**1. Vector fields and the Euler characteristic**

It is a classical fact that a compact manifold admitting a nowhere vanishing vector field satisfies . One way to prove this is to note that the local flows generated by the vector field are homotopic to the identity, but have no fixed points for small (since the vector field is nonvanishing). By the Lefschetz fixed point theorem, we find that the Lefschetz number of , which is , 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 -dimensional oriented Riemannian manifold , the Euler characteristic can be computed as the index of the elliptic operator

from even-dimensional differential forms to odd-dimensional ones. Here is exterior differentiation and 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

In fact, can be defined on the entire space , and there it is *self-adjoint* (consequently with index zero).

It follows that

The elements in are precisely the *harmonic* differentials (in fact, is a square root of the Hodge Laplacian ), and by Hodge theory these represent cohomology classes on . It follows thus that

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 (or a perturbation thereof) to show that its index is zero.

Theorem 1Let be a vector field on with no zeros. Then .

**2. Clifford algebras**

To see this, it is more convenient to use Clifford multiplication rather than exterior multiplication. Associated to the Riemannian bundle there is a bundle of Clifford algebras, , such that the fiber at each is the Clifford algebra . This is a bundle of -graded vector spaces (even algebras), and as a bundle of -graded vector spaces, it is isomorphic to the exterior algebra of the cotangent bundle.

In fact, for any inner product space , there is an action of the Clifford algebra on the exterior algebra , by which a vector acts by the operator . Here is the operator of wedging with and is contraction with . This structure defines a map

which makes the exterior algebra into a -module. To check this, one simply has to check that the operators satisfy the Clifford commutation relations, which is straightforward.

A useful fact is:

Proposition 2The exterior algebra is a free -module of rank 1, generated by the unit element of the exterior algebra.

In this way, we can actually *identify* the Clifford algebra with the exterior algebra. Associated to the natural -grading of the Clifford algebra is the natural -grading of the exterior algebra.

What if we did this globally? For a manifold , we find that there is an isomorphism of -graded vector bundles

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

**3. The symbol of **

The symbol of the de Rham complex of differential operators

is given by the *Koszul complex*. Namely, on a cotangent vector , one has the complex

where each map is exterior multiplication by . This is precisely the symbol of the de Rham complex, at a cotangent vector . The operator 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 is, at a cotangent vector ,

But acts by left Clifford multiplication by , since . It follows that the symbol of , on a cotangent vector , is just given by left Clifford multiplication by . 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

is given by left Clifford multiplication by , at a cotangent vector .

**4. Symmetries**

Now, our goal is to show that if one has a nowhere zero vector field , then

Atiyah’s idea is to use the vector field to generate a symmetry of —or rather its symbol, which anyway is the only thing that affects the index. Namely, we have the operator of *right* Clifford multiplication, on the exterior algebra bundle (identified with the Clifford algebra). Then has the property that it commutes with left Clifford multiplication, and so the symbol of the elliptic operator

is just left Clifford multiplication by at a cotangent vector .

We have seen, however, that left Clifford multiplication by is precisely the symbol of and, in particular,

have the same symbol. They thus have the same index, which implies

this means that . This proves that .

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.

April 10, 2012 at 7:15 pm

[…] 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 […]