5
$\begingroup$

I already posted my question on mathstackexchange

For a separable Hilbert space $H$ it is known that the unitary group $U(H)$ is contractible, both for the norm topology (Kuiper's theorem) and for the strong operator topology (Dixmier and Douady, 1963). Note that the strong operator topology coincides with the compact open topology in this case.

Does somebody know whether contractibility of the unitary group of at least some special pre Hilbert spaces equipped with the strong operator topology or the compact open topology still holds?

The example I have in mind is the underlying algebraic Fock space $\bigwedge PH \oplus (PH^\perp)^*$ of the Fock space.

Thanks in advance, StW

$\endgroup$
3
  • $\begingroup$ Is the algebraic Fock space locally contractible? $\endgroup$ Commented Jun 16, 2015 at 9:21
  • 2
    $\begingroup$ All locally convex topological vector spaces are locally contractible: use straight line homotopy. $\endgroup$ Commented Jun 16, 2015 at 11:33
  • $\begingroup$ You say: Yes, because the algebraic Fock space is locally convex. But in general, a pre Hilbert space does not even need to be locally pathconnected. $\endgroup$ Commented Jun 17, 2015 at 7:56

1 Answer 1

7
$\begingroup$

The proof of contractibility in the strong(=weak=compact-open) topology is very easy:

Identify $H$ with $L^2([0,1])$.
The path $\{u_t\}$ that connects any unitary $u$ to the identity element is then given by:

$$u_t:L^2([0,1])= L^2([0,t])\oplus L^2([t,1])\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$ $$\qquad\qquad\qquad \xrightarrow{\varphi_t u \varphi_t^{-1}\,\oplus\, id} L^2([0,t])\oplus L^2([t,1])=L^2([0,1]) $$

were $\varphi_t:L^2([0,1])\to L^2([0,t])$ is the unitary given by the obvious reparametrization.


To answer your question, I would say that it's a case-by-case analysis:

Examples:

For an infinite algebraic direct sum of Hilbert spaces, I can mimic the above argument and make it work: use $\bigoplus^\infty L^2([0,1])$ and do the same shrinking to $\bigoplus^\infty L^2([0,t])$.

For the $n$-th algebraic tensor power of a Hilbert space, I can also make it work. Again, it's a matter of finding a 1-parameter family of embeddings of the pre-Hilbert space into itself that "shrinks it to zero". Think about $L^2([0,1]^n)$ shrinking to $L^2([0,t]^n)$ and note that it preserves the subspace given by the $n$-th algebraic tensor power of $L^2([0,1])$.

For exterior powers, it's the same as for tensor powers.

For your particular example of the algebraic Fock space, that same strategy also works, and so the unitary group is strong-contractible. Think of $\mathcal F$ as given by by $\bigoplus_{n=0}^\infty\bigwedge^n (L^2([0,1]))$ and note that you can shrink it to $\bigoplus_{n=0}^\infty\bigwedge^n (L^2([0,t]))$. Once again, this shrinking procedure preserves the "algebraic" subspace of the Hilbert space $\mathcal F$.

$\endgroup$

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.