Virasoro Algebra

Conformal Invariance in 2D

To begin, recall that in two dimensions, the conformal transformations are generated by holomorphic and anti-holomorphic transformations. At the infinitesimal level, let $\ell_n := -z^{n+1} \partial_z$ be a basis of holomorphic vector fields. These satisfy the Witt algebra
$$[\ell_m, \ell_n] = (m-n)\ell_{m+n}.$$
Similarly, we can define $\bar{\ell}_m = -\bar{z}^{n+1} \partial_{\bar{z}}$, and in addition to the Witt algebra these new generators satisfy $[\bar{\ell}_m, \ell_n]=0$.

Now, we could try to define a 2D conformal quantum field theory to be a unitary representation of the Witt algebra (or rather, of two copies of the Witt algebra, since we have both holomorphic and anti-holomorphic vector fields--but nevermind that). But this is too naive.

Central Extensions

Recall that in quantum mechanics, states are represented by vectors in some Hilbert space $\mathcal{H}$. However, the state $|\phi\rangle$ and $\alpha|\phi\rangle$ are physically equivalent for any non-zero complex number $\alpha$. The reason, of course, is that the expectation value of an operator $\mathcal{O}$ is defined to be $\langle \phi|\mathcal{O}|\phi\rangle / \langle \phi|\phi\rangle$, and such expressions are invariant under rescaling in $\mathcal{H}$.

Thus,  a symmetry group $G$ for a theory does not necessarily act via a map $G \to U(\mathcal{H})$. It suffices to have a projective representation $G \to PU(\mathcal{H})$. Let $\mathfrak{g}, \mathfrak{pu}$ be the Lie algebras of $G$ and $PU$, respectively. A projective representation gives a map
$$\mathfrak{g} \to \mathfrak{pu}.$$
Since $PU$ is a quotient of $U$, we have a short exact sequence
$$0 \to \mathbb{C} \to \mathfrak{u} \to \mathfrak{pu} \to 0.$$
Now let $\hat{\mathfrak{g}}$ be defined as
$$\hat{\mathfrak{g}} = \{ (\xi, \eta) \in \mathfrak{u}\oplus\mathfrak{g} \ | \ \pi(\xi) = \rho(\eta) \}$$
This comes with a natural projection $\hat{\mathfrak{g}} \to \mathfrak{g}$. If we suppose that the projective representation $\rho$ is faithful, then the kernel of this map is exactly $\mathbb{C}$. Hence, a faithful projective representation of $\mathfrak{g}$ yields a short exact sequence of Lie algebras
$$0 \to \mathbb{C} \to \hat{\mathfrak{g}} \to \mathfrak{g} \to 0.$$
We have obtained a central extension of $\mathfrak{g}$.

Virasoro Algebra

Finally, we can define the Virasoro algebra. It has generators $L_n$ and $c$, with defining relations
$$[L_m, L_n] = (m-n) L_{m+n} + \frac{c}{12}(m^3-m) \delta_{m+n,0}, [c, L_n] = 0.$$
The generator $c$ acts as a scalar in any irreducible representation, and its value is called the central charge. The factor of $1/12$ is entirely conventional. Now, the amazing fact is the following.

Theorem. Up to equivalence, the Virasoro algebra is the unique non-trivial central extension of the Witt algebra.

Proof sketch. This is essentially just a calculation. Any central extension has to be of the form
$$[L_m, L_n] = (m-n) L_{m+n} + A(m,n) c$$
for some function $A(m,n)$. If we make the replacement $L_m \mapsto L_m + a_m c$, then we have
$$[L_m, L_n] = (m-n) L_{m+n} + \left( A(m,n) + (m-n) a_{m+n} \right) c$$
Taking $n = 0$, we have
$$[L_m, L_0] = m L_{m} + \left( A(m,0) + m a_{m} \right) c$$
Hence for $m\neq0$ we can take $a_m = m^{-1} A(m,0)$. Having done this, we are now free to assume that $A(m,0) = 0$ for all $m$. Then we may apply the Jacobi identity to deduce that $A(m,n)=0$ except possibly for $m=-n$, so that $A(m,n)$ can be written in the form $A(m,n) = A_m \delta_{m+n, 0}$. Finally, another application of the Jacobi identity yields a simple recurrence relation for the coefficients $A_m$, and it is easily seen that every solution of this recurrence is proportional to $m^3-m$.

Now we can take our (preliminary, and still too naive) definition of a quantum conformal field theory to be a unitary representation of the Virasoro algebra.

Stress-Energy Tensor and OPE

The operator $L_0$ behaves like the Hamiltonian of the theory, and the Virasoro relations show that $L_n$ for $n>0$ act as lowering operators. Hence, in a physically sensible representation, the vacuum vector $|\Omega\rangle$ will be annihilated by $L_n$ for all $n > 0$. Unitary requires $L_n^\dagger = L_{-n}$, so additionally we have $\langle \Omega|L_n = 0$ for $n < 0$. Hence

$$\langle \Omega | L_m L_n | \Omega \rangle = 0 \ \textrm{unless}\ n \leq 0, m \geq 0$$

Now define the stress-energy tensor to be the operator-valued formal power series

$$T(z) = \sum_n \frac{L_n}{z^{n+2}}$$
We can consider the vacuum expectation of the product $T(z) T(w)$. By the above remarks, many terms in the expansion will vanish. In fact, it is a straightforward (but tedious!) exercise to check the following.

Theorem. The stress-energy tensor satisfies the operator product expansion
$$T(z) T(w) \sim \frac{c/2}{(z-w)^4} + \frac{2 T(w)}{(z-w)^2} + \frac{\partial_w T(w)}{z-w}$$
where $\sim$ denotes that the left- and right-hand sides are equal up to the addition of terms with vanishing vev and/or regular as $z \to w$.