Talk:Unitary representation

definition of complete reducibility

by definition, a unitary representation π is a mapping from G to U(H). then, what is an "invariant subspace" under π? or is it meant that π(A) is completely reducible for all A in G? in this case it would make sense to talk about "invariant subspace" under π(A).

An 'invariant subspace' for the action of G is a subspace invariant under π(g) for all g in G. Algebraist 22:04, 11 February 2008 (UTC)[reply]

"See also Unitary representation of a real Lie algebra"

is such a letdown! 131.111.55.75 (talk) 00:46, 8 May 2008 (UTC)[reply]

But going around in circles is such fun!--CSTAR (talk) 02:35, 8 May 2008 (UTC)[reply]

Smooth and analytic vectors

The sentence on compatibility with smooth/analytic structures of a Lie group is wrong and misleading. The correct statement is that smooth vectors are dense (because convolution by smooth functions of compact support gives a dense set of smooth vectors) and that analytic vectors are dense (by the argument of Nelson, Garding and Roe Goodman using the heat kernel of an elliptic element in the universal enveloping algebra). More precisely they form a common core for Lie algebra elements. There seems to be no article on smooth vectors or analytic vectors (see e,g, Garth Warner's book) on WP, which is unfortunate, because I needed them for the article on Zonal spherical functions. Perhaps the material exists in some spectral theory article. I will check this and modify this article appropriately. Mathsci (talk) 15:32, 16 June 2008 (UTC)[reply]

Your recent edit to the article on "Unitary Representations"

Smooth vectors are dense in H by a classical argument of Lars Gårding, since convolution by smooth functions of compact support yields smooth vectors.

Can you explain (preferably within the article) how it is possible to convolve with functions when all we have is an abstract Hilbert space? What are the functions and how are they related to the vectors of the Hilbert space? 84.227.20.67 (talk) 17:10, 12 August 2015 (UTC)[reply]

I thought you had problems with compact support. I just made a blue link for it, that's all. YohanN7 (talk) 17:54, 12 August 2015 (UTC)[reply]

Fair enough. In any case I found the answer here: http://math.stackexchange.com/questions/398881/a-technical-problem-on-constructing-smooth-vectors-in-a-representation. The averaging is done against a compactly supported smooth function on the group. It produces an "approximate identity" operator which can then be applied to a vector. The result is a smooth vector. 84.227.20.67 (talk) 20:08, 12 August 2015 (UTC)[reply]

I suppose a reference to Lars Gårding's original result wouldn't out of place. If someone knows which it is and where to find it, ... YohanN7 (talk) 20:20, 12 August 2015 (UTC)[reply]