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Given a decomposition $H=H_1\otimes H_2$ of a Hilbert space $H$ into the tensor product of the Hilbert spaces $H_1$ and $H_2$ and a *-isomorphism $U: B(H_0)\to B(H)$, where $H_0$ is another Hilbert space, the equation $$J(A):=U(A)\otimes 1~~~~~~(1)$$ defines an injective *-homomorphism $J: B(H_0)\to B(H)$.

Is the converse always true? I.e., is every injective *-homomorphism $J: B(H_0)\to B(H)$, where $H_0$ and $H$ are Hilbert spaces, of the form (1) for some decomposition of $H$? If not, which conditions characterize these homomorphisms?

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    $\begingroup$ What is $H_0$? Is your question the following: does a $\ast$-homomorphism like that imply that $H$ splits as a tensor product? That's true. $\endgroup$
    – Mateusz Wasilewski
    Oct 25, 2019 at 9:50
  • $\begingroup$ I improved my question. Why is it true (reference?) $\endgroup$
    – Arnold Neumaier
    Oct 25, 2019 at 10:47
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    $\begingroup$ It follows from the fact that all representations of the algebra of compact operators are multiples of the canonical identity representation, and the restricted representation uniquely determines the corresponding representation of the algebra of bounded operators (representations extend uniquely from ideals). A good reference is Arveson's book "An invitation to C*-algebras" (see page 15 for the uniqueness of the extension and Corollary 1. on page 20 for the classification of representations of compact operators). $\endgroup$
    – Mateusz Wasilewski
    Oct 25, 2019 at 11:15
  • $\begingroup$ @MateuszWasilewski: Is there also a direct proof, avoiding representation theory? $\endgroup$
    – Arnold Neumaier
    Oct 25, 2019 at 11:28
  • $\begingroup$ Not that I know of, but maybe there is one. But, as a consequence, this direct approach would imply the classification of representations, so I don't think it can be much simpler. $\endgroup$
    – Mateusz Wasilewski
    Oct 28, 2019 at 8:43

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