12.10 Supersymmetric proof of index theorem

The proof of the index theorem in its simplest setting will be given in the present section by making use of the supersymmetric quantum mechanics developed in the previous section.

12.10.1 The index

Let us consider vector bundles $E_{\pm }\stackrel{\pi }{\to }M$ , $E=E_{+}\oplus E_{-}$ and let $\mathcal{D}$ be an elliptic differential operator acting as

$$\mathcal{D}:\Gamma (M,E^{+})\to \Gamma (M,E^{-}).$$

It is possible, by using the fibre norm, to define the adjoint of $\mathcal{D}$ as

$${\mathcal{D}}^{†}:\Gamma (M,E^{-})\to \Gamma (M,E^{+}).$$

Assuming that $\mathcal{D}$ is Fredholm, the index

$$\mathrm{Ind}\mathcal{D}=\dim \ker \mathcal{D}-\dim \ker {\mathcal{D}}^{†}\text{\hspace{1em}(12.161)}$$

is well defined.

Theorem 12.4

The number ind $\mathcal{D}$ is invariant under a ‘small’ deformation of $\mathcal{D}$ .

Proof

Note, first, that ${\mathcal{DD}}^{†}$ and ${\mathcal{D}}^{†}\mathcal{D}$ are non-negative and, hence, it follows that

$$\ker \mathcal{D}=\ker {\mathcal{D}}^{†}\mathcal{D}\ker {\mathcal{D}}^{†}=\ker \mathcal{D}{\mathcal{D}}^{†}.$$

Let $\{{\varphi }_n\}$ be the orthonormal set of eigensections of ${\mathcal{D}}^{†}\mathcal{D}:\Gamma (M,E^{+})\to$ $\Gamma (M,E^{+})$ ..

$$({\mathcal{D}}^{†}\mathcal{D}){\varphi }_n={\lambda }_n{\varphi }_n.$$

Define ${\psi }_n\equiv \mathcal{D}{\varphi }_n/\sqrt{{\lambda }_n}$ for ${\lambda }_n>0$ , namely ${\varphi }_n\in (\ker \mathcal{D}{)}^{\perp }$ . Then we find that ${\psi }_n$ is an eigensection with the same eigenvalue ${\lambda }_n$ , namely ${\psi }_n\in (\ker {\mathcal{D}}^{†}{)}^{\perp }$ since

$$(\mathcal{D}{\mathcal{D}}^{†}){\psi }_n=\mathcal{D}({\mathcal{D}}^{†}\mathcal{D}{\varphi }_n)/\sqrt{{\lambda }_n}={\lambda }_n\mathcal{D}{\varphi }_n/\sqrt{{\lambda }_n}={\lambda }_n{\psi }_n.$$

Note also that $\{{\psi }_n\}$ is an orthonormal eigensection,

$$⟨{\psi }_n|{\psi }_m⟩=\frac{1}{\sqrt{{\lambda }_n{\lambda }_m}}⟨{\varphi }_n|{\mathcal{D}}^{†}\mathcal{D}|{\varphi }_m⟩=\frac{{\lambda }_m}{\sqrt{{\lambda }_n{\lambda }_m}}{\delta }_{nm}={\delta }_{nm}.$$

Thus, it follows that there is a natural isomorphism between $(\ker \mathcal{D}{)}^{\perp }$ and $(\ker {\mathcal{D}}^{†}{)}^{\perp }$ . Note, however, that there exists no such isomorphism between $\ker \mathcal{D}$ and $\ker {\mathcal{D}}^{†}$ . Suppose $N$ states in $\ker \mathcal{D}$ obtain non-vanishing eigenvalues as a result of a small perturbation of the operator $\mathcal{D}$ and $\dim \ker \mathcal{D}$ decreases by $N$ . Then it follows from this observation that the same number of states must also leave $\ker {\mathcal{D}}^{†}$ . Otherwise $(\ker \mathcal{D}{)}^{\perp }$ is no longer isomorphic to $(\ker {\mathcal{D}}^{†}{)}^{\perp }$ . Similary, if $\dim \ker \mathcal{D}$ increases by $N$ , $\dim \ker {\mathcal{D}}^{†}$ must also increase by $N$ to keep the pairing properties of $(\ker \mathcal{D}{)}^{\perp }$ and $(\ker {\mathcal{D}}^{†}{)}^{\perp }$ . Therefore, $\mathrm{ind}\mathcal{D}$ is invariant under small perturbations of $\mathcal{D}$ .

Theorem 12.5

Let $\mathcal{D}$ be a Fredholm differential operator. Then its index is given by

$$\mathrm{ind}\mathcal{D}=\mathrm{Tr}{\mathrm{e}}^{-\beta {\mathcal{D}}^{†}\mathcal{D}}-\mathrm{Tr}{\mathrm{e}}^{-\beta {\mathcal{DD}}^{†}}\text{\hspace{1em}(12.162)}$$

where $\beta >0$ is a real constant. In fact, the index is independent of $\beta$ .

Proof

The traces in (12.162) are over $\{{\varphi }_n\}$ and $\{{\psi }_n\}$ , respectively. Let $\{{\varphi }_i^0\}$ and $\{{\psi }_i^0\}$ be orthonormal eigensections of $\ker \mathcal{D}$ and $\ker {\mathcal{D}}^{†}$ , respectively, and $1\le i\le \dim \ker \mathcal{D}$ and $1\le j\le \dim \ker {\mathcal{D}}^{†}$ . Then it follows that

$$\begin{array}{rl}& {\mathrm{Tre}}^{-\beta {\mathcal{D}}^{†}\mathcal{D}}-{\mathrm{Tre}}^{-\beta {\mathcal{DD}}^{†}}\\ & =\sum _{{\lambda }_n\ne 0}⟨{\varphi }_n|e^{-\beta {\mathcal{D}}^{†}\mathcal{D}}|{\varphi }_n⟩-\sum _{{\lambda }_n\ne 0}⟨{\psi }_n|e^{-\beta \mathcal{D}{\mathcal{D}}^{†}}|{\psi }_n⟩\\ & +\sum _i⟨{\varphi }_i^0|{\varphi }_i^0⟩-\sum _j⟨{\psi }_j^0|{\psi }_j^0⟩\\ & =\sum _{{\lambda }_n\ne 0}{\mathrm{e}}^{-\beta {\lambda }_n}\left(⟨{\varphi }_n|{\varphi }_n⟩-⟨{\psi }_n|{\psi }_n⟩\right)+\sum _{i}1-\sum _{j}1\\ & =\dim \ker \mathcal{D}-\dim \ker {\mathcal{D}}^{†}\\ & =\mathrm{ind}\mathcal{D}.\end{array}$$

Since the summations over $i$ and $j$ are independent of $\beta$ , $\mathrm{ind}\mathcal{D}$ thus defined is independent of $\beta$ .

The trace that appears in theorem 12.5 is identified with the heat kernel. Let $E=E_{+}\oplus E_{-}$ and define a differential operator acting on $E$ by (cf equation (12.79))

$$\mathrm{i}Q\equiv \left(\begin{array}{cc}0& {\mathcal{D}}^{†}\\ \mathcal{D}& 0\end{array}\right):E\to E.\text{\hspace{1em}(12.163)}$$

Moreover, define a ‘Hamiltonian’ and a matrix $\Gamma$ by

$$H=(\mathrm{i}Q{)}^2=\left(\begin{array}{cc}{\mathcal{D}}^{†}\mathcal{D}& 0\\ 0& \mathcal{D}{\mathcal{D}}^{†}\end{array}\right)\Gamma =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right).\text{\hspace{1em}(12.164)}$$

Since $Q$ thus defined is anti-Hermitian, the operator $H$ is Hermite and nonnegative. The index of $\mathcal{D}$ is rewritten in a compact form by making use of $\Gamma$ as

$$\mathrm{ind}\mathcal{D}=\mathrm{Tr}\Gamma {\mathrm{e}}^{-\beta H}.\text{\hspace{1em}(12.165)}$$

Let $M$ be a spin manifold, for which the second Stiefel-Whitney class $w_2(M)$ is trivial. Accordingly, the $\mathrm{SO}(k)$ principal bundle over $M$ may be lifted to the $\mathrm{SPIN}(k)$ principal bundle as

$$\begin{array}{rl}& \\ & \begin{array}{ccc}\mathrm{SO}(k)& \to & \mathrm{SPIN}(k).\end{array}\\ & \downarrow \pi \\ & M\end{array}$$

Let $E=\Delta (M)$ be this spin bundle. Then, associated with $\Delta (M)$ is a Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\delta }^{\mu \nu }$ . Let us define the chirality operator

$${\gamma }_{2n+1}\equiv {\mathrm{i}}^n{\gamma }_1{\gamma }_2\dots {\gamma }_{2n}.\text{\hspace{1em}(12.166)}$$

5 The operator $Q$ will be identified with the supercharge later.

It follows from ${\gamma }_{2n+1}^2=1$ that the eigenvalues of ${\gamma }_{2n+1}$ are restricted to be $\pm 1$ , which we call chirality.

Exercise 12.10

Use the Clifford algebra to show that

$${\gamma }_{2n+1}^2=1\{{\gamma }_{\mu },{\gamma }_{2n+1}\}=0.$$

The set of sections $\Gamma (M,\Delta )$ for an even $k$ is not an irreducible representation of $\mathrm{Spin}(k)$ but can be decomposed into two subspaces according to the chirality as

$$\Gamma (M,\Delta )=\Gamma (M,{\Delta }^{+})\oplus \Gamma (M,{\Delta }^{-})\text{\hspace{1em}(12.167)}$$

where ${\psi }_{\pm }\in \Gamma (M,{\Delta }^{\pm })$ satisfy ${\gamma }_{2n+1}{\psi }_{\pm }=\pm {\psi }_{\pm }$ . We assign the fermion number $F=0$ to sections in $\Gamma (M,{\Delta }^{+})$ while $F=1$ for those in $\Gamma (M,{\Delta }^{-})$ . Then the $\Gamma$ defined in (12.164) can be written as

$$\Gamma =(-1{)}^F.\text{\hspace{1em}(12.168)}$$

It is clear that the operator $Q$ flips the chirality and hence $\{Q,\Gamma \}=0$

Let $Q$ be the Dirac operator on $M$ and let $\Gamma ={\gamma }_{2n+1}$ . In fact, it follows from exercise 12.11 that $\{Q,{\gamma }_{2n+1}\}=0$ and ${\gamma }_{2n+1}$ is identified with $(-1{)}^F$ . When $\Gamma$ is diagonalized as in (12.164), the chirality eigensections are expressed as¹

$${\psi }_{+}=\left(\begin{array}{c}{\psi }_{+}\\ 0\end{array}\right){\psi }_{-}=\left(\begin{array}{c}0\\ {\psi }_{-}\end{array}\right).\text{\hspace{1em}(12.169)}$$

It should be then clear that $\mathcal{D}:\Gamma (M,{\Delta }^{+})\to \Gamma (M,{\Delta }^{-})$ and ${\mathcal{D}}^{†}:\Gamma (M,{\Delta }^{-})\to$ $\Gamma (M,{\Delta }^{+})$ are identified with $D$ and $D^{†}$ , respectively, in (12.79). Accordingly, the index of the Dirac operator is defined as

$$\mathrm{ind}Q=\dim \ker D-\dim \ker D^{†}.\text{\hspace{1em}(12.170)}$$

Physicists often call the sections in $\ker D$ and $\ker D^{†}$ zero modes. Then, the index of the Dirac operator is the difference between the number of positive and negative chirality zero modes. This index has a path integral expression as we see in the next subsection.

12.10.2 Path integral and index theorem

Let us consider a Dirac operator $Q$ on a $2n$ -dimensional spin manifold $M$ . We employ Euclidean time $(t\to -\mathrm{i}t)$ from now on.

Let $H=(\mathrm{i}Q{)}^2=\frac{1}{2}{g}_{\mu \nu }{p}^{\mu }{p}^{\nu }$ be the Hamiltonian corresponding to $Q$ . Then the index of the Dirac operator has a path integral expression

$$\begin{array}{rl}\mathrm{ind}Q& =\mathrm{Tr}\Gamma {\mathrm{e}}^{-\beta H}=\mathrm{Tr}(-1{)}^F{\mathrm{e}}^{-\beta H}\\ & ={\int }_{\mathrm{PBC}}\mathcal{D}x\mathcal{D}\psi {\mathrm{e}}^{-{\int }_0^{\beta }\mathrm{d}tL}\end{array}\text{\hspace{1em}(12.171)}$$

where the Lagrangian $L$ has been introduced in (12.154),

$$L=\frac{1}{2}{g}_{\mu v}(x){\dot{x}}^{\mu }{\dot{x}}^v+\frac{1}{2}{g}_{\mu v}(x){\psi }^{\mu }\frac{\mathrm{D}{\psi }^v}{\mathrm{D}t}\text{\hspace{1em}(12.172)}$$

and PBC stands for the boundary condition in which the path integral is over functions satisfying a periodic boundary condition over $[0,\beta ]$ . The factor $(-1{)}^F$ disappears if the anti-periodic boundary condition for the fermionic variables is changed into a periodic one. This can be seen from the following observation. In the path integral formalism, the trace with $(-1{)}^F$ is (see section 1.5)

$$\begin{array}{rl}\mathrm{tr}(-1{)}^F{\mathrm{e}}^{-\beta H}& =\sum _n⟨n|(-1{)}^F{\mathrm{e}}^{-\beta H}|n⟩\\ & =\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta ⟨-\theta |(-1{)}^F{\mathrm{e}}^{-\beta H}|\theta ⟩{\mathrm{e}}^{-{\theta }^{\ast }\theta }\end{array}\text{\hspace{1em}(12.173)}$$

where $F=c^{†}c$ is the Fermion number operator. By noting that

$$|\theta ⟩=|0⟩+|1⟩\theta (-1{)}^F|\theta ⟩=|0⟩-|1⟩\theta =|-\theta ⟩$$

this integral is cast into the form

$$\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta ⟨\theta |{\mathrm{e}}^{-\beta H}|\theta ⟩{\mathrm{e}}^{-{\theta }^{\ast }\theta }.\text{\hspace{1em}(12.174)}$$

Thus, by eliminating $(-1{)}^F$ , we have to change the boundary condition to a periodic one.

This path integral is evaluated in the rest of this section to show that it reduces to a topological index obtained from the Dirac $\hat{A}$ -genus.

The SUSY transformation in Euclidean time is obtained by the replacement $t\to -\mathrm{i}t$ in (12.140) as

$$\delta x^{\mu }=\mathrm{i}ϵ{\psi }^{\mu }\delta {\psi }^{\mu }=-\mathrm{i}ϵ{\dot{x}}^{\mu }.$$

As was shown in the previous subsection, the index is independent of $\beta$ and, hence, we may consider the limit $\beta \downarrow 0$ in computing the trace. By rescaling the time parameter as $t=\beta s$ , we cast the action into the form

$$\begin{array}{rl}& {\int }_0^{\beta }\mathrm{d}t\left[\frac{1}{2}{g}_{\mu \nu }(x){\dot{x}}^{\mu }{\dot{x}}^{\nu }+\frac{1}{2}{g}_{\mu \nu }(x){\psi }^{\mu }\frac{\mathrm{D}{\psi }^{\nu }}{\mathrm{D}t}\right]\\ & ={\int }_0^1\mathrm{d}s\left[\frac{1}{\beta }\frac{1}{2}{g}_{\mu v}(x)\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}s}\frac{\mathrm{d}{x}^v}{\mathrm{d}s}+\frac{1}{2}{g}_{\mu v}(x){\psi }^{\mu }\frac{\mathrm{D}{\psi }^v}{\mathrm{D}s}\right].\end{array}\text{\hspace{1em}(12.175)}$$

Thus, any path with $\dot{x}\ne 0$ has an exponentially small contribution to the path integral in the limit $\beta \downarrow 0$ . Accordingly, the contributions to the path integral come only from paths $x(t)=\text{constant}$ in this limit. Clearly, these paths satisfy the periodic boundary condition.

The periodic boundary condition forces us to take the set of loops in $M$ , which we will denote as $L(M)$ , as the configuration space of the bosonic coordinates. To apply the saddle point method to the evaluation of the path integral, we have to find the set $\mathcal{M}$ of the extrema of the action, namely the solutions of the classical Euler-Lagrange equations

$$-g_{\lambda \mu }(x)\frac{\mathrm{D}{\dot{x}}^{\mu }}{\mathrm{D}t}+\frac{1}{2}{R}_{\mu \nu \lambda \rho }{\psi }^{\mu }{\psi }^{\nu }{\dot{x}}^{\rho }=0\text{\hspace{1em}(12.176)}$$

$$\frac{\mathrm{D}{\psi }^{\mu }}{\mathrm{D}t}=\frac{\mathrm{d}{\psi }^{\mu }}{\mathrm{d}t}+{\dot{x}}^{\lambda }{\Gamma }_{\lambda v}^{\mu }{\psi }^v=0.\text{\hspace{1em}(12.177)}$$

It is instructive to outline the derivation of these equations since the anti-commutativity of Grassmann numbers and the symmetries of the Riemann tensor are fully utilized. The Euler-Lagrange equation for ${\psi }^{\mu }$ is

$$\begin{array}{rl}0& =\frac{\partial L}{\partial {\psi }^{\rho }}-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial {\dot{\psi }}^{\rho }}\right)\\ & =\frac{1}{2}{g}_{\rho v}\frac{\mathrm{D}{\psi }^v}{\mathrm{D}t}-\frac{1}{2}{g}_{\kappa v}{\psi }^{\kappa }{\dot{x}}^{\lambda }{\Gamma }_{\lambda \rho }^v+\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{\rho v}{\psi }^v\right)\\ & =\frac{1}{2}\left[g_{\rho \nu }\frac{\mathrm{D}{\psi }^{\nu }}{\mathrm{D}t}-g_{\kappa \nu }{\dot{x}}^{\lambda }{\Gamma }_{\lambda \rho }^{\nu }{\psi }^{\kappa }+\left({\partial }_{\lambda }{g}_{\rho \nu }\right){\dot{x}}^{\lambda }{\psi }^{\nu }+g_{\rho \nu }{\dot{\psi }}^{\nu }\right].\end{array}$$

By multiplying both sides by $g^{\mu \rho }$ and summing over $\rho$ , we have

$$\begin{array}{rl}0& =\frac{\mathrm{D}{\psi }^{\mu }}{\mathrm{D}t}-g^{\mu \rho }{g}_{\kappa \nu }{\dot{x}}^{\lambda }{\Gamma }_{\lambda \rho }^{\nu }{\psi }^{\kappa }+g^{\mu \rho }\left({\partial }_{\lambda }{g}_{\rho \nu }\right){\dot{x}}^{\lambda }{\psi }^{\nu }+{\dot{\psi }}^{\mu }\\ & =\frac{\mathrm{D}{\psi }^{\mu }}{\mathrm{D}t}+{\dot{\psi }}^{\mu }+{\dot{x}}^{\lambda }\left[g^{\mu \rho }\left({\partial }_{\lambda }{g}_{\rho v}\right)-g^{\mu \rho }{g}_{v\kappa }{\Gamma }_{\lambda \rho }^{\kappa }\right]{\psi }^v=2\frac{\mathrm{D}{\psi }^{\mu }}{\mathrm{D}t}\end{array}$$

which proves (12.177). Here, use has been made of the identity

$$\begin{array}{rl}& g^{\mu \rho }\left[\left({\partial }_{\lambda }{g}_{\rho v}\right)-\frac{1}{2}\left({\partial }_{\lambda }{g}_{v\rho }+{\partial }_{\rho }{g}_{v\lambda }-{\partial }_v{g}_{\lambda \rho }\right)\right]\\ & =g^{\mu \rho }\frac{1}{2}\left({\partial }_{\lambda }{g}_{\rho \nu }+{\partial }_{\nu }{g}_{\lambda \rho }-{\partial }_{\rho }{g}_{\nu \lambda }\right)={\Gamma }_{\nu \lambda }^{\mu }\end{array}$$

in the square brackets in the second line above.

Let us prove the equation of motion for $x^{\mu }$ next. We find

$$\begin{array}{rl}& \frac{\partial L}{\partial x^{\mu }}-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial {\dot{x}}^{\mu }}\right)\\ & =\frac{1}{2}\left({\partial }_{\mu }{g}_{\alpha \beta }\right){\dot{x}}^{\alpha }{\dot{x}}^{\beta }+\frac{1}{2}\left({\partial }_{\mu }{g}_{\alpha \beta }\right){\psi }^{\alpha }\frac{\mathrm{D}{\psi }^{\beta }}{\mathrm{D}t}+\frac{1}{2}{g}_{\alpha \beta }{\psi }^{\alpha }{\dot{x}}^{\lambda }{\partial }_{\mu }{\Gamma }_{\lambda \kappa }^{\beta }{\psi }^{\kappa }\\ & -\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{\mu \nu }{\dot{x}}^{\nu }+\frac{1}{2}{g}_{\alpha \beta }{\psi }^{\alpha }{\Gamma }_{\mu \kappa }^{\beta }{\psi }^{\kappa }\right)\\ & =-\left[g_{\mu \nu }{\ddot{x}}^{\nu }+\frac{1}{2}\left({\partial }_{\lambda }{g}_{\mu \nu }+{\partial }_{\nu }{g}_{\mu \lambda }-{\partial }_{\mu }{g}_{\nu \lambda }\right){\dot{x}}^{\nu }{\dot{x}}^{\lambda }\right]\end{array}$$

$$\begin{array}{rl}& +\frac{1}{2}\left[g_{\alpha \beta }{\partial }_{\mu }{\Gamma }_{\lambda \kappa }^{\beta }-{\partial }_{\lambda }{g}_{\alpha \beta }{\Gamma }_{\mu \kappa }^{\beta }-g_{\alpha \beta }{\partial }_{\lambda }{\Gamma }_{\mu \kappa }^{\beta }\right]{\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }\\ & +\frac{1}{2}{g}_{\alpha \beta }{\dot{x}}^{\lambda }{\Gamma }_{\lambda \gamma }^{\alpha }{\psi }^{\gamma }{\Gamma }_{\mu \kappa }^{\beta }{\psi }^{\kappa }+\frac{1}{2}{g}_{\alpha \beta }{\psi }^{\alpha }{\Gamma }_{\mu \kappa }^{\beta }{\dot{x}}^{\lambda }{\Gamma }_{\lambda \nu }^{\kappa }{\psi }^{\nu }\\ & =-g_{\mu \nu }\frac{\mathrm{D}{\dot{x}}^{\nu }}{\mathrm{D}t}+\frac{1}{2}[g_{\alpha \beta }{\partial }_{\mu }{\Gamma }_{\lambda \kappa }^{\beta }-g_{\alpha \beta }{\partial }_{\lambda }{\Gamma }_{\mu \kappa }^{\beta }-{\partial }_{\lambda }{g}_{\alpha \beta }{\Gamma }_{\mu \kappa }^{\beta }\\ & +g_{\gamma \beta }{\Gamma }_{\lambda \alpha }^{\gamma }{\Gamma }_{\mu \kappa }^{\beta }+g_{\alpha \beta }{\Gamma }_{\mu \gamma }^{\beta }{\Gamma }_{\lambda \kappa }^{\gamma }]{\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }\\ & =-g_{\mu \nu }\frac{\mathrm{D}{\dot{x}}^{\nu }}{\mathrm{D}t}+\frac{1}{2}\left({\partial }_{\mu }{\Gamma }_{\lambda \kappa }^{\beta }-{\partial }_{\lambda }{\Gamma }_{\mu \kappa }^{\beta }+{\Gamma }_{\mu \kappa }^{\beta }{\Gamma }_{\lambda \kappa }^{\gamma }\right){\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }\\ & +\frac{1}{2}\left(g_{\gamma \beta }{\Gamma }_{\lambda \alpha }^{\gamma }-{\partial }_{\lambda }{g}_{\alpha \beta }\right){\Gamma }_{\mu \kappa }^{\beta }{\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }.\\ & +\frac{1}{2}\left[g_{\alpha \beta }{\partial }_{\mu }{\Gamma }_{\lambda \kappa }^{\beta }-{\partial }_{\lambda }{g}_{\alpha \beta }{\Gamma }_{\mu \kappa }^{\beta }-g_{\alpha \beta }{\partial }_{\lambda }{\Gamma }_{\mu \kappa }^{\beta }\right]{\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }\end{array}$$

The last term of the last line of this equation is written as

$$\begin{array}{rl}& [g_{\gamma \beta }\frac{1}{2}{g}^{\gamma \nu }({\partial }_{\lambda }{g}_{\nu \alpha }+{\partial }_{\alpha }{g}_{\nu \lambda }-{\partial }_{\nu }{g}_{\lambda \alpha })-{\partial }_{\lambda }{g}_{\alpha \beta }]{\Gamma }_{\mu \nu }^{\beta }{\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }\\ & =-\frac{1}{2}\left({\partial }_{\lambda }{g}_{\alpha \beta }+{\partial }_{\beta }{g}_{\lambda \alpha }-{\partial }_{\alpha }{g}_{\lambda \beta }\right){\Gamma }_{\mu \nu }^{\beta }{\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }\\ & =-{\Gamma }_{\alpha \lambda \beta }{\Gamma }_{\mu \kappa }^{\beta }{\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }\\ & =-g_{\alpha \beta }{\Gamma }_{\lambda \beta }^{\beta }{\Gamma }_{\mu \kappa }^{\beta }{\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }\end{array}$$

from which we obtain

$$\begin{array}{rl}0& =-g_{\mu \nu }\frac{\mathrm{D}{\dot{x}}^{\nu }}{\mathrm{D}t}+\frac{1}{2}\left({\partial }_{\mu }{\Gamma }_{\lambda \kappa }^{\beta }-{\partial }_{\lambda }{\Gamma }_{\mu \kappa }^{\beta }+{\Gamma }_{\mu \gamma }^{\beta }{\Gamma }_{\lambda \kappa }^{\gamma }-{\Gamma }_{\lambda \gamma }^{\beta }{\Gamma }_{\mu \kappa }^{\gamma }\right){\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }\\ & =-g_{\mu \nu }\frac{\mathrm{D}{\dot{x}}^{\nu }}{\mathrm{D}t}+\frac{1}{2}{R}_{\alpha \kappa \mu \lambda }{\psi }^{\alpha }{\psi }^{\kappa }{\dot{x}}^{\lambda }.\end{array}$$

Equation (12.176) follows by renaming dummy indices.

Let us come back to the study of the solutions of the equations of motion (12.176) and (12.177). Clearly, the pair $x=$ constant and $\psi =$ constant is one of solutions. Therefore, $x_p:t\mapsto p\in M$ is always contained in the solutions, which may be written as $M\subset \mathcal{M}$ . Equation (12.176) reduces to the geodesic equation when $\psi =0$ but not necessarily so in general. When the fundamental group ${\pi }_1(M)$ is non-trivial, there exist non-contractible geodesics in general. Their contributions to the path integral, however, vanish exponentially as $\exp (-c/\beta )$ as $\beta \downarrow 0$ and, hence, are negligible.

Before we proceed to the proof of the index theorem, we need to explain the saddle point method. Let us start with a simple example. Consider the integral

$$Z={\int }_{-\infty }^{\infty }\frac{\mathrm{d}x}{\sqrt{2\pi \hslash }}{\mathrm{e}}^{-f(x)/\hslash }.$$

The function $f(x)$ is assumed to have only one minimum at $x=x_0$ and that $f(x)\to \infty$ as $x\to \pm \infty$ . Let us consider the asymptotic expansion of the integral $Z$ when the limit $\hslash \to 0$ is taken. Put $x=x_0+\sqrt{\hslash }y$ and expand $f(x)$ at $x_0$ . Taking $f^{\prime }(x_0)=0$ into account, we obtain the expansion

$$f(x)=f(x_0)+\frac{1}{2!}\hslash y^2{f}^{\prime \prime }(x_0)+\frac{1}{3!}{\hslash }^{3/2}{y}^3{f}^{(3)}(x_0)+\frac{1}{4!}{\hslash }^2{y}^4{f}^{(4)}(x_0)+\dots .$$

If this expansion is substituted into $Z$ , we have

$$\begin{array}{rl}& Z={\mathrm{e}}^{-f(x_0)/\hslash }{\int }_{-\infty }^{\infty }\frac{\mathrm{d}y}{\sqrt{2\pi }}\\ & \times \exp \left[-\frac{1}{2}{y}^2{f}^{\prime \prime }(x_0)-\left(\frac{1}{3!}{\hslash }^{1/2}{y}^3{f}^{(3)}(x_0)+\frac{1}{4!}\hslash y^4{f}^{(4)}(x_0)+\dots \right)\right].\end{array}$$

Let us define the moment of $y$ by

$$⟨y^n⟩=\frac{\int \frac{\mathrm{d}y}{\sqrt{2\pi }}{y}^n{\mathrm{e}}^{-y^2{f}^{\prime \prime }(x_0)/2}}{\int \frac{\mathrm{d}y}{\sqrt{2\pi }}{\mathrm{e}}^{-y^2{f}^{\prime \prime }(x_0)/2}}.$$

Then we finally obtain the expansion of $Z$ as

$$Z=\frac{{\mathrm{e}}^{-f(x_0)/\hslash }}{\sqrt{f^{\prime \prime }(x_0)}}⟨\exp \left[-\frac{1}{3!}{\hslash }^{1/2}{y}^3{f}^{(3)}(x_0)-\frac{1}{4!}\hslash y^4{f}^{(4)}(x_0)\dots \right]⟩.$$

One might think that one will get terms of order $\mathrm{O}({\hslash }^{1/2})$ if $⟨\dots ⟩$ is expanded. However, this is not the case since $⟨y^3⟩=0$ and one has $⟨\dots ⟩=1+O(\hslash )$ in reality. In the proof of the following index theorem, the parameter $\hslash$ is replaced by $\beta$ . The index is, however, independent of $\beta$ and we conclude that terms of order $\mathrm{O}(\beta )$ vanish and, hence, we need to take only the extrema of the action and the second-order fluctuations thereof into account.

Exercise 12.11

Use the previous expansion to prove the Staring formula

$$n!\simeq \sqrt{2\pi n}{\mathrm{e}}^{-n}{n}^n\text{\hspace{1em}(12.178)}$$

for $n\gg 1$

Let us come back to SUSYQM. We take the second-order fluctuation around the solutions of the classical equations of motion in evaluating $Z$ . The principal contribution to the path integral comes from the solution $x=x_0$ and $\psi ={\psi }_0$ . We employ the Riemann normal coordinate based at $x=x_0$ to make our life easier. This is to take a coordinate system in which the metric tensor satisfies conditions

$$g_{\mu \nu }(x_0)={\delta }_{\mu \nu }\frac{\partial }{\partial x^{\lambda }}{g}_{\mu \nu }(x_0)=0.$$

Thus, we have $g\equiv \det g=1$ . We define the fluctuations in this coordinate system as

$$x^{\mu }(t)=x_0^{\mu }+{\xi }^{\mu }(t)$$

$${\psi }^{\mu }(t)={\psi }_0^{\mu }+{\eta }^{\mu }(t).$$

7 Of course, this choice does not imply that the Riemann tensor vanishes in general.

Note here that $\mathrm{d}{x}^{\mu }=\mathrm{d}{\xi }^{\mu }$ , $\mathrm{d}{\psi }^{\mu }=\mathrm{d}{\eta }^{\mu }$ . The second-order expansion of the action is now written as

$$S_2={\int }_0^{\beta }\mathrm{d}t\left[\frac{1}{2}\frac{\mathrm{d}{\xi }^{\mu }}{\mathrm{d}t}\frac{\mathrm{d}{\xi }^{\mu }}{\mathrm{d}t}+\frac{1}{2}{\eta }^{\mu }\frac{\mathrm{d}{\eta }^{\mu }}{\mathrm{d}t}+\frac{1}{2}{\tilde{\mathcal{R}}}_{\mu v}(x_0){\xi }^{\mu }\frac{\mathrm{d}{\xi }^v}{\mathrm{d}t}\right]\text{\hspace{1em}(12.179)}$$

where we have put

$${\tilde{\mathcal{R}}}_{\mu \nu }(x_0)=\frac{1}{2}{R}_{\mu \nu \rho \sigma }(x_0){\psi }_0^{\rho }{\psi }_0^{\sigma }.$$

Needless to say, the zeroth-order action $S_0=S(x_0,{\psi }_0)$ vanishes identically.

Let us evaluate the index

$$\mathrm{ind}Q=\int \mathcal{D}\xi \mathcal{D}\eta {\mathrm{e}}^{-S_2}\text{\hspace{1em}(12.180)}$$

using the second-order action $S_2$ . Here we have taken the translational invariance of the path integral measure $\mathcal{D}x\mathcal{D}\psi =\mathcal{D}\xi \mathcal{D}\eta$ . Taking the periodic boundary condition of $\xi$ , $\eta$ into account, their Fourier expansions are given by

$${\xi }^{\mu }=\frac{1}{\sqrt{\beta }}\sum _{n=-\infty }^{\infty }{\xi }_n^{\mu }{\mathrm{e}}^{2\pi int/\beta }$$

$${\eta }^{\mu }=\frac{1}{\sqrt{\beta }}\sum _{n=-\infty }^{\infty }{\eta }_n^{\mu }{\mathrm{e}}^{2\pi int/\beta }.$$

The fluctuation operator for $\xi$ in $S_2$ is

$$-{\delta }_{\mu \nu }\frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}+{\tilde{\mathcal{R}}}_{\mu \nu }\frac{\mathrm{d}}{\mathrm{d}t}$$

while that for $\eta$ is

$${\delta }_{\mu \nu }\frac{d}{dt}.$$

We have to consider the zero modes ${\xi }_0^{\mu }$ and ${\eta }_0^{\mu }$ , for which $n=0$ , separately in the following Gaussian integrals. Taking these into account, we write

$$\begin{array}{rl}& \mathrm{ind}Q=\mathcal{N}\int \prod _{\mu =1}^d\frac{\mathrm{d}{\xi }_0^{\mu }}{\sqrt{2\pi }}\mathrm{d}{\eta }_0^{\mu }{\left[{\mathrm{Det}}_{\mathrm{PBC}}^{\prime }\left({\delta }_{\mu \nu }\frac{\mathrm{d}}{\mathrm{d}t}\right)\right]}^{1/2}\\ & \times {\left[{\mathrm{Det}}_{\mathrm{PBC}}^{\prime }\left(-{\delta }_{\mu \nu }\frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}+{\tilde{\mathcal{R}}}_{\mu \nu }(x_0)\frac{\mathrm{d}}{\mathrm{d}t}\right)\right]}^{-1/2}\\ & =\mathcal{N}\int \prod _{\mu =1}^d\frac{\mathrm{d}{\xi }_0^{\mu }}{\sqrt{2\pi }}\mathrm{d}{\eta }_0^{\mu }{\left[{\mathrm{Det}}_{\mathrm{PBC}}^{\prime }\left(-{\delta }_{\mu v}\frac{\mathrm{d}}{\mathrm{d}t}+{\tilde{\mathcal{R}}}_{\mu v}(x_0)\right)\right]}^{-1/2}(12.181)\end{array}$$

where $\prime$ indicates that the zero modes are omitted while $\mathcal{N}$ is the normalization factor, which takes care of the ambiguities associated with the ordering of Grassmann numbers. Let us evaluate this factor now.

Since $\mathrm{ind}Q$ is independent of $\beta$ , we put $\beta =1$ for simplicity. We also simplify our calculation by choosing the metric to be $g_{\mu \nu }={\delta }_{\mu \nu }$ . Then the fermion and boson parts separate completely. The fermionic part is evaluated, by noting $H_{\mathrm{fermion}}=0$ , to yield

$$\begin{array}{rl}\mathrm{Tr}{\gamma }_{2n+1}& ={\int }_{\mathrm{PBC}}\mathcal{D}\psi {\mathrm{e}}^{-\frac{1}{2}{\int }_0^1\psi \cdot \dot{\psi }\mathrm{d}t}\\ & ={\mathcal{N}}_f{\mathrm{Det}}_{\mathrm{PBC}}^{\prime }({\delta }_{\mu \nu }{\partial }_t{)}^{1/2}\int \mathrm{d}{\psi }_0^1\dots \mathrm{d}{\psi }_0^{2n},\end{array}$$

where ${\psi }_0^{\mu }$ is the zero mode. The determinant is evaluated as follows. First, note that the argument in section 1.5 shows that the determinant is, in fact,

$${\mathrm{Det}}_{\mathrm{PBC}}^{\prime }\left({\partial }_t+\omega \right)=\underset{\epsilon \to 0}{\lim }{\mathrm{Det}}^{\prime }\left(\left(1-\epsilon \omega \right){\partial }_t+\omega \right)$$

where we have introduced the harmonic oscillator frequency $\omega$ , which will be set to zero at the end of the calculation. The ‘partition function’ is

$$\begin{array}{rl}\mathrm{tr}(-1{)}^F{\mathrm{e}}^{-\beta H}& =2\sinh (\beta \omega /2)\\ & ={\mathrm{e}}^{\beta \omega /2}{\mathrm{Det}}_{\mathrm{PBC}}^{\prime }((1-\epsilon \omega ){\partial }_t+\omega ).\end{array}\text{\hspace{1em}(12.182)}$$

Therefore, the determinant in the limit $\omega \to 0$ is

$${\mathrm{Det}}_{\mathrm{PBC}}^{\prime }\left({\partial }_t\right)=\underset{\omega \to 0}{\lim }{\mathrm{e}}^{-\beta \omega /2}2\sinh (\beta \omega /2)=1.\text{\hspace{1em}(12.183)}$$

Thus, we finally obtained

$$\mathrm{Tr}{\gamma }_{2n+1}={\mathcal{N}}_f\int \mathrm{d}{\psi }_0^1\dots \mathrm{d}{\psi }_0^{2n}.\text{\hspace{1em}(12.184)}$$

We insert

$${\gamma }_{2n+1}={\mathrm{i}}^n{\gamma }_0^1\dots {\gamma }_0^{2n}=(2\mathrm{i}{)}^n{\psi }_0^1\dots {\psi }_0^{2n}$$

further in the trace. Since $\mathrm{Tr}{\gamma }_{2n+1}^2=\mathrm{Tr}I=2^n$ , we obtain

$$\mathrm{Tr}{\gamma }_{2n+1}^2=2^n={\mathcal{N}}_f\int \mathrm{d}{\psi }_0^1\dots \mathrm{d}{\psi }_0^{2n}(2\mathrm{i}{)}^n{\psi }_0^1\dots {\psi }_0^{2n}={\mathcal{N}}_f(-2\mathrm{i}{)}^n$$

which leads to

$${\mathcal{N}}_f={\mathrm{i}}^n.$$

Next, we evaluate the normalization factor ${\mathcal{N}}_b$ of the boson part. If we employ imaginary time in (1.101) to obtain $⟨x,1|x,0⟩=(2\pi {)}^{-1/2}$ , we have

$$\int \mathcal{D}{x}^{\mu }{\mathrm{e}}^{-\frac{1}{2}{\int }_0^1{\dot{x}}^{\mu }{}^2}={\mathcal{N}}_b\frac{1}{{\mathrm{Det}}^{1/2}(-{\delta }_{\mu \nu }{\partial }_t^2)}\int \prod _{\mu =1}^{2n}\frac{\mathrm{d}{x}^{\mu }}{\sqrt{2\pi }}=(2\pi {)}^{-n}\int \prod _{\mu =1}^{2n}\mathrm{d}{x}^{\mu }.$$

The determinant is evaluated using the $\zeta$ -function regularization as in section 1.4. The eigenvalue of $-{\mathrm{d}}^2/\mathrm{d}{t}^2$ with the periodic boundary condition is ${\lambda }_n=(2n\pi /\beta {)}^2$ and then

$${\mathrm{Det}}_{\mathrm{PBC}}^{\prime }\left(-\frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}\right)=\prod _{n\in \mathbb{Z},n\ne 0}{\left(\frac{2\pi n}{\beta }\right)}^2.$$

The spectral $\zeta$ -function is

$${\zeta }_{-{\mathrm{d}}^2/\mathrm{d}{t}^2}(s)=\sum _{n\in \mathbb{Z},n\ne 0}^{\infty }{\left[{\left(\frac{2n\pi }{\beta }\right)}^2\right]}^{-s}=2{\left(\frac{\beta }{2\pi }\right)}^{2s}\zeta (2s)$$

from which we find

$$\begin{array}{rl}{\zeta }_{-{\mathrm{d}}^2/\mathrm{d}{t}^2}^{\prime }(0)& =4\log (\beta /2\pi ){\mathrm{e}}^{2s\log (\beta /2\pi )}\zeta (2s)+4{\mathrm{e}}^{2s\log (\beta /2\pi )}{\zeta }^{\prime }(2s){|}_{s=0}\\ & =4\left[\log \left(\beta /2\pi \right)\zeta (0)+{\zeta }^{\prime }(0)\right]=-2\log \beta .\end{array}$$

Therefore, the determinant is

$${\mathrm{Det}}_{\mathrm{PBC}}^{\prime }\left(-\frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}\right)=\exp \left[-{\zeta }_{-{\mathrm{d}}^2/\mathrm{d}{t}^2}^{\prime }(0)\right]={\beta }^2.\text{\hspace{1em}(12.185)}$$

By putting $\beta =1$ , we find ${\mathrm{Det}}_{\mathrm{PBC}}^{\prime }(-{\mathrm{d}}^2/\mathrm{d}{t}^2)=1$ . Thus, we have obtained the normalization factor

$${\mathcal{N}}_b=1.$$

Putting these results together, we have shown that $\mathcal{N}={\mathcal{N}}_f{\mathcal{N}}_b={\mathrm{i}}^n$ . Accordingly, the index is expressed as

$$\mathrm{ind}Q={\mathrm{i}}^n\int \prod _{\mu =1}^d\frac{\mathrm{d}{\xi }_0^{\mu }}{\sqrt{2\pi }}\mathrm{d}{\eta }_0^{\mu }{\left[{\mathrm{Det}}_{\mathrm{PBC}}^{\prime }\left(-{\delta }_{\mu v}\frac{\mathrm{d}}{\mathrm{d}t}+{\tilde{\mathcal{R}}}_{\mu v}(x_0)\right)\right]}^{-1/2}.\text{\hspace{1em}(12.186)}$$

Let us evaluate the functional determinant in (12.186). Since the Fermi variables are contained only in ${\tilde{\mathcal{R}}}_{\mu \nu }(x_0)$ and this is Grassmann-even, we pretend this part is a commuting number for the time being. The anti-symmetry of the Riemann tensor implies that ${\tilde{\mathcal{R}}}_{\mu \nu }(x_0)$ satisfies ${\tilde{\mathcal{R}}}_{\mu \nu }=-{\tilde{\mathcal{R}}}_{\nu \mu }$ . Therefore, it is possible, in an even-dimensional manifold $M$ , to block-diagonalize ${\tilde{\mathcal{R}}}_{\mu \nu }$ in the form

$${\tilde{\mathcal{R}}}_{\mu v}=\left(\begin{array}{ccccc}0& y_1& & & \\ -y_1& 0& & & \\ & & \ddots & & \\ & & & 0& y_n\\ & & & -y_n& 0\end{array}\right).\text{\hspace{1em}(12.187)}$$

Let us concentrate on the first block. The operator

$$-{\delta }_{\mu \nu }\frac{\mathrm{d}}{\mathrm{d}t}+{\tilde{\mathcal{R}}}_{\mu \nu }(x_0)$$

is real and, hence, the eigenvalues are made of complex conjugate pairs. Let us express the determinant of this block in terms of the product of these complex eigenvalues. We find

$$\begin{array}{rl}\stackrel{\prime }{\det }\left(\begin{array}{cc}-\frac{\mathrm{d}}{\mathrm{d}t}& y_1\\ -y_1& -\frac{\mathrm{d}}{\mathrm{d}t}\end{array}\right)& ={\mathrm{Det}}^{\prime }\left(\frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}+y_1^2\right)=\prod _{n\ne 0}\left(y_1^2-(2\pi n/\beta {)}^2\right)\\ & ={\left[\prod _{n\ge 1}{\left(\frac{2\pi n}{\beta }\right)}^2\prod _{n\ge 1}\left[1-{\left(\frac{y_1\beta }{2\pi n}\right)}^2\right]\right]}^2\\ & ={\left(\frac{\sin \beta y_1/2}{y_1/2}\right)}^2.\end{array}\text{\hspace{1em}(12.188)}$$

Now the index is expressed as

$$\mathrm{ind}Q={\mathrm{i}}^n\int \prod _{\mu =1}^{2n}\frac{\mathrm{d}{\xi }_0^{\mu }}{\sqrt{2\pi }}\mathrm{d}{\eta }_0^{\mu }\prod _{j=1}^n\frac{y_j/2}{\sin \beta y_j/2}.\text{\hspace{1em}(12.189)}$$

The product with respect to $j$ is written as

$$\frac{1}{{\beta }^{d/2}}\det {\left(\frac{\beta \tilde{\mathcal{R}}/2}{\sin \beta \tilde{\mathcal{R}}/2}\right)}^{1/2}.$$

Note that any Taylor expansion with respect to $\tilde{\mathcal{R}}$ terminates at finite order since ${\tilde{\mathcal{R}}}^p=0$ for $p>d/2$ .

We have evaluated the contributions of the second-order fluctuations around a particular pair $x_0,{\psi }_0$ so far. Now we need to take the contributions coming from all the solutions to the classical equations of motion into account. We have noted before that the set $\mathcal{M}$ of the solutions of the equations of motion contains the constant solution $(x_0,{\psi }_0)$ as a subset and that the contributions from nonconstant solutions are exponentially small as $\beta \downarrow 0$ . Therefore, we neglect all periodic solutions except for constant solutions. If we note the expansion

$$x^{\mu }=x_0^{\mu }+\frac{1}{\sqrt{\beta }}{\xi }_0^{\mu }+\dots$$

we find that the integral over $x_0$ is equivalent with that over ${\xi }_0/\sqrt{\beta }$ , namely $\mathrm{d}{x}_0^{\mu }=\mathrm{d}{\xi }_0^{\mu }/\sqrt{\beta }$ . This argument is also applied to the Grassmannian zero mode

and we find $\mathrm{d}{\psi }_0^{\mu }=\sqrt{\beta }\mathrm{d}{\eta }_0^{\mu }$ . In summary, the index is now written as

$$\mathrm{ind}Q={\mathrm{i}}^n\int \prod _{\mu =1}^{2n}\frac{\mathrm{d}{x}_0^{\mu }}{\sqrt{2\pi }}\mathrm{d}{\psi }_0^{\mu }\frac{1}{{\beta }^{d/2}}\det {\left(\frac{\beta \tilde{\mathcal{R}}/2}{\sin \beta \tilde{\mathcal{R}}/2}\right)}^{1/2}.\text{\hspace{1em}(12.190)}$$

We make the following change of variables to erase the apparent $\beta$ -dependence of the index,

$${\psi }_0^{\mu }=\frac{{\chi }_0^{\mu }}{\sqrt{2\pi \beta }},\mathrm{d}{\psi }_0^{\mu }=\sqrt{2\pi \beta }\mathrm{d}{\chi }_0^{\mu }.$$

Substituting

$$\beta {\tilde{\mathcal{R}}}_{\mu \nu }=\frac{1}{2\pi }\frac{1}{2}{\mathcal{R}}_{\mu \nu \rho \sigma }{\chi }_0^{\rho }{\chi }_0^{\sigma }$$

into the integrand, we obtain

$$\mathrm{ind}Q={\mathrm{i}}^n\int \prod _{\mu =1}^{2n}\mathrm{d}{x}_0^{\mu }\mathrm{d}{\chi }_0^{\mu }\det {\left(\frac{\frac{1}{2}\frac{1}{2\pi }\frac{1}{2}{\mathcal{R}}_{\mu \nu \rho \sigma }(x_0){\chi }_0^{\rho }{\chi }_0^{\sigma }}{\sin \frac{1}{2}\frac{1}{2\pi }\frac{1}{2}{\mathcal{R}}_{\mu \nu \rho \sigma }(x_0){\chi }_0^{\rho }{\chi }_0^{\sigma }}\right)}^{1/2}.\text{\hspace{1em}(12.191)}$$

This is the Atiyah-Singer index theorem for the Dirac operator.

Let us rewrite the previous theorem in a more familiar form. Note that only terms of order $2n$ in $\chi$ in the integrand yield non-vanishing contributions upon integration over $\prod \mathrm{d}{\chi }_0^{\mu }$ . Note also that $\prod \mathrm{d}{x}_0^{\mu }$ is just an ordinary volume element. Then define the curvature two-form

$${\mathcal{R}}_{\mu v}=\frac{1}{2}{R}_{\mu v\rho \sigma }\mathrm{d}{x}^{\rho }\wedge \mathrm{d}{x}^{\sigma }.\text{\hspace{1em}(12.192)}$$

Then note that $\mathcal{R}/\sin \mathcal{R}$ is even in $\mathcal{R}$ and, hence, the integral is non-vanishing only when $n$ is even, that is only when $d$ is a multiple of four. If this is the case, the factor ${\mathrm{i}}^n$ takes only $\pm 1$ and we can formally replace the integrand as

$${\mathrm{i}}^n\frac{\mathcal{R}}{\sin \mathcal{R}}\to \frac{\mathcal{R}}{\sinh \mathcal{R}}.$$

The reader should verify the first few terms. Then the index is now written in the well-known form as

$$\mathrm{ind}Q={\int }_M\det {\left(\frac{\frac{1}{2}\frac{1}{2\pi }\mathcal{R}}{\sinh \frac{1}{2}\frac{1}{2\pi }\mathcal{R}}\right)}^{1/2}.$$

We, moreover, define the $\hat{A}$ -genus. Since $\mathcal{R}$ is anti-symmetric, it can be block-

diagonalized as

$$\frac{1}{2\pi }{\mathcal{R}}_{\mu \nu }=\left(\begin{array}{ccccc}0& x_1& & & \\ -x_1& 0& & & \\ & & \ddots & & \\ & & & 0& x_n\\ & & & -x_n& 0\end{array}\right).$$

Then define the $\hat{A}$ -genus of $M$ by

$$\hat{A}(M)=\prod _{j=1}^n\frac{x_j/2}{\sinh x_j/2}\text{\hspace{1em}(12.193)}$$

where the RHS is defined by its formal expansion with respect to $x_j$ .

In summary, we have proved the Atiyah-Singer index theorem in the simplest setting (the spin complex).

Theorem 12.6 · Index theorem for a spin complex

The index of a Dirac operator defined in $M$ is

$$\mathrm{ind}Q={\int }_M\hat{A}(M).\text{\hspace{1em}(12.194)}$$

12.1 In the text, we dealt only with compact manifolds. The extension of the AS index theorem to non-compact manifolds is the Callias-Bott-Seely index theorem (Callias 1978, Bott and Seely 1978). Here we consider the simplest case studied by Hirayama (1983). Consider a pair of operators

$$L\equiv \frac{1}{\mathrm{i}}\frac{\mathrm{d}}{\mathrm{d}x}-\mathrm{i}W(x)L^{†}\equiv \frac{1}{\mathrm{i}}\frac{\mathrm{d}}{\mathrm{d}x}+\mathrm{i}W(x)$$

where $W(+\infty )=\mu$ and $W(-\infty )=\lambda$ .

(a) Show that ${\mathrm{Spec}}^{\prime }{L}^{†}L={\mathrm{Spec}}^{\prime }L{L}^{†}$ , where the prime indicates that the zero eigenvalues are omitted.
(b) Show that

$$J(z)\equiv \mathrm{tr}\left(\frac{z}{L^{†}L+z}-\frac{z}{L{L}^{†}+z}\right)=\frac{1}{2}\left(\frac{\mu }{({\mu }^2+z{)}^{1/2}}-\frac{\lambda }{({\lambda }^2+z{)}^{1/2}}\right).$$

Endnotes

Unplaced source note

The integrations over ${\xi }_0$ and ${\eta }_0$ are equivalent with those over $x_0$ and ${\psi }_0$ .

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：8e5daa17552aefc71a9622c9ffb168c6218e6eafc3c723bc46da387abf893f4d

Footnotes

1. Note the slight abuse of notations. The symbols ${\psi }_{\pm }$ have been used to denote sections in $\Gamma (M,S)$ as well as those in $\Gamma (M,{\Delta }^{\pm })$ . ↩