Virasoro Algebra from the Free Boson without Regularization

Usual physics derivations of the Virasoro algebra from the free boson in two dimensions usually use some sort of regularization procedure to compute the central charge. Following these notes I'd like to give a purely algebraic calculation of the central charge.


Let $A = \mathbf C[x_1, x_2, \dots]$ be the polynomial algebra in countably many generators. For an integer $k > 0$, define a $k$-linear operator $a_k$ on $A$ by
\[ a_k = \frac{\partial}{\partial x_k}, \ k > 0 \]
Similarly, for $k < 0$ we define $a_k$ by multiplication:
\[ a_{-k} = k x_k, k > 0 \]
For $k = 0$, we define $a_0$ to be multiplication by some fixed complex number (which by abuse of notation we also denote by $a_0$).

Lemma. We have the commutation relation $[a_m, a_n] = m \delta_{m+n}$ as linear operators on $A$.

For any monomial in the $a_k$, we define normal ordering $::$ to be the monomial obtained by reordering the terms so that the indices are increasing. (Mathematical interpretation: it is a section of the quotient map from the tensor algebra in the $a_k$ to the symmetric algebra, defined by lexicographic order.) For example,
\[ :a_j a_k:\ = \left\{ \begin{array}{rr} a_j a_k, & j \leq k \\ a_k a_j, & j > k \end{array} \right. \]

Next we formally define a set of operators $L_k$ by
\[ L_k = \frac{1}{2}\sum_j :a_j a_{k-j}: \]

Proposition. The $L_k$ are well-defined as linear operators on $A$.

Proof. For sufficiently large $|j|$, at least one of $j$ or $k-j$ is positive, and hence $:a_j a_{k-j}:$ contains a differentiation (on the right). Since any element $f \in A$ is annihilated by all but finitely many of the differentiation operators $\partial_j$, the formal expression $L_k f$ contains only finitely many non-zero terms, and hence is well-defined.

Lemma. As operators on $A$, we have $[a_k, L_n] = k a_{k+n}$.

Theorem. As operators on $A$, we have
\[ [L_m, L_n] = (m-n) L_{m+n} + \frac{1}{12} (m^3-m) \delta_{m+n} \]

Proof. Fix $m,n$. For the sake of simplicity we will assume $m \neq n$ and $mn \neq 0$. (The other special cases can be treated by similar arguments.) By the same argument as the proof of the preceding proposition, for any fixed element $f \in A$ there exists some $N \gg 0$ such that
\[ [L_m, L_n]f = [L_m^N, L_n] f \]
where $L_m^N$ is the truncated operator
\[ L_m^N = \frac{1}{2}\sum_{|j| < N} :a_j a_{m-j}: \]
Let us compute (noting that since $m \neq 0$, $:a_j a_{m-j}: = a_j a_{m-j}$)
\begin{align}
  [L_m^N, L_n] &= \frac{1}{2} \sum_{|j| < N} [a_j a_{m-j}, L_n] \\
  &= \frac{1}{2} \sum_{|j| < N} (m-j) a_j a_{m+n-j} + \frac{1}{2} \sum_{|j| < N}j a_{n+j} a_{m-j}
\end{align}
Denote the two sums above by $S_1$ and $S_2$. It is clear that these should be related to the operator $L_{m+n}$, but to see the exact relation we will have to normal order the terms. Let's start with $S_1$. Note that $a_j a_{m+n-j}$ is already normal ordered, unless $j > m+n-j$. Hence
\begin{align}
  S_1 &= \frac{1}{2} \sum_{|j| < N} (m-j) :a_j a_{m+n-j}: +  \frac{1}{2} \sum_{m+n\lt2j\lt2N} (m-j) [a_j, a_{m+n-j}] \\
&= \frac{1}{2} \sum_{|j| < N} (m-j) :a_j a_{m+n-j}: +  \frac{\delta_{m+n}}{2} \sum_{m+n\lt2j\lt2N} j(m-j) \\
&= \frac{1}{2} \sum_{|j| < N} (m-j) :a_j a_{m+n-j}: +  \frac{\delta_{m+n}}{2} \sum_{0\lt j\lt N} j(m-j)
\end{align}

Similarly, we normal order the terms in $S_2$:
\begin{align}
  S_2 &= \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: -\frac{1}{2} \sum_{m-n\lt2j\lt2N}j [a_{m-j}, a_{n+j}] \\
&= \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: -\frac{\delta_{m+n}}{2} \sum_{m\lt j\lt N}j (m-j) \\
&= \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: -\frac{\delta_{m+n}}{2} \sum_{m\lt j\lt N}j (m-j) \\
&= \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: -\frac{\delta_{m+n}}{2} \sum_{m\lt j\lt N}j (m-j)
\end{align}

Hence we have
\[ S_1 + S_2 = \frac{1}{2} \sum_{|j| < N} (m-j) :a_j a_{m+n-j}: +  \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: + \frac{\delta_{m+n}}{2} \sum_{0\lt j\leq m} j(m-j)  \]
Now that everything is normal ordered, we can take the limit $N \to \infty$ without fear. After a simple cancellation, and explicitly summing the last term using well-known formulas for sums of powers of integers, we obtain:
\[ [L_m, L_n] = (m-n) L_{m+n} + \frac{1}{12} (m^3-m) \delta_{m+n} \]
For the usual conventions of the Virasoro algebra, this shows that this representation corresponds to central charge $c=1$.

Remark. If we tried to take the limit $N \to \infty$ in each of the terms $S_1, S_2$ separately, before taking their sum, we would obtain a formal infinite constant $\sum_{j} j(m-j)$. In any physics textbook, the author will simply zeta-regularize this sum to obtain a fininte result. However, the above calculation shows that this is not necessary. By taking care that each term in the expression $S_1+S_2$ was normal-ordered, before taking the limit, we obtain only finite constants with no need to regularize. Zeta regularization certainly has its uses (for example in rigorous definitions of functional determinants), but as the above calculation shows, it can also be an unnecessary crutch that obscures the underlying mathematical phenomena.

Remark. There is an analogous calculation which shows that one obtains a Virasoro representation from affine Lie algebras. Physically, this corresponds to the WZW model. Roughly, the generators of the affine Lie algebra behave as an infinite set of harmonic oscillators, similar to the Heisenberg algebra above. Sometime in the future I may write a sequel to this post giving the details of this calculation.

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 \(z \to w\).