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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
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 mht04:48, 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 salad11:05, 24 July 2007 (UTC)reply
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.