Talk:Gelfand–Naimark–Segal construction

Latest comment: 17 years ago by Functor salad in topic Non-degeneracy

The article doesn't define what a cyclic representation is, though it defines a cyclic vector and its relation to a cyclic representation...

how about one with a cyclic vector, :-)? Mct mht 07:51, 9 July 2007 (UTC)Reply

Non-degeneracy

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Hello,

I was confused at first, because I thought that non-degeneracy means that no vector should vanish under all  . But I think that this is equivalent to the definition given in the article, since

 

So the intersection of the kernels is trivial iff the union of the images is dense, or am I missing something? Functor salad 19:26, 19 July 2007 (UTC)Reply

that doesn't look right. why does the last equality hold? non-degeneracy means the Hilbert space is as small as can be. your condition that all operators   be injective is much more restrictive. take a full concrete algebra of bounded operators on some Hilbert space. this is already a nondegenerate representation but fails to satisfy your requirement. Mct mht 04:48, 24 July 2007 (UTC)Reply
Hi,
I didn't require all   to be injective , just that for all nonzero  , there exists an   such that   (This is already satisfied if at least one of the   is injective.)
The last equality holds because the union of something over all   is the same as the union over all  , since   is a bijection from   to itself. Functor salad 11:05, 24 July 2007 (UTC)Reply
ok, you're right. if there is some v that vanishes under all  , then span{v} violates non-degeneracy. Mct mht 17:33, 24 July 2007 (UTC)Reply
late late comment: what you wanna say is
 
where the V denotes the linear span. Mct mht 16:24, 13 October 2007 (UTC)Reply
That's the same thing as what I said, since the orthogonal subspace is a linear subspace anyway. Functor salad 20:43, 13 October 2007 (UTC)Reply

States and representations

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Equivalence

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The above shows that there is a bijective correspondence between positive linear functionals and cyclic representations. Two cyclic representations πφ and πψ with corresponding positive functionals φ and ψ are unitarily equivalent if and only if φ = α ψ for some positive number α.

This is definitely wrong. If u is a unitary element of A then for   the representation   is equivalent to   but in general   will not be a multiple of  . --- Matthias Lesch, Bonn.