Fredholm Theory — Mathematics Knowledge System

Overview

Fredholm theory studies a class of bounded linear operators called Fredholm operators, which are characterised by having finite-dimensional kernel and cokernel, and closed range. The Fredholm index (dim ker T - dim coker T) is a robust topological invariant. The Fredholm alternative states: either the equation Tx = y has a unique solution for every right-hand side, or the homogeneous equation Tx = 0 has nontrivial solutions (and the inhomogeneous equation has solutions only when y satisfies orthogonality conditions). Fredholm theory is central to the analysis of elliptic PDEs.

Intuition

A Fredholm operator is 'almost invertible': it has a finite-dimensional kernel (failure of injectivity) and finite-dimensional cokernel (failure of surjectivity). The index measures the net failure: index = dim ker - dim coker. For compact perturbations of the identity (I - K with K compact), the Fredholm alternative says exactly: either (I-K) is invertible, or it has a finite-dimensional kernel of the same dimension as the cokernel. This is the precise infinite-dimensional version of the fact that for a square matrix, rank-nullity forces dim ker = dim coker.

Formal Definition

Definition

A bounded linear operator T: X -> Y between Banach spaces is Fredholm if (1) ker T is finite-dimensional, (2) range T is closed in Y, and (3) coker T = Y/range(T) is finite-dimensional. The Fredholm index is ind(T) = dim ker(T) - dim coker(T) = dim ker(T) - codim range(T). The set Fred(X,Y) of Fredholm operators is open in B(X,Y) and the index is locally constant.

\mathrm{ind}(T) = \dim\ker T - \dim\mathrm{coker}\, T

Fredholm index

\mathrm{coker}\, T = Y / \mathrm{range}(T) \cong (\ker T^*)^*

Cokernel

\mathrm{ind}(T + K) = \mathrm{ind}(T) \quad \text{for } K \text{ compact}

Index stability under compact perturbations

I - K \text{ Fredholm of index } 0 \text{ for } K \text{ compact}

Compact perturbations of identity

Notation

Notation	Meaning
$\ker T$	Kernel (null space) of T
$\mathrm{coker}\, T$	Cokernel: Y/range(T)
$\mathrm{ind}(T)$	Fredholm index of T
$\mathrm{Fred}(X,Y)$	Set of Fredholm operators from X to Y

Theorems

Theorem 11.1: Fredholm Alternative

Let K: X -> X be a compact operator on a Banach space X. Then either (I - K) is bijective (and has a bounded inverse), or ker(I - K) is nontrivial (has positive dimension). In the second case, dim ker(I-K) = dim ker(I-K*) < infinity, and range(I-K) = {y in X : f(y) = 0 for all f in ker(I-K*)}.

Theorem 11.2: Stability of Fredholm Index

The Fredholm index is locally constant: if T is Fredholm and S has sufficiently small norm, T + S is Fredholm with ind(T+S) = ind(T). The index is constant on connected components of Fred(X,Y).

Theorem 11.3: Atkinson's Theorem

A bounded operator T: X -> Y is Fredholm if and only if it is invertible modulo compact operators: there exists S: Y -> X with I - ST and I - TS compact.

Theorem 11.4: Index Additivity

If T: X -> Y and S: Y -> Z are Fredholm, then ST: X -> Z is Fredholm and ind(ST) = ind(S) + ind(T).

Worked Examples

1
K is a Hilbert-Schmidt operator (kernel K(s,t) = 1 for s <= t, 0 otherwise, in L^2([0,1]^2)). Hence K is compact.
2
Since K is compact, I - K is a compact perturbation of the identity, hence Fredholm by Theorem 11.1.
3
For compact perturbations of the identity, ind(I-K) = 0.
4
We can verify directly: ker(I-K) = {f : f = Kf}. If f = Kf, then f(t) = integral_0^t f(s)ds, so f'(t) = f(t) and f(0) = 0, giving f = 0. So ker = {0}.
5
By Fredholm alternative (index 0 and trivial kernel), I-K is bijective: unique solution for every right-hand side.

✓ Answer

I - K is Fredholm of index 0 with trivial kernel, hence bijective. The Volterra integral equation always has a unique L^2 solution.

Practice Problems

Hardproof writing

Prove that if T is Fredholm of index k and K is compact, then T + K is Fredholm of index k.

Mediumfree response

State the Fredholm alternative for the equation (I - K)u = f where K is compact, and explain its physical interpretation for integral equations.

MediumMultiple choice

A Fredholm operator of index 0:

Is always bijectiveHas dim ker = dim coker, so is bijective iff it is injectiveHas trivial kernelIs always a compact perturbation of the identity

Common Mistakes

Common Mistake

Thinking Fredholm index 0 implies the operator is bijective.

Index 0 means dim ker = dim coker. The operator is bijective iff additionally ker = {0} (equivalently, it is injective). For example, the zero operator on R^2 has index 0 but is not bijective.

Common Mistake

Confusing the Fredholm alternative with saying 'either uniqueness or existence'.

The Fredholm alternative says: either both existence and uniqueness hold for all right-hand sides, or neither holds for generic right-hand sides (solutions exist only on a proper subspace). The Fredholm alternative identifies exactly when solutions exist in the second case.

Quiz

Historical Background

Ivar Fredholm studied integral equations of the form f(s) - lambda integral K(s,t) f(t) dt = g(s) in 1900, observing a fundamental dichotomy. Hilbert and Riesz reformulated and generalised this in the framework of infinite-dimensional linear algebra. The abstract theory was developed by Atkinson (1951) and others. The Atiyah-Singer index theorem (1963) vastly generalised the Fredholm index to elliptic differential operators on manifolds, connecting it to topological invariants.

1900
Fredholm studies integral equations and discovers the Fredholm alternative
Ivar Fredholm
1918
Riesz formalises Fredholm theory for completely continuous operators
Frédéric Riesz
1951
Atkinson characterises Fredholm operators in terms of invertibility modulo compact operators
F. V. Atkinson
1963
Atiyah-Singer index theorem generalises the Fredholm index
Michael Atiyah, Isadore Singer

Summary

A Fredholm operator has finite-dimensional kernel and cokernel, and closed range.
The Fredholm index ind(T) = dim ker(T) - dim coker(T) is a topological invariant stable under compact perturbations.
Fredholm alternative: for compact K, either (I-K) is bijective, or it has finite-dimensional kernel and the equation (I-K)u=f requires orthogonality conditions on f.
Atkinson's theorem: T is Fredholm iff it is invertible in the Calkin algebra B(X)/K(X).
The Atiyah-Singer index theorem extends the Fredholm index to elliptic differential operators, connecting analysis to topology.

References

BookRudin, W. — Functional Analysis (2nd ed.), Chapter 5
BookBrezis, H. — Functional Analysis, Sobolev Spaces and PDEs, Chapter 6
WebsiteWikipedia: Fredholm operator