Representations of simple C$^*$-algebras

Question

I am reading Further Representations of the Canonical Commutation Relations by Florig and Summers, and am a bit stumped by footnote 4 on page 5:

Actually, I will copy the relevant text because it disappears off the bottom of the page (or at least it does when I view it on my laptop).

The fact that the CCR-algebra is simple can be seen as the correct generalization of the Stone–von Neumann uniqueness theorem to the case of infinitely many degrees of freedom. Indeed, since $\mathcal{A}(H, \sigma)$ is simple, all of its representations are isomorphic. When $H$ is finite-dimensional, this isomorphism is unitarily implementable, entailing the result in [20].

The object $\mathcal{A}(H,\sigma)$ which is mentioned is a C$^*$-algebra.

I have found only one mention of isomorphic representations for C$^*$-algebras, which is Definition 2 in the following document:

http://www.math.ru.nl/~tcrisp/teaching/2017-Cstar-reps/notes/2017-09-25-Cstar-reps-notes.pdf

And my problem is that I can’t make any progress with proving the implication hinted at by the footnote, that if $A$ is a C$^*$-algebra, then

$$\text{$A$ simple} \implies \text{all of its representations are isomorphic} \,.$$

Well, to me this is what the footnote is hinting at, but it could well be the case that the author just means that this is the case for $\mathcal{A}(H,\sigma)$, so I guess I should also be thinking about possible counter examples for the general case. However I’d be the first to admit that my knowledge of representation theory isn’t particularly great yet…

As usual, I would appreciate any hints/suggestions etc. I’m really quite interested by this footnote, as it seems to say that Slawny’s Theorem implies the Stone–Von Neumann Theorem which is something I’ve not seen anywhere else.

Answer 1

It is not true that simplicity implies isomorphism of representations (it may be true for your algebra, that I don't know what it is).

I cannot immediately think of an elementary example, but here are a couple.

• Take the free groups $\mathbb F_2$ and $\mathbb F_3$ and their reduced C$^*$-algebras. The canonical homomorphism $\mathbb F_3\to\mathbb F_2$ lifts to a $*$-epimomorphism $C_r^*(\mathbb F_3)\to C_r^*(\mathbb F_2)$. Compare with the identity representation $C_r^*(\mathbb F_3)\to C_r^*(\mathbb F_3)$ and the fact that $C_r^*(\mathbb F_3)$ and $C_r^*(\mathbb F_2)$ are not isomorphic (this is due to Pimsner-Voiculescu if I'm not wrong), and that both are simple.

• Take the Cuntz algebras $\mathbb O_2$ and $\mathbb O_3$. It is well-known that they are not isomorphic, and $\mathbb O_3$ embeds in $\mathbb O_2$ (all exact algebras embed in it). So again we get two representations of $\mathbb O_3$ (one into itself, one into $\mathbb O_2$) that are not isomorphic. And Cuntz algebras are simple.

Answer 2

[Edited after helpful comments by Jendrik]

Let $\mathcal{A}$ be a $C^*$-algebra, let $\pi_i: \mathcal{A} \to B(H_i)$ be representations.

Say that $\pi_1$ and $\pi_2$ are isomorphic representations if the images $\pi_1(\mathcal{A})$ and $\pi_2(\mathcal{A})$ are isomorphic as $C^*$-algebras. The representations are unitarily equivalent if there exists a unitary $U: H_1 \to H_2$ such that $\pi_2(x) = U \pi_1(x) U^\dagger$ for all $x\in\mathcal{A}$.

The paper is concerned with $\mathcal{A}(H,\sigma)$, where $(H,\sigma)$ is a symplectic vector space. One can define $\mathcal{A}(H,\sigma)$ as the universal $C^*$-algebra with generators $\{W(f)\,|\, f\in H\}$ subject to the "canonical commutation relations in Weyl form" $W(f)W(g)=e^{-i\sigma(f,g)} W(f+g), W(f)^*=W(-f)$.

True statements:

• For general $C^*$-algebras: unitary equivalence implies isomorphism of representations.

This is obvious.

• $\mathcal{A}$ is simple if and only if all representations are equivalent.

This is elementary. If $\mathcal{A}$ is simple, it does not contain any non-trivial closed ideals. This means that if $\pi$ is a a non-trivial homomorphism from $\mathcal{A}$ into another $C^*$-algebra $\mathcal{B}$, then $\ker\pi=\{0\}$. Thus $\pi$ is faithulful and therefore $\pi(\mathcal{A})$ is isomorphic to $\mathcal{A}$ as a $C^*$ algebra. Conversely, assume that $\mathcal{A}$ is not simple. Let $I\subset\mathcal{A}$ be a non-trivial closed ideal, then the quotient map $\pi: \mathcal{A} \to \mathcal{A}/I$ has kernel $I$, so $\pi(\mathcal{A})$ is not isomorphic to $\mathcal{A}$.

• $\mathcal{A}(H,\sigma)$ is simple.

This is sometimes called Slwany's theorem (e.g. in Petz, A invitation to the algebra of the CCRs). C.f. Petz' book or Theorem 5.2.8 of Bratelli-Robinson for a proof.

• If $\dim H < \infty$, then all representations of $\mathcal{A}(H,\sigma)$ are unitarily equivalent.

This is the Stone-von Neumann Theorem.

• If $H$ is infinite-dimensional, there are unitarily inequivalent (but equivalent, in the above sense) representations.

This is well-known in quantum field theory. See e.g. the presentation of Shale's theorem in Chapter 9 of the book by Petz.

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[Version before edit. I guess I should delete this, but not sure what the etiquette is on Stackexchange.]

I don't think anything deep is happening.

Let $\mathcal{A}$ be a simple $C^*$-algebra, $\mathcal{B}$ another $C^*$-algebra and $\pi:\mathcal{A}\to\mathcal{B}$ a non-trivial homomorphism. Because $\ker\pi$ is a closed ideal, simplicity of $\mathcal{A}$ implies that $\ker\pi=\{0\}$. Thus $\pi(\mathcal{A})$ is isomorphic to $\mathcal{A}$. In particular, any two representations are isomorphic.

So what about the two sketches in Martin's answer suggesting that the claim is wrong? I'm not sure I understand what he had in mind. However, here's why I can't see how they contradict what I'm saying above:

• I don't see how to construct the map $C^*_r(\mathbb{F}_3)\to C^*_r(\mathbb{F}_2)$. If one defines $\phi:\mathbb{F}_3\to\mathbb{F}_2$ be just setting the third generator and its inverse to $\mathbb{1}$, then the resulting map $\ell_2(\mathbb{F}_3) \to \ell_2(\mathbb{F}_2)$ does not seem bounded.
• Unless it is claimed that the embedding of $\mathbb{O}_3\to\mathbb{O}_2$ is onto, I don't see how $\mathbb{O}_2$ and $\mathbb{O}_3$ not being isomorphic has any bearing on the problem.