M721: Index Theory

Ruminations of a Graduate Class

Examples of Atiyah-Singer Index Theorem

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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


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).


Written by topoclyb

January 13, 2013 at 7:37 pm

Posted in Uncategorized

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