Naming isomorphism of bounded operators on Hilbert spaces coming from isomorphism of Hilbert spaces

31 Views Asked by At

Suppose I have an (isometric) isomorphism of separable Hilbert spaces $i:\mathcal{H}_1 \to \mathcal{H}_2$, then I can define $I:\mathcal{B}(\mathcal{H}_1) \to \mathcal{B}(\mathcal{H}_2)$ given by $$ I(A) y = I(A) i(i^{-1}(y)) = i(Ai^{-1}(y)) $$ where $y \in \mathcal{H}_2, A \in \mathcal{B}(\mathcal{H}_1)$.

Do the maps $I$ and $I(A)$ have names?

1

There are 1 best solutions below

0
On BEST ANSWER

This operation is commonly called "conjugation by $i$". Very generally, $bab^{-1}$ is referred to as "$a$ conjugated by $b$" in any context in which the expression makes sense (e.g. elements of a group, or functions where $b$ is a bijection). (Beware that depending on context, "$a$ conjugated by $b$" could also sometimes refer instead to $b^{-1}ab$, but usually there is little risk of confusion in context.)

Other names you might see are simply something like "the isomorphism $\mathcal{B}(\mathcal{H}_1)\to\mathcal{B}(\mathcal{H}_2)$ induced by $i$" or "transport of structure along $i$" (which refers more generally to taking some structure on $\mathcal{H}_1$ and turning it into a structure on $\mathcal{H}_2$ by replacing elements of $\mathcal{H}_1$ with the corresponding elements of $\mathcal{H}_2$ via $i$).