9
$\begingroup$

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 $\phi(A)\subseteq A?$

$\endgroup$

1 Answer 1

5
$\begingroup$

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 $g$.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer. Why is $M(A) = C(\beta \mathbb{N})$? $\endgroup$ Apr 8 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 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.