# 4 Proving Bott Periodicity II

It remains to show that the green and red arrows are homotopy equivalences. Note that the loop space involves only the identity component. Let $L_\blacksquare ^{k - 1}$ and $G_\blacksquare ^{k - 1}$ denote the components of $L^{k - 1}$ and $G^{k - 1}$ respectively that contain $J_{k - 1}$. Then the fact that the green arrows are homotopy equivalences follow from the following two lemmas:

We have a fiber sequence

$\Omega _{k - 1} \to L_\blacksquare ^{k - 1} \to G_\blacksquare ^{k - 1},$$L_\blacksquare ^{k - 1}$ is contractible.

First observe that $L^{k - 1}_\blacksquare \to G^{k - 1}_{\blacksquare }$ has a local section by restricting the local section of $\mathcal{A} \to \mathcal{A}/\mathcal{K}$, averaging, and then applying unitary retract. So $L_\blacksquare ^{k -1 } \to G_{\blacksquare }^{k - 1}$ is a fiber bundle. So it suffices to show that the fiber through $J_{k - 1}$ is $\Omega _{k - 1}$. It is clear that it is given by $\Omega _{k - 1} \cap L^{k - 1}_{\blacksquare }$. So we want to show that $\Omega _{k - 1} \subseteq L^{k - 1}_{\blacksquare }$.

Since $L_{k - 1}$ is connected (and in fact contractible, by Schur's lemma and Kuiper's theorem), it suffices to show that if $J \in \Omega _{k - 1}$, then $J$ is conjugate to $J_{k - 1}$ by an element in $L_{k - 1}$. Equivalently, we want to show that when $J$ acts as $J_{k - 1}$, there are infinitely many copies of each irrep.

We only have to consider the case when $C_{k - 1}$ is not simple. In this case, we know the projections onto the subspaces spanned by each irrep are given by $\frac{1 \pm w}{2}$. Thus, the image is finite dimensional iff $\frac{1 \pm w}{2}$ is compact. But the $w(J) \equiv w(J_{k - 1})$ modulo compact operators. Since $w(J_{k - 1})$ is not compact, the same is true for $w(J)$.

This would imply the result, since both $L_k$ and $L_{k - 1}$ are contractible.

The second map is given the action $A \mapsto AJ_{k - 1} A^{-1}$, and since $L_{k - 1}$ is connected, the image lies in $L_{\blacksquare }^{k - 1}$. It follows from [4, Theorem 6] that $L_{k - 1} \to L^{k - 1}_\blacksquare$ has a local section, which implies that the orbit of $J_{k - 1}$ open. Since the same would be true for the orbits of all other elements in $L^{k - 1}$ under the conjugation action, the orbit of $J_{k - 1}$ is closed, hence a component, i.e. it is $L^{k - 1}_\blacksquare$. Since the stabilizer of $J_{k - 1}$ is $L_k$, we would be done.

It remains to show that the red arrow is a homotopy equivalence. To do so, we first translate the spaces involved by $J_{k - 1}$, so that the map between them is simply given by exponentiation. Define

$\tilde{F}^k_* = F^k_* J_{k - 1},\quad \tilde{\Omega }_{k - 1} = J_{k - 1} \Omega _{k - 1}.$Then we can characterize $\tilde{F}^k_*$ as the operators $B$ such that

$B$ is Fredholm and skew-Hermitian;

$B$ commutes with $J_1, \ldots , J_{k - 2}$ and anti-commutes with $J_{k - 1}$;

$\| B\| = 1$ and $BB^* = 1$ modulo compact operators.

If $k \equiv -1 \pmod4$, then $J_1 \cdots J_{k - 2} B$ is not essentially definite.

Similarly, $\tilde{\Omega }_{k - 1}$ consists of operators $S$ such that

$S$ is orthogonal;

$S$ commutes with $J_1, \ldots , J_{k - 2}$ and $(J_{k - 1} S)^2 = -1$

$S \equiv -1$ modulo compact operators.

We now want to show that $\exp \pi : \tilde{F}_*^k \to \tilde{\Omega }_{k - 1}$ is a homotopy equivalence. Dropping the tildes, we consider filtrations on $F_*^k$ and $\Omega _{k - 1}$ as follows: We define $\Omega _{k - 1}(n)$ to be those operators $T$ such that $I + T$ has rank $\leq n$, and $F_*^k(n) = (\exp \pi )^{-1} (\Omega _{k - 1}(n))$.

For any $m$ and $a \in \Omega _{k - 1}(m), b \in F^k_*(m)$, the natural inclusion induces bijections

$\begin{aligned} \lim _{n \to \infty } \pi _k(\Omega _{k - 1}(n), a) & \to \pi _k(\Omega _{k - 1}, a)\\ \lim _{n \to \infty } \pi _k(F_*^k(n), b) & \to \pi _k(F_*^k, b). \end{aligned}$Here orthogonality forces the eigenvalues to lie on the unit circle, and the fact that $I + T$ is compact means $-1$ is the only point in the essential spectrum.

Thus, given any angle $\theta$, we can apply a spectral deformation to shrink all the points that are at most $\theta$ away from $-1$ to $-1$, and what is left is something in $\Omega _{k - 1}(n)$ for some $n$, since there are only finitely many eigenvalues at an angle $> \theta$ away from $-1$. By picking a spectral deformation that maps the unit circle to the unit circle and invariant under complex conjugation, this ensures the operator stays orthogonal and real. If we further choose it so that it is odd, then this preserves the set of operators that commute and anti-commute with it. So we stay inside $\Omega _{k - 1}$.

In fact, given any compact subset of $\Omega _{k - 1}$, applying such a spectral deformation will send it to something lying in $\Omega _{k - 1}(n)$ for some large $n$. Thus, given any $a \in \Omega _{k - 1}(m)$, we can pick $\theta$ so that all the eigenvalues of $T$ are either $-1$ or $>\theta$ away from $-1$. Then the associated spectral deformation fixes $a$, but retracts any compact subset (and in particular, any image of maps from spheres) into one of the $\Omega _{k - 1}(n)$.

In the case of $F_k^*(m)$, the spectrum instead looks like

Now it suffices to show that $\exp \pi : F^k_*(m) \to \Omega _{k - 1}(m)$ is a homotopy equivalence for all $m$. For convenience, set $H^k(m) = F^k_*(m) \setminus F^k_*(m - 1)$ and $D^k(m) = \Omega _{k - 1}(m) \setminus \Omega _{k - 1}(m - 1)$. We shall show that $\exp \pi : H^k(m) \to D^k(m)$ is a fiber bundle with contractible fiber for all $m$, and then use the following standard result in homotopy theory:

Let $(Y, Y_1, Y_2)$ and $(Z, Z_1, Z_2)$ be excisive triads, and $f: Y \to Z$ a map of triads. If

$f|_{Y_1}: Y_1 \to Z_1,\quad f|_{Y_2}: Y_2 \to Z_2,\quad f|_{Y_1 \cap Y_2} \to Z_1 \cap Z_2$are all weak homotopy equivalences, then so is $f$.

We shall take

$\begin{aligned} Y & = F^k_*(m) & Y_1 & = F^k_*(m - 1)\\ Z & = \Omega _{k - 1}(m) & Z_1 & = \Omega _{k - 1}(m - 1). \end{aligned}$By a simple spectral deformation argument, we can show that $F^k_*(n - 1)$ is the deformation retract of some open neighbourhood of $F_*^k(n)$ (add to $F_*^k(n - 1)$ those operators whose eigenvalue closest to $-1$ has small real part, then retract those to $-1$), and similarly for $\Omega _{k - 1}(n - 1)$ in a compatible way. We can then $Y_2$ and $Z_2$ to be these neighbourhoods.

By induction, we can conclude that $f|_{Y_1}$ is a homotopy equivalence. Then the fact that $\exp \pi$ restricts to a fiber bundle with contractible fiber shows that the other two maps are homotopy equivalences.

$\exp \pi : H^k(m) \to D^k(m)$ is a fiber bundle with contractible fiber.

Essentially, the “information lost” by passing from $T \in H^k(m)$ to $\exp \pi T \in D^k(m)$ is how the spectrum is distributed between $\pm i$.

To make this concrete, observe that since $A \in D(n)$ is such that $I + A$ has constant rank $n$, it follows that $\ker (I + A)$ is a Hilbert space subbundle $\mathcal{H}$ of $D(n) \times H$ with orthogonal complement $\mathcal{H}^\perp$ an $n$-dimensional vector bundle.

Given $T \in H^k(m)$, it acts on $\mathcal{H}$ and $\mathcal{H}^\perp$ disjointly. Its action on $\mathcal{H}^\perp$ is completely determined by $\exp \pi T$ (by linear algebra), and the action of $T$ on $\mathcal{H}$ is orthogonal, skew-Hermitian, etc. Thus, local triviality of $\mathcal{H}$ and $\mathcal{H}^\perp$ implies $\exp \pi : H^k(m) \to D^k(m)$ is a fiber bundle whose fiber consists of operators $T: H' \to H'$ (where $H'$ is any fiber of $\mathcal{H}$) that are

Orthogonal and square $-1$;

Commuting with $J_1, \ldots , J_{k - 2}$ and anti-commuting with $J_{k - 1}$;

If $k \equiv -1 \pmod4$, then $J_1 \cdots J_{k - 2} T$ is not essentially definite.

Equivalently, if we replace $B$ by $BJ_{k - 1}$ (thus reversing the previous shift), this is the number of choices of $J_k$ such that each irrep appears with infinite multiplicity. Thus, this is given by the quotient of the space of unitary $C_{k - 1}$-automorphisms of $H'$ by the centralizer of $J_k$, i.e. by the space of $C_k$-automorphisms of $H'$. Both of these are contractible. So we are done.