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I'm reading the following fragment in the paper "Notes on compact quantum groups":

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While I'm familiar with the multiplier algebra (constructed via double centralizers) and its universal property in terms of essential ideals, I'm a little bit unsure why one can extend the map $$\mathcal{B}_0(\mathcal{H}) \otimes A \to \mathcal{B}_0(\mathcal{H}) \otimes A \otimes A: x \mapsto x \otimes 1$$ to a map $$M(\mathcal{B}_0(\mathcal{H}) \otimes A) \to M(\mathcal{B}_0(\mathcal{H}) \otimes A \otimes A)$$

Here, the tensor product is the minimal one.

Does every map $*$-morphism $A \to B$ between $C^*$-algebras extend to a $*$-morphism $M(A) \to M(B)$?

Thanks in advance for any reference/input/links.

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    $\begingroup$ There is a general result that says that you get your extension provided your morphism $\phi:A\to B$ satisfies $\overline{\phi(A)B}=B$. Since this holds in your case, your extension exists. $\endgroup$
    – Ruy
    Commented Nov 3, 2020 at 2:25
  • $\begingroup$ @Ruy Do you have a reference? $\endgroup$
    – user167952
    Commented Nov 3, 2020 at 8:48
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    $\begingroup$ This is proven if I remember correctly in the book by Lance, "Hilbert $C^*$-modules: A toolkit for operator algebraists". $\endgroup$
    – Ulrich Pennig
    Commented Nov 3, 2020 at 10:54
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    $\begingroup$ This is Proposition 2.5 in Lance's book, but it takes a bit of interpretation. You need to take the Hilbert module $E$ mentioned there to be $E=B$, and then you need to know that $\mathcal L(E)$ is the multiplier algebra of $B$ (actually this is Lance's definition of the multiplier algebra). $\endgroup$
    – Ruy
    Commented Nov 3, 2020 at 11:51

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The answer is no in general: take $B = C[0,1]$, let $A$ be the subalgebra consisting of continuous functions which vanish at $0$, and let $\phi: A \to B$ be the inclusion map. This doesn't extend to a $*$-homomorphism from $M(A) \cong C_b(0,1]$ into $B$. (There's no continuous function from $[0,1]$ into the Stone–Čech compactification of $(0,1]$ which is the identity on $(0,1]$.)

However, as Ruy points out in a comment, we are okay if $\overline{\phi(A)B} = B$. To see this, put $B$ inside some $B(K)$ and regard $\phi$ as a $*$-homomorphism from $A$ into $B(K)$. Since $A$ is an ideal of $M(A)$, this extends to a $*$-homomorphism $\tilde{\phi}$ from $M(A)$ into $B(K)$. Does it map into $M(B)$? We must show that $\tilde{\phi}(a)b \in B$ for any $a \in M(A)$ and $b \in B$. But if $b = \phi(a_0)b_0$ for some $a_0 \in A$ and $b_0 \in B$, then $\tilde{\phi}(a)b = \phi(aa_0)b_0 \in B$. Since $\overline{\phi(A)B} = B$, we are done by continuity.

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  • $\begingroup$ Can't we get away with slightly less: say $\phi(u_\lambda)$ converges strictly to a projection on $B$, where $(u_\lambda)$ is an approximate identity for $A$? $\endgroup$
    – David Roberts ♦
    Commented Nov 3, 2020 at 5:59
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    $\begingroup$ @user839372 $M(A)$ is the completion of $A$ in the strict topology, so the extension you seek exists precisely when the original $*$-homomorphism is strictly continuous. You don't need the concrete representation of $B$ at all, I presume that was just motivation. $\endgroup$
    – David Roberts ♦
    Commented Nov 3, 2020 at 9:39
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    $\begingroup$ Every representation of an ideal extends to a representation of the ambient algebra, so this is why it is useful to represent $B$. But you may also get away without it: by Cohen-Hewitt every $b$ in $B$ may be written on the nose as $b=φ(a_1)b_1$, so if $m$ is a multiplier you may define the extension by $\tildeφ(m)b=φ(ma_1)b_1$. $\endgroup$
    – Ruy
    Commented Nov 3, 2020 at 11:33
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    $\begingroup$ @user839372 Ruy answered this, but: I put $B$ in $B(K)$ because $A$ is an ideal of $M(A)$, and it is a general fact that any $*$-homomorphism from an ideal of a C*-algebra into $B(K)$ extends to the whole algebra. $\endgroup$
    – Nik Weaver
    Commented Nov 3, 2020 at 13:32
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    $\begingroup$ You are welcome! $\endgroup$
    – Nik Weaver
    Commented Nov 3, 2020 at 16:21

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