# Fredholm Operators

One way of obtaining K-theory classes is to take a ${\displaystyle 2}$-term complex of vector bundles

${\displaystyle E_{1}{\xrightarrow {d}}E_{2}}$
we get get a K-theory class ${\displaystyle [E_{1}]-[E_{2}]=[ker(d)]-[Im(d)]}$. In fact all K-theory classes occur like this if we allow ${\displaystyle E_{i}}$ to be infinite dimensional. This motivates the study of Fredholm operators.

Fix ${\displaystyle V}$ to be a separable Hilbert space over ${\displaystyle \mathbb {C} }$. [A vector space ${\displaystyle V}$ with an inner product that induces a complete metric on V].

Definition 13.2

Let ${\displaystyle T:V\rightarrow V}$ be a bounded linear map (the image of the unit ball is bounded).

1. We say that ${\displaystyle T}$ is Fredholm if both ${\displaystyle Ker(T)}$ and ${\displaystyle Coker(T)}$ are finite dimensional.
2. The index of a Fredolm operator ${\displaystyle T}$ is defined ${\displaystyle index(T)=dim(Ker(T))-dim(Coker(T))}$.
3. ${\displaystyle \mathbf {\mathcal {F}} :=q\{}$ Fredholm operators ${\displaystyle :V\rightarrow V\}}$.

Exercise 13.5

Consider the Hilbert space ${\displaystyle L^{2}(\mathbb {Z} _{+})=\{(a_{0},a_{1},...)\in \mathbb {C} ^{\infty }:\sum |a_{i}|^{2}<\infty \}}$. The maps

${\displaystyle T:(a_{0},a_{1},...)=(0,a_{0},a_{1},...)}$
${\displaystyle T^{-1}:(a_{0},a_{1},...)=(a_{1},a_{2},....)}$
have index ${\displaystyle -1,1}$ respectively.

In fact ${\displaystyle Index(T^{n})=n}$${\displaystyle \forall n\in \mathbb {Z} }$${\displaystyle T^{-1}}$ is a left inverse of ${\displaystyle T}$. The operator ${\displaystyle T^{n}T^{-n}-T^{-n}T^{n}}$ is the finite rank operator:


${\displaystyle (a_{0},a_{1},...)\mapsto (a_{0},a_{1},...,a_{n-1},0,...)}$

## Constructing Index bundles

The theory of Fredholm operators is closely related to K-theory: Suppose that we had a smoothly varying family of Fredholm operators ${\displaystyle T_{x}}$ over a compact, connected space ${\displaystyle X}$. Then such a family gives a K-theory class in ${\displaystyle X}$, and all K-theory classes arise in this way.

Definition 13.3

Let ${\displaystyle f:X\rightarrow {\mathcal {F}}}$ be a family of Fredholm operators ${\displaystyle \{T_{x}\}}$. Then we may associate a canonically defined K-theory class ${\displaystyle Ind(T)\in K(X)}$.

Proof

The map ${\displaystyle Index:{\mathcal {F}}\rightarrow \mathbb {Z} }$ is continuous so ${\displaystyle Ind(T_{x})=dim(Ker(T_{x}))-dim(Coker(T_{x}))}$ is constant. In the simplified situation where ${\displaystyle dim(Ker(T_{x}))}$ and hence ${\displaystyle dim(Coker(T_{x}))}$ are also constant we can give an explicit description of the index bundles.

${\displaystyle E=\bigsqcup _{x\in X}Ker(T_{x}),F=\bigsqcup _{x\in X}Coker(T_{x})}$
inherit the structure of vector bundles from the family. Define ${\displaystyle Ind(T)=E-F}$.

Now for the general situation consider the map ${\displaystyle H_{n}:(a_{0},a_{1},...)\mapsto (0,...,a_{n},...)}$ (assuming for simplicity that ${\displaystyle V=L^{2}(\mathbb {Z} _{+})}$). Fredholm operators are invertible up to compact operators, so for each ${\displaystyle x\in X}$${\displaystyle \exists N:Im(H_{N}T_{x})=Im(H_{N})}$. By compactness of ${\displaystyle X}$ there is a global choice of ${\displaystyle N}$ that works for all ${\displaystyle x}$. Now the family ${\displaystyle \{H_{N}T_{x}\}}$ has Cokernel of constant dimension so apply the above construction. [HARD BIT: prove this is independent of ${\displaystyle n}$].

Exercise 13.6

Show that all K(X) classes arise this way by considering the sequence

${\displaystyle 0\rightarrow Ker\rightarrow L^{2}(E)\rightarrow E\rightarrow 0.}$
[Here ${\displaystyle L^{2}(E)}$ is the trivial vector bundle of ${\displaystyle L^{2}}$ sections of ${\displaystyle E}$. The second map is ${\displaystyle (x,s)\mapsto s(x)}$]

Exercise 13.7

Fredholm operators and winding numbers.

Let ${\displaystyle V=L^{2}(S^{1})}$ be the space of square integrable (complex) functions on ${\displaystyle S^{1}}$. All such functions have a Fourier series

${\displaystyle f(z)=\sum _{k\in \mathbb {Z} }a_{k}z^{k}}$
Define the Hardy space ${\displaystyle {\mathcal {H}}^{2}}$ to be the subspace of functions with ${\displaystyle a_{k}=0}$ for all ${\displaystyle k<0}$.

Let ${\displaystyle g:S^{1}\rightarrow \mathbb {C} }$ be a continuous function. Define an operator ${\displaystyle T_{g}:{\mathcal {H}}^{2}\rightarrow {\mathcal {H}}^{2}}$. Note that for ${\displaystyle f\in L^{2}}$${\displaystyle f*g\in L^{2}}$. Define ${\displaystyle T_{f}(g)}$ to be the projection of ${\displaystyle f*g}$ to ${\displaystyle {\mathcal {H}}^{2}}$ by killing the negative Fourier coefficients.

${\displaystyle T_{g}}$ is Fredholm ${\displaystyle \Leftrightarrow }$${\displaystyle Im(g)\subset \mathbb {C} ^{\star }}$. The nice fact about these Fredholm operators is that the index is the winding number of ${\displaystyle g}$ around ${\displaystyle 0}$.