[Nakahara GTP] 13.2 Abelian anomalies

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

We work in an even-dimensional manifold $M$ ($\dim M=m=2l$) with a Euclidean signiture. We assume our system is non-chiral, namely, the gauge field couples to the right and left components in the same way. (but SM?...) Our convention is $$\gamma^{\mu\dagger}=\gamma^\mu\quad \{\gamma^\mu,\gamma^\nu\}=2\delta^{\mu\nu}\quad \gamma^{m+1}=(i)^l\gamma^1\cdots\gamma^m\\ \gamma^{m+1\dagger}=\gamma^{m+1}\quad (\gamma^{m+1})^2=+I.$$ The Lie group generators $\{T_\alpha\}$ satisfy $$T^\dagger_\alpha=-T_\alpha\quad [T_\alpha,T_\beta]=f_{\alpha\beta}^\gamma T_\gamma \quad \mbox{tr}(T^\alpha T^\beta)=-\frac{1}{2}\delta^{\alpha\beta}.$$

13.2.1 Fujikawa’s method 

Fujikawa's way reveals the topological and geometrical nature of the problem most directly. This method is equivalent to the heat kernel proof of the relevant index theorem.

Let $\psi$ be a massless Dirac field interacting with an external non-Abelian gauge field $\mathcal{A}_\mu$. The effective action $W[\mathcal{A}]$ is given by $$\begin{align}e^{-W[\mathcal{A}]}=\int \mathcal{D}\psi \mathcal{D}\bar{\psi}e^{-\int dx\, \bar{\psi}i\nabla\!\!\! /\psi}\end{align}$$ where $i\nabla\!\!\!\! / =i\gamma^\mu \nabla_\mu=i\gamma^\mu(\partial_\mu+\omega_\mu+\mathcal{A}_\mu)$, with $\omega_\mu=\frac{1}{2}\omega_{\mu\alpha\beta}\Sigma^{\alpha\beta}$ being the spin connection of the background space. We compactly the space in such a way that the geometry (the spin connection) plays toe role. For example, this can be achieved by compactifying $\mathbb{R}^4$ to $S^4=\mathbb{R}^4\cup \{\infty\}$, for which the Dirac genus $\hat{A}(TM)$ is trivial; see example 12.5. If this is the case, the spin connection is irrelevant and may be dropped from $i\nabla\!\!\!\! /$. The classical action $\int dx\bar{\psi}i\nabla\!\!\!\! / \psi$ is invariant with respect to the chiral rotation, $$\begin{align}\psi\rightarrow e^{i\gamma^{m+1}\alpha}\psi,\ \bar{\psi}\rightarrow \bar{\psi}e^{i\gamma^{m+1}\alpha}.\end{align}$$ We expand $\psi$ and $\bar{\psi}$ as $$\begin{align}\psi=\sum_i a_i\psi_i\quad \bar{\psi}=\sum_i \bar{b}_i\psi^\dagger_i\end{align}$$ where $a_i$ and $\bar{b}_i$ are anti-commuting Grassmann variables, $$\{a_i,a_j\}=0\quad \{\bar{b}_i,\bar{b}_j\}=0\quad \{a_i,\bar{b}_j\}=0$$ and $\psi_i$ is an eigenvector of the Dirac operator $$\begin{align}i\nabla\!\!\!\! /\psi_i=\lambda_i\psi_i.\end{align}$$ Since $i\nabla\!\!\!\! /$ is Hermitian, $\lambda_i$ is real. Since $M$ is compact, $\psi_i$ can be normalized as $$\langle \psi_i|\psi_j\rangle= \int dx\, \psi^\dagger_i(x)\psi_j(x)=\delta_{ij}.$$ Now the path integrals over $\psi$ and $\bar{\psi}$ are replaced by those over $a_i$ and $\bar{b}_i$.

Consider an infinitesimal chiral transformation, $$\begin{align}\psi(x)\rightarrow (x)+i\alpha(x)\gamma^{m+1}\psi(x),\\ \bar{\psi}(x)\rightarrow \bar{\psi}(x)+i\bar{\psi}(x)\alpha(x)\gamma^{m+1}.\end{align}$$ As usual, we take $\alpha=\alpha(x)$ to be $x$-dependent. Under this change, the classical action transforms as $$\begin{align}\int dx\, \bar{\psi}i\nabla\!\!\!\! /\psi\rightarrow \int dx\, (\bar{\psi}+i\bar{\psi}\alpha \gamma^{m+1})i\nabla\!\!\!\!(\psi+i\alpha\gamma^{m+1}\psi)\nonumber\\ =\int dx\, \bar{\psi}i\nabla\!\!\!\! /\psi+i\int dx [\alpha \bar{\psi}\gamma^{m+1}i\nabla\!\!\!\! /\psi+\bar{\psi}i\nabla\!\!\!\! /(\alpha \gamma^{m+1}\psi)]\nonumber\\ \int dx\, \bar{\psi}i\nabla\!\!\!\! \psi-\int dx\, [\alpha\bar{\psi}\gamma^{m+1}\gamma^\mu(\partial_\mu+\mathcal{A}_\mu)\psi\nonumber\\ +\bar{\psi}\gamma^\mu(\partial_\mu+\mathcal{A}_\mu)(\alpha\gamma^{m+1}\psi)]\nonumber\\ =\int dx\, \bar{\psi}i\nabla\!\!\!\! /\psi+\int dx\, \alpha(x)\partial_\mu j^\mu_{m+1}(x)\end{align}$$ where we have used the anti-commutation relations $\{\gamma^\mu,\gamma^{m+1}\}=0$ and $$\begin{align}j^\mu_{m+1}(x)\equiv \bar{\psi}(x)\gamma^\mu \gamma^{m+1}\psi(x)\end{align}$$ is the chiral current. This is the higher-dimensional analogue of $j^\mu_5$ defined previously. If (7) were the only change caused by (5,6), naive application of the Ward-Takahashi relation would imply the conservation of the axial current $\partial_\mu j^\mu_{m+1}=0$. In quantum theory, however, we have an additional change, namely the change of the path integral measure. Define the chiral-rotated fields by $$\begin{align}\psi'=\psi+i\alpha\gamma^{m+1}\psi=\sum a'_i\psi_i,\\ \bar{\psi}'=\bar{\psi}+i\bar{\psi}\alpha\gamma^{m+1}=\sum \bar{b}'_i\psi^\dagger_i.\end{align}$$ Now the measure changes as $$\begin{align}\int \prod_i da_i\, d\bar{b}_i\rightarrow \int\prod_i da'_i\, d\bar{b}'_i.\end{align}$$ From the orthonormality of $\{\psi_i\}$, we find that $$\begin{align}a'_i=\langle \psi_i|\psi'\rangle=\langle \psi_i|(1+i\alpha\gamma^{m+1})\psi\rangle \nonumber\\ \sum_j \langle \psi_i|(1+i\alpha \gamma^{m+1})\psi_j\rangle a_j\equiv \sum_j C_{ij}a_j\end{align}$$ where $$\begin{align}C_{ij}=\langle \psi_i|(1+i\alpha \gamma^{m+1})\psi_j\rangle=\delta_{ij}+i\alpha \langle \psi_i|\gamma^{m+1}\psi_j\rangle.\end{align}$$ The measure in terms of the new variables is $$\begin{align}\prod da'_j=[\det C_{ij}]^{-1}\prod da_i=\exp (-\mbox{tr}\ln C_{ij})\prod da_i\nonumber\\ =\exp [-\mbox{tr}\ln (I+i\alpha\langle \psi_i|\gamma^{m+1}\psi_j\rangle)]\prod da_i\nonumber\\ \approx \exp (-\mbox{tr}i\alpha \langle \psi_i|\gamma^{m+1}\psi_j\rangle)\prod da_i\nonumber\\ =\exp \left( -i\alpha \sum_i \langle \psi_i|\gamma^{m+1}\psi_i\rangle \right)\prod da_i\end{align}$$ where the inverse of the determinant appears since $a_i$ and $a'_i$ are Grassmann variables. As for $\bar{b}_i\rightarrow \bar{b}'_i$, we have $$\begin{align}\bar{b}'_i=\sum_j \bar{b}_j\langle \psi_j| (1+i\alpha \gamma^{m+1})|\psi_i\rangle =\sum_j C_{ji}\bar{b}_j.\end{align}$$ The Jacobian for the change $\bar{b}_i\rightarrow \bar{b}'_i$ agrees with (14). Thus, the measure transforms under the chiral rotation (9,10) as $$\begin{align}\prod_i da_i\, d\bar{b}_i\rightarrow \prod_i da'_i\, d\bar{b}'_i\, \exp \left( -2i\int dx\, \alpha(x)\, \sum \psi^\dagger_n(x)\gamma^{m+1}\psi_n(x)\right).\end{align}$$ Now the effective acrtion has two expressions: $$\begin{align}e^{-W[\mathcal{A}]}=\int \prod_i da_i\, d\bar{b}_i\, \exp \left( -\int dx\, \bar{\psi}i\nabla\!\!\!\!/ \psi\right)\nonumber\\ =\int \prod_i da'_i\, d\bar{b}'_i\, \exp \left( -\int dx\, \bar{\psi}i\nabla\!\!\!\!/ \psi -\int dx\, \alpha(x)\partial_\mu j^\mu_{m+1}(x)-2i\int dx\, \alpha(x)\mathbf{A}(x)\right)\end{align}$$ where $$\begin{align}\mathbf{A}(x)\equiv \sum_i \psi^\dagger_i(x)\gamma^{m+1}\psi_i(x).\end{align}$$ Since $\alpha(x)$ is arbitrary, we have $$\begin{align}\partial_\mu j^\mu_{m+1}(x)=-2i\mathbf{A}(x).\end{align}$$ Thus, naive conservation of an axial current does not hold in quantum theory. This non-conservation of the current $j^\mu_{m+1}$ is called the Abelian anomaly (or chiral anomaly or axial anomaly). 

How is this related to the topology? Let us look at the Jacobian (16) and assume that $\alpha(x)$ is independent of $x$. The integral in (16) is not well defined and must be regularized. We introduce the Gaussian cut-off (heat-kernel regularization) as $$\begin{align} \int dx\, \mathbf{A}(x)=\int dx \sum_i \psi^\dagger_i(x)\gamma^{m+1}\psi_i(x)\exp [-(\lambda_i/M)^2]|_{M\rightarrow \infty}\nonumber\\ =\sum \langle \psi_i|\gamma^{m+1}\exp [-i(i\nabla\!\!\!\! / M)^2]|\psi_i\rangle |_{M\rightarrow \infty}.\end{align}$$ In (20), $1/M^2$ corresponds to the 'time' parameter $t$ in the previous chapter and $M\rightarrow \infty$ implies $t\rightarrow \infty$. Let $|\psi_i\rangle$ be an eigenstate of $i\nabla\!\!\!\! /$ with non-vanishing eigenvalue $\lambda_i$. Among the eigenstates, there exists a state $|\psi_i\rangle^\chi\equiv \gamma^{m+1}|\psi\rangle$ with eigenvalue: $-\lambda_i$: $$i\nabla\!\!\!\!/ |\psi_i\rangle^\chi =i\nabla\!\!\!\! \gamma^{m+1}|\psi_i\rangle =-\gamma^{m+1}i\nabla\!\!\!\! |\psi_i\rangle \\ =-\lambda_i\gamma^{m+1}|\psi_i\rangle=-\lambda_i|\psi_i\rangle^\chi$$ where use has been made of the anti-commutation relation $\{\gamma^{m+1},i\nabla\!\!\!\! /\}=0$. Since $i\nabla\!\!\!\! /$ is a Hermitian operator, eigenvectors which belong to different eigenvalues are orthogonal, hence $\langle \psi_i|\psi_i\rangle^\chi=\langle \psi_i|\gamma^{m+1}|\psi_i\rangle=0$. This shows that $$\langle \psi_i|\gamma^{m+1}\exp [-(i\nabla\!\!\!\!//M)^2]|\psi\rangle=\langle \psi_i|\gamma^{m+1}|\psi_i\rangle \exp [-(\lambda_i/M)^2]=0.$$ Thus, the contribution to the RHS of (20) comes only froim the zero-energy modes. Let $|0,i\rangle$ be the zero-energy moides of $i\nabla\!\!\!\!/$, ($1\le i\le n_0$). They are not in an irreducible representation of the spin algebra and should be classified according to the eigenvalue of $\gamma^{m+1}$. We write $$\begin{align}\gamma^{m+1}|0,i\rangle_\pm=\pm |0,i\rangle_\pm.\end{align}$$ Then, (20) becomes $$\begin{align}\int dx\, \mathbf{A}(x)=\sum \langle \psi_i|\gamma^{m+1}\exp [-(i\nabla\!\!\!\! //M)^2]|\psi_i\rangle|_{M\rightarrow \infty}\nonumber\\ \sum_{i,+} \langle 0,i|0,i\rangle_+-\sum_{i,-}\langle 0,i|0,i\rangle_-\nonumber\\ =\nu_+-\nu_-=\mbox{ind}i\nabla\!\!\!\! /_+\end{align}$$ where $\nu_+$ ($\nu_-$) is the number of zero-energy modes with positive (negative) chirality ($\nu_++\nu_-=n_0$) and $i\nabla\!\!\!\!/_+$ is defined by $$i\nabla\!\!\!\! /=\begin{pmatrix}0&i\nabla\!\!\!\! /_-\\ i\nabla\!\!\!\!/_+&0\end{pmatrix}\quad i\nabla\!\!\!\! /_-=(i\nabla\!\!\!\!_+)^\dagger.$$ The Atiyah-Singer index theorem now comes into the problem.

To show that (22), indeed, represents an integral of the relevant Chern character, we first note that $$\begin{align}(i\nabla\!\!\!\! /)^2=-\gamma^\mu\gamma^\nu \nabla_\mu \nabla_\nu=-\{\delta^{\mu\nu}+\frac{1}{2}[\gamma^\mu,\gamma^\nu]\}\frac{1}{2}[\{\nabla_\mu,\nabla_\nu\}+\mathcal{F}_{\mu\nu}]\nonumber\\ =-\nabla_\mu \nabla^\mu -\frac{1}{4}[\gamma^\mu,\gamma^\nu]\mathcal{F}_{\mu\nu}\end{align}$$ where use has been made of the relation $[\nabla_\mu,\nabla_\nu]=\mathcal{F}_{\mu\nu}$. Then $$\begin{align}\mathbf{A}(x)=\sum_i \langle \psi_i|x\rangle\langle x|\gamma^{m+1}\exp [(\nabla^2 +\frac{1}{4}[\gamma^\mu,\gamma^\nu]\mathcal{F}_{\mu\nu})/M^2]|\psi_i\rangle |_{M\rightarrow \infty}.\end{align}$$ Let us take $m=4$ for definiteness. We introduce the plane wave basis as $$\langle x|\psi_i\rangle =\int \frac{d^4k}{(2\pi)^4}\langle x|k\rangle \langle k|\psi_i\rangle.$$ Then (24) becomes $$\begin{align}\mathbf{A}(x)=\int \frac{dk}{(2\pi)^4}\int \frac{dk'}{(2\pi)^4}\sum_i \langle \psi_i |k'\rangle \langle k'|x\rangle \nonumber\\ \times \left. \gamma^{m+1}\exp [(\nabla^2+\frac{1}{4}[\gamma^\mu,\gamma^\nu]\mathcal{F}_{\mu\nu})/M^2]\langle x|k\rangle\langle k|\psi_i\rangle \right|_{M\rightarrow \infty,\ y\rightarrow x}\nonumber\\ =\int \frac{dk}{(2\pi)^4}\mbox{tr}\gamma^{m+1}\exp [(-k^2+\frac{1}{4}[\gamma^\mu,\gamma^\nu]\mathcal{F}_{\mu\nu})/M^2]_{M\rightarrow \infty}\end{align}$$ where use has been made of completeness property $$\sum_i \langle k|\psi_i\rangle\langle \psi_i|k'\rangle =(2\pi)^4\delta^4 (k-k').$$ In (25), we have replaced $\nabla^2$ be the symbol $-k^2$ since the residual terms containing $\mathcal{A}$ do not survive in the limit $M\rightarrow \infty$. If we put $\tilde{k}^\mu\equiv k^\mu/M$. (25) becomes $$\mathbf{A}(x)=\mbox{tr}[\gamma^5 \exp (\frac{1}{4}[\gamma^\mu,\gamma^\nu]\mathcal{F}_{\mu\nu}/M^2)]M^4\int \frac{d\tilde{k}}{(2\pi)^4}\exp (-\tilde{k}^2).$$ We expand the first exponential and use $$\mbox{tr}\gamma^5=\mbox{tr}\gamma^5\gamma^\mu\gamma^\nu=0\quad \mbox{tr}\gamma^5\gamma^\kappa\gamma^\lambda\gamma^\mu\gamma^\nu=-4\epsilon^{\kappa\lambda\mu\nu}\\ \int d\tilde{k}\exp (-\tilde{k}^2)=\pi^2$$ to obtain $$\begin{align}\mathbf{A}(x)=\frac{1}{2}\mbox{tr}\left[ \gamma^5\frac{1}{4^2}\{[\gamma^\mu,\gamma^\nu]\mathcal{F}_{\mu\nu}\}^2\right] \frac{1}{16\pi^2}\nonumber\\ =\frac{-1}{32\pi^2}\mbox{tr}\epsilon^{\kappa\lambda\mu\nu}\mathcal{F}_{\kappa\lambda}(x)\mathcal{F}_{\mu\nu}(x).\end{align}$$ Note that the higher-order terms in the expansion of the exponential vanish in the limit $M\rightarrow \infty$. The anomalous conservation law (19) now becomes $$\begin{align}\partial_\mu j^\mu_5=\frac{1}{16\pi^2}\mbox{tr}\epsilon^{\kappa\lambda\mu\nu}\mathcal{F}_{\kappa\lambda}\mathcal{F}_{\mu\nu}\nonumber\\ =\frac{1}{4\pi^2}\mbox{tr}[\epsilon^{\kappa\lambda\mu\nu} \partial_\kappa (\mathcal{A}_\lambda\partial_\mu\mathcal{A}_\nu+\frac{2}{3}\mathcal{A}_\lambda\mathcal{A}_\mu\mathcal{A}_\nu)].\end{align}$$ This is regarded as a local version of the AS index theorem. Let us write (27) in terms of the field strength $\mathcal{F}=\frac{1}{2}\mathcal{F}_{\mu\nu}dx^\mu\wedge dx^\nu$. We easily verify that $$\begin{align}\nu_+-\nu_-=\int_M dx\, \partial_\mu j^\mu_{m+1}=\int_M\mbox{ch}_2(\mathcal{F}).\end{align}$$ This is the index theorem for a twisted spinor complex with trivial background geometry ($\hat{A}(TM)=1$).

For $\dim M=m=2l$, we have the following identity: $$\begin{align}\nu_+-\nu_-=\int_Mdx\, \partial_\mu j^\mu_{m+1}=\int_M\mbox{ch}_l(\mathcal{F})=\int_M\frac{1}{l!}\mbox{tr}\left(\frac{i\mathcal{F}}{2\pi}\right)^l.\end{align}$$