April 15, 2024

[Francesco CFT] Conformal Anomaly

This article is one of CFT/TFT.

This is most important topic for me.

I have to understand at most as possible.

The central charge also arises when a conformal field theory is defined on a curved two-dimensional manifold. The curvature introduces a macroscopic scale in the system, and the expectation value of the trace of the energy-momentum tensor, instead of vanishing, is proportional to both the curvature $R$ and the central charge $c$: $$\begin{align}\langle T^\mu_\mu(x)\rangle_g=\frac{c}{24\pi}R(x)\end{align}$$ This quantum breaking of scale invariance is called the trace anomaly.

General Approach - Ward Identity for Scale Tranformation

Now we demonstrate Eq. (1) for the trace anomaly for a free boson. We consider the generating functional $$\begin{align}Z[g]=\int [d\varphi]_ge^{-S[\varphi,g]}=e^{-W[g]}\end{align}$$ where $S[\varphi,g]$ is the action of a free scalar field in a background metric $g_{\mu\nu}$: $$\begin{align}S[\varphi,g]=\int d^2x\, \sqrt{g}g^{\mu\nu}\partial_\mu\varphi\partial_\nu\varphi=-\int d^2x\, \sqrt{g}\varphi\Delta\varphi\end{align}$$ We have introduced the Laplacian operator $\Delta$: $$\begin{align}\Delta\varphi=\frac{1}{\sqrt{g}}\partial_\mu(\sqrt{g}g^{\mu\nu}\partial_\nu\varphi)\end{align}$$ Under a local scale transformation of the metric $g_{\mu\nu}\rightarrow (1+\sigma(x))g_{\mu\nu}$, the action varies according to $$\begin{align}\delta S[\varphi,g]=-\frac{1}{2}\int d^2x\, T^{\mu\nu}\delta g_{\mu\nu}=-\frac{1}{2}\int d^2x\, \sigma(x)T^\mu_\mu\end{align}$$ where $\sigma(x)$ is infinitesimal. Consequently, the variation of the connected vacuum functional $W[g]$ is $$\begin{align}\delta W[g]=-\frac{1}{2}\int d^2x\, \sigma(x)\langle T^\mu_\mu(x)\rangle\end{align}$$ According to the Ward identities previously derived, this variation vanishes in flat space, since $\langle T^\mu_\mu(x)\rangle=0$ This is no longer true on an arbitrary manifold.

Reparametrize the Functional Space

To see this, we define the functional measure $[d\varphi]$ in a fashion more suited to an arbitrary metric. We proeceed by analogy with integration on a general manifold of dimension $d$: the line element is then $ds^2=g_{\mu\nu}dx^\mu dx^\nu$, and the volume element is $d\Omega=\sqrt{g}dx^1\cdots dx^d$. If a coordinate system can be found such that $g_{\mu\nu}=\eta_{\mu\nu}$, then $\sqrt{g}=1$ and the integration measure simplifies accordingly. In the space of field configurations, the analog of the metric is defined in a reparametrization invariant way: $$\begin{align}(\varphi_1,\varphi_2)=\int d^2x\, \sqrt{g}\varphi^*_1\varphi_2\end{align}$$ and the line element is simply $$\begin{align}\lVert \delta\varphi\rVert^2=(\delta\varphi,\delta\varphi)\end{align}$$ In order to diagonalize this "functional metric", we introduce a complete set of orthonormal functions $\{\varphi_i\}$ (i.e., such that $(\varphi_m,\varphi_n)=\delta_{mn}$) and express any general field configuration as $\varphi=\sum_nc_n\varphi_n$. The line element thus reduces to $$\begin{align}\lVert \delta\varphi\rVert^2=\sum_n(\delta c_n)^2\end{align}$$ which allows us to define the functional integration measure as $$\begin{align}[d\varphi]=\prod_n dx_n\end{align}$$

Diagonalize Laplacian

Of all possible complete sets $\{\varphi_n\}$, the most useful is the set of normalized eigenfunctions of the Laplacian, with eigenvalues $-\lambda_n$: $$\begin{align}\Delta\varphi_n=-\lambda_n\varphi_n\end{align}$$ The action of a configuration specified by the expansion coefficients $c_n$ is then simply $$\begin{align}S[\varphi,g]=\sum_n \lambda_n c^2_n\end{align}$$ which means that the modes $\varphi_n$ decouple. However, all is not trivial since the eigenfunctions $\varphi_n$ and the eigenvalues $\lambda_n$ depend on the background metric $g_{\mu\nu}$. The vacuum functional may be written as $$\begin{align}Z[g]=\int \prod_n\left\{ dc_ne^{-\lambda_nc^2_n}\right\}=\prod_n \sqrt{\frac{2\pi}{\lambda_n}}\end{align}$$

Infrared Regularization

We must be cautious here, since the Laplacian always has a zero-mode $\varphi_0=const.$ with vanishing eigenvalue. Such a mode is a source of divergence in the vacuum functional. To fix this "infrared" problem, we "compactify" the field $\varphi$: We assume that $\varphi$ takes its values on a circle, such that the values $\varphi$ and $\varphi+a$ are equivalent. The circumference $a$ can be chosen very large, and taken to infinity at the end of the calculation. Then the range of integration of $c_0$ is no longer the whole real axis, but the segment $[0,a\sqrt{A}]$, where $A$ is the area of the manifold. This follows from the normalization condition $(\varphi_0,\varphi_0)=A\varphi_0^2=1$ and the condition $0< c_0\varphi_0<a$. The above expression for the vacuum functional is then replaced by $$\begin{align}Z[g]=a\sqrt{A}\prod_{n\ne 0}\sqrt{\frac{2\pi}{\lambda_n}}\end{align}$$ The connected functional $W[g]$ is then $$\begin{align}W[g]=-\ln a-\frac{1}{2}\ln A+\frac{1}{2}\mbox{Tr}'\ln \frac{-\Delta}{2\pi}\end{align}$$ where $\mbox{Tr}'$ indicates a trace taken over all nonzero modes. We then use the following representation of the logarithm: $$\begin{align}\ln B=-\lim_{\epsilon\rightarrow 0}\int^\infty_\epsilon \frac{dt}{t}(e^{-Bt}-e^{-t})\end{align}$$ in order to write $$\begin{align}W[g]=-\ln a-\frac{1}{2}\ln A-\frac{1}{2}\mbox{Tr}'\left\{\int^\infty_\epsilon\frac{dt}{t}(e^{t\Delta}-e^{-2\pi t})\right\}\end{align}$$ (we have scaled $t\rightarrow 2\pi t$). Form now on we keep $\epsilon$ finite and shall send it to zero at the end of the calculation.

We now perform an infinitesimal local scale transformation. The variation of the metric is $\delta g_{\mu\nu}=\sigma g_{\mu\nu}$, and that of the Laplatican is $\delta\Delta=-\sigma\Delta$. The variation of the second term of (17) is $$\begin{align}\delta(-\frac{1}{2}\ln A)=-\frac{\delta A}{2A}=-\frac{1}{2A}\int d^2x\sqrt{g}\sigma \end{align}$$ and that of the trace in Eq. (17) is $$\begin{align}\frac{1}{2}\mbox{Tr}'\left\{\int^\infty_\epsilon dt\, \sigma\Delta e^{t\Delta}\right\}=\frac{1}{2}\mbox{Tr}'\left\{\int^\infty_\epsilon dt\, \sigma\frac{d}{dt}e^{t\Delta}\right\}=-\frac{1}{2}\mbox{Tr}'(\sigma e^{\epsilon\Delta})\end{align}$$ In the second equality, we used the property that all nonzero eigenvalues of $\Delta$ are negative, so that only the lower-bound of the integral over $t$ contributes. Since $$\begin{align}-\frac{1}{2A}\int d^2x\sqrt{g}\sigma=-\frac{1}{2}(\varphi_0,\sigma\varphi_0)=-\frac{1}{2}(\varphi_0,\sigma e^{\epsilon \Delta}\varphi_0)\end{align}$$ we may combine the two variations into a single expression: $$\begin{align}\delta W[g]=\frac{1}{2}\mbox{Tr}(\sigma e^{\epsilon\Delta})\end{align}$$ This expression contains the contribution of the zero-mode, hence $\mbox{Tr}'$ has been replaced by $\mbox{Tr}$.

9801056.pdf (arxiv.org)

Heat Kernel Expansion

To preceed, we introduce the heat kernel $$\begin{align}G(x,y;t)=\begin{cases}\langle x|e^{t\Delta}|y\rangle& (t\ge 0)\\ 0& (t<0)\end{cases}\end{align}$$ Since the eigenvalues of $\Delta$ can be arbitrarily negative, the expression $e^{t\Delta}$ has meaning only for $t\ge 0$. In terms of this kernel, the variation of $W[g]$ is $$\begin{align}\delta W[g]=-\frac{1}{2}\int d^2x\sqrt{g}\sigma(x)G(x,x;\epsilon)\end{align}$$ The crucial point here is the shot-time behavior of the diagonal kernel, which can be shown to be $$\begin{align}G(x,x;\epsilon)=\frac{1}{4\pi \epsilon}+\frac{1}{24\pi}R(x)+O(\epsilon)\end{align}$$ (this result is prove later). It follows that $$\begin{align}\delta W[g]=-\frac{1}{8\pi\epsilon}\int d^2x\sqrt{g}\sigma(x)-\frac{1}{48\pi}\int d^2x\sqrt{g}\sigma(x)R(x)\end{align}$$ In the limit $\epsilon\rightarrow 0$, the first term seems problematic, being infinite. The origin of this divergence lies in the assumed finite size of the manifold and has nothing to do with curvature. To fix it, we add to the origin action the following $\varphi$-independent counterterm: $$\begin{align}S_1[g]=\mu \int d^2x\sqrt{g}\end{align}$$ which is simply equal to $\mu A$. Under a local scale transformation it underoges the following variation: $$\begin{align}\delta S_1[g]=\mu\int d^2x\sqrt{g}\sigma(x)\end{align}$$ By suitably choosing $\mu$ to be equal to $-1/8\pi \epsilon$, the variation of the counterterm action $S_1$ cancels the divergent term in (25). The second term in (25) cannot be eliminated in the same way. Indeed, if we add a second counterterm of the form $$\begin{align}S_2[g]=\int d^2x\sqrt{g}R(x)\end{align}$$ we find that it is proportional to the Euler characteristics $\chi$, a topological invariant that depends only on the number of handles of the manifold. Therefore, it is invariant under a local scale transformation, and cannot cancel the rest of the variation $\delta W[g]$. Then, the equivalence of (6) and (25) implies that the trace of the energy-momentum tensor does not vanish, according to (1), with the value $c = 1$ appropriate for a free boson.

Relation with Central Charge

In order to relate the trace anomaly to the central charge figuring in the OPE of the energy-momentum tensor or, equivalently, in the two-point function $\langle T_{\mu\nu}(x)T_{\rho\lambda}(y)\rangle$, we proceed as follows. We use the "conformal gauge", a coordinate system in which the metric tensor is diagonal: $$\begin{align}g_{\mu\nu}=\delta_{\mu\nu}e^{2\varphi(x)}\end{align}$$ In two dimensions it is always possible to find such a system, at least locally. In terms of the field $\varphi$, the determinant $\sqrt{g}$ and the curvature are $$\begin{align}\sqrt{g}=e^{2\varphi}\quad \sqrt{g}R=\partial^2\varphi\end{align}$$ Since a local scale transformation amounts to a local variation of the field rp, the corresponding variation of the connected functional $W[g]$ is $$\begin{align}\delta W[g]=-\frac{c}{24\pi}\int d^2x\, \partial^2\varphi\delta\varphi\end{align}$$ where $c$ is some constant, equal to unity in the case of a free boson, as argued above. This implies that $$\begin{align}W[g]=\frac{c}{48\pi}\int d^2x\, (\partial\varphi)^2\end{align}$$ up to terms independent of $\varphi$. In terms of the Green function $K(x,y)$ of the Laplatican, this is $$\begin{align}W[g]=-\frac{c}{48}\int d^2x\, d^2y\, \partial^2\varphi(x)K(x,y)\partial^2\varphi(y)\end{align}$$ This follows from the defining property $\partial^2_x K(x,y)=\delta (x-y)$ and integration by parts. The natural extension of the above to an arbitrary coordinate system is $$\begin{align}W[g]=-\frac{c}{48\pi}\int d^2x\, d^2y\, \sqrt{g(x)}\sqrt{g(y)}R(x)K(x,y)R(y)\end{align}$$ where $K(x,y)$ now satisfies $$\begin{align}\sqrt{g(x)}\Delta_x K(x,y)=\delta (x-y)\end{align}$$ The above expression for $W[g]$ can used to calculate the two-point function of the energy-momentum tensor (the Schwinger function): $$\begin{align}\langle T_{\mu\nu}(x)T_{\rho\lambda}(y)\rangle=\frac{\delta^2 W}{\delta g_{\mu\nu}(x)\delta g_{\rho\lambda}(y)}\end{align}$$ Without a detailed calculation, it is by now clear that the Schwinger function will be proportional to $c$, which confirms that the central charge and the coefficient of the trace anomaly are one and the same thing.

The Heat Kernel

We show that the heat kernel $G(x,y;t)$ defined in (22) has the short-time behavior given in (24) for $x=y$.

Definition of Heat Kernel

From the definition of the heat kernel, we see that it satisfies the equations $$\begin{align}\frac{\partial}{\partial t}G(x,y;t)=\Delta_x G(x,y;t)&\quad \Delta=\frac{1}{\sqrt{g}}\partial_\mu \sqrt{g}\partial^\mu\\ G(x,y;0)=\frac{1}{\sqrt{g}}\delta(x-y)\nonumber & \end{align}$$ These two equations may be combined into $$\begin{align}(\partial_t-\Delta_x)G(x,y;t)=\frac{1}{\sqrt{g}}\delta(x-y)\delta (t)\end{align}$$ The equivalence of this single equation with Eq. (37) may be first considering the case $t>0$, and by integrating the above over $t$ from $-\epsilon$ to $\epsilon$, where $\epsilon$ is infinitesimal, remembering that $G(x,y;t)=0$ if $t<0$. The heat kernel is then the Green function for the diffusion equaltion: $$\begin{align}G(x,y;t)=\langle x,t|(\partial_t-\Delta)^{-1}|y,0\rangle\end{align}$$

General Curved Spacetime

We know the (normalized) solution to this equation in flat infinite space: $$\begin{align}G_0(x,y,t)=\frac{1}{4\pi t}\exp -\frac{(x-y)^2}{4t}\end{align}$$ We now with to find the small $t$ behavior of $G(x,x;t)$ on a general curved manifold. Physically, $G(x,x;t)$ is the probability that a random walker will diffuse from $x$ back to $x$ in a time $t$. If $t$ is small the diffusion cannot go very far, and we can restrict our attention to the immediate neighborhood of $x$. To this end we write $y=x+\delta x$ and use a locally inertial frame at $x$, with $g_{\mu\nu}(x)=\eta_{\mu\nu}$ and $\partial_\lambda g_{\mu\nu}(x)=0$: $$\begin{align}g_{\mu\nu}(x+\delta x)\sim \eta_{\mu\nu}+\frac{1}{2}C_{\mu\nu\rho\lambda}\delta x^\rho \delta x^\lambda\end{align}$$ where the constants $C_{\mu\nu\rho\lambda}$ are symmetric under the interchanges $\mu\leftrightarrow \nu$ and $\rho\leftrightarrow \lambda$. It is then a simple exercise to show that $$\begin{align}\Delta(x)\sim \partial_\mu\partial^\mu+a^{\mu\nu}\partial_\mu\partial_\nu+b^\mu\partial_\mu\end{align}$$ wherein $$\begin{align}a^{\mu\nu}=-\frac{1}{2}{C^{\mu\nu}}_{\rho\lambda}\delta x^\rho \delta x^\lambda\quad \mbox{and}\quad b^\mu=\frac{1}{2}({{{C^\rho}_\rho}^\mu}_\lambda-{C^{\mu\rho}}_{\rho\lambda})\delta x^\lambda\end{align}$$

Short-Time Expansion

The heat kernel ten becomes $$\begin{align}G(x,y;t)=\langle x,t|\frac{1}{A-B}|y,0\rangle\end{align}$$ where $$\begin{align}\begin{matrix}A=\partial_t-\partial_\mu\partial^\mu\\ B=a^{\mu\nu}\partial_\mu\partial_\nu+b^\mu\partial_\mu\end{matrix}\end{align}$$ A perturbative solution for $G(x,y;t)$ is obtained by expanding $$\begin{align}\frac{1}{A-B}=\frac{1}{A}+\frac{1}{A}B\frac{1}{A}+\frac{1}{A}B\frac{1}{A}B\frac{1}{A}+\cdots\end{align}$$ To first order, this yields $$\begin{align}G(x,y;t)=\langle x,t|A^{-1}|y,0\rangle+\int d\tau dz\, \langle x,t|A^{-1}|z,\tau\rangle \langle z,\tau |BA^{-1}|y,0\rangle\\ =G_0(x,y;t)+\int^t_0d\tau \int d^2z\, G_0(x,z,t-\tau)\nonumber\\ \times \left\{a^{\mu\nu}(z)\frac{\partial^2}{\partial z^\mu\partial z^\nu}+b^\mu(z)\frac{\partial}{\partial z^\mu}\right\} G_0(z,y;\tau)\nonumber\end{align}$$ The range of the $\tau$ integration follows from the vanishing of $G_0(x,y;t)$ for $t<0$. One checks that the low $t$ behavior of the $n$-th order contribution in perturbation theory is $t^{n-1}$. We are thus justified in keeping only the first-order contributions. Substitution of the explicit form (40) of $G_0(x,y;t)$ and (43) of $a^{\mu\nu}$ and $b^\mu$ yields $$\begin{align}G(x,x;t)=\frac{1}{4\pi t}+\frac{1}{24\pi}\left({C^{\mu\lambda}}_{\mu\lambda}-{{{C^\mu}_\mu}^\lambda}_\lambda\right)+O(t)\end{align}$$

Ricci Curvature

On the other hand, it is straightforward to show that the scalar curvature is given by $$\begin{align}R(x)=\left({C^{\mu\lambda}}_{\mu\lambda}-{{{C^\mu}_\mu}^\lambda}_\lambda\right)\end{align}$$ Therefore, the short-time behavior of the heat kernel on a curved manifold is given by $$\begin{align}G(x,x;\epsilon)=\frac{1}{4\pi \epsilon}+\frac{1}{24\pi}R(x)+O(\epsilon)\end{align}$$ Even if this result is obtained in a specific local inertial frame, the relation of the curvature with the short-time heat kernel is coordinate independent.

Exercise

5.10

Demonstrate in detail the expressions (42) and (43) for the Laplacian in a locally inertial frame near the origin.

Sol)

5.11 Heat kernel on a sphere

The Laplacian operator on a sphere of radius r embedded in three-dimensional space is $\Delta=-\mathbf{L}^2/r^2$, where $\mathbf{L}$ is the angular momentum operator of quantum mechanics.

a) Show that the heat kernel $G(x,x;t)$ is given by $$\begin{align}G(x,x;t)=\frac{1}{r^2}\sum_{l,m}|Y_{l,m}(x)|^2e^{-tl(l+1)/r^2}\end{align}$$ where $x$ stands for the angular coordinates $(\theta,\varphi)$. The spherical harmonics $Y_{l,m}(\theta,\varphi)$ are eigenfunctions of $\mathbf{L}^2$ and $L_z$: $$\begin{align}\mathbf{L}^2Y_{l,m}=l(l+1)Y_{l,m}\quad L_z Y_{l,m}=mY_{l,m}\end{align}$$

b) By setting $x=0$ (the north pole $\theta=0$) and using Euler's summation formula, show explicitly that $$\begin{align}G(0,0;t)=\frac{1}{4\pi t}+\frac{1}{12\pi r^2}+\cdots\end{align}$$ This result agrees with Eq. (24) (or (50)), since the scalar curvature $R$ of a sphere of radius $r$ is $R=2/r^2$.