2.2 Kuiper's theorem
We shall take a short break from Fredholm operators, and prove the following extremely important theorem:
(Kuiper, 1964, [uiper196519)
Let be a separable infinite-dimensional (real or complex) Hilbert space. Then for any compact space , we have . In particular, is weakly contractible.
To prove this, we first establish a useful “move”.
Let be a fixed space, and continuous functions. Then the maps
are homotopic as maps .
The mental picture we should have in mind is that we are allowed to “rotate”
from the top-left corner to the bottom-right corner. The proof, unsurprisingly, is a literal rotation.
for going from to .
If is such that there is an infinite-dimensional subspace with for all , then is homotopic to the constant map .
In the decomposition
, the matrix of
for some invertible matrix . We can linearly homotopy to kill off the on the bottom left corner. We then write as the infinite sum of infinite-dimensional Hilbert subspaces, so that the matrix now looks like this:
Using the fact that , the previous lemma tells us we have homotopies
We now apply the lemma again, but “bracketing” in a different way, to obtain
So the idea is to pick some subspace and modify until the is the identity on . In general, suppose we have another map , and we want to modify so that for all . The key idea is that we can do so as long as and have orthogonal images, in which case we just do a simple rotation, using to identify the two subspaces we want to rotate. I must emphasize that the actual result and proofs are much easier and more intuitive than how I wrote them down, but I couldn't find a better way to do so.
Suppose , , and are such that for all and is an isomorphism onto its image. Then is homotopic to a map such that .
. Then the map
is homotopic to the identity. Indeed, we can achieve this if and weren't there by a rotation, and we just have to conjugate that homotopy by . Then take .
Thus, if we can decompose such that , then the theorem follows by performing two swaps — first homotope so that , and then swap the image back to the identity on .
We may assume the image of is contained in a finite-dimensional subspace of .
This is a special case of the following more general fact: Let
be a normed vector space and
an open set. Suppose
are points in
are such that
. Then there is a deformation retract of
into the simplicial complex spanned by
. Indeed, pick a partition of unity
Then we can use the deformation
and the hypothesis ensures this remains in .
There exists an orthogonal decomposition such that for all , and and are both infinite-dimensional.
Pick a vector
, put it in
. The span of all
is finite-dimensional, so we can put finitely many vectors in
. Pick any vector orthogonal to the vectors we have chosen and put it in
Next, pick that is orthogonal to everything chosen so far, and also so that is orthogonal to everything chosen so far for all . This is possible since these conditions only exclude a finite-dimensional subspace. Keep going on countably many times, and afterwards if there is anything left, put it in .