Fredholm index

Let $P$ be a Fredholm operator.The index of $P$ is defined as

$index(P)$$=$$dim\ker (P)-dimcoker(P)$

$=$$dim\ker (P)-dim\ker (P^{*}).$

Note: this is well defined as $\ker (P)$ and $\ker (P^{*})$ are finite-dimensionalvector spaces, for $P$ Fredholm.

• $index(P^{*})=-index(P)$.

• $index(P+K)=index(P)$ for any compact operator $K$.

• If $P_1:{\mathscr{H}}_1\to {\mathscr{H}}_2$ and $P_2:{\mathscr{H}}_2\to {\mathscr{H}}_3$are Fredholm operators, then $index(P_2{P}_1)=index(P_1)+index(P_2)$.

• If $t\to P_t$, $t\in [0,1]$ is a norm continuous path of Fredholm operators,then $index(P_t)=index(P_0)$.

Fredholm operators of the form $\mathrm{𝑖𝑛𝑣𝑒𝑟𝑡𝑖𝑏𝑙𝑒}+\mathrm{𝑐𝑜𝑚𝑝𝑎𝑐𝑡}$ have index zero.