In the $AdS_3/CFT_2$ correspondence, the central charge of the dual CFT2 is universally given by $$ c = \frac{3\ell}{2G} $$ This is independent of the matter in the bulk of AdS3. Is it also universal in the higher dimensional analogues $AdS_d/CFT_{d-1}$ or does it depend on details of the matter in the bulk, etc.?

In particular, what is the central charge for $d=4$?

  • $\begingroup$ Any thoughts??? $\endgroup$ – Prahar Jun 17 '14 at 8:00

The expression you've written derives from a proposal, due to Ryu and Takayanagi (See: http://arxiv.org/abs/hep-th/0603001), to calculate entanglement entropy of a CFT by making use of a gravity dual.

Define a CFT in $D$ space-time dimensions. Take a spatial slice, a ball of radius $R$ for example, and calculate the entanglement entropy associated with it. You'll get the following behaviours:

$S_{D=2}(R)=\frac{c}{3}\log \frac{R}{\epsilon}$

$S_{D=3}(R)=k_1 \frac{R}{\epsilon} -F$

$S_{D=4}(R)=k_2 \frac{R^2}{\epsilon^2} -a \log \frac{R}{\epsilon}$

where $c$, $F$, $a$ are universal terms. For the case of $D=2$ it's the central charge.

Now if this CFT admits a holographic dual then there's another expression for the entanglement entropy:

$S(R) = \frac{A_{\gamma(R)}}{4G_N}$

where $G_N$ is Newton's constant and $A_{\gamma(R)}$ is the area of a minimal surface $\gamma(R)$ in $AdS_{D+1}$ whose boundary is exactly the boundary of the spatial region you're computing $S(R)$ for. Like this picture: enter image description here

The intuition for why this is true is that tracing out a region is like inducing a horizon beyond which you don't know what's going on. Then this holographic expression is like the Bekenstein-Hawking formula.

Matching expressions can give you a formula like you desire. An important thing to note though is that this $a$ for the case of $D=4$ isn't always the "$c$" central charge. (See my comment for a bit more on this) They differ in cases where your CFT is dual to a higher curvature gravity theory.(http://arxiv.org/abs/hep-th/9904179) This is related to its appearance in the trace anomaly.

This $a$ quantity does fulfill a role in generalizing the $c$-theorem of $D=2$ CFT to higher dimensions though; it's an RG monotone. This line of inquiry has been pursued by Rob Myers and Aninda Sinha. (See: http://arxiv.org/abs/1011.5819 and http://arxiv.org/abs/1006.1263)


As promised this will provide a formula for a 'central charge', or combination of them, in any number of dimensions which is just a function of constants and the AdS radius $R$.

For $D=4$ this yields: $a=\frac{\pi^{3/2}}{4 G^{(5)}_N} \frac{R^3}{\Gamma(\frac{3}{2})}$

  • $\begingroup$ Even though it is interesting in its own right, I doubt that this answer's Prahar's question. It is not about finding expressions for quantities obeying c/F/a-theorems, but explicitly about the central charge, which also exists in higher dimensions than 2. $\endgroup$ – Frederic Brünner Jun 18 '14 at 21:01
  • $\begingroup$ Well I guess I'd like to know what he meant by 'central charge'. If you think about the trace anomaly in $D=4$ then $\vev{T_{\mu}^{\mu}} \propto -a E_4 + c W^2$ where $E_4$ is the Euler density and $W$ is Weyl tensor. I think both are colloquially referred to as 'central charges' but $a$ is the RG monotone and in many cases ($N=4$ SYM ...) they're the same. Neither appear in anything like the Virasoro algebra and $a$ is what appears as the logarithmic piece of the entanglement entropy. $\endgroup$ – SM Kravec Jun 18 '14 at 23:04
  • $\begingroup$ @SMKravec - Thank you for that answer. However, as pointed out by Frederic, it is not what I'm looking for. You can take your favourite definition of the central charge in higher dimension. For example, in $d=4$, we have $a$ and $c$. In $d=3$, we also refer to the central charge as the constant appearing in the 2-point function of stress-tensor. Either way, the question I'm asking is if there is a universal constant that one can attribute to the dual CFT that is independent of the precise bulk d.o.f. like there is in $d=2$. $\endgroup$ – Prahar Jun 19 '14 at 3:36
  • $\begingroup$ Yes for $D=4$ it's $a$. I wrote the formula I believe you seek. This method gives you the formula you wrote for $D=2$. I don't know if one exists for the 'other' central charge $c$ since it doesn't appear in the entanglement entropy. See arxiv.org/pdf/0905.0932v2.pdf section 3.5 if this is mysterious to you. $\endgroup$ – SM Kravec Jun 19 '14 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.