Fredholm Operators

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

$$E_{1}{\xrightarrow {d}}E_{2}$$

$$E_{1}{\xrightarrow {d}}E_{2}$$

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

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

Definition 13.2

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

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

 

Exercise 13.5

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

$$T:(a_{0},a_{1},...)=(0,a_{0},a_{1},...)$$

$$T:(a_{0},a_{1},...)=(0,a_{0},a_{1},...)$$

$$T^{-1}:(a_{0},a_{1},...)=(a_{1},a_{2},....)$$

$$T^{-1}:(a_{0},a_{1},...)=(a_{1},a_{2},....)$$

have index $-1,1$$-1,1$ respectively.

 

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

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

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

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 $T_{x}$$T_{x}$ over a compact, connected space $X$$X$. Then such a family gives a K-theory class in $X$$X$, and all K-theory classes arise in this way.

Definition 13.3

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

 

Proof

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

$$E=\bigsqcup _{x\in X}Ker(T_{x}),F=\bigsqcup _{x\in X}Coker(T_{x})$$

$$E=\bigsqcup _{x\in X}Ker(T_{x}),F=\bigsqcup _{x\in X}Coker(T_{x})$$

inherit the structure of vector bundles from the family. Define $Ind(T)=E-F$$Ind(T)=E-F$.

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

 

Exercise 13.6

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

$$0\rightarrow Ker\rightarrow L^{2}(E)\rightarrow E\rightarrow 0.$$

$$0\rightarrow Ker\rightarrow L^{2}(E)\rightarrow E\rightarrow 0.$$

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

 

Exercise 13.7

Fredholm operators and winding numbers.

 

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

$$f(z)=\sum _{k\in \mathbb {Z} }a_{k}z^{k}$$

$$f(z)=\sum _{k\in \mathbb {Z} }a_{k}z^{k}$$

Define the Hardy space ${\mathcal {H}}^{2}$${\mathcal {H}}^{2}$ to be the subspace of functions with $a_{k}=0$$a_{k}=0$ for all $k<0$$k<0$.

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

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