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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.

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  • $\begingroup$ 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
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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 $\alpha^{-1}\in\overline{\operatorname{Inn}}(A)$.

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  • $\begingroup$ I think your argument is ok. Note that automorphisms are isometric, not just "norm decreasing". $\endgroup$ – Martin Argerami Jun 5 '17 at 16:47
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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: $$ \lVert \alpha^{-1}(a)-\beta_n^{-1}(a) \rVert = \lVert \beta_n(\alpha^{-1}(a))-a\rVert \to 0 \qquad ( a \in A). $$

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