# Identifying the multiplier $C^*$-algebra $M(C_0(X) \otimes B)$

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?

• 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. Nov 8 '20 at 1:45
• @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.
– user167952
Nov 8 '20 at 9:48
• @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).
– user167952
Nov 8 '20 at 12:13
• For my purposes though, the version you quoted is also good because I can assume $X$ is compact anyway.
– user167952
Nov 8 '20 at 12:14

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.

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.
• Thanks a lot! This is very nice.
– user167952
Nov 8 '20 at 22:56
• 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 $\|F(x_\alpha)b-F(x)b\|_B = \| \pi_{x_\alpha}(v)b- \pi_{x}(v)b\|_B \to 0$?
– user167952
Jan 21 at 9:53
• I don't understand: where is your $F$ from? Jan 21 at 10:33
• The $F$ you define in the surjectivity: $F(x) = \pi_x(L)$
– user167952
Jan 21 at 10:35
• 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 $F(x_\alpha) b = \pi_{x_\alpha}(L(f\otimes b)) \rightarrow \pi_x(L(f\otimes b)) = F(x)b$ Jan 21 at 10:55