Entanglement Entropy in $T\overline{T}$-Deformed CFT

TL;DR

This work analyzes entanglement in a $T\overline{T}$-deformed CFT on a cylinder by applying conformal perturbation theory to leading order in $\mu$, and compares with a holographic description in AdS$_3$ with a finite-radius cutoff. The main finding is a nonzero leading correction to entanglement entropy at finite temperature, $$\delta S(A) = -\frac{\mu\pi^4 c^2 l}{9 \beta^3}\coth\left(\frac{\pi l}{\beta}\right)$$, while no correction appears at leading order in the finite-size case; the corresponding Rényi corrections are computed as well. On the gravity side, the BTZ-cutoff geometry reproduces the field-theory result via the RT formula with the identified relation between $\mu$ and the bulk cutoff, $\mu = \frac{6R^4}{\pi c r_c^2}$ and $\epsilon = \frac{R^2}{r_c}$. In the zero-temperature finite-size setup dual to global AdS$_3$, the leading entanglement entropy remains the conventional CFT result, with no $\mu$-correction at this order. The results support Verlinde's finite-cutoff holographic conjecture for $T\overline{T}$-deformed holographic CFTs and motivate further nonperturbative explorations of the duality and Rényi entropies in this context.

Abstract

In this paper, we study the entanglement entropy of a single interval on a cylinder in two-dimensional $T\overline{T}$-deformed conformal field theory. For such case, the (Rényi) entanglement entropy takes a universal form in a CFT. We compute the correction due to the deformation up to the leading order of the deformation parameter in the framework of the conformal perturbation theory. We find that the correction to the entanglement entropy is nonvanishing in the finite temperature case, while it is vanishing in the finite size case. For the deformed holographic large $c$ CFT, which is proposed to be dual to a AdS$_3$ gravity in a finite region, we find the agreement with the holographic entanglement entropy via the Ryu-Takayanagi formula. Moreover, we compute the leading order correction to the Rényi entropy, and discuss its holographic picture as well.

Figures (2)

• Figure 1: $e^{\frac{2 \pi (x+i \tau )}{\beta }}$ runs on the orange circle clockwise, and $(e^{\frac{2 \pi (x+i \tau )}{\beta }}-1)$ is represented by the red arrow. When moving the head of the arrow around the orange circle once: in (a) the argument of the red arrow is added by $2\pi$, so $\log \left(e^{\frac{2 \pi (x+i \tau )}{\beta }}-1\right)|_{\tau=0}^{\tau=\beta}=2\pi i$; in (b) the argument of the red arrow doesn't change, so $\log \left(e^{\frac{2 \pi (x+i \tau )}{\beta }}-1\right)|_{\tau=0}^{\tau=\beta}=0$.
• Figure 2: Again $e^{\frac{2 \pi (x+i \tau )}{\beta }}$ runs on the orange circle clockwise. $(e^{\frac{2 \pi (x+i \tau )}{\beta }}-1)$ is represented by the red arrow, and $(e^{\frac{2 \pi (x+i \tau )}{\beta }}-e^{\frac{2 \pi l}{\beta }})$ is represented by the blue arrow. When moving the heads of the arrows around the orange circle once: in (a) the arguments of both arrows won't change, which means $g(x)=0$; in (b) the argument of the red arrow is added by $2\pi$, while the argument of the bule arrow doesn't change, so we have $g(x)=2\pi i$; in (c) the arguments of both arrows are added by $2\pi$, but their contributions cancle each other, so $g(x)=0$.