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Let U be a unitary transformation from Hilbert space H to Hilbert space K.

How do you call a *-homomorphism f from B(H) to B(K), defined by $f(a) = U a U^{-1}$?

I'm interested both in a symbol, which can be used in formulas, and a name for it, which can be used in texts or in speech.

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Probably, one can denote this f by $U_*$, or something like that. It would be great if it was so, because then it would be possible not to invent a new letter for this homomorphism for every U. –  Fiktor Aug 13 '10 at 18:18

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This homomorphism is often denoted $\mathrm{Ad}(U)$ in the case where $H=K$. The map $\mathrm{Ad}:\mathcal{U}(H)\to\mathrm{Aut}(B(H))$ is a surjective homomorphism from the unitary group of $H$ to the $*$-automorphism group of $B(H)$ with kernel $\mathbb{T}I$. For example, this notation is used in Davidson's C*-algebras by example and in Raeburn and Williams's Morita equivalence and continuous trace C*-algebras. The same notation is often used for the corresponding automorphism of the algebra of compact operators on $H$, even though in that case it is not an inner automorphism unless $H$ is finite dimensional.

I don't know whether the same notation is in common use for the case where $H\neq K$.

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If $H=K$ and $U$ is a unitary, then $f$ is called an inner automorphism of $\mathbb B(H)$, because it is given by conjugation with a unitary element of your algebra $\mathbb B(H)$ (these are the "obviously existing automorphism").

Some notation (for the group case) is proposed here. At least for rings this notation isn't standard, however.

The above naming doesn't nicely generalise to your more general situation and I don't think there's a common name for isomorphisms given by conjugation with a unitary transformation.

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Probably I'm abusing terminology, but by "U is unitary" I mean the following. U is a bounded operator such that it has bounded inverse and $U^{-1}=U^*$. –  Fiktor Aug 13 '10 at 18:12
    
You were right. I should say "unitary transformation" instead of "unitary operator" in order to express what I mean. The question was corrected. –  Fiktor Aug 13 '10 at 18:27

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