# The Atiyah-Singer Index formula (INCOMPLETE)

[Let ${\displaystyle E}$,${\displaystyle F}$ be vector bundles of rank ${\displaystyle m}$ and ${\displaystyle n}$ respectively over ${\displaystyle X}$ a manifold of dimension ${\displaystyle k}$. A a rank${\displaystyle \mathbf {k} }$differential operator is a map ${\displaystyle D:\Gamma (E)\rightarrow \Gamma (F)}$ satisfying the following: About each point choose an open set such that ${\displaystyle E}$ and ${\displaystyle F}$ are trivial. ${\displaystyle s\in \Gamma (E)}$ and ${\displaystyle Ds\in \Gamma (F)}$ may be expressed locally as ${\displaystyle s:U\rightarrow \mathbb {R} ^{m}}$, ${\displaystyle Ds:U\rightarrow \mathbb {R} ^{n}}$. We say that ${\displaystyle D}$ is a differential operator of order ${\displaystyle k}$ if locally ${\displaystyle Ds=As}$ where A is a matrix with coefficients of the form:

${\displaystyle \sum f_{i_{1},...,i_{j}}{\frac {\partial ^{j}}{\partial x_{i_{1}}...x_{i_{j}}}}}$
and ${\displaystyle j\leq k}$.

Exercise 13.8
• The Laplace operator. ${\displaystyle \Delta =\sum {\frac {\partial ^{2}}{\partial x_{i}^{2}}}}$ acting on complex functions is a differential operator of degree ${\displaystyle 2}$.
• The exterior derivative. ${\displaystyle d:\Omega ^{p}\rightarrow \Omega ^{p+1}}$ is a differential operator of degree ${\displaystyle 1}$ (exercise).]

Suppose that D is a differential operator of degree ${\displaystyle k}$. Now for each 1-form ${\displaystyle \omega \neq 0}$ we define ${\displaystyle \sigma (D)(\omega )\in Hom(E,F)}$ called the symbol of ${\displaystyle D}$. This measures the top-order behaviour of the operator and may be described locally:

We form a new matrix ${\displaystyle A^{P}}$ from ${\displaystyle A}$ above by forgetting all differentials of order ${\displaystyle , now write the 1-form ${\displaystyle \omega }$ in local coordinates ${\displaystyle \omega (p)=(v_{1},...,v_{k})}$. Replace each differential ${\displaystyle {\frac {\partial }{\partial x_{i}}}}$ in the matrix ${\displaystyle A^{p}}$ with ${\displaystyle v_{i}\in \mathbb {R} }$. This matrix gives a linear map ${\displaystyle E_{p}\rightarrow V_{p}}$ and globally we have a bundle homomorphism ${\displaystyle \sigma (D)(\omega )\in Hom(E,F)}$ which is called the symbol.

Exercise 13.9

Consider the cotangent bundle ${\displaystyle \pi :T_{X}^{\star }\setminus 0\rightarrow X}$, we may pull back the bundles to obtain ${\displaystyle \pi ^{\star }(E),\pi ^{\star }(F)}$. Rearrange the above to show that the symbol can be expressed as a bundle morphism ${\displaystyle \sigma (D):\pi ^{\star }(E)\rightarrow \pi ^{\star }(F)}$.

Definition 13.4

${\displaystyle D}$ is called an Elliptic operator if ${\displaystyle \sigma (D)(\omega )}$ is a bundle isomorphism for all ${\displaystyle \omega \in \Omega _{X}^{1}\setminus 0}$.

Exercise 13.10

For the Exterior derivative we may check that ${\displaystyle \sigma (d)(\alpha ):\Omega ^{k}\rightarrow \Omega ^{k+1}=\cdot \mapsto \alpha \wedge \cdot }$.

By computing the rank of ${\displaystyle \Omega ^{k}}$ we see that this can't be an isomorphism in all but the most trivial cases, therefore ${\displaystyle d}$ is not an elliptic operator. Define ${\displaystyle d^{\star }:\Omega ^{k+1}\rightarrow \Omega ^{k}}$ by

${\displaystyle fdx_{1}...dx_{p}\mapsto \sum _{i}{\frac {\partial f}{\partial x_{i}}}dx_{1}...{\hat {dx_{i}}}...dx_{p}}$
. Then ${\displaystyle d^{\star }}$ is the formal adjoint of d. ${\displaystyle Ker(d)=coker(d^{\star })}$ and ${\displaystyle coker(d)=ker(d^{\star })}$, and ${\displaystyle d+d^{\star }:\Omega ^{\star }\rightarrow \Omega ^{\star }}$ is an elliptic operator (this is seen by computing the symbol: ${\displaystyle \sigma (d+d^{\star })(a)(\cdot )=a\wedge \cdot +a\lrcorner \cdot }$). Note that ${\displaystyle (d+d^{\star })(\Omega ^{2\star })\subset \Omega ^{2*+1}}$, restricting the operator in this way ${\displaystyle Ker(d+d^{\star })=\oplus H^{2\star }(X)}$, ${\displaystyle Coker(d+d^{\star })=\oplus H^{2\star +1}(X)}$.

It turns out that an elliptic complex is always a Fredholm operator (${\displaystyle \Gamma (E),\Gamma (F)}$ considered as trivial bundles over ${\displaystyle X}$) and we see in this case ${\displaystyle Index(D)=\sum dim(H^{2i})-\sum dim(H^{2i+1})=e(X)}$ (exercise).

Elliptic operators are invertible up to lower order operators. Using the compact Rellich lemma (which states that ${\displaystyle L_{k}^{2}\rightarrow L_{k-1}^{2}}$ is a compact embedding) shows that

${\displaystyle L^{2}(E){\xrightarrow {D}}L^{2}(F)}$
is Fredholm. Index(D) depends only on the homotopy class of the map ${\displaystyle T_{X}^{*}\setminus 0\rightarrow Iso(E,F)}$ given ${\displaystyle \omega \mapsto \sigma (D)(\omega )}$ and can provide information about the topology of ${\displaystyle X}$.

Consider ${\displaystyle \pi ^{\star }(E)-\pi ^{\star }(F)\in K(T^{\star }(X))}$. Subtracting ${\displaystyle \rho ^{\star }(F)}$ to get a class ${\displaystyle {\tilde {\sigma }}\in {\tilde {K}}(X)}$.

We may embed ${\displaystyle X\hookrightarrow N_{X}\hookrightarrow \mathbb {R} ^{n}}$. Suppose that ${\displaystyle N_{X}}$ was trivial ${\displaystyle N_{X}=X\times \mathbb {R} ^{m}}$ (${\displaystyle m=codim(X)}$).

Now there exists an operator on the trivial bundle ${\displaystyle \mathbb {R} \times \mathbb {C} }$ over ${\displaystyle \mathbb {R} }$. Whose symbol over ${\displaystyle T^{\star }\mathbb {R} \cong \mathbb {R} ^{2}}$, gives the Bott class ${\displaystyle b\in {\tilde {K}}(\mathbb {R} ^{2})}$. We want to form ${\displaystyle D\boxplus B^{\boxplus m}\in {\tilde {K}}(T^{\star }X\times R^{2m})={\tilde {K}}(T^{\star }(N_{X}))}$.