a morphism
A tedious computation shows that this morphism can be described as follows:
N o t a t i o n : Let G and H be groups, and U be an H-set G. If X is a G-set, if (Z,f) and (T, g) are objects of Du(X), [ denote by Z~T their pullback over X
Z~T= {(z,t) 9 a • T If(z) = g(t)} I denote by z~t the couple (z, t) of Z~T. I denote by f~g the map from Z~T to X defined by (f~g)(z~t) = f(z) = g(t) Definition: [[ M and N are Maekey functors for H, if X is a G-set, if (Z, f ) is an object of D u ( X ) , if m 9 M(G\U.Z), and if n E TCu(N)(X), I define a sequence Ox(a(y,f),n)(w,g ) indexed by the objects (T,g) of Du(X), such that Ox(a(y,l),n)(r,g) C
(M~N)(G\U.T), by setting Ox(a(rd),n)(r,g) = [M* ( G'u'(z~t) ~ (m) | n(zbrd~g)] "~ ( ~,). \ GAt.z ] J(G\U.(ZbT),( ~:~, )) L e m m a 10.5.1: The previous equality defines a bilinear m a p • r
Ox :
u(N)(X) --,
Proof: First I must check that Ox is well defined. So let (S, a) be a u-disjoint H-set over G\U.Z, and let m 9 M(S). Then
\
G.u.z ] M.(a)(m) | n(z~r,/~g) (G\U(Z~r)'(G2:(,:~O))
10.5. R I G H T A D J O I N T S AND TENSOR P R O D U C T
243
Let S', ct' and/3 filling the cartesian square ! O~
s'
S
O'
, G\U.(&T)
,
\ a.~.~ ] G\U.Z
Then M* (a.,,.(z~t)'~ < G.u.z ] M,,(cz) = M,,(a')M*(/3), and
....
[M* (/3)(m)| N*(a')(n(zbT,]~))] (S',(G2:[:~'))o~')
But (S', a') is a u-disjoint H-set over G\U.(Z~T): indeed, if c~'(s') = G.u.(z~t) and if (u, s') E U CH S', then
( G.~.(z~t) a.~.z
(J) : ~9(~') = a.~,.z
and (u,]3(s')) 6 U OH S, which is impossible if (S,c~) is . - d i s j o i n t .
T h e n as n
S u ( N ) ((G\U.(Z~T)), I have
N*(a')(n(z~rj~)) = 0 so
Ox(M.(a)(m)(z,j),n) = O. Now if c~: ( Z , f ) --+ ( Z ' , f ' ) is a morphism in "Du(X), then
ox (M.(a\ U.~)(.~)(~, s,), n)(~,~) . . . . ....
M* ( a.u.(z'~O
It is clear that the squares
~T
U.(~T)
Z~T
, ZgT
(z:') 1
1
,
Z'
U.Z
Z
,
U.(Z~T)
O:
,
)
U.(Z'~T) \~.z'] U.Z'
U.(~
are cartesian, and as U.c~ is injective on the orbits of G, lemma 9.3.3 shows that the square
G\U.(Z~T)
a\u.z
, O\U.(Z'~T)
a \ u . ~I
a\u.z'
C H A P T E R 10. A D J U N C T I O N A N D G R E E N F U N C T O R S
244 is cartesian. Then
ox (M.(a\u.~)(,~)(~,j,), ~)(~. ) ....
But the morphism
\ a.~.~(~) ) is a morphism in D u ( X ) : indeed, the morphism U . ( ~ T ) is injective on the orbits of G on U.(ZhT): if x E G, and if
then xu = u, xt = t and c~(xz) = c~(z). Thus xz = z, since U.c~ is injective on the orbits of G. It follows that
and then
ox ( M.( G\ v.~ )(,~)(z,,s,), ~) (~,~) .... "'"
: [M" (a.~.(z~)) (,~) | \
G.~z.z ]
r*(ZbT'Ibg).I(G\U.(Zi~T),(~:(:~,)))
Ox(m(z,f),n)
Finally, the map Ox is well defined. Now if (S, c~) is a u-disjoint H-set over G\U.T, and if m E M ( S ) , then I fill the cartesian square
~1 s'
/ O~
[/G'~(z~')'/
, a\U.(Z~T)
'
S
\ a.~.t G\U.T
~
)
so that
9
\
a.,~.z ] ('~) | N ( )((z~r,i~)) (s,,~)
But (S', a ' ) is as before a u-disjoint H-set over G\U.(Z~T), and then N*(c~')(n(z~Tj~g)) = O~ so
(M+N)*(~) (0x(~(z,j), ~)(T,g)) = o
I0.5. RIGHT ADJOINTS AND TENSOR PRODUCT Finally if
c~: (T', g') --~ (T, g)
is a morphism in
G\U,(Z~.)
O\U.(Z~T')
O\ U.o~
G\U.(Z~oO,
then as the square
O\U.(Z~T)
)
G\ U.T' is cartesian, setting r =
Z)u(X),
245
G\ U.T
I have
( M Q N )"( G\ U.a) (Ox(m(z,.f}, n )(?,~)) ....
Now the morphism
U.(Z~a)
is injective on the orbits of G, thus *
n
and moreover
{t G.uz ] =
M*(r
M"
(\ C.u.t ]
Fir~a]Iy
(M+N)*tG\U.a) (Oxtm(zj), n)(T,g)) = Ox(m(z,l), n)(T,,,,) which proves that the image of follows.
L e m m a 10.5.2: T h e m a p s t o Tgu(M@N).
Ox
is contained in
Ox define
T@(M~N)(X),
and the lemma 9
a bilinear m o r p h i s m f r o m s
Tiu(N)
Proof: I must check the three conditions of the proposition 1.8.3. Let then r : X --* be a morphism of G-sets. If (Z', f ' ) is an object of l)u(X'), if m 6 M(G\U.Z'), and if n e 7"{u(M)(X'), then to compute s162 I fill the cartesian square
X'
Z X
a
, Z' ~ X'
and then
s162
= M*(G\U.a)(m)(z,])
In those conditions
Ox ( s (M )*(r )(m(z,j,} ), 7~u(N)* (q~)(r0)(:r,g)
. . . .
= M* \{G'u(z~t)G.u.z ]1 M*(G\U.a)(m) | 7~u(M)*(r162
CHAPTER 10. ADJUNCTfON AND GREEN FUNCTORS
246
.... [M* (G.u.(zOt)~ (m)@n(ZbT,.$(/~g)) ] \ C.,,.a(z) ]
(aW.(zm,(~:t:~ ') ))
On the other hand
(nu(M@N)*(~)Ox,(m(z,j,), n))(r,9) = Ox,(rn(z,j,), n)(T,gg) . . . .
'
(G.u.(z'~t)) ( , ~ ) e
= [M* \ a . u . /
rt(Z'~T'f'~("~g)) ]
(lo.s)
But the diagram a
Z[T
, Z
, Z'
~ X
> X'
'l
T
g
l"
is composed of two cartesian squares. Thus the square
Z[T
-+ Z'
T
,
X'
is cartesian. It follows that in equality (10.8), I have Z~T = Z'~T, and with this (c.~.(zb0 identification, the map \(G.~.(z'b0'~ a.~.z, j becomes the map t,a.~.a(z))" Those remarks prove that
Ox ( s
9 (~)(rn(z,./,)),T@(N) .
(r
))
= 7~u(M|~
.
(r
which is condition iii) of proposition 1.8.3. Now if m E M(G\U.Z), then
s so that for an object
M).( (o)(m(z,/) ) = m(z.'gs)
(T,g) of Du(X'), and n E nu(N)(X'), I have
Ox, ( s (M). ( ~ )(m(zj) ), n)(T,g) . . . .
'"
G.~.~ ]
(c\u.(ar),(~:(:.~') )~
10.5. R I G H T A D J O I N T S AND T E N S O R P R O D U C T
247
where the pull-back product Z~T fills the cartesian square
(zl )
ZtT
,
T (c)
Z
, X' ~f
I must compare this element with
To compute this element, I fill the cartesian square O/
Z'
> T
X
> X'
so that
/
.
.
\
Moreover
0x (~(=,s), nu(Nl'(r 9
..
=
]
( G.u.(zhz') ~ ..--/M[ \ G.u.z ](m)|162
,
,,
J (GW.(=b='),( ~ ~:V~ ) ) )
On the other hand This gives M*
a ~ ztz' '( )
'
m
G.~.~
=
/ ('~) |
Then the following diagram is composed of two cartesian squares
(
ztz'~ ZI ]
Z~Z'
Z f
Og
, Z'
,
,
, X'
X
T
...
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
248
It follows that that in (C), I can identify
Z[T with Z[Z'. With this identification, I
}lave
(r
=r
k G.u.t ]
\ C.u.a(z')
so that I have
which is condition i) of proposition 1.8.3. Finally for condition ii) I must compare E' and
= Ox,
(m(z,~), ~ u ( N ) . ( r
E" = Tiu(M@N).(r
when (Z, f) is an object of Du(X') and object (T,g) of IDu(X'), I have
(s162
n)
m E M(G\U.Z), and r~ E Ru(N)(X). For an
] (a\u(Z~T,(C2:t:~O))
E~T,g)= [M* {G'u'(z~t)~ \ C.u.z ] (m) | T@(N).(r To compute this expression, I fill the cartesian square
S
Ol
, Z~T
(c,) X
,
X'
so that
= N.( G\U.a)(n(s,~))
T@(N).( r Thus
E~T'g)= [ M* ( Gu(zlt)'~
1
which is also
On the other hand, to compute
s162
I fill the cartesian square a
S'
,
Z
(c~) X
I X'
10.5.
RIGHT ADJOINTS
AND TENSOR PRODUCT
249
so that
s162
= M*(G\U.a)(m)(s,,b)
Let Then E
,, ( T , g ) = ( I ~ u ( M @ N ) , ( r
)) (T,9)
To compute this expression, I fill the cartesian square a
S"
!
T
(c~) X
X'
Then
E"(T,a) = (M+N).(G\U.a')(F(s,,,b,)) Moreover
G.~.s,
]
(G\u.(s, ts,,), (* ;~:.?,~")))
Hence
a.~.s,
J (~\u.(s,bs,,),( S:!:,:;:'; ) )
In those conditions, the set 5" identifies with the pull-back SPaS " of S' and S" over X: indeed, if s'[ls" E S'[IS", then b(s') = b'(s"), so
Cb(s') = f a ( s ' ) = ga'(s") = Cb'(s') and then I have the element a(s')~a'(s") of Z~T. Moreover
(f~) (a(~,)~,(~,,)) = fa(~,) = ~b(~ p) Then as (C1) is cartesian, there exists a unique s E S such that
~(s') = ~(~)
~(~')~'(~") = ~(~)
So I have a map s'~s" ~-~ s from S'~S" to S. Conversely, if s E S, and if a(s) = z~t, then
(f~g)(z~t) = f ( z ) = (f~g)a(s) = r As (C2) is cartesian, there exists a unique s' E S' such that b(~') = ~ ( ~ )
a(~') = z
Similarly, a.s
(f~g)(z~t) = g(t) = (f~g)a(s) = r
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
250
and as (C3) is cartesian, there exists a unique s" C S" such that b'(s") = 9(s)
a'(s") = t
Moreover b(s') = b'(s"), so I have the element s'~s" of S'~S". This correspondence s ~ s'~s" is clearly inverse of the previous one. As they are morphisms of G-sets, I have S ~- S'~S". W i t h this identification, I have
Ct(,S'~S") = a( s')~a' (s" )
Z(~'~") = b(~') = b'(~")
and I can write E~T,g) as
EIT'g) = [M*(G\U'a)M* (Gu'('z~t)G.u.z \
(m)
=, (
))
Moreover
(G.u.(z~t)~ (G.u.(a(s')~a'(s"))) . . . .
a.u.z
]
=
G.u.z ]
. . . .
C.~,.~(~') =
(G\U.~) \
C.u.~' ]
Then
(G.u.(z~t)~ M* (G.u.(s'~s")~ M*(G\U.a)M* \ G.u.z ] (m) = \ G.u.s' ] M*(G\U.a)(m) whence E~T,e) = E"(T,g), SO E t This proves l e m m a 10.5.2.
10.6
=
and condition
E',
ii)
of proposition 1.8.3 holds. 9
n u ( M ) as / 2 u ( A ) - m o d u l e
Let G and H be groups, and U be a G-set-H. If A is a Green functor for H , then s is a Green functor for G. If M is an A-module, then 7~u(M) is a Mackey functor for G. I have moreover a bilinear morphism
s
x nu(M) -~ nu(A@M)
that I can compose with the morphism
n~j(A@M) -, n~(M) deduced of the product
A@M -* M.
Now I obtain a product
f-,u(A )@Tiu( M) ~ nu( M)
10.6. R u ( M ) AS L u ( A ) - M O D U L E
251
L e m m a 10.6.1: Let X be a G-set, let (Z, f) be an o b j e c t of Du(X), a n d a C A(G\U.Z). I f m E Tgu(M)(X), t h e a b o v e p r o d u c t is such t h a t for a n y o b j e c t (T, g) of 7?u(X)
(a(zj).m)(r,g) =
M*
{G.u.(z~')]
{G.u.(z~t)]
w h e r e t h e p r o d u c t in t h e right h a n d side is the p r o d u c t "." for t h e Am o d u l e M. Proof: This follows from the definition, and from the formula
G.u.z
] (a) |
J (G\U.(Z~T),( ~:~uzt ))
L e m m a 10.6.2: Let (Z, f) a n d (Z', f') be o b j e c t s of T~u(X). I f a E A ( G \ U . Z ) a n d a'E A(G\U.Z'), t h e n in s I have ,
'
a(zj)'a(z"s')
\
G.u.z
]
\
G.u.z'
] (a)(z~z,,S~I,)
Proof: Indeed, by definition
() x (a(z,]) xx
' =s a(z,s).a(z,,s,)
x a(z,j,)) . . . .
.... L;u(A)* (xxx) (A*(~u'Z')(ax a')(z.z,,s.S')) To compute this expression, I fill the cartesian square i
P
X
~
(')
Z.Z'
~ XxX
ZX
It is then clear that P is isomorphic to the pull-back of Z and Z' over X, with
i(z~z') = z.z'
j(z~z') = f ( z ) = f'(z')
In other words, the map j identifies with f~f'. Then ,
a(zj)'a(z'J')
:
A.
\ G.u.z.z'
/
,
A* \ G . u . z G.u.z'] (a x a )(z~z',I~I') . . . . = "
'
"
A* ( C . u . ( z ~ z ' )
~ (a x a')(z~z'jby')
\a.u.zG.u.z']
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
252 Moreover
\
G.u,.z
] (a).m* \
G.u2.z'
] (a')
G.u2.z'
] (a') . . . .
~ I have and then denoting by ~b the map (c.~.(z~z') \G .... c.~.z,),
a(z,]).a~z,j, ) . . . . . . . . A*(r
G.u,.z "'"
]
\
= A. (G.u.(z~z')~ (a).A.(G'u'(z~z')) , G.u.z ] \ G.u.z' (a )(zbz',]~]')
which proves the lemma.
9
Proposition 10.6.3: T h e a b o v e p r o d u c t
s turns
A )~Tiv( M) ~ T~u(M)
T~u(M) into an s
P r o o f : The product "." satisfies by construction the three conditions of proposition 1.8.3. Then the associated product " x " is bifunctorial. It suffices then to check that it is associative and unitary, which is equivalent to check that the product "." itself is associative and unitary. Let (Z, f) and (Z', f ' ) be objects of l)u(X), and let a E A(G\U.Z) and a' E A(G\U.Z'). If m E Tiu(M)(X), then for any object (T, g) of I)u(X), I must compute
v = (a~z,,~.r
. . . .
J A" \
. . . .
Moreover, identifying
G.u.z ](a)'(a(z"f')'m)(zhT,f~gl
Z'~(Z~T) and (Z'~Z)~T with Z'~Z~T
,
(a(z,j,).m)(Z~Tjbg) = M. \ G.u.(z~t) ) A*
G.u.z'
] (a ).m(z'~zbT,f'b$~g)
I set
, ] (a).m(z'bzbT,f'bfbg)
E = A * " ""(C'u'~z'~z~t)~ " "
\
G.u.z'
and
\ G.u.z ] (a) Then
=
A* (G.u.(z~t)~ (a).(a~z,,l,).m))(Z~T,lbg) F.M.
k C.~.z )
(a.~.(z'~z~t)~
G.u.(z~t) ) (E) . . . .
10.6. Ru(M) AS Lu(A)-MODULE
253
.... M.\ G.,,.(~b,)J[A"\
c.~.(zhO ]
(~).~]
As
\ a.u.(z~t) ] (F) = A* \
C.u.z
] (a)
I have
A. (G.u.(z'~z~t) "] .(G.u.(z'Oz~t))(a).A.(G.u.(z'~z~t)' ] 9 \ C.u.(z~t) ] (F).E:A G.u.z \ G.u.z' ](a').m(z,bz~T,f,~f~g) Moreover
\
G.u.z
] (a).A* \ .
:~
c.~.(~z,)
~ \
.
.
C.~.~' ) (a') . . . .
.
c.~.~ )t~.~, t c.~.(~,) )~' \ c.~,.~, ](~) ....
"'
\ G.u.(z[z') ] [A"
(a).A*
Setting
F' = A* { G.u.(z'~z~t)
( G.u.(z~z')'~
( G.u.(z~z')'] (a')) .m(z,~z~T,],~]~s)
\ G.u.(z~z,) ) (A* \ c.~,.~ ] (a).A* \ c.,~.z, ]
this gives
P= M. \ C.u.t ] M* \ G.u.(z~t) ] (F') o1"
P = M. \
c.~.t
/ ( F')
Setting
=
\ G.u.(z[lz,) ) (A* \ G.~.~ )(a].A* \ G.~.z' ]
I have P = M.
'~t)~
(c.~.(~ c.~.~
....
j[E'.m~,~,~,~,~,]
M. \
. . . .
C.~,.~ ] M. \a.u.(~'~t)]
Moreover
(G.u.(z'bz~t)'~ [E'.m(z,~z~Tj,~ybg)] : M* I G'u'(z~zgt)'~ [E'.m(z,~Z~T,I,W~g)]....
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
254 Finally
\ c.~,.(z,~z~),] (E') . . . . "'" =
\ G.u,(z~z') ] (A*
\ c.~,.z ] (a).A" \ c.~,.z, ]
]
As m 6 g u ( M ) ( X ) , and as the map \(zbz'~Q , ' ~ t / is bijective, hence a morphism in 7)u(X), I have also M*
(c.~.(~b~'~t)) \ O.u.(zgzbt)
(m(z'~zbr,1'~jbg)) = m(ZbZ'br.S~'bg)
Finally, setting
\
C.~,.z
(,~).A'\
C.~.~' }
this gives
\ c.~,.(z~') ] (a).mi~z,~r,i~i,~g ) and since a ( z d ) . a ( z , ~,) = a
(z~z'd~]')
I have
p : ((a(z,~.alz, g,)).q~,g ) which proves associativity of the product. Moreover the unit ex of (s is equal to
As the square _
(:)
X
, U/H
X
,
i
(:) is trivially cartesian, I have also
~ _- ~-~.(x))~,/>~)~<~,,~, _- ~, ( % % ~<~.,~) Then if m 6 T@(M)(X), I have, for an object (T,g) of :Du(X)
10.7. Lu(A)-MODULES AND RIGHT ADJOINTS
255
As the square
Id
T
X
, T
, X
Id
is trivially cartesian, I have X~T = T, and then
since A* (a:.t) (CA) is the unit of the ring (A(G\U.T),.). So the product is unitary, and this completes the proof of the proposition. .. k
10.7
s
/
and right adjoints
Let G and H be groups, and U be a G-set-H. If A is a Green functor for H, I have defined s which is a Green functor for the group G, and the functor N ~ NoUIA from s to A-Mod: if N is an s and Z is an H-set, then
(:v o u,A)(z) = :,~,.(~) •
(N o u ) ( z )
where the product in the right hand side is the product of the s
o U-module
NoU. L e m m a 10.7.1: Let n E (N o U)(Z). T h e n d e n o t i n g by P~\u.(uoHz) the u n i q u e m o r p h i s m f r o m G\U.(U OH Z) to *, I have
w h e r e the p r o d u c t in the right h a n d side is the p r o d u c t "." for the s m o d u l e N. P r o o f : Indeed by definition
( ~ ~(U ( ~A,.(CA) xUTt = X *,~.,Z,~A,.,CA, • n) where the product in the right hand side is is the product of the s Let X = U OH Z. Then 5A,.(CA) E s and n E N ( X ) . Thus
'
\
uH
N.
]
So
\ u g ] -IA'~ But if x -= ('u, z) E X, then
(~)
CHAPTER 10 ADJUNCTION AND GREEN FUNCTORS
256 and then
N*~5 U ~N* ( u H z ) Moreover
uH /
'
Hence
~(
) / .,S~(A)
~ ~H ) (~A,./cA)) : ~(A)"
To compute this expression, I use the cartesian square
UoHZ
, U/H
UOHZ
) U/H
and then
Z~IA)'/e[z)e~IA)" (~H x~ (~../cA)) = '
\
uH
/
...
=
.
O H
oHZ,Tr
which proves the lemma.
Z
9
P r o p o s i t i o n 10.7.2: L e t G a n d H be g r o u p s , a n d U b e a G - s e t - H . I f A is a G r e e n f u n c t o r for H, t h e n t h e c o r r e s p o n d e n c e M ~ 7r is a f u n c t o r f r o m A-Mod to s w h i c h is r i g h t a d j o i n t t o t h e f u n c t o r N ~ N o UIA. P r o o f : I already know that the correspondence M ~ g u ( M ) is a functor between the categories of Mackey functors. Moreover, if r : M --+ M' is a morphism of A-modules, then the morphism R u ( r is given for a G-set X, an element m E 7r and an object (T,g) of Du(X) by
nu(r
= Ca\v.r(~(r,~))
Then if (Z, f) is an object of Du(X) and if m E M(G\U.Z), I have
n~(r
= ~>~w.~((~(~,s).~)(~,~)) . . . .
10.7. Lu(A)-MODULES AND RIGHT ADJOINTS
257
As r is a morphism of Mackey functors, it is also
G.u.t ] r
[A" \ G.u.z ) (a).m(Z~T,lbg)]
Finally as r is a morphism of A-modules, I have
TgU(r )( a(zj).m )(r,g) . . . . 9
\
c.u.z
]
(a(z,j).T~u(r
. . . .
which proves that T@(r
. . . .
s
is a morphism of
Let 0 be a m o r p h i s m o f A-modules from NOUIA to M. As NOUIA is a direct summand of N o U as Mackey fnnctor, I have a morphism O' of Mackey functors from N o U to M, defined on the H-set Z by
0 (m) =
• m)
By adjunction, it correspond to 0' a morphism r of Mackey functors from N to as follows: if X is a C-set and (r,g) an object of 7)u(X), and if
Tiu(M), defined n E N(X), then
r
= O~\u.TN.(vT)N*(gx)(n) 6 M ( G \ U . T )
Thus ~(n)(T,~) = 0a\U.T(AA,.(CA) • The previous lemma shows that setting T' =
AA,.(eA) X U N.(vT)N*(gx)(n) But as N is an
=
N.(.~)N*(gx)(n))
G\U.T
A*(pc\u.(Uo.T,))(CA)(Uo.T,,~r,).N.(~'T)N*(gx)(n)
s
A*(pa\u.(Uo,T,))(C a)Wo,T,,~,).N.(,T)N*(gx )(n) . . . .
Let
dr
be the map from T to 2F defined by
T
l/T
T
, UoHT ~
, UoHT' PT
is cartesian. So
dT(t) = (t,gg(t)).
The square
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
258
....
A*( G\ U.~'T)A*(pG\u.(UoHT'))(CA )(T,dr) . . . . ....
A*(PG\U.T)(eA)(T,dT)
This gives finally
~(n)(T,g) = Oa\u.TN,(vT)(A (PG\U.T)(~A)(T,dT).N
(~X)(/7))
I must show that ~b is a morphism of s from N to g u ( M ) . So let (Z, f) be an object of 9 and a E A(G\U.Z). Then by definition of the product on Tgu(M), I have
(o
~.,,, j[A. , ~.,,.~ j <~/+~,,~<~..,~,]
Moreover
I set =
*
sV*
so that
(~{~"~~)(~,g/: M, < c.~.t / L"4. < o.~.z J As 0 is a morphism of A-modules, I have
A* {a.u.(z~t) where the product .t~ is the product of the A-module NOUIA. Setting Z' = G\U.(Z~T), and
a': ,X~,~,A*(C.~,.(~t)~C;..<,.z S (a) I have
Then by definition
d Y E : (W o v)*
(z,)
_-'~' (a' •
E) : (N o U)"
(z,) z'~'
X*(~,,z,)(a' • E)
where the product in the right hand side is the product of the s if (u, z') C U OH Z', then
ZIZ !
It follows that
dYE=N ~(\(~, (~' ~')~') ) (~' • ~')0,,
E) : a'.B
N. But
10.7. Lu(A)-MODULES AND RIGHT ADJOINTS
259
where the product in the right hand side is the product "." of the s Now setting
N.
E' = A*(pa\u.(ZbT))(CA)(Z~T,dz~r).N*((f~g )x ) (n) N,(~'ZbT)(E').
I have E =
But as N is an
s
aI.E = N.(~'ZbT)(s Moreover
To compute
s
I use the cartesian square
Z[T
lYZbT ~
U OH Z'
<
, Z~T
~ U OH Z'
which shows that
{C.,,.(z~t))
s As
= A*(G\U.uz~T)A*(~z,)A* \ G.u.z ] (a)(ZbT,dz~T)
A (G\[ UZ~T)A (r;z)
is the identity (by an adjnnction argument), it follows that
Cu(A)*(uz~r)(a')= A* (G.u.(z~t)~ Then
s "'"
,)
= A. (G.~L.(z~) ,
\ G.u.z
:... (a)(Z~T,dz~r).A*(pa\cr(ZbT))(SA)(ZbT,dz~r).N*((f~g)X)(n)
Moreover, the square
Z[T
Id
Id I
?z~T
Z~T
dzbT is cartesian, because
dz~T is
, Z~T ) zbr
injective. Then by lemma 10.6.2
\ G.u.z / (a)(ZhT'dz~r)'A (Pa\u.(Z~T))(~-A)(Z~T,dz~r)=A ' l SO
(7.u.z J
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
260 As moreover the square
Z~T
, Z
dz~r ~_
! dz
Z~T
, Z
is cartesian, I have
A*(G.
Moreover (S~glx = ~x o U )' so
. . . . N*
a(z,dz).N*(fx)(n)
Finally, I have
(a(z,/).n)(T,s)= M.
G.u.t ] OG\U.(ZIT)N.(L'ZtT)N"
[a(z,dz).N'(fx)(n)]
As 0 is a morphism of Mackey functors, I have also M.\ Moreover
G.u.t
(
N. U ott \
] OG\U'(Z~T)= OG\U'TN* U ~ \
a.u.t
/ ] N.(z/Z~T) = N.(UT)N.
This gives finally
The square
ZiT T
. . . . . . -~ Z --~ g
X
G.u.t
It
10.7. L u ( A ) - M O D U L E S A N D R I G H T A D J O I N T S
261
being cartesian by definition, I have
and then (a(zd).n)(r,g) = Oa\u.TN.(uT)N*(g)N.(f)[a(z,az).N*(fx)(n)] Let p be the projection from X onto X. Then f x = f o p, and
....
s
*
(P)( Tt )
Moreover
N*(g) [s
dz)).N*(p)(n)] . . . .
9.. = Eu(A)*(g)Eu(A), (f) (a(z,az)). N*(g)N*(p)(n) = ... .... Finally s
Eu(A)*(g)gu(A).(f)(a(z,dz)).N*(gx)(n)
= af~ z, and the square
<) Z~T
~ Z
dT(t)] I .,
Y
is cartesian, Then
gu(A)*(g)(a~dz) = A* (G.u.(z~t)~ whence
\ a.~.z ] (a)(zh~,,~;%;X I must compare this element with ~(a(z,]).n)(T,g). By definition of ~b, I have
r
= 0o\u.TN.(~T)(A*(pcw.~)(~A)(~,~).N*(g~)(a(~j).~))
But
N*(gz)(a(zd).n) = s
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
262
Let P , a and fl fill the cartesian square O/
P
,
T
Z
, X
gx Then
s
= A*(G\U.a)(a)(p,z)
and I can write
r
=
Oa\u.TN.(vT) [A*(pc\u.r)(cA)(T,dr).A*(G\U.a)(a)(p,z).N*(gx)(n)]
To compute this product by lemma i0.6.2, I fill the cartesian square
T~P
O, !
T
, P
,
T
dz so that
A*(Pa\v.T )(e A)(T,dT).A*(G\ U.a)( a )(P,Z) . . . . "'"
\
G.u.t )(eA).A*
a*(G\U.a)(a)(The, dr~)
Then the square OLO~t
ThP
, Z
T
, X g-xdT
is composed of the two previous cartesian squares. It is then cartesian, which proves that T~P identifies with Z~T. W i t h this identification, I have fl'(z~t) = t, and a'a(z~t) = z. Then
r
= Oa\u.TN.(vT)
[
A* \(G'u'(z~t)]G.u.t J (a)(Z~T,(~;}'O)) "N ( g x ) ( )
which proves finally that
~(a(z,~).~) = a(z,1).r and ~b is a morphism of
s
Conversely, being given a morphism of L~u(A)-modules ~b from N to T~u(M), I have a morphism of s o U-modules ~b o U from N o U to TZu(M) o U. Composing
10.7. L u ( A ) - M O D U L E S A N D R I G H T A D J O I N T S
263
this morphism with the morphism of Mackey functors from 7~u(M) o U to M, I obtain a morphism of Mackey functors from N o U to M. If I know that this morphism is compatible with the action of A, then it will be entirely determined by its restriction to N o UIA: indeed, the image of an element is then equal to the image of its product by AA,.(eA). In other words, the proposition will be proved if I know that the co-unit morphism rb : 7 r
o U --* M
is compatible with the action of A. But q) is defined for an H-set Z and an element rn E ( ~ u ( M ) o U)(Z) = "gu(M)(U oH Z) by
(~z(rn) = M,(~z)(rn(uo.z,~z)) E M ( Z ) Let a r A(Z). Then a acts on (T@(M) o U)(Z) through the morphism A : A s A ) o U. In other words
a.rn = •A,a(a).Um I already observed that AA,z(a).Um = AA,z(a).rn, this product being the product of nu(M) an Cu(A)-module. Then =
But ,~A,Z(a) = A*(r?z)(a)(uo,Z,~z), so setting T = U oil Z, I have
(/kA,Z(a).m)(uOHZ,~z) = (A*(rlz)(a)(T,,z).rn)(T,,z) To compute this product, I must build TIT: but the square
T
Id
T
~ T
, T 7CZ
is cartesian since 7rz is injective. Thus T~T = T, and then
(A*(~z)(a)(T,,~z).rn)(T,,~z) : M . ( I d ) [A*(Id)A*(rlz)(a).m(T~T,~z~,~z)] . . . . ....
In those conditions
42z(a.rn) = M,(rlz ) (A*(rlz)(a).rn(T,,~z)) = a.M.(rlz)(m(T,,~z) ) = a.•z(rn) which proves that r is compatible with the action of A, and completes the proof of the proposition. 9
C H A P T E R I0. A D J U N C T I O N A N D G R E E N F U N C T O R S
264
10.8 10.8.1
Examples and applications Induction
and
restriction
Let G be a group, and H be a subgroup of G. Let U be the set G, viewed as a G-set-H, and V be the set G, viewed as an H-set-G. Then I know that if A is a Green functor for G, the functors A o U and s are both equal to the restriction of A to H. It is not clear a priori that the products defined on A o U and s are the same. But they are: the isomorphism
s
~- R e s ~ A ( X )
comes from the fact that the category D r ( X ) identifies with H - s e t S x , which has a final object (X, Id). Moreover, the set H \ V . X identifies with I n d e X . If a and a' are elements of A ( I n d ~ X ) , then
But obviously X ~ X = X , so
a(X,ld).a~X,ld) = a,a{x,id) and the product of L;v(A) ~s the product of Res~A. A similar argument shows that if B is a Green functor for H, then the products defined on s ~- I n d ~ B ~- B o V are the same. 10.8.2
The
case
U/H
=.
P r o p o s i t i o n 10.8.1: L e t G a n d H b e g r o u p s , a n d U b e a G - s e t - H . following conditions are equivalent:
The
1. T h e g r o u p H is t r a n s i t i v e on U, i.e. U / H = .. 2. F o r a n y G r e e n f u n c t o r A for H, t h e m o r p h i s m
/~A : A ~ s
oU
is u n i t a r y . 3. T h e f u n c t o r A H Cu(A) f r o m G r e e n ( H ) t o G r e e n ( G ) is l e f t a d j o i n t t o t h e f u n c t o r B H B o U. P r o o f : I recall the formulae
AA,.(r
= A*(Pa\u)(r
r
= A*(pa\u.(U/H)2)(eA)((U/H)2,H)
10.8. EXAMPLES AND APPLICATIONS
265
Moreover, as 7to is the diagonal injection from U/H to (U/H) 2, it is a morphism from (U/H, rr.) to ((U/H) 2, Id) in the category Z)u(U/H). So in s I have also AA,.(~A) = A.(rc~ If 1) holds, then U/H ~_ (U/H) ~ _~., and in this case
~A,.(e A) = ec~(a)o~ so 2) holds. s
Conversely if 2) holds, then for any Green functor A for H and any N, I have
NoUIA = N o U Then HomA(A, N o UFA) ~_ (N o U)(.) = N(U/H) "z_ ttomLv(A)(s
N) ~ N ( . )
Taking for A the Burnside functor b, and for N the module f~g(b) ~- bv/u, this gives
b(U/H) ~_ b((V/H) ~) But (U/H) 2 is the disjoint union of its diagonal, isomorphic to U/H, and of its complement C in (U/H) 2. Then the above isomorphism shows that b(C) = 0. But the only set X such that b(X) = 0 is the empty set. Then U/H ~- (U/H) 2, and U/H ~ . , so 1) holds. Now if 3) holds, then as s is the unit of the adjunction, it is a unitary morphism of Green functors. So 2) holds. And if 2) holds, then to prove 3), I must prove that the unitary morphisms of Green functors from s to B are sent by adjunction to unitary morphisms of Green flmctors from A to B o U, and conversely. But if g5 is a unitary morphism of Green functors from s to B, then 05o U is a unitary morphism of Green functors from s o U to B o U. Composing this morphism with hA, which is a unitary morphism of Green functors by hypothesis, I obtain the morphism from A to B o U which is associated to r hence this is a unitary morphism of Green functors. To prove the inverse correspondence, it suffices similarly to prove that the co-unit of the adjunction
0 :s
o U) -, B
is a unitary morphism of Green functors. But if X is a G-set, as U/H = . , the objects of 9 are just the G-sets over X. Let (Z, f) be such a set, and b e (BoU)(G\U.Z) (furthermore the set U.Z is equal to U x Z in this case). Then
Ox(b(z,f)) = B.(f)B*(l,z)(b) The unit of s
o U) is by definition
scv(BoU) = (BoU)*(pa\u)(S,oC~)(u/udd) = B*(UoHpa\u)B*(p~s/H)(SB)(.,Id) . . . . ....
B*(PUoMa\U))(SB)(.,Id)
So o.(c~(sov))
= B.(Id)B*(,.)B*(pVo,(G\v))(~,)
= B'(p.)(c,)
= ~
C H A P T E R 10. A D J U N C T I O N AND GREEN F U N C T O R S
266
Thus O is unitary. It remains to see that it is a morphism of Green functors. So let Y be a G-set, let (T, g) be a set over Y, and let b' E B(G\U.T). Then i b(z,]) x b(T,g ):
B.
[
U
t~z,r)(b x b )(Z.TJ.g)
But Z.T = Z • T and f.g = f x g. Then !
• b(T,g))
Ox•
=
B . ( f • g)B * ( . z •
*
(az,U T)(b x b')
Let moreover u be any element of H. Then
SO
Finally
Ox•
• b~T,g)) = B . ( f x g)B'(v'z x .T)(b x b') . . . . ....
B.(f)B*(v'z)(b) x B.(g)B*(V'T)(b') = @x(b(z,y)) x Oy(b(T,g))
Thus O is a unitary morphism of Green functors. The proposition follows.
10.8.3
Adjunction and Morita contexts
Let G and H be groups, and U be a G-set-H. Let A be a Green functor for H, and M be a module-A. If N is a Mackey functor for H, then /-/(M, N) has a natural structure of A-module (see proposition 6.4.2), defined as follows: if X and Y are H-sets, if a E A ( X ) and
r E TI(M, N ) ( Y ) = HomM~k(H)(M, Ny) then a x r is the morphism from M to Nx• defined for an H-set Z and m E M ( Z ) by (a • r = Czxx(m x a) e N ( Z • X • Y) = N x • The Mackey functor s has also a structure of module-/:v(A). a Mackey functor for G, the functor ~ ( s is an s module " H ( s
is an s
Then if P is So the
By restriction along the morphism
AA : A ~ s
o U, I obtain an A-module 7Y(s P) o UIA. On the other hand, the functor 7-t(M, P o U) is also an A-module. I have built morphisms of Mackey functors from ~ ( M , P o U) to 7-l(s P) o U. It is natural to ask if these morphisms are compatible with the product of A.
Proposition 10.8.2: L e t G and H b e g r o u p s , a n d U b e a G - s e t - H . L e t A b e a G r e e n functor for H, and M b e a m o d u l e - A . I f P is a M a c k e y f u n c t o r , t h e n t h e m o r p h i s m s 1-t(M,P o U) ~ " H ( s o U i n d u c e isomorphisms of A-modules
TI( M, P o U)
~- ) "H(s
M), P) oUla
10.8. E X A M P L E S AND APPLICATIONS
267
Proof." Let O be the morphism from "H(M, P o U) to ~ ( s P) o U. If I prove that O is compatible with the product of A, then as the module in the left hand side is unitary, the image of O will be contained in ~ ( s P) o UIA. So let X and Y be H-sets. Let a E A ( X ) and
r E I~(M, P o U)(Y) = gomMack(H)(M, (P o U)y) Then let
= O(c~)E [ ~ ( s
( Y ) = HomM=ck(a)(s
If S is a G-set, and (T,g) is an object of Du(S), and if rn E M(G\U.T), then the image |162 is given by the following diagram
.
.~+,~>
• M~o< P(r • (Cro. Y)) '~ • ~ ~
P((x • (Vo. Y))
In other words, denoting by 7 the functor from H - s e t to G-set$u/H defined by -y(X) = U OH X, and co its left adjoint, defined by co(T) = G\U.T, I have
U ~s(rn(T,g)) = P.(gs • Id~(z))P*(u(r.g) x Id.~(y))P.(5~(r),y)r
)
To simplify this expression, I filI the cartesian square
C~
r
,
~(co(r) • r )
r • 7(Z)
, ~co(r) • ~(Y)
U(r,g) • Ida(y) Let (t, u, y) E T x 3'(Y). If u' E U is such that gu(t) = u'H, then
Let (ul,G.u2.t,,y,) e 7(w(T) x Y). Then
6wU(T),y(ttl,a'U2"tl, ~]1) =
((?/1, a'tt2"tl), (~1, Yl))
Those images are equal if and only if
(~t', ~.~".t): (Ul, a.tt2.tl)
(tt, y) : (ttl,Yl)
This is equivalent to say that there exists h, h ~ C H such that
ul = u~h
G.u~.tl = G.u~h.t
uh ~ = Ul
htyl = y
It follows in particular that u~H = ulH = uH. Moreover uh' = u'h, thus (ul, ~.t~2.tl, Yl)
:
(uh', G.ulh.t, yl)
=
(tth', G.uhl.t, yl) = (u, G.u.t, h'yl)
=
(tt, G.u.t, y)
CHAPTER 10. ADJUNCTION AND GREEN F U N C T O R S
268
Then the element (t, u, y) is in T.7(Y). If C~T,y is the map from T."/(Y) to 3' (w(T) x Y) defined by ~r,r(t,~,y)
=
(~, a.~.t,y)
I have C~T,y(t, u, y) = (ul, G.u>tl, yt). If moreover/~r,y is the inclusion from T.~/(Y) into T x "~(Y), then the square
sT,y
T x "~(Z) l/(T,g ) X
• Y)
, ")a(T) x ~f(Y) Id~(z)
is commutative. The above argument shows that the associated morphism s from T."/(Y) to F is surjective. As 13z,y factors through s, the morphism s is also injective. Thus F is isomorphic to T."/(Y). In those conditions, I can write
~Ss(m(T,g)) = P.(gs • Id-r(Y))P.(•T,Y)P*(aT,Y)r As the composite map
(gs x Id,(y)) o/3T,y is nothing but (g.rry)sx~(y), I have finally
On the other hand, if Z is an H-set, and if m' C M ( Z ) , then
(a x r
= r215
• a)
It follows that
O(a x r
= P.((g.lrxxy)s•215215162215
x a)
(10.9)
The image of a under An is the element
b= AA,x(a)= A'(~x)(a)(~(X)#x, C (f~u(A) o U ) ( X ) = Then, by definition of the product on ? f ( s a • ~ = 7t~s
s
I have
P)
•
where the product in the right hand side is the product of ~ ( s module. Then (a x r
= P*(Ids x 5U,y)~bSx.r(x)(m(T,.q) X b)
Moreover
rn(T,g) x b : M *(~r,~(x)) u (rn x A*(~x)(a))(T.~(x),g.~x) : ...
5S
s (10.10)
10.8.
269
EXAMPLES AND APPLICATIONS M*/nu [ T,7(X))~M*~r-* [ l c ~ ( T ) x ~ x ) ( m x a)(r.,(x),g.,~) = M * ( r ) ( m
....
x
a)(T..y(X),g.r~x)
where I denote by r the composite map r = (Ida(T) x fiX)
U o XT,,~(X)
Finally, (a • ~b)s(mlr,~)) . . . .
9..
xy)P.
g.rrx.rcy)Sx.,(x)x,(y) P (C~T..,(X)y)r
(r)(m X a)
As q~ is a morphism of Mackey functors, I have O~(f.-r(x))M*(r) = (P o U)*(r x Idz)(o~(r)• SO
(a x =
P*(z~•
~)~('~(r,~/)
U
....
((g.,-~.,-Y)~•165
*
(,~.~(x),y)(PoU) (,.•162215 *
I must compare this expression with (10.9). I have already observed that ~ ( x ) . - ~ ( Y ) ~_ ~ ( x • Y)
Indeed, the map p : 7 ( X x Y) ~ 3'(X).7(Y) defined by p(,,,x,v) = (~,x).(~,v)
is an isomorphism. Now it follows from the injectivity of 5xV,g that the square P
T."/(X x Y )
(g'TrXxY)Sx~r(XxY) 1 S x ",/(X
)
T.-~(X).-~(Y)
[ (g.;rx.;ry )sx,,/(X)x-ffy) x Y)
)
Ids x
s • -r(x) • -y(y)
5X,Y u
is cartesian. Thus
and then
(~ • ~)s('~(r,~/) . . . .
. . . . P, ((g.~x •215
P*(p)P*(<~.;(x),~)(P o v)*(~ • I~)r215
Let t . ( u , x , y ) E T."/(X x Y), Then
Thus
• a)
270
C H A P T E R 10. A D J U N C T I O N A N D G R E E N F U N C T O R S
Then U
As r = (Ida(T) • ~x) o tCT,~(X) u , and as r/X ( G . u . ( u , z )) = z, this gives
Finally (a • r
= P.((g.~rxxy)s•215162
x a)
The comparison with (10.9) shows that a • ~ = O(a • r
=.
• O(r
hence O is compatible with the product of A. In particular
Conversely, let r U ( e u ( M ) , P) o U(Y) I can apply fornmla (10.10) in the case X = . , and a = CA: this gives b = A*(rl.)(ea)(~l.),~.l SO
(CA X @)S(m(T,g)) = P * ( I d s x ~~176
x b)
Moreover m(T,g) • b = M * ( r ) ( m • a)(T.~(.),g.~.) Here m • a = m, and T.'y(.) = T . ( U / H ) ~_ T. The map g.~r. identifies then with the map t E T H (g(t),gu(t)) r S x ( U / H ) 2 Now the map r is r = (Ida(T) x rlo) o ~uT,-~(.) Then identifying w (T.7(.)) with w(T), the map r becomes the identity of w ( T ) , and then (CA • O)s(rn(y,g)) = P * ( I d s • 5.,v)~s• Moreover m(T,g.,.) = s where dr is the map from T to T x ( U / H ) de~ned by dT(t) = ( t , g u ( t ) ) . morphism of Mackey functors, I have ~s•163
= P . ( f • Id,(y))~T
and then (CA z r
= P*(Ids x 5.y)P.(f
z Id~(v))r
As ~ is a
10.8. EXAMPLES AND APPLICATIONS
271
It is easy to see that the square T.7(Y)
/~T,Y
~
T • ~,(Y)
(9.r~y)sx~(z) I
[ f • Id.~(y)
T • .y(Z)
~ T • ( U / H ) • -~(Y)
IdT • ~Uy is cartesian. Finally, this gives
The question is now to know if there exists ~b E ~ ( M ,
(~A • ~ ) s ( . ~ , ~ ) )
P o U)(Y) such that
= P, ((g.~)~•
This relation must hold for any T and g. This forces the equality
P'(9~,~)~(~,~)
:
P*(~r,Y)~{r)(~)
In the case T = ~/(X) for an H-set X, and rn =
(10.11)
M*(Tix)(m'), this relation gives
P*(/L+x.),r)~.~(x)( M*(qx )(m')c~cxl,~x) ) = P*(a.y(x)y )~.~(x)M*(rlx )(rn') As 0 is a morphism of Mackey functors, I have
r
) = P*(U oH (71x • Idg))Ox
Thus
P*(13.y(x),y)~b.,(x)(M*(rlx)(m')(,(x),,,x)) = P*(a.~(xly)P*(e OH (fix x Idr))(ox(m') The map a.~(x),y is a map fi'om ~/(X).'~(Y) to "~(w(7(X)) x Y). I can compose the previous equality with P*(p), where p is the isomorphism from 7 ( X x Y) to "/(X).7(Y). Then, as
this gives
c)x(m') = P*(p)P*(/3~(x)y )'g,~(x)(M*(rlx )(m')(~(x),~)) Moreover, the map fl~.(x),Y o p is the inclusion from 3'(X • Y) into 7(X) x "),(Y), that is (~u X,Y" Thus for any X and any m E M(X), I have
which proves at least unicity of & (I know that | is a split monomorphism). I observe that this eqnMity can be written as
~x(m) = P*(sU,y)~UoHX/~M,X(m)
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
272
In this form, It is easy to check that this equality defines a morphism of Mackey functors fiom M to Po U. Moreover, relation (10.11) holds: indeed, with this definition of qS, I have
P*(C~Ty)6~(T)(m ) = P*(C~T,y)IP*((5~(T),y)t',..,~(T)(:}I*(rl~(T))(rn)(W(T),~:~(T))) Moreover, if
t.(u, y) C T.7(Y), then
Thus 5~iT)y o at,y is the restriction to
T.7(Y) of u(r,g) • Ida0, ), that is
5wU(T),y 00~T, r = (1/(T,9)
X
/d,(v))
o
flT,Y
In those conditions,
P*(aT,Y)O~(T)(m) = P*(flT,y)P*(u(T,g) • [d.yo'))g'.).~(T)(M*(Tlw(T))(m)(.-r~(r),r~(T))) As ~b is by hypothesis a morphism of Mackey t\mctors from s
P* ( U(T,g) • I G(r) )g'.~,(T) = I~'Ts
to
PUoY, I have
]~ )* ( IIT )
So
P*(aXy )CD~(x)(m) = P*(flX,y )~f.,fs M)*(UT)( M'(rl~(r))(ra )(~(T),~(rl)) As the square
T
I/T
dT I T
, .~(T)
~(T) ,-yco(T) YT
is cartesian, it follows that
f2u(M)*(ur)( m*(rl,,,(r))(m)(~,(T),,~(r))) = M*(G\uT)M*(rl~o('r))(m)(T, av) = m(T,dr) Then n*(~r,y)r
= P*(flr,~')~r(.~(r,<~))
which proves (10.11), hence that c x ~b = | So the image of 0 is exactly "H(s
o UIA, and the proposition follows. ,,
P r o p o s i t i o n 10.8.3: Let G a n d H b e g r o u p s , a n d U h e a G - s e t - H . If A is a G r e e n f u n c t o r for H, if M is a m o d u l e - A a n d N an A - m o d u l e , t h e n
s
~- ~U(~I)@s163
)
P r o o f : This result follows from the previous one by adjunction: if N is an A-module,
then by proposition 6.6.2, I have
HomA(N,~f(M, PoU)) : HomM~ok(H)(M~aN, P o U ) : HomM=~k(~)(s
10.8. EXAMPLES AND APPLICATIONS
273
On the other hand HomA (N, ~(f-.u(M), P)o UIA) = Hom&;(A)(.s ....
~(s
P)) . . . .
nomM.~(c) (s163
P)
The comparison of these two equalities now gives the claimed isomorphism.
R e m a r k : Similar tedious computations show that if N is an A-module, then ~ ( P , ~ u ( N ) ) o U~A --~ ~ ( P o U, N 1
s
o QA ~- M ~ ( P o U)
as A-modules.
P r o p o s i t i o n 10.8.4: Let G and H be groups, and U be a G-set-H. If A and B are G r e e n f u n c t o r s for H, if (P,(2, q~,~P) is a surjective M o r i t a c o n t e x t for A and B, if @ (resp. 0') is t h e i s o m o r p h i s m from s163 to s (resp. f r o m s163 to Z:u((2~BP)), t h e n t h e 4-tuple (s163163163 is a surjective M o r i t a c o n t e x t for s and
s
Proof: This is clear, since if P~AQ ~- B, then
Moreover, the isomorphisms of proposition 10.8.3 are natural enough to give here isomorphisms of bimodules. 9 Taking for U the set H, viewed as (l)-set-H, then the fnnctor s is the restriction functor to the trivial subgroup, that is the ewlua.tion functor at the trivial subgroup.
N o t a t i o n : If M is a Mackey functor for the group G, and H is a subgroup of G, I set
M(H) = (!tesa~(H)M)H(H/H) = M ( H ) / ~
t~V(L)
LcH
LcH
IrA is a Green functoz3 then A.(H) is a unitary., ring for the product " ,, of (Res~.G(H)A and A(H) is a two-sided ideal of (A(H), .). c; A )H Now the rings .4(H) are evaluations at the trivial group of functors (ResNc(H) which are the functors L:u(A) for suitable sets U: indeed, let U = H \ G , viewed as a N(~(H)/H-se t -G' by
~.Hx.g
=
Hnxg
where n ~-+ r7 is the projection from Nc~(H) to Na(H)/H. Then for any G-set X, I tl a v e
U oG X = X H
274
CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS
Indeed, in (H\G) oe X, I have (Hy, x ) = (H, gx), and (H,x) 6 (H\G) oa X if and only if z E X H. Moreover (H,x) = (H,x') if and only if x = z'. Thus if M is a Mackey functor for G, I have s
c, H = (ResNc(HtM)
This gives in particular the following lemma: L e m m a 10.8.5: Let H be a s u b g r o u p of G, and A, B and C b e G r e e n funct o r s for G. I f M is a B - m o d u l e - A , and N is an A - m o d u l e - C , t h e n
~- (Reslv~(H)M)
"@{F{esNCG(H}A)H(Res%G(H)N) t/
as (Res~va(H)B)H-modules-(Res~c(H)C)". Now evaluating this isomorphism at the trivial Nc;(H)/H-set, I have the following consequence: P r o p o s i t i o n 10.8.6: Let G and H be groups. I f A and B are G r e e n f u n c t o r s for G, if (P,(2,~, tit) is a surjective M o r i t a c o n t e x t for A and B, t h e n for any s u b g r o u p H of G, t h e 4-tuple (P(H), ~)(H), ~, ~) is a s u r j e c t i v e M o r i t a c o n t e x t for A(H) and B(H).
Chapter 11 The simple modules 11.1
Generalities
Let A be a Green functor for the group G. Theorem 3.3.5 states the equivalence between the category of A-modules and the category of representations of C.4. I can in particular apply to A - M o d the generalities on representations of categories proved in [3]. In particular, the simple A-modules can be described as follows: P r o p o s i t i o n 11.1.1: L e t A be a G r e e n f u n c t o r for t h e g r o u p G. 1. If X is a G - s e t , a n d V is a s i m p l e m o d u l e for t h e a l g e b r a EndeA(X) = A(X2), t h e n t h e m o d u l e Lx,v defined b y
Lx,v(Y) = A(YX) •A(x2) V has a u n i q u e m a x i m a l s u b m o d u l e
Jx,v(Y) = {~i ai|
'Vr
Jx,v,
defined by
A(XY), ~-~.(r
a,).vi =O }
i
The quotient
Sx,y = Lx,y/Jx,v
is t h e n a s i m p l e m o d u l e .
2. C o n v e r s e l y , if S is a s i m p l e A - m o d u l e , a n d if X is a G-set such t h a t S(X) r O, t h e n t h e m o d u l e V = ,q'(X) is a s i m p l e A ( X 2 ) - m o d u l e , a n d S is i s o m o r p h i c to Sx,v. P r o o f : The first assertion follows from [3]. The second is not stated there explicitly,
but it is a consequence of the following argument: if W is a non-zero A(X2)-submodule of S(X), then by adjunction there is a non-zero morphism from Lx,w to S, which is onto since S is simple. Then as Lx,w(X) = W, it follows that l/V = S(X), thus S(X) is a simple A(X2)-module. Then ,5" is a simple quotient of Lx,s(x), which has a unique simple quotient Sx,s(x). Thus S _~ ,5'x,s(x). "
11.2
C l a s s i f i c a t i o n of t h e s i m p l e m o d u l e s
In the special case of a functor on CA, I have moreover the notion of minimal subgroup: a subgroup H of G is called minhna/for the functor F on CA if F(G/H) r O, but
CItAPTER 11. THE SIMPLE MODULES
276
if F(G/K) = 0 for any subgroup K of C of strictly smaller order. Such a subgroup always exists if F is non-zero, since F is additive. Let S be a simple A-module, and II be a minimal subgroup for S. Let K a subgroup of G such that 1Is < iH[, and ~) be an element of A((G/H) 2) which factors through G/K in Ca, i.e. such that.
:
oG/K/7
for
C A(r
x
(r
/9
• IC,/H))
Then setting P = G/H, and V = S(F), I have the following equality in Lr,v(F) = V for v E V 1A(r~) | r = r | v = (c~ %./~,- ~3) q) v = Sr.v(a)(/7 @ v) But as H is minimal for S, the element/7 @ v of S(K) is zero. So the element 4.v is zero. Thus the module V is annihilated by any endomorphism of F which factors by a set G/Is such that IKI < IHI. N o t a t i o n : Let A be a Green [unctor, and H be a subg~'oup of G. I denote by ]~(H)
the quotient of A((G/H) 2) by the two-,sided ideal generated by the elements of the fo~m ~ oa/s, 17, fo~ IKI < IHI. With this definition, I have the following classification Proposition
11.2.1: L e t A b e a G r e e n f u n c t o r for t h e g r o u p G.
1. If S is a s i m p l e A - m o d u l e , and H is a m i n i m a l s u b g r o u p for S, t h e n V = S(H) is a s i m p l e A ( H ) - m o d u l e , a n d S is i s o m o r p h i c t o Sc./Hy.
2. C o n v e r s e l y , if H is a s u b g r o u p of G, a n d V is a s i m p l e / i ( H ) - m o d u l e , t h e n Sa/~,v is a s i m p l e A - m o d u l e , t h e g r o u p H is m i n i m a l for Sc,,/u,v, a n d m o r e o v e r Sa/H,v(H) ~- V. 3. L e t H b e a s u b g r o u p of G, and V b e a s i m p l e / i ( H ) - m o d u l e . I f X is a G - s e t s u c h t h a t Sa/H,v(X) r 0, t h e n X u r ~. I n p a r t i c u l a r , t h e m i n i m a l s u b g r o u p s of Sc,/H.v are t h e c o n j u g a t e s of H in G.
4. L e t H a n d K b e s u b g r o u p s of G. If V is a s i m p l e A ( H ) - m o d u l e , a n d if W is a s i m p l e / t ( K ) - m o d u l e , t h e n t h e m o d u l e s Sc,/s,v and ,5'c,/r,,w are i s o m o r p h i c if and o n l y if t h e r e e x i s t s x E G s u c h t h a t K = XH a n d
W ~V. R e m a r k s : a) I have identified a A(H)-module V with the associated . 4 ( ( g / H ) 2 ) module. b) The notation W _~ ~V means that V maps to W by the isomorphism c:~,. fi'om G/H to G / K in CA, deduced from the conjugation c~: : G/H --+ G/K defined by
c~(g.H ) = 9z-I.K. P r o o f : I have already proved assertion 1). For assertion 2), I already know that if V is a simple /t(H)-modnle, viewed as an A((G/H)2)-module, then ScUH,v is a
i1.2. CLASSIFICATION OF THE SIMPLE MODULES
277
simple A-module such that SH,v(H) "z_ V. It remains to show that H is minimal for Sa/u.v. So let K be a subgroup of G such that IKI < Igl. Let a | v be an element of La/n,v(K). Then a | v E Ja/n,v(K): indeed, if r C A ( ( G / H ) x ( G / K ) ) , then r oa/K a factors through G / K , so annihilates V. Thus a | v E Ja/H,v(K), and
Sc/I~,V(K) = O. Under the hypothesis of assertion 3), there exists a G A ( X x (G/H)) and v C V such that a | v is non-zero in Sa/n,v(X). In particular, there exists an element E A ( ( G / H ) x X ) such that 6 ox a has non-zero action on V, so does not factor through sets G / K such that IK[ < IH[. But L e m m a 11.2.2: L e t X a n d Y be G-sets, a n d a C A ( X Y ) . L e t
b = A.
xy yxy
(a) C A ( Y X Y ) = Hornc~(Y, Y X )
c=
x
.
E Hornc~(YX, X )
T h e n a = c ovx b in Ca. Admitting this lemma for a while, I see in particular that a factors in CA through (G/H) x X , and so does r a. As
II
(a/H) • x
n
xEG\X g~H\G/Gx I see that there exists x and g E G such that [H N G~I = Igl, and then gx C X H, which proves that X H r O. Then if K is another minimal subgroup of Sa/H,v, and if W = Sc/~,v(K), I know that Sa/H,V ~-- Sc/K,w. The group K has fixed points on G/H, and H has fixed points on G / K , so K is conjugate to H. This proves assertion 3). Under the hypothesis of assertion 4), the group K is minimal for SC/H,V, so it is conjugate to H. I can then suppose H = K, and the proposition follows then from the fact that V ~ Sa/H,v(H) ~- S G / H , w ( H ) ~ W . P r o o f of l e m m a 11.2.2: With the notations of the temma, I have
c o y x b = A . ( X l y l x 2 Y 2 1 A , ( Xlylx2y2 / \ xly2 / \XlYlX2YlX2y2/
(cxb)
But
cxb=A.
.... A.(
ylZlX2y2 IA*(yl~lX2Y21(a)
kxlylxly2x2y2/ As the square
k
x2y2 /
(&) XY
~
YX~Y k xlylxl y2x2y2 /
(Xy)2
(
XlYlX2y2 i
\xWlx2ylx2Y2
(XY)3
CHAPTER 11. THE SIMPLE MODULES
278 is cartesian, I have
A * ( xlylx2y2 I A . ( ylx~x2y2 / = A . ( xy ) A * ( xy ) \XlYlX2YlX2y2] XlYlXlY2X2~]2] xyxy yxxy and then e~
xly 2
]
xl]xy
~xxy
....
A.
\
xy
x2y 2 /
....
xy
which proves the lemma.
11.3
9
The structure of algebras A(H)
Having classified the simple A-modules, 1 have still to describe the structure of the algebras /i(H): Notations: Let K be a group acting by automorphisms on ~n R-algebra A. I denote by A | K the tensor product A | RK, with the following multiplication
(a | k).(a' | k') = ak(a') Q kk' If H is a subgroup of G, [ denote by NG(H) the quotient No(H)/H. P r o p o s i t i o n 11.3.1: Let A be a Green functor for the group G. If H is a s u b g r o u p of G, then A(H) identifies with 5.(H) | No(H), by the m a p
c~ E A((G/H) 2) ~
~ ~(H)
A*(iH,~,.)(c0 | n
w h e r e iH,n,H : G/H -'~ (G/H) 2 maps gH to (gH, gnH). Proof: By definition, I have
A ( ( G / H T ) ) ~-
0 A(H n ' H ) xEH\G/H
the isomorphism being obtained by the maps
c~ E A((G/H) ~) ~
~ A'(iH,~,H)(a) xEH\G/H
where ill,z, H is the map
iH,x,H : G/(H ~ ~H) ~ (G/H) 2 defined by
ig,~,H(g(H M ~H)) = (gH, gxg) The inverse isomorphism maps the element fl E A(H N ~H) to A.(ig,.,g)(fl). But if x • No(H), then m.(iH,x,u)(fl) factors through G/(H N ~H):
11.3. THE STRUCTURE OF ALGEBRAS A(H)
279
L e m m a 11.3.2: L e t X a n d Y b e G - s e t s . I f f a n d g a r e m o r p h i s m s o f G - s e t s f r o m X t o Y, a n d if a a n d b a r e e l e m e n t s o f A(X), t h e n
A. g(x)x (a) oxA.
(x) xf(x)
g(x)f(x)
(b) z A .
P r o o f : It suffices to compute
(x)
x )) (b) ....
A. g(x)x (a) ox A. xf(x ...zA.
y, xY2) A.(y,xy2) [A. YlY2
\ylxxy2
The cartesian square
~(axb)]
x,x2
\g(xl)XlX2f(x2) ]
(x) XX
X
~
X2
\g(xl)xlx2f(z2)] YXY
( ylxy2 } YX~Y \ylxxy2
shows that
A.(ylxy2 IA.(\g(x,)x,x2f(x2)] ZlX 2 ~ ( X )A.(Z) kY,XXy2/ = A. g(x)xf(x) xx
so that
A. g(x)x (a)oxA. xf(x)
( b ) = A . \ YlY2 / A.
....
A.
g(x)xf(x)
g(x)f(x
=..
(a.b)
which proves the lemma. Taking
xx
9
Y = G/H and X = G/(H N XH), and f : G/(H
r~XH) --~ G/H
f(u(HN~:H)) =uH
g: G/(H A XH) ~ G/H a = 9 ~ A ( H n ~H)
this l e m m a proves that A.(iH,~:,H)(fl) factors through G/(H A xH) in CA. Its image in A.(H) is then zero if x ~ Na(H). If 7r denotes the canonical projection from A((G/H) 2) o n t o / i ( H ) , there is a surjective morphism 0
O:
(~ A(H) ---+f4(H) ne~Va(H)
CHAPTER 11. THE SIMPLE MODULES
280
which maps the element/3 of the component n of the left hand side to ~rA.(iH,n,H)(/3). If K is a proper subgroup of H, and if ~ = tH(y), for 7 ~ A(I(), then denoting by 7rH the projection from G / K to G/H, I have
A.(iH,~,H)(/~) z A.(iH,,~,H o ~rH)(?) L e m m a 11.2.2 shows that this element factors through G / K , and its image under ~r is zero. Finally I have a surjective morphism 0
0: 9
X(H)+A(H)
~Na(H) Denoting by ~ | n the element/~ of the n component, I have
0(8 | n) = 7rA.( iH,~,H)(/3) If n and n' are elements of Na(H), and if fl a n d / 3 ' are elements of A(H), then
A.(iH,~,H)(/~) O~/H A.(i.,~,,.)(~') = A.(ig,n~,,H)(/~.~l~ ') Indeed, setting X = G/H, and denoting by x ~ xn the map from X to X which sends gH to gnH, I have
d.(ig,,~,H)(/~) Oa/H d.(iu,n, U)(/~') . . . . \
XlX2
As the square
/
\ZlX2X2X3 /
X l X l r t X2 Z2 n!
(/~ X /~1)
(x) X Xf t
X
x
--
--~
X2
1
[
X3
xlx2
~ X4 ZlX2X3
I
XlX2X2X3 /
is cartesian, I have
A xlx2x3
xlx
\XlX2X2X3]
)
X l X l ~ X2 X2 rtt
( A.
z
x x zn
xnn'
)A,( x ) x xn
whence
The right hand side can also be written as A.
(/~ x •') . . . . X Xn?'l I
...
XX
"
\X 1 X2rt/'
*
XX
x
xn
(•'
= A.(iH,,~n,,H)(/~.~/~ ')
11.3. THE STRUCTURE OF ALGEBRAS A(H)
281
It follows that 0 is a morphism of algebras from A.(H) | Na(H) to /~(H). This morphism is moreover unitary, since the unit of A(H) | Na(H) is e~-]-ffN 1, which is mapped to
Conversely, if a E A((G/H)e),
and if n e -No(H) I can consider the element
d*(ig,,~,u)(a) of 7.(g). If a factors through Y = G/I<, then setting X = G/H, there exists b E A ( X Y ) and c E A ( Y X ) such that a = boy c. Then for n E No(H), I have
A*(iH,n,H)(a) = A* ( x ) A. (xlyx2~ A* ( x~yx2 ~ ( b • c) xn '. xlx2 / \xlyyx~/ As the square
XY
<11 X
)
XYX
~
X2
l
is cartesian, I have also
As
X Y = (G/H) x (G/K) ~-
II G/(H N gK) geH\GIK
then for any c~ r A(XY), the element A. ( 7 ) ( c ~ ) i s a linear combination of elements of the form tHHn~Kag. If the order of K is strictly smaller than the order of H, the groups Y N g I ( are proper subgroups of H, and it follows that the image of A. ( 7 ) (c0 in J.(H) is zero. So I have a morphism r f r o m / ~ ( H ) to A(H) | -Na(H), defined by
n6No(H)
Moreover r
| n) =
~
A*(iH,n',H)A.(iH,n,H)(/3) | n'
n'6NG(H)
The only non-zero product A*(iH,n,H)A.(iH,~.H) corresponds to n' = n, and it is equal to the identity of A(H). It follows that r is the identity, so that 0 is injective, hence an isomorphism. Then r is the inverse isomorphism, which proves the proposition. 9
CHAPTER 11. THE SIMPLE MODULES
282
11.4
T h e s t r u c t u r e of s i m p l e m o d u l e s
I know that the isomorphism classes of simple modules can be indexed by the conjugacy classes of couples (H,V), where H is a subgroup of G, and V is a simple ft.(H) | Na(H) module. G N o t a t i o n : I will denote by SHy or SHy (instead of SO/H,V) the module associated
to the couple (H, V). The module V can be viewed as 7VG(H)-module, by the homomorphism n
eC/H | n from No(H) to A(H) | NG(H). Then denoting by [Y] the permutation module associated to a set Y, with the set Y as a basis, I have the P r o p o s i t i o n 11.4.1: L e t H b e a s u b g r o u p of G, a n d V b e a s i m p l e A(H) @ ~ T g ( H ) - m o d u l e . I f X is a G-set, t h e n IX H] is a JVG(H)-module, a n d
S~,v(X) ~- Horn(IX'], V)~G<m I t is t h e set of m o r p h i s m s of N u ( H ) - m o d u l e s f r o m [X H] t o V w h i c h f a c t o r through a projective module. I f f : X -+ Y is a m o r p h i s m of G-sets, t h e n for c~ r SH,v(X) a n d y E y S , I have
SH,v,(f)(cO(y ) = ~ c~(x) zEX H ](x)=y I f g : Y --+ X is a m o r p h i s m of G-sets, t h e n for a E SH,v(X) a n d y E yH, I have =
F i n a l l y if a C A(X) a n d f E SHy(Y), t h e n a • f is t h e m o r p h i s m [(X • y)H] = IX u • yH] t o V d e f i n e d b y
from
(a x f)(x,y)= (A*(m~)(a)| l ) . f ( y ) w h e r e mr is t h e m o r p h i s m of G-sets f r o m G/H to X d e f i n e d b y m~(uH) = ux. P r o o f : I will first prove the isomorphism
&,v(x)
Horn(IX'l,
~
and the other formulae will follow. m
11.4.1
The isomorphism
SHy(X) ~
Hom([XH],v)~ ~(H)
First I observe that X H identifies with the set of morphisms of G-sets from G/H to X, by the map sending z to m~. Let a | v be an element of LG/H,v(X). Then a 6 A ( X • (G/H)). If x 6 X H, then m• C A((G/H) • X), and I can consider the
-o
(
)
product m r x a, which is an element of A (G/H) 2
Its image rr(m* ox a) is then in
11.4. THE STRUCTURE OF SIMPLE MODULES
A(H). This
Na(H),
is identified with A(H) |
283
which acts on V. Now I can consider
the map ),~,v : x ~ X ~ ~
7~(m;
ox a).v
Proposition 11.3.1 shows that this is also
7r(rn*~ox a).v =
E
[A*(iH,~,H)(m* Ox a)| n] .v
nE/~G(H)
But 1 can view a as an element of
AG/H(X),
and then
m x* ox a = A'a~H (m~)(a) = A* (m~ • I d)( a) Thus Aa,v(X) =
E
[A*(iH,.,H)A*(m~ x Id)(a) | n] .v . . . .
nEI~G(H) . . . .
ux unH )
neNG(H)
Now let g~,~ be the linear map from
IX H] to
V defined by
ux uH (a) | l .v E V Then
TrN~ ~(H)(g.,.)(z) = neNG(H)
This can also be written as
TrNla(H)(g~,.)(x)
=-
~
[~A*
(un_lxuH)uH
(a)|
nE~ra(H) or
TrNa(H)(ga,.)(X) =
~ A* (unH) A* (un_lx uH) (a)Qn .v ne~Va(H)
whence finally
TrNl C(H)(ga,.)(x) =
~
A* (ux unH) (a)|
.v
This formula proves that A~,v = Tr~C(H)(ga,v), hence that A~,. is a morphism of JVa(H)-modules from [X H] to V, which factors by a projective module. I obtain that way a morphism
~: A(x • (G/H)) |
V -~ ~om([X"l, V ) ~ G(")
Moreover, ifbEA((G/H)2),then /~aOG/Hb,v z /~aw(b).v
a | v H Ao,~
CHAPTER 11. THE SIMPLE MOD ULES
284 Indeed, for x E X H
(" m~ o x a o a / . b) .v =
~(~;
ox a)~(b).~
The morphism ~ is then a morphism from LG/H,v(X) to Hom([XH], v)N1 G(H), which passes down to the quotient SH,v(X): indeed, if ~ a~ | v~ is an element of Ja/H,v(X), then for any ~b E A((G/H) x X ) , I have ~(r
ox ai).vi = ~ 7r(r o x ai).vi = 0
i
i
It follows that for any x E
XH E 7r(m*z ~ ai).vi = 0 i
so that ~ i ~a,,~, = 0. Thus I have a morphism, that I still denote by ~, from SHy(X) to Hom([XH], V)~ c(H). Conversely, let f E Hom([XH], V)~ a(u), expressed as
f = TrNIG(H)(g)
(11.1)
where g E Hom([XH], V). If x E X H, then m~,. E A ( X • (G/H)), and I set
#(f) = ~
m~,. | g(x) E SHy(X)
(11.2)
x6X n
Then # ( f ) does not depend on the choice of the element g in (11.1): this is equivalent to say that if TrNa(H)(g) = 0, then the expression (11.2) is zero. In SH,v(X), this means that for any r E A((G/H) • X ) , I have
(r ox ,,,~..).g(x) = 0 x6X H
Let g be any element of Homn([XH], V). Then
xEX H
~EX H
But setting Y = G/H, I have
'
\ YlY2 /
\ylXlXly2/
Moreover
r • rex,. = r • A.
m~(y)y
(e) = A. ~kylxlmx(Y2)Y2]
\
ylXl
11.4. THE STRUCTURE OF SIMPLE MODULES
285
As the square
YlY2 (
y2 ~
YlY2
> YXY ( ylxly2 \ylzlm=(y2)y2 ) \ylXlXly~]
is cartesian, I have
A*( Y~xlY2 I \ylxlxly2 /
(
ylXlY2
~
~
YlY2
~A*(
YlY2
and this gives ~boxm~,.= A,(ylxly2)
\ YlY2 / \YlI=
\ylm=(y=)/ (~) = A*(Id x m=)(@)
Finally Z
[A*(iH,.,H)A*(Id • m=)(@) | n] .g(x) . . . .
(r
zEX H
xEX H
~eR~(H) =~x-
'
~H ~x
9
@) | ~ . s
( ~ ) @ Tt . g ( r t - l x )
....
....
.~Ra(H)
xEX H
Thus if TrlNG(H)(g) = 0, then the right hand side of (11.2) is zero9 It follows that # ( f ) is well defined by equations (119 and (11.2). So I have a morphism from Hom([XH], V)~ G(H) to SHy(X). Moreover, if a | v E SHy(X), then as ~ = Tr~G(H)(9~,~), I have ~(Ao,~) = ~
.~=,. e go,v(=)
x~.k"H
If ~b 9 A((G/H) • X ) , the previous computation shows that I have
286
C H A P T E R 11. T H E S I M P L E M O D U L E S
9
,9
z
xEX H
[ ""1 A* ( u H ux
]
(~b) | 1 .Aa,v(X) . . . .
? (,x%),~
E
xs H
....
nENG(H) =
But
E xEX H neNG(H)
uH
uH "uH
"'"
uH
"uH
u l H UlX U2X u2nH.] (~ x a) . . . .
uH ug
"'"
= A* ( u H
uH ux ux "unH) (r x a)
This gives E (r OX lT~x,*)*ga,v(X) : xEX H
A* E xEX H n6/~G(H)
u H ux ux u n H
(~ x a ) |
.v
On the other hand
(~ ox a).v
=
E
[A*(i.,.,.)(r
o~
a) | ~] .v
Moreover A ( ' . , . , g ) ( ~ OX a) = A (ZH,.,H)A. \ u , H "u,H /
\u,H x x u,H/(r
As the square
G/H x X uH x
u H x "~ u H x "unH ) ) (G/H) x X x (G/H)
GIH
,
\ u l H u2H / (GIH) 2
uH
('uH un, H ) is cartesian, I have A.(iHnH)A. ''
('ul H X "u2gx~ : A . ( ' u S x~ A* ( u g x ) \ u l H u2H / \ uH / \uH x unH/
x a)
11.4. THE STRUCTURE OF SIMPLE MODULES
287
and then
(uHX']A.( u I I x ) A * ( ~lHX'u2g ) = A. \ uH ] \ u H x unH] 2u~H x x usH ( r
A*(iH,n,H)(r
....
( u H x ) A* ( uH x "~ . . . . A. \ uH ] \ u H x x unH] (~b x a) This gives
(r
E
ox a).~ :
A, 2(UHuHX'~]A* ( u H x x unH ] (r x a) | n .v
,ae~Va(H) But the map sending (uH, x) E (G/H) x X to g-~x 6 H \ X induces a bijection
G\((G/H) • X) ~_ H \ X Then the map c~ defined by
~:u(HnG~)e
II a/(Hna~)~(u(HnC~)) zeH\X
=(gH, gx) e ( O / H ) •
is an isomorphism of G-sets. In particular,
IdA((a/H)•
= Y~ A.(cr~)A*(~) x~H\X
Then (r
=
Z
A.
(u")A.(ax)A (a~)A (uHxxunH)(~b xa)| uH
x
9
9
uH
x
....
xEH\X
[A . \{u(HNG~,)]A. (u(HNG~) ~ ].v uH " ] \uHuxuxuna] (r • 1 7 4
E neNc(H) xEH\X
= tH But A. \(u(HnG~) ~H ~H ] ( a ) = 0 for a n y a . HnG~" Thus i f H ~: G~, then A , k Now I can restrict the summation to x E H \ X H = X H, and then
(~20X a).tl
:
E
[A,
uH
,,.
uHuxuxunH
)
]
(r • 1 7 4
.v . . . .
zEX H
....
[-<
~ A* uH ,~Zc(H) uHuxuxunH xEX H
Finedly, for any ~, e A((G/H) x X ) , I have
xEX ~
)
(~ x a ) |
]
.v
C H A P T E R 11. THE S I M P L E M O D U L E S
288 which proves that in S H y ( X ) , I have
:zE X H
or that # o A is the identity. Conversely, if f E Hom([XH], V ) ~ G(H), i.e. if f is of the form
f = Tr~c(H)(g) then
#(f)=
~
m~,.Qg(x)
xEX ~
so that for y E X H, I have
A o #(f)(y) = neJVc(/4) zEX H Moreover
uy u n H
(m~,,) = A* \ u y u n H
A. \ u x u H /
For g i v e n n , x a n d y , let Q ..... 9, a .... ~ and b.... y be such that the square ar~,m ,xj
Q .... ~
>
C/H
G/H
,
x • (G/H)
uy u n H is cartesian. Then if Q~,y is non-empty, there exists u and u' in G such that ux = u'y and u H = u'nH. These equalities imply that
uH.x = {ux} = {u'y} ~ u'nH.x = {u'nx} uH
uH
s o y : nx. Thus if y ~ nx, then Q .... y : 0, and the product A* ( ~ u ~ - , ) A. (=~,H) is zero. And if y = nx, then Q .... y identifies with G / H , the m a p b,~,,:,y being the identity, and the map a .... y being right multiplication by n. So I can write . . . .
-e~a(H)
t
\
--~
.e~o(H)
j
~eRa(H)
....
T~lVC(")(g)(~) = f ( y )
which proves that I o # is also the identity, hence that )~ and # are mutual inverse isomorphisms.
11.4. THE STRUCTURE OF SIMPLE MODULES 11.4.2
The
A-module
289
o f SH,V
structure
Let f : X ~ Y be a morphism of G-sets, and h E Hom([ZH], V)~ ~(H), expressed as
h = Tr~G(H)(k). Then #(f)=
~
m~,.|
x6X H
so that
SH,V,.(I)@(f)) = ~
m.(f x Id)(m~,.) | k(x)
x6X H
Under the isomorphism A, this element is sent to the map which sends y 6 yH to the element
~ex"
uy uH
A . ( f • Id)(m~,.) |
.k(x)
neNG(H)
Moreover
A . ( f • Id)(m~,.) : A , ( f • Id)A,
(r
: A.
(Ca~H)
Let Q~,~,y, a .... y and b.... y such that the square an,x,y
G/H
, g • (G/H)
uy unH is cartesian. If Q .... y r 0, then there exists u and u' in G such that uy = u'f(x) and unH = u'H. These equalities imply that
unH.f(x) = {unf(x)} = u'H.f(x) = {u'f(x)} = {uy} uH
uH
Thus if y r nf(x), then Q .... y---- O, and the product A" (~yuH)A. (,d(~)uH)is zero. If y = nf(x), then Q,:,y ~- G/H, the map b~,~,y is the identity and the map a .... y is right multiplication n. This gives e ----
E nENG(H) xeX H, f(nz)=y
IA*
uH n I .k(x) = ( u n H ) (Ea/H| ""
H
(~7-~ | ~).k(~-'x) =
xEX H nENc(H)
~ h(x) ](x)=y x6xH
](~)=y So I have proved that
SH,y,.(f)(h)(y) = H xEX H
f(~)=y
h(x)
290
CHAPTER
11.
THE SIMPLE MODULES
Now if g is a morphism of G-sets from Y to X, then S;/,y~(h) =
~
A*(g • Id)(m~,.) 0 k(x)
x~X H
The image under A of this element is the map sending y E yH to
E
uy u H
A*(g • I d ) ( m x , , ) @ n
.k(x) . . . .
xE X J4
~Va(H)
= ~o,(h)(g(y))
~e_x H n~Nc(H)
= ~g(y)
ug(y) u H
So I have
s;~,v(g)(h)(y)
=
hg(:y)
Finally, let X and Y he G-sets, and
h = Tr~G(H)(k) E Hom([rH], V) Then
#(f)
=
~
my,.|
yEY H
Thus, if a E A ( X ) , I have a • #(f)
(a • . ~ , . ) 0 k(y)
--yEY H
The image under element
A
of this element is tile map sending z C Z H = X H X y H to the
y6Y H
Moreover
. . . . A*(
x,,H )
""
\uz unH/
x uy u H )
Let Q ..... y, a .... ~ and bn,z,~ such that, the square an,z,y
Q .... ~
--~
X•
"
uy u H ]
GIN
, uz u n H
z •
(a/H)
11.5. THE SIMPLE GREEN FUNCTORS
291
is cartesian. If z = (x0, Y0), and if Q~,z,y ~r 0, then there exists u and u' in G such that uz = (UXo, uyo) = (x, u'y) and unH = u'H. In those conditions
u'H.y = {u'y} = {uyo} = unH.y = {uny} Thus if y0 r ny, the set Q,~,z,y is empty, and the product A* ( . . . . H) A. \x ~ ~HJ zero9 and if Y0 = ny, then Q,~,z,~ ~- G/H, the map b~,~,y is the identity, and the map an,z,y is defined by a,~,~,y(uH) = (UXo, uH) It follows that
e=
E ~e~ra(H)
[A* (uxUoHuH) A* (x:H)(a)@rt]
nC/VG(H) L
....
\ j
(A*(rn~:o)(a) | 1)
u/ x
./r
....
0
• (CG/H@rt).~(l't-lyo) .... ~Na(H) 9
, .
z
...
....
| l).h(>)
which proves that
(a • h)(xo, yo) = (A*(mxo)(a) @ 1).h(yo) and completes the proof of the proposition.
11.5
T h e simple Green functors
The previous results give a new proof of a theorem of Th6venaz (see[13]) on simple Green functors. The notion of simple Green functor relies on the notion of functorial ideal, that Th~venaz defines as follows (see [13] 1.8): a functorial ideal I of the Green functor A for the group G is a sub-Mackey functor such that for any subgroup H, the module I(H) is moreover a two-sided ideal of A(H). It is equivalent to say that the Green functor structure of A passes down to the quotient A / I . Translating this definition in terms of the product x, I see that for any G-sets X and Y, I must have I ( X ) • A(Y) C_ I ( X x Y)
a ( x ) • I(Y) g I ( X • Y)
Conversely, if I is a sub-Mackey functor of A such that this condition holds for any X and Y, then the product • of A passes down to the quotient A / I . In other words, a functorial ideal is nothing but an A-submodule-A of A. Thus a Green functor A is a simple Green functor if and only if A is simple as an A-module-A. But the A-modules-A are also the A@A~ Thus if A is a simple Green functor, then there exists a subgroup H of G and an AQA~ | JVa(H)-module V such that
~,A~A~ A "~ ~H,V
CHAPTER 11. THE SIMPLE MODULES
292
as AQA~ (the exponent A(~A ~ recalls that the simple module SHy is a simple module for this Green functor). Then H is the unique minimal subgroup of A up to conjugation, and V = A(H) = ~.(H) is a simple A~A~ | Na(H)-module. Now
(A~A~ g ~ AH~(A~ H ~ AH~)(AH) ~ This follows from lemma 10.8.5, since moreover for U = H \ G as a .Nc(H)-set-G, I have U/G = . , and so
(Res~va(H)b)H = s
= bu/G = b
This isomorphism is now clearly an isomorphism of Green functors, and it follows from evaluation at H the isomorphism of algebras
A@A~
~_ A(H) | A(H) ~ = A(H) | A(H) ~
Now say that A(H) is a simple A(H) | A(H) ~ | JVa(H)-module, is equivalent to say that d ( Y ) is a _Na(H)-algebra, o1" an algebra with an action of -Na(g), having no proper two-sided ideal invariant by Na(H). Moreover, if A is non-zero, then A(*) # 0. As SH@g4~ ~ Hom({.l, v)N1o(H) I see that T r ~ ( H ) ( A ( H ) ) has to be non-zero. As it is a two-sided invariant ideal of A(H), I must then have %
/
A(H) :
which proves that A(H) is moreover a projective/Va(H)-algebra. Conversely, if B is a Nc(H)-algebra, then I have a Green functor FPB, such that for any subgroup K of NG(H), the ring FPB(H) is equal to B H, the restrictions being inclusions, and the transfers being relative traces. In those conditions, for any /Va(H)-set Z, I have FPB(Z) ~-- H o m ~ ( H ) ( [ Z ], B). If B is a projective _NG(H)-algebra, having no proper two-sided invariant ideal, then let G NG(H)
A = IndNG(H)InfK%(H)FPB
Then for any G-set X, I have
A(X) = Hom~G(HI([X"], B) and as B is projective, it is also
A(X)
=
Hom([XH], B)~ c(H)
In those conditions, the group H is a minimal subgroup for A, and A(H) = A(H) = B is a simple A(H) | A(H) ~ | Na(H)-module, so that
A
~- S A~gA~
H,B
as A-modules-A. Thus A is a simple Green functor. So I must find, for a group K, which are the algebras B with a K-action which are projective and have no proper two-sided invariant ideal. Let ] be any maximal
11.5. THE SIMPLE GREEN FUNCTORS
293
(proper) two-sided ideal of B as an algebra.. Such an ideal exists by Zorn's lemma. Let L be the stabilizer of [ in K. Then for any k 5 K , the image k(I) of I by k is also a m a x i m a l two sided ideal of B. In particular
f-) k(tl = 0 kEK
since it is a two-sided invariant ideal. Let P be a subset of K of maximal cardinality such that
J = (~ k(i) # o kEP
Then P :~ K , and there exists ko C K - P. Replacing P by kolP, I can suppose k0 = 1 ~ P , and then
I N J =O Moreover, the m a x i m a l i t y of I implies I + J = / 3 , thus
B=I| Then J identifies with B/I, which is a simple B-module-B. Thus J is a minimal two-sided ideal of B. Then J I is contained in J V/I = 0. So I is contained in the right annihilator of J , which is a two-sided ideal. As J # 0, the annihilator of J is equal to I. On the other hand the sum
keKIL is a non-zero two-sided invariant ideal of B, so it is equal to B. As J.k(J) is contained in g A k ( J ) , which is a sub-two-sided ideal of J , I have J.k(g) = 0 if k ~ L. Thus if (jk)keK/n is a family of elements of J such that
Z
k(jk) = 0
kEK/L
then for any kl 6 K , I have
o = kl(J) E
k, ( E
keK/L
= 1(J.5 1)
\keK/L
]
Then Jkl is in the annihilator of J , hence in I, hence in I C~J. Thus Jkl = 0, and B =
k(J)
(9 keK/L
In particular, the unit of B decomposes as IB :
E
ek
ke1,/L for some ek E k(J). If kl E K/L, I have
keh'lL
CHAPTER 11. THE SIMPLE MOD ULES
294
and unicity of the decomposition implies ek = k(el). Moreover
el ~---el Z
k(el) z e~
kEK/L and for any b E B, I have
elb =
elb
E
kEK/L
k(Cl):
Z
kEK/L
elbk(el)=elbel
elbk(el) G J.k(J) = 0 if k ~ L. The same argument shows then that be1 = etbel, so that el is central in B. Then J = elB is a simple algebra with an
because
action of L, and B is isomorphic to the induced K-algebra B _~ I n d 2 J =
RK |
J
with product defined by
(k|174
0 if k - l k ' ~ L k @ j.(k-lk')(j ') otherwise
Now say that B is projective means that there exists elements jk, for k E that / \
kl EK
K/L, such
kl ( ~ k(jk)) =1t3= ~_, k(el) \kEK/L / kEK/L
The product by el gives el=
Y~ ~l]C(Jk) z E l(jk): Zrf ( E jk) kl EK kEh'/L \keK/L I
kEK/L klk6L
IEL
It follows that J is projective as L-algebra. Conversely, if
el = TrL(j) = ~ l(j) IEL then
T@'(j) = TrI~-TrL(j) = Tr~L'(el) = 1B and B is projective. Similarly, if J is a simple algebra on which L acts, then IndLKJ is a K-algebra without proper two-sided invariant ideals: if U is such an ideal, and if el is the unit of J, then V is the direct sum of k(el)V, for k E h'/L, and k(el)U is isomorphic to elU, which is a two-sided ideal of J. If elU = 0, then U = 0, and if elU = J, then U=B. Thus the algebra B is isomorphic to I n d , J , where L is a subgroup of K, and J is a simple algebra with an action of L, the algebra J being moreover projective as L-Mgebra. Finally, I observe that the group L and the L-algebra J are entirely determined if B is known as K-algebra: indeed, if/1 is a maximal two-sided ideal of B, then k(el)I1
11.6. SIMPLE FUNCTORS AND ENDOMORPHISMS
295
is isomorphic to elk-l(/1), which is a two-sided ideal of J, hence equal to zero or J, and I,= @ k(el)J kEK/L k(e~ )11 r
As I1 is maximal, there exists a unique ko C K/L such that h =
@
k(
l)J
keK/L-{ko}
The maximal two-sided ideals of B are then conjugate by K. The group L is the stabilizer of a maximal two-sided ideal of B, and the algebra J is the unique simple quotient of B: the couple (L, J) is then unique up to conjugation by K. Going back to the case of a simple Green functor A, and observing that induction and inflation of algebras commute with the functor FP_, it follows that there exists a subgroup M of Na(H), containing H (such that M/H is the above group n), and a simple algebra S (equal to J) on which M/H acts projectively, such that a M p A ~- IndMInfM/HF S So the previous discussion gives a new proof, under slightly weaker hypothesis, of the following theorem:
P r o p o s i t i o n 11.5.1: ' T h 6 v e n a z [13] T h e o r e m 12.11) 1. Let A be a s i m p l e G r e e n f u n c t o r for G. T h e n t h e r e exists a s u b g r o u p M of G, a n o r m a l s u b g r o u p H of M, and a s i m p l e a l g e b r a S on which M / H acts p r o j e c t i v e l y , such t h a t A ~_ IndMInfM/HFPs G M T h e t r i p l e (M,H,S) is unique up to c o n j u g a t i o n b y G (and up to i s o m o r p h i s m of M/H-algebras for S). 2. C o n v e r s e l y , if H_~ M are s u b g r o u p s of G, if S is a s i m p l e a l g e b r a on which M/H acts p r o j e c t i v e l y , t h e n IndaMInfM/HFPs is a s i m p l e G r e e n functor.
11.6
Simple functors and endomorphisms
Let G be a group and A be a Green functor for G. In this section I will study the G relations between the simple A-modules SH, v and the functors - 8 - and 7 - / ( - , - ) . First I observe that proposition 11.4.1 gives a definition of S H,V c for any J.(H) | Na(H)-module V (non-necessarily simple). Moreover, the structure of Mackey functor G y depends only on the restriction of V to NG(H) ~- b(H) | N a ( H ) : in other of SH, words
aesbS ,v = SH,Res~rO(H)V a Similarly, I can define:
296
CHAPTER 11. THE SIMPLE MODULES
Definition: Let H be a subgroup of G, and V be an A(H)| [denote by FP~, V = FPH,v the A-module defined like SHy, but with any homomorphisms, and not only those which factor through a projective module: if X is a G-set, then FPg, v ( X ) = HomK~G(H)([xH], V) ff f : X ~ Y is a morphism of G-sets, then for a 6 FPI~,v(X) and y 6 yH, [have FP~,v.(f)(cO(y ) =
~
a(x)
x6X H
f(~)=y If g : Y ~ X is a morphism of G-sets, then for c~ E FpGH,v(X) and y E yH, I have
FP~,~*(g)(9)(y) = ~g(y) Finally, ira E A ( X ) and f E FP~,v(Y ), then a x f is the morphism from [(X • y)H] = [X H x yH] to V defined by (a x m)(x, y) = (A*(m~)(a) | 1).f(y) where mx is the morphism of G-sets from G / H to X defined by m~(uH) = ux. I define duMly FQaH,V for a G-set X by F QH,v(X) a = If f : X ---+Y is
a
[x']
v
|
morphism of G-sets, then FQ~,v.(f)(x | v) = f(x) | v
If g : Y --+ X is
a
morphism of G-sets, then G
*
FQ~,v (f)(x | vl =
E
Y| v
y6Y H
g(y)=x Finally, ira E A(X), and y | v C FQ~,v(Y ), then a x (y|
= ~
(x,y)|
xEX H
It is easy to see that these definitions turn FPI~,V and FQ~,v into A-modules. The notation comes from the fact that if A is the Burnside functor, then FP~,v and FQH, VG identify respectively with the functors denoted by Ind~%(H)InfN:((HH))FPv
IndGxc(H)Inf~7:((SH))FQv
by Th6venaz and Webb. The following lemma is a way to recover this isomorphism: L e m m a 11.6.1: L e t M be an A-module. Then, for any subgroup H of G, the m o d u l e s _~r(H) and M__(H) are A(H) | Moreover HomA(FQGH,v, M) = Hom~(H)|
) (V, M ( H ) )
HomA(M, FPg, v) = Hom~(H)|
) (M(H), V)
11.6. SIMPLE FUNCTORS AND ENDOMORPHISMS
297
A(H) | Nc(H) on M(H) induces an action of
P r o o f : It is clear that the action of A(H) | N o ( H ) on M ( H ) , since
H H (tH(a) | n).m = tH(a).n.rn = th-(a.rK(n.m))
Similarly, if m E M ( H ) , and if K C H, then H (a = tK
|
/-/
= 0
because = n. /,.~(m) : O. So M(H) is also an J,(H) | .Nc(H)-module. Moreover if r is a morphism from FQH,V a to M, then r is a morphism of A(H) | N c ( H ) - m o d u l e s from F Q ~ , v ( H ) to M(H). But it is clear that FQH,v(H) is isomorphic to V, and that FQ~,v(K ) = 0 if K is a proper subgroup of H. Then the image of r is contained in M ( H ) , and r is a morphism of A(H) | Na(H)modules from V to M_M_(H). Conversely, if ~b is a morphism of A(H) | NG(H)-modules from V to M ( H ) , and if X is a a-set, I define a morphism ~bx from FQ~,v(X ) to M(X) by setting
Cx(x o
=
It is easy to see that this defines a morphism of A-modules from first adjunction of the lemma follows. The second one follows from a dual proof.
a to M. The FQH,v 9
L e m m a 11.6.2: L e t H be a s u b g r o u p of G, a n d V be a ATc(H)-module. I f K is a s u b g r o u p o f G, t h e n
S~,v(K) : 0 = SaH,v(K) if It* r
H
S~,v(H) : V : S~,v(H) P r o o f : By definition, as
(G/H) H = Na(H)/H, I have
S~4,v(G/H) = Hom([(G/H)"], V)~ a(H) = Horn(R-No(H), V)i~G(H) -~ V Then v E V is associated to the morphism
from from
RNc(H) to V. Moreover, if X is a G-set, if x0 E X, and if 5~0,v is the morphism IX H] to V defined by '
f 0 if
X
l,
x =
v
if
#
then
nWVc(H) n.v o :x
Xo xo
CHAPTER 11. THE SIMPLE MODULES
298
Let rn~ 0 the morphism of G-sets from C/H to X defined by rn~:o(GH) = gxo. It follows from proposition 11.4.1 that ~elV~(H) nxo:x
thus Tr[a(s4)(5~o,, ) = Sa,v.(m~o). If I( is a subgroup of G, then
S~4,v(I<) = Hom([(G/Ii)U], V)~ c(H) is zero if H is not contained in K modulo G, because then argument proves moreover that
S~,v(K ) = ~
(G/K) s = 0. The previous
tH.Su,v(H ) ,, a
~:EG H~CA
-
It follows that = 0
if Ix" #G H. It is clear moreover that S~,v(H ) = S~I,v(H) = V. A dual argument proves the assertions on S~,v(K), and the lemma follows. " P r o p o s i t i o n 11.6.3: L e t G be a g r o u p , a n d M a n d N b e M a e k e y f u n c t o r s for G. * I f f o r a n y s u b g r o u p H of G, one of t h e m o d u l e s ~r(H) or N ( H ) is z e r o , t h e n M@N = O. . I f for a n y s u b g r o u p H of G, one of t h e m o d u l e s M(H) or N ( H ) is zero, t h e n N(M, N) = 0 P r o o f : Indeed, I have already seen that if M and N are Mackey functors for G, then
M~N(H) " M(H) | N(H) If one of the modules ~r(H) or iV(H) is zero for any H, then M@N(H) = 0 for any H. If M@N # 0, there is a minimal subgroup H for MQN. Then
0 # (M~N)(H) ~- (M~N)(H) = 0 This contradiction proves the first assertion of the proposition. The second one follows similarly from the equality
n u n ( M , N) = n(s
~u(N))
for U = H\G, viewed as a IVa(H)-set-G (see proposition 10.1.2), which proves that
7{(i, N)(H) = H o m . (_~(H), N ( H ) ) Then if 7{(M, N) ~ 0, for any minimal subgroup H of 7-/(M, N), I have
0 r 7{(i, N)(H) = 7t(M, N)(H) = Hom.(-~r(H), N__(H)) so the two modules
M(H) and _N(H) are non-zero.
,,
11.6. SIMPLE FUNCTORS AND ENDOMORPHISMS
299
Proposition 11.6.4: L e t H a n d K b e s u b g r o u p s o f G. I f V is a Na(H)m o d u l e , a n d W a Na(K)-module, a n d if H a n d K a r e n o t c o n j u g a t e in G, then G
~
SH,VeSK,
G
G
W ~- 0
~t~(SH,V,
G
S K , w ) ~--- 0
M o r e o v e r , if K = H, t h e n G
~
G
F
G
G
SH,VeSH,W ~-- QH, v . . w
G
G
H(X~,v, SK,W) ~-- FP~,Hom.(V,W)
as M a c k e y f u n c t o r s . P r o o f : The first assertion follows from proposition 11.6.3 and from lemma 11.6.2. As moreover the module S~,v(K ) is zero for any proper subgroup of H, it is clear that the same is true for (SH,vQS~,w)(Ii), 6' ^ a - and that moreover
~- SH, v(H)eSH,w(H ) ~-- ViW" and lemma 11.6.1 shows that there exists Then (S~4,v+S~,w)(H) = (SH,V| a ^ a a a ^ a a morphism
r
: x @ (v | w) E FQH,v+w(X) ~
Conversely, if X and Y are G-sets, it is easy to check that the correspondence which maps r = TrFa(H)(r E S~,v(X) = Hom([XH], V) Na(H)
r E S~I,w(Y) = Hom([yH], W)F a(H) to
Ox,y(r162 = ~
(x,y)|
(r
E FQ~,v(X x Y)
| r
xEX H
yEY H
is well defined, and bifunctoriah indeed, if r = TrF~(H)(r OX,Y( r r
=
iX, y) @ (r
E
I have also
| ~)O(n--lx)) . . . .
xEX ~/
yEY H
nE-NG(H)
E
i
. . . .
xEX H
xEX H
yE y ti nEI~G(H)
yEY H
~eRG(H)
E
(x,y)e(n-lr174162
(x,v)|162174162
z E X .4
xE X ~I
yEy tt
yEY H
nERo(H) It follows that Oxy(r r is independent of the choice of r such that r = Tr N~ If moreover f : X --+ X' and g : Y -+ Y' are morphisms of G-sets, then for x' E X', I have
r
xEX H
f(=)=~'
CHAPTER 11. THE SIMPLE MODULES
300 Setting r
=
~
Co(X)
x6X H
f(~):~'
I have SHa,v.(f)(r = TrNa{H)(r G
, and
then
= Z
G
,
i Xl
,
sG
)) ....
x'6x'H y'6y 'H
....
(x',y')|174
~
= FQaH,v.(f xg)Ox,Y(C,r
(x,y)e(x • (:xg)(~,y)=(:~',r
Now if f is a morphism from x' C X'
X'
to X and g is a morphism from Y' to Y, then for
G * SH,V (f)(C)(x')
= r
Setting C~ = C0 o f, I have S~,v*(f)(C) = Trfc(H)(C'o),
E
(x',y')|
(x',y')6x'H •
....
~
~
so X/ @
I
....
'H
(x',y')|174
= FQ~,v*(f xg)Oi,r(m,n)
(~,y)e(x xY)~ (~%')e(x'xY') H (: xg)(~',r
Ox,y FQH,v|
is bifunctorial, and induces a bilinear morphism from SGH,v, S~,,w to G ^ G This gives a morphism 0 from SH,V@SH, W t o FQH,v6w , defined for a G-set X by
Thus G
ex:
[r | r
9
(S~,y@S~,w)(x) ~ ~
f(y)
| (r
y6Y H
where Co is a morphism
[yH]
from
to
V such that
Then (I) and 0 are mutual inverse isomorphisms: if G x | (v | w) 9 FQH,v|
then ~(x) = [5~ | 5.o](a/H,~) Let 5o,~ be the map from
Na(H)
to V defined by
5o,.(n) = Then 5. = 00(x|174
v if n = l 0 otherwise
Tr~C(H)(5o,~), and =
E n6/VG(H)
rn~(n)|174174174174174
| r
9
FQaH,V|
i1.6. SIMPLE FUNCTORS AND ENDOMORPHLY'MS
301
So O~ is the identity,. Conversely, if (Y, f) is a G-set over X, then
ox([r o <(~,s)): ~
,r(:,j) e (~o(~j) o ~,(,,j))
y~yH
where r
is such that Tr~G(s)(r
= r
Thus
(1~.3) y6Y H
Let r = ~ e Z ~ ~q,G mv.(m~)(6r
. Then if Y0 ~ y H I have
H
r
: ~ Siv.(,,,,,)(ar
:
~
yEY
~,,0(.~,(,~). . . .
,qEY ' t
n6Nc;(tt) m v n)=90
....
Z
,,,o(y) =
y6Y t!~
Z
,,,~o(,-b0) = r
n6NG(H)
ny:yo
'H,V ( ',)( Co(y))" Then -,G
~
/
yEY H
Moreover if n 6 NG(H)
&f,v (-,..~)(~')(,.):
V,(,,v)
&,~(,J
So yEY H
As moreover .fm~ = rn](,v), this expression is equal to the right hand side of (11.3), thus 0 0 is also the identity, proving that ~G
5
~<;
~%~H,VQg'gH,W "~
F
G
QH,V|
To prove that ~(, c;c; H,t....qG H,W)) ~- FP[j.Hom~,(V,W)' I wi]] use the following lemma: L e m m a 11.6.5: L e t H be a s u b g r o u p of G, a n d V b e an A(H) | NG(H)m o d u l e . I f X is a G-set, t h e n Homz,([Xn], V) is an 7,(tt)Q , ~ G ( H ) - i n o d u l e by : (,, r
a n d t h e r e is all i s o m o r p h i s m of A - m o d u l e s
(S~,v)X ~-- '-"H,HomR([Xq.V)
CHAPTER 11. TIlE SIMPLE MODULES
302 Proof: Indeed, if 4) e then
S~v(X) and ifa@n and a'~Sn' are elements of A ( H ) Q , ~ o ( H ) ,
. . . .
thus HomR([xH], V ) i s an A ( H ) C ' !VG(lI)-module. Moreover. if Y is an H-set : sg,
O"x) = Hom([ 'H • x ' ] ,
The structure of N a ( H ) - m o d u l e of
Hom1{([xU],V) is
v))
such that
Hom([Y H • xH], V)~ a(Ir ~_ Hom([yH], HomR([xHI, V))~Va(~r) So
,G
,"
(S~,v)x(Y) ~ 5H,,om,~([XH3,V)(~ ) It is easy to see that these isomorphisms are compatible with the structures of Amodules of both sides, and the lemma follows. 9 It follows in particular that = HomM ~k(a)( 5H,V, H,HomR (IXH].W)) ~
Then I use the following lemma: L e m m a 11.6.6: Let H be a subgroup of G, and V and W be iVc(H)-modules. Then G ,G 5H,W) HomM~k(o)( SH,v, ~_ Homg~(H)(~/; W)
then OH Proof: Indeed, if 0 is a morphism of Mackey functors from SHy ,a to ,SH,W, ,a is a morphism of Na(H)-modules from ,5'~,v(H ) = V to S~,w(H ) = W. Conversely, if ~/, is a morphism of .Na(H)-modules from V to W, and if X is a G-set, I define a. morphism ~bx from S~,v(X) to Ssa w(X) by fx: ~ r
Hom([XH],V)~ G(H) ~
~o
It is clear indeed that if ~ factors through a projective module, so does bb~. It is easy to see that these correspondences are inverse to each other, and this proves the lemma. 9 In those conditions
?-t( S~,v, Sg,w ) (X ) ~_I-Iom~vG(H)(V, HomR( [xH], W)) = Honwa(H) ( V | [X H], W) . . . . ....
HomK%(H)([xH],
MotoR(V, W ) ) =
[~'PH,HomR(V,w)(X)
Those isomorphisms are moreover natural in X, and this gives an isomorphism of Mackey functors
7-((,fH,V , SH,W ) ~-- F PIt,Hom:a(v,w)
11.6.
SIMPLE FUNCTORS
303
AND ENDOMORPHISMS
and completes the proof of the proposition.
9
R e m a r k s : 1) In the case when V and W are A(H) @ NG(H)-modnles, then the module Hom g(H)(V, W) becomes a Nc;(H)-module, and then I have the isomorphism of Mackey functors l PH,Hom~(H)(V,W )
,
2) Let H be a subgroup of G, and V be a simple A(H) Q Nc,(H)-modnle. Then Iet B
~C
~,C
G
The Green [unct, or B admits H as a minimal subgroup, and B ( H ) ~- EndRV. This algebra is simple if and only if R is a. field k, and l/ is finite dimensional over k. Now ~B| ~ the functor /~ is a simple Green functor if and only if it is isomorphic to ,. H,E=a~V This is equive~ient to say that for a.ny (-,'-set X, I ha.re
The case X = 9 shows that this condition holds if" and only if (Endk l/)Nc.(Ju) = ( E n d k V ) f G'(H) i.e. if the kAWc,(H)-module V is projective. In those conditions, denoting by V ~ = Homk( V, k) the dual of V, I have V~ kV -~ Endk V if and only if V is finite dhnensional over k, and '-q,c; t t , V o ~ ~qc, k'~ H , V ]ho 9 Then -,G o < ~,O (bH,V) r v ~
c,G A q, G DH,Vo'JO'-tI,V
~
G
FQHyo|
~
S
G
QH,EnakV
Moreover, it is clear that F(~)~ E,a v ( t f ) = End~.V = FPI~Endkv(H). The morphism F QH,E,dkV ---+ b- P~',S,d~V which follows from lemma 11.6.1 factors a , through 5H,~:na~v, which is ~ quotient of ["Qu,s~a~v and a submodule of FPU,E,~d~V. It is an isomorphism if and only if ,
k
'
"
-,C1'
G
,
q,G
~ F
FQH,Endk~ ~ ~ H.EndkV --
G
P~I,EndkV
which is equivalent to say that Endj,.V is projective as k/Vc,(H)-module. Finally, those considerations prove the following proposition: P r o p o s i t i o n 11.6.7: L e t R = k b e a field. L e t C be a g r o u p , a n d A b e a G r e e n f u n c t o r for G. L e t m o r e o v e r It be a s u b g r o u p of G, and V b e a s i m p l e A(H) 4) ~ c ; ( H ) - l n o d u l e . T h e f o l l o w i n g c o n d i t i o u s are e q u i v a l e n t : 1. T i l e G r e e n f u n c t o r
,G ~tr q,C,' ~(5HV,
is s i m p l e .
2. A s M a c k e y f u n c t o r s
3. T h e m o d u l e V is finite d i m e n s i o n a l o v e r k a n d p r o j e c t i v e as kNG(H)module.
304
('tfAPTER i1. THE SIMPLE MODULES
I will say then that S~, v is endosimple. The number of endosimple modules is related to Alperin's conjecture in the following way: let .4 = b~ be tile in-part of the Burnside functor, for a prime number in: if X is a C-set. then bp(X) is the Grothendieck group of the full subcategory of G-sets over X, formed of G-sets }" such tha.t for any .y E Y, the stabilizer of y in G is a in-group. Then [~(tt) is zero if H is not a in-group, and isomorphic to h otherwise. Let k be an algebraically closed field of characteristic p. Then if ~~) is a in-subgroup of G, and V is a simple NG(Q) module (hence finite dimensional over k), the bp,G module ,SH, v is endosimple if and only if the module V is simple and projective. Thus Alperin's conjecture (in its global form) can be expressed by saying that the number of endosimple bp-modules is equal to the number of simple/,'G-modules.
Chapter 12 Centres 12.1
The centre of a Green functor
Definition: Let G be a group, and A be a Green fimctor for G. I call centre of A, and I denote by Z ( A ) , the commutant of A in A: if X is a G-set, then
Z(A)(X)
= {o~ E A ( X )
I v Y , v/~ c A ( Y ) , c~ x /3 = e~ x ~ fl}
W i t h this definition, t h e functor Z(A) is commutative.
a,
sul>Creen functor of A, which is clearly
Definitions: If A and B are Green flmctors fox" the group G, then the direct sum A | B of A and B is the direct sum of A and B as Mackey flmetors, with the product defined for O-sets X and Y, and elements a E A ( X ) , b E B ( X ) , c E A(}") and
(z ~ B(Y) bx (,~ r
b) • (~ ~ d) = (a • c) r (t, • d) c ( A ~ B ) ( X
• Y)
The unit of A @ t3 is the element CA @ CB of ( A @ B )( . ). If M is an A-module, and if z E Z ( A ) ( . ) , I denote by z x M the A-module defined for a G-set X by
(~ • M)(X)
= ~ • M(X)
1Te is an idempotent of Z ( A ) ( . ) , I denote by e x A the subfunctor of A defined for a O-set X by
(~
• A)(X)
-
~: • A ( X )
c_ A ( X )
Then e x A is a sub-Green functor of A (the inclusion being not unitary in general), with e = e x CA C (e X A ) ( . ) as unit. Moreover, if M is an A-module, then ( x M is a e x A-module T h e m o d u l e z • M is an A - m o d u l e , because if X and Y are G-sets, if c~ E A ( X ) and rn E M ( Y ) , t h e n o~ x ~ x m = (<~ x ~ :~) • m
( z x ,~) x '.~
If e is an i d e m p o t e n t of Z(A)(e), the functor e • M is an e x A module, since (e x ~) x (e x , ~ )
W i t h these definitions:
= e x ( . x ~ c) x m = ~ x ~ x a x b =
~ x <, x . ,
('HAPTER 12. CENTRES
306
L e m m a 12.1.1: L e t A, B, and C be G r e e n f u n c t o r s for t h e g r o u p G. T h e n A is i s o m o r p h i c to B @ C if a n d o n l y if t h e r e e x i s t s o r t h o g o n a l i d e m p o t e n t s e a n d f o f Z ( A ) ( . ) , s u c h t h a t e + f = <.1, a n d B _~ e • A a n d C ~ f • A. P r o o f : Indeed, the unit of B is in Z ( B H)C)( 9 b r B ( X ) and c E C ( X ) , I have cB x
since for a G-set X and elements
(b@c) = (es @ 0) x (b<"bc) = (CB x b) @ 0 = b
and similarly c) x eB = (b x ~ ) + : : ~ 0 = b
Then if A ~ B r C, the image e of ~s and the linage f of go' in A ( I ) a.re ort.hogonal idempotents, and e + f = CA. Converse 9 it" c a~(t .f are ort.hogonaI idempotents of Z ( A ) ( . ) such that e + f = e_4, then it is clear that the maps a C A ( X ) ~ (e
x
a) ~ (.f x ~) r (~: x A ) ( X )
r (.f x A ) ( X )
define an isomorphism from A to (e x A) I~t (.f x A). The main interest of this lemma is that since Z(A) is a Green functor, there is always a. (unique) unitary morphism of Green functors from b to Z(A). Thus any decomposition of Cb as a sum of orthogona.1 idempotents of b( 9 hldttces a decomposition of A as a direct sum of Green fnnctors. L e m m a 12.1.2: L e t G and H b e g r o u p s , a n d A b e a G r e e n f u n c t o r for G. L e t i b e t h e i n j e c t i o n f r o m Z(A) i n t o A. 9 If U is a G - s e t - H , t h e n
9 If V is an H - s e t - G , t h e n
Proof: Indeed, if X and Y are H-sets, if
z c (Z(A)o U)(X) -- Z(A)(U o. X) and if a r (A o U)(Y), then theh' product for A o U is
Shnilarly, in ( A o U) ~
z xV~
(}Sz)
,u
= ( A o U). k~:~/ A*(ay, x ) ( . x z)
B u t a • z = A* ( (,~,y)(.~,~,z)'~ ( z x a ), and moreover
o@, x
o
Uon
12.1.
307
T H E C E N T R E OF A G R E E N F U N C T O R
Thus z x U~
= A.
U OH
my
and the first assertion follows. Similarly, for the second assertion, if X and Y are H-sets, and (Z, f ) and (T,g) are H-sets respectively over X and Y, and if z C Z ( A ) ( H \ V . Z ) and a E A ( H \ i < T ) ,
then in Z . v ( A ) ( X • Y), ~ h~,~e Z(ZJ) • a(Tm) = A*(t~;,T)(Z • a)(z.TJm)
On the other hand
But a • z = A* (\ aG.va.z . , , . t aG.vl.t ..... )~ (z • a), and k G.v2.z G.vx.t )
T,Z =
Z,T
0
Thus \xY) A s (t~) zt is bijective, in s
• Y), l h a v e
~/[oreover
s
xy
Finally as
~
~y
o
tz
=
,
,ave
z(zd) x~ a(Tm) = A*(nVT)( z x a)(z.T#mt = Z(zd) x a(r,~l
which proves the second assertion of the lemnm.
s
9
I have already used an application of these lemmas, when I have built for an N the module N o UI,~ = )<<.(CA) • (=V o U): indeed, the element
XA,.(SA) is in Z ( s
o U)(.).
N o t a t i o n : I will denote by rc = ~rR(G) the set of 1)rime factors of the order of G which are not invertible in 19. I will suppose for simplicity that l{ is contained in A(o) (otherwise I can replace R by its image in A ( - ) ) . Then the Burnside ring b(.) = bR(G) with coefficieui,s in
CHAPTER 12. (_:ENTRES
308
Fg has a family of idempotents f ~ indexed by rr-perfect subgroups of G (see theorem (9.3) of Thavena~ and Webb [~5], or chapter 5.4 of Benson's book [1]): the primitive idempotents of bq(G) are indexed by the subgroups L of G, up to conjugation, and given by the following formulae of Gluck (see [7]) 1
e ~ - INn(L) I Z
hCL
II"I~]I;,GG/I'~
The idempotent c~ is characterized by the fact that for any G-set X, I have
~'.x -- I X H I 4 gI1 and in particular I(ea)~'~ I is zero if H and Ix are not conjugate in G, and equal to 1 otherwise. Then the idempotent .f~ corresponding to a ,-r-perfect subgroup H of G is given by
fD =
E
~:2
O=(L)=II L rood.Nee(H)
where O~(L .) is the smallest normal subgroup N of L such that the quotient L/N is a solvable ~r-group. Now a theorem of Dress (see [1] 5.4.7 and 5.4.8) shows th&t the idempotents f~, as H runs through a system of representatives of the conjngacy classes of rr-perfect subgroups of G (i.e. subgroups H such that H = O~(H), or equivalently subgroups having no non-trivial p-quotient, for p r 7r), are mutual orthogonal idempotents of sum GIG in bR(G). It follows that for any Green functor A for G, I have A ~_ @ ,l';i • A H
where the sum runs on 7r-perfect subgroups of G up to conjugation by G. In this expression I also denote by f~; the image of the element .f~ of b(G) in Z(A)(.). o l'Na(H) , a.nd it, seems natural to compa,re the Now observe that f ~ . . . .I.~A . Nc,(Htau Green fnnctors ,f~ x A and f ~ ( H ) x Reset (H ) A 9 It is actually easier to compare their associated categories: ~,Nc,(H)
L e m m a 12.1.8: L e t B = f ~ x A a n d C = .~H
F = G/No(H). 1. F o r a n y
G
• ResNG(H)A. L e t m o r e o v e r
Then G-set
X
a a H).~" ) (ReSNa(H),4)( [{esNc;(
A(V x X)
Moreover, if Y is a G-set, if a E A(I" x X) a n d fl E A(E x G product a x T fl f o r t h e f u n e t o r H.esN,:;(fi)A is given b y
c~ x~ fl = A* ( ?x!/ l (c~ x d ) \ ?x"/y /
Y), t h e n
their
12.1.
309
T H E C E N T R E OF A G R E E N F U N C T O R
2. For a n y G-set X
3. T h e c o r r e s p o n d e n c e 7~ m a p p i n g t h e G-set X to Res~G(H)X , a n d t h e e l e m e n t u 6 B ( Y X ) to fNHC(H) X U is a fully faithful f u n c t o r f r o m Cs to Cc. a n y o b j e c t of Cc is i s o m o r p h i c to a d i r e c t s u m m a n d of a n o b j e c t in t h e i m a g e of 7~.
4. M o r e o v e r ,
Proof: Set N = N o ( H ) . The first assertion is clear, since for any G-set X (Res~A)(Res~X) = A(Ind~vRes~X) Moreover Ind~ResCNX ~- F x X, and assertions 1) and 2) follow from the definition of the product for the functor Res~A. Now if u E A ( Y X ) , then A* ( ~ ) (u) E A(rYX), and
\?Tyx
\ yx /
\77yx
\ ~yx ]
\ TYX l
This shows that Tt(u) 6 C(ReSCNY x Res~vX). Now if X, Y and Z are G-sets, if u 6 A ( F Y X ) and v E A ( F Z Y ) , then the composition of ~ ( v ) and 7r is given by
\ 7 zx I
\Tzyyx] ....
A.(?zYX]A \ 7zx I
*(
Tzyx ] (fZ • v • fZ • u)
?zy~yx/
As f ~ is in the center, this is also
\ 7zx /
•
\~zyyx]
\ yzx ]
ButA'(<)(SZ•
....
"77
\zyyx]
N N = f N , SO = f~l.f~l
k "Tzx /
zyyx
\ zx i
kzyyx]
C H A P T E R 12. C E N T R E S
310 Moreover
Ts
x 1A(X2)) = fN • fr~ X 1A(X2) = f ~ X fH • A.
(s) =
ZZ
Furthermore = f~.A
(f~) = fHN .ResNfn a C,
But ResN.fH a c is the sum of the f ~ , where K runs through the different conjugates of H in G which are contained in N. So fN • f ~ = fN, and
which it the unit of C(Res~X2). So 7s is a flmctor from CB to Co. Note that the previous argument shows that
Now if X and g are G-sets, and if v 6 A(F • Y x X), then define
\yXl The equalities
show that Ts and 8 are mutual inverse bijections between B ( Y X ) and C ( / ~ ( Y ) R ( X ) ) , so the functor R is fully faithful. Finally, as any N-set Y is a sub-N-set of ResNIndNY a c , it is a direct summand in (Jm of a.n object in the image of Ts 9 C o r o l l a r y 12.1.4: T h e functors
M ~ fN~(U) • ReSNa(H) M c
and
L
indic(H) L
are m u t u a l inverse e q u i v a l e n c e s of c a t e g o r i e s b e t w e e n B - M o d and C - M o d .
P r o o f i If C is an R-additive category, define the category C in the following way: the objects of C are the couples (X,i), where X is an object of C, and i ~nd idempotent endoinorphism of X in C. A morphism in C fi'om ( X , i ) to (Y,j) is a morphism f from X to Y in C, such that f o i = f = j o f. In other words Homd((X, i), ( Y , j ) ) = j o Homc(X, Y) o i The composition of morphisms is the composition in C. The category C is obtained by "adding direct summands" of objects of C. It is also an /~-additive category (the direct sum of ( X , i ) and (Y,j) is of course (X @ Y,i @j)).
12.1. THE CENTRE OF A GREEN FUNCTOR
311
There is a canonical functor X ~ (X, ldx) from C to C. So if/~ is an additive functor from d to R - M o d , it gives by composition an additive functor fi'om C to R - M o d . Conversely, if F is an additive functor fi'om C to R - M o d , then F admits a unique extension/~ to C, defined by
i) = F(,)(F(x)) This is because if (X, i) is an object of C, then i : X --+ X defines two morphisms in d i+: (X,i) ~ (X, Idx)
i- : (X, Idx) ---, (X,i)
Moreover i- o i + is equal to i, which is the identity morphism of (X, i) in C. So (X, i) is a direct summand of (X, Idx) in C. This shows that the categories of representations of C and d are equivalent. Now if 9 : C - 4 / ) is a functor between additive categories, it induces a functor ~ : d -~ 75. It is clear moreover that if ~ is fully faithfld, then ~5 is fully faithfuh indeed, the bijection F : Homc(X, Y) --* Horny (F(X), F ( Y ) ) restricts to a bijection/" from I-Iornd ((X, i), (IG j)) = j o Home(X, Y ) o i t o
,om
F(Y,j)) : F(j)o Homo(F(X), F(Y)) o
If moreover any object Y o f / 9 is a direct summand of an object F ( X ) in the image of F, therl there are morphisms ~: P(X) ~ Z
fl: Z ~ F ( X )
such that c~ofl = Idv. I f i is any idempotent in Endv(Y), then / 3 o i o c ~ is an idempotent endomorphism of F(X). So it is equal to F(7), for a unique 3' E Endc(X). This implies that "y is an idempotent. Now the morphisms
are mutual inverse isomorphisms in D. So /~' is essentially surjective, hence it is an equivalence of categories. Thus in the situation of the lemma, the functor ~ is an equivalence of categories: this implies in particular that 7-4 induces an equivalence of categories between the categories of representations of CB and Ct). It remains to check that this equivalence is as stated in the corollary, which is clear since if L is a C-module, then for any G-set X fHc x (Indaa(~jL)(X) = f~' x L(Res~a(H)X ) . . . .
....
f ~ x fN~(H) • L(ResGva(H)X ) = fNG(H) X L(Res~%(H)X ) = L(Res~va(H)X)
which proves that f,~ x Ind~G(H)L = IndNc;(H)L.
9
L e m m a 12.1.5: L e t H b e a n o r m a l s u b g r o u p of G. If A is a G r e e n f u n c t o r for G s u c h t h a t f~/ x A = A, t h e n flc'/h' • A H = AH~ a n d t h e c o r r e s p o n d e n c e X ~ X H i n d u c e s an e q u i v a l e n c e of c a t e g o r i e s f r o m Ca t o CAH.
CHAPTER 12. CENTRES
312 P r o o f : Let Y be a (G/H)-set. Then
AM(y) = A(In~/,~')/ ~ A . ( f ) ( A ( Z ) ) (z,:) where (Z, f ) runs through the G-sets over In~/HY such that Z H = O. Now since f ~ G is the unit of A, and as R e sGK f ~, = 0 if H Z K, it follows that A(K) = 0 if H ~ K , so A(Z) = 0 if Z H = O. Thus AH(y) = A(In~/HY ). Let ix denote the injection from In~/H(X H) into X. Then
AH(x") = A(In~/HX n) The unit A(X) --* (In~/HAH)(x)
=
AH(x")is
tlae map A'(ix).
Now as
X = (In~/HX H) ~l Z with Z H = 0, it follows thai, A(Z) = 0, and the map A*(ix) induces an isomorphism
A(X) ---+AH(X~). Now it is clear that the correspondence
X H Xg
f C A(YX) ~ A*(iyx)(f) r A H ( y H x H)
is a fully faithful functor from CA to CAH. As any (G/H)-set Y is isomorphic to (Inf~a/HY) H, this functor is essentially surjective, hence it is an equivalence of categories. Finally, the unit of A H is f ~ / H because .f~ - In~/Hf]/H is a linear combination of G/If, for H ~ K , so it acts by zero on A. 9 C o r o l l a r y 12.1.6: L e t H b e a n o r m a l s u b g r o u p of G. If A is a G r e e n f u n c t o r for G s u c h t h a t fHa X A = A, t h e n t h e f u n c t o r s
M ~ MH
and
L ~ In~/NL
are m u t u a l i n v e r s e e q u i v a l e n c e s of c a t e g o r i e s b e t w e e n A - M o d and A H - M o d . P r o o f : This follows from lemma 12.1.5, and from the definition
(In~/HL)(X) = L(X H) So In~/g is an equivalence of categories, and its inverse is the left adjoint functor
M~M
H.
9
L e m m a 12.1.7: L e t H b e a s u b g r o u p of G, and M b e a M a e k e y f u n c t o r for N o ( H ) . I f Z is a n e l e m e n t o f b(.), t h e n
G Na(H) ~-- ,tn(]NG(H ,C Z • IndNc(y)InfNa(H)M )Inf N~(H) ~Va(H)[IzH X M) P r o o f : I f Z i s a G - s c t , I k n o w t h a t Z=b,(~.)b'(~.)(cb). L is a Mackey functor for G, then for I 6 L(X) I have
L. (y) L, (y)
Thus if X is a G set, aad
12.1, THE CENTRE OF A GREEN FUNCTOR
313
G N Set N = No(H) and N = Na(H). If L = tndNInfvM , I have L(X) = M(XH), and setting Z ' = Z H and X ' = X H, the maps L*(":) arid L. (;~) are respectively equal
to M" ( ~ ' ) of..
zlm #
zig I
and M. ( ~ , ) . z#x !
M(X H) via Z H, for
It follows that the image of L. ( ~ ) L* ( ~ ) i s
the image
), that
,n o her woras, the e,ement any Z 6 b(G). It follows that
on
(Z x Ind~InfNN-M)(X) = Z " • M ( X H) = indNinf~v( g a N H X M)(X) This proves the lemma.
9
L e m m a 12.1.8: L e t H be a z,-perfect n o r m a l s u b g r o u p bR(G/H), I have
of G.
Then
in
= fC,,/H ,1
{ ,cG~H (G / H
P r o o f : Let /7 = K / H be a ~r-perfect subgroup of G/H. If va~s~ "a/
c l H L) I (eLa) " = "teg G/H'ClS-f"eC'" h' = I(In~/H eU ) It Z,)
Thus L / H is conjugate of ~7 in G/H. Then L is conjugate to L' in G, and then O'(L) = H is conjugate in G to O"(L ') = K. Thus H = K, and the lemma follows. 9 C o r o l l a r y 12.1.9: I f H is a rc,-perfect n o r m a l s u b g r o u p of G, t h e n for a n y A-module M and any AN-module L
(f~s • M) H ~- flWH • MH
f~ x InfaalHL ~- Ing/s_s(f~ C(") x L)
Finally, corollary 12.1.4, 12.1.6 and 12.1.9 show that if A is a Green functor for G, and H is a rr-perfect subgroup of G, then the functor G
NG(H)
L ~ IndNc(H)Inf~c,(H)L is an equivalence of categories from ,fl~G(H) • (Resg-a(H)A)H-Mod to fHc' x A M o d . The inverse equivalence is the adjoint functor
(Res,%.(H)M)" In particular, the unit morphism a
N~(H)
f~ • A ~ IndNa(H)In~a(H)(f~
X (Res~vc(H)A)H)
is an isomorphism of A-modules. As it is a morphism of Green functors, it is an isomorphism of Green functors. The following lemma gives a characterization of the Green functors .f~ • A as the functors which are projective relative to the set E~(G) of solvable rr-subgroups of G (i.e. relative to the set I]H G/H, where H runs through E~(G)):
CHAPTER 12. CENTRES
314
P r o p o s i t i o n 12.1.10: L e t G be a g r o u p , a n d A be a G r e e n f u n c t o r for G over R. L e t Ir t h e set of p r i m e f a c t o r s of IGI w h i c h are n o t i n v e r t i b l e in R. T h e following c o n d i t i o n s are e q u i v a l e n t : 1. T h e f u n c t o r A is p r o j e c t i v e r e l a t i v e to t h e set of solvable re-subgroups. 2. T h e i d e m p o t e n t fla a c t s as t h e i d e n t i t y on A, i.e. A = ,f~ x A. P r o o f i It is clear that 2) implies 1), because if A = f a • A, then <4 is the image in A(.) of fla. This element is a sum of el, for L E E,(G), hence a linear combination of G / K , for K C_ L, thus t( E E~(G). Then the identity map of A(.) = A(G) is a a for K E E.(G) Thus A is projective relative to the set linear combination of tic-rio of solvable rr-subgroups. Conversely, if A is projective relative to the set of solvable ~r-suhgroups of G, then eA can be written as
~A
~
=
t G~,- (~,")
Then
t~.:(rl,.f, .oeK) IqEE~(G)
and it is easy to see that for any subgroup Ix" of G G a=f[ Resr,,fl and that f#"
=
K / K if I(
is
re-solvable. Thus .f~
=
r
and 2) holds.
.,
Finally, I have proved the following proposition: P r o p o s i t i o n 12.1.11: L e t G be a g r o u p , a n d A be a G r e e n f u n c t o r for G over R. L e t rr t h e set o f p r i m e f a c t o r s of
I01 w h i c h
are n o t i n v e r t i b l e in /2.
T h e n t h e r e is an i s o m o r p h i s m of G r e e n f u n c t o r s
A~-(~S~
x A
H
w h e r e t h e s u m m a t i o n r u n s t h r o u g h a set of r e p r e s e n t a t i v e s of t h e c o n j u g a c y classes of 7r-perfect s u b g r o u p s of G. I f H is such a s u b g r o u p , t h e n
J • a
,
,a
,
~Nc(H)
{ ~Fc(H)
•
H)
T h e f u n e t o r fife(H) x (Res~o(H)A)n is p r o j e c t i v e r e l a t i v e t o t h e set of s o l v a b l e re-subgroups, a n d t h e f u n c t o r s G No(H) M ~-+ IndNG(y)InfFrc(H)M
G H L ~-~ S~c,(H) x (ReSNa(H)L)
are m u t u a l inverse e q u i v a l e n c e s of c a t e g o r i e s b e t w e e n (f~ x A ) - M o d a n d /1No(H) X (ResaNa(s)A)U-Mod.
12.2. THE FUNCTORS (A
315
R e m a r k s : 1) In the case of Mackey functors (i.e. the case A = b), this is essentially theorem 10.1 of Th4venaz and Webb ([15]). 2) In the case 7r = 0, i.e. when the order of G is invertible in R, the solvable ~rsubgroups are trivial. In those conditions, I have fac = ~a[G/1. Thus if f ~ • A -- A, then the maps
a 6 A(H) ~ rlHa 6 A(1) H
b E A(1) H ~-~ 1-~--tHb
114[
are mutual inverse isomorphisms between A and FPA(1). Thus if the order of G is invertible in R, then the isomorphism of the proposition gives A ~- ( ~ Indff'c(") I n f N o ( HN~.~/F ) G~[ ) P~'H'() H
This is theorem 13.1 of Th&enaz (see [13]).
12.2
T h e f u n c t o r s (A
12.2.1
Another analogue of the centre
The definition of the centre of a Green functor given in the previous section is certainly natural, and mimics the definition of the centre of a ring. However, one can also define the centre of a ring from its category of modules, as the endomorphism ring of the identity functor ([1] prop. 2.2.7). If A is a Green functor for the group G, then the endomorphisms ring of the identity functor of A - M o d is just a ring, isomorphic to the centre of A(ft2). I would like to have a similar construction, using the category A - M o d , and producing a Green functor (A- The idea is then to mimic the construction of the Green functors 7-/(M, M), setting for a subgroup H of G
CA(H) = Endf~nct(Res~) where EndFu~,(Res~) denotes the endomorphisms of the restriction functor from A - M o d to R e s ~ A - M o d : such an endomorphism is determined by giving, for any A-module M, a morphism OM of Res~A-modules from R e s ~ M to itself, such that if ~b : M --+ N is a morphism of A-modules, the square Resr~M
Resin
OM
0~
, Res~M
,
t
Res/~N
is commutative. With this definition, it is clear that if A is a Green functor on R, then (A(H) has a natural R-algebra structure. It is natural to try to turn (A into a Green functor. First I observe that if 2- denotes the identity functor of A - M o d , and 2-G/H the functor M H MO/H from A - M o d to A - M o d , then
~A(H) ~-- Homf~n~t(Z, Za/H)
CHAPTER 12. CENTRES
316
Indeed, I know that 2-a/H(M) ~- Ind~Res~M, flmctorially in M, and that Ind/~ is right adjoint to Rest. The above isomorphism follows then from the next lemma: L e m m a 12.2.1: L e t C a n d D be c a t e g o r i e s , a n d F b e a f u n c t o r f r o m C t o 7) h a v i n g a r i g h t a d j o i n t F'. T h e n if g is a c a t e g o r y , if A is a f u n c t o r f r o m $ to C, a n d B is a f u n c t o r f r o m g t o 19, I have HomF~n~t(F o A, B) _~ HomElier(A, F ' o B) P r o o f i Indeed, a morphism of functors from F o A to B is determined by giving, for any object X of g, a morphism Ox from F o A(X) to B(X), such that if q~ is a morphism in 5" from X to Y, the square
OX
F o A(X)
, B(X)
F o A(f) [
I B(f)
f o A(Y)
Oy
, B(z)
is commutative. Denoting by u ~-+ u ~ the bijection
Hom•(F(Z),T)
~ Homc(Z,F'(T))
deduced of the adjunction, I obtain for any X a morphism 0) from A(X) to F o B ( X ) . Moreover the square
A(X)
0)
, F' o B(X)
A(f) l A(Y)
[FoB(f) O{
, F'o B(Y)
is commutative, because its diagonal h : A(X) --+ F' o B(Y) is both equal to (B(f) o
Ox) ~ and to (0y o ( F o A)(f)) ~ Thus the maps 0~c define a morphism of functors from A to F' o B. There is an obvious inverse construction, which gives the isomorphism of the lemma. 9 Now for any G-set X, I can define the endofunctor Zx of A - M o d by I x ( M ) = Mx. Then, for any G-set X, I define
(A(x) : Hom~.~,(Z, Zx) If f : X ~ Y is a morphism of G-sets, then f determines a natural transformation from I x to 2-z: if M is an A-module, and Z is a G-set, then the maps
M,(Idz •
Mx(Z) --~ M y ( Z )
define a natural transformation from Mx to Mz, which is itself natural in M. Similarly, the maps
M*(Idz x f) : M y ( Z ) --+ Mx(Z)
12.2. THE FUNCTORS (A
317
define a natural transformation from I T to 2-x. This gives by composition morphisms ( J , , ( f ) : (A(X) -~ (A(Y)
:r
r
---' (A(X)
Finally, if c~ r CA(X) and /9 r r then for any A-module M, I have a morphism aM : M ~ Mx, and a morphism ~M : M ~ My. I set then (o, x ,3)M = <,,,1 x a M
the product in the right hand side being computed in 7-((M, M). In other words, if Z is a G-set, I have for nz E M ( Z )
M. CZYz I \ zx!t /
(a x 3)M,z(,,,)
Then if 0 : M ~ N is a morphism of Mackey functors, I have (c~ x 3)~v,zOz = X . \ ~.W / o C~N,Zr o ~N,Z o Oz('~) . . . .
....
N.
zx~
:~r9]
....
Oz.,x'y o AI.
which proves that a x/3 r r
z.W
o a,~i,n. o ~ M , z ( m ) = Ozxy o (c~ x ,J)M,z
x Y).
L e m m a 12.2.2: T h e a b o v e definitions turn CA into a G r e e n functor, such that
(A(*) = EndF,~,~o~(Z) ~- Z(A(fl2)) Proof: The verifications to make are not difficult, and similar to those made for the functors "H(M,M). The equality CA(e) ~-- Z(A(fi2)) foIlows from the fact that the category A - M o d is equivalent to A ( f F ) - M o d . Thus r identifies to the endomorphisms of the identity functor in this category, hence to the centre of the algebra A(W).
Let X be a G-set, and ~ E r
Then for any A-module A'/, I have a morphism
c~M : M --~ Mx, such that if q5 is a. morphism of A - m o d u b s from !1,1 to N, the square (~A,I M
>
Mx
(c) N
--
~
oN
is conmmtative.
N\'
C H A P T E R 12. C E N T R E S
318
In particular, if Y is a G-set, I can take M = Ay. Then HomA(A},-, M) ~_ M ( Y ) (see proposition 3.1.3): the element m 9 M ( Y ) is associated to the morphism 4,~ : Av --* M, defined for any G-set Z by
a 9 A v ( Z ) = A ( Z x Y) ~ a %-,)) = M.
M*
zyy
It follows in particular that the morphism aAy : Av ~ ( A v ) x = A ( X Y ) is determined by an element ay 9 A x y ( Y ) = A ( Y X Y ) Let n 9 N ( Y ) . I must have a commutative square
Ay
N
OZ,4y
CtN
~ Axv
,
Nx
Then for any G-set Z and any a E A v ( Z ) = A ( Z Y ) , I must have
The case Z = Y and a = 1A(V~) gives then
Conversely, if this equality is used to define a n y for any N and any Y, then the commutativity of the square (C) is equivalent to
fo any Y and any m r M ( Y ) . But
As r is a morphism of Mackey functors, the right hand side is
N. ( y ' x Y 2 1 N * ( YlxY2 t r k ylx } kylXy2y2/
• m)
As r is a morphism of A-modules, it is also
N. (yIxy2 / N* ( ylxy2 k ylx /
)(ayx
~ y ( w i ) ) = ay oy @y(fil)
kylxy2y2
Thus the square (C) is commutative. The only condition on the elements a r comes then from the fact that the morphism c~v is determined twice by the previous argument: I must have for any Z and any a 9 Av(Z) = A(ZY) Cry(a) = a Oy ay = az OZ a
12.2.
T H E F U N C T O R S (a
319
Finally, if those conditions are satisfied, then the morphisms C~N,Z are morphisms of A-modules: indeed, if a E A ( Z ) and ~ E N ( Y ) , then x o,:,,~,-(,~)
= a x (,:,~,-o~,-,~)
whereas But
z~ zyy
a x n = A.
As A.
(7) (,,)
azy ozy (a x n) =
....
(A.
A*
(a) o y . .
A( YY), have
( azy ozy A.
o,.,,
zyy
(zY)A*(;Y)(a)oyay)oyn zyy
. . . .
=A.(ZY)A'(:J)(a)~ zyy ....
g,yOy1q.) ~ . . .
a • (ay oy n)
So I have If f : X ~ X ' is a morphism of G-sets, then the sequence (ay) determines a morphism from 2- to 2-x,, defined by composition for an A-module 3J and a G-set Z by
M.(Idz x f) M(z)
, ~,~x(z) = M ( Z X )
, A,I(ZX')
In particular, if M = Az and Z = Y, the image of 1A(Z2) in A r ( Y X ' ) = A ( Y X ' Y ) is
Ar,.(Idz x f)(ay) = A.(ldz x f x Idy)(ay) In other words, if s is the sequence ( a r ) , | have
(A,.(f)(s)r = A . ( l d v x f x I d r ) ( a y ) It is clear similarly that if s' = ( @ ) is a sequence defining an element of (A(X'), then G ( f ) ( s ' ) r = A*(ldy x
,f x [ d r ) ( @ )
If s = ( a r ) defines the element a C (A (X), and t = (br) defines the element/3 E G ( Y ) , then for any A-module M, any G-set Z and any m E M ( Z ) , I have
\ zxy /
\ zxy / ....
M. ( z Y x l (azz ozv b z o z ,n) \zxy/
Moreover as bz E A ( Z Y Z ) , I have azy
Ozy bz = bz Oz az
CHAPTER I2. CENTRES
320 Thus
(~ • 3)M,Z('~) = M. (\ :zwY : r ) ( b z o z a z o z m ) In particular, if M = Am and m = la(z_~), this equality shows that (~ x /~ is defined by the sequence .~ • t such that
A* ( z~itl:rz21(bz ~
(sxt)y=
Finally, I have proved the following proposition: P r o p o s i t i o n 12.2.3: Let A b e a G r e e n f u n c t o r for t h e g r o u p G, a n d X b e a G - s e t . T h e n ~.4(X) i d e n t i f i e s w i t h t h e s e t of s e q u e n c e s .s = (.~<), i n d e x e d b y t h e G - s e t s , s u c h t h a t st. C A ( Y X Y ) a n d a Oy ,gy = *~Z OZ (/
for a n y G - s e t s Y a n d Z a n d a n y a C A(ZY). I f f : X ~ X ' is a m o r p h i s m
of G - s e t s , if s E CA(X) a n d s' C ( a ( X ' ) , t h e n
CA,.(f)(S)y = A.(ldy • f • Idy)(.sy)
= A*(Idy • f • Idz)(Jy)
r
I f s E CA(X) and t 6 (A(Y), t h e n for any Z
(s x t)z = A. (Z~:rz~) (tzoz ~z) \ zl ;r.7tz2
Let X and Y be G-sets, such that X divides a multiple of Y in CA. This is equivalent to say t h a t there exists elements a{ E A ( X Y ) and fl~. E A ( Y X ) , for 1 < i < n, such
that 1A(X~) : ~ ~i oy fl~ i
T h e n let Z be a G-set, and s E (A(Z). I cart write 8 X ~- 3 X OX 1 A ( X 2) = Z
3 X OX O:i Oy /3 i = E i
Q" Oy 8 y Oy fli
i
This f o r m u l a shows that s x is d e t e r m i n e d by sy. T h e element sv is such t h a t .syogtt = u oy sy for any u E A(Y2). Conversely, if T is a G-set such that any G-set divides in CA a mnltiple of T, if I choose an element .s E A(TZT), such that s OT u = zL OT s for any u C A(T2), and if I choose for any X elements ai,x E A ( X T ) and fli.x C A(TX), for 1 < i < rzx, such that 7tX
1A(x~) = ~ c~{..
I can se~ ~zX
sx = ~ c~,x oT s OT 5<X C A ( X Z X ) i=1
i2.2.
THE FUNCTORS (A
321
This element does not depend on the choices of c~i,x and di,x: indeed, if !
1A(X~) = ~ C~j,X o7c fl'j,X d=l
then
rzX I
/
SX Ox O~j,X = E
Cti,X OT S 0 T fli,X OX Ctj,X
i=1
As fl~,x ox ct},x r A(T2), it is also nX
SX OX O~j,X :
~
Ozi,X OT fli,X OX dtj,X ~
S :
1A(X2) Ox as
OT S :
~ j , X OT 8
i=1
It follows that 1
!
/
/
SX Ox Ctj,X 0 1 / ' f l j , X = (~j,X 0 T S 0 T flj,X
and summing over j
~k
6X =
X;'Ozj,X
0T 8 0T
9'j,X
j=l
Moreover, the sequence s x is an element of (A(Z): indeed, if X and Y are G-sets, and if a E A ( X Y ) , then rzy
a = )__s a or c,j,~,. OT/3j,r
j=l Thus ,SX 0 X a =
v--,L ~ l
OT 5 o T fli,x o x a Oy O~j,y 0 T flj,y
As fli,x ox a %, cU,y C A(T2), it is also ~ly
8xOxO,
E
1
O : i , X O T ~ i , x O x ( I O Y d t j y O T SOY flj,Y = E
a~176
T S O Y flj,Y = (tOy S y
j=l
Then I see that (A(Z) identifies with the set of elements s in A ( T Z T ) such that s OT u = 'u o f s for any u E A(T2). But
and I have also A( T 2) ~- J'{A ( AT, AT)('). Moreover, the product of u r ?-{A ( AT, AT)(') and s r "HA(AT, AT)(Z) for the functor J-(A(AT, AT) is precisely zt OTS. It follows that CA is the commutant of J-((AT, AT)(.) in "}-{(AT, AT). Now say that any X divides a multiple of T in CA is equivalent to say that AT is a progenerator of A - M o d . A natural question is then to know if, whenever P is a. progenerator of A - M o d , I have
CHAPTER
322
12. C E N T R E S
First it is clear that if a E ( A ( X ) , then o'p is a element of Hom~ (P, Px" ) = ~.4( P, P ) ( X ) . I o b t a i n tha~. wa.y a n a t u r a l m o r p h i s m (.4(X) ~ 7~A(P, P ) ( X ) , which induces a morphism 0 : ( . ~ -~ ~.~(P, P) If f E E n d a ( P ) = ~ a ( P , P ) ( * ) , then the square (rp
P
~ Px
P
~ t~ 0 p
has to be c o m m u t a t i v e . This expresses exactly the fact that the image of (9 is contained in the c o m m u t a n t of 7-LA(P, P ) ( . ) . Let p(O he a direct s u m of copies of P. It. is easy to see that the c o m m u t a t i v i t y of tile squares
P
(.'r p
)
X
'
Px
where fi is the projection on the c o m p o n e n t of index i E I, forces at,!n = (c~p) (t). T h e n if a p = 0, I have an(~ 0 for any i. As any A - m o d u l e M is a quotient of some p(I); for a suitable (I), the c o m m u t a t i v i t y of tile square
p(1)
_C~P(1)> [)~I)
M
~
Mx
OCM
forces t h e n C~Mo ~r = 0, thus a M z 0 if (7 is surjective. T h e n ct = 0, which proves that @ is injective. Convers~e.lv,, if ct E HomA(P, P x ) is such that all the squares (~
P
,1 P
are c o m m u t a t i v e , then setting ap(•
-
'
Pv
--
)
Pv
I fX
= (c~)(/), it is easy to see that all the squares
,1
lsx
p(d)
p(3) --
O'p(J)
)
X
(c)
12.2.
THE FUNCTORS
CA
323
obtained for index sets I and d and a morphism f : p(S) __+ p ( a ) are commutative. In particular, if L is the image of f, I see that ap(,l(L) C L x . So if M is an arbitrary A-module, and ~r a surjective morphism from p(a) to M, then there exists a unique morphism C~M such that the square P( J)
Rid)
_~
OI
~. OX
M
C*M
~
Mx
is commutative. The commutativity of the squares (C) and the projectivity of p(1) shows that C~M does not depend on (d) or on ~r. It is then clear that the c
(A "" C~A(P,P)(EndA(P)) In p a r t i c u l a r , if T is a G-set such t h a t any G-set d i v i d e s a m u l t i p l e of T in CA, t h e n for any G-set Z CA(Z) = {s ff A(TZT)
12.2.2
IV. E
A(T2), .s OT u = u or s}
E n d o m o r p h i s m s o f t h e restriction f u n c t o r
Let H be a subgroup of G. The first definition of ~A(H) is
where Reset is the restriction functor from A - M o d to Res,~A-Mod. Setting
A-CG
KCH
the evaluation M ~-+ ]IJ(Q) is an equivalence of categories from A - M o d to A(Q 2) G 4M o d , and the evaluation M' ~-+ M'(gt') is an equivalence of categories from iRes~, M o d to (Res~A)(f2'2)-Mod. Moreover = G Qt2 (Res~a)(fY ~) A(IndH ) and the product of the elements c~ and fl of (Res~A)(~t '2) is defined by ' fl = (Res~A). e of~,
w'w2c% / /
(ResHA) c; *
~lw2w3 I /, / /
(c* x' fl)
where the products o' and x' are those of the Green functor Res~A. Thus
c'~oa , f l = A .
(
In H \
c0~w~ ]
)(
,,,
\w,w2co2w3)]
C H A P T E R 12. CENTRES
324
G #2 2 defined by where 5 is the map from Ind~FY 4 to (IndHf~)
031'022'023'024) =
0,q'031'032)' . . . (g' . ~'3'CJ'I /)
It follows that
,
'
'
# / , ! (g,~o~,02~,~)
,'
\ (~,~,02~) ) ,/
'~ (~x/3)
(~,02~,02~)(g,02~,02~)] /
l
I
#
l
Let i be the map fi'om Inds~fl ~2 to f~2 defined by
i(g,J~,J~)-
(,qJ1,g022)
where the expression g03~ is equal to 9hK if 03' = DK C fY. The map i is injective: indeed, if
i(g,03',,03;) = i(.q', 02'/, 03;') then g02~ = g'03'[ and as 02~ and w'/are contained in H, I have g'tl = gH. Then there exists h E H such that g#
= gh
02'1 =
]702'1;
02; =
]7027
#1 ## ! i Then (g # ,02~,022)= (g'h,03~,022) = (g,%,022), and i is injective. It follows that A.(i) is an injective morphism of (Res~A)(fi '2) into A ( ~ ) . over
' ~ ,, , , A* ( A.(i)(a o~,/3) = A . . ( (g'02''cc~'a~3). # # ~,
\
~02,,~02~ )
More-
~ (o, • ;~)
(~,~,03~,02~)",,
(g,02,,~)(~,02~,~,~)) /
/
/
,1
On the ol, her hand A . ( i ) ( o ~ ) o f ~ n , ( i ) ( f f ] ) : A . (021022~ k
021023
/
A* ( 021032023 ~ (n.(i)(o~) x n.(i)(/_~)) : . , \0210220220"13/ !
' 9 9
k 031033 /
#
\021022~'2023/
,t
l
, ~/ #
,/
,
,1
glCOlglO.:2g2~3g2",.
/]
Let (C) be the square
1G ,'~;3 n<~.~
~
(g,03~, 02,1,~;) (g, 02~,02~)(j, 02~I,021)J
, (Ind~fY~) 2
( 02,022023 I
~
k 031C02C02cO3/
This ~qu~re ~s cartesian: h~deed, if (~o1,~.,~,~:~) ~nd t(~,~.,,<~)(.<~,02~.~':,)) that (w~, 022,02~, w3)
~ g2a3, ~ ff203~) = ( 9 t % ~, .(k 022,
a,'e
snch
12.2, THE FUNCTORS (A
325
t = g202;. As 02; and 023 ! are contained in H, this forces gltt = g2H, and there then 91022 exists h E H such that 91 = g2h. Then %' = ha;;, and I
!
(g l h
(g2'cd3'024) =
--1
/
Ind/%'*.
Let then u = (g, 02'1,02;,h-~o2~) E
P
/
I t--1
' had2,024) = [ g 1 , 0 2 2 , "
!\ 024)
I have
((g'<'<'<))(,,) (9, ~ , 021)0,4, 02.~))
= ((gl, ' 021,
and =
I
/
I
I
,
,
,
=
/
(
On the other hand, the map i = 9
I
....
0,~,~,%) O,q,~;)Cg.%.~i) I
l
,
!
)
is injective: indeed, if I!
II
II
then there exists h E H such that
g' = gh -1
J(=
h02t1
oJ; = h02;
0,23 = h02;
and then I
It
II
II
(g ,021,022,023)
=
l
I
f
( g h - 1 , hcu1, h022, h023) :
(g, 021, ' %', % ')
So the square (C) is cartesian, and then
A.(i)(~) oa A.(i)(fl) . . . .
"
\
g021wi
/
\(g ~ 4 ) 0 , 4 , 4 ) / % • ! . . . . i(~ o~, fi)
....
It follows that A. (i) is a morphism of algebras (non-unitary in general) from Res~rA(fY 2) to A(f~). In terms of generators, if K and L are subgroups of H, if x E H, and if iK,~-,L is the map from H / ( K N XL) to fy2 defined by
then i o Ind~iK#,L is the map
g(K
n
~L) e a / ( I (
n
~L) ~ (gK, gxL) C
as
It follows that
d *~)(~ ~.,,,KKNXL .A, a,Kn=L xr~L K L F[znL)~ -~ tKnXL~a,Kn~LXTK=nL tn other words, the morphism A.(i) is just the inclusion a " "fy2x] = (ResHA)/
@ K,LCH xeli~H/L
I have finally proved the
A(K
rn ~L)
~-~
@ K,LC_G xeK\G/L
A(K
n
~L) = A(a 2)
CHAPTER 12. CENTRES
326
L e m m a 12.2.5: L e t H be a s u b g r o u p of G. T h e n t h e i n c l u s i o n i f r o m Ind~fY 2 into f~2 i n d u c e s an i n j e c t i v e m o r p h i s m of a l g e b r a s
A,(i)
=
(Res~)(A)(fY 2) ~-~ A(f~ ~)
R e m a r k : This is the only point on which I disagree with Th6venaz and Webb (see [15] 5.3 to 5.4): this lemma shows in particular that the morphism b.(i) from the Mackey algebra, of H to the Mackey algebra of G is injective. If M is an A-module, then KCH
NCH
-
In other words, the restriction functor can be translated in terms of the algebras B = A(Q 2) and C = (Res~A)(fY 23 as
(Res~M)(W) = A,(i)(lc) oa M(f~) Let e be the idempotent A.(i)(1c) of B. Then e oa B is a C-module-B, and
aes~M(a') ~_ (~ o~ B) O~ M(n) The restriction functor is then given by tensoring with a bimodule, and then I can apply the following lemma: L e m m a 12.2.6: L e t A a n d B b e R - a l g e b r a s , a n d M be an A - m o d u l e - B . T h e n
E n d F ~ t ( M @B - ) ~-- EndA| I n p a r t i c u l a r , if f : A --~ B is a m o r p h i s m of a l g e b r a s , if e = .f(1A), a n d M = eB, t h e n eM @B -- is t h e f u n c t o r Resj of r e s t r i c t i o n a l o n g f , a n d
E n d f ~ M a e s s ) ~-- {5 ~ ~B~ IV(,
~
A, bf(a)
=
f(a)b)
P r o o f i An endomorphism r of the functor M | - is determined by giving, for any B-module L, a morphism of A-modules eL from M @B L to itself, such that for any morphism of B-modules ~b : L --+ L', the square
r M|
, M|
IM|
M@B~ t M @B L'
)
M | L'
eL' is commutative. In particular CB is an endomorphism of M | B, which is isomorphic to M by the map 0 defined by O(m | b) = rnb. Then ~ = OCBO-1 is an endomorphism of M. Moreover HomB(B, L) _~ L for any L, the e l e m e n t / E L defining the morphism #l : b ~-+ bl from B to L. The commutativity of the square
M@BB~_M
~B
~ M|
I
M
@t? #I ~
~ M @B #t
4.
M@L
1 '
cs
M|
(c)
12,2. THE F U N C T O R S s
327
forces then
eL(re|
= (M|174
Thus ~ determines r
(M@pz)((I)(m)@l)= O ( m ) |
(M|
Moreover, if a E A and b E B, I mnst have
The case L = B now implies that ~ is an endomorphism of M as A-module-B. Conversely, being given r I can define (I)L by qSc(rn| = (I)(m)| This definition makes sense, because for b C B Cc(,~b | l) = ~(mb) | Z = ~(m)b O l = ~(m) O bl = CL(-~ | bl) If moreover 9 is a morphism of A-modules, then for any a E A, I have CL(a,~ | l) = ~(am) | 1 = a~(n~) | l = aCL(,~ | l) Thus qSc is a morphism of A-modules from M to M x . With this definition of eL, it is clear that all squares (C) are commutative. This proves the first assertion of the ]emma. For the second assertion, I must find the endomorphisms of eB as an A-module-B. Such an endomorphism r is entirely determined by the image of e E eB, since
r Then I must have r
= r
~) = ~5(e) = r
and also
E eBe. Moreover, if a C A, then
So r
f(a)e = f(a)f(1A) = f(a) = f(1A)f(a) = ef(a) Conversely, if p E |
commutes with any element in the image of f, then setting
r
= pb
I obtain an endomorphism of eB as A-module-B, since
r
= f3(f(a)ebb') = r
= pf(a)bb' = f(a)pbb' = a.(pb).b'
This proves the lemma.
"
This lemma applies to the morphism A.(i) from (tles~/A)(ft a) to A(f~2), and this gives the P r o p o s i t i o n 12.2.7: Let A be a Green functor for G, and H be a subgroup of G. T h e n s identifies w i t h t h e set of e l e m e n t s in
4A(n~)t~: I(CH
which c o m m u t e to all the e l e m e n t s
t~'~:nxrfi1,n* C,~Xr~xnC for K,L C H, for a C A(K) and x E H. In particular, the ring (CA(I),.) is i s o m o r p h i c to the centralizer of ,4(1) in A(1) | G.
328
CHAPTER
12.
CENTRES
Proof: The first assertion is only a reformulation of the previous results: if z E (~
t~[)A(fl2)( E
KCH
t~[)
NCH
and if z commutes with all the t~:, for K _C H, then Z
=
E
~K . L ~KZ~L =
K,LCH
.h.L Ix" IN .L ~lggL z = Z ~ ~I( z = E LKZ~ hh-,LC_H KC_H KCH
thus z C L, tK ~ ~ K, and the first assertion holds. The second one follows from the case H = {1}: the algebra t l A ( n ~ ) f i is formed of the elements t] ~\l,a:rr] for a 6 A(1) and x r G. Hence it is isomorphic to A ( 1 ) @ G . The image of A(fl '2) corresponds to the elements such that x = 1. Thus it is A(1), and the proposition follows. " Let z E ( A ( H ) .
~K ~i, Say that z commutes Then z = Z~,cH zK, where zK = ~Kz~K.
with all the ~K I,Kn~: L ~ KN~L,a x r L K # n L is equivalent to say that K
L
I~"
L
(12.1)
ZKI~Kr3zL)~KcT~L,aXrI
which gives in particular, for L C K C_ H, for a C A ( K ) and x E H zKt~
t Lh ZL
r Li4 ZK = Z L ? h ~L
ZN.~K,a ~ /~IC,aZI.2
XZNTs -1 = Zx K
Conversely, these relations imply equalities (12.1). A computation similar to the one made to identify the functors H ( M , N) (see proposition 1.4.1) shows then the following proposition:
Proposition 12.2.8: Let A b e a G r e e n f u n c t o r for G. I f H is a s u b g r o u p o f G, t h e n ( A ( H ) i d e n t i f i e s w i t h t h e s e t of s e q u e n c e s (zi,-) i n d e x e d b y t h e K 2 )t K, N and subgroups of H, such that z~- E ti,,A(f~ zKt~- : t LK ZL
I~ f LK ZK = ZLT L
Zl,[/~lq,a = )~B.',aZK
X Z K X -1 = Zxlx"
for any subgroups K D L o f H , a n d a n y a G A ( K ) a n d x G H. I f H ' C H a n d z = (zh) G ( A ( H ) , t h e n
r.",(~)K = ~, i f H C_ H ' and ~ = (~,,-) ~ G ( H ) ,
(t HH ~z)i, .=
~
W~" C_ H '
then ~.r~ Kn~HXZKxnHX
-1 r KI,n ~ H
VK C __ H '
xC=K\H'/H
F i n a l l y if g C G a n d z = (zK) E ( A ( H ) , t h e n (9,7.)12 = 9(Zh'g )
The product
V J; C 99
z . z ' o f t w o e l e m e n t s o f ~A(H) is g i v e n b y
(~.~')~, = ~ ~,,'
VI(CH _
12.2.
T H E F U N C T O R S CA
12.2.3
Induction
329
and inflation
Let G and H be groups, and U be a G-set-H. If A is a Green functor for G, and X is an H-set, then an element a E CA(U OH X) is a sequence (ay), indexed by the G-sets, such that ay E A ( Y • (U OH X ) • Y). Then if T is an H-set, I set
bT = A 9(@,X,T)(aUo,T) u E (A o U ) ( T X T ) L e m m a 12.2.9: If U/H = i, the above construction is a unitary m o r p h i s m of Green functors f r o m ~A o U t o ~AoU. Proof: First, if U / H = o, and if X and Y are G-sets, then
U o . ( X • Y) ~_ (u o . x ) ( u o . Y) = (U o . x ) • (U oH Y) so that the maps FUxy are bijective. Then the maps
~u T , X , T = ((SU,x
U
x
IdUoHT ) o @X,T
are also bijective. Let then S be an H-set, and ~ E (A o U)(TS). Then denoting by o' and x ' the products o and x for the functor A o U, I have
bTO'~/3 = (AoU). (t,xt2s'~ (AoU)* ( t,xt2s ,. , , . . , . . . . (AoU). \ tlxs / (AoU)*
....
( \tlXt2t23J
Moreover, setting/3' = A.(@U,s)(/3), I have 9
U
A (@,X,T)(aUosT)•
*
U
*
U
/
*
U
= A (@,X,T)(aUouT)X A ( @ , s ) ( / 3 ) = A (@,X,T,T,S)(aUouT X/3')
Setting
T' = U
OH T
X'
= U OH X
S' = U o H S
it is then clear that
( '1x'28 "~ . u
( t'lx't'2s' "~
(A o U)* \ t , x t 2 t ~ s / A (@,X,'r,T,s) = A*(5~X,T,s)A*
tlx,t,2t,2s, ]
Similarly
(A o U). (l~lX*2s~A*(~SUxTS) = (A o U). (~1x'28"~ A . ( ~ T u, X , T , s -1 ) \ tlxt2
/
' ' '
\
tlxs
/
....
= A.@Uxs-')A. (t~x't'2s"~ 9
. .
,
,
\
so that finally
bT OtT /3 = A *(~ST,X,S)(a U T,
0 T' /~')
As a E ~A(X'), and as /3' E A(T'S'), I have also t * O t / b r o T fl = A ( ~ S T , X , S ) ( / 3 0 S , a s )
~;~'~'/
C H A P T E R 12. C E N T R E S
330 The same argument applied to fl o~ bs shows that !
bT OT /3 = ~ os bs hence that the sequence bT defines an element of (AoU(X). Now the l e m m a follows from proposition 8.4.1, which says that the correspondence
X ~-* U on X
a C A ( Y X ) ~-* A.(Su,y)(a) C A ( ( U oH X ) x (U OH Y ) )
is a functor from CAoU to CA.
9
In some cases, the morphism from (A o U to (Aog is an isomorphism:
P r o p o s i t i o n 12.2.10: Let L~_K be subgroups of G, and A be a Green funetor for the group K / L . T h e n , a, (indl~.inf~/LA ~-- ind alx.lnIi(/L(A
Proof: I will show that if G and H are groups, if U is a G-set-H such that U / H = 9 and such that G acts freely on U, then (A o U ~-- (AoV This assertion is equivalent to the proposition, because with the hypothesis on U, there exists a subgroup P of G x H such that
p:(P) = G
k , ( P ) = {1}
U -~ (G x H ) / P
In those conditions, if K = P2(P) and L = k2(P), I have an isomorphism 0 from K / L on G, and if Z is an H-set, I have
VonX
~ - 0 ( ( a e d Z ) L)
Then if G = K / L and 0 = Id, and if A is a Green functor for G, I have
A o U = Ind1(Inf1~/LA H I, W i t h the notations of l e m m a 12.2.9, if br = 0 for any T, then ay = 0 for any Y: the hypothesis implies indeed that ay = 0 if Y is of the form U ou T, since A.(SU, x,T) is an isomorphism. But if Y is a G-set, I have the morphism
uy : Y --* g oH ( G \ U . Y ) defined in this case by .y(y)
=
for an arbitrary element u C U (indeed U.Y = U x Y if U / H = . ) . If moreover G acts freely on U, then vy is injective: indeed, if
12.2.
THE F U N C T O R S (A
331
then there exists h E H and g ff G such that
Whence gy = y' and gu = uh = u. Then g = 1 and y~ = y. Thus any G-set is a subset of a set of the form U OH T. divides an object of this type. But I know that if Y divides Y' then ay, determines ay. In those conditions, I have then ag morphism from (A o U to (AoU is injective. It is also surjective: if b C (Aou(X), and if T is an H-set,
Then in CA, any object in CA, and if a E (A(X), = 0 for any Y, and the I have an element
br E ( A o U ) ( T X T ) I have thus an element b~ : A.(~T,X,T)(bT) r A((U OH T) • (U oH X ) • (U oH T)) and it is easy to see by proposition 8.4.1 that if a E A ( ( U OH T) x (U OH T')), then setting
o / = A*(fU, T,)(v~) r (A o U)(TT') I have /
0
U
U
/
U
t
bT UonTa = A.(@,X,T)(bT)OUoHT A*(~T,T,)(OI ) = A*(@,X,T')(bT~ U 9 ..
t I
=
U
or,
bT,)
i
t
) ....
U
t
=
=
If Y is a G-set, then Y maps into U OH (G\U.Y) via uy, and I set for T = G \ U . Y 9
!
ay = uy OUoffT bT OUoHT Yy,. If Y' is another G-set, if T' = G\U.Y', and if a r A ( Y Y ' ) , then ely Oy OL ~ Y y OUolr T bT OUoI4T b'y,. Oy OZ Oy~ lYyr OUoHT~ IZyl,.
But uy,. oy a oy, @, r A ( ( U OH T) x (U OH T')). So this gives a y Oy O~ = b'y OUoHT lYy,. Oy O~ Oy, lly, OUoHT, b !T' OU~
lYy, . = OL Oy, a y ,
So the sequence ay defines an element of (A(U OH X). If moreover Y = U OH T, then for T ' = G \ U . Y , I have ay :
Uy OUoHT, bT, OUoHT, 12y..
But uy,, is an element of
A((u o. r') • y) : A((v o. r') • (u o. r)) Then a y = l]y OUoHT, Py.. OUoHT bT =
b~T
and then 9
U
A ((~T,X,T)(auo;_.:T) =
which proves the proposition.
A
*
U
!
( ~ T , X , T ) ( b T ) = bT
9
CHAPTER 12. CENTRES
332
12.3
Examples
Let A be a Green functor for G. If M is an A module, if Y is a G-set, and if z C CA(Y), then z induces by definition a morphism zM of A-modules from M to My. It is clear that I obtain that way a morphism of Mackey functors from (A to ~ ( M , My). The construction of the product on (A, by the formulae (~ •
= o:M • Z .
where the product of the right hand side is the product of 7~(M, M), shows that this morphism is compatible with the product. It is moreover clearly unitary. Thus M is a (A-module, and the image of ~A in 7-{(M, M) is contained in 7 ~ A ( M , M) (since z M is a morphism of A-modules). Thus M is an A(~(a-module. P r o p o s i t i o n 12.3.1: L e t A be a G r e e n f u n c t o r for t h e g r o u p G. I f M is an
A - m o d u l e , a n d if X, Y a n d Z are G-sets, t h e n t h e p r o d u c t
aEA(X), b ~ ( A ( Y ) ' r n E M ( Z ) H a x b x m = a x M * ( z Y ) ( b z ~ t u r n s M i n t o an A ~ ( A - m o d u l e .
In particular, the primitive idempotents of CA( ' ) = Z (A(gt 2)) lead to a decomposition of A-modules in blocks: the block j is formed of the A-modules M such that j x M = M.
12.3.1
The
functors
FPB
Let B be a G-algebra, and A = G-set X by
FPB be the fixed points functor on B, defined for a
FP~(X) = nom~([X], B) Then if X and Y are G-sets, the module FPB(XY) identifies with the set of matrices re(x, y) indexed by X x Y, with coefficients in B, which are invariant by G, i.e. such that for any g C G A similar computation as in proposition 4.5.2 shows that if Z is a G-set, if" p(y,z) is a matrix in FPB(YZ), then the product m oy p is obtained as a product of matrices, i.e. (.~ oy p)(~, ~) : X~ "~(~, y)p(y, ~) yEY
where the expression in the right hand side is computed in the algebra B. Let X be a G-set, and rn E (A(X). Then for any G-set Y, I have an element my C A(YXY), that is a "matrix" my(y,x, y') indexed by Y X Y , and invariant by G, i.e. such that for any g E G, I have
-~y(gy, g~, gy') = g.~y (y,., y')
12.3. E X A M P L E S Say t h a t rn E
333
~A(X) iS
equivalent to say that for any G-sets Y and Z, for any
a E A ( Y Z ) , and for any (a:,y,z) E X Y Z , I have y'ffY
#EZ
First I will t~ke Z = G/1. I set m(:r,z) = ma/l(1,x,z)
tor
x E X, z E G = (;/1
so t h a t mo/l(z', x, z) = ~'rn(zt-lx, z'-lz). Similarly, if & is a m a p from Y to B, and if I set I o b t a i n an element of A ( Y • ( G / l ) ) , and any element of A ( Y • ( G / l ) ) i s of this form. In particular, if r = 1 for y = y0 and r = 0 otherwise, equality (12.2) becomes ?Tty(Y'X'ZyO) = E zt'TL(zI-lx'~ZI-Iz) z'EG zt-1Y=YO or for y' = zyo
7Tty(y,',C,y') :
E z'E(;
z'~lYt(zt--lx;zt--lz)
(12.3)
zt--1 ~I=Z-- I 9 t
T h e n taking Y = G/1, this equality forces
,~G/, (> ,~, v') = ~-~(v -1., y-l,/) = y~,-, ~.~(z-lv,v-lx, z-,v,y-lz) which can also be written as
m(y -1 x, y-iv,) = ~'-' =.m(z-~v,y-lz, z-~y,y-lz) This relation must hold for any x C X, and any y, S , z and z' in G. Changing x to yx and z to y'z, this gives
g,z(x, y- ly,) = z?,,.(z-lx, z- l y- , ytz ) or changing f u r t h e r m o r e y' to yy' .~.(.~, ~') = - ' r e ( z - l . , ~ - l v ' ~ ) which gives finally, changing z to z -1 and y' to y
In those conditions, equation (12.3) becomes
zJEG
~.l--I 9--Z--1 7:Ji
C H A P T E R 12. CENTRES
334 which gives (y
y')
m(z,g) 9EG 9y=y'
Using this relation in equation (12.2) gives
y'EY 9EG gy=yt
z~CZ 9EG gzt=z
or Z ?Tt(X,g)(;t(g~],Z) = Z (/(Y'g-Iz)?T/(X' g) 9EG
9EG
that is 9EG
9r (;
Let again Y = Z = G/1. Let b E /3, and r be the function from Y to /~' which is equal to b in Y0 and to 0 elsewhere. If a(y, z) = ~r equation (12.4) gives TYt(X, zyoy-1).Zb = gy~ b.fft(x, zy0y -1 )
This equation gives for Y = Yo = 1
Conversely, it is clear that this relation implies equation (12.4). Thus CA(X) is isomorphic to the set of matrices re(z, z), indexed by X x G, and satisfying the following conditions
Zm(x, y) = ~(~z, ~y) V:~ ~ X, W, y ~ G
Then let r'n(z) = ~ e a re(z, z) | z 9 B | G. The second condition is equivalent to the fact that r'n(x) commutes with all the elements of B, I have then a map
7~ : X ---, CBoc(B) The first condition shows that for z E G
(1 e z),a(x)(1 o z -1) = ~ z~(~, y) | z~ = }2 - 4 % zy) e z~ = ,~,(~x) yEG yEG In other words, if C = CB| with its natural structure of G-Mgebra (induced by the morphism 9 ~ 1 | g from G to B (x) G), I see that rh is just a morphism of RG-modules from IX] to C. On the other hand, the product in CA of the element rn E CA(X) by the element 0 ' r CA(X') is defined by tZZ2)
( o • ,.')y = A. kzlx~'z~
I
( ~ y or my)
12.3. EXAMPLES
335
Then if m corresponds to the matrix re(x, z), and m' to the matrix m'(x', z), I have
(,~•
= ~ ~ ( y , ~ ' , v )-~(y , x , y ) = y" EY
~
m'(~',g).~(~,g')
. . . .
gY=Y" yy,,=yl
....
F~
~
~'(x',g).~(x,g')
9"C=0 9,g~EG g,,y=yt gig=g,,
It follows that
,./(~:',g).~(x,J) og'g
~T.~.,(~:,s) = ~ g,g~EG
But
,~(x).~,(x') =
E
g
I
/
l
g,gSEG
=
I
I
I
g,glEG
Then (A(X) ~-- Hom~([X], C), and it is clear that this isomorphism induces an isomorphism of Green functors from (a to FIX:. Now let b C Z(B)(X). Then b is a morphism of RG-modnles from IX] to B, such that for any Y, and any morphism c of RG-modules from Y to B
The case Y = G/1 shows then that b(x) E Z(B), and it follows an isomorphism of Green functors Z(FPB) ~-- FPz(B). Finally, I have the following proposition:
P r o p o s i t i o n 12.3.2: Let B be a G - a l g e b r a , a n d C = CB|
Z(FPs) ~- FPz(B) as G r e e n functors. In particular @p.(*) ~-
Then
@e. ~-- FPc
Z(B | G) and
T h u s t h e centre of Y o s h i d a algebra (see p r o p o s i t i o n 4.5.2) is i s o m o r p h i c to t h e centre of J~G. R e m a r k : If the algebra B is an interior algebra (see [11]), then the algebra C is isomorphic to Z(B)G: indeed, if 9 ~ P(g) is a morphism from G to the group of invertible elements of/5', and if g acts on B by conjugation by 0(9), then say that ~g % ~ g is in C is equivalent to say that for any g, the element %p(g) -~ is in Z(B), hence that Eg(%p(9)-l)g E Z(B)G. This correspondence is moreover an isomorphism of algebras C ~- Z( B)G.
12.3.2
The blocks of Mackey algebra
Let R = k be a field of characteristic p > 0 and A = bp the p-part of the Burnside functor, with coefficients in k. Then A(1) ~ k, and (A(1) ~-- kG. On the other hand, (A(G) is the centre of the p-part #e(G) of the Mackey algebra (i.e. the piece of the Mackey algebra corresponding to the central idempotent fla, or the subalgebra of the Mackey algebra formed of elements ~px'1rKp~, t for any subgroups H and K of G, any
336
CHAPTER
I2.
CENTRES
element z of G, and any p-subgroup P of 1t N ~K). The lq(G)-modules are exactly the Mackey functors which are pr@ective relative to p-subgroups. Then I have the map rla : Z(#I((-/)) --4 (k.(;) c" = Z~:G. It is a morphism of algebras, it is unitary, and surjective: indeed, Za'(; is generated as ~'-module by the elements of the form for x E G. Let for K C_ C a ( x ) It is easy to see that this defines an eleme,tt of (.a (Cc;(.~')). Moreover G G
Thus r~ induces a surjection f r o m
-
,,r 111 rd>.(~.)( r
Z(/ll(("))
to Z~'(;. The kernel of this surjection
is moreover nilpotent, because if e is an idempotent of the kernel, then 7'~;(e') = 0. The module e#l(G) is then a projective Mackey functor, which is projective relative to p-subgroups, and equal to zero at {1}. Hence it is zero (see [15[ corollary 12.2) and It follows that rla induces a bijection between the blocks of kG and those of #~(G) (see [15] Theorem 17.1).
Bibliography [1] D. Benson. Representations and cohornology L volmne 30 of Cambridge studies in advanced mathematics. Cambridge University Press, 1991.
[2] S. Bouc. Construction de foncteurs entre categories de G-ensembles. J. of Algebra , 183(0239):737 825, 1996. [3] S. Bouc. Foncteurs d'ensembles munis d'une double action. 183(0238):664-736, 1996.
J. of Algebra,
[4] C. Curtis and I. Reiner. Methods of representation theory with applications to ,finite groups and orders, volume 1 of Wiley classics library. Wiley, 1990. [5] A. Dress. Contributions to the theory o.f induced representations, volume 342 of Lecture Notes in Mathematics, pages 183 240. Springer-Verlag, 1973. [6] L. G. Lewis, Jr. The theory of Green functors. Unpublished notes, 1981. [7] D. Gluck. Idempotent formula for the Burnside ring with applications to the p subgroup simplicial complex, lllinois J. Math., 25:63--67, 1981. [8] J. Green. Axiomatic representation theory for finite groups. J. Pure Appl. Algebra, 1:41-77, 1971. [9] H. Lindner. A remark on Mackey functors. Manuscripta Math., 18:273-278, 1976. [10] S. Mac Lane. Categories for the working m.athematician, volume 5 of Graduate tezts in Mathematics. Springer, 1971. [11] L. Puig. Pointed groups and construction of characters. Math. Z., 176:265 292, 1981. [12] H. Sasaki. Green correspondence atld transfer theorems of Wielandt type for G fimctors. J. Algebra, 79:98-120, March 1982. [13] J. Th~venaz. Defect theory for maximal ideals and simple ftlnctors. J. of Algebra, 140(2):426-483, July 1991. [14] J. Th~venaz and P. Webb. Simple Mackey functors. In Proceedings of the 2nd International group theory conference Bressanone 1989, volume 23 of l?end. Circ. Mat. Palertoo, pages 299 319, 1990. Serie II. [15] J. Th~venaz and P. Webb. The structure of Mackey functors. Math. 5'oc., 347(6):1865-1961, June 1995.
Trans. Amer.
BIBLIOGRAPHY
338
[16] P. Webb. A split exact sequence fin' Mackey functors. Commerzt. Math. Helv., 66:34 69, 1991. [17] T.
Yoshida. On G-functors (II): J.Math.Soc.Japan, 35:179-190, 1983.
Hecke operators
and
G-functors.
Index (LI), 5o
.v, 181 A - M o d , 48 A(X2), 81 A ~ 127 A | K, 278 Av.j, 115 CM(L), 143 CM(O~), 143 FP~,v, 296 F OH,v, a 296 FM, 71 G.s.t, 229
OH, 167
5U X1,...,Xn 171
~lz, 185 G-set~x, 53 A(H), 276
d, a o
~,
G\U.Y, 18a G\U.c~, 183 GreenR(G), 47 H| 97 Lx,v, 275 M o U, 168 M@BN, 146 Ms-+My, 8 M o, 130 MF, 72 MackR(G), 7 N
o
UIA , 237, 256
S.T, 229 S r 206 S H,v, a 282 Sx,v, 275 Sr 206 T r 193 Tr 193 U.Y, 183 U[X], 173 Z(A), 305 Z~T, 242 [m @ n]K, 14 [m@
n](y,r
fl, 84 f Q , 112 E~(G), 313
21s
~ , 218 ~uS,T~ 230 Az, 199 lq (G), 335 >disjoint, 188 u(yj), 185 ~N(K), 218 v-perfect, 308 rrR(G), 307 ~z, 199 M ( H ) , 273
M(H), 27a ~ ( H ) , 27s r XUy 168 x ~ 127 • U, 171 xH, 181 U• 231 CA, 134 5AoU, 171 s 232 X, 194 (A(H), 315
(A(X), 316
16
a.m, 49 a ox m, 72 a oz a', 65 a x b, 46 a x r n , 47,76 b, 52 b 5 a, 125 e~, 308
INDEX
340
f.,f*, 68 fHa, 308 kl(L), 183 k2(L), 183 rn(g.~), t94 pI(L), 183
p~(L), ls3
q(L), 183 Du(X), 186 ~(M, N), 9
~A(M, N),
145
Zx, 316
~:(M,,..., ,%; P), s 193 s 196
29
Qv(M), 188 7~(M), 2o7
T~u(O), 209 8u(M), 205 adjunction between ~ and ~ , 38 between @B and "HA, 148 co-unit, 185 unit, 185 algebra associated to a Green functor, 84, 99, 164 Alperin's conjecture, 304 balanced, 99, 154 bifunctoriality, 46, 47 bimodule, 141 construction, 15a structure on 7-{(M, N), 142 biproducts, 68 biset, 167 composition with, 168 blocks of Mackey algebra, aa5 Burnside functor as Green functor, ,55 as initial object, 57 as Mackey functor, 52 as unit, 59 cartesian product in CA, 103 in CA • CA, 109
category Du(X), 186 adding direct snmmands to a, 310 associated to a Green functor, 67 equivalence of, 79, 310 .312, 314 representation of, 71, 81 centre, 305 of Yoshida algebra, a35 coinduction. I80 coinfiation, 168, 217 commutant, l,t3 commutative Green functor, 110, 129, 305 commute, 136 composition and associated categories, 173 and Green functors, 170 and modules, 175 and tensor product, 168 with a biset, 168 direct summand in CA, 82 divides in C.4, 82 dual of a module, 130 embedding of a.Lgebras, 112 endosimple A module, 304 equivalence of categories, 79, 84, 112, 129, 141, 155, 310-312, 314 of the definitions of (-been functors, 48 examples of algebras A(fT~), 94, 95 of composition with a biset, 168 of Frobenius morphisms, 227 of functors (,4, 3a2 of functors s and •u(M), 215 of Green functors s 264 finitely generated module over a. Green functor, 114 Frobenius nlorl)hisn~s, 223 functor from G-sets to CA, 68 of evaluation, 81
INDEX functorial ideal, 291 functoriality of M@BN, 147
of U ~ U oa X, 177 of "H(M, N) and MQN, 24 of 7fA(M, N), 146 generators and relations for A(f~2), 85, 88 for the Mackey algebra, 7 Green functors and solvable re-subgroups, 314 centre, 305 composition with a biset, 170 definition in terms of G-sets, 46 definition in terms of subgroups, 41 direct sum, 305 tensor product, 134 identification of ~A, 320, 323, 327 of "]~A(M,_N), 145 of internal homomorphisms, 11 of tensor product (first), 15 of tensor product (second), 23 injective on the orbits, 177, 186, 192 left adjoint and Morita contexts, 273 and tensor product, 272 toM~MoU, 197 to Z ~-+ U OH Z, 183 Lindner construction, 94 Mackey algebra, 7 anti-automorphism, 10 axiom, 5 functors composition with a biset, 168 definition as modules, 7 definition in terms of G-sets, 6 definition in terms of subgroups, 5 internal homomorphisms, 9 tensor product of, 9 module dual, 130
341 of finite type, 114 over a Green functor, 41, 47 examples, 61 over the Burnside functor, ,59 right, 129 Morita context, 100, 154 equivalence, 99 of algebras A(X2), 99, 100 theory, 1,55, 160 morphism n-linear, 29 universal property, 29 bilinear, 37, 127 Probenius, 223 of bimodules, 141 of Green functor, 41 of Mackey functors, 5, 6 of modules over a Green functor, 42 multiple of in CA, 82 natural transformation, 316 non-commutative tensor product, 164 opposite Green fimctor, 127 product OH, 167 8 , 125 • for A~B, 134 • for CA, 317 x for ~ ( M , M), 123 • for s 231 progenerator, 115, 117, 118, 155, 161, 321 projective G-algebra, 292 projective relative to, 102 solvable ~r-subgroups, 314 relative projectivity, 100 representation of a category, 71, 81 residual rings, 274 restriction, 5 for M~N, 15, 23 for ~ ( M , N), 10
342 right adjoint to M ~ - + M o U , 210 right modules, 129 same stabilizers, 100 simple Green functors, 291 classification, 295 simple modules, 275 classification, 276 structure, 282 source algebra, 114 surjective Morita context, 154 tensor product 7~-fold, 25 universal property, 29 and composition with a biset, 168 and left adjoints, 227 and residues, 298 and right adjoints, 242 associativity, 38 commutativity, 38 of "algebras" over b(ft2), 164 of Green functors, 134 universal property, 137 of modules over Green functors, 140 of simple modules, 298 Th~venaz's theorem, 295 trace, see transfer transfer, 5 for M Q N , 15, 23 for 7-{(M,N), 10 unitary Green functor, 46 morphism of Green functors, 41, 43 Yoshida algebra, 95 centre, 335
INDEX
