From the negative answer to this question we know that C$^*$-algebras that are isomorphic after tensoring with $M_n$ for all $n\geq 2$ need not be isomorphic. So what happens when we strengthen this?

In the following, if $\mathcal A\subset B(\mathcal H)$ and $\mathcal B\subset B(\mathcal K)$ then the minimum tensor product $\mathcal A\otimes_\textrm{min} \mathcal B$ is the closure of the algebraic tensor product $\mathcal A\odot \mathcal B$ in the $B(\mathcal H\otimes \mathcal K)$ norm.

Question 1: Let $\mathcal A,\mathcal B$ be C$^*$-algebras such that for all C$^*$-algebras $\mathcal C$ non-isomorphic to $\mathbb C$
$$\mathcal A\otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C $$ Does this imply $\mathcal A\simeq \mathcal B$?

Secondly, can one accomplish this with a single C$^*$-algebra.

Question 2: Does there exist a C$^*$-algebra $\mathcal C$ not isomorphic to $\mathbb C$ such that for all C$^*$-algebras $A,B$ we have $$ \mathcal A \otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C \ \Rightarrow \ \mathcal A \simeq\mathcal B\ ? $$

One can also ask these questions with the max tensor product.

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    $\begingroup$ That "set" is in fact a proper class because there are $C^\ast$-algebras of arbitrarily large cardinality. You could introduce a bound on the cardinality to work around that. $\endgroup$ Aug 22, 2015 at 20:39
  • $\begingroup$ @JohannesHahn Of course you are right, I'll fix that. $\endgroup$ Aug 22, 2015 at 22:57

2 Answers 2


The answer to question 1 is `yes': Let $A$ and $B$ be any $C^*$-algebras. Let $N$ be a simple $C^*$-algebra of such high cardinality that it does not embed into either $A$ or $B$. Then take $C:=N\oplus\mathbb{C}$.

We have $$ A\otimes C \cong (A\otimes N) \oplus A, \quad B\otimes C \cong (B\otimes N) \oplus B. $$ Let $\varphi\colon (A\otimes N) \oplus A\to (B\otimes N) \oplus B$ be an isomorphism. The choice of $N$ ensures that no quotient of $A\otimes N$ embeds into $B$. Thus, $\varphi$ maps the summand $A\otimes N$ to $B\otimes N$. Applying the same argument to the inverse of $\varphi$, we see that $\varphi$ is given by an isomorphism $A\otimes N\cong B\otimes N$ and an isomorphism $A\cong B$.

This also answers question 2 if one puts some restrictions on the cardinality (or better: density character) of the $C^*$-algebras. For instance, for separable $C^*$-algebras, one could use the test-algebra $C:=\mathcal{R}\oplus\mathbb{C}$, where $\mathcal{R}$ is the hyperfinite II-1 factor.


I think the solution could be $\mathcal{C}= \mathbb{C} \oplus \mathbb{C}$. Then you should show $A \oplus A \cong B \oplus B$ ($=B \otimes \mathcal{C}$) implies $A \cong B$.

Let $\pi:A \oplus A \rightarrow B \otimes B$ be the isomorphism. Consider the 4 projections $p_1:= \pi(1 \oplus 0)$, $p_2:=\pi(0 \oplus 1)$, $q_1 := 1 \oplus 0$, $q_2 := 0 \oplus 1$ in the center of (the multiplier algebra of) $B \oplus B$.

Note that $\pi(A \oplus 0) = p_1 \cdot (B \oplus B)$. By refining those 4 projections you see that $$B \oplus B = (W \oplus X) \oplus (Y \oplus Z)$$ where $A \cong X\oplus Y \cong W \oplus Z$, the two copies of $A$.

Hence $W \oplus X = Y \oplus Z$. Even if the sitiuation is as in the beginning, by refining you could proceed further.

It appears to me that at the end $A \cong B$, as the double of $A$ is the double of $B$ means that $A$ has the same components as $B$ under the infinite refinements.


Even it is wrong, let me still give an essentially same but more transparent argument:

Write $A$ as an (infinite) direct sum $A = \bigoplus_i A_i$, where each $A_i$ has no further direct summands (=inessential ideals). Similarly $B=\bigoplus_j B_j$. Then it is clear that $A$ and $B$ share the same components (=summands).

Just as I said above.


It is clear that it is not possible to write $A$ as a direct sum as claimed in general.

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    $\begingroup$ I don't think this works. According to the answer to this question there exist non-homeomorphic compact Hausdorff spaces $X$ and $Y$ such that $X\coprod X \cong Y\coprod Y$. $\endgroup$
    – Nik Weaver
    May 18, 2018 at 18:48
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    $\begingroup$ See also this question. $\endgroup$
    – Nik Weaver
    May 18, 2018 at 19:01

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