Talk:Gelfand-Naimark-Segal construction

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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)

[edit] Non-degeneracy

Hello,

I was confused at first, because I thought that non-degeneracy means that no vector should vanish under all $π(x)$ . 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

π(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

π(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

π(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