2.2 Kuiper's theorem

We shall take a short break from Fredholm operators, and prove the following extremely important theorem:

To prove this, we first establish a useful “move”.

The mental picture we should have in mind is that we are allowed to “rotate” $S$ from the top-left corner to the bottom-right corner. The proof, unsurprisingly, is a literal rotation.

We pick

$$H_t = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix} \begin{pmatrix} S & 0 \\ 0 & I \end{pmatrix} \begin{pmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{pmatrix} \begin{pmatrix} T & 0 \\ 0 & I \end{pmatrix}$$

for $t$ going from $0$ to $\frac{\pi }{2}$.

In the decomposition $H = H_0^\perp \oplus H_0$, the matrix of $f(x)$ looks like

$$\begin{pmatrix} Q & 0 \\*& I \end{pmatrix}.$$

for some invertible matrix $Q = Q(x)$. We can linearly homotopy $\tilde{f}$ to kill off the $*$ on the bottom left corner. We then write $H_0$ as the infinite sum of infinite-dimensional Hilbert subspaces, so that the matrix now looks like this:

Using the fact that $I=Q^{-1}Q$, 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 $H_0 \subseteq H$ and modify $f$ until the $f$ is the identity on $H_0$. In general, suppose we have another map $g(x): H_0 \to H$, and we want to modify $f$ so that $f(x)|_{H_0} = g(x)$ for all $x$. The key idea is that we can do so as long as $g(x)$ and $f(x)$ have orthogonal images, in which case we just do a simple rotation, using $g(x) f(x)^{-1}$ 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.

For each $x$, decompose $H = f(x)H_0 \oplus g(x) H_0 \oplus H_x$ for some $H_x$. Then the map

$$\begin{aligned} \varphi _x: f(x) H_0 \oplus g(x) H_0 \oplus H_x & \to f(x) H_0 \oplus g(x)H_0 \oplus H_x\\ f(x) a \oplus g(x) b \oplus c & \mapsto f(x) b \oplus (-g(x)a) \oplus c. \end{aligned}$$

is homotopic to the identity. Indeed, we can achieve this if $f(x)$ and $g(x)$ weren't there by a rotation, and we just have to conjugate that homotopy by $f(x) \oplus g(x) \oplus I_{H_x}$. Then take $\tilde{f}(x) = \varphi _x^{-1} f(x)$.

Thus, if we can decompose $H = H_0 \oplus H_1 \oplus H_2$ such that $f(x)(H_0) \perp H_2$, then the theorem follows by performing two swaps — first homotope $f$ so that $f(x)(H_0) = H_2$, and then swap the image back to the identity on $H_0$.

This is a special case of the following more general fact: Let $V$ be a normed vector space and $U \subseteq V$ an open set. Suppose $x_1, \ldots , x_n$ are points in $U$ and $\varepsilon _i > 0$ are such that $B(x_i, 3 \varepsilon _i) \subseteq U$. Then there is a deformation retract of $U_* = \bigcup _{i = 1}^n B(x_i, \varepsilon _i)$ into the simplicial complex spanned by $x_1, \ldots , x_n$. Indeed, pick a partition of unity

$$\begin{aligned} \psi _j(x) & = \max (\varepsilon _i - \| x - x_j\| , 0),\\ \phi _j(z) & = \frac{\psi _j(z)}{\sum _{k = 1}^N \psi _k(z)}. \end{aligned}$$

Then we can use the deformation

$$g_t(x) = (1 - t) x + t \sum _{j = 1}^N \phi _j(x) x_j,$$

and the hypothesis ensures this remains in $\mathrm{GL}(H)$.

Pick a vector $a_1$, put it in $H_1$. The span of all $f(x)(a_1)$ is finite-dimensional, so we can put finitely many vectors in $H_2$ so that $f(x)(a_1) \in H_1 \oplus H_2$ for all $x$. Pick any vector orthogonal to the vectors we have chosen and put it in $H_3$.

Next, pick $a_2$ that is orthogonal to everything chosen so far, and also so that $f(x)(a_2)$ is orthogonal to everything chosen so far for all $x$. 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 $H_2$.