Let $A$ be a (non-unital) $C^*$-algebra with multiplier $C^*$-algebra $M(A)$. Let $\phi: M(A) \to M(A)$ be a $*$-automorphism. Is it true that $\phi$ is automatically strictly continuous (on bounded subsets)?

Some remarks/observations:

(1) If $A = B_0(H)$, then this is true because $*$-automorphisms of $B(H) = M(B_0(H))$ are automatically strict (as they are given by conjugation with a unitary).

(2) If $A$ is separable, this is true due to a result by Woronowicz which says that $$A= \{x \in M(A): xM(A) \mathrm{\ is \ separable}\}$$ so that we can reconstruct $A$ from its multiplier $C^*$-algebra $A$.

(3) I tried to see what happens in the commutative case, so $A=C_0(X)$. Then a $*$-automorphism of $M(A) = C_b(X)= C(\beta X)$ corresponds to a homeomorphism $\beta X \to \beta X$. I have hope that if a counterexample exists, then a smart example of such a homeomorphism can lead to a counterexample.

(4) The following question seems to be related: Does a strict $*$-automorphism $\phi: M(A) \to M(A)$ preserve the subalgebra $A$, i.e. do we have $\phi(A)\subseteq A?$


1 Answer 1


I think that the answer is no.

Let $\mu$ be a non-trivial homeomorphism of $\beta \bf N$ with distinct points $y,z\in \beta\bf N\setminus \bf N$ such that $\mu(y)=z$ and $\mu(z)=y$. Set $A=\{f\in C(\beta{\bf N} ): f(y)=0\}$. Then $M(A)=C(\beta {\bf N})$. Let $\phi: M(A)\to M(A)$ be given by $\phi(f)(x)=f(\mu(x))$ $(f\in C(\beta {\bf N}), x\in \beta {\bf N})$.

Let $(f_{\alpha})$ be a bounded approximate identity for $A$. Then $(f_{\alpha})$ converges strictly to $1\in M(A)$ but $(\phi(f_{\alpha}))$ does not converge strictly to $\phi(1)=1$. To see this, take $g\in A$ such that $g(z)=1$. Then $\phi(f_{\alpha})(z)g(z)=\phi(f_{\alpha})(z)=f_{\alpha}(\mu(z))=f_{\alpha}(y)=0$ but $\phi(1)(z)g(z)=1$. Hence $\phi(f_{\alpha})g$ does not converge in norm to $g$.

  • $\begingroup$ Thanks for your answer. Why is $M(A) = C(\beta \mathbb{N})$? $\endgroup$
    – J. De Ro
    Apr 8, 2022 at 15:07
  • 3
    $\begingroup$ @QuantumSpace It is a general theorem about multiplier algebras of C$^*$-algebras that if $B\subseteq A\subseteq M(B)$ and $A$ is an ideal in $M(B)$ then $M(B)$ is the multiplier algebra of $A$ as well (follows from the approximate identity in $B$, if I remember). So taking $B=C_0({\bf N})$ we get $M(A)=M(B)=C(\beta {\bf N})$. $\endgroup$ Apr 8, 2022 at 15:59

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