I get an extra term when I try to derive $\nabla\!\!\!\!/\ ^2=\nabla_\mu \nabla^\mu+\frac{1}{4}[\gamma^\mu,\gamma^\nu]F_{\mu\nu}$

Question

The equation in the title can be found in Nakahara's book, page 507 (chapter 13, Anomalies in Gauge Field Theories) and in Fujikawa's book, pages 69 and 78 (chapter 5, The Jacobian in Path Integrals and Quantum Anomalies). However, when I try to derive this equation, using the trick $BA=\frac{1}{2}\{B,A\}+\frac{1}{2}[B,A]$, I get an extra term: \begin{align*} &(\gamma^\mu\gamma^\nu)(\nabla_\mu\nabla_\nu)=\frac{1}{2}\big(\{\gamma^\mu,\gamma^\nu\}+[\gamma^\mu,\gamma^\nu]\big)\frac{1}{2}\big(\{\nabla_\mu,\nabla_\nu\}+[\nabla_\mu,\nabla_\nu]\big)&\\ &=\underbrace{\frac{1}{4}\{\gamma^\mu,\gamma^\nu\}\{\nabla_\mu,\nabla_\nu\}+\frac{1}{4}\{\gamma^\mu,\gamma^\nu\}[\nabla_\mu,\nabla_\nu]}_{\frac{1}{2}\{\gamma^\mu,\gamma^\nu\}\nabla_\mu\nabla_\nu=\eta^{\mu\nu}\nabla_\mu\nabla_\nu=\nabla_\mu\nabla^\mu}+\frac{1}{4}[\gamma^\mu,\gamma^\nu]\{\nabla_\mu,\nabla_\nu\}+\frac{1}{4}[\gamma^\mu,\gamma^\nu]\underbrace{[\nabla_\mu,\nabla_\nu]}_{=F_{\mu\nu}}&\\ &=\nabla_\mu\nabla^\mu+\underbrace{\frac{1}{4}[\gamma^{\mu}\gamma^{\nu}]\{\nabla_\mu,\nabla_\nu\}}_{\text{extra term}}+\frac{1}{4}[\gamma^{\mu},\gamma^{\nu}]F_{\mu\nu}& \end{align*}

Is the "extra term" zero for some reason or is $\nabla^\mu$ defined in a different way (I assumed $\nabla^\mu=\eta^{\mu\nu}\nabla_\nu$)?

Answer 1

Let $P\to M$ be a principal $G$-bundle with a representation $\rho\colon G\to\mathrm{GL}(V)$.

The "extra term" is the endomorphism $O\colon C^{\infty}(M,\Delta\otimes V)\to C^{\infty}(M,\Delta\otimes V)$ defined by \begin{equation} O(\psi)=[\gamma^k,\gamma^l]\cdot\{\nabla_k,\nabla_l\}(\psi)=\sum_{k,l}[\gamma^k,\gamma^l]\cdot\{\nabla_k,\nabla_l\}(\psi). \end{equation} We prove $O=-O$ (which is equivalent to $O=0$):

Since $[A,B]=-[B,A]$ and $\{A,B\}=\{B,A\}$, \begin{equation} O(\psi)=[\gamma^k,\gamma^l]\cdot\{\nabla_k,\nabla_l\}(\psi)=-[\gamma^l,\gamma^k]\cdot\{\nabla_l,\nabla_k\}(\psi)=-O(\psi) \end{equation} for all $\psi\in C^{\infty}(M,\Delta\otimes V)$.

(To be precise, we would have to replace $\gamma^\mu$ by $\gamma^\mu\otimes 1$.)