# [1805.06286] [Chakrabortya, Giveona, Itzhakib, Kutasov] Entanglement beyond AdS ## A Different $T\overline{T}$ Duality - **Different sign** from H. Verlinde et al [1611.03470], or Silverstein et al [1909.13808]. - **Different dual geometry:** $\mcal{M}_3\times \mcal{N}$, AdS now sits in the *interior* while near the boundary we have a linear dilaton background. In other words, - IR: small $U,\ \mcal{M}_3 = \mrm{AdS}_3$ - UV: large $U,\ \mcal{M}_3 = \mathbb{R}^{1,1}\times \mathbb{R} _\phi$ No longer a cutoff geometry, except a $U_{\max}$ manually introduced for regularization. - There is a **minimum entanglement length** $L_{\min}$: as the peak of the minimal surface $U_0\to\infty$, $L$ does not shrink to zero but is bounded from below: $L\to L_{\min}$. See Fig 1. - The notation is slightly misleading: note that $U_{\max}$ does not correspond to $L_{\min}$; $U_{\max}$ is manually introduced, while $L_{\min}$ can be found as $U\to\infty$, _without any cutoff!_ ## C-function Cutoff-independent C-function (per CGIK's convention; Silverstein does not have this extra pre-factor of 3): $$ C(L) = 3L\,\pdv{S_{EE}(L)}{L} \tag{CGIK, 1.2} $$ Field theory calculation, similar to Donnelly & Shyam, [1806.07444]: - Deformation parameter: $\lambda$, or $(-\mu)$ in [1806.07444]. Trace flow equation (1.3) should be independent of $(\mop{sign} \lambda)$. Solution of $T_{ab}$ is given by (2.2). - $\pdv{\,\log Z}{L}$ is related to $T^a_a$; with Kutasov et al's convention, we get: $$ L\,\pdv{\,\log Z}{L} = -\frac{2L^2}{\lambda} \pqty{ \sqrt{1 - \frac{c\lambda}{3L^2}} - 1 }, \\ \pdv{\,\log Z}{L} = -\frac{2}{\lambda} \pqty{ \sqrt{L^2 - \frac{c\lambda}{3}} - L }, \tag{DS, 2.3$'$} $$ The square root is real iff. $$ L \ge L_0 = \sqrt{\frac{c\lambda}{3}} \label{eq:ft_length_bound} $$ > **Conventions:** > > - [1806.07444] Donnelly & Shyam: $L = 2r,\ \mu = -2\pi\lambda$ > - [1909.13808] Silverstein et al: $L = 2r$ - Compare (cf. PengXiang's note): $$ L_0 = \sqrt{\frac{6k\cdot 2\alpha'}{3}} = 2\sqrt{k\alpha'}, \\ L_{\min} = \frac{\pi}{2}\sqrt{k\alpha'} = 1.57 \sqrt{k\alpha'}, $$ If I've got their conventions right, then **they don't match quantitatively**. _But I am confused about conventions so I could be wrong!_ Note: $L_{\min}$ is the lower bound produced from holography (RT) by Kutasov et al. Here $k$ is the level of the _worldsheet_ $\mathrm{SL}(2,\mathbb{R})$ current algebra. > **Correspondence:** $c = 6k,\ \lambda = 2\alpha'$. _Why?_ I have not yet understood this, might need to check Kutasov's original: [1701.05576]. - A convenient substitution for the integral: $$ L = L_0 \cosh x = \sqrt{\frac{c\lambda}{3}} \cosh x, $$ $$ \begin{aligned} \log Z &= -\frac{2}{\lambda} \frac{c\lambda}{3} \int \dd{x}\,\sinh x \cdot \pqty{ \sinh x - \cosh x } \\ &\sim - \frac{c}{3} \pqty{ -x - \frac{1}{2} e^{-2x} } \\ &\sim \frac{c}{3}\, \cosh^{-1} \frac{L}{L_0} + \frac{L^2}{\lambda}\,\pqty{ 1 - \sqrt{1 - \pqty{\frac{L_0}{L}}^{\!\!2}} }, \end{aligned} $$ $$ S = \pqty{ 1 - \frac{L}{2} \frac{\pd}{\pd L} } \log Z = \frac{c}{3}\,\cosh^{-1} \frac{L}{L_0}, $$ $$ C(L) = 3L\,\pdv{S}{L} = \frac{c}{\sqrt{1 - \frac{c\lambda}{3L^2}}} $$ **Note:** $\log Z$ and $S$ might differ by an $L$-independent "constant", but that is eliminated in $C(L)$ due to the $L$-derivative. ## RT calculation Here we check the RT calculation, with the help of ADS (Apolo, Detournay & Song) [1911.12359] when it comes to confusing conventions. **From now on, we default to the convention of ADS [1911.12359]** The metric, B-field and dilation - at finite temperature ($T_u, T_v$) - in string frame - in static patch are given by ADS (3.13). Setting $T_u = T_v = 0$, the asymptotically AdS part is reduced to: $$ \dd{s}^2 = \ell^2 \pqty{ \frac{\dd{r}^2}{4r^2} + \frac{r\dd{u}\dd{v}}{1 + 2\lambda r} },\\ e^{2\Phi} = \frac{k}{p} \frac{1}{1 + 2\lambda r}\, e^{-2\phi_0}, \\ u,v = \varphi \pm t/\ell, \tag{ADS, 3.13$'$} \label{eq:staticMetric} $$ This should correspond to CGIK (2.3). The finite temperature version should roughly correspond to CGIK (5.1). The $\lambda$ in \eqref{eq:staticMetric} is the string theory deformation parameter; by ADS (2.43) and ADS Section 2.3, we have: $$ \lambda = \frac{k}{\ell^2} \mu, \quad k = \ell^2/\ell_s^2 \tag{ADS, 2.43$'$} $$ $$ \lambda = \mu / \ell_s^2 \tag{ADS, 3.38$'$} $$ Moreover, we have $c = 6pk$ by ADS (3.7). ### Conventions By comparing the $\abs{w} = 1$ spectrum: $$ E = - \frac{R}{2\mu} \Bqty{ 1 - \sqrt{ 1 + \frac{4\mu}{R} E(0) + \frac{4\mu^2}{R^4} J(0)^2 } }, \tag{ADS, 3.39$'$} $$ With the one in Silverstein et al (LLST) [1909.13808]: $$ \begin{aligned} E_S &= \frac{2R}{\lambda_S} \Bqty{ 1 - \sqrt{ 1 - \frac{\lambda_S}{R^2} \pqty{ \Delta + \bar{\Delta} - \frac{c}{12} } + \frac{\lambda_S^2}{4R^4} \pqty{ \Delta - \bar{\Delta} }^2 } } \\ &= \frac{2R}{\lambda_S} \Bqty{ 1 - \sqrt{ 1 - \frac{\lambda_S}{R} E(0) + \frac{\lambda_S^2}{4R^4} J(0)^2 } }, \end{aligned} \tag{LLST, 2.7$'$} \\ E(0) = \pqty{ \Delta + \bar{\Delta} - \tfrac{c}{12} } / R,\quad J(0) = \pqty{ \Delta - \bar{\Delta} }, $$ $\lambda_S$ is the deformation parameter in Silverstein et al's convention. We have a correspondence: $$ E_S = E_{\mathrm{ADS}}, \quad \lambda_S = -4\mu_{\mathrm{ADS}} $$ _Thank Luis & Wei for help with this!_ With ADS's convention, the lower bound \eqref{eq:ft_length_bound} can be expressed as: $$ L_0 = \sqrt{\frac{6k\cdot 4\mu}{3}} = \sqrt{8k\mu} $$ With the conventions sorted out, let's try to compute the geodesic with \eqref{eq:staticMetric}. ### Poincaré Patch **Again, we default to the convention of ADS [1911.12359]** At first glace \eqref{eq:staticMetric} may resemble the metric in some "polar" coordinates, but in fact it's more closely related to the Poincare coordinates. Recall that the standard BTZ black hole comes from a quotient of the Poincare patch. To see this more explicitly, we can set $\ell = 1$ and take $r = \frac{1}{z^2}$, which gives: $$ \dd{s}^2 = \frac{\dd{z}^2}{z^2} + \frac{\dd{u}\dd{v}}{z^2 + 2\lambda},\\ e^{2\Phi} = \frac{k}{p} \frac{z^2}{z^2 + 2\lambda}\, e^{-2\phi_0}, \tag{ADS, 3.13$''$} \label{eq:staticMetricPoincare} $$ _Again thank Luis for help with this!_ ### RT with dilaton The RT proposal needs to be generalized in the presence of dilatons; an updated proposal was introduced by R & T themselves in Section 7.7 of [hep-th/0605073]; see Section 3.3 of Peng-Xiang's note. This is the proposal adopted by Kutasov et al. More specifically, the "generalized area" of the RT surface in (bulk) 3D is modified by including a dilaton factor: $$ L_E = \int \dd{\tau}\,e^{-2\Phi} \sqrt{ g_{\mu\nu} \pd_\tau X^\mu \pd_\tau X^\nu } = \int \dd{\tau} \sqrt{ G_{\mu\nu} \pd_\tau X^\mu \pd_\tau X^\nu } $$ Which we would like to minimize. Here $g_{\mu\nu}$ is the string-frame metric, while $G_{\mu\nu}$ is the Einstein-frame metric; $$ G_{\mu\nu} = e^{-4\Phi} g_{\mu\nu},\\ $$ Basically $L_E$ is the Einstein-frame length; this $e^{-2\Phi}$ is the same as the one in the _spacetime effective action:_ $\cdots\int e^{-2\Phi}\,\mathcal{R}\cdots$ We have: $$ \dd{s}^2_E = \pqty{ \frac{z^2 + 2\lambda}{z^2} }^{\!\!2} \pqty{ \frac{\dd{z}^2}{z^2} + \frac{\dd{u}\dd{v}}{z^2 + 2\lambda} } $$ The minimal surface / geodesic $(t,z,\varphi)_\tau$ at the $t = 0$ slice is constrained as follows: $$ t = 0 = u - v, \\ p_\varphi = \mathrm{const.} = \pqty{ \frac{z^2 + 2\lambda}{z^2} }^{\!\!2} \frac{1}{z^2 + 2\lambda}\, \dot{\varphi},\\ p_\mu p^\mu = \pqty{ \frac{z^2 + 2\lambda}{z^2} }^{\!\!2} \pqty{ \frac{\dot{z}^2}{z^2} + \frac{\dot{\varphi}^2}{z^2 + 2\lambda} } = 1 $$ This $p_\varphi$ is the conserved (angular) momentum, not to be confused with the parameter $p = 1$. This can be reduced to: $$ \dot{\varphi} = \frac{z^4}{z^2 + 2\lambda}\, p_\varphi,\\ \dot{z}^2 = \frac{z^6}{(z^2 + 2\lambda)^3} \pqty{ z^2 + 2\lambda - p^2_\varphi z^4 } $$ Hence, $$ \Bqty{z'(\varphi)}^2 = \frac{1}{z^2 p^2_\varphi} - \frac{z^2}{z^2 + 2\lambda} $$ w.l.o.g, let's assume that entanglement entropy peaks at $\varphi = 0, z = z_0$; we have $z'(0) = 0$, and we can express $p_\varphi$ in terms of the more geometric parameter $z_0$: $$ p_\varphi = \frac{\sqrt{z_0^2 + 2\lambda}}{z_0^2} $$ $$ \Bqty{z'(\varphi)}^2 = \frac{z_0^2}{z^2} \frac{z_0^2}{z_0^2 + 2\lambda} - \frac{z^2}{z^2 + 2\lambda} $$ The **coordinate length** $L$ of the entanglement region at the boundary is given by: $$ \begin{aligned} L &= \int \dd{\varphi} \\ &= 2\int_0^{z_0} \abs{\dv{\varphi}{z}} \dd{z} \\ &= 2\int_0^{z_0} \frac{1}{\sqrt{ \frac{z_0^2}{z^2} \frac{z_0^2}{z_0^2 + 2\lambda} - \frac{z^2}{z^2 + 2\lambda} }} \dd{z}, \quad z = \frac{1}{w} \\ &= 2\int_{w_0}^{\infty} \frac{ \sqrt{1 + 2\lambda w^2} }{ \sqrt{ \frac{w^2}{w_0^2} \frac{ 1 + 2\lambda w^2 }{1 + 2\lambda w_0^2} - 1 } } \frac{\dd{w}}{w^2} \\ &= \frac{2}{w_0} \int_{1}^{\infty} \frac{ \sqrt{1 + 2\lambda w_0^2 y^2} }{ y^2 \sqrt{ y^2 \frac{ 1 + 2\lambda w_0^2 y^2 }{1 + 2\lambda w_0^2} - 1 } }\, \dd{y} \\ \end{aligned} $$ As $w_0 \to \infty$, we have: $$ L \ge L_{\min} = \frac{2}{w_0} \sqrt{2\lambda w_0^2} \int_{1}^{\infty} \frac{y}{y^2 \sqrt{y^4 - 1}}\, \dd{y} = \sqrt{8\lambda}\,\frac{\pi}{4} = \sqrt{\frac{\lambda}{2}}\,\pi $$ Note that: - The general formula for $L$ agrees with Kutasov's (3.3), with $2\lambda w^2 \leftrightarrow kU^2$, and $\ell = 1$. More explicitly, with $z = \frac{1}{w}$, or $r = w^2$, we have: $$ \dd{s}^2 = \ell^2 \pqty{ \frac{\dd{w}^2}{w^2} + \frac{\dd{u}\dd{v}}{ w^{-2} + 2\lambda } },\\ e^{2\Phi} = \frac{k}{p} \frac{w^{-2}}{w^{-2} + 2\lambda}\, e^{-2\phi_0}, $$ Here we've restored the AdS radius $\ell$, $$ \ell = \ell_s \sqrt{k} = \sqrt{k\alpha'} $$ - With $\lambda = 1/2$, we recover Kutasov's result: $L_{\min} = \frac{\pi}{2} \sqrt{k\alpha'}$. Why set $\lambda = 1/2$ ? Because with this choice the boundary (string-frame) metric has a trivial overall factor: $$ \dd{s}^2 \sim \ell^2\cdot \dd{u}\,\dd{v} $$ This is the normalization chosen by Kutasov et al in their calculations. - With $\ell = 1$, $\lambda = k\mu$, and we have: $$ L_{\min} = \frac{\pi}{\sqrt{2}} \sqrt{k\mu} $$ Recall that from the field theory side, we have $L_0 = \sqrt{8k\mu}$, therefore: $$ L_{\min} / L_0 = \pi / 4 \ne 1 $$ Indeed the two lower bounds do NOT agree with each other. This implies that we need to modify the RT proposal, at least in the UV ($z \to 0$) region, since the UV behavior determines the lower bound of the entanglement region. --- $$ \begin{aligned} 4G_N S &= \int \frac{z^2 + 2\lambda}{z^2} \sqrt{ \frac{\dd{z}^2}{z^2} + \frac{\dd{\varphi}^2}{z^2 + 2\lambda} } \\ &= 2\int_0^{z_0} \frac{z^2 + 2\lambda}{z^2} \dd{z} \sqrt{ \frac{1}{z^2} + \frac{1}{z^2 + 2\lambda} \pqty{\dv{z}{\varphi}}^{\!\!-2} }, \quad z = \frac{1}{w} \\ &= 2\int_{w_0}^{\infty} \frac{ 1 + 2\lambda w^2 }{ \sqrt{ 1 - \frac{w_0^2}{w^2} \frac{ 1 + 2\lambda w_0^2 }{1 + 2\lambda w^2} } } \frac{\dd{w}}{w} \\ &= 2\int_1^{\infty} \frac{ 1 + 2\lambda w_0^2 y^2 }{ \sqrt{ 1 - \frac{1}{y^2} \frac{ 1 + 2\lambda w_0^2 }{1 + 2\lambda w_0^2 y^2} } } \frac{\dd{y}}{y} \\ \end{aligned} $$ Again this agrees with CGIK (3.4) with $2\lambda w^2 \leftrightarrow kU^2$.