Show that $A \otimes B = A \odot B$ if $A$ is a finite-dimensional $C*$-algebra.

Question

Let $A$ and $B$ be $C^*$-algebras. If $A$ is finite-dimensional, then I want to show that the algebraic tensor product $A \odot B$ is complete for the minimal $C^*$-norm on it (in fact, the choice of norm does not matter since $A$ is nuclear).

Attempt:

I know that $$A \cong M_{n_1}(\mathbb{C}) \oplus \dots \oplus M_{n_k}(\mathbb{C})$$

And thus $$A \odot B \cong (A \odot M_{n_1}(\mathbb{C}))\oplus \dots \oplus (A\odot M_{n_k}(\mathbb{C})) $$

so I believe it suffices to show that $A \odot M_{n_1}(\mathbb{C})\cong A \odot \mathbb{C}^{n^ 2}$ is complete.

Then I'm stuck.

Maybe I'm overcomplicating this though....

Answer 1

You wrote as if $B$ is the finite-dimensional one, so I'll stick with that.

Let $\gamma$ be a C$^*$-norm on $A\otimes M_n(\mathbb C)$ (one certainly exists, because we can represent $A\subset B(H)$ and then $A\otimes M_n(\mathbb C)$ can be represented in $B(H\otimes\mathbb C^n)$).

For any $k$, the map $a\longmapsto \gamma(a\otimes E_{kk})$ defines a C$^*$-norm on $A$. As a C$^*$-algebra admits a unique C$^*$-norm, we get that $\gamma(a\otimes E_{kk})=\|a\|$. Even if $k\ne j$, $$ \gamma(a\otimes E_{kj})^2=\gamma(a^*a\otimes E_{jj})=\|a^*a\|=\|a\|^2. $$ Now suppose that we have a Cauchy sequence $\{x_n\}\subset A\otimes M_n(\mathbb C)$. We can write $$x_n=\sum_{k,j} a_n(k,j)\otimes E_{k,j}.$$ For any $k,j$, \begin{align} \|a_n(k,j)-a_m(k,j)\|^2&=\gamma((a_n(k,j)-a_m(k,j))\otimes E_{11})\\[0.3cm] &= \gamma((1\otimes E_{11})(x_n-x_m)(1\otimes E_{11}))\\[0.3cm] &\leq\gamma(x_n-x_m). \end{align} So each sequence $\{a_n(k,j)\}_n$ is Cauchy on $A$. As $A$ is a C$^*$-algebra, there exist $a(k,j)\in A$ with $\lim_na_n(k,j)=a(k,j)$. We want to show that $x_n\to\sum_{k,j} a(k,j)\otimes E_{k,j}$. For this \begin{align} \gamma(x_n-\sum_{k,j} a(k,j)\otimes E_{k,j}) &=\gamma(\sum_{k,j} (a_n(k,j)-a(k,j))\otimes E_{k,j})\\[0.3cm] &\leq \sum_{k,j} \gamma((a_n(k,j)-a(k,j))\otimes E_{k,j})\\[0.3cm] &=\sum_{k,j} \|a_n(k,j)-a(k,j)\|\\[0.3cm] &\to0, \end{align} where it is essential in this last step that the sum is finite. So $A\otimes M_n(\mathbb C)$ is complete with respect to $\gamma$, and so it is a C$^*$-algebra. This in particular shows that $\gamma$ is necessarily the norm described in the second paragraph.

Answer 2

The point is that $A \odot B$ is closed in $A \otimes B$. To see this, choose a basis $\{e_i\}_{i=1}^n$ for $A$ and suppose that $\{x_k\}_{k=1}^\infty$ is a sequence in $A \odot B$ that converges to $x \in A \otimes B$. Then we can write $$x_k = \sum_{i=1}^n e_i \otimes b_i^k$$ and since $x_k \to x$ in $A \otimes B$, applying the slice map $\tau_j \otimes \iota: A \otimes B \to B$ where $\tau \in A^*$ satisfies $\tau(e_i) = \delta_{i,j}$, we see that $$b_j^k = (\tau_j \otimes \iota)(x_k) \to (\tau_j \otimes \iota)(x).$$ Hence, $$x=\sum_{i=1}^n e_i \otimes (\tau_i \otimes \iota)(x) \in A \odot B.$$