Analytic index

Remarks

If $E$ and $F$ carry Hermitian metrics and $M$ carries a smooth positive density, then $\operatorname{coker} P$ is isomorphic to $\ker P^*$, where $P^*$ is the formal adjoint, so $\operatorname{ind} P = \dim \ker P - \dim \ker P^*$. Let $m$ be the order of $P$ and let $H^s$ denote Sobolev spaces. For every real $s$, the extension $P \colon H^{s}(M; E) \to H^{s - m}(M; F)$ is a Fredholm operator. Its Fredholm index equals $\operatorname{ind} P$. The index depends only on the homotopy class of the principal symbol of $P$ through elliptic symbols.

Example: on $S^1 = \R / 2\pi\Z$, let $P = d/d\theta$ on complex functions. The kernel is the constant functions, of dimension $1$. The image is the functions with $\int_0^{2\pi} f \, d\theta = 0$, so the cokernel has dimension $1$, and $\operatorname{ind} P = 0$.

Non-example: the zero operator $C^\infty(S^1) \to C^\infty(S^1)$, viewed as an operator of order $1$, is not elliptic, since its principal symbol vanishes. Its kernel and cokernel are infinite-dimensional, so it has no analytic index.