# First introduction to Fredholm operators.

In this post I introduce the notion of a Fredholm operator between two Hilbert spaces, following part 1 of Booss & Bleecker’s book Topology and Analysis. These operators end up providing some very nice bridges between topological ideas and functional analytic ideas. For example the Atiyah-Jänich theorem says that the space of Fredholm operators on the complex Hilbert space represents the K functor in topological K-theory. It is also at the heart of index theory (see the Atiyah-Singer index theorem), where some deep analogies between ideas coming from the study of PDEs and ideas from geometry are established. In the hope of understanding this all one day, let’s start with those Fredholm operators.

We will place ourselves in the context of a complex Hilbert space having a countable orthonormal basis. Note that all such Hilbert spaces are isometrically isomorphic to . We will denote the Banach algebra of bounded linear operators on . A **Fredholm** operator is an operator with closed range and finite dimensional kernel and cokernel. We will denote the set of Fredholm operators by . The **index** of a Fredholm operator is defined as

.

For example, the situation for an operator between two finite dimensional spaces, the situation is quite simple: We have since and .

These operators originally appeared in the theory of integral equations. The condition of being Fredholm means that the equation has a finite dimensional space of solutions and that there are a finite number of linear relations we can impose on to make sure that is solvable (because has to be in the orthogonal complement of the cokernel).

We can also phrase the index of in terms of its adjoint. Recall that by Riesz’s representation theorem, for each we have the adjoint operator that satisfies, for all ,

such that is an isometry. It is an easy exercice to check that and thus . In fact, this holds for any bounded linear operator with closed range. We can then give an alternative definition: An operator is **Fredholm** if its image is closed and both and are finite dimensional. The **index** is then . Here is another example:

Proposition:The operator for a finite-rank operator (ie. an operator with finite dimensional image) is Fredholm with index 0.

**Proof:** We first show that is Fredholm. Let . We proceed in two steps: (1) show that and then (2) that . For (1), if , then for , we have

so . For (2), first note that is orthogonal to . To see this, take and . We get

as wanted.

Now define a linear map by . This is well defined since . Since , this map is injective and (2) follows.

Consider now the following diagram:

Clearly the rows are exact. To see that it is commutative and that the columns are well-defined, it is enough to show that . to see this, let and . Then

so . Now since is finite dimensional, we have and since , it’s index is also 0. Computing the alternating sum of the dimensions we get from the snake lemma, we finally conclude that so that .

QED

Fredholm operators are often encoutered in the study of PDE’s and more classically in the theory of integral equations. Since for , the equation is solvable for iff . In particular, if the index of is 0, i.e. , then we get the so-called **Fredholm alternative** : Either (1) the inhomogeneous equation has a unique solution for every or (2) the homogeneous equation has linearly independant solutions and there are such that if for then has a solution.

For example, if for some closed interval, then it can be shown that for ,

is a Fredholm operator of index 0. It follows that the equation

has a solution if and only if is -orthogonal to every solution of the homogeneous adjoint equation .

This principle can also be seen to be behind the fact that the laplace equation can be solved for on a compact manifold iff .

The fact that the integral equation above gives a Fredholm operator of index 0 is a consequence of the compactness of and of the

**Theorem **(Riesz, 1918): For any compact operator , the operator is Fredholm with index 0.

In the next post, I will prove this after introducing compact operators and then discuss a close relation between compact and Fredholm operators given by Atkinson’s theorem.