Fredholm index ($\mathrm{ind}T=\dim \ker T-\dim \mathrm{coker}T$)

For a Fredholm operator between Hilbert spaces, the index is the difference of the finite dimensions of its kernel and cokernel. Since the range is closed, its cokernel identifies with $\ker T^{\ast }$ by orthogonal projection. An invertible operator has index zero. Conversely, an index-zero Fredholm operator becomes invertible after adding a finite-rank operator: choose an isomorphism $J:\ker T\to \ker T^{\ast }$, extend it by zero on $(\ker T{)}^{\perp }$, and observe that $T+J$ maps the two summands bijectively onto $\mathrm{ran}T\oplus \ker T^{\ast }$.

An equivariant Fredholm operator between unitary representations of a compact group has finite-dimensional invariant kernel and cokernel. Their formal difference is its index in the representation ring of a compact group. If domain and target decompose into irreducibles with finite multiplicities $m_{\pi },m_{\pi }^{\prime }$, Schur's lemma makes each intertwining block a matrix on the multiplicity spaces. Its rank cancels between kernel and cokernel, giving ${\mathrm{ind}}_{G}T={\sum }_{\pi }(m_{\pi }-m_{\pi }^{\prime })[\pi ]$, with finitely many nonzero terms.

Near a Fredholm operator, split the domain into its kernel and orthogonal complement and the target into its range and orthogonal complement. The large invertible block stays invertible under small norm perturbations. Invertible block elimination reduces the perturbed operator to that block and a map between the finite-dimensional kernel and cokernel spaces. Its index is their dimension difference regardless of its rank. Thus the Fredholm set is open and its index is locally constant.