Question about $C_0(X)$-algebras and $C_b(X)$.

Question

Let $X$ be a locally compact Hausdorff space. Denote by $C_0(X)$ its C*-algebra of continuous functions that vanish on infinity and by $C_b(X)$ its C*-algebra of bounded functions. Now, let $A$ be a C*-algebra and $M(A)$ its multiplier algebra. Denote by $\mathcal{Z}M(A)$ its center.

A structure of "$C_0(X)$-algebra" in $A$ is a $*$-morphism $\pi:C_0(X) \rightarrow \mathcal{Z}M(A)$ that has "non-degeneracy", meaning that $$\pi(C_0(X))\cdot A = \mathrm{span}\{ \pi(f)\cdot a : f \in C_0(X), a \in A\} $$ is dense in $A$.

Being somewhat careful, one can see that this is actually equivalent to the fact that if $\{u_{\lambda}\}$ is an approximate unit of $C_0(X)$ then $\{\pi(u_{\lambda})\}$ is an approximate unit of $A$, in the sense that $$ \lim_{\lambda \to \infty} \| \pi(u_{\lambda}) a - a \| = 0$$

My question now is, how can you extend $\pi$ to a $*$-morphism $$ \hat{\pi}:C_b(X) \rightarrow \mathcal{Z}M(A)? $$

I've seen this result quoted somewhere but I haven't been able to prove it.

Thanks in advance.

Answer 1

It all works a bit more generally than that.

Proposition: Let $A$ and $B$ be C*-algebras and suppose $\pi: A \to M(B)$ is a $*$-homomorphism that is "nondegenerate" in the sense you described. That is, $\mathrm{span}\{ \pi(a) b : a \in A, b \in B\}$ is dense in $B$. Then $\pi$ extends uniquely to a $*$-homomorphism $\overline \pi : M(A) \to M(B)$.

How does is this extension defined? Well...

• Note that, to determine a multiplier $y \in M(B)$, it is enough to say what is $yb \in B$ for each $b \in B$. In fact, by linearity and continuity of left multiplication, it suffices to know $yb$ for a set of $b \in B$ with dense linear span.
• By the above observation, given $x \in M(A)$, to determine $\overline \pi(x)$ as a multiplier of $B$, it suffices to say what is $\overline \pi(x) \pi(a) b$ for all $a \in A$, $b \in B$ (since the elements $\pi(a)b \in B$ have dense linear span). However, since $\overline \pi$ is a homomorphism extending $\pi$, we have $$\overline \pi(x) \pi(a) b = \overline \pi(xa)b = \pi(xa)b.$$

Thus the definition of $\overline \pi$ is forced on you. It remains to check the formula \begin{align*} \overline \pi(x) \pi(a) b = \pi(xa)b && a \in A, B \in B \end{align*} defines an element $\overline \pi(x) \in M(B)$ and that the assigmnent $x \mapsto \overline \pi(x)$ is a $*$-homomorphism, but this is mostly routine. Do you think you can take it from here?

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Some references...

• Proposition 2.5 in the book "Hilbert C*-Modules: A Toolkit for Operator Algebraists" by Lance.
• Lemma 1.1 in the paper "Representations of crossed products by coactions and principal bundles" by Landstad, Phillips, Raeburn and Sutherland.