Clifford Algebras and Bott Periodicity — Proving Bott Periodicity II

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 Lk1L_\blacksquare ^{k - 1} and Gk1G_\blacksquare ^{k - 1} denote the components of Lk1L^{k - 1} and Gk1G^{k - 1} respectively that contain Jk1J_{k - 1}. Then the fact that the green arrows are homotopy equivalences follow from the following two lemmas:

Lemma

We have a fiber sequence

Ωk1Lk1Gk1, \Omega _{k - 1} \to L_\blacksquare ^{k - 1} \to G_\blacksquare ^{k - 1},

Lemma

Lk1L_\blacksquare ^{k - 1} is contractible.

Proof of first lemma

Let LkL_ k be the subgroup of LL consisting of matrices that commute with J1,,JkJ_1, \ldots , J_ k.

First observe that Lk1Gk1L^{k - 1}_\blacksquare \to G^{k - 1}_{\blacksquare } has a local section by restricting the local section of AA/K\mathcal{A} \to \mathcal{A}/\mathcal{K}, averaging, and then applying unitary retract. So Lk1Gk1L_\blacksquare ^{k -1 } \to G_{\blacksquare }^{k - 1} is a fiber bundle. So it suffices to show that the fiber through Jk1J_{k - 1} is Ωk1\Omega _{k - 1}. It is clear that it is given by Ωk1Lk1\Omega _{k - 1} \cap L^{k - 1}_{\blacksquare }. So we want to show that Ωk1Lk1\Omega _{k - 1} \subseteq L^{k - 1}_{\blacksquare }.

Since Lk1L_{k - 1} is connected (and in fact contractible, by Schur's lemma and Kuiper's theorem), it suffices to show that if JΩk1J \in \Omega _{k - 1}, then JJ is conjugate to Jk1J_{k - 1} by an element in Lk1L_{k - 1}. Equivalently, we want to show that when JJ acts as Jk1J_{k - 1}, there are infinitely many copies of each irrep.

We only have to consider the case when Ck1C_{k - 1} is not simple. In this case, we know the projections onto the subspaces spanned by each irrep are given by 1±w2\frac{1 \pm w}{2}. Thus, the image is finite dimensional iff 1±w2\frac{1 \pm w}{2} is compact. But the w(J)w(Jk1)w(J) \equiv w(J_{k - 1}) modulo compact operators. Since w(Jk1)w(J_{k - 1}) is not compact, the same is true for w(J)w(J).

Proof of second lemma

We claim that there is a fiber sequence

LkLk1Lk1. L_ k \to L_{k - 1} \to L_\blacksquare ^{k - 1}.

This would imply the result, since both LkL_ k and Lk1L_{k - 1} are contractible.

The second map is given the action AAJk1A1A \mapsto AJ_{k - 1} A^{-1}, and since Lk1L_{k - 1} is connected, the image lies in Lk1L_{\blacksquare }^{k - 1}. It follows from [ 4 , Theorem 6 ] that Lk1Lk1L_{k - 1} \to L^{k - 1}_\blacksquare has a local section, which implies that the orbit of Jk1J_{k - 1} open. Since the same would be true for the orbits of all other elements in Lk1L^{k - 1} under the conjugation action, the orbit of Jk1J_{k - 1} is closed, hence a component, i.e. it is Lk1L^{k - 1}_\blacksquare . Since the stabilizer of Jk1J_{k - 1} is LkL_ 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 Jk1J_{k - 1}, so that the map between them is simply given by exponentiation. Define

F~k=FkJk1,Ω~k1=Jk1Ωk1. \tilde{F}^ k_* = F^ k_* J_{k - 1},\quad \tilde{\Omega }_{k - 1} = J_{k - 1} \Omega _{k - 1}.

Then we can characterize F~k\tilde{F}^ k_* as the operators BB such that

  1. BB is Fredholm and skew-Hermitian;

  2. BB commutes with J1,,Jk2J_1, \ldots , J_{k - 2} and anti-commutes with Jk1J_{k - 1};

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

  4. If k1(mod4)k \equiv -1 \pmod4, then J1Jk2BJ_1 \cdots J_{k - 2} B is not essentially definite.

Similarly, Ω~k1\tilde{\Omega }_{k - 1} consists of operators SS such that

  1. SS is orthogonal;

  2. SS commutes with J1,,Jk2J_1, \ldots , J_{k - 2} and (Jk1S)2=1(J_{k - 1} S)^2 = -1

  3. S1S \equiv -1 modulo compact operators.

We now want to show that expπ:F~kΩ~k1\exp \pi : \tilde{F}_*^ k \to \tilde{\Omega }_{k - 1} is a homotopy equivalence. Dropping the tildes, we consider filtrations on FkF_*^ k and Ωk1\Omega _{k - 1} as follows: We define Ωk1(n)\Omega _{k - 1}(n) to be those operators TT such that I+TI + T has rank n\leq n, and Fk(n)=(expπ)1(Ωk1(n))F_*^ k(n) = (\exp \pi )^{-1} (\Omega _{k - 1}(n)).

Lemma

For any mm and aΩk1(m),bFk(m)a \in \Omega _{k - 1}(m), b \in F^ k_*(m), the natural inclusion induces bijections

limnπk(Ωk1(n),a)πk(Ωk1,a)limnπk(Fk(n),b)πk(Fk,b).\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}

Proof

We only do the case of Ωk1\Omega _{k - 1}. The other is similar. Given an element TΩk1T \in \Omega _{k - 1}, its spectrum looks roughly like this:

\begin{tikzpicture} 
      \draw [gray] circle [radius=1];
      \fill (-1, 0) circle [radius=0.08];
      \foreach \rot in {3, 5, 8, 12, 17, 23, 30} {
        \pgfmathsetmacro\scal{0.01 * ln(\rot)};
        \fill [rotate=\rot] (-1, 0) circle [radius=\scal];
        \fill [rotate=-\rot] (-1, 0) circle [radius=\scal];
      }
    \end{tikzpicture}

Here orthogonality forces the eigenvalues to lie on the unit circle, and the fact that I+TI + T is compact means 1-1 is the only point in the essential spectrum.

Thus, given any angle θ\theta , we can apply a spectral deformation to shrink all the points that are at most θ\theta away from 1-1 to 1-1, and what is left is something in Ωk1(n)\Omega _{k - 1}(n) for some nn, since there are only finitely many eigenvalues at an angle >θ> \theta away from 1-1. 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 Ωk1\Omega _{k - 1}.

In fact, given any compact subset of Ωk1\Omega _{k - 1}, applying such a spectral deformation will send it to something lying in Ωk1(n)\Omega _{k - 1}(n) for some large nn. Thus, given any aΩk1(m)a \in \Omega _{k - 1}(m), we can pick θ\theta so that all the eigenvalues of TT are either 1-1 or >θ>\theta away from 1-1. Then the associated spectral deformation fixes aa, but retracts any compact subset (and in particular, any image of maps from spheres) into one of the Ωk1(n)\Omega _{k - 1}(n).

In the case of Fk(m)F_ k^*(m), the spectrum instead looks like

\begin{tikzpicture} 
      \fill (0, 1) circle [radius=0.08];
      \fill (0, -1) circle [radius=0.08];
      \draw [gray] (0, 1) -- (0, -1);
      \foreach \rot in {3, 5, 8, 12, 17, 23, 30} {
        \pgfmathsetmacro\scal{0.008 * ln(\rot)};
        \pgfmathsetmacro\y{1 - 0.015 * \rot};
        \fill (0, \y) circle [radius=\scal];
        \fill (0, -\y) circle [radius=\scal];
      }
    \end{tikzpicture}

Now it suffices to show that expπ:Fk(m)Ωk1(m)\exp \pi : F^ k_*(m) \to \Omega _{k - 1}(m) is a homotopy equivalence for all mm. For convenience, set Hk(m)=Fk(m)Fk(m1)H^ k(m) = F^ k_*(m) \setminus F^ k_*(m - 1) and Dk(m)=Ωk1(m)Ωk1(m1)D^ k(m) = \Omega _{k - 1}(m) \setminus \Omega _{k - 1}(m - 1). We shall show that expπ:Hk(m)Dk(m)\exp \pi : H^ k(m) \to D^ k(m) is a fiber bundle with contractible fiber for all mm, and then use the following standard result in homotopy theory:

Lemma

Let (Y,Y1,Y2)(Y, Y_1, Y_2) and (Z,Z1,Z2)(Z, Z_1, Z_2) be excisive triads, and f:YZf: Y \to Z a map of triads. If

fY1:Y1Z1,fY2:Y2Z2,fY1Y2Z1Z2 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 ff.

We shall take

Y=Fk(m)Y1=Fk(m1)Z=Ωk1(m)Z1=Ωk1(m1).\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 Fk(n1)F^ k_*(n - 1) is the deformation retract of some open neighbourhood of Fk(n)F_*^ k(n) (add to Fk(n1)F_*^ k(n - 1) those operators whose eigenvalue closest to 1-1 has small real part, then retract those to 1-1), and similarly for Ωk1(n1)\Omega _{k - 1}(n - 1) in a compatible way. We can then Y2Y_2 and Z2Z_2 to be these neighbourhoods.

By induction, we can conclude that fY1f|_{Y_1} is a homotopy equivalence. Then the fact that expπ\exp \pi restricts to a fiber bundle with contractible fiber shows that the other two maps are homotopy equivalences.

Theorem

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

Proof

Visually, the exponential map does something like this:

\begin{tikzpicture} 
      \fill [blue!30!red] (0, 1) circle [radius=0.08];
      \fill [blue!30!red] (0, -1) circle [radius=0.08];
      \draw [gray] (0, 1) -- (0, -1);
      \foreach \rot in {12, 17, 23, 30} {
        \pgfmathsetmacro\scal{0.01 * ln(\rot)};
        \pgfmathsetmacro\y{1 - 0.015 * \rot};
        \pgfmathsetmacro\col{\rot*2};

        \fill [blue!\col!red] (0, \y) circle [radius=\scal];
        \fill [blue!\col!red] (0, -\y) circle [radius=\scal];
      }

      \draw [->] (1, 0) -- (3, 0);
      \begin{scope}[shift={(5, 0)}]
        \draw [gray] circle [radius=1];
        \fill [blue!30!red] (-1, 0) circle [radius=0.08];
        \foreach \rot in {12, 17, 23, 30} {
          \pgfmathsetmacro\col{\rot*2};
          \pgfmathsetmacro\scal{0.01 * ln(\rot)};
          \fill [blue!\col!red, rotate=\rot] (-1, 0) circle [radius=\scal];
          \fill [blue!\col!red, rotate=-\rot] (-1, 0) circle [radius=\scal];
        }
      \end{scope}
    \end{tikzpicture}

Essentially, the ``information lost'' by passing from THk(m)T \in H^ k(m) to expπTDk(m)\exp \pi T \in D^ k(m) is how the spectrum is distributed between ±i\pm i.

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

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

  1. Orthogonal and square 1-1;

  2. Commuting with J1,,Jk2J_1, \ldots , J_{k - 2} and anti-commuting with Jk1J_{k - 1};

  3. If k1(mod4)k \equiv -1 \pmod4, then J1Jk2TJ_1 \cdots J_{k - 2} T is not essentially definite.

Equivalently, if we replace BB by BJk1BJ_{k - 1} (thus reversing the previous shift), this is the number of choices of JkJ_ k such that each irrep appears with infinite multiplicity. Thus, this is given by the quotient of the space of unitary Ck1C_{k - 1}-automorphisms of HH' by the centralizer of JkJ_ k, i.e. by the space of CkC_ k-automorphisms of HH'. Both of these are contractible. So we are done.