# Tensoring with a CAR-algebra

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.

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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.

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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. –  Gabor Szabo Jul 4 '13 at 20:47
Thanks, Gabor! What is the reference for this? –  Ulrich Pennig Jul 5 '13 at 14:02
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. –  Gabor Szabo Jul 5 '13 at 16:43

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.

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