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)

States and representations

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.