It is known that if $\varphi \colon B(H) \to B(K)$ is a unital non-zero normal $*$-homomorphism (for Hilbert spaces $H$ and $K$), then there exists a Hilbert space $K'$ and a unitary $U' \colon K \to H \otimes K'$ such that $\varphi(a) = U^* (a \otimes 1) U$ for all $a \in B(H)$.

More well-known is the corollary is that any unital normal $*$-isomorphism between type I von Neumann algebras is inner; i.e. of the form $a \mapsto U^*aU$ for some unitary $U$.

I can't find a reference for it though. (I'm only looking for a reference, I already have a proof.)

  • $\begingroup$ I unfortunately don't have a reference, but I assume you want the word injective in there somewhere. $\endgroup$ – Josh Keneda Mar 10 at 23:04
  • $\begingroup$ I forgot to write non-zero. Then it’s automatically injective as its kernel is a ultraweakly closed two sided ideal and thus equal $z B(H)$ for some central projection $z$. Only choice is $z=0$ $\endgroup$ – westerbaan Mar 11 at 8:09
  • $\begingroup$ Ah, my mistake. For some reason I thought you were trying to characterize certain unital normal completely positive maps $\phi: B(H) \rightarrow B(K)$, and I knew something was missing. Turns out I just suck at reading. Carry on. :) $\endgroup$ – Josh Keneda Mar 11 at 8:35
  • 2
    $\begingroup$ See Theorem 5.5. of Chapter IV in Takesaki's book. $\endgroup$ – MaoWao Mar 11 at 15:37
  • 1
    $\begingroup$ Well, $\phi$ is a normal $\ast$-homomorphism onto its image, and the image is a von Neumann algebra. $\endgroup$ – MaoWao Mar 14 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.