Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be a $C^*$-algebra. The group $Aut(A)$ of $\ast$-automorphisms of $A$ is usually equipped either with the pointwise norm topology, i.e. the topology generated by the semi-norms $\lVert \varphi \rVert_a = \lVert \varphi(a) \rVert$ or with the uniform topology, i.e. the one which it earns by inclusion into the Banach space of bounded linear operators from $A$ to $A$. Moreover, let $U(A) = U(M(A))$ be the unitary group of the multiplier algebra equipped with the strict topology. Now, the map $Ad : U(A) \to Aut(A)$ is continuous if $Aut(A)$ carries the pointwise norm topology. It induces a continuous bijection $U(A) / Z(U(A)) \to Inn(A)$ onto the inner automorphisms, which is not a homeomorphism unless the $C^\ast$-algebra is a continuous trace $C^{\ast}$-algebra. So, my questions are

Is the induced bijection $U(A)/Z(U(A)) \to Inn(A)$ a homeomorphism, if $Aut(A)$ carries the uniform topology?

Is there a natural topology on $Aut(A)$, which induces a homeomorphism $U(A)/Z(U(A)) \to Inn(A)$?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

If $Aut(A)$ carries the uniform topology but $U(A)/Z(U(A))$ the topology induced by the strict one, then the bijection is not continuous.

For example, let $A=C_0(N,M_2(C))$ be the $C^*$-algebra of all $2\times 2$-matrix-valued functions on the naturals vanishing at infinity, and define for each $n \in N \cup \{\infty\}$ a unitary $U_n \in U(A)$ such that for $k \leq n$, $U_n(k)$ is $0$ on the diagonal and $1$ off the diagonal and for $k > n$, $U_n(k)$ is the identity.Then the sequence of the $U_n$ converges strictly to $U_{\infty}$. But, if $\chi_{m} \in A$ denotes the function that is equal to the identity in $M_2(C)$ on $\{0,\ldots,m\}$ and $0$ thereafter, then $\|(Ad_{U_n}- Ad_{U_{n+1}})(\chi_m)\|$ is constant non-zero for $m>n$ independently of $n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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