# Is a $*$-automorphism $M(A) \to M(A)$ automatically strictly continuous?

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 $$\phi(A)\subseteq A?$$

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 $$g$$.
• Thanks for your answer. Why is $M(A) = C(\beta \mathbb{N})$? Apr 8, 2022 at 15:07
• @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 $M(A)=M(B)=C(\beta {\bf N})$. Apr 8, 2022 at 15:59