[Nakahara GTP] 12.6 Spin complexes

   This article is one of Manifold, Differential Geometry, Fibre Bundle.

 

12.6.1 Dirac operator 

Let us consider a spin bundle $S(M)$ over an $m$-dimensional orientable manifold $M$. We shall denote the set of sections of this bundle by $\Delta(M)=\Gamma(M,S(M))$. We assume that $m=2l$ is an even integer. The spin group $\mbox{SPIN}(m)$ is generated by $m$ Dirac matrices $\{\gamma^\alpha\}$, which satisfy $$\begin{align}\gamma^{\alpha\dagger}=\gamma^\alpha \\ \{\gamma^\alpha,\gamma^\beta\}=2\delta^{\alpha\beta}.\end{align}$$ Throughout this chapter we assume that the metric has the Euclidean signature. The Clifford algebra is generated by $$ 1;\ \gamma^\alpha;\ \gamma^{\alpha_1}\gamma^{\alpha_2}(\alpha_1<\alpha_2);\ \cdots;\\ \gamma^{\alpha_1}\cdots\gamma^{\alpha_k}(\alpha_1<\cdots<\alpha_k);\ \cdots;\ \gamma^1\cdots\gamma^{2l}.$$ The last generator is of particular importance and we define $$\begin{align}\gamma^{m+1}\equiv i^l\gamma^1\cdots\gamma^m.\end{align}$$ Our convention is such that $(\gamma^{m+1})^2=I$ and $(\gamma^{m+1})^\dagger=\gamma^{m+1}$. It can be shown from the general theory of the Clifford algebra that the $\gamma^x$ are represented by $2^l\times 2^l$ matrices with complex entries. It is convenient to take a representation of $\{\gamma^x\}$ such that $\gamma^{m+1}$ is diagonal, $$\begin{align}\gamma^{m+1}=\begin{pmatrix}\mathbf{1}&0\\0&-\mathbf{1}\end{pmatrix}\end{align}$$ where $\mathbf{1}$ here is the $2^{l-1}\times 2^{l-1}$ unit matrix.

Example 12.4. Form $m=2$, we take $$\gamma^0=\sigma_2\quad \gamma^1=\sigma_1\quad \gamma^3=i\gamma^0\gamma^1=\sigma_3$$ $\sigma_\alpha$ being Pauli matrices, $$\sigma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}\quad \sigma_2=\begin{pmatrix}0&-i\\i&0\end{pmatrix}\quad \sigma_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ For $m=4$, we may take $$\gamma\beta=\begin{pmatrix}0&i\alpha^\beta\\ -i\bar{\alpha}^\beta & 0\end{pmatrix}\quad \alpha^\beta=(I_2,-i\vec{\sigma}),\ \bar{\alpha}^\beta=(I_2,i\vec{\sigma})\\ \gamma^5=-\gamma^0\gamma^1\gamma^2\gamma^3=\begin{pmatrix}I_2&0\\ 0&0I_2\end{pmatrix}.$$ 

A Dirac spinor $\psi\in\Delta(M)$ is an irreducible representation of the Clifford algebra but not that of $\mbox{SPIN}(2l)$. Irreducible representation of $\mbox{SPIN}(2l)$ are obtained by separating $\Delta(M)$ according to the eigenvalues of $\gamma^{m+1}$. Since $(\gamma^{m+1})^2=I$, the eigenvalues of $\gamma^{m+1}$, called the chirality, must be $\pm1$. Then $\Delta(M)$ is separated into two eigenspaces $$\begin{align}\Delta(M)=\Delta^+(M)\oplus \Delta^-(M)\end{align}$$ where $\gamma^{m+1}\psi^\pm=\pm\psi^\pm$ for $\psi^\pm\in \Delta^\pm(M)$. The projection operators $\mathcal{P}^\pm$ onto $\Delta^\pm$ are given by $$\begin{align}\mathcal{P}^+\equiv \frac{1}{2}(I+\gamma^{m+1})=\begin{pmatrix}\mathbf{1}&0\\ 0&0\end{pmatrix}\\ \mathcal{P}^-\equiv \frac{1}{2}(I-\gamma^{m+1})=\begin{pmatrix}0&0\\0&\mathbf{1}\end{pmatrix}.\end{align}$$ Thus, we may write $$\begin{align}\psi^+=\begin{pmatrix}\psi^+\\ 0\end{pmatrix}\in\Delta^+(M),\quad \psi^-=\begin{pmatrix}0\\ \psi^+\end{pmatrix}\in\Delta^-(M).\end{align}$$ The reader should verify that $\mathcal{P}^++\mathcal{P}^-=\mathbf{1}$, $(\mathcal{P}^\pm)^2=\mathcal{P}^\pm$, $\mathcal{P}^+\mathcal{P}^-=0$, $\mathcal{P}^\pm\psi^\pm=\psi^\pm$ and $\mathcal{P}^\pm \psi^\mp=0$.

The Dirac operator in a curved space is given by (section 7.10) $$\begin{align}i\nabla\!\!\!\! /\psi\equiv i\gamma^\mu \nabla_{\partial/\partial x^\mu}\psi=i\gamma^\mu (\partial_\mu+\omega_\mu)\psi\end{align}$$ where $\omega_\mu=\frac{1}{2}i\omega_\mu^{\alpha\beta}\sum_{\alpha\beta}$ is the spin connection and $\gamma^\mu=\gamma^\alpha e_\alpha^\mu$. Let us prove that $i\nabla\!\!\!\! /$ is elliptic. Let $f$ be a function defined near $p\in M$ such that $f(p)=0$ and $i\gamma^\mu\partial_\mu f(p)=i\gamma^\mu\xi_\mu\equiv i\xi\!\!/$. Take a section $\tilde{\psi}\in\Delta(M)$ such that $\tilde{\psi}(p)=\psi$. From (12.1.4), we have $$\sigma(i\nabla\!\!\!\! /,\xi)\psi=i\nabla\!\!\!\! /(f\tilde{\psi})|_p=(i\nabla\!\!\!\! /f)\tilde{\psi}|_p=i\xi\!\!\!/\psi$$ which shows that $$\begin{align}\sigma(i\nabla\!\!\!\! /,\xi\!\!\! /)=i\xi\!\!\! /.\end{align}$$ If we note that $\xi\!\!/\xi\!\!/=\xi_\alpha\xi_\beta\gamma^\alpha \gamma^\beta=\xi^\mu\xi_\mu$, we find that (10) is invertible for $i\xi\!\!\! /\ne 0$, hence $i\nabla\!\!\!\! /$ is an elliptic operator.

It can be shown that $\{\gamma^\alpha\}$ is taken in the form $$\begin{align}\gamma^\beta =\begin{pmatrix}0&i\alpha_\beta\\ -i\bar{\alpha}_\beta&0\end{pmatrix}\quad \alpha^\dagger_\beta=\bar{\alpha}_\beta\end{align}$$ see example 12.4 for $m=2$ and $4$. Then (9) becomes $$\begin{align}i\nabla\!\!\!\! /=\begin{pmatrix}0&D^\dagger\\ D&0\end{pmatrix}\end{align}$$ where $$\begin{align}D\equiv \bar{\alpha}^\beta e_\beta^\mu(\partial_\mu+\omega_\mu)\quad D^\dagger \equiv -\alpha^\beta e_\beta^\mu (\partial_\mu + \omega_\mu).\end{align}$$ Hence, $D^\dagger$ is, indeed, the adjoint of $D$ (note that $\partial_\mu+\omega_\mu$ is anti-Hermitian). For $$\begin{pmatrix}\psi^+\\ 0\end{pmatrix} \in \Delta^+(M)$$ we have $$i\nabla\!\!\!\! /\begin{pmatrix}\psi^+\\ 0\end{pmatrix}=\begin{pmatrix} 0&D^\dagger\\ D&0\end{pmatrix}\begin{pmatrix} \psi^+\\ 0\end{pmatrix}=\begin{pmatrix}0\\ D\psi^+\end{pmatrix}$$ while for $$\begin{pmatrix}0 \\ \psi^-\end{pmatrix}\in \Delta^-(M)$$ we have $$i\nabla\!\!\!\! /\begin{pmatrix}0\\ \psi^-\end{pmatrix}=\begin{pmatrix}D^\dagger \psi^-\end{pmatrix}.$$ Hence, $D=i\nabla\!\!\!\! / \mathcal{P}^+:\Delta^+(M)\rightarrow \Delta^-(M)$ and $D^\dagger=i\nabla\!\!\!\! /\mathcal{P}^-:\Delta^-(M)\rightarrow \Delta^+(M)$. Now we have a two-term complex $$\begin{align}\Delta^+(M)\stackrel{D}{\rightarrow}\Delta^-(M)\end{align}$$ called the spin complex. The analytical index of this complex is $$\begin{align}\mbox{ind}D=\dim\ker D-\dim\ker D^\dagger =\nu_+-\nu_-\end{align}$$ where $\nu_+$ ($\nu_-$) is the number of zero-energy modes of chirality $+$ ($-$).

Let us apply the AS index theorem to this case. Without getting into the details of the Clifford algebra and the spin complex, we simply write down the result. The AS index theorem for the spin complex (14) is $$\begin{align}\nu_+-\nu_-=\int_M\mbox{ch}(\Delta^+(M)-\Delta^-(M))\left. \frac{\mbox{Td}(TM^\mathbb{C})}{e(TM)}\right|_{vol}\nonumber\\ =\int_M\hat{A}(TM)|_{vol}\end{align}$$ where $\hat{A}$ is the Dirac genus defined by (11.94). Since $\hat{A}$ contains only $4j$-forms, $\nu_+-\nu_-$ vanishes unless $m=0\ \mbox{mod} 4$. Of course, this does note necessarily imply $\nu_+=\nu_-=0$. The proof of (16) will be given (16) will be given later in sections 12.9 and 12.10.

12.6.2 Twisted spin complexes

In physics, a spinor field may belong to a representation of a group $G$. For example, the quark field in QCD belongs to the $\mathbf{3}$ of $SU(3)$. A spinor which belongs to a representation of $G$ is a section of the product bundle $S(M)\otimes E$, where $E$ is an associated vector bundle of $P(M,G)$ in an appropriate representation. The Dirac operator $D_E:\Delta^+(M)\otimes E\rightarrow \Delta(M)^-\otimes E$ in this case is $$\begin{align}D_E=i\gamma^\alpha e_\alpha^\mu(\partial_\mu+\omega_\mu+\mathcal{A}_\mu)\mathcal{D}_+\end{align}$$ where $\mathcal{A}_\mu$ is the gauge potential on $E$. The AS index theorem for this twisted spin complex is $$\begin{align}\nu_+-\nu_-=\int_M \hat{A}(TM)\mbox{ch}(E)|_{vol}.\end{align}$$

For $\dim M=2$, we have $$\begin{align}\nu_+-\nu_-=\int_M\mbox{ch}_1(E)=\frac{i}{2\pi}\int_M\mbox{tr}\mathcal{F}\end{align}$$ while for $\dim M=4$, $$\begin{align}\nu_+-\nu_-=\int_M[\mbox{ch}_2(E)+\hat{A}_1(TM)\mbox{ch}_0(E)]\nonumber\\ =\frac{-1}{8\pi^2}\int_M \mbox{tr}\mathcal{F}^2+\frac{\dim E}{192\pi^2}\int_M \mbox{tr}\mathcal{R}^2.\end{align}$$

Example 12.5. Let $$M=T^{2l}=S^1\times \cdots\times S^1.$$ Then we find $$\hat{A}(TM)=\hat{A}\left( \oplus^{2l}_1 TS^1\right) =\prod^{2l}_1 \hat{A}(TS^1)=1.$$ We also have $\hat{A}(TS^{2l})=1$. Accordingly, the index of these bundles is $$\begin{align}\nu_+-\nu_-=\int_M \mbox{ch}(E)|_{vol}.\end{align}$$

Example 12.6. Let us consider the monopole bundle $P(S^2,U(1))$. If $\mathcal{A}$ is the local gauge potential, the field strength is $\mathcal{F}=d\mathcal{A}$. The index theorem is $$\begin{align}\nu_+-\nu_-=\frac{i}{2\pi}\int_{S^2}\mathcal{F}=-\frac{1}{2\pi}\int_{S^2}F\end{align}$$ where $\mathcal{F}=iF$. As was shown in section 10.5, the RHS represents the winding number $\pi_1(U(1))=\mathbb{Z}$ and analytical information (the LHS) is now expressed in a topological way (the RHS).

Let $P(S^4,SU(2))$ be the instanton bundle. Expression (21) reads as $$\begin{align}\nu_+-\nu_-=\int_{S^4}\mbox{ch}_2(\mathcal{F})=\frac{-1}{8\pi^2}\int_{S^4}\mbox{tr}\mathcal{F}^2.\end{align}$$ The RHS represents the instanton number $k\in\pi_3(SU(2))=\mathbb{Z}$. Note that $k>0$ if $\mathcal{F}=\star \mathcal{F}$ while $k<0$ if $\mathcal{F}=-\star\mathcal{F}$. It can be shown that $\nu_-=0$ ($\nu_+=0$) if $k>0$ ($k<0$). For example, let $\mathcal{F}$ be self-dual. Suppose $\psi^-\in\ker D^\dagger=\ker DD^\dagger$. From (13), we find that $$DD^\dagger\psi^-=[(\partial_\mu+\mathcal{A}_\mu)^2+2i\bar{\sigma}_{\mu\nu}\mathcal{F}^{\mu\nu}]\psi^-=0$$ where $\bar{\sigma}_{\mu\nu}\equiv (1/4i)(\alpha^\mu \bar{\alpha}^\nu-\alpha^\nu\bar{\sigma}^\mu)$. It is easily verified that $\bar{\sigma}^{\mu\nu}$ is anti-self-dual ($\bar{\sigma}^{\mu\nu}=-\star \bar{\sigma}^{\nu\mu}$) and hence $\bar{\sigma}_{\mu\nu}\mathcal{F}^{\mu\nu}=0$. Since $(\partial_\mu+\mathcal{A}_\mu)^2$ is a positive-definite operator, it has no normalizable bound states. This verifies that $\ker D^\dagger=\varnothing$.