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. (more…)