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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 $U(A)/Z(U(A)) \to Inn(A)$?

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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 $n$.

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