[Nakahara GTP] 13.3 Non-Abelian anomalies

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

In the last section we considered the chiral current which is a gauge singlet. Now we turn to the study of the gauge current $j^\mu_\alpha$ where $\alpha$ is the gauge index. Here we consider a chiral theory in which the gauge field $\mathcal{A}$ couples only to the left-handed Weyl fermion $\psi$. Suppose $\psi$ transforms in a complex representation $\mathbf{r}$ of the gauge group $G$. For example, suppose $\psi$ belongs to a $\mathbf{3}$ of $SU(3)$. The effective action $W_\mathbf{r}[\mathcal{A}]$ is given by $$\begin{align}e^{-W_\mathbf{r}[\mathcal{A}]}=\int \mathcal{D}\psi \mathcal{D} \bar{\psi}\exp \left( -\int dx\, \bar{\psi}i\nabla\!\!\!\! /_+\psi\right)\end{align}$$ where $$\begin{align}i\nabla\!\!\!\! /_+=i\gamma^\mu (\partial_\mu+\mathcal{A}_\mu)\mathcal{P}_+\quad \mathcal{P}_\pm=\frac{1}{2}(1\pm \gamma^{m+1}).\end{align}$$ The gauge current is $$\begin{align}j^\mu_\alpha=i\bar{\psi}\gamma^\mu T_\alpha \mathcal{P}_+\psi.\end{align}$$ Let $v=v^\alpha T_\alpha$ be an infinitesimal gauge transformation parameter, $g=1-v$ under which we have $$\begin{align}\mathcal{A}_\mu\rightarrow (1+v)(\mathcal{A}_\mu+d)(1-v)=\mathcal{A}_\mu-\mathcal{D}_\mu v\end{align}$$ where $\mathcal{D}_\mu v\equiv \partial_\mu v+[\mathcal{A}_\mu,v]$ is the covariant derivative for a field in the adjoint representation. The effective action transforms as $$\begin{align}W_r[\mathcal{A}]\rightarrow W_r[\mathcal{A}-\mathcal{D}v]\nonumber\\ W_r[\mathcal{A}]-\int dx\, \mbox{tr}\left( \mathcal{D}v\frac{\delta}{\delta\mathcal{A}}W_r[\mathcal{A}]\right)\nonumber\\ =W_r[\mathcal{A}]-\int dx\, \mbox{tr}(\partial_\mu v^\alpha+f_{\alpha\beta\gamma}A_\mu^\beta v^\gamma)\frac{\delta}{\delta A_\mu^\alpha}W_r[\mathcal{A}]\nonumber\\ =W_r[\mathcal{A}]+\int dx\, \mbox{tr}\left( v^\alpha \mathcal{D}\frac{\delta}{\delta \mathcal{A}}W_r[\mathcal{A}]_\alpha\right).\end{align}$$ Since $$\frac{\delta}{\delta A_\mu^\alpha}W_r[\mathcal{A}]=\langle i\bar{\psi}\gamma^\mu T_\alpha \frac{1}{2}(1+\gamma^{m+1})\psi\rangle_\mathcal{A}=\langle j^\mu_\alpha\rangle$$ we obtain $$\begin{align}W_r[\mathcal{A}-\mathcal{D}v]-W_r[\mathcal{A}]=\int dx\, \mbox{tr}(v^\alpha \mathcal{D}_\mu \langle j^\mu \rangle_\alpha).\end{align}$$  

We are naively tempted to regard (1) as $\det (i\nabla\!\!\!\! /)=\prod \lambda'_i$, $\lambda_i$ eing the 'eigenvalue' of $i\nabla\!\!\!\! /$. A subtlety arises here: $i\nabla\!\!\!\! /_+$ maps sections of $S_+\otimes E$ to those of $S_-\otimes E$, where $E$ is the vector bundle associated with the $G$ bundle and $S_\pm$ are spin bundles with chirality $\pm$. Accordingly, the equation $i\nabla\!\!\!\!_+\psi=\lambda\psi$ is meaningless. To avoid this difficulty, we formally introduce a Dirac spinor $\psi$ and define $$\begin{align}e^{-W_r[\mathcal{A}]}=\int \mathcal{D}\psi\mathcal{D}\bar{\psi}\exp \left( -\int dx\, \bar{\psi}i\hat{D}\psi\right)\end{align}$$ where $i\hat{D}$ is defined by $$\begin{align}i\hat{D}\equiv i\gamma^\mu (\partial_\mu+i\mathcal{A}_\mu\mathcal{P}_+)=\begin{pmatrix}0&i\partial\!\!\! /_-\\ i\nabla\!\!\!\! /_+&0\end{pmatrix}\end{align}$$ where we have diagonalized $\gamma^{m+1}$. In (8), the gauge field $\mathcal{A}$ couples only to the positive chirality field. Now the eigenvalue problem $i\hat{D}\psi_i=\lambda \psi_i$ is well defined. Note that $i\hat{D}$ is not Hermitian and $\lambda_i$ is a complex number in general. Moreover, we need to introduce right and left eigenfunctions separately by $$\begin{align}i\hat{D}\psi_i=\lambda_i\psi_i\nonumber\\ \chi^\dagger_i(i\hat{D})=\lambda_i\chi^\dagger_i\quad (i\hat{D})^\dagger\chi_i=\bar{\lambda}_i\chi_i.\end{align}$$ Since $\int \chi^\dagger_i\psi_jdx=0$ for $i\ne j$, we may choose an orthonormal basis, $$\begin{align}\int \chi^\dagger_i\psi_jdx=\delta_{ij}.\end{align}$$ It should be noted that the eigenvalue $\lambda_i$ is not gauge invariant. This follows from the observation that $$\begin{align}g(i\hat{D}(\mathcal{A}^g))g^{-1}=gi\gamma^\mu[\partial_\mu+g^{-1}(\mathcal{A}_\mu+\partial_\mu)g\mathcal{P}_+]g^{-1}\nonumber\\ =i\hat{D}(\mathcal{A})-i\partial\!\!\!/ gg^{-1}+i\partial\!\!\! /gg^{-1}\mathcal{P}_+\ne i\hat{D}(\mathcal{A}).\end{align}$$ If the equality were to hold in (11), $g^{-1}\psi_i$ would satisfy $i\hat{D}(\mathcal{A}^g)g^{-1}\psi_i=\lambda_ig^{-1}\psi_i$ when $i\hat{D}(\mathcal{A})\psi_i=\lambda_i\psi_i$. Then $\mbox{Spec}i\hat{D}(\mathcal{A})$ would be gauge invariant. Although individual eigenvalues are not gauge invariant, the absolute value of the product of eigenvalues of $i\hat{D}$ is gauge invairnat. In fact, $$\begin{align}\det (i\hat{D})\det ((i\hat{D})^\dagger)=\det (i\hat{D}(i\hat{D})^\dagger)\nonumber\\ =\det \begin{pmatrix}(i\partial\!\!\!/_-)(i\partial\!\!\! /_+)& 0\\ 0& (i\nabla\!\!\!\! /_+)(i\nabla\!\!\!\! /_-)\end{pmatrix}\nonumber\\ =\det (i\partial\!\!\! /_-i\partial\!\!\!/_+)\det (i\nabla\!\!\!\! /_+i\nabla\!\!\!\!/_-)\end{align}$$ where $i\partial\!\!\!_+=(i\partial\!\!\!/_-)^\dagger$ and $i\nabla\!\!\!\!/_-=(i\nabla\!\!\!\!_+)^\dagger$. This is simply the Dirac determinant (up to an irrelevant factor $\det (i\partial\!\!\!/_-i\partial\!\!\!/_+)$), $$\begin{align}[\det(i\nabla\!\!\!\! /)]^2=\det \begin{pmatrix}i\nabla\!\!\!\!/_-i\nabla\!\!\!\!/_+&0\\ 0&i\nabla\!\!\!\!/_+i\nabla\!\!\!\!/_-\end{pmatrix}=[\det(i\nabla\!\!\!\!/_+i\nabla\!\!\!\!/_-)]^2\end{align}$$ where $i\nabla\!\!\!\!/$ is given by $$\begin{align}i\nabla\!\!\!\!/=\begin{pmatrix}0&i\nabla\!\!\!\!/_-\\ i\nabla\!\!\!\!_+&0\end{pmatrix}.\end{align}$$ The Dirac determinant is gaug invariant, hence so if $|\det (i\hat{D})|$. It then follows that $\mbox{Re}W_\mathbf{r}[\mathcal{A}]$ is gauge invariant since $$\exp (-W_r[\mathcal{A}])\exp (-\bar{W_\mathbf{r}[\mathcal{A}]})=\det (i\hat{D})\det((i\hat{D})^\dagger)\propto \det (i\nabla\!\!\!\!/_+i\nabla\!\!\!\!/_-)$$ is gauge invairnat. Therefore, only the imaginary part of $W_\mathbf{r}[\mathcal{A}]$, that is the phase of $\det (i\hat{D})$, may gain an anomalous variation under gauge transformations. 

The anomaly  may be computed by evaluating the Jacobian as before. The functional measure is taken to be $\prod_ida_i\, d\bar{b}_i$. We consider an infinitesimal gauge transformation, $$\begin{align}\mathcal{A}\rightarrow \mathcal{A}-\mathcal{D}v\quad \psi\rightarrow \psi+v\psi_+\quad \bar{\psi}\rightarrow \bar{\psi}-\bar{\psi}_-v\end{align}$$ where the gauge transformation rotates the positive chirality part only. The Jacobian factor is $$\begin{align}\int dx\, \mbox{tr}v(x)\sum_n (n|x\rangle \gamma^{m+1}\langle x|n\rangle\end{align}$$ where $\langle x|n\rangle=\psi_n(x)$ and $(n|x\rangle=\chi_n^\dagger(x)$.  This integral is ill dfined and must be regularized. As before, we employ the Gaussian regulator, $$\begin{align}\int dx\, \lim_{M\rightarrow,\ x\rightarrow y}\mbox{tr}\, v(x)\sum_n(n|y\rangle \gamma^{m+1}\langle x|e^{-(i\hat{D})^2/M^2}|n\rangle\nonumber\\ =\int dx\, \lim_{M\rightarrow,\ x\rightarrow y}\mbox{tr}\, v(x)\gamma^{m+1}e^{-i\hat{D}_x)^2/M^2}\delta(x-y)\end{align}$$ where use has been made of the completeness relation $$\begin{align}\sum_n|n\rangle (n|=I.\end{align}$$ It follows from (6) and (17) that $$\begin{align}\int dx\, v^\alpha \mathcal{D}_\mu \left(\frac{\delta}{\delta A_\mu^\alpha}W_r[\mathcal{A}]\right)=\int dx\lim_{M\rightarrow,\ x\rightarrow y}\mbox{tr}[v\gamma^{m+1}e^{-(i\hat{D}_x)^2/M^2}\delta(x-y)].\end{align}$$ In the present case $W_\mathbf{r}$ really changes under (15). The trace may be written as $$\begin{align}\mbox{tr}[v\gamma^{m+1}e^{-(i\hat{D}_x)^2/M^2}]=\mbox{tr}[v(\mathcal{P}_+-\mathcal{P}_-)e^{-(i\partial\!\!/_-i\nabla\!\!\!/_+)-(i\nabla\!\!\!/_-i\partial\!\!/_+)/M^2}]\nonumber\\ =\mbox{tr}[vP_+e^{(i\partial\!\! /i\nabla\!\!\!/)/M^2}]-\mbox{tr}[vP_-e^{(i\nabla\!\!\!/i\partial\!\!/)/M^2}].\end{align}$$ (20) can be evaluated in the plane wave basis, which is straightforward but tedious. We derived the non-Abelian anomaly from a topological viewpoint in the next section. For $m=4$, the anomalous variation is $$\begin{align}W_r[\mathcal{A}-\mathcal{D}v]-W_r[\mathcal{A}]=\int dx\, v^\alpha\mathcal{D}_\mu \langle j^\mu\rangle_\alpha\nonumber\\ =\frac{1}{24\pi^2}\int dx\, \mbox{tr}\{ v^\alpha T_\alpha \epsilon^{\kappa\lambda\mu\nu}\partial_\kappa [\mathcal{A}_\lambda\partial_\mu\mathcal{A}_\nu+\frac{1}{2}\mathcal{A}_\lambda\mathcal{A}_\mu\mathcal{A}_\nu]\}\nonumber\\ =\frac{1}{24\pi^2}\int \mbox{tr}\{vd[\mathcal{A}d\mathcal{A}+\frac{1}{2}\mathcal{A}^3]\}.\end{align}$$ The anomalous divergence of the gauge current is $$\begin{align}\mathcal{D}_\mu\langle j^\mu\rangle_\alpha=\frac{1}{24\pi^2}\mbox{tr}\{T_\alpha \epsilon^{\kappa\lambda\mu\nu}\partial_\kappa [\mathcal{A}_\lambda\partial_\mu\mathcal{A}_\nu+\frac{1}{2}\mathcal{A}_\lambda\mathcal{A}_\mu\mathcal{A}_\nu]\}.\end{align}$$ This should be compared with 13.2. There are two differences between these results: the two-thrids in front of $\mathcal{A}^3$ is replaced by a half and the overall factor is different.