Alot of the material I'm reading lately seems to mention Fredholm operators and the 'Fredholm alternative' and operators being 'Fredholm of index $0$'.

Can someone give me a high level overview of what's the reason for caring whether an operator is Fredholm or not? What does it enable us to do with the operator? Is being Fredholm of index $0$ a good thing or a bad thing? Would be prefer an operator to be, say, Fredholm of index $2$ for example?


A Fredholm operator $T$ is an operator for which the solutions of the nonhomogeneous linear problem $Tx = y$ can be described using "finitely many pieces of data" just like in the finite dimensional case even though the operator acts on a possible infinite dimensional space. More explicitly, if $T$ is Fredholm then $\ker(T)$ is finite dimensional and so we can (at least theoretically) find a finite basis $v_1, \dots, v_n$ for $\ker T$. Since the cokernel is also finite dimensional, we can find finitely many linear functionals $\varphi^1, \dots, \varphi^k$ such that $y \in \operatorname{Im}(T)$ if and only if $\varphi^1(y) = \dots \varphi^k(y) = 0$. Then:

  1. The equation $Tx = y$ has a solution if and only if $\varphi^1(y) = \dots = \varphi^k(y) = 0$.
  2. If the equation $Tx = y$ has a solution, it has a finite affine space of solutions given by $x_0 + \left< v_1, \dots, v_n \right>$ where $x_0$ is some arbitrary particular solution to the inhomogeneous problem (that is, $Tx_0 = y$).

The index of $T$ is $n - k$. The important thing about the index is that unlike the quantities $k$ and $n$ alone, it is invariant under compact perturbations and it is a continuous map on the (open) set of Fredholm operators. In fact, the connected components are precisely in bijection with the index.

The fact that $k$ and $n$ are not invariant along families but $n - k$ is can already be seen in the finite dimensional case. If $T \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ is a linear map, it is automatically Fredholm of index zero. When $\lambda$ runs over the complex numbers, the kernel of $T - \lambda I$ is usually trivial and $T - \lambda I$ is onto unless $\lambda$ hits an eigenvalue and then the dimension of the kernel jumps up (by the geometric multiplicity of $\lambda$) while the dimension of the image jumps down (by the same quantity, a consequence of the rank-nullity theorem) so the index stays zero.

In the infinite dimensional case, we don't have the rank-nullity theorem and in general there is no relation between the dimensions of the kernel of $T$ and the image of $T$ (which is usually infinite dimensional). However, for Fredholm operators we have the next best thing - for a Fredholm operator $T \colon X \rightarrow Y$ of index $L$, the relation between the dimensions of the kernel and the cokernel is described by $L$ and if $X = Y$ and $T - \lambda I$ stays Fredholm for all $\lambda$ then the quantity by which the dimension of the kernel of $T - \lambda I$ jumps up when $\lambda$ hits an eigenvalue and the "dimension of the image jumps down" is also given by $L$. If $L = 0$, the jumps balance out. If $L > 0$, $T - \lambda I$ can't be injective. If $T - \lambda I$ is surjective, we know that the dimension of the kernel is precisely $L$. In general, the dimension of the kernel can jump from $L$ to $L + r$ (with $r > 0$) but then the "dimension of the image will jump down" by $r$ to compensate (more precisely, the dimension of the cokernel will also jump up from $0$ to $r$). If $L < 0$, $T - \lambda I$ can't be surjective and a similar analysis follows.

For me, the most important examples of Fredholm operators come from elliptic operators. In that context, the fact that an elliptic operator is Fredholm is tied closely with elliptic regularity. Consider for example a second order elliptic operator $L \colon H^2(\Omega) \cap H^1_0(\Omega) \rightarrow L^2(\Omega)$ where $\Omega \subseteq \mathbb{R}^n$ is nice enough. In that context, the fact that $L$ is Fredholm is pretty much equivalent to the estimate

$$ ||u||_{H^2(\Omega)} \leq C \left( ||Lu||_{L^2(\Omega)} + ||u||_{L^2(\Omega)} \right) $$

which lies at the basis of elliptic regularity theory that guarantees that the solutions to the equation $Lu = v$ are as regular as $v$ and $L$ allows. The inequality above implies in a purely formal way (without knowing anything about PDEs, only a little about Sobolev spaces) that $L$ is semi-Fredholm (has finite dimensional kernel and closed image) and if you know that $L$ is (semi-)Fredholm, you can deduce the inequality above.

Finally, another thing worth noting is that while the dimensions of the kernel and cokernel of a Fredholm operator are often impossible to compute explicitly, there are many methods to compute the index which is a much coarser invariant. For example, determining the multiplicity of a specific eigenvalue for the Laplacian with Dirichlet boundary conditions on a domain in $\mathbb{R}^n$ with $n \geq 2$ is usually impossible but since the Laplacian is self-adjoint, the index is zero. The index does give you some information in some cases. For example, a Fredholm operator of index $2$ must have kernel of dimension at least $2$ so if you want many solutions to $Lu = 0$, you might prefer to have a positive index. You might not be able to determine the dimension of the kernel but if you can determine the index, you will at least have a lower bound.

  • $\begingroup$ Excellent explanation of Fredholm operators, the best I've seen thanks! Is there a slight mix-up in paragraph 4? For $L = \text{dim ker}(T) - \text{dim coker}(T)$ to be greater than zero shouldn't it read - If $L > 0$, $T - \lambda I$ can't be injective, the dimension of the kernel can jump up by $\textbf{r + L}$ but then the "dimension of the image will jump down" by $\textbf{r}$ to compensate (more precisely, the dimension of the cokernel will also jump up by $\textbf{r}$). $\endgroup$ – csss Jan 24 '17 at 15:30
  • $\begingroup$ I just posted a related question if you get a chance to take a look: math.stackexchange.com/questions/2112065/… $\endgroup$ – csss Jan 24 '17 at 17:00
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    $\begingroup$ @csss: You're right, I've edited my answer. Thanks for the comment and the error report. $\endgroup$ – levap Jan 24 '17 at 19:46

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