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If an automorphism $\alpha$ of a C*-algebra $A$ is inner then whenever $A$ is a subalgebra of another C*-algebra $B$, $\alpha$ obviously extends to $B$.

Is the converse true: if an automorphism $\alpha$ of $A$ is such that whenever $A \subset B$ then $\alpha$ extends to an automorphism of $B$, is $\alpha$ necessarily inner?

The analogous question for groups has a positive solution, see: Are the inner automorphisms the only ones that extend to every overgroup?

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This is by no means a full answer, but Kishimoto has shown in Theorem 4.1 of his paper "Universally weakly inner one-parameter automorphism groups" that for an automorphism $\alpha$ of a separable $C^*$-algebra the following statements are equivalent:

(1) $\alpha$ is extendible in every irreducible representation
(2) $\alpha$ is universally weakly inner

If I read the last property correctly, then this means that there is a $u \in A^{**}$ such that $\alpha(a) = uau^*$.

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    $\begingroup$ On the other hand, in "Outer automorphisms and reduced crossed products of simple C*-algebras", Kishimoto shows that for simple separable C*-algebras, universally weakly inner automorphisms are conjugation by a unitary in the multiplier algebra. Together with Theorem 4.1 above, this answers Aaron's question for separable, unital, simple C*-algebras. $\endgroup$ – Leonel Robert Jan 6 '15 at 21:47

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