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 $H$ a complex Hilbert space having a countable orthonormal basis. Note that all such Hilbert spaces are isometrically isomorphic to $l^2$. We will denote $B$ the Banach algebra of bounded linear operators on $H$. A Fredholm operator is an operator $T \in B$ with closed range and finite dimensional kernel and cokernel. We will denote the set of Fredholm operators by $F$. The index of a Fredholm operator $T \in F$ is defined as

$\text{ind } T := \dim \ker T - \dim \text{coker } T$.

For example, the situation for $T : V \to V'$ an operator between two finite dimensional spaces, the situation is quite simple: We have $\text{ind } T = \dim V - \dim V'$ since $\dim V = \dim \ker T + \dim \text{Im } T$ and $\dim V' = \dim \text{coker }T + \dim \text{Im }T$.

These operators originally appeared in the theory of integral equations. The condition of being Fredholm means that the equation $Tu = 0$ has a finite dimensional space of solutions and that there are a finite number of linear relations we can impose on $v$ to make sure that $Tu = v$ is solvable (because $v$ has to be in the orthogonal complement of the cokernel).

We can also phrase the index of $T$ in terms of its adjoint. Recall that by Riesz’s representation theorem, for each $T \in B$ we have the adjoint operator $T^* \in B$ that satisfies, for all $u,v \in H$,

$\langle u,T^*v \rangle = \langle Tu, v \rangle$

such that $T \mapsto T^*$ is an isometry. It is an easy exercice to check that $\text{im } T = (\ker T^*)^{\perp}$ and thus $\text{coker } T = \ker T^*$. In fact, this holds for any bounded linear operator with closed range. We can then give an alternative definition: An operator $T \in B$ is Fredholm if its image is closed and both $\ker T$ and $\ker T^*$ are finite dimensional. The index is then $\text{ind } T = \dim \ker T - \dim \ker T^*$. Here is another example:

Proposition: The operator $Id + P$ for $P : H \to H$ a finite-rank operator (ie. an operator with finite dimensional image) is Fredholm with index 0.

Proof: We first show that $Id + P$ is Fredholm. Let $h = \text{im }P$. We proceed in two steps: (1) show that $\ker (Id + P) \subset h$ and then (2) that $\dim \text{coker}(Id + P) \leq \dim h$. For (1), if $u \in \ker(Id + P)$, then for $w \in (\ker P^*)$, we have

$0 = \langle u + Pu, w \rangle = \langle u, w + P^*w \rangle = \langle u, w \rangle$

so $u \in (\ker P^*)^{\perp} = h$. For (2), first note that $\text{coker }(Id + P) = \ker (Id + P^*)$ is orthogonal to $\ker P$. To see this, take $u \in \ker(Id + P^*)$ and $w \in \ker P$. We get

$0 = \langle u + P^*u, w \rangle = \langle u, w + Pw \rangle = \langle u,w \rangle$

as wanted.

Now define a linear map $\varphi : \text{coker }(Id + P) \to \text{im }P$ by $\varphi(v + \text{im}(Id + P)) = Pv$. This is well defined since $P(\text{im}(Id + P)) \subset \text{im}(Id + P)$. Since $\text{coker}(Id + P) \perp \ker P$, 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 $(Id + P)(h) \subset h$. to see this, let $u \in h = (\ker P^*)^{\perp}$ and $w \in \ker P^*$. Then

$\langle u+Pu, w \rangle = \langle u,w \rangle + \langle u, P^*w \rangle = 0$

so $(Id + P)u \in (\ker P^*)^{\perp} = h$. Now since $h$ is finite dimensional, we have $\text{ind}(Id + P)|_h = \dim h - \dim h = 0$ and since $(Id+P)_{H/h} = Id_{H/h}$, it’s index is also 0. Computing the alternating sum of the dimensions we get from the snake lemma, we finally conclude that $\dim \ker(Id + P) = \dim\text{coker}(Id + P)$ so that $\text{ind }(Id + P) = 0$.

QED

Fredholm operators are often encoutered in the study of PDE’s and more classically in the theory of integral equations. Since $\text{im }T = (\ker T^*)^{\perp}$ for $T \in F$, the equation $Tu = v$ is solvable for $u$ iff $v \perp \ker T^*$. In particular, if the index of $T$ is 0, i.e. $\dim \ker T = \dim \ker T^*$, then we get the so-called Fredholm alternative : Either (1) the inhomogeneous equation $Tu = v$ has a unique solution for every $v \in H$ or (2) the homogeneous equation $Tu = 0$ has $k (> 0)$ linearly independant solutions and there are $w_1, \ldots, w_k$ such that if $\langle v, w_j \rangle = 0$ for $j = 1, \ldots, k$ then $Tu = v$ has a solution.

For example, if $K \in L^2(I \times I)$ for $I$ some closed interval, then it can be shown that for $\phi \in L^2(I)$,

$\phi \mapsto \phi + \int_I K(x,y)\phi(y)dy$

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

$\phi(x) + \int_I K(x,y) \phi(y) dy = h(x)$

has a solution if and only if $h(x)$ is $L^2$-orthogonal to every solution $\psi(x)$ of the homogeneous adjoint equation $u(x) + \int_I \overline{K(y,x)}u(y)dy = 0$.

This principle can also be seen to be behind the fact that the laplace equation $\Delta f = g$ can be solved for $f$ on a compact manifold iff $\int g = 0$.

The fact that the integral equation above gives a Fredholm operator of index 0 is a consequence of the compactness of $\phi \mapsto \int K(x,y) \phi(y)dy$ and of the

Theorem (Riesz, 1918): For any compact operator $Q \in K$, the operator $Id + Q$ 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.