How Virasoro constraints imply conformal symmetry?

Question

On one of the string theory notes I read that the conformal symmetry of the bosonic string theory is implied by virasoro constraints. Could someone explain what is the relationship between conformal symmetry and the virasoro constraints?

Answer 1

In bosonic string theory, we apply reparametrisation invariance and Weyl invariance to put the worlsheet metric in conformal gauge, i.e. $h_{\mu\nu}=\eta_{\mu\nu}$. However, there exists a residual symmetry, which in light cone-coordinates is $\sigma^+\to f(\sigma^+)$ and $\sigma^-\to f(\sigma^-)$. Moreover, in light cone coordinates, the conservation laws of the energy-momentum tensors read $\partial_-T_{++}=0$ and $\partial_+T_{--}=0$. These facts imply an infinite number of conserved charges:

$$ L_f=2T\int_0^l d\sigma \,f(\sigma^+)T_{++}(\sigma^+)$$

For a closed string, taking into account the periodic boundary conditions, one chooses $f$ in the form $f(\sigma^{\pm})=\exp(\frac{2\pi i}{l} m\sigma^{\pm})$. The Virasoro generators are defined to be the corresponding charges at $\tau=0$:

$$ L_n=-\frac{l}{4\pi^2}\int_0^ld\sigma\, e^{-\frac{2\pi i}{l}n\sigma}T_{--}$$

$$ \bar{L}_n=-\frac{l}{4\pi^2}\int_0^ld\sigma\, e^{\frac{2\pi i}{l}n\sigma}T_{++} $$

The Poisson brackets are

$$ \{ L_m,L_n\}=-i(m-n)L_{m+n}, \hspace{3mm} \{ \bar{L}_m,\bar{L}_n\}=-i(m-n)\bar{L}_{m+n}, \hspace{3mm}\{ \bar{L}_m,L_n\}=0$$

This algebra is calles Virasoro algebra. If we replace Poisson brackets by Lie brackets, a realisation of the Virasoro algebra is furnished by the vector fields

$$\bar{L}_n=e^{\frac{2\pi i}{l} n\sigma^+}\partial_+, \hspace{3 mm} L_n=e^{\frac{2\pi i}{l} n\sigma^-}\partial_- $$

If we define the variable $z=e^{\frac{2\pi i}{l} n\sigma^-}$, then $L_n=iz^{n+1}\partial_z$, which are the generators of the conformal group in two dimensions.