Fredholm index

Remarks

Example: the right shift $S$ on $\ell^2$ has $\ker S = \set{0}$ and $\operatorname{coker} S$ of dimension $1$, so $\operatorname{ind} S = -1$. The left shift $S^*(x_1, x_2, \ldots) = (x_2, x_3, \ldots)$ is onto with a kernel of dimension $1$, so $\operatorname{ind} S^* = 1$. Non-example: the diagonal operator $D(x_n) = (x_n / n)$ on $\ell^2$ is not Fredholm, so it has no index. For a linear map $T \colon X \to Y$ between finite-dimensional spaces, the rank–nullity theorem gives $\operatorname{ind} T = \dim X - \dim Y$. For Fredholm operators $T \colon X \to Y$ and $S \colon Y \to Z$, the product $ST$ is Fredholm and $\operatorname{ind}(ST) = \operatorname{ind} S + \operatorname{ind} T$. If $K \colon X \to Y$ is a compact operator, then $T + K$ is Fredholm and $\operatorname{ind}(T + K) = \operatorname{ind} T$. The Fredholm operators form an open subset of the bounded operators, and the index is constant on its connected components.