3
$\begingroup$

Let $A$ and $B$ be two unital infinite-dimensional simple separable nuclear $C^{\ast}$-algebras and let $C$ be a CAR-algebra. When does $A\otimes C \simeq B\otimes C$, imply $A\simeq B$?

The answer is clear if $A$ and $B$ absorb the CAR-algebra $C$. Although, I don't know what type of algebras can absorb a CAR-algebra.

$\endgroup$

2 Answers 2

3
$\begingroup$

The CAR-algebra is isomorphic to the UHF-algebra $M_{2^{\infty}}$, i.e. the infinite tensor products of $M_2(\mathbb{C})$. This is explained in Example 1.2.6 in the book Classification of Nuclear, Simple $C^*$-algebras" by Rordam. As an infinite UHF-algebra it is what is called strongly self-absorbing. This is a definition of Toms and Winter, which can be found here. In particular, it implies that $M_{2^{\infty}} \otimes M_{2^{\infty}} \cong M_{2^{\infty}}$, i.e. it absorbs itself. Unfortunately, I don't know if there is a classification of $C^*$-algebras that absorb some infinite UHF-algebra. For other strongly self-absorbing algebras, there are such results. For example, a separable, simple and nuclear $C^*$-algebra $A$ is purely infinite if and only if $A \otimes \mathcal{O}_{\infty} \cong A$, where $\mathcal{O}_{\infty}$ is the infinite Cuntz algebra (Theorem 7.2.6 in Rordam's book).

If $A$ satisfies $A \otimes \mathcal{O}_{\infty} \otimes \mathbb{K} \cong A$ and similarly for $B$ there might be a chance that you can reduce your first question to one about $KK$-groups using the classification results by Phillips, in particular Theorem 4.1.3 in the paper A classification theorem for nuclear purely infinite simple $C^*$-algebras.

$\endgroup$
3
  • $\begingroup$ There is certainly no known classification of C*-algebras observing the universal UHF algebra. (And in particular none for any other infinite one!) In fact some recent results in classification show that such a classification in the case of separable, simple, nuclear, unital C*-algebras would lead to $\mathcal{Z}$-stable classification, which is still an open problem. $\endgroup$ Jul 4, 2013 at 20:47
  • $\begingroup$ Thanks, Gabor! What is the reference for this? $\endgroup$ Jul 5, 2013 at 14:02
  • $\begingroup$ I have never bothered to look up the details for this because it seems to be hard to understand. But I guess the introduction of arxiv.org/abs/1307.1342 can give you a good idea, and there you can also find references. $\endgroup$ Jul 5, 2013 at 16:43
3
$\begingroup$

It looks like $A$ and $B$ should be pretty close to being classifiable if one wants to have such an implication. Since the CAR algebra $M_{2^\infty}$ absorb the Jiang-Su algebra $\mathcal{Z}$ tensorically, you always have $A\otimes M_{2^\infty} \cong (A\otimes\mathcal{Z})\otimes M_{2^\infty}$.

In particular, one should assume $\mathcal{Z}$-stability of the C*-algebras in question. But $\mathcal{Z}$-stable, unital, simple, separable, nuclear C*-algebras are conjectured to being classified by the Elliott invariant.

So one maybe has to look for $K$-theoretic obstructions to this implication. By the Künneth formula, one has $$K_*(A\otimes M_{2^\infty}) = K_*(A)\otimes_{\mathbb{Z}}K_0(M_{2^\infty}) = K_*(A)\otimes_{\mathbb{Z}} \mathbb{Z}\left[\frac{1}{2}\right].$$

One could go in two directions: For one, the implication is trivially true when the C*-algebras $A$ and $B$ absorb $M_{2^\infty}$ tensorially. In $K$-theory, this means by the above formula that the $K$-groups of $A$ and $B$ have to be uniquely $2$-divisible, i.e. multiplying by $2$ yields a group automorphism.

On the other hand, one can ask that $A$ and $B$ be as 'far away' from being $M_{2^\infty}$-stable as possible. On the level of $K$-groups, being very far away from uniquely $2$-divisibility could mean that you have no $x\in K_*(A)$ with $2x=[1_A]_{K_*}$ and also no torsion of powers of $2$ is allowed. I don't know if this is actually enough, but it could be a starting point. I doubt this is a class that can overall be described nicely, though.

Some nice immediate subclass is the set of all unital infinite dimensional UHF-algebras arising as unital subalgebras of $$ \mathcal{U}_{>2} = \bigotimes_{p\neq2\; \text{prime}} M_{p^\infty} .$$

I hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.