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Is there an accessible proof for the following fact?

If $A=C_0(X)$ with $X$ locally compact Hausdorff and $B$ is a $C^\ast$-algebra then $M(A\otimes B)$ is the set of bounded strictly continuous functions $X \to M(B)$.

Denote the set of bounded strictly continuous functions by $C_b^s (X, M(B))$.

Thanks to the hint in the comments, we can say the following:

Given $x \in X$, there is a mapping $$\pi_x: C_0(X) \otimes B \to B: f \otimes b \mapsto f(x)b$$ which extends to a map $$\pi_x: M(C_0(X) \otimes B) \to M(B)$$ and this allows us to define $$M(C_0(X) \otimes B) \to C_b^s(X,M(B)): L \mapsto (x \mapsto \pi_x(L))$$

Why is this an isomorphism of $C^*$-algebras, i.e. why is it injective and surjective?

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    $\begingroup$ Where did you find this "fact"? It's not correct as it is written. Probably, you mean bounded and strictly continuous functions from $X$ into $M(B)$, instead of those from $\beta X$ (which forces to have compact ranges)? Evaluation at $x\in X$ gives rise to the surjective *-homomorphism $\pi_x\colon C_0(X)\otimes B \to B$, which extends on the multipliers. $\endgroup$
    – Narutaka OZAWA
    Commented Nov 8, 2020 at 1:45
  • $\begingroup$ @NarutakaOZAWA Thanks! I read it in an answer of this thread: mathoverflow.net/questions/239720/… But I guess that contained a mistake then. I edited the question. $\endgroup$
    – user167952
    Commented Nov 8, 2020 at 9:48
  • $\begingroup$ @NarutakaOZAWA I happen to have found a reference to the fact I initially claimed: It's in Bruce Blackadar's book on operator algebras p147 ex. II 7.3.12 (iv). $\endgroup$
    – user167952
    Commented Nov 8, 2020 at 12:13
  • $\begingroup$ For my purposes though, the version you quoted is also good because I can assume $X$ is compact anyway. $\endgroup$
    – user167952
    Commented Nov 8, 2020 at 12:14

1 Answer 1

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Indeed, Blackadar appears to have made a mistake here. But the reference he gives is good, and seems to give both a correct statement, and a proof:

Akemann, Charles A.; Pedersen, Gert K.; Tomiyama, Jun Multipliers of C∗-algebras. J. Functional Analysis 13 (1973), 277–301.

MR470685 DOI: 10.1016/0022-1236(73)90036-0

See Corollary 3.4.

Edit: How I would approach this. Firstly, understand carefully the proof that $C_0(X) \otimes B \cong C_0(X,B)$. Much the same ideas are used for the multiplier algebra case. In particular, for $f\in C_0(X), b\in B$ we identify $f\otimes b$ with the continuous map $X\rightarrow B; x\mapsto f(x) b$.

I would look at $\Phi: C^b_{str}(X, M(B)) \rightarrow M(C_0(X,B))$ defined by pointwise multiplication:

  • First show this is well-defined. This is easy, as for $F\in C^b_{str}(X, M(B))$ we have that $F(f\otimes b) \in C_0(X,B)$. Then copy the proof that $C_0(X) \otimes B \cong C_0(X,B)$ to show that $F$ does multiply $C_0(X,B)$ into itself.
  • $\Phi$ is clearly injective.
  • To show $\Phi$ is surjective, argue as in the OP: given $L\in M(C_0(X,B))$ we define $F(x) = \pi_x(L)$. Then $X\rightarrow B; x \mapsto \pi_x(L) f(x) b = \pi_x(L(f\otimes b))$ is continuous, for each $f\in C_0(X)$ and $b\in B$. This is enough to show that $F$ is continuous, for the strict topology
  • Checking that $\Phi$ is a $*$-homomorphism is routine.
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  • $\begingroup$ Thanks a lot! This is very nice. $\endgroup$
    – user167952
    Commented Nov 8, 2020 at 22:56
  • $\begingroup$ Hi. I'm afraid I don't quite see why $F$ is continuous. I took a net $x_\alpha \to x$ and I want to show that $F(x_\alpha) \to F(x)$ in the strict topology on $M(B)$. Thus we must show $F(x_\alpha)b \to F(x)b$ in the $B$-norm and $bF(x_\alpha) \to bF(x)$ in the $B$-norm, but I can't see why this is true. In particular, my follow up question is: is there a quick way to see $\|F(x_\alpha)b-F(x)b\|_B = \| \pi_{x_\alpha}(v)b- \pi_{x}(v)b\|_B \to 0$? $\endgroup$
    – user167952
    Commented Jan 21, 2021 at 9:53
  • $\begingroup$ I don't understand: where is your $F$ from? $\endgroup$
    – Matthew Daws
    Commented Jan 21, 2021 at 10:33
  • $\begingroup$ The $F$ you define in the surjectivity: $F(x) = \pi_x(L)$ $\endgroup$
    – user167952
    Commented Jan 21, 2021 at 10:35
  • $\begingroup$ Continuity follows from the equation I wrote: $F(x) f(x) b = \pi_x(L(f\otimes b))$. If $(x_\alpha)$ converges to $x$ then choose f to be identically 1 on a neighbourhood of $x$. Then $L(f\otimes b)$ is is $C_0(X)\otimes B = C_0(X,B)$ and so continuous, so $F(x_\alpha) b = \pi_{x_\alpha}(L(f\otimes b)) \rightarrow \pi_x(L(f\otimes b)) = F(x)b$ $\endgroup$
    – Matthew Daws
    Commented Jan 21, 2021 at 10:55

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