Minimal tensor product of $B(H)$ and $C(G)$

Question

Let $H$ be a finite dimensional vector space, and $G$ be a compact group. Let $B(H)$ be the bounded operators on $H$, let $C(G)$ be the complex valued continuous functions on $G$, and let $C(G;B(H))$ be the $B(H)$ valued continuous functions on $G$. Let $B(H)\otimes C(G)$ be the minimal tensor product of $B(H)$ and $C(G)$.

Question: Show that $B(H)\otimes C(G)=C(G;B(H))$?

I got stuck in this problem while reading the note 'Compact Quantum Groups and Their Representation Categories' by Neshveyev and Tuset. Thanks in advance.

Answer 1

The natural map here would be $$ \gamma:\sum_k T_k\otimes f_k\longmapsto \big( g\longmapsto \sum_k f_k(g)\,T_k\big). $$ This map is clearly a $*$-homomorphism, if it is well-defined. For that, $\sum_kT_k\otimes f_k=0$, if and only if there exist coefficients $c_{kj}$ such that $$ \sum_m c_{mk}T_m=0,\qquad \sum_m c_{kn}f_n=f_k\qquad \text{ for all }k. $$ Then $$ \sum_kf_k(g)T_k=\sum_k\sum_n c_{kn}f_k(g)T_k=\sum_nf_k(g)\sum_k c_{kn}T_k=0. $$ So $\gamma$ is well-defined. Next we check that $\gamma$ is injective. Suppose that $\sum_kf_k(g)T_k=0$ for all $g\in G$. Let $S_1,\ldots,S_r$ be a basis of $\operatorname{span}\{T_k:\ k\}$. Then there exists coefficients such that, for all $k,n$,
$$ T_k=\sum_m\alpha_{km}S_m,\qquad S_n=\sum_j\beta_{nj}T_j. $$ Then $$ 0=\sum_kf_k(g)T_k=\sum_k\sum_m\alpha_{km}f_k(g)S_m=\sum_m\Big(\sum_k\alpha_{km}f_k(g)\Big)S_m. $$ As the $S_m$ are linearly independent and $g$ is arbitrary, $$\sum_k\alpha_{km}f_k=0.$$ Then $$ \sum_k f_k\otimes T_k=\sum_k\sum_m\alpha_{km}(f_k\otimes S_m) =\sum_k\Big(\sum_k\alpha_{km}f_k\Big)\otimes S_m=0. $$ So $\gamma$ is injective. Being an injective $*$-homomorphism, it is isometric.

All that remains is to show that $\gamma$ is surjective. Let $f\in C(G,B(H))$ and fix $\varepsilon>0$. Since $G$ is compact and $f$ is continuous, $f(G)$ is compact. So there exists $T_1,\ldots,T_r\in f(G)$ such that $f(G)\subset \bigcup_kB_\varepsilon(T_k)$. Let $g_k\in G$ with $f(g_k)=T_k$, and $V_k=f^{-1}(B_\varepsilon(T_k))$. These are open sets that cover $G$. Let $h_1,\ldots,h_r$ be a partition of unity relative to $V_1,\ldots,V_r$. That is, $\sum_k h_k=1$, $0\leq h_k\leq1$, continuous, and $\operatorname{supp}h_k\subset V_k$. Let $f_k=f\,h_k$. Then, taking into account that $h_k$ is nonzero only inside $V_k$, \begin{align} \Big\|f(g)-\sum_k h_k(g)T_k\Big\| &=\Big\|\sum_k h_k(g)\,\big(f(g)-f(g_k)\big)\Big\| \leq\varepsilon\,\Big\|\sum_kh_k(g)\Big\|=\varepsilon. \end{align} The above shows that the image of $\gamma$ is dense. Being isometric, $\gamma$ extends to a $*$-isomorphism.

Answer 2

Another perspective:

We have the following chain of well-known isomorphisms:

$$B(H)\otimes C(G)\cong M_n(\mathbb{C})\otimes C(G)\cong M_n(C(G))\cong C(G, M_n(\mathbb{C}))\cong C(G, B(H)).$$

If you calculate this explicit composition, you easily see that it agrees with the map in the answer of Martin Argerami.

Also, the 'minimal' tensor product doesn't matter here. $C(G)$ is nuclear, so there is a unique $C^*$-norm on every algebraic tensor product you form.