# Modern form of Brown-Henneaux formula

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

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 $$c=\frac{3l}{2G}$$