Unitarizable representations and amenable operator algebras Yemon - - PowerPoint PPT Presentation

Unitarizable representations and amenable

• perator algebras

Yemon Choi Lancaster University QOP Network Meeting Lancaster University, 25th September 2014

0 / 21

Today’s menu

1

Unitarizable representations

2

Amenable operator algebras?

3

Non-unitarizable representations

4

Unitarizable representations (reprise)

5

Open questions

0 / 21

Today’s menu

1

Unitarizable representations

2

Amenable operator algebras?

3

Non-unitarizable representations

4

Unitarizable representations (reprise)

5

Open questions

0 / 21

Unitarizable representations and fixed points What do I mean by “a representation (θ, Γ, A)”? In this talk: a discrete group Γ, a unital C∗-algebra A, and a HM θ : Γ → Ainv. We say the representation is bounded if supx∈Γ θ(x) < ∞.

1 / 21

Unitarizable representations and fixed points What do I mean by “a representation (θ, Γ, A)”? In this talk: a discrete group Γ, a unital C∗-algebra A, and a HM θ : Γ → Ainv. We say the representation is bounded if supx∈Γ θ(x) < ∞. A (bounded) representation (θ, Γ, A) is unitarizable, or similar to a ∗-representation, if there exists s ∈ Ainv such that sθ(x)s−1 ∈ U(A) for all x ∈ Γ. We say that s is a similarity element for θ.

1 / 21

Let A+

inv = Ainv ∩ A+. Then Γ acts on A+ inv, as follows;

θ+(x) : h → θ(x)hθ(x)∗ . Exercise (θ, Γ, A) is unitarizable if and only if θ+ : Γ A+

inv has a fixed point.

2 / 21

Let A+

inv = Ainv ∩ A+. Then Γ acts on A+ inv, as follows;

θ+(x) : h → θ(x)hθ(x)∗ . Exercise (θ, Γ, A) is unitarizable if and only if θ+ : Γ A+

inv has a fixed point.

Exercise Suppose Γ is finite and let θ : Γ → Ainv be a HM. Show that the action θ+ : Γ A+

inv has a fixed point. (Hint: average over orbits.)

Thus every finite subgroup of Ainv is similar to a subgroup of U(A).

2 / 21

Example 1. Let ε > 0 and consider x = 1 −1

• ,

y = 1 ε −1

• .

These give a pair of representations θx, θy : Z/2Z → (M2)inv. θ+

x and θ+ y act on (M2)+ inv, but have no common fixed point. Hence

there is no s ∈ (M2)inv which simultaneously unitarizes θx and θy.

3 / 21

A theorem of Day and Dixmier Theorem (Day, 1950; Dixmier, 1950) Let Γ be an amenable discrete group and M a von Neumann algebra. Then every bounded representation (θ, Γ, M) is unitarizable. The case Γ = Z, M = B(H) was proved by Sz.-Nagy (1947) and contains the essential ideas for the general case.

4 / 21

A theorem of Day and Dixmier Theorem (Day, 1950; Dixmier, 1950) Let Γ be an amenable discrete group and M a von Neumann algebra. Then every bounded representation (θ, Γ, M) is unitarizable. The case Γ = Z, M = B(H) was proved by Sz.-Nagy (1947) and contains the essential ideas for the general case. Theorem (Pisier, 2007) If Γ is a discrete non-amenable group, then there is some von Neumann algebra M and some bounded, non-unitarizable rep (θ, Γ, M). Unknown if we can always take M = B(H)!

4 / 21

Today’s menu

1

Unitarizable representations

2

Amenable operator algebras?

3

Non-unitarizable representations

4

Unitarizable representations (reprise)

5

Open questions

4 / 21

Amenable Banach algebras Quick definition: a Banach algebra A is amenable if it has a bounded approximate diagonal, i.e. a bounded net (mα) ∈ A ⊗ A satisfying a · mα − mα · a → 0 and aπ(mα) → a for each a ∈ A. Example 2. [Johnson, 1972] If Γ is a discrete amenable group, then ℓ1(Γ) is amenable. In particular, ℓ1(Γ) is amenable whenever Γ is abelian.

5 / 21

Amenable Banach algebras Quick definition: a Banach algebra A is amenable if it has a bounded approximate diagonal, i.e. a bounded net (mα) ∈ A ⊗ A satisfying a · mα − mα · a → 0 and aπ(mα) → a for each a ∈ A. Example 2. [Johnson, 1972] If Γ is a discrete amenable group, then ℓ1(Γ) is amenable. In particular, ℓ1(Γ) is amenable whenever Γ is abelian. Some hereditary properties if A is amenable and φ : A → B is a HM with dense range, then B is amenable; if A is a Banach algebra, J is a closed ideal in A, and J and A/J are both amenable, then so is A.

5 / 21

Example 3. [Johnson, 1972] C(X) and K(H) are amenable. Both examples are closures of HM’ic images of ℓ1(Γ), for some choice of amenable Γ.

6 / 21

Example 3. [Johnson, 1972] C(X) and K(H) are amenable. Both examples are closures of HM’ic images of ℓ1(Γ), for some choice of amenable Γ. Remark Johnson went on to show (1972) that every GCR (i.e. Type I) C∗-algebra is amenable. (In fact, strongly amenable.) Also, the algebras On, 2 ≤ n ≤ ∞, are amenable (but not strongly amenable). [Rosenberg, 1977] None of these proofs need the word “nuclear”

6 / 21

Amenable operator algebras? (“Operator algebra” = norm closed subalg of B(H).) We might wish to study amenable operator algebras. But how can we find examples?

7 / 21

Amenable operator algebras? (“Operator algebra” = norm closed subalg of B(H).) We might wish to study amenable operator algebras. But how can we find examples? Question. Let A be an amenable operator algebra. Must A be isomorphic to (the underlying Banach algebra of) some C∗-algebra? In the finite-dimensional setting, the answer is YES, by Wedderburn’s

• theorem. This was pushed further by Gifford in his PhD thesis.

7 / 21

Amenable operator algebras? (“Operator algebra” = norm closed subalg of B(H).) We might wish to study amenable operator algebras. But how can we find examples? Question. Let A be an amenable operator algebra. Must A be isomorphic to (the underlying Banach algebra of) some C∗-algebra? In the finite-dimensional setting, the answer is YES, by Wedderburn’s

• theorem. This was pushed further by Gifford in his PhD thesis.

Theorem (Gifford, 1997/2006) Amenable, closed subalgebras of K(H) are isomorphic to C∗-algebras. In full generality, this question resisted attempts over many years. . .

7 / 21

A question and idea of Ozawa Let Q(H) := B(H)/K(H) denote the Calkin algebra and q : B(H) → Q(H) the quotient HM. Question. Is there a bounded, non-unitarizable rep (θ, Z, Q(H))? What if we replace Z by some other discrete abelian group?

8 / 21

A question and idea of Ozawa Let Q(H) := B(H)/K(H) denote the Calkin algebra and q : B(H) → Q(H) the quotient HM. Question. Is there a bounded, non-unitarizable rep (θ, Z, Q(H))? What if we replace Z by some other discrete abelian group? The point of Ozawa’s question: suppose Γ is abelian; then each bounded rep (θ, Γ, Q(H)) gives an amenable A ⊂ B(H); if A is isomorphic to a C∗-algebra then θ is unitarizable.

8 / 21

A question and idea of Ozawa Let Q(H) := B(H)/K(H) denote the Calkin algebra and q : B(H) → Q(H) the quotient HM. Question. Is there a bounded, non-unitarizable rep (θ, Z, Q(H))? What if we replace Z by some other discrete abelian group? The point of Ozawa’s question: suppose Γ is abelian; then each bounded rep (θ, Γ, Q(H)) gives an amenable A ⊂ B(H); if A is isomorphic to a C∗-algebra then θ is unitarizable. So, a bounded non-unitarizable (θ, Γ, Q(H)) gives rise to an amenable

• perator algebra not isomorphic to any C∗-algebra.

8 / 21

Details Given θ : Γ → Q(H) define B = lin{θ(x) : x ∈ Γ}. B is amenable. Let A = q−1(B). There is a short exact sequence 0 → K(H) → A

q

− → B → 0 By hereditary properties, A is an amenable operator algebra.

9 / 21

Details Given θ : Γ → Q(H) define B = lin{θ(x) : x ∈ Γ}. B is amenable. Let A = q−1(B). There is a short exact sequence 0 → K(H) → A

q

− → B → 0 By hereditary properties, A is an amenable operator algebra. Now suppose A is also isomorphic to a C∗-algebra. Then there exists R ∈ B(H)inv such that RAR−1 is a self-adjoint subalgebra of B(H). Put s := q(R). Then sBs−1 is a commutative and self-adjoint subalgebra of Q(H). Observe: if x ∈ Γ, then sθ(x)s−1 is normal with spectrum contained in T, hence is unitary. So s unitarizes θ.

9 / 21

Today’s menu

1

Unitarizable representations

2

Amenable operator algebras?

3

Non-unitarizable representations

4

Unitarizable representations (reprise)

5

Open questions

9 / 21

A construction of Farah and Ozawa Theorem (see arXiv:1309.2415v1) There is a set T of bounded HMs Z⊕c → Q(ℓ2), with |T| = 2c, such that T is parametrized by certain “1-cocycles” Let θ ∈ T; then (θ, Z⊕c, Q(ℓ2)) is unitarizable iff θ corresponds to an “inner” cocycle.

10 / 21

A construction of Farah and Ozawa Theorem (see arXiv:1309.2415v1) There is a set T of bounded HMs Z⊕c → Q(ℓ2), with |T| = 2c, such that T is parametrized by certain “1-cocycles” Let θ ∈ T; then (θ, Z⊕c, Q(ℓ2)) is unitarizable iff θ corresponds to an “inner” cocycle. But |{inner cocycles}| ≤ Q(ℓ2) = c < 2c = |T|. Therefore there is a bounded, non-unitarizable (θ, Z⊕c, Q(ℓ2))

10 / 21

Hence, by Ozawa’s ingenious observation, our infamous question has the answer NO. Corollary (Farah, Ozawa, ibid.) There is an amenable closed A ⊂ B(ℓ2) not isomorphic to any C∗-algebra. A surprising feature is that A is “locally very nice”:

11 / 21

Hence, by Ozawa’s ingenious observation, our infamous question has the answer NO. Corollary (Farah, Ozawa, ibid.) There is an amenable closed A ⊂ B(ℓ2) not isomorphic to any C∗-algebra. A surprising feature is that A is “locally very nice”: for every countable subset X ⊂ c, the rep (θ, Z⊕X, Q(ℓ2)) is unitarizable. Thus A = lim − →X AX, where each AX is separable, amenable and similar to a C∗-algebra. Moreover, similarity elements sX exist with supX sX

• s−1

X

• < ∞.

11 / 21

Sharpening the construction By making various technical modifications, the method of Farah–Ozawa can be refined to give an even more striking example.

12 / 21

Sharpening the construction By making various technical modifications, the method of Farah–Ozawa can be refined to give an even more striking example. Theorem (C.–Farah–Ozawa, 2014) Let C = ℓ∞/c0. There is a bounded HM θ : (Z/2Z)⊕ℵ1 → C ⊗ M2 which is not unitarizable (inside C ⊗ M2). This gives rise to A ⊂ ℓ∞ ⊗ M2 which has density character ℵ1, and is amenable, but not isomorphic to any C∗-algebra.

12 / 21

Sharpening the construction By making various technical modifications, the method of Farah–Ozawa can be refined to give an even more striking example. Theorem (C.–Farah–Ozawa, 2014) Let C = ℓ∞/c0. There is a bounded HM θ : (Z/2Z)⊕ℵ1 → C ⊗ M2 which is not unitarizable (inside C ⊗ M2). This gives rise to A ⊂ ℓ∞ ⊗ M2 which has density character ℵ1, and is amenable, but not isomorphic to any C∗-algebra. Remark Any amenable closed subalgebra of ℓ∞ is automatically self-adjoint (corollary of Sheinberg, 1977), hence isomorphic to some C0(X).

12 / 21

Some details of the construction Let Γ = (Z/2Z)⊕ℵ1. We will construct two commuting, bounded representations θx, θy : Γ → C ⊗ M2 which cannot be simultaneously unitarized. Then θx × θy : Γ × Γ → (ℓ∞/c0) ⊗ M2 is the desired bounded but non-unitarizable representation. All the real work takes place inside C = ℓ∞/c0 = C(βN \ N).

13 / 21

We can find F, G ⊂ 2N, with |F| = |G| = ℵ1, such that (q(1J))J∈F ∪ (q(1K))K∈G is a family of non-zero, pairwise-orthogonal projections in ℓ∞/c0. We can also arrange for the following condition to hold. “Gap condition” For each partition N = X ⊔ Y, either there exists J ∈ F such that X ∩ J is infinite, or there exists K ∈ G such that Y ∩ K is infinite.

14 / 21

Pick two involutions x, y ∈ M2 such that θ+

x , θ+ y have no common fixed

• point. (We saw easy examples earlier!)

For each J ∈ F and K ∈ G, define involutions in ℓ∞ ⊗ M2 by xJ = 1J ⊗ x + 1N\J ⊗ I2 and yK = 1K ⊗ y + 1N\K ⊗ I2. Define Θx : Γ → (ℓ∞/c0) ⊗ M2 by Θx(eJ) = (q ⊗ id)(xJ), and define Θy

• similarly. These representations of Γ are bounded, and their ranges

commute, as required. We get actions Θ+

x , Θ+ y : Γ ((ℓ∞/c0) ⊗ M2)+ inv.

It suffices to show these actions have no common fixed point.

15 / 21

Suppose Θ+

x and Θ+ y have a common fixed point, say q(s) for some

positive invertible s = (sn) ∈ ℓ∞ ⊗ M2. A little work shows that limn∈J dist(sn, Fix(θ+

x )) = 0, for all J ∈ F;

limn∈K dist(sn, Fix(θ+

y )) = 0, for all K ∈ G.

16 / 21

Suppose Θ+

x and Θ+ y have a common fixed point, say q(s) for some

positive invertible s = (sn) ∈ ℓ∞ ⊗ M2. A little work shows that limn∈J dist(sn, Fix(θ+

x )) = 0, for all J ∈ F;

limn∈K dist(sn, Fix(θ+

y )) = 0, for all K ∈ G.

But our choice of x and y turns out to force inf

n dist(sn, Fix(θ+ x )) + dist(sn, Fix(θ+ y )) = δ > 0.

From these we get X, Y ⊆ N, with X ∪ Y = N, such that |X ∩ J| < ∞ for all J ∈ F , |Y ∩ K| < ∞ for all K ∈ G. This contradicts the gap condition. So no such s exists.

• 16 / 21

Today’s menu

1

Unitarizable representations

2

Amenable operator algebras?

3

Non-unitarizable representations

4

Unitarizable representations (reprise)

5

Open questions

16 / 21

A limit to our methods We found non-unitarizable reps (θ, Γ, Q(H)), but required uncountable Γ. Recall: Ozawa’s original question was might there be a non-unitarizable (θ, Z, Q(H))?

17 / 21

A limit to our methods We found non-unitarizable reps (θ, Γ, Q(H)), but required uncountable Γ. Recall: Ozawa’s original question was might there be a non-unitarizable (θ, Z, Q(H))? Answer: NO. Theorem (C.–Farah–Ozawa, ibid.) Let Γ be a countable amenable group. Then every bounded representation (θ, Γ, Q(H)) is unitarizable. The same is true if we replace the Calkin algebra by certain other algebras, e.g.

n Mn/ Mn or (ℓ∞/c0) ⊗ Mn or ultraproducts of a

sequence of C∗-algebras.

17 / 21

Some comments on the proof We need to find a fixed point of the action θ+ : Γ A+

inv.

θ+(x)(h) := θ(x)hθ(x)∗ (x ∈ Γ, h ∈ A+

inv).

A standard theme: when looking for a fixed point of a (semi)group action, try to take an “average over an orbit”.

18 / 21

Some comments on the proof We need to find a fixed point of the action θ+ : Γ A+

inv.

θ+(x)(h) := θ(x)hθ(x)∗ (x ∈ Γ, h ∈ A+

inv).

A standard theme: when looking for a fixed point of a (semi)group action, try to take an “average over an orbit”. So now suppose Γ has a Følner sequence (Fn). Put hn = 1 |Fn|

• y∈Fn

θ(y)θ(y)∗ . Then for any x ∈ Γ, θ(x)hnθ(x)∗ − hn ≤ |Fn|−1|xFn△Fn| θ2 → 0 , so (hn) is an “asymptotically invariant” sequence in A+

inv.

18 / 21

The key point If A has a certain “countable saturation property”, tools from the metric model theory of C∗-algebras allow us to construct the desired h from the sequence (hn). (These tools are an axiomatic version of ideas used by G. K. Pedersen to study derivations from separable C∗-algebras into corona algebras.)

19 / 21

Today’s menu

1

Unitarizable representations

2

Amenable operator algebras?

3

Non-unitarizable representations

4

Unitarizable representations (reprise)

5

Open questions

19 / 21

Commutative amenable operator algebras? Theorem (C., 2013) Let A be a closed, commutative subalgebra of a finite von Neumann

• algebra. If A is amenable, then it is isomorphic to C0(X) for some X.

Recently this was significantly improved:

20 / 21

Commutative amenable operator algebras? Theorem (C., 2013) Let A be a closed, commutative subalgebra of a finite von Neumann

• algebra. If A is amenable, then it is isomorphic to C0(X) for some X.

Recently this was significantly improved: Theorem (Marcoux–Popov, 2013 preprint) Let A be a closed, commutative subalgebra of B(H). If A is amenable, then it is isomorphic to C0(X) for some X. In both papers, one uses amenability to show that the Gelfand transform A → C0(ΦA) is bounded below. (From there the rest is a standard application of Sheinberg’s theorem.)

20 / 21

Questions Question. Let A be a separable closed subalgebra of B(H). If A is amenable, must it be isomorphic to a C∗-algebra? Question. What about amenable subalgebras of e.g. the CAR algebra? The final question was suggested to me by S. A. White. Question. Let A be a weak*-closed, “Connes-amenable” subalgebra of B(H). Must it be isomorphic to a von Neumann algebra?

21 / 21