Let me recall the following fact:

If $A$ is a $C^*$-algebra and $\pi: A \to \mathcal{B}(\mathcal{H})$ is a faithful non-degenerate representation, then we can explicitely realise the multiplier algebra $M(A)$ as bounded operators on $H$. More precisely, $M(A) \cong M_\pi(A)$ where $$M_\pi(A) = \{T \in \mathcal{B}(\mathcal{H})\mid T\pi(A) \subseteq \pi(A), \ \pi(A) T \subseteq \pi(A)\}.$$

Let $A$ be a $C^*$-algebra and $\mathcal{H}$ be a Hilbert space. The canonical $*$-morphism $$\mathcal{B}_0(\mathcal{H}) \otimes A \to \mathcal{B}_0(\mathcal{H}) \otimes A \otimes A: x \otimes a \mapsto x \otimes a \otimes 1$$ is non-degenerate, so extends uniquely to a strict $*$-morphism between the multiplier algebras. Let $\pi_u$ be a faithful and non-degenerate representation of $A$ on the Hilbert space $\mathcal{K}$. Let $\text{id}$ be the identity representation of $\mathcal{B}_0(\mathcal{H})$. I want to show that the following diagram commutes:

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I can see that we have two maps $$\mathcal{M}(\mathcal{B}_0(\mathcal{H}\otimes A) \to \mathcal{M}(\mathcal{B}_0(\mathcal{H}\otimes A \otimes A)$$ such that $\mathcal{B}_0(\mathcal{H})\otimes A \ni x \otimes a \mapsto x \otimes a \otimes 1$. Somehow I want to invoke uniqueness of the extension that these maps must agree everywhere but for that I need to show that the composition along the two isomorphisms and the map $T \mapsto T \otimes 1$ (first down, then go right, then go up) is strictly continuous, and I can't see why this holds.

Question: What is the quickest way to see that this diagram commutes?

  • $\begingroup$ I have (of course) left an answer, which does not use "strict continuity" (except perhaps in heavy disguise). But thinking more about your question, isn't it clear that the "down" isomorphisms are strictly continuous (in either direction) and that $T\mapsto T\otimes 1$ is strictly continuous? $\endgroup$ Nov 29 '20 at 11:18
  • $\begingroup$ Maybe yes, I just did a long walk and I also thought this should not be hard to show. But your answer offers a nicer perspecive though! $\endgroup$
    – user167952
    Nov 29 '20 at 11:21

This is just a definition chase. Let me state a more general result: let $\newcommand{\mc}{\mathcal} C,D$ be $C^\ast$-algebras, let $\pi_C: C\rightarrow\mc B(H_C)$ be a faithful $*$-representation, and similarly $\pi_D$ on $H_D$. Let $\phi:C\rightarrow D$ be a non-degenerate $*$-homomorphism with extension $\tilde\phi:M(C)\rightarrow M(D)$. Similar remarks would apply to $\phi:C\rightarrow M(D)$.

Let $\alpha:\mc B(H_C)\rightarrow\mc B(H_D)$ be a $*$-homomorphism such that the following diagram commutes:$\require{AMScd}$

$$ \begin{CD} C @>\phi>> D \\ @V{\pi_C}VV @VV{\pi_D}V\\ \mc B(H_C) @>\alpha>> \mc B(H_D) \end{CD} $$

Then I claim also the following diagram commutes:

$$ \begin{CD} M(C) @>{\tilde\phi}>> M(D) \\ @V{\tilde\pi_C}VV @VV{\tilde\pi_D}V\\ \mc B(H_C) @>\alpha>> \mc B(H_D) \end{CD} $$

Here $\tilde\pi_C:M(C)\rightarrow\mc B(H_C)$ is the extension map, which sends $M(C)$ to $M_{\pi_C}(C) = \{ T\in\mc B(H_C) : T\pi_C(a), \pi_C(a)T\in\pi_C(C) \ (a\in C) \}$.

To show the claim, let $x\in M(C), c\in C$ so $\phi(xc) = \tilde\phi(x)\phi(c)$ and hence $$ \alpha(\tilde\pi_C(x)) \pi_D(\phi(c)) = \alpha(\tilde\pi_C(x)) \alpha(\pi_C(c)) = \alpha(\pi_C(xc)) = \pi_D(\phi(xc)) = \tilde\pi_D(\tilde\phi(x)) \pi_D(\phi(c)). $$ As $\phi$ and $\pi_D$ are non-degenerate both $\{ \phi(c)d : c\in C, d \in D \}$ is linear dense in $D$, and $\{ \pi_D(\phi(c)) \pi_D(d) \xi : c\in C, d\in D, \xi\in H_D \}$ is linearly dense in $H_D$. So we have that $$ \alpha(\tilde\pi_C(x)) \pi_D(\phi(c)) \pi_D(d) \xi = \tilde\pi_D(\tilde\phi(x)) \pi_D(\phi(c)) \pi_D(d) \xi $$ for all $c,d,\xi,x$ and hence $$ \alpha(\tilde\pi_C(x)) = \tilde\pi_D(\tilde\phi(x)) \qquad (x\in M(C)), $$ as required.

In your case, set $C=\mc B_0(H)\otimes A$ and $D=\mc B_0(H)\otimes A\otimes A$ with $\phi(c) = c\otimes 1$ and $\alpha(T)=T\otimes 1$, and $\pi_C, \pi_D$ the obvious maps to $H\otimes K$ and $H\otimes K\otimes K$, respectively. The middle row of your commutative diagram follows by restriction.

Notice that I did not assume that $\alpha$ was "continuous" in any sense, but that the first diagram commutes imposes conditions on $\alpha$. That $\pi_D$ and $\phi$ are non-degenerate shows that $\alpha$ is non-degenerate in the sense that $\{ \alpha(T)\xi : T\in\mc B(H_C), \xi\in H_D \}$ is linear dense in $H_D$. Indeed, also $\pi_C$ is non-degenerate, and this shows that also $\pi_C:C\rightarrow M(\mc B_0(H_C))$ is non-degenerate, so $\{ \pi_C(c) t : c\in C, t\in\mc B_0(H_C) \}$ is linearly dense in $\mc B_0(H_C)$. Thus, if $\alpha$ is at least bounded, \begin{align*}\newcommand{\lin}{\operatorname{lin}} &\overline{\lin}\{ \alpha(t)s : t\in \mc B_0(H_C), s\in \mc B_0(H_D) \} \\ &= \overline{\lin}\{ \alpha(\pi_C(c)t)s :c\in C,t\in \mc B_0(H_C), s\in \mc B_0(H_D) \} \\ &= \overline{\lin}\{ \pi_D(\phi(c)) \alpha(t) s:c\in C,t\in \mc B_0(H_C), s\in \mc B_0(H_D) \}. \end{align*} However, this does not seem to necessarily be all of $\mc B_0(H_D)$. Thus, if we consider $\mc B(H_C)$ as $M(\mc B_0(H_C))$, and restrict $\alpha$ to $\mc B_0(H_C)$ say giving $\alpha_0:\mc B_0(H_C) \rightarrow \mc B(H_D)$, I do not see why necessarily $\alpha_0$ is non-degenerate; equivalently, whether $\alpha$ is strictly continuous. It is in your case, of course.

  • $\begingroup$ Thanks! I believe this is exactly what I'm looking for! I will accept the answer once I worked through it! $\endgroup$
    – user167952
    Nov 29 '20 at 11:19

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