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Consider a collection of $C^*$-algebras $\{A_i\}_{i \in I}$. We can form the direct sum $$\bigoplus_{i \in I}^{c_0} A_i:= \left\{(a_i)_{i \in I} \in \prod_{i\in I} A_i: \lim_{i \in I} \|a_i\| = 0\right\}$$ which is an ideal in the $C^*$-algebra $$\bigoplus_{i \in I}^{\ell^\infty} A_i:= \left\{(a_i)_{i \in I} \in \prod_{i\in I} A_i: \sup_{i \in I} \|a_i\| <\infty\right\}.$$ Given a $C^*$-algebra $A$, denote its multiplier algebra by $M(A)$. One possible realisation of the multiplier algebra is by setting $M(A):= \mathcal{L}_A(A)$, the adjointable operators when we view $A$ as a (right) Hilbert $C^*$-module over itself.

I believe I have proven that $$M\left(\bigoplus_{i \in I}^{c_0}A_i\right) \cong \bigoplus_{i \in I}^{\ell^\infty} M(A_i)\cong M\left(\bigoplus_{i \in I}^{\ell^\infty}A_i\right)$$ but I can't find a reference for this statement. So, is my assertion true?


Here is a proof sketch: We use the implementation of the multiplier $C^*$-algebra as adjointable operators. We then have natural maps $$M\left(\bigoplus_{i \in I}^{c_0}A_i\right) \to \bigoplus_{i \in I}^{\ell^\infty} M(A_i): t \mapsto (\iota_i ^* t\iota_i)_{i \in I}$$ where $\iota_i: A_i \hookrightarrow \bigoplus_{i \in I}^{c_0} A_i$ is the inclusion map and $$\bigoplus_{i \in I}^{\ell^\infty} M(A_i) \to M\left(\bigoplus_{i \in I}^{c_0}A_i\right) : (t_i)_{i \in I} \mapsto [(a_i)_i \mapsto (t_i(a_i))]$$ These are easily checked to be $*$-isomorphisms that are inverse to each other, and this establishes the isomorphism $M\left(\bigoplus_{i \in I}^{c_0}A_i\right) \cong \bigoplus_{i \in I}^{\ell^\infty} M(A_i)$. The other isomorphism is shown similarly.

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    $\begingroup$ The isomorphism you sketch is correct. The one you are not sketching is incorrect. There are in general a lot more multiplier on $\ell^\infty-\bigoplus_i A_i$ than on $c_0-\bigoplus_i A_i$. $\endgroup$
    – Jamie Gabe
    Aug 24, 2021 at 7:51
  • $\begingroup$ @Jamie Gabe Thanks. Where does the proof break down in the $\ell^\infty$-case? $\endgroup$
    – Andromeda
    Aug 24, 2021 at 8:04
  • $\begingroup$ In your sketch proof, I do not (immediately) see why: (a) $\iota_i^* t \iota_i$ need be adjointable; and (b) why are the two maps mutual inverses? It seems like more argument is needed, and it's here the $\ell^\infty$ case would be different. $\endgroup$ Aug 24, 2021 at 8:39
  • $\begingroup$ Sorry, more to the point, what is $\iota_i^*$? Because $\iota_i$ is a map between Hilbert $C^*$-modules over different algebras, so I don't really know what adjointable or the adjoint would mean in such a situation. $\endgroup$ Aug 24, 2021 at 10:29
  • $\begingroup$ @MatthewDaws Yes, you are right. We should replace it by the projection onto the $i$-th component. $\endgroup$
    – Andromeda
    Aug 24, 2021 at 10:55

1 Answer 1

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Let's try to flesh out your "sketch". Set $A=c_0-\oplus_i A_i$ and consider this as a Hilbert $C^*$-module over itself. Let $\iota_i:A_i\rightarrow A$ be the inclusion, and $\jmath_i:A\rightarrow A_i$ the left inverse to $\iota_i$. These are both non-degenerate $*$-homomorphisms, and so extend to unital $*$-homomorphisms $\overline\iota_i:M(A_i)\rightarrow M(A)$ and $\overline\jmath_i:M(A)\rightarrow M(A_i)$. Also, notice that by definition of the $A$-valued inner-product on $A$, $$ (b|\iota_i(a)) = \iota_i\big( (\jmath_i(b)|a) \big) \qquad (b\in A, a\in A_i). $$

Given $T\in M(A)$ set $T_i = \overline\jmath_i(T)\in M(A_i)$. By definition, $T_i(\jmath_i(a)) = \jmath_i(T(a))$ for $a\in A$, and as $\jmath_i$ is a $*$-homomorphism, also $T_i^*(\jmath_i(a)) = \jmath_i(T^*(a))$. For $a\in A_i$ and $b=(b_j)\in A$, $$ (T\iota_i(a)|b) = (\iota_i(a)|T^*(b)) = \iota_i\big( (a|\jmath_i(T^*(b))) \big) = \iota_i\big( (a|T_i^*(\jmath_i(b))) \big) $$ while $$ (\iota_i T_i(a)|b) = \iota_i\big( (T_i(a)|\jmath_i(b)) \big) = \iota_i\big( (a|T_i^*(\jmath_i(b))) \big). $$ Thus $T\iota_i = \iota_iT_i$ for each $i$.

As the linear span of the images of the $\iota_i$ are dense in $A$, it now follows that $$ T(a) = \sum_i T\iota_i\jmath_i(a) = \sum_i \iota_i\big( T_i\jmath_i(a) \big), $$ and the isomorphism $M(A) \cong \ell^\infty-\oplus_i M(A_i)$ now follows.


I actually find arguing using Hilbert $^*$-modules a bit cumbersome. Instead, let $A_i\subseteq\mathcal B(H_i)$ non-degenerately, for each $i$, for some suitable Hilbert space $H_i$. Set $H = \oplus_i H_i$, so naturally $A$ acts non-degenerately on $H$. Set $B=\ell^\infty-\oplus_i A_i$, so also $B$ acts non-degenerately on $H$. I'll now consider $B$, but much the same argument works for $A$.

We know that $M(B)\subseteq B''\subseteq \mathcal B(H)$, and indeed $$ M(B) = \big\{ T\in B'' : Tb, bT\in B \ (b\in B) \big\}. $$ Given $b=(b_i)\in B$ and $\xi=(\xi_i)\in H$, by definition, $b(\xi) = (b_i(\xi_i))$. As such, with $p_i\in\mathcal B(H)$ the projection onto the factor $H_i$, we see that $b p_i = p_i b$ so $p_i\in B'$.

Thus, any $T\in M(B)\subseteq B''$ commutes with each $p_i$. By linearity and continuity, there is $(T_i) \in \ell^\infty-\oplus_i \mathcal B(H_i)$ with $T(\xi) = (T_i(\xi_i))$ for each $\xi\in H$. Using the inclusions $A_i\rightarrow B$, we can now show that each $T_i\in M(A_i)$.

We have hence showed that $M(B) \cong \ell^\infty-\oplus_i M(A_i)$. [This seems surprising to me, but I believe this 2nd argument.]

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  • $\begingroup$ Thanks! So the second isomorphism holds after all, but my approach did not work. $\endgroup$
    – Andromeda
    Aug 24, 2021 at 21:07
  • $\begingroup$ This gives nice examples of inclusions $A \subsetneq B$ where $M(A) = M(B)$. $\endgroup$
    – Andromeda
    Aug 24, 2021 at 21:09
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    $\begingroup$ Oh, I was wrong in saying that $\ell^\infty-\oplus A_i$ had a different multiplier algebra! Yikes! You're absolutely right, Matt! One can also deduce this from the fact that if $A \subseteq B \subseteq M(A)$ are inclusions of two-sided closed ideals (so that the composition is the canonical inclusion $A\subseteq M(A)$), then $M(A) = M(B)$. This is an easy consequence of the universal property of multiplier algebras. One can apply this to $A = c_0-\oplus A_i$ and $B = \ell^\infty-\oplus_Ai$ once one has convinced themselves that $M(A) = \ell^\infty -\oplus M(A_i)$. $\endgroup$
    – Jamie Gabe
    Aug 25, 2021 at 9:22
  • $\begingroup$ @JamieGabe That's a very nice observation, and a much easier way to get the result! $\endgroup$ Aug 25, 2021 at 10:31

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