Let $A$ be a C$^{*}$-algebra. Let $\operatorname{Aut}(A)$ denote the set of all $*$-isomorphisms from $A$ onto itself. We call an element $\alpha \in \operatorname{Aut}(A)$ an inner automorphism if there is a unitary element $u$ in the unitization $\widetilde{A}$ of $A$ such that $\alpha(a)=uau^{*}$ for all $a$ in $A$.

An $\alpha\in \operatorname{Aut}(A)$ is called approximately inner if for every finite subset $F$ of $A$ and every $\epsilon>0$, there is an inner automorphism $\beta$ such that $\|\alpha(a)-\beta(a)\|<\epsilon$. We denote the set of approximately inner automorphisms of $A$ be $\overline{\operatorname{Inn}}(A)$.

I am trying to prove that $\overline{\operatorname{Inn}}(A)$ is a normal subgroup of $\operatorname{Aut}(A)$.

I know that the set of all inner automorphisms of $A$ form a normal subgroup of $\operatorname{Aut}(A)$, and using this I was able to show that $\overline{\operatorname{Inn}}(A)$ is closed under composition and that it is a normal subgroup assuming it is a group.

I have been struggling to show that if $\alpha\in\overline{\operatorname{Inn}}(A)$, then $\alpha^{-1}\in\overline{\operatorname{Inn}}(A)$.

This is all I need to complete the proof. I am having trouble determining what inner automorphism to approximate $\alpha^{-1}$ with for a given $F$ and $\epsilon$. Any help is very much appreciated.

Thank you.

  • $\begingroup$ I know very little about this subject, but in a $C^*$-algebra the inverse map from $A$ to $A$ is continuous, right? Could that be useful? $\endgroup$ – Matthew Leingang Jun 5 '17 at 13:42
  • $\begingroup$ The inversion map is indeed continuous, but I'm not sure if this is going to help in general since the inner automorphism which we approximate $\alpha^{-1}$ by should be close to $\alpha^{-1}$ on all finite subsets of $A$. $\endgroup$ – ervx Jun 5 '17 at 13:48

I think I got it.

Suppose $\alpha\in\overline{\operatorname{Inn}}(A)$. Let $F\subset A$ be finite and let $\epsilon>0$. Note that $\alpha^{-2}(F)\subset A$ is finite. Thus, there is a $\beta\in\operatorname{Inn}(A)$ such that for all $a\in F$ $$ \|\alpha(\alpha^{-2}(a))-\beta(\alpha^{-2}(a))\|<\epsilon. $$ Since $\operatorname{Inn}(A)$ is a normal subgroup of $\operatorname{Aut}(A)$, $\beta\alpha^{-2}=\alpha^{-2}\gamma$ for some $\gamma\in\operatorname{Inn}(A)$. Thus, for all $a\in F$, $$ \|\alpha^{-1}(a)-\alpha^{-2}(\gamma(a))\|<\epsilon. $$ Now, $*$-homomorphisms are norm-decreasing and, hence, $$ \|a-\alpha^{-1}(\gamma(a))\|=\|\alpha(\alpha^{-1}(a)-\alpha^{-2}(\gamma(a)))\|\leq\|\alpha^{-1}(a)-\alpha^{-2}(\gamma(a))\|<\epsilon $$ for all $a\in F$.

Again, by the normality of $\operatorname{Inn}(A)$, there is a $\delta\in\operatorname{Inn}(A)$, such that $\alpha^{-1}\gamma=\delta\alpha^{-1}$. Therefore, for all $a\in F$, $$ \|a-\delta(\alpha^{-1}(a))\|<\epsilon. $$ Thus, since $\delta^{-1}$ is norm-decreasing, we have for all $a\in F$: $$ \|\alpha^{-1}(a)-\delta^{-1}(a)\|=\|\delta^{-1}(a-\delta(\alpha^{-1}(a)))\|\leq\|a-\delta(\alpha^{-1}(a))\|<\epsilon. $$ Whence, $\delta^{-1}$ is the inner automorphism that approximates $\alpha$ within $\epsilon$ on $F$, so that $\alpha^{-1}\in\overline{\operatorname{Inn}}(A)$.

  • $\begingroup$ I think your argument is ok. Note that automorphisms are isometric, not just "norm decreasing". $\endgroup$ – Martin Argerami Jun 5 '17 at 16:47

One can shorten your proof a little bit:

Let $\beta_n$ be a net of inner automorphisms with $\beta_n \to \alpha$ pointwise. Then

$$ \lVert \alpha(\alpha^{-1}(a)) - \beta_n(\alpha^{-1}(a) \rVert = \lVert a - \beta_n (\alpha^{-1}(a)) \rVert \to 0 \qquad ( a \in A). $$ Using this and the fact that injective $*$-homomorphisms are isometric, we get: $$ \lVert \alpha^{-1}(a)-\beta_n^{-1}(a) \rVert = \lVert \beta_n(\alpha^{-1}(a))-a\rVert \to 0 \qquad ( a \in A). $$


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.