# C$^*$-algebras isomorphic after tensoring with $M_n(\mathbb C)$

In 1977, Joan Plastiras gave a striking example of two non $*$-isomorphic C$^*$-algebras $\mathcal A$ and $\mathcal B$ such that $$\mathcal A \otimes M_2(\mathbb C) \simeq \mathcal B\otimes M_2(\mathbb C)$$ (http://www.ams.org/journals/proc/1977-066-02/S0002-9939-1977-0461158-9/S0002-9939-1977-0461158-9.pdf).

My question is this: can you find two non $*$-isomorphic C$^*$-algebras $\mathcal A,\mathcal B$ such that $$\mathcal A\otimes M_n(\mathbb C) \simeq \mathcal B\otimes M_n(\mathbb C), \ n=2,3?$$ Note that Plastiras' example does not satisfy this condition. Furthermore, can you find non $*$-isomorphic $\mathcal A, \mathcal B$ such that $$\mathcal A\otimes M_n(\mathbb C) \simeq \mathcal B\otimes M_n(\mathbb C), \ \forall n\geq 2$$ or does this condition imply that $\mathcal A\simeq \mathcal B$?

Such examples do exist. In the paper "Stability of C*-algebras is not a stable property, Doc. Math. J. DMV , 2, (1997), 375-386.", Rordam gives examples of simple C*-algebras $A$ such that $M_2(A)$ is a stable C*-algebra but $A$ is not. Now take the pair $A$ and $B=M_2(A)$. These C*-algebras are not isomorphic, since $A$ is not stable, but $B$ is. On the other hand, $M_n(A)$ is isomorphic to $M_n(B)$ which is isomorphic to $A\otimes \mathcal K$ for all $n\geq 2$. (Once $M_n(A)$ is stable for some $n$, it remains stable for all larger $n$. This is proven in the same paper.)

Part of the hard work in Rordam's paper is to find an example as above with $A$ simple. At the heart of Rordam's construction is the following non-simple example: There exists a countable generated Hilbert C*-module $H$ over $C(X)$, with $X=\prod_{i=1}^\infty S^2$ (an infinite product of 2-spheres) such that $H$ is not isomorphic to $\ell^2(C(X))$ but $H\oplus H$ is. Then one can choose $A=K(H)$ and $B=K(H\oplus H)$. $A$ is not stable since this is equivalent to $H\cong \ell^2(C(X))$ but $M_2(A)\cong B$ is stable.

This is only a comment: It seems to me that it should be possible to find (hermitian) complex vector bundles $\eta_1$ and $\eta_2$ over some space $X$ such that $\eta_1^{n\oplus}\cong \eta_2^{n\oplus}$ for all $n\geq 2$ but $\eta_1\ncong \eta_2$. Then $A=\mathrm{End}(\eta_1)$ and $B=\mathrm{End}(\eta_2)$ are candidate examples. From $\eta_1^{n\oplus}\cong \eta_2^{n\oplus}$ for all $n\geq 2$ we can already deduce that $M_n(A)\cong M_n(B)$ for all $n\geq 2$. If we choose them with some care (they can't be line bundles) then we should also have that $A\ncong B$.

• This is a fantastic example. It raises the question for which subsequences $\{n_i\}_{i\geq 2}$ of $\mathbb N$ do there exist non-isomorphic C$^*$-algebras $A$ and $B$ such that $M_n(A) \simeq M_n(B)$ if and only if $n\in\{n_i\}_{i\geq 2}$. One obvious condition is that if $n\in\{n_i\}_{i\geq 2}$ then so are all multiples of $n$. – Chris Ramsey Aug 8 '15 at 17:37

This is not an answer, but too long for a comment and just an idea for an approach.

Kirchberg algebras (unital, purely infinite, separable, nuclear) in the UCT class are classified by a triple $(G_0, g_0, G_1)$ where $G_0$ is the $K_0$ group, $G_1$ is the $K_1$ group and $g_0$ is the image of the identity in $G_0.$ Every pair of countable abelian groups and $g_0\in G_0$ appears as the invariant of an actual Kirchberg algebra in the UCT class. (See for example Rordam's book Classification of nuclear C*-algebras)

Now let $A$ be a Kirchberg algebra. Then $M_n\otimes A$ of course has the same $K$-groups but the distinguished element gets multiplied by $n.$

If one works in this class of algebras, your question now becomes:

Does there exist a countable abelian group $G$ and elements $x,y\in G$ such that $\phi(x)\neq y$ for every automorphism of $G$ but for every $n\geq 2$ there is an automorphism $\phi_n$ such that $\phi_n(nx)=ny$?

I don't know of any countable abelian groups with this property (I'm sure no finitely generated one has this property for example) so this approach may be useless.

• Thanks, this is a great place to look for examples. I've added the gr.group-theory tag to hopefully get someone who can answer this question. – Chris Ramsey Aug 5 '15 at 15:34