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 and denote the components of and respectively that contain . Then the fact that the green arrows are homotopy equivalences follow from the following two lemmas:
We have a fiber sequence
Let be the subgroup of consisting of matrices that commute with .
First observe that has a local section by restricting the local section of , averaging, and then applying unitary retract. So is a fiber bundle. So it suffices to show that the fiber through is . It is clear that it is given by . So we want to show that .
Since is connected (and in fact contractible, by Schur's lemma and Kuiper's theorem), it suffices to show that if , then is conjugate to by an element in . Equivalently, we want to show that when acts as , there are infinitely many copies of each irrep.
We only have to consider the case when is not simple. In this case, we know the projections onto the subspaces spanned by each irrep are given by . Thus, the image is finite dimensional iff is compact. But the modulo compact operators. Since is not compact, the same is true for .□
We claim that there is a fiber sequence
This would imply the result, since both and are contractible.
The second map is given the action , and since is connected, the image lies in . It follows from [4, Theorem 6] that has a local section, which implies that the orbit of open. Since the same would be true for the orbits of all other elements in under the conjugation action, the orbit of is closed, hence a component, i.e. it is . Since the stabilizer of is , 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 , so that the map between them is simply given by exponentiation. Define
Then we can characterize as the operators such that
is Fredholm and skew-Hermitian;
commutes with and anti-commutes with ;
and modulo compact operators.
If , then is not essentially definite.
Similarly, consists of operators such that
commutes with and
modulo compact operators.
We now want to show that is a homotopy equivalence. Dropping the tildes, we consider filtrations on and as follows: We define to be those operators such that has rank , and .
For any and , the natural inclusion induces bijections
We only do the case of . The other is similar. Given an element , its spectrum looks roughly like this:
Here orthogonality forces the eigenvalues to lie on the unit circle, and the fact that is compact means is the only point in the essential spectrum.
Thus, given any angle , we can apply a spectral deformation to shrink all the points that are at most away from to , and what is left is something in for some , since there are only finitely many eigenvalues at an angle away from . 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 .
In fact, given any compact subset of , applying such a spectral deformation will send it to something lying in for some large . Thus, given any , we can pick so that all the eigenvalues of are either or away from . Then the associated spectral deformation fixes , but retracts any compact subset (and in particular, any image of maps from spheres) into one of the .
In the case of , the spectrum instead looks like
Now it suffices to show that is a homotopy equivalence for all . For convenience, set and . We shall show that is a fiber bundle with contractible fiber for all , and then use the following standard result in homotopy theory:
Let and be excisive triads, and a map of triads. If
are all weak homotopy equivalences, then so is .□
We shall take
By a simple spectral deformation argument, we can show that is the deformation retract of some open neighbourhood of (add to those operators whose eigenvalue closest to has small real part, then retract those to ), and similarly for in a compatible way. We can then and to be these neighbourhoods.
By induction, we can conclude that is a homotopy equivalence. Then the fact that restricts to a fiber bundle with contractible fiber shows that the other two maps are homotopy equivalences.
is a fiber bundle with contractible fiber.
Visually, the exponential map does something like this:
Essentially, the ``information lost'' by passing from to is how the spectrum is distributed between .
To make this concrete, observe that since is such that has constant rank , it follows that is a Hilbert space subbundle of with orthogonal complement an -dimensional vector bundle.
Given , it acts on and disjointly. Its action on is completely determined by (by linear algebra), and the action of on is orthogonal, skew-Hermitian, etc. Thus, local triviality of and implies is a fiber bundle whose fiber consists of operators (where is any fiber of ) that are
Orthogonal and square ;
Commuting with and anti-commuting with ;
If , then is not essentially definite.
Equivalently, if we replace by (thus reversing the previous shift), this is the number of choices of such that each irrep appears with infinite multiplicity. Thus, this is given by the quotient of the space of unitary -automorphisms of by the centralizer of , i.e. by the space of -automorphisms of . Both of these are contractible. So we are done.□