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 \(z \to w\).
