2
$\begingroup$

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 $H$).

Now, in the non-equivariant case, Kuiper's theorem states that $\text{Gl}(H)$ is (weakly) contractible. Is this this also true for $\text{Gl}_{G}(H)$?

$\endgroup$
1
  • 1
    $\begingroup$ If $G$ is compact so that $H$ splits up as a topological sum of isotypical components, then each isotypical component has to have infinite multiplicity. Under that assumption it seems to me that one can apply Kuiper's result to each of the components. $\endgroup$ Oct 17, 2011 at 15:59

1 Answer 1

7
$\begingroup$

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.

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.

$\endgroup$
2
  • $\begingroup$ Thank you very much - is it at least true that $\text{GL}_{S^1}(H)$ is path-connected ($H$ a complex Hilbert space as above)? $\endgroup$ Oct 18, 2011 at 9:36
  • 2
    $\begingroup$ @ J. Fabian Meier: assuming $G$ acts unitarily on $H$, the group $GL_G(H)$ is indeed path-connected. Indeed it is the invertible group of the commutant of $G$ in $B(H)$, which is a von Neumann algebra, and any von Neumann algebra $M$ has path-connected invertible group. [Recall the argument: by the polar decomposition, the invertible group retracts on the unitary group; and the unitary group is path-connected because, using a measurable branch of the logarithm on $S^1$, we may express any unitary as $e^{iH}$, for some self-adjoint $H\in M$.] $\endgroup$ Oct 18, 2011 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.