The Virasoro algebra \begin{equation} [L_m,L_n]=(m-n) L_{m+n} +\frac{c}{12} (m^3-m) \delta_{m+n,0} \end{equation} of the stress energy tensor $T$ was said to follow from the witt algebra of the local conformal transformation $z\rightarrow z+\epsilon_n$ where $\epsilon_n=-z^{n+1}$ , and $\ell_n=-z^{n+1} \partial _z$ was the generator of the local conformal transformation with the Witt algebra \begin{equation} [\ell_m ,\ell_n]=(m-n) \ell_{m+n} \end{equation}

However, the $L_n=\frac{1}{2\pi i}\oint dz z^{n+1} T(z)$ came from the mode expansion of $T(z)$ and the Virasoro algebra followed from the definition of the quasi primary fields ($T(z)T(w)$ expansion). It was not clear where did the generator of the local conformal transformation $\ell_n$ come to play, i.e. $L_n$ seemed to have nothing to do with $\ell_n$.

How did the two copies of the Witt algebra become two copies of the Virasoro algebra in the CFT? Especially, how does the global subgroup $\ell_0,\ell_{\pm 1}$ influence $L_n$? Was there necessary a bijection, or was it simply an algebraic coincidence?

Related: Virasoro Algebra vs Witt Algebra


1 Answer 1


$L_n$ seemed to have nothing to do with $\ell_n$.

There exists a specific relation between these two. The symmetry of the action is reflected on the correlation functions (which are the 'observables' of a theory) through Ward Identities. The general form of these identities are: $$\frac{\partial}{\partial x_{\mu}}\langle j^{\mu}(x) \Phi(x_1)\Phi(x_2)... \rangle = \sum \delta(x-x_i) \langle\Phi(x_1)...G\Phi(x_i)... \rangle$$

In the above expression, $j^{\mu}$ is the current associated with the symmetry and $G\Phi(x)$ denotes the variation of the field under the symmetry transformations.

Now, the modes of the stress tensor generates the conformal symmetry. What this means is that all currents associated to conformal symmetry can be constructed from the stress tensor.

For example, we know that the current associated with the translation (along $x^{\nu}$) symmetry is $(j^{\mu}_{Trans})^{\nu} = T^{\mu\nu}$. Similarly, the other symmetry generators are $$(j_{Lorentz}^{\mu})^{\rho\sigma} = T^{\mu\nu}x^{\rho}- T^{\mu\rho}x^{\sigma}$$ $$j^{\mu}_{Dilatation} = T^{\mu}_{\,\,\,\nu} x^{\nu}$$

This is the reason that modes of stress tensor $L_n$ generates conformal symmetry. Why the modes have non-zero central charge is another matter and that has to do with breaking of conformal symmetry at quantum level (see the related question).


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.