1. Vector fields and the Euler characteristic

It is a classical fact that a compact manifold ${M}$ admitting a nowhere vanishing vector field satisfies ${\chi(M) = 0}$. One way to prove this is to note that the local flows ${\phi_\epsilon}$ generated by the vector field are homotopic to the identity, but have no fixed points for ${\epsilon }$ small (since the vector field is nonvanishing). By the Lefschetz fixed point theorem, we find that the Lefschetz number of ${\phi_\epsilon}$, which is ${\chi(M)}$, must vanish.

There is another way of proving this theorem, which uses the theory of elliptic operators instead of the Lefschetz fixed-point theorem. On any ${n}$-dimensional oriented Riemannian manifold ${M}$, the Euler characteristic can be computed as the index of the elliptic operator $\displaystyle D = d + d^* : \Omega^{even}(M) \rightarrow \Omega^{odd}(M)$

from even-dimensional differential forms to odd-dimensional ones. Here ${d}$ is exterior differentiation and ${d^*}$ the formal adjoint, which comes from the metric. One way to see this is to observe that the elliptic operator thus defined is just a “rolled up” version of the usual de Rham complex $\displaystyle 0 \rightarrow \Omega^0(M) \rightarrow \Omega^1(M) \rightarrow \dots.$

In fact, ${d + d^*}$ can be defined on the entire space ${\Omega^\bullet(M)}$, and there it is self-adjoint (consequently with index zero).

It follows that $\displaystyle \mathrm{index}D = \dim \ker D - \dim \mathrm{coker }D = \dim \ker (d + d^*)|_{\Omega^{even}(M)}- \dim \ker (d + d^*)|_{\Omega^{odd}(M)}.$

The elements in ${\ker d + d^*}$ are precisely the harmonic differentials (in fact, ${d + d^*}$ is a square root of the Hodge Laplacian ${dd^* + d^* d}$), and by Hodge theory these represent cohomology classes on ${M}$. It follows thus that $\displaystyle \mathrm{index} D = \dim H^{even}(M) - \dim H^{odd}(M).$

Atiyah’s idea, in his paper “Vector fields on manifolds,” is to use the existence of a nowhere vanishing vector field to get a symmetry of ${D}$ (or a perturbation thereof) to show that its index is zero.

Theorem 1 Let ${X}$ be a vector field on ${M}$ with no zeros. Then ${\chi(M) =0}$.

2. Clifford algebras

To see this, it is more convenient to use Clifford multiplication rather than exterior multiplication. Associated to the Riemannian bundle ${TM}$ there is a bundle of Clifford algebras, ${\mathrm{Cl}(TM)}$, such that the fiber at each ${x \in M}$ is the Clifford algebra ${\mathrm{Cl}(T_x M)}$. This is a bundle of ${\mathbb{Z}/2}$-graded vector spaces (even algebras), and as a bundle of ${\mathbb{Z}/2}$-graded vector spaces, it is isomorphic to the exterior algebra ${(T^*M)^\bullet}$ of the cotangent bundle.

In fact, for any inner product space ${V}$, there is an action of the Clifford algebra ${\mathrm{Cl}(V)}$ on the exterior algebra ${\bigwedge^\bullet V}$, by which a vector ${v}$ acts by the operator ${e_v - \iota_v}$. Here ${e_v}$ is the operator of wedging with ${v}$ and ${\iota_v}$ is contraction with ${v}$. This structure defines a map $\displaystyle \mathrm{Cl}(V) \otimes \bigwedge^\bullet V \rightarrow \bigwedge^\bullet V,$

which makes the exterior algebra ${\bigwedge^\bullet V}$ into a ${\mathrm{Cl}(V)}$-module. To check this, one simply has to check that the operators ${e_v - \iota_v}$ satisfy the Clifford commutation relations, which is straightforward.

A useful fact is:

Proposition 2 The exterior algebra ${\bigwedge^\bullet V}$ is a free ${\mathrm{Cl}(V)}$-module of rank 1, generated by the unit element ${1}$ of the exterior algebra.

In this way, we can actually identify the Clifford algebra with the exterior algebra. Associated to the natural ${\mathbb{Z}/2}$-grading of the Clifford algebra is the natural ${\mathbb{Z}/2}$-grading of the exterior algebra.

What if we did this globally? For a manifold ${M}$, we find that there is an isomorphism of ${\mathbb{Z}/2}$-graded vector bundles $\displaystyle \mathrm{Cl}(T^* M) \simeq \bigwedge^\bullet T^* M.$

We thus can talk about Clifford multiplication of forms, just as honestly as exterior multiplication.

3. The symbol of ${D}$

The symbol of the de Rham complex of differential operators $\displaystyle 0 \rightarrow \Omega^0(M) \stackrel{d}{\rightarrow} \Omega^1(M) \stackrel{d}{\rightarrow} \dots$

is given by the Koszul complex. Namely, on a cotangent vector ${v \in T^*_x M}$, one has the complex $\displaystyle 0 \rightarrow \mathbb{R} \stackrel{e_v}{\rightarrow} T^*_x M \stackrel{e_v}{\rightarrow} \wedge^2 T^*_x M \rightarrow \dots$

where each map is exterior multiplication by ${v}$. This is precisely the symbol of the de Rham complex, at a cotangent vector ${v \in T^*_x M}$. The operator ${D}$ was obtained by “rolling up” the de Rham complex, namely by taking the differential in the de Rham complex and adding it to its adjoint. All this is compatible with passage to symbols (except one picks up a sign from taking the adjoint), and we find that the symbol of ${D}$ is, at a cotangent vector ${v}$, $\displaystyle \bigwedge^{even} T^*_xM \stackrel{e_v - e_v^*}{\rightarrow} \bigwedge^{odd} T^*_x M.$

But ${e_v - e_v^*}$ acts by left Clifford multiplication by ${v}$, since ${e_v^* = \iota_v}$. It follows that the symbol of ${D}$, on a cotangent vector ${v}$, is just given by left Clifford multiplication by ${v}$. ${D}$ is a special (and relatively easy) case of the so-called “Dirac operator,” which makes sense on any Clifford bundle.

In a similar manner, we find that the symbol of the formal adjoint $\displaystyle D^* = d + d^*: \Omega^{odd}( M) \rightarrow \Omega^{even}(M)$

is given by left Clifford multiplication by ${v}$, at a cotangent vector ${v}$.

4. Symmetries

Now, our goal is to show that if one has a nowhere zero vector field ${X}$, then $\displaystyle \mathrm{index} D = 0.$

Atiyah’s idea is to use the vector field ${X}$ to generate a symmetry of ${D}$—or rather its symbol, which anyway is the only thing that affects the index. Namely, we have the operator ${R_X}$ of right Clifford multiplication, on the exterior algebra bundle ${\bigwedge^\bullet T^* M}$ (identified with the Clifford algebra). Then ${R_X}$ has the property that it commutes with left Clifford multiplication, and so the symbol of the elliptic operator $\displaystyle R_X D R_X^{-1}: \Omega^{odd} (M) \rightarrow \Omega^{even}(M)$

is just left Clifford multiplication by ${v}$ at a cotangent vector ${v}$.

We have seen, however, that left Clifford multiplication by ${v}$ is precisely the symbol of ${D^* = d+ d^*: \Omega^{odd}( M) \rightarrow \Omega^{even}(M) }$ and, in particular, $\displaystyle R_X D R_X^{-1}, D^*$

have the same symbol. They thus have the same index, which implies $\displaystyle \mathrm{index}D = \mathrm{index} R_X D R_X^{-1} = \mathrm{index} D^* = - \mathrm{index}D;$

this means that ${\mathrm{index} D = 0}$. This proves that ${\chi(M) = 0}$.

In the rest of the paper, Atiyah uses more sophisticated versions of this idea to prove a number of congruences on the signature of manifolds admitting fields of tangent planes.