# M721: Index Theory

## Examples of Atiyah-Singer Index Theorem

Let’s begin with some notations.

Let $D:C^\infty (E) \rightarrow C^\infty (F)$ be an elliptic differential operator, where $E$ and $F$ are vector bundles over a closed oriented manifold $X^n$. Suppose $E, F$ and $TX$ have smooth inner product structures. Let $D(X):=D(T^*X)$ and $S(X):=S(T^*X)$ be the disk and sphere bundle of the cotangent bundle, respectively. Let $\pi: D(X)\rightarrow X$ be the projection. Let $[\sigma_D]=[\pi^*E, \pi^*F; \sigma_D]\in K(D(X),S(X))$ be the associated symbol class. Let $\textrm{ch} ([\sigma_D])\in H^{2*}(D(X),S(X);\Bbb{Q})$ be its Chern character. Let $\textrm{td} (X)$ denote the pullback of the Todd class $\pi^*\textrm{td} (T^*X\otimes \Bbb{C})\in H^{2*}(D(X);\Bbb{Q})$. Let $[D(X)]\in H_{2n}(D(X),S(X))$ be the fundamental class.

Definition.  The topological index of $D$ is defined to be $\textrm{t-ind} \ D:=(-1)^n\langle \textrm{ch}[\sigma_D]\textrm{td} (X),[D(X)]\rangle$.

Atiyah-Singer Index Theorem.  $\textrm{ind} \ D=\textrm{t-ind} \ D$, where $\textrm{ind} \ D:=\textrm{dim}_{\Bbb{C}}(\textrm{ker} D)-\textrm{dim}_{\Bbb{C}}(\textrm{cok} \ D)$ is the analytical index.

Example. [Point Case] Let $X$ be a point. Then $E$ and $F$ are finite dimensional vector spaces. Any non-trivial differential operator $D$ is a linear map between them, and hence of order 0. Thus, $\sigma_D=D$. Note that $D(X)=X, S(X)$ is empty set. Recall the definition of Chern character, then we have $\textrm{ch} (E)=\textrm{dim} E$ and $\textrm{ch} (F)=\textrm{dim} F$. Since $S(X)$ is empty set, then $[\sigma _D]=[E]-[F]$ and hence $\textrm{ch} ([\sigma _D])=\textrm{dim} E-\textrm{dim} F$. Recall the definition of Todd class, we then have $\textrm{td} (X)=1$. Therefore, by the definition of topological index, we have $\textrm{t-ind} \ D=\textrm{dim} E-\textrm{dim} F$. By Atiyah-Singer, $\textrm{dim} (\textrm{ker} D)-\textrm{dim} (\textrm{cok} \ D)=\textrm{dim} E-\textrm{dim} F$.

Example [$S^1$ Case] Let $X=S^1=\Bbb{R}/2\pi\Bbb{Z}$, $E=F=S^1\times\Bbb{C}$, $D=\frac{d}{dx}: C^\infty(S^1,\Bbb{C})\rightarrow C^\infty(S^1,\Bbb{C}).$ Then $D$ is a first order elliptic operator. We now claim that $\textrm{ind} \ D=0$. We give four different proofs.
Proof 1. We compute $\textrm{ind} \ D$ directly. Note that $\textrm{ker} \frac{d}{dx}=\{constant functions\}$ and that $\textrm{cok} \ \frac{d}{dx}\xrightarrow{\approx}\Bbb{C}$ via $[f]\mapsto\int_{S^1}f$.
Proof 2. Since $E=F$, we have $[\sigma_D]=[\pi^*E,\pi^*F;\sigma_D]=[\pi^*E\cup_{\sigma_D}\pi^*F]-[\pi^*F\cup_{\textrm{id}}\pi^*F]=[\pi^*F\cup_{\textrm{id}}\pi^*F]-[\pi^*F\cup_{\textrm{id}}\pi^*F]=0.$
Proof 3. $\textrm{ind} \ \frac{d}{dx}=\textrm{ind} \ i\frac{d}{dx}=0$, since $i\frac{d}{dx}$ is self-adjoint.
Proof 4. We will show that the topological index vanishes whenever $n$ is odd. See next example.

Example [Odd Dimensional Case, Theorem 13.12 in Lawson-Michaelson]
We will show that the topological index of any elliptic differential operator vanishes whenever $n$ is odd.

We want to show that $\textrm{t-ind} \ D =-\textrm{t-ind} \ D$, where $D$ is an elliptic differential operator of order $m$. Consider the diffeomorphic involution $c: TX\rightarrow TX$ given by $c(v)=-v$. Since
$\textrm{t-ind}\ D=-\textrm{ch} ([\sigma_D])\textrm{td} (X)[D(X)]$

$\ \ \ \ \ \ \ \ \ \ \ \ =-\textrm{ch} ([\sigma_D])\textrm{td} (X)c_*c_*[D(X)]$

$=-c^*(\textrm{ch} ([\sigma_D])\textrm{td} (X)) c_*[D(X)]$

$=-(\textrm{ch} (c^*[\sigma_D]))\textrm{td} (X) (-[D(X)])$,
it suffices to show $c^*[\sigma_D]=[\sigma_D]$. In fact, $c^*[\sigma_D]=[\pi^*E,\pi^* F;(-1)^m\sigma_D]=[\pi^*E,\pi^* F;\sigma_D]=[\sigma_D]$, since $\sigma_D$ is homotopic to $-\sigma_D$ via $e^{i\pi t}D, t\in [0,1]$.
Next, we need to introduce the Thom Isomorphism to talk about the de Rham operator.

Let $E$ be an oriented $\Bbb{R}^k$-vector bundle over $X$, with inner product on each fiber. We now give the notion of Thom class and Thom space.

Definition. $u(E)\in H^k(D(E)),S(E))$ is a Thom Class of $E$ if it restricts to a generator of $H^k(D^k,S^{k-1})\approx H^k(\pi^{-1}\{x\}\cap D(E),\pi^{-1}\{x\}\cap S(E))$ on each fiber. The quotient $D(E)/S(E)$ is called the Thom space of $E$, and denoted by $\textrm{Th} (E)$.

Thom Isomorphism Theorem. The composition $H^*(X)\xrightarrow{\approx}H^*(D(E))\xrightarrow{\smallsmile u}H^{*+k}(D(E),S(E))$ is an isomorphism.

We denote the composition by $\textrm{-}\smallsmile u : H^*(X)\xrightarrow{\approx}H^{*+k}(D(E),S(E))=\widetilde{H}^{*+k}(\textrm{Th}(E))$ and denote its inverse by $\pi_!$.

Remark.  $\pi_!$ has the following two other interpretations.

(1). Integration over the fiber. $\pi_! : \Omega^{i+k}(E)\rightarrow\Omega^i(X)$, where $E$ is a $\Bbb{R}^k$ bundle over $X$.
Let $\tau\in\Omega^{i+k}(E)$ be given and choose $x\in X$ and $v_1,...,v_i\in T_xX$. Associated to these data is a form $\tau^{x,v_1,...,v_i}\in\Omega^k(E_x)$, defined as follows. Given $e\in E_x$ and a basis $e_1,...,e_k\in T_e E_x$, choose lifts $\widetilde{v_l}\in T_e E_x$ such that $d\pi (e)\widetilde{v_l}=v_l$, for each $l\leq i$, and define

$(\tau^{x,v_1,...,v_i})_e(e_1,...,e_k):=\tau_e(\widetilde{v_1},...,\widetilde{v_i},e_1,...,e_k).$ Now $\pi_! \tau\in\Omega^k(X)$ is defined by

$(\pi_!\tau)_x(v_1,...,v_i):=\int_{E_x}\tau^{x,v_1,...,v_i}.$

Integrating over the fibers will give the second formulation of the topological index, which is the next theorem. The factor $(-1)^{\frac{n(n+1)}{2}}$ compensates for the difference between the orientation on $TX$ induced by the one on $X$, and the canonical orientation on $TX$ inherited from its almost complex structure.

(2) . Also, we can use the second interpretation of $\pi_!$ to give that formulation.

$\pi_!$ is the composition (Poincare duality) $\ \circ\ \pi_*\ \circ$ (Poincare-Lefschetz duality),

$H^{*+k}(D(E),S(E))\rightarrow H_{n-*}(D(E))\xrightarrow{\pi_*} H_{n-*}(X)\rightarrow H^*(X).$
Then, we compute
$\textrm{t-ind}\ D=(-1)^n \textrm{ch}[\sigma_D]\textrm{td} (X)[D(X)]$
$=(-1)^{\frac{n(n+1)}{2}}\textrm{td} (X) \pi_*(\textrm{ch} ([\sigma_D])\smallfrown [D(X)])$
$= (-1)^{\frac{n(n+1)}{2}}\textrm{td} (X) \pi_!(\textrm{ch} ([\sigma_D])\smallfrown [X])$
$= (-1)^{\frac{n(n+1)}{2}}\pi_!( \textrm{ch} ([\sigma_D]))\textrm{td} (X)\ [X]$.

Theorem.  $\textrm{t-ind} \ D=(-1)^{\frac{n(n+1)}{2}}\pi_!( \textrm{ch} ([\sigma_D]))\textrm{td} (X)\ [X]$.

We will then give the third formulation of the topological index. To do this we need the notion of Euler class.

Definition. The Euler class of an oriented $\Bbb{R}^k$-bundle over $X$, denoted by $latexe(E)$, is the image of the Thom class $u$ under the following isomorphism: $H^k(D(E),S(E))\xrightarrow{i^*}H^k(D(E))\xrightarrow{\approx}H^k(X)$. We may denote the composition by $i^*$.

Theorem. [Gysin Sequence]  To any bundle $E$ as above there is associated an exact sequence of the form $\cdots\rightarrow H^i(X)\xrightarrow{\smallsmile e(E)} H^{i+k}(X)\xrightarrow{\pi^*} H^{i+k}(S(E))\rightarrow H^{i+1}(X)\xrightarrow{\smallsmile e}\cdots.$

Definition. The Euler characteristic of $X$ is defined to be $\chi (X):=\langle e(TX),[X]\rangle.$

From now on, we assume $n=2m$.

We want to analyze $\pi_!\textrm{ch}([\sigma_D])$ to give the third formulation of the topological index. For details please see Lawson-Michaelson, P258, Theorem 13.13.

Since –$\smallsmile u$ and $\pi_!$ are inverse to each other, we have $(\pi_!\textrm{ch}([\sigma_D]))\smallsmile u(TX)=\textrm{ch}([\sigma_D])$. Applying $i^*$ to both sides, we then get $(\pi_!\textrm{ch}([\sigma_D]))\smallsmile e(TX)=i^*\textrm{ch}([\sigma_D])=\textrm{ch} (i^*\sigma_D])=\textrm{ch}(E)-\textrm{ch}(F).$ Thus we can write $\pi_!\textrm{ch}([\sigma_D])=\frac{\textrm{ch}(E)-\textrm{ch}(F)}{e(TX)}$, if $e(TX)\neq 0$.

Theorem. $\textrm{t-ind} \ D=(-1)^{\frac{n(n+1)}{2}}\frac{\textrm{ch}(E)-\textrm{ch}(F)}{e(TX)}\textrm{td} (X)\ [X]$, if $e(TX)$ is not zero.

Now, we are trying to apply this formula to the de Rham operator.

Example. [de Rham operator]  Let $\Omega^k_\Bbb{C}=C^\infty(\Lambda^k\ T^*X\otimes \Bbb{C}).$ Then we already know that $(\Omega_\Bbb{C}^\bullet, d)$ is an elliptic complex, that $d: \Omega_\Bbb{C}^{even}\rightarrow\Omega_\Bbb{C}^{odd}$ is an elliptic operator and that $\textrm{ind} \ d=\Sigma(-1)^i\ \textrm{rk}(H^i_{DR}(X))$. We want to use the above theorem to show that $\textrm{t-ind}\ d=\Sigma (-1)^i \ \textrm{rk}( H^i(X)),$ the Euler charateristic.

For a complex vector bundle $E$, by the splitting principle, we can write $E$ as $\oplus_i L_i$. Then $\Sigma_0^n \Lambda^i E=\otimes_1^n\Lambda L_i=\otimes_1^n (\Bbb{C}\oplus L_i)$. It follows that $\textrm{ch} (\Sigma_0^n \Lambda^i E)=\prod_1^n (1+e^{x_i}).$ Similarly, we have $\textrm{ch} (\Sigma_0^n (-1)^i\Lambda^i E)=\prod_1^n (1-e^{x_i}).$

Back to our example, applying the real splitting principle to $T^*X$, we compute

$\textrm{ch} (\Sigma (-1)^i \Lambda^i T^*X\otimes\Bbb{C})=\prod_1^n (1-e^{x_i})=\prod_1^n (1-e^{-x_i}),$

since $x_i=-x_{i+m}$ for $1\leq i\leq m$. Note that $\textrm{td} (X)=\prod_1^n \frac{x_i}{1-e^{-x_i}}$, and that $e=\prod_1^m x_i$, then by the theorem above, we obtain

$\textrm{t-ind} \ d=(-1)^{\frac{n(n+1)}{2}}\frac{\prod_1^n (1-e^{-x_i})}{\prod_1^m x_i}\prod_1^n \frac{x_i}{1-e^{-x_i}} [X]=(-1)=\prod_1^m x_i [X]=\chi (X).$