Existence of conditional expectations map onto subalgebras

Question

Let $B\subset A$ be an inclusion of $C^*$ - algebras. I am having confusions on the existence of a conditional expectation $E: A \to B$. I could see that in general an inclusion need not have any conditional expectation. I couldnt get an example towards this.

1. But if we consider the inclusion of tracial von Neumann algebras, there exists a unique trace preserving faithful normal conditional expectation.

2. Is the same true for inclusion of finite von Neumann algebras ? I knw there exists a unique centre valued trace but can we find a scalar valued trace? Because the construction of the conditional expectation for a tracial von Neumann algebras are via the GNS construction. Is the same true for finite von Neumann algebras also?

3. For a $C^*$ algebra $A$, with $B$ is a Cartan subalgebra, then also there exists a faithful conditional expectation.

Is there any unified theorem which tells the existence of a conditional expectation(faithful) for a $C^*$ - algebra inclusion?

Answer 1

As you mentioned, there are many concrete examples of $C^*$-algebra embeddings that admit a conditional expectation.

In general, for a given embedding $A\subset B$ this is a difficult question, it depends a lot on the type of embedding.

But there are situations where we always find conditional expectations: this is the case, for instance, if $A$ is an injective $C^*$-algebra, as for example $A=B(H)$ (the $C^*$-algebra of all bounded operators); this is due to Arverson's extension theorem. There are many other examples of injective $C^*$-algebras: for instance, the bidual von Neumann algebra of a nuclear $C^*$-algebra is always injective, so for instance, $\ell^\infty(X)$ is injective for any set $X$.

A von Neumann algebra $M\subset B(H)$ is injective if and only if there is a conditional expectation $B(H)\to M$. From this one can give several examples of embeddings that do not admit a conditional expectation: just take any non-injective von Neumann algebra $M$. For instance, if $G$ is a discrete group, its von Neumann algebra $W^*_r(G)$, generated by the image left regular representation $\lambda\colon G\to B(\ell^2(G))$, is injective if and only if $G$ is amenable. So, if $G$ is not amenable (e.g. a non-commutative free group), then the embedding $W^*_r(G)\subset B(\ell^2(G))$ does not admit a conditional expectation.

More generally, we can also ask about existence of weak conditional expectations: for an embedding $A\subset B$, this means that the bidual embedding $A''\subset B''$ has a (normal) conditional expectation.

If for fixed $A$, all the embeddings $A\subset B$ admit a weak conditional expectation, then $A$ is said to have the WEP (Weak Expectation Property).

All nuclear $C^*$-algebras have the WEP. It turns out that a $C^*$-algebra $A$ has the WEP if and only if the algebraic tensor product $A\odot C^*(\mathbb{F})$ has a unique $C^*$-norm, see Brown-Ozawa's book ``$C^*$-algebras and finite dimensional approximations'', for instance. Here $\mathbb{F}$ denotes a non-commutative free group.

Also, it follows that the celebrated Connes embedding conjecture is equivalent to the fact that $C^*(\mathbb{F})$ has the WEP. And this conjecture has been announced to be false recently.

To have a concrete example of a non-WEP $C^*$-algebra, one can also look at the reduced $C^*$-algebra $C^*_r(\mathbb{F_2})$ of the free group $\mathbb{F}_2$ on two generators (see again the book of Brown-Ozawa, exercise 13.2.2). In fact, this exercise shows that every exact $C^*$-algebra that has the WEP must be nuclear.