Talk:Gelfand–Naimark–Segal construction

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)

Non-degeneracy
Hello,

I was confused at first, because I thought that non-degeneracy means that no vector should vanish under all $$\pi(x)$$. But I think that this is equivalent to the definition given in the article, since
 * $$\bigcap_x \ker \pi(x) =

\bigcap_x (\operatorname{im } \pi(x^*))^\bot = \left(\bigcup_x \operatorname{im } \pi(x^*)\right)^\bot = \left(\bigcup_x \operatorname{im } \pi(x)\right)^\bot $$ 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)


 * 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 $$\pi(x)$$ 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)


 * Hi,
 * I didn't require all $$\pi(x)$$ to be injective, just that for all nonzero $$v \in H$$, there exists an $$x \in A$$ such that $$\pi(x)v \neq 0$$ (This is already satisfied if at least one of the $$\pi(x)$$ is injective.)
 * The last equality holds because the union of something over all $$x \in A$$ is the same as the union over all $$x^* \in A$$, since $$\ ^*$$ is a bijection from $$A$$ to itself. Functor salad 11:05, 24 July 2007 (UTC)


 * ok, you're right. if there is some v that vanishes under all $$\pi(x)$$, then span{v} violates non-degeneracy. Mct mht 17:33, 24 July 2007 (UTC)


 * late late comment: what you wanna say is


 * $$\bigcap_x \ker \pi(x) =

\left(\bigvee_x \operatorname{im } \pi(x)\right)^\bot, $$


 * where the V denotes the linear span. Mct mht 16:24, 13 October 2007 (UTC)


 * 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)

Equivalence

 * 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 $$\psi(a):= \varphi(u^*au) $$ the representation $$ \pi_\psi $$ is equivalent to $$ \pi_\varphi $$ but in general $$\psi$$ will not be a multiple of $$\varphi$$. --- Matthias Lesch, Bonn.