Almost every paper mentioning Brown and Henneaux's matching of asymptotic symmetries of AdS$_3$ with the Virasoro algebra of a $1{+}1$-dimensional CFT summarizes their results in the formula $$c=\frac{3R}{2G},$$ whereby the central charge $c$ is expressed in terms of the AdS radius $R$ and the gravitational constant $G$.

However, Brown and Henneaux's original paper doesn't contain this formula. Instead, it derives an asymptotic algebra of the form \begin{equation} \{J[L_n],J[L_m]\}^\star = (n-m) J[L_{n+m}] + 2\pi i \sigma R n (n^2 - 1) \delta_{n,-m} \end{equation} for charges $J[L_n]$ (on page 222), which looks very close to the standard Virasoro algebra \begin{equation} [L_n,L_m] = (n-m) L_{n+m} + \frac{c}{12} n (n^2 - 1) \delta_{n,-m} \ . \end{equation} But I don't see how this form would relate the central charge in any way to $G$ nor figure out where the factor $3/2$ comes from. Can anyone help me figure out the missing steps?


1 Answer 1


Computing the Poisson bracket gives $$I=i\{L_{m}^{(+)},L_{n}^{(+)}\}=\frac{il}{\kappa}\int_{0}^{2\pi}d\phi e^{imx^{+}}\left(e^{inx^{+}}\partial_{+}L_{+}+2L_{+}\partial_{+}e^{inx^{+}}-\frac{1}{2}\partial_{+}^{3}e^{inx^{+}} \right),$$

where $\kappa=8\pi G$ and $L_{m}^{(\pm)} = \frac{l}{\kappa}\int_{0}^{2\pi}d\phi\ L_{\pm}\left(x^{\pm}\right)e^{imx^{\pm}}$. Integrating by-parts on $\partial_{+}$ gives us $$I=\frac{il}{\kappa}\int_{0}^{2\pi}d\phi\ \left(\underbrace{-i(n+m)e^{i(n+m)x^{+}}L_{+} + 2inL_{+}e^{i(n+m)x^{+}}}_{=-i(m-n)L_{+}e^{i(m+n)x^{+}}} - \frac{1}{2}(-i)n^{3}e^{i(n+m)x^{+}} \right).$$

Now, the integral representation of the discrete-delta funciton is given by: $\delta(n) = \frac{1}{2\pi}\int_{0}^{2\pi}e^{int}dt$ see here. Thus, we have $$I = (m-n)L^{(+)}_{(m+n)} + m^{3}\delta_{(m+n),0}\frac{l}{8G},$$

where the central charge is $m^{3}\delta_{(m+n),0}\frac{l}{8G} = \frac{c}{12}m^{3}\delta_{(m+n),0}$ with the Brown-Henneaux central charge $c=\frac{3l}{2G}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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