This is the second post devoted to describing some of the ideas in Atiyah’s paper “Vector fields on manifolds.” Last time, we saw that one could prove the classical vanishing of the Euler characteristic on a manifold admitting a nowhere zero vector field using the symmetries of the de Rham complex. In this post, I’ll describe how analogous methods lead to some of the deeper results in the paper.

1. The case of a field of planes

One of the benefits of Atiyah’s idea of using symmetries of differential operators is that it gives us a host of other results, which are not connected with the Lefschetz fixed-point theorem.

For instance:

Theorem 3 Let ${M}$ be a compact manifold admitting an oriented two-dimensional subbundle ${F \subset TM}$. Then ${\chi(M)}$ is even.

The proof of this result starts off as before. Yesterday, we observed that the Euler characteristic of a Riemannian manifold ${M}$ can be obtained as the index of the elliptic operator

$\displaystyle D = d + d^* : \Omega^{even}(M) \rightarrow \Omega^{odd}(M).$

The operator ${D}$ (obtained by “rolling up” the de Rham complex, whose index is precisely ${\chi(M)}$) is a map of the global sections ${\bigwedge^{even} T^*M \rightarrow \bigwedge^{odd} T^* M}$. As we saw yesterday, the symbol of this operator ${D}$ is precisely given by left Clifford multiplication. In other words, the symbol of ${D}$ at a cotangent vector ${v \in T_x^* M}$ is precisely given by left Clifford multiplication ${L_v}$ by ${v}$ on ${\bigwedge^{even} T^*_x M = \mathrm{Cl}^0(T^*_x M)}$.

The method Atiyah uses to construct symmetries of ${D}$ is to use the simple observation that left and right Clifford multiplication commute. This enables him to construct an operator ${I}$ commuting with ${D}$ such that ${I^2 = - 1}$, thus—approximately—endowing the kernel and cokernel of ${D}$ with a complex structure.

Proof: Let ${F \subset T^* M}$ be a two-dimensional vector subbundle; we identify the tangent and cotangent bundles via a Riemannian metric. We can define a global section of ${\mathrm{Cl}(F) \subset \mathrm{Cl}(T^*M)}$ as follows. If ${e_1, e_2}$ is a local oriented orthonormal frame for ${F}$, then consider the Clifford product ${s = e_1 . e_2}$. This is well-defined: it doesn’t depend on the orthonormal frame. Consequently, ${s}$ is a global section of ${\mathrm{Cl}(F)}$.

Observe that the Clifford product ${s^2}$ is identically ${e_1 e_2 e_1 e_2 =-1}$ by the anticommutation relations.

Now consider the ${\mathbb{Z}/2}$-graded operator

$\displaystyle R_s: \mathrm{Cl}(T^*M) \rightarrow \mathrm{Cl}(T^* M)$

given by right Clifford multiplication by ${s}$. Then ${R_s^2 \equiv -1}$. Since the symbol of ${R_s}$ is rightClifford multiplication by ${s}$, we find that the symbol of ${D}$ commutes with ${R_s}$.

The idea is to use ${R_s}$ to obtain a complex structure on the kernel and cokernel of ${D}$. However, ${R_s}$ itself does not commute with ${D}$, only to first order. But instead we can consider

$\displaystyle D' = \frac{1}{2}(D + R_s D R_s^{-1}).$

Then ${D'}$ has the same symbol as ${D}$ and honestly commutes with ${R_s}$. Consequently, the kernel and cokernel of ${D'}$ are endowed with complex structures, which gives

$\displaystyle \mathrm{index} D' \in 2 \mathbb{Z}.$

However, since ${D}$ and ${D'}$ share the same symbol, we have

$\displaystyle \mathrm{index} D' = \mathrm{index} D = \chi(M) \in 2 \mathbb{Z} .$

$\Box$

2. The signature operator

There are a number of subtler results proved in the paper using the same method: one has a given geometric invariant as an index of an elliptic operator. The symbol of this elliptic operator commutes with right Clifford multiplication. Given the existence of fields of tangent planes (or something like that), one uses right Clifford multiplication to produce an action of a Clifford algebra on the kernel and cokernel of an “averaged” version of this elliptic operator, which still has the symbol. But the dimensions of representations of a Clifford algebras satisfy divisibility relations, and these give divisibility relations on the relevant geometric invariant.

An example is the so-called signature operator, whose index gives the signature. Let ${M}$ be a compact, oriented manifold of dimension ${4k}$. The signature of ${M}$, ${\sigma(M)}$, is the signature of the quadratic form on ${H^{2k}(M; \mathbb{R})}$ given by the cup product followed by integration over ${M}$. It is a crucial observation that ${\sigma(M)}$ is actually the index of an elliptic operator constructed out of ${TM}$.

Let us assume that ${M}$ is endowed with a Riemannian structure, as usual. Then on ${2k}$-dimensional forms, one has the Hodge star

$\displaystyle \ast : \Omega^{2k}(M) \rightarrow \Omega^{2k}(M).$

It has the property that

$\displaystyle \int \omega \wedge \ast \omega = \int (\omega, \omega) d\mathrm{vol} \geq 0,$

where ${(\cdot, \cdot)}$ is the wedge product metric on ${\wedge^{2k} T_x^* M}$. The integral is even positive if ${\omega \not\equiv 0}$. It follows in particular that the ${1}$-eigenspace for ${\ast}$ consists of forms ${\omega}$ such that ${\int \omega \wedge \omega > 0}$, while the ${-1}$-eigenspace for ${\ast}$ consists of forms ${\omega}$ such that ${\int \omega \wedge \omega < 0}$. Since ${\ast^2 =1}$ on ${\Omega^{2k}}$, we get a complete decomposition of ${\Omega^{2k}(M)}$.

In fact, the action of ${\ast}$ preserves the harmonic ${2k}$-dimensional forms ${\mathcal{H}^{2k}}$, which form a space isomorphic to ${H^{2k}(M; \mathbb{R})}$. It follows that we get a decomposition

$\displaystyle \mathcal{H}^{2k} = \mathcal{H}^{2k}_+ \oplus \mathcal{H}^{2k}_-$

of the harmonic ${2k}$-forms into positive and negative eigenspaces for ${\ast}$. Moreover,

$\displaystyle \sigma(M) = \dim \mathcal{H}^{2k}_+ - \dim \mathcal{H}^{2k}_-$

since the cup square is positive definite on ${\mathcal{H}^{2k}_+}$ and negative definite on ${\mathcal{H}^{2k}_-}$.

The idea now is to find an elliptic operator whose kernel is precisely ${\mathcal{H}^{2k}_+}$ and whose cokernel (or whose adjoint’s kernel) is precisely ${\mathcal{H}^{2k}_-}$. In fact, the elliptic operator will precisely be ${ d + d^*}$, but on a different splitting of ${\mathrm{Cl}(T^* M) = \bigwedge^\bullet T^* M}$. Let ${w}$ be the volume form, so in the Clifford algebra,

$\displaystyle w = e_1 \dots e_{4k}$

for ${e_1, \dots, e_{4k}}$ a local orthonormal frame. Then ${w}$ has the property that, in the Clifford algebra,

$\displaystyle w^2 = (e_1 \dots e_{4k})^2 = (-1)^{s}$

where ${s}$ is the number of times we have to move an ${e_i }$ past an ${e_j}$ (with ${i < j}$); it is thus the number of pairs ${(i, j) \in \left\{1, 2, \dots, 4k\right\}}$ with ${i < j}$. So

$\displaystyle s = \frac{4k(4k-1)}{2} = 2k (4k-1).$

Thus the sign is ${1}$: ${w^2 = 1}$.

In particular, the operator ${L_w}$ of left multiplication by ${w}$ has the property that ${L_w^2 = 1}$: ${L_w}$ is an involution. For our purposes, we need the involution ${\tau = (-1)^q L_w}$ (which just takes care of a sign). It defines a splitting

$\displaystyle \mathrm{Cl}(T^* M) = \mathrm{Cl}(T^* M)_+ \oplus \mathrm{Cl}(T^* M)_-$

into positive and negative eigenspaces for ${\tau}$. The operator ${d + d^*}$ anticommutes with ${L_w}$ (and thus ${\tau}$); this is because ${d + d^*= \sum e_i . \nabla_{e_i}}$ and the product ${e_1 \dots e_{4k}}$ is parallel and anticentral (with respect to the Levi-Civita connection).

Consequently, ${d + d^*}$ defines an operator

$\displaystyle D^+ : \mathrm{Cl}(T^* M)_+ \rightarrow \mathrm{Cl}(T^* M)_-$

(or, rather, on the space of sections of these bundles). This is analogous to the operator that gave the Euler characteristic, except we have used a different ${\mathbb{Z}/2}$-grading of the Clifford (or exterior) bundle.

In view of the next proposition, we can call ${D^+}$ the signature operator.

Proposition 4 The index of ${D^+ = D |_{\mathrm{Cl}(T^*M)_+}}$ is precisely the signature ${\sigma(M)}$.

Proof: Since ${d + d^* }$ is self-adjoint, the index of ${D^+}$ is precisely the difference

$\displaystyle \dim \ker (d + d^*)|_{\mathrm{Cl}(T^* M)_+} - \dim \ker (d + d^*)|_{\mathrm{Cl}(T^* M)_-} .$

In other words, we have to look at the space of all harmonic forms in ${\mathrm{Cl}(T^* M)_+}$ and the space of harmonic forms in ${\mathrm{Cl}(T^*M)_-}$, and subtract the dimensions. The claim is that we will get precisely

$\displaystyle \sigma(M) = \dim \mathcal{H}^{2k}_+ - \dim \mathcal{H}^{2k}_-$

using the previous notation.

A potential problem is that the harmonic forms in ${\mathrm{Cl}(T^* M)_+}$ include ${p}$-forms for ${p \neq 2k}$, which must be counted in the index. However, these don’t contribute. If ${\mathcal{H}_p}$ denotes harmonic ${p}$-forms, then ${\tau}$ acts on

$\displaystyle \mathcal{H}_p \oplus \mathcal{H}_{4k-p}$

by switching the two factors (if ${p \neq 2k}$). This means that the positive and negative eigenspaces of ${\tau}$ on ${\mathcal{H}_p \oplus \mathcal{H}_{4k-p} }$ have equal dimension (they are the graphs of ${\tau}$ and ${-\tau}$, respectively). It follows that, when computing the index of ${D^+}$, we need only look at ${2k}$-forms after all.

But, on ${2k}$-forms, ${\tau}$ is precisely the Hodge star ${\ast}$. Thus the splitting induced by ${\tau}$ on ${2k}$-dimensional forms is precisely the splitting described earlier, and the index of ${D^+}$ can be computed as ${\dim \mathcal{H}_{2k}^+ - \dim \mathcal{H}_{2k}^-}$. As we have seen, this is ${\sigma(M)}$. $\Box$

3. Divisibility conditions

Finally, I’ll describe the divisibility conditions imposed on the signature of manifolds admitting lots of independent vector fields.

Theorem 5 Let ${M}$ be a manifold admitting ${r}$ linearly independent vector fields. Then ${\sigma(M)}$ is divisible by ${a_r}$, where ${a_r}$ is the dimension of an irreducible representation of the Clifford algebra ${\mathrm{Cl}(\mathbb{R}^r)}$.

Proof: In fact, we suppose given ${r}$ linearly independent vector fields ${X_1, \dots, X_r}$ on ${M}$; we can assume them orthonormal at each point. These define operators ${R_i}$ of right Clifford multiplication by ${X_i}$ on the Clifford bundle ${\mathrm{Cl}(T^*M)}$. Since the ${R_i}$ commute with ${\tau}$, we get an induced action of the ${R_i}$ on the splitting

$\displaystyle \mathrm{Cl}(T^* M) = \mathrm{Cl}(T^* M)_+ \oplus \mathrm{Cl}(T^* M)_-$

induced by ${\tau}$.

Moreover, the ${R_i}$ commute with the symbol of ${D^+}$ because, again, left Clifford multiplication commutes with right Clifford multiplication. This suggests that a suitable average of ${D^+}$ will be equivariant for the group generated by the ${R_i}$, and this is indeed the case.

In fact, the ${R_i}$‘s satisfy the relations

$\displaystyle R_i R_j = - R_j R_i, \ \text{for } i \neq j, \quad R_i^2 = -1.$

They thus generate a finite group ${G}$ containing ${\pm 1}$. The group algebra of ${G}$, when quotiented by the relation that the element ${-1 \in G}$ is identified with ${-1 \in \mathbb{C}}$, is precisely the Clifford algebra ${\mathrm{Cl}(\mathbb{R}^r)}$.

The average

$\displaystyle D' = \frac{1}{|G|} \sum_{\sigma \in G} \sigma D^+ \sigma^{-1}$

is thus a ${G}$-equivariant operator. We find that

$\displaystyle \ker D', \mathrm{coker } D'$

are representations of ${\mathrm{Cl}(\mathbb{R}^r)}$, and in particular have dimensions divisible by ${a_r}$.

Thus

$\displaystyle \mathrm{index} D^+ = \mathrm{index } D' \in a_r \mathbb{Z},$

where, again, we use the fact that ${D^+, D'}$ have the same symbol. $\Box$

By the classification of Clifford algebras, the ${a_r}$ grow in ${r}$ fairly rapidly. For ${r =1 , 2, \dots, 8}$, one has ${a_r = 1, 2, 4, 4, 8, 8, 8, 8}$, and by the periodicity thereafter ${a_{r+8} = 16 a_r}$. Thus, one gets fairly strong divisibility conditions on the signature.

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