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Jan 24, 2020 at 8:56 vote accept Diego Martinez
Jan 23, 2020 at 17:50 answer added Yemon Choi timeline score: 5
Jan 23, 2020 at 17:44 comment added Robert Israel Hmm, I take that back. I think it is not enough for $V$ to have closed range.
Jan 23, 2020 at 17:27 comment added Yemon Choi @RobertIsrael If the cp map sends 1 to 1 then as Diego points out, the V constructed in Stinespring's representation is an isometry. But I think that by choosing suitable integral operators on C[0,1] one can get positive operators that send 1 to 1 yet do not have closed range.
Jan 23, 2020 at 16:59 comment added Diego Martinez @RobertIsrael if you ever find the time and energy to provide a sketch of the proof that'd be great, because that'd mean that unital cp maps, for instance, do have closed range. Thank you as well
Jan 23, 2020 at 16:58 comment added Diego Martinez @YemonChoi that, erm, makes a lot of sense. What I meant was whether injective cp maps of norm 1 are isometric, but that is also false. Thank you for pointing it out.
Jan 23, 2020 at 16:43 comment added Robert Israel In the Stinespring case, you want $\mathcal B = B(H)$ where $H$ is a Hilbert space, $\pi$ is a *-homomorphism into $B(K)$ for some Hilbert space $K$, and $V: K \to H$. I think in this case you should have closed image if $V$ has closed range.
Jan 23, 2020 at 16:15 comment added Yemon Choi Based on the uiquity of counterexamples (i.e. cp maps with non-closed range) I don't think "under which conditions to cp maps have closed range" is sufficiently focused to make a good MO question: they have closed range when, erm, they have closed range.
Jan 23, 2020 at 16:14 comment added Yemon Choi You should be able to easily find counterexamples for commutative A and B, when cp is the same as usual positivity by Naimark's theorem (see Effros-Ruan or Paulsen's book). I think this is a useful exercise for you to figure out.
Jan 23, 2020 at 16:00 review First posts
Jan 23, 2020 at 18:20
Jan 23, 2020 at 15:58 history asked Diego Martinez CC BY-SA 4.0