Fredholm operator

A closed operator is Fredholm when it has closed range and finite-dimensional kernel and cokernel. It is upper semi-Fredholm when only closed range and finite kernel are required, and lower semi-Fredholm when only closed range and finite cokernel are required.

For a smooth Hilbert-space map $F$ with $F(0)=0$ and Fredholm derivative $L$, split the source into $\ker L$ and a complement and the target into $\mathrm{im}L$ and a finite-dimensional cokernel. The implicit function theorem solves the range component uniquely as a graph over $\ker L$. The remaining equation is a smooth map $\kappa :\ker L\to \mathrm{coker}L$ with $\kappa (0)=d\kappa (0)=0$. Thus the local zero set is ${\kappa }^{-1}(0)$. Compact symmetry groups permit equivariant splittings.

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A bounded operator is Fredholm if and only if it has a bounded two-sided inverse modulo compact operators. For the forward implication invert its restriction from the orthogonal kernel complement to its closed range and extend by zero. Conversely a parametrix makes the identity compact on the kernel, forcing finite dimension. A failure of a lower bound on the kernel complement would yield unit vectors $x_n$ with $T{x}_n\to 0$; the compact error then forces a norm-convergent subsequence, whose nonzero limit lies both in the kernel and its complement. This proves closed range; applying the other parametrix identity to the adjoint gives finite cokernel.

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.

For a Fredholm operator $L$ on a Hilbert space, invert its restriction from the kernel complement to its closed range, extending by zero on the range complement. The resulting bounded $B$ has $BL=I-P$ and $LB=I-Q$ with finite-rank projections. Adding a compact operator $K$ gives two-sided inverses modulo compact errors. Such a parametrix forces finite kernel, a lower bound on the kernel complement, closed range and finite adjoint kernel; the latter is the cokernel. Repeating with $-K$ proves the equivalence without assuming self-adjointness.

Several essential spectra distinguish failure of upper and lower semi-Fredholm properties. For $T=A-zI$, these failures are detected at zero by the positive self-adjoint operators $T^{\ast }T$ and $T{T}^{\ast }$.

For a bounded linear operator on an infinite-dimensional complex Hilbert space, define the essential spectrum by failure of the Fredholm property for $T-\lambda I$. The Atkinson theorem identifies it with the spectrum in the Calkin algebra, hence it is compact and nonempty. On a finite-dimensional Hilbert space it is empty. For a bounded real diagonal operator on a separable infinite-dimensional space it equals the subsequential cluster set of the diagonal entries, including values of infinite multiplicity.

Let $A$ be a closed operator and let $V:D(A)\to H$ be compact when the domain carries its graph norm. If $A+V$ is closed on the same domain, compact perturbation invariance of Fredholm operators applied to the maps $D(A)\to H$ proves invariance of the Fredholm essential spectrum. For self-adjoint operators this is also the essential spectrum obtained by deleting isolated finite-multiplicity eigenvalues. A bounded potential tending to zero at spatial infinity is relatively compact with respect to a second-order constant-coefficient operator: local Rellich-Kondrachov compactness theorem gives compactness on bounded intervals and the small tails give compactness globally.

For bounded $L$ and compact $K$, use the Fredholm convention ${\Sigma }_{\mathrm{e}}(L)=\{z:L-zI\text{ is not Fredholm}\}$. Applying compact perturbation invariance of Fredholm operators to every shift proves the equality. This general theorem needs no self-adjointness; other definitions of essential spectrum for nonnormal operators must be distinguished.

For a bounded self-adjoint operator $L$ on a complex Hilbert space, defineThis equals the spectrum of a bounded operator after removing its isolated eigenvalues of finite multiplicity. To see isolation when the kernel is finite and the shifted range is closed, restrict the shifted operator to its kernel complement: it is invertible there by image-kernel orthogonality for an adjoint. A Neumann-series perturbation leaves nearby nonzero shifts invertible on that complement and on the finite kernel. The closed range must concern the shifted operator, not the original one.

If $L$ is bounded and self-adjoint on a complex Hilbert space and $K$ is compact and self-adjoint, thenA singular Weyl sequence for $L$ stays singular for $L+K$, because compact operators send weak convergence to norm convergence and therefore $K{f}_n\to 0$. Applying the same argument with $-K$ proves the reverse inclusion. Finite-multiplicity isolated eigenvalues can move under such perturbations; the essential spectrum of a bounded self-adjoint operator is unchanged.

The discrete spectrum consists of isolated eigenvalues of finite algebraic multiplicity.