Theorem of Kuiper for Hilbert spaces with group action - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:22:41Z http://mathoverflow.net/feeds/question/78347 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78347/theorem-of-kuiper-for-hilbert-spaces-with-group-action Theorem of Kuiper for Hilbert spaces with group action J. Fabian Meier 2011-10-17T15:39:10Z 2012-01-14T20:07:44Z <p>Let $H$ be an infinite dimensional separable complex Hilbert space with Lie group action (I am mostly interested in the case $S^1$). Let $\text{Gl}_{G}(H)$ be the space of invertible, bounded and equivariant linear maps (from $H$ to $H$).</p> <p>Now, in the non-equivariant case, Kuiper's theorem states that $\text{Gl}(H)$ is (weakly) contractible. Is this this also true for $\text{Gl}_{G}(H)$?</p> http://mathoverflow.net/questions/78347/theorem-of-kuiper-for-hilbert-spaces-with-group-action/78351#78351 Answer by Alain Valette for Theorem of Kuiper for Hilbert spaces with group action Alain Valette 2011-10-17T16:26:55Z 2011-10-17T16:32:28Z <p>Assume that $G$ acts on $H$ through a unitary irreducible representation. Then by Schur's lemma, $GL_G(H)$ is $\mathbb{C}^\times$, which is of course not contractible.</p> <p>For $G=S^1$: consider the left regular representation on $L^2(S^1)$. Then $GL_G(H)$ is the multiplicative group of bounded sequences with values in $\mathbb{C}^\times$, which is not contractible.</p>