# Virasoro algebra Virasoro algebra is the unique central extension of the Witt algebra. The generators satisfy $ \left[L_{m}, L_{n}\right]=(m-n) L_{m+n}+\frac{c}{12}\left(m^{3}-m\right) \delta_{m+n, 0} $ where $c$ is the [[0033 Central charge|central charge]]. ## The algebra of 2d CFT In 2d CFT, the local [[0028 Conformal symmetry|conformal symmetry]] becomes two copies of the Virasoro algebra: - $\left[L_{n}, L_{m}\right]=(n-m) L_{n+m}+\frac{c}{12} n\left(n^{2}-1\right) \delta_{n+m, 0}$ - $\left[L_{n}, \bar{L}_{m}\right]=0$ - $\left[\bar{L}_{n}, \bar{L}_{m}\right]=(n-m) \bar{L}_{n+m}+\frac{\bar{c}}{12} n\left(n^{2}-1\right) \delta_{n+m, 0}$ - usually $c = \bar{c}$ ## The global subalgebra Subalgebra involving $L_{\pm 1}$ and $L_0$ agree with those of the global transformation $l_\pm$ and $l_0$. See [[0028 Conformal symmetry|conformal symmetry]]. ## Algebra from [[0030 Operator product expansion|OPE]] \[*The following notes are taken from [[Rsc0006 Tong String theory]]*.\] We need to compute: $\left[L_{m}, L_{n}\right]=\left(\oint \frac{d z}{2 \pi i} \oint \frac{d w}{2 \pi i}-\oint \frac{d w}{2 \pi i} \oint \frac{d z}{2 \pi i}\right) z^{m+1} w^{n+1} T(z) T(w)$ ![[0032_contours.png|400]] The trick is to first fix $w$ and do the $z$ integral: - the contour becomes ![[0032_trick.png|350]] - using [[0030 Operator product expansion|OPE]], we have $\begin{aligned}\left[L_{m}, L_{n}\right] &=\oint \frac{d w}{2 \pi i} \oint_{w} \frac{d z}{2 \pi i} z^{m+1} w^{n+1} T(z) T(w) \\ &=\oint \frac{d w}{2 \pi i} \operatorname{Res}\left[z^{m+1} w^{n+1}\left(\frac{c / 2}{(z-w)^{4}}+\frac{2 T(w)}{(z-w)^{2}}+\frac{\partial T(w)}{z-w}+\ldots\right)\right] \end{aligned}$ - expand $z^{m+1}$ around $w$: $z^{m+1}=w^{m+1}+(m+1) w^{m}(z-w)+\frac{1}{2} m(m+1) w^{m-1}(z-w)^{2}+\frac{1}{6} m\left(m^{2}-1\right) w^{m-2}(z-w)^{3}+\ldots$ - then the residue theorem picks up products at order $1/(z-w)$, of which there are three, so $\left[L_{m}, L_{n}\right]=\oint \frac{d w}{2 \pi i} w^{n+1}\left[w^{m+1} \partial T(w)+2(m+1) w^{m} T(w)+\frac{c}{12} m\left(m^{2}-1\right) w^{m-2}\right]$ - doing the remaining integral gives $\left[L_{m}, L_{n}\right]=(m-n) L_{m+n}+\frac{c}{12} m\left(m^{2}-1\right) \delta_{m+n, 0}$ ## Related - [[0030 Operator product expansion]] - [[0114 Celestial OPE]] - [[0073 AdS3-CFT2]] - [[0085 Asymptotic symmetry of AdS3]]