The Kadison Similarity Problem

Similarity Problem

10. The Kadison

Summary: This chapter concentrates on the (still open) Kadison Similarity problem (= Problem 0.2). We show that this problem is equivalent to several seemingly different questions which appeared in other contexts: Arveson's hyperreflexivity problem and Dixmier's invariant opemtor range problem. Finally, we describe the author's recent work on the notion of "length" of an opemtor algebm, which also leads to one more reformulation of Kadison's question.

general form, the Kadison similarity problem asks whether any bounded homomorphism u from a C*-algebra A into B(H) is similar to a homomorphism (= "orthogonalizable"). This is the same as Problem 0.2 from the introduction. Indeed, this can be In its most

reduced to the unital

Recall first how ideal in

A

and if

A and A in C

follows.

C*-algebra

is the unit in

e

we

case as

a

A is embedded into its unitization

A

we

A + Ce,

Ila + AeJJX

<

1Ilall, JAI

2

any bounded

homomorphism

u:

A

--+

B(H)

JAI

can

B (H) by setting ii(a + \e) homomorphism ii: , orthogonalizable, u (being its restriction to A) also --

Remark. With the

passing that

Ilull

> 1

if

Also since

a

preceding notation,

since A has

u: 0, + Ae

moreover

A is

for any

a

an

in

have

max

Thus,

have

A:

hence -

an

lliill

A is

a

we

have

Ilall.

be extended to

u(a)

+ AL

a

bounded

Clearly,

if ii is

is.

Ilull

<

lliill

in its unit ball

approximate unit :5

=

+

:5

21jull

we

+ 1. Note in

have

necessarily

31jull.

*-homomorphism

it is

completely contractive,

from

which it is easy to deduce that

IJUIlcb :5 lliillcb ::_ 21julIcb Thus

we

have

proved

lem reduces to the unital

Quite curiously,

that the above

+ I <

general

31julIcb-

formulation of Kadison's

the Kadison

probEl

case.

problem

turns out to be

equivalent

to

a

number

of questions which come up in various different contexts and which make its solution even more desirable. We have already seen one such equivalence:

10. The Kadison

Theorem 10.1.

derivation

The Kadison

problem (=

This is due to the end of

Problem

Similarity

similarity problem (= 0.2) are equivalent.

Kirchberg [Ki]. We have included

Problem

detailed

a

Problem

0.2)

169

and the

proof above,

at

7.

equivalent form of the preceding problem which arose algebras". This is a certain class of non-selfadjoint subalgebras of B(H), of which the simplest example is the algebra of upper triangular matrices in M,, or in B(H). More precisely, a nest on a Hilbert space H is a family C of subspaces of H, totally ordered under inclusion and such that for any subfamily (Ei)iE, in C we have We

turn to

now

in Arveson's

theory

an

of "nest

nEi

EC

UEjEC.

and

iEI

iEI

Given such

a

nest

all operators T in

C,

one

T(E)

V E (=- C The

subalgebra A(C)

associates to it the

c

B(H)

formed of

such that

B(H)

C E.

resulting algebras are called "nest algebras". For instance, the algebra of triangular matrices in M,, corresponds to the nest (Ei, 1 < i < n) defined

upper

by Ei

=

span[el.... ei],

(ei) denotes the canonical basis of fn2. These algebras were introduced by Ringrose, motivated by Kadison and Singer's earlier work on "triangular algebras". We refer the reader to [Dad] for a detailed exposition of this subject. The most striking result of this theory is probably Arveson's distance formula which we will now explain (this was briefly mentioned in the Notes at the end of 4). Let A c B(H) be a subalgebra, we denote by Lat (A) the set of all closed where

subspaces

E C H which

operator T in

B(H)

we

are

A-invariant

d(T, A) Then it is easy to

see

-

an

equality, namely

supf 11 (1

(10.2) This led him to an

algebra

=

inff JIT

-

-

PE)TPE I I

PE)TPEII I

The basic result of Arveson's

becomes

a

in

A).

Given

an

all I

<

a

E

Al.

subspace

E

we

have

d(T, A),

B(H)

supf 11 (1

(10.1)

aE C E for all

that for any A-invariant

I (I whence for any T in

(i.e.

set

study

-

theory

E

c

Lat(A)J

<

is that if A is

for any T in

PE)TPE 111

B(H)

E E

we

a

this distance formula for

nest

algebra then (10.1)

have

Lat(A) I

is called "reflexive" if it satisfies the

d(T, A).

=

d(T, A).

general classes implication

more

of

algebras:

Similarity Problem

10. The Kadison

170

it is called T in

VE c

c E

(T(E)

if

"hyperreflexive"

Lat(A)) there is

moreover

=#-

T c-

A;

constant K such that for any

a

B(H) d(T, A)

KsupJJJ(1

-.5

-

PE)TP-pil I

Lat(A)J.

E E

1 for nest algebras. The first examples hyperreflexive algebras were given by Kraus and Larson [KrL]. These are not selfadjoint. In the selfadjoint case, von Neumann's bicommutant theorem implies that every von Neumann algebra is reflexive. The problem whether they are hyperreflexive was raised by Arveson in [Ar2] and has

As

just

we

saw, this holds with K

=

of reflexive but not

remained open since: Problem 10.2. Is every

Neumann

von

Theorem 10.3. Kadison's mann

problem hyperreflexive.

is

algebra

Proof: We have

already

M is inner, then M' is

is

us now a

<

von

KsupJJJ(I

where the supremum runs A fortiori, if we let 6T(X)

over

=

d(T,M) a

check the

fortiori

we

:5

xT

-

B(H)

M c

and any T in

P)TP111:5 KsupJJJTP

-

PT111

all projections P in M'. Tx, we find -

<

1i

x

E

M11.

have

when

ir:

A

case we

--*

B(H)

may

M

=

as

is

well

A' in

an

isometric

assume

(10.3)

we

J6T11cb(A,B(H))

embedding

that A c

obtain, !

a

B(H)

can

(v), but restricted to the a C*-algebra A. Indeed,

of

and

fortiori, for

4KI16T: A"

--+

exchange the roles of

ir

is the inclusion map.

any T in

B(H)

B(H)11.

But since the unit ball of A is dense in that of A" in the weak operator

find

116T: A" and

we

(10.4)

we

116TIleb,.5 4KI16TIl

Then, taking

we

algebra

Assume that every M argument, we find a universal

for any inner derivation 6T: M' -+ B(H). Since we M' and M, we obtain the situation of Theorem 7.21 in that

Neumann

Neu-

M and its commutant,

algebra

2KSUPJJJ6T(X)JJ I JJXJJ

(10.3)

case

von

converse.

sum

Neumann

on a von

every

have

d(T, M)

whence

positive solution iff

simple direct

constant K such that for any we

a

hyperreflexive. Exchanging

only hyperreflexive. Then, by

B(H)

has

that if every derivation

seen

if part. Let

obtain the

algebra hyperreflexive?

--+

B(H)11

=

116T:

A

obtain

116T116:5 4KI16TIl

-4

B(H)11

topology

10. The Kadison

for any 7r-inner 6T: A

--*

B(H).

Similarity Problem

complete the proof,

To

it remains

that the assumption that 7r: A --+ B(H) is isometric is very easy: given a fixed inclusion A C B(h) and a a:

A

a

0

0

a(a )

)

representation (=*-homomorphism)

7r(a) This is 0

an

0)

isometric

we

can

to show

only

be removed. This

(possibly

not

faithful)

consider

--+

B(H,)

we

E

B (H ED

H,).

any T in B (H,), if

representation and for

171

we

let 0

have

T

0

0

ir(a)O

07r(a)

-

=

0

(0 o,(a)T-To,(a))

which shows that

16TIl

=

and

11boll

116T116

=

116011cb.

(10.4) holds for a faithful 7r, it also holds if 7r is replaced by an arbitrary representation (=*-homomorphism) a. Thus we obtain the condition (v) in Theorem 7.21 for any C*-algebra A, which is equivalent to the Kadison 11 problem. We now turn to a topic which was in vogue many years ago, namely "operator ranges". An operator range is a linear subspace V of a Hilbert space H which is the range of bounded operator T: H --> H, i.e. we have V T(H). Typically we will be dealing with subspaces which are not closed, and Foias introduced the term "paraclosed subspaces" to designate operator ranges. In the early 70's, this subject was studied by several authors, among whom Fbias [Foi2], Fillmore and Williams [FiW] and Voiculescu [Vo]. The following problem, due to Dixmier, (whose 1949 thesis is devoted to a related topic) has circulated for a long time: This shows that if

=

invariant operator range problem.) Let V C H be an B(H) be a unital C*-algebra. Assume that V is

(The

Problem 10.4.

operator range and let A C invariant under

A'

(=

Dixmier, whom in

a

aV C V for any

A, meaning

the commutant of A) such that I

consulted, kindly

letter addressed to

von

find another operator

T

wrote

me

in A. Is there

an

operator T in

V?

back that he asked this question

Neumann in 1950.

T(H)

Remark. Of course, if V problem is to show that if V =

T(H)

a

=

=

and

T(H)

in A' with the

that V is not assumed closed

T

E

A' then V is invariant under A. The

is invariant with T E

B(H),

then

we can

range. Again, we should emphasize the solution is trivial: we can take

same

(otherwise,

T the orthogonal projection onto V which clearly is in A' if V is closed and invariant). Following an idea of Foias ([Foi2]) we will show that this problem is equivalent

for

to the Kadison

one.

10. The Kadison

172

Proof: a

problem (= Problem 10.4) has a similarity problem (= Problem 10.2) also does.

The invariant operator range

Theorem 10.5.

solution iff the Kadison

positive

Problem

Similarity

Following [Foi2], we will associate to an invariant operator homomorphism. u: A -- B(H). So we assume V

bounded unital

C*-subalgebra. injective. Fix

Let K

A and k

a

E

T(H)

with

C

=

=

Ilk. By the

that aTk

=

V for any a in A where A c B(H) is a unital V and of course TIK is ker(T)J-. Note that T(K)

such that aV

B(H)

T E

range V C H

aT(K)

K. Since closed

c

T(K)

a unique k in K such correspondence k --+ i defines

there is

the

graph theorem,

bounded linear operator which we denote by u(a). It is easy to check that u(a) e B(K) is a unital homomorphism. Moreover, one more application of the closed graph theorem shows that a --* u(a) is bounded. Thus we obtain the a

a

--+

announced

homomorphism. associated

u(a))

Tu(a),

aT

=

hence if 0

to V. Note that

K

TIK:

=

u(a)

VaE,A but in this formula

Then there is

an

0-1 is we

H

--+

Now

S-'ir(a)S

=

(by

have

definition of

0-'aO

=

K such that

have 0-'aO

we

have

we

priori unbounded.

invertible S: K Thus

homomorphism.

a

--+

assume u -+

a

7r(a)

orthogonalizable.

Su(a)S-'

=

and if we let

x

=

is

a

OS-': K

x7r(a) ax for any a in A. But since A is a C*-algebra and 7r(a*) x7r(a)* x7r(a*) a*x, hence taking adjoints lr(a)x* x*a, and 7r(a)*, xx*a. But we can also write finally (after left multiplication by x) x7r(a)x* H

we

have

=

have

we

=

=

=

=

x7r(a)x* Thus

the x

we

have

(x7r(a))x*

xx*: H

proved that

(xx*)'I'.

for

same

=

=

(ax)x*

=

axx*.

A, which implies polar decomposition of

H is in the commutant

--+

A moment of

thought

on

the

shows that

T(H) and since

x

=

OS-' with S invertible

x(K) Thus

we

We have thus shown that lem 10.4.

Conversely,

u:

A

-+

a

=

=

have

T(H).

.

.

.

be

a

bounded unital

notation. Let

(6,,)

B(H,,)

-+

be

u(A) j

T

c

A'.

a

one

to Prob-

+

homomorphism. Assume H separable H, and let H,, C H

dense sequence in +

u(A) ,.

defined by u.,,(a) hence, by Theorem 7. 11,

B(H.,,) set

,

able. Whence S,,: H,, -4

with

positive answer to Problem 0.2 implies positive answer to Problem 10.4.

homomorphism u,,: A ,, as cyclic 1,

admits

A

O(K)

T(H)

H,,

7r,,:

clearly

assume a

B(H)

simplicity of be defined by

for

The

we

x(K)

conclude V

Let

=

=

H,, invertible (and bounded) and

such that

u,,(.)

=

S,-,1,7r,,(-)S.,,.

u(a)IH,,, obviously

=

u,, a

is orthogonaliz*-homomorphism

We may

as

well

assume

10. The Kadison

JIS,,11

1. Let

B(ft)

ii: A

Hi 6) H2 (D

be defined

We have

(a)

is

1,

=

S-' (a),

:

F1

Problem

`R: A

B

--

173

(ft)

and

by

EDS,

=

and let

...

Similarity

(a)

injective,

where here

ax,,(a)

=

its

-1

and

i (a)

u,,(a).

=

r( )

is invariant under

image

and

have

we

priori unbounded. Suppose for a moment that we can show that i is orthogonalizable. Then it is easy to see that u itself must be orthogonalizable. Indeed, we have 11U11cb IP16 hence SUP JIU,,Ilcb u is orthogonalizable by Corollary 4.4 (one can also prove it directly). Thus to complete the proof it suffices to show that any homomorphism u: A B(H) of the form u(a) H bounded, injective and with 0-17r(a)0 with 0: H V O(H) invariant under 7r(A), is orthogonalizable. If we assume a positive solution to Problem 10.4, this is easy: we can find 0 in 7r(A)' such that O(H) V. Let Ho (ker0)_L. Note that Ho is invariant =

is

a

=

=

--+

--+

=

=

_

=

=

7r(A). 7r(a)Oo

under

Let

have

=

Oo: Ho

--+

0o7r(a)jH,,

H be the restriction of 0 to

for any

a

7r(a) But V

=

O(H)

=

0-10o7r(a)IH000

W(H)

10.

But

in A

=

hence,

on

Ho which is injective. We 00 we have

the range of

Oo7r(a)IHOOO 1.

is also the range of 00. Hence we have now 00 10: H -* Ho and 0-100: Ho

u(a) ---

=

H

0-17F(a)O

are

bounded

by the closed graph theorem (note here that, of course, Ho is closed), hence u is orthogonalizable. This completes the proof that "yes" to Problem 10.4 implies 11 44yes" to Problem 0.2. Remark. Note that

we

may

as

well

assume

in Problem 10.4 that A is

a von

already know this for Problem 0.2 (by passing bidual). See [On] for more details on this particular point. mann

algebra,

since

The results in for the

(SP)

Chapter 4, and notably Theorem 4.3 provide ample

Neu-

to the

motivation

following:

Generalized the

we

Similarity

Problem.

following property (SP) For any H, every bounded

Which unital operator

algebras A have

?

unital

homomorphism

u:

A

--+

B (H) is

c.

b.

Loosely speaking, this property (SP) could be described as "automatic complete boundedness" in analogy with the field of automatic continuity for homomorphisms between Banach algebras (see [DW]). The Kadison problem is equivalent to showing that every C*-algebra satisfies

(SP). [P9],

In

for

an

show that this property (SP) is closely linked to a notion of "length" operator algebra. Related results appear in [P10-11-12]. The rest of this we

chapter The

is

an

outline of all these results.

"length"

that

we

have in mind is

analogous

to the

following

situation:

a unital semi-group S and a unital generating subset B C S, it is natural to say that B generates S with length < d if any x in S can be written as

consider

a

product

x

=

b1b2

...

bd with each bi in B. Our main idea

can

be illustrated

10. The Kadison

174

in

Similarity Problem

rather transparent way

a

simple model of semi-groups

the above

on

fol-

as

lows. Assume that B generates S with length < d. Then any homomorphism 7r: S -- B(H) (i.e. 7r(st) 7r(s)-7r(t) and 7r(l) 1) which is bounded on B with =

=

sup

11-7r(b) 11

be bounded

< c must

11 7r(s) 11A

the whole of S with sup

on

SES

bEB

Conversely, assume that we know that morphisms ir: S -- B(H) satisfy, for some sup

11 7r (b) 11

:5

c =:>.

for

sup

11 7r (s) I

<

n

>

0, all homo-

following implication:

> 1, the

c

and

> 0

some a

rw'.

SES

bEB

Then it is rather easy to see that B (=integral part of a), so that we can

necessarily generates S with length replace a by [a] and r, by 1.

<

[a]

Definition. An operator algebra A C B(H) is said to be of length < d if there is a constant K such that, for any n and any x in Mn (A), there is an integer N =

ad-1 E MN(C)i N(n, x) and scalar matrices ao E Mn,N(C), al E MN(O)) in MN(A) satisfying ad E MN,n(O) together with diagonal matrices Dj,..., Dd -

-

.,

I

x

=

aoDlaID2... Ddad

d

d

fj Had 11 IlDill We denote of A

(so

KlIxII.

the smallest d for which this holds and

by t(A)

that A has

Equivalently,

<

1

0

length

<

d is indeed the

may reformulate this

we

same as

we

t(A)

using infinite

:!5

usual

k(A)

=

Mn(A) UM,,(A)

c

Mn+,(A)

be the

via the

mapping

completion of the

the norm, then it is easy to check that

f(A)

x

--+

"length"

d). if

matrices:

we

view

0

(0 0), X

as

call it the

and if

we

let

union with the natural extension of <

d iff any

x

in

IC(A)

can

be written

as x

=

aODjaj

with ai in K(C) and Di diagonal in by the open mapping theorem.) We start with

Proposition

an

...

Ddad

IC(A). (The

constant K

automatically

exists

easy observation.

10.6. Assume

We have then for any

f(A)

=

d <

homomorphism

oo

with

d, K

.1rom A to

u

as

in the above definition.

B(H)

JIU116:5 KIJUIld. Proof: Consider

x

in

M,,(A).

Let u,,:

M,,(A)

-+

M,,(B(H))

denote the usual

SUP111un 111 n > 11. Let homomorphism (= 10 u). Recall IJUlIcb unit ball of M,,(A). Consider a factorization of the above form: =

x

with ai "scalar" and Di

=

aoD,

"diagonal".

...

Ddad

We have then

x

be in the

10. The Kadison

Un(X)

aOUN(Dl)al

...

Similarity

Problem

175

UN(Dd)ad

hence

11Un(x)11 but

clearly

since the

Di's

are

diagonal JJUN(Di)JJ

HUnW11 which

yields (recalling

the

fl 11aill 11 JJUN(Di)JJ

:5

<

11ull IlDill

hence

11ulld H 11aill fl IlDill

:5

meaning of t(A)

d)

<

JJU116:5 K11U11d. El

The main result concerning this sufficient for (SP), it is also necessary: Theorem 10.7. A unital operator over, let

d(A) (here

of course

u

=

infla

denotes

an

algebra A

13

> 0

"length"

K V

satisfies

t(A)

(SP)

<

00

is not

iff t(A) <

oo.

only

More-

1JU116:5 KilullI

u

unital

arbitrary

is that

homomorphism from A

to

B(H)),

then

d(A) and the infimum

defining d(A)

We refer the reader to

[P91

=

t(A)

is attained. or

[P16]

for the

proof C*-algebra, let us momentaxily some K, all bounded (not necB(H) satisfy IJUIleb < Kilulld. By the

Remark. If A is any (not necessarily unital) denote by do(A) the smallest d such that, for

essarily unital)'homomorphisms u: A -+ remark on A which opens this chapter, it is easy to check that do(A) d(.X). On the other hand, we have (this is a particularly easy case of (10.6) below) f(A) (A). Thus (by Theorem 10.7) we obtain the equality do(A) f(A) for any C*-algebra A. A C (as unital C*-algebras) hence Now, if A happens to be unital, then clearly d(A) d(A). Thus we have do(A) d(A) in this case. Therefore, there is no need to distinguish do(A) and d(A), at least when A is a C*-algebra, so that from now on we set d(A) do (A) for any C*-algebra (unital or not). Thus we can now write d(A) t(A) for any C*-algebra A. Remark 10.8. Let us briefly return to the derivation problem. By Theorem =

=

=

=

=

=

=

7.21 and

(7.26)*,

if

we

have

11611cb

(10.4) for all

7r

and all 7r-derivations 6: A

of

a.

u as

in Theorem 10.7. Therefore

This leads

us

to

a

conjecture

:

a11611

B(H)

JJUJJcb

(10.5) for all

-+

<

<

then

we

have

HU11a

t(A)

is less

or

equal

to the

integral part

Similarity Problem

10. The Kadison

176

Conjecture. For any infinite dimensional C*-algebra A, the best possible (10.4) is always an integer. Let KA be the best K such that:

V

:5 K IlUlld(A). We have

IJUJIcb

u

example of an infinite dimensional C*-algebra A for which KA a

a

But

> 1.

we

in

no

have

suspect:

Conjecture. If A

B (E2) E)

=

too then KA

(here d(A)

> 1

It is easy to see that the property (SP) passes to if I c A is a closed two-sided ideal then

f(AII) and also that f (A) :5

maxjt(I), f(AII)}. f(A)

(10.6) To show

there is

in I

x

aai

particular,

aia

-

Let

(i) (ii)

(D

...

of infinite direct

us now

Kadison

-4

below).

quotients. More precisely,

t(A)

If A is

a

C*-algebra,

we

have

maxf (I), f(A/I)}.

ED

&

.

A,,,)

sums

review the

have xai -+ a in A.

we

and aix

x

x

--+

and

moreover

0 for any

for all finite sets

t(Ai case

see

: (A) we merely use the fact (due to Arveson, see [Wa]) that "quasi-central approximate unit" in I, i.e. a net (ai) in the unit ball

(quasi-centrality)

The

=

3,

t(I)

a

of I such that for any In

<

=

.

=

.

,

A,, of operator algebras

maxjt(Ai) 11

is discussed in

cases

for which

a

< i <

we

have

nf.

[P12].

positive solution

is known for the

problem:

A is commutative.

C*-algebra, denoted by K, of all compact operators on t2, or its unitization, denoted by k. More generally, by Theorem 7.16, A can be any nuclear C*-algebra. (iii) A B(H) or more generally A can be any C*-algebra without tracial. states, see Corollary 7.14 above. IC 0 B with B an arbitrary C*-algebra (here again A has no tracial (iv) A A is either the

=

=

states)(V)

A is

a

Hi-factor with Murray and

for instance when A is the so-called

product We will

preceding

of 2

2 matrices with normalized

trace).

give the degree (=length) of the algebras appearing in'the 1, so from now dim(A) < oo, we have clearly d(A)

now

list. When 1 <

on we assume

(i) (ii) (iii) (iv)

x

Neumann's property F (see [C4]) hyperfinite III-factor (= infinite tensor von

dim(A)

If A is commutative, If A

=

If A

=

If A

=

IC, A

=

B(H), k (9 B

k

or

then

=

oo.

=

d(A)

if A is

d(A)

with B

arbitrary C*-algebra )

=

=

2.

nuclear, also d(A)

=

2.

3.

arbitrary unital C*-algebra (or then 2 <

d(A)

< 3.

if A

=

IC 0 B with B

10. The Kadison

If A is

(v)

< 3 is

(ii)

and

(i)

Notes:

Hi-factor with property P then d(A)

a

are

=

Similarity

Problem

177

3.

(see [C2]).

due to J. Bunce and E. Christensen

(iii)

In

proved in [Hl] while > 3 is proved in [P9] (see below for more on this). essentially in [111]. Finally, concerning (v), Christensen proves in [C4] that

(iv) is d(A) < 44,

but this bound is reduced to < 5 in

[P12],

d(A)

and

:! 3 appears in

the very recent preprint [C8J. It is also mentioned in [P12] that (as pointed out by N. Ozawa) Anderson's construction in [And] remains valid on any 11, factor,

yielding d(A) > 3 for any II, factor A by the same argument as in [P9]. sharp contrast, the conjecture is still open when A is the reduced C*-algebra

thus In

of the free group with > 2 generators,

EX

M=

or even

when A

M with

=

=fx=(X,,)Jx,,EM,, SUPIIX,,11<001.

n>1 00

C*-algebra f,,o (IN; M) (formed of all bounded famcounterexample to the Kadison conjecture. The evidence for this lies in the following result from [P12] and its corollary: we feel the property appearing in this corollary is "too strong to be true" but unfortunately we

We

strongly suspect

(Xi)iE]N

ilies

M)

in

that the

is

a

cannot prove this!

Proposition 10.9. Let A be an operator algebra. following assertions are equivalent.

(i) Welave (ii) There is

(IN, A))

an

integer d

> 1.

The

d.

<

that, for

constant K such

a

Fix

any n, there is

an

integer N

=

N(n)

and scalar matrices of norm 1

Mn,N (0)

aO G

such that for any

,

x

MN (C))

al E

in

Mn(A)

...

there

i

ad- 1 E

are

MN (C)

diagonal

7

ad G

matrices

MN,n M Dl,..., Dd

in

d

rl I I Di 11

MN (A) with

:5 K I I x I I and satisfying

1

x

CorollarylO.10. then there is

a

=

aoDlalD2... Ddad.

If the C* -algebra

(IN; M)

constant K such that for any

satisfies the Kadison conjecture, there is an integer N N(n)

n

=

and scalar matrices of norm I aO E

such

Mn,N (C)

that, for

any

,

al E

integer

MN (0)

m

and any

ad- 1 E

x

MN (C)

,

ad E

in the unit ball of

MN,n M

Mn(M,,)

there

are

d

diagonal

matrices

Dl,..., Dd

in

MN (M,,,) with

rl 11 Dill

:5 K and satisfying

1

x

The class of

algebras

that of "amenable Banach

=

with

aoDlalD2

...

Ddad-

d(A)(= f(A)) equal to 2 is closely related to e.g. [M]). A von Neumann algebra

algebras" (see

10. The Kadison

178

M C

with a

B(7i) JJPJJ

Similarity

is called amenable

=

Problem

(= injective)

if there is a projection P: B(H) M deep paper [Co], it is known (see [CE]) that amenable, see [H7]) iff for every representation B(H), the von Neumann algebra M, 7r(A)" injective). This motivates the following --+

1. Thanks to Connes'

C*-algebra A

is nuclear

(= *-homomorphism)

(,

A

7r:

=

generated by,7r

is amenable

Definition. A

C*-algebra is called semi-nuclear if for any representation,7r: A von Neumann algebra 7r (A)", the generated alge-

B(H) generating a semi-finite bra ir(A)" is injective. Theorem 10.11.

([P9])

For

C*-algebra A, d(A)

a

< 2

implies that A

is semi-

nuclear. It is

an

problem whether

open

if A is either the reduced

or

A nuclear

general semi-nuclear =: , nuclear. However, C*-algebra of a discrete group G, then

in

the full

A semi-nuclear

#:

-*

G amenable.

The preceding result shows that d(B(H)) > 2, since otherwise B(H) would semi-nuclear, which contradicts the main result in [And]. Hence, we have d(B(H)) > 3. Actually, using the length t(B(H)) instead, we can obtain a very simple proof that d(B(H)) 3, as follows. Proof that (B(H)) < 3: This very direct proof comes from [P12]. Fix n > 1. Let W, and W2 be any two n x n unitary matrices such that be

=

V i,j

JW2(iii)l

JW1(iJ)J

Note that there

are

associated to the

examples

many

of such

(suitably normalized)

n- 1/2

:=

matrices, for. instance the

Fourier transform

on

group of order

there

are

matrix

any commutative

n. Then, for any x in the unit ball of Mn (B (H)) (with dim H diagonal matrices D1, D2, D3 also in the unit ball of Mn (B (H))

=

00)

such

that x

The

proof of this orthogonal ranges

is very so

=

DlWlD2W2D3-

simple. Let Si,

i

1,

=

.

.

.

,

n

be isometries

on

H with

that

V iJ

Si*Si

=

6ii I

-

Then let

D, (i, i) and

=

Si*

and

D3

Sj

moreover

D2 (k, k)

=

n

7 W, (i, k) Sixij Sj* W2 (k, ij

The announced properties

are now

but

a

simple verification.

El

10. The Kadison

Similarity Problem

179

Proposition 10.12. The Kadison similarity problem has a positive answer for all unital C*-algebras A iff there is an integer do such that (A) :! do for any C*-algebra A. By Proposition 10.6, if we have a bound for the length, the Kadison conjecture is immediate. Conversely, if there are C*-algebras Ad with length tending to infinity with d, then clearly the length of their direct sum (in the f,, sense) must be infinite, so this direct sum must fail the Kadison conjecture. El Unfortunately, up to now, the highest known value of t(A) for a C*-algebra is 3, but we conjecture that there are examples of arbitrarily large length. However, in the non-self-adjoint case, we have recently been able to prove the following. Prooh

Theorem 10.13. For any integer d > operator algebra Ad such that

i(Ad)

1, there

=

is

a

(non-self-adjoint)

unital

d.

algebras with arbitrarily large finite length? example with 2 < t(A) < oo is known. However, algebras is proved in [P9] that any proper uniform algebra A must satisfy t(A) > 2. It also unknown whether there are Q-algebras (= quotients of uniform algebras)

Problem. Are there uniform For uniform

with 2 <

t(A)

<

no

it is

A

oo.

Remark.The above similarity problems are also intimately connected to several perturbation problems for von Neumann algebras posed in the 70's by Kadison and Kastler

[KK],

as

follows.

M, N be two von Neumann algebras in B (H). Let d(M, N) denote their Hausdorff distance, and suppose d(M, N) < c (this means that any point of the unit ball of M is at distance < c from some point in the unit ball of N, and vice versa). The Kadison-Kastler problem can then be stated like this: if c > 0 I < 6 (6) is small enough, does there exist a unitary U in B (H) such that U ? N U*MU that with 6(E) --* 0 when e 0, and such A positive solution to the Kadison problem (Problem 0.2) would almost surely allow a breakthrough on this problem too. Indeed, there are some important contributions of E. Christensen ([C5, C6, C7], see also [Ph]) showing that, under be solved some supplementary assumptions, the Kadison-Kastler problem can Kadison the N problem. for von Neumann algebras M, verifying Let

-

=

--

Note in

particular, that

d(M,, (M), M,, (N))

if both

rather

well,

t(M) we

<

oo

find for

and

t(N)

some

< oo,

then

we can

estimate

constant C:

supfd(M,,(M),M,,(N)) Jn>1J
length generated by

to the closed

convex

is

a

hull of the union

U34=lCj

where

10. The Kadison

180

Cj

Assuming

=

Similarity Problem

jX1X2

that this

another Banach

holds, algebra 0.

let

...

u:

Xj

I

B

F-+

Xk E C

Vk

0 be

a

continuous

homomorphism

into

It is then easy to check that d

Ilull

:5 K

E MaXIIIUIB1 Ili IIUjB2 11 jj j=1

where K is

Thus if

a

(depending only

constant

and

B,,3

all unital

u are

Ilull In the

:5

we

on

obtain

d and the size of the subset

(since

all the

C).

norms are now >

1)

dKMaXfIIUIB, Ili IJUIB2 111d.

direction, assuming again B1, B2 and B all unital, let algebra generated by B, and B2, which we assume is alg(Bl, B2) dense in B. Assume that every unital homomorphism u: alg(BI, B2) --+ 3 into an arbitrary unital Banach algebra 13 such that IIUIB1 11 < oo and IIUJB2 11 < C)O is bounded and satisfies actually converse

denote the

Ilull where K and

<

K(maxf IIUIB, 11) IIUJB2 11 J)C'

independent of u and 3. (see [P9, 8]) that B1, B2 generate B with length at most equal to the integral part of a. For example, let IC(A) be as defined before Proposition 10.6. We may clearly identify IC(A) with a subalgebra of the algebra M,,,) (A) c B( 2(H)) of bi-infinite matrices with entries in A, equipped with the operator norm. Taking B IC(A), we may consider the subalgebra B, c B formed of all the diagonal matrices with entries in A and we let B2 k(C)It is then easy to check that f(A) < d implies that B1, B2 generate B with length < 2d + 1. Conversely, if B1, B2 generate B with length < m, then (A) < [ '+]. 2 a >

0

are

Then it follows

==

=

Remark. The

slight discrepancy appearing here comes from the fact that in products appearing in the subset Cd we do not specify that the first term of the product must lie in B2 or B, while in the corresponding definition of f(A) the analogous term must be in B2. This difficulty can be circumvented: one should then consider homomorphisms u: alg(BI, B2) 1 0 such that 11 U1 B2 11 and study the inequality Ilull :5 KIIUIB1 11"- See [PIO] for more variations on this the

---+

=

theme. The

case

kind, for

study of f(A) suggests examining

instance the

pair B,

=

IC(Al), B2

many other

=

)C(A2)

examples of the same A, c A, A2 c A

where

subalgebras. In particular, we may consider the case where A is product of two unital C*-algebras C1, C2: namely we take 1 (9 C2. All these cases are studied A C, (9 1 and A2 C1 Omax C2 with A, in [P10], to which we refer the reader for several illustrative examples and more information. See also [NW] for a discussion of the length of crossed products. are

two closed

the maximal tensor =

=

=

Notes and Remarks

on

Chapter

10

chapter (as in the preceding one) most references are given in the text. Theorem 10.3 is surely well known to specialists. Theorem 10.5 is perhaps new. The results on the length all come from the series of papers [P9, P10, P11, P12] to which we refer the reader. More recent results along this line appear in [P13, P14]. See also Le Merdy's papers [LeM5, LeM6]. In particular, [LeM6] develops a notion of length, specially adapted for the "dual case"' (i.e. for dual operator algebras, normal homomorphisms, and so on). For all the possible variants of length, the main open problems are the construction of examples of arbitrarily large length. We have been able to produce such examples among (non-self-adjoint and non-commutative) operator algebras ([P11j), but the same question remains open for uniform algebras, or Q-algebras, or even for commutative operator algebras (perhaps this last class can be treated by the same general strategy, as in [P11]). Note that it is conceivable that, like the disc algebra (see 9), all proper uniform algebras have infinite length. In [P9], we show that they all have length > 3, and the disc algebra case implies that the ball algebra or the polydisc algebra all have infinite length, but a general argument is lacking. See [Kis2] for a partial result in this direction. In this