# 2D Liouville Stress-Energy tensor

I am working on 2D Liouville field theory and trying to follow mostly Harold Erbin's note on 2d quantum gravity and Liouville theory.

I have a really simple question:

One consider the Euclidean Liouville action which is given by

$S_L = \frac{1}{4\pi}\int d^2\sigma\sqrt{h}\left(h^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi+QR\phi+4\pi\mu e^{2b\phi}\right)$

Now, in order to get the e.o.m as well as the stress-energy tensor, we have to vary the action, which yields

$$\delta_hS_L = \frac{1}{4\pi}\int d^2\sigma\sqrt{h}\delta h^{\mu\nu}\left[-\frac{1}{2}h_{\mu\nu}\left(h^{\rho\sigma}\partial_{\rho}\phi\partial_{\sigma}\phi+QR\phi+4\pi\mu e^{2b\phi}\right)\\+\left(\partial_{\mu}\phi\partial_{\nu}\phi+QR_{\mu\nu}\phi+Q(h_{\mu\nu}\Delta\phi-\nabla_{\mu}\nabla_{\nu}\phi)\right)\right] ,$$ while the variation w.r.t. $\phi$ gives, $$\delta_{\phi}S_L = \frac{1}{4\pi}\int d^2\sigma\sqrt{h}\delta \phi\left(-2\Delta\phi+QR+8\pi\mu be^{2b\phi}\right)$$The equation of motion for $\phi$ are given in the usual way and one obtain $$QR[h]-2\Delta\phi=-8\pi\mu b e^{2b\phi}$$ If one consider the flat metric, then this reduces to $$\partial_{\mu}\partial^{\mu}\phi=4\pi\mu b e^{2b\phi}$$

The stress energy tensor is computed in the usual way using $$T_{\mu\nu} = -\frac{4\pi}{\sqrt{h}}\frac{\delta S}{\delta h^{\mu\nu}}$$ In the notes, it is claimed that this gives $$T_{\mu\nu}= -\left(\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}h_{\mu\nu}h^{\rho\sigma}\partial_{\rho}\phi\partial_{\sigma}\phi\right)+Q(-h_{\mu\nu}\Delta\phi+\nabla_{\mu}\nabla_{\nu}\phi)+2\pi\mu be^{2b\phi}h_{\mu\nu}$$

$\textbf{Question 1:}$ Why does the contribution proportional to $R$ vanishes? I have the feeling that this is the stress energy tensor for flat space but not for curved space of do they vanish in every case?

Then, we go to complex plane (section 6.5.2). The metrix is given by $ds^2 = dzd\bar{z}$. This means that

$$g_{zz}=g_{\bar{z}\bar{z}}=0, g_{z\bar{z}}=g_{\bar{z}z}=\frac{1}{2}$$ Also, the complex coordinates derivative are easily found to be $\partial_{z}= \frac{1}{2}(\partial_0-i\partial_1)$.

$\textbf{Question 2:}$ I don't fully understand how to transform the e.o.m as well as the stress-energy tensor in those new coordinates. The result should be:

$$\partial\bar{\partial}\phi = 4\pi \mu be^{eb\phi}$$ $$T(z) = T_{zz} = -(\partial\phi)^2+Q\partial^2\phi+2\pi\mu e^{eb\phi}$$

Can someone give me a hint please?

## 1 Answer

First, I should say that I am the author of the notes; numbering refers to the online version of the notes.

## Question 1

The formula (6.27) for the energy-momentum tensor follows from the general variation of the action (6.22a). The terms in the bracket match $T_{\mu\nu}$ up to the term $$Q \phi \, G_{\mu\nu} = Q \phi \left( R_{\mu\nu} - \frac{R}{2} \, h_{\mu\nu} \right),$$ but this term is identically zero in two dimensions. The reason is that the Riemann tensor has only one non-vanishing component, and thus the Ricci tensor must be proportional to $R$. In the notes, this is briefly (and badly) explained in (4.11) and below.

## Question 2

To transform to complex coordinates you have two possibilities:

1. Label the components of every tensor with complex coordinates, and expand in components using the explicit form of the metric.
2. Use the explicit formulas (tensor transformation, or chaining rule) to relate between the Cartesian and complex components.

I find 1) simpler in general. For example the Laplacian: $$\Delta = h^{\mu\nu} \partial_\mu \partial_\nu = \underbrace{h^{zz} \partial_z \partial_z}_{= 0} + \underbrace{h^{\bar z \bar z} \partial_{\bar z} \partial_{\bar z}}_{= 0} + 2 h^{z \bar z} \partial_z \partial_{\bar z} = 4 \, \partial \bar\partial$$ Hence the equation of motion becomes $$\partial \bar\partial \phi = \pi\mu \, \mathrm{e}^{2 b \phi}$$ (there are typos for the factors in my notes, I was not careful with the conventions used for complex coordinates). For the energy-momentum tensor: $$T_{zz} = - \bigg( \partial_z \phi \partial_z \phi - \underbrace{h_{z z} (\partial_\mu \phi}_{= 0} )^2 \bigg) + Q \big( \underbrace{h_{zz} \Delta \phi}_{= 0} + \partial_z \partial_z \phi \big) + \underbrace{2 \pi b \mu \, h_{z z} \, \mathrm{e}^{2 b \phi}}_{= 0},$$ and this simplifies to $$T = - (\partial \phi)^2 + Q \, \partial^2 \phi.$$ This is the most common form of the energy-momentum tensor. The mu term in my notes are a mistake: I have not been careful with h_{zz} and derivative indices. On the other hand what is true is that the trace is equal to (after simplifying vanishing term) $$h^{\mu\nu} T_{\mu\nu} = 2 h^{z \bar z} T_{z \bar z} = T_{z \bar z} = Q (- h_{z \bar z} 4 \partial \bar\partial \phi + \partial \bar\partial \phi) + 2 \pi b \mu \, h_{z \bar z} \, \mathrm{e}^{2 b \phi} = - Q \, \partial \bar\partial \phi + \pi b \mu \, \mathrm{e}^{2 b \phi}.$$ At the classical level we have $Q = 1/b$ which implies that the energy-momentum is traceless on-shell. This is also why the mu term cannot appear in $T$, since it is the only conformal-breaking term, and it must thus appears in the trace only (except if it can be removed by the eom).

• Thanks for your answer! I think you miss a $\phi$ in the left hand side of the e.o.m. I have another remaining question. When you compute the Stress energy tensor in complex coordinate, the last term with the exponential is multiplied by a $h_{\mu \nu}$ in the initial form (6.27). How does it survived the complex coordinate transformation as $h_{zz}$ is 0 and makes the two other terms vanish? – Ezareth Apr 9 '18 at 9:58
• This was a typo, there is no mu term in $T$; I fixed the above explanation. – Harold Apr 9 '18 at 12:16
• Thank you very much for all the details and the explanation! Helped a lot! – Ezareth Apr 10 '18 at 21:20