Virasoro Generators, Descendants, and Central Charge

Radial quantization turns stress-tensor modes into more than notation. In two dimensions, the holomorphic stress tensor $T(z)$ generates an infinite-dimensional algebra of local conformal transformations. A primary field is the starting point of a representation of this algebra, and its descendants are obtained by acting with the negative modes $L_{-1},L_{-2},\ldots$.

The quantum stress algebra has a central extension: the stress tensor has a singular OPE with itself that contains a c-number fourth-order pole. Its coefficient is written as $c/2$, where $c$ is the central charge. It measures the short-distance strength of stress-tensor fluctuations, controls the central extension of the Virasoro algebra, and fixes the Schwarzian term in the transformation law of $T$ introduced in the preceding lessons.

Required background. Primary fields, cylinder maps and mode expansions supplies radial ordering, local descendants, the state–operator map and the stress-mode convention. We use its plane identity vacuum and counterclockwise contours. The free-field checks also use Wick contraction, including fermionic interchange signs.

Stress-tensor modes as local generators

Virasoro modes and central-charge normalization

We work on the Euclidean plane in one holomorphic sector, with radially ordered OPEs and integer stress modes. The antiholomorphic sector has generators $\bar L_n$ and central charge $\bar c$. We normalize modes by

$$L_n={1\over 2\pi i}\oint dz\,z^{n+1}T(z), \qquad T(z)=\sum_{n\in\mathbb Z}L_n z^{-n-2}.$$

A primary field $\mathcal O$ of weights $(h,\bar h)$ obeys

$$T(z)\mathcal O(w,\bar w) \sim {h\mathcal O(w,\bar w)\over (z-w)^2} +{\partial_w\mathcal O(w,\bar w)\over z-w},$$

and similarly for $\bar T(\bar z)$. The central charge is normalized by

$$T(z)T(w) \sim {c/2\over (z-w)^4} +{2T(w)\over (z-w)^2} +{\partial T(w)\over z-w}.$$

The definition

$$L_n={1\over 2\pi i}\oint dz\,z^{n+1}T(z)$$

says that $L_n$ is a contour integral of the conserved holomorphic current associated with the vector field

$$v_n(z)=z^{n+1}\partial_z.$$

The three modes $L_{-1},L_0,L_1$ correspond to translations, dilatations/rotations, and special conformal transformations on the plane. The modes with all other $n$ generate local conformal transformations that are not globally well-defined on the Riemann sphere but are perfectly meaningful as contour operations around operator insertions.

For descendant fields, retain the notation of Lesson 22’s local modes:

$$(L_n^{(a)}\mathcal O)(a) =\frac{1}{2\pi i}\oint_{C_a}dz\,(z-a)^{n+1}T(z)\mathcal O(a).$$

Here $C_a$ surrounds only $a$. This is a coefficient in a radially ordered OPE, not generally the commutator of an origin mode with a displaced field. At the origin,

$$T(z)\mathcal O(0) = \sum_{n\in\mathbb Z} z^{-n-2}(L_n^{(0)}\mathcal O)(0).$$

For a primary field, the OPE is

$$T(z)\mathcal O(0) \sim {h\mathcal O(0)\over z^2} +{\partial\mathcal O(0)\over z} +\text{regular terms}.$$

Comparing coefficients gives

$$\begin{aligned} (L_0^{(0)}\mathcal O)(0)&=h\mathcal O(0),\\ (L_{-1}^{(0)}\mathcal O)(0)&=\partial\mathcal O(0),\\ (L_n^{(0)}\mathcal O)(0)&=0\qquad(n>0). \end{aligned}$$

The positive local modes annihilate a primary insertion. Under the state–operator map, this is why primary states are called highest-weight states: the positive state modes lower the $L_0$ eigenvalue, and a primary state cannot be lowered further inside its conformal family.

The negative modes generate descendants. The first one is special:

$$L_{-1}^{(0)}\mathcal O=\partial\mathcal O.$$

But the next one is not simply a second derivative. The field

$$(L_{-2}^{(0)}\mathcal O)(0) ={1\over 2\pi i}\oint_0 dz\,{1\over z}T(z)\mathcal O(0)$$

is the finite-part stress descendant, generally different from the translation descendant $(L_{-1}^{(0)})^2\mathcal O=\partial^2\mathcal O$. They need not be independent: null relations can identify combinations. The local definition and its state correspondence are developed in Di Francesco, Mathieu and Sénéchal 1997, § 6.6.1, pp. 177–178, Eqs. 6.147–6.155.

Primary states and descendant towers

Use the plane identity vacuum with $\langle0|0\rangle=1$. Regularity of $T(z)|0\rangle$ at the origin gives

$$L_n|0\rangle=0\qquad(n\ge-1).$$

This includes invariance under the three global generators; global invariance alone is not the stated regularity condition for every positive mode. Radial quantization converts a local operator into a state:

$$|\mathcal O\rangle=\mathcal O(0)|0\rangle.$$

At the origin, the state of $(L_n^{(0)}\mathcal O)(0)$ is $L_n|\mathcal O\rangle$. It is not generally $[L_n,\mathcal O(0)]|0\rangle$, because $\mathcal O(0)L_n|0\rangle$ need not vanish for $n\le-2$. A primary with holomorphic weight $h$ gives a state satisfying

$$L_0|\mathcal O\rangle=h|\mathcal O\rangle, \qquad L_n|\mathcal O\rangle=0\quad(n>0).$$

The descendants are

$$L_{-n_1}L_{-n_2}\cdots L_{-n_k}|\mathcal O\rangle, \qquad n_i>0.$$

The level is

$$N=n_1+n_2+\cdots+n_k.$$

The commutator with $L_0$, derived below from the stress OPE, is

$$[L_0,L_{-n}]=nL_{-n},$$

so a level-$N$ descendant has holomorphic weight $h+N$:

$$L_0\left(L_{-n_1}\cdots L_{-n_k}|\mathcal O\rangle\right) =(h+N)L_{-n_1}\cdots L_{-n_k}|\mathcal O\rangle.$$

To avoid counting different orderings twice, choose $n_1\ge n_2\ge\cdots\ge n_k\ge1$. Commutators express another ordering as a linear combination of ordered products. For example,

$$L_{-1}L_{-2}|h\rangle =L_{-2}L_{-1}|h\rangle+L_{-3}|h\rangle.$$

The first levels of the formal Verma module are listed below. Each row has $L_0$ eigenvalue $h+N$.

Ordered descendants in a formal Verma module

Level N	Operators acting on the primary state	Formal count p(N)
0	$\mathbf1$	1
1	$L_{-1}$	1
2	$L_{-2}$, $L_{-1}^2$	2
3	$L_{-3}$, $L_{-2}L_{-1}$, $L_{-1}^3$	3

The Poincaré–Birkhoff–Witt ordering gives $p(N)$, the number of partitions of $N$, as the dimension of level $N$ in the formal Verma module. A physical representation can be a quotient in which null vectors and their descendants are identified with zero. It can therefore contain fewer independent states. For example, $L_{-1}|0\rangle=0$ in the identity-vacuum representation. The displayed level equation still holds for a zero vector, but only a nonzero surviving state is an eigenvector. See Di Francesco, Mathieu and Sénéchal 1997, § 6.2.2, pp. 157–158, Eqs. 6.26–6.38, and § 7.1, pp. 202–205 for the ordered products and null-submodule quotient.

Why the stress tensor is special

For an ordinary primary field of weight $h$, the stress-tensor OPE has only a second- and first-order pole. Since $T$ has weight $2$, one might guess

$$T(z)T(w) \stackrel{?}{\sim} {2T(w)\over (z-w)^2} +{\partial T(w)\over z-w}.$$

For the positive active contour action inherited from Lesson 21, a quadratic differential without an anomalous term would obey

$$\delta_\epsilon T =\epsilon\partial T+2(\partial\epsilon)T.$$

Quantum mechanically this is incomplete. The product $T(z)T(w)$ has a c-number singularity:

$$\boxed{ T(z)T(w) \sim {c/2\over (z-w)^4} +{2T(w)\over (z-w)^2} +{\partial T(w)\over z-w}. }$$

The first term is proportional to the identity operator, rather than another nontrivial local field. The coefficient $c$ is the central charge. The three poles have different jobs in the commutator $[L_m,L_n]$ calculated below.

Contributions of the three singular stress-OPE terms

Pole order	Coefficient in the OPE	Contribution after both contour integrals
4	$c/2$ times the identity	$\frac{c}{12}m(m^2-1)\delta_{m+n,0}$
2	$2T(w)$	$2(m+1)L_{m+n}$
1	$\partial T(w)$	$-(m+n+2)L_{m+n}$

In the plane conformal identity vacuum, with no additional insertion or background,

$$\langle T(z)\rangle=0.$$

The OPE fixes the singular part of the two-point function. In this vacuum, global covariance and regularity at infinity exclude an additional regular holomorphic term, giving

$$\boxed{ \langle T(z)T(w)\rangle={c/2\over (z-w)^4}. }$$

These are plane-vacuum statements; a local OPE alone would not fix a regular addition in another state or geometry. The normalized OPE and vacuum correlator appear in Di Francesco, Mathieu and Sénéchal 1997, § 5.4, pp. 135–136, Eqs. 5.121–5.122. Thus $c$ is the normalization of the stress-tensor two-point function after the stress tensor itself has already been normalized by the Ward identity. It is not removed by rescaling $T$, because rescaling $T$ would also rescale the generator of conformal transformations and spoil the standard transformation law of every operator.

A more invariant way to say the same thing is this: the stress tensor is not a primary field when $c\ne0$. It is quasiprimary, because the global modes $L_{-1},L_0,L_1$ still act on it as expected, but under general local conformal transformations it acquires an anomalous c-number term. Lesson 24 studies this Schwarzian term in more detail.

The central term from contour algebra

The Virasoro commutator follows from the $T T$ OPE inside radially ordered matrix elements. Subtract two counterclockwise origin-centered contours, one just outside $w$ and one just inside it. Both surround zero; the intervening annulus contains no other insertion, pole or cut. Their difference is a small counterclockwise contour around $w$, excluding zero, as in Lesson 22’s contour comparison. This exclusion matters for negative Laurent powers. On a common domain of such radial matrix elements, start with

$$[L_m,L_n] = {1\over(2\pi i)^2}\oint_0 dw\,w^{n+1} \oint_w dz\,z^{m+1}T(z)T(w),$$

where the inner contour around $w$ computes the singular part of the OPE. Insert

$$T(z)T(w) \sim {c/2\over (z-w)^4} +{2T(w)\over (z-w)^2} +{\partial T(w)\over z-w}.$$

The double-pole term gives

$${1\over2\pi i}\oint_w dz\,{z^{m+1}2T(w)\over (z-w)^2} =2(m+1)w^mT(w).$$

The simple-pole term gives

$${1\over2\pi i}\oint_w dz\,{z^{m+1}\partial T(w)\over z-w} =w^{m+1}\partial T(w).$$

Multiplying by $w^{n+1}$ and integrating around the origin,

$${1\over2\pi i}\oint dw\, \left[2(m+1)w^{m+n+1}T(w)+w^{m+n+2}\partial T(w)\right].$$

The second term is integrated by parts:

$${1\over2\pi i}\oint dw\,w^{m+n+2}\partial T(w) =-(m+n+2){1\over2\pi i}\oint dw\,w^{m+n+1}T(w).$$

Thus the noncentral contribution is

$$\bigl(2m+2-m-n-2\bigr)L_{m+n} =(m-n)L_{m+n}.$$

For the fourth-order pole,

$${1\over2\pi i}\oint_w dz\,{z^{m+1}\over (z-w)^4} ={1\over 3!}\partial_w^3 w^{m+1} ={m(m^2-1)\over6}w^{m-2}.$$

Multiplying by $c/2$ and integrating over $w$ gives

$${c\over 12}m(m^2-1)\delta_{m+n,0}.$$

Therefore

$$\boxed{ [L_m,L_n] =(m-n)L_{m+n} +{c\over12}m(m^2-1)\delta_{m+n,0}. }$$

This is the Virasoro algebra. The contour construction and residue calculation are developed in Di Francesco, Mathieu and Sénéchal 1997, § 6.1.2, pp. 154–155, Eqs. 6.13–6.18, and § 6.2.1, pp. 155–157, Eqs. 6.21–6.25. The central term vanishes for $m=-1,0,1$, so the global conformal subalgebra remains

$$[L_0,L_{-1}]=L_{-1}, \qquad [L_0,L_1]=-L_1, \qquad [L_1,L_{-1}]=2L_0.$$

The infinite-dimensional extension is quantum mechanically projective: the symmetry generators close up to a central c-number.

A normalized current algebra and positivity

A chiral current provides a simpler example of the same mechanism. Let $J(z)$ be a bosonic holomorphic U(1) current with

$$J(z)J(w)\sim\frac{k}{(z-w)^2}, \qquad J_n=\frac{1}{2\pi i}\oint_0 dz\,z^nJ(z).$$

The current level $k$ is defined by this normalization. The same nested-contour argument gives

$$\begin{aligned} [J_m,J_n] &=\frac{k}{2\pi i}\oint_0dw\,w^n\,\partial_w w^m\\ &=km\,\delta_{m+n,0}. \end{aligned}$$

For the free boson of Exercise 3, $J=i\partial\varphi$ has $k=1$: the factor $i^2$ cancels the minus sign of $\partial\varphi\partial\varphi$. The unit oscillator-current realization is described in Di Francesco, Mathieu and Sénéchal 1997, § 6.3.1, pp. 160–161, Eqs. 6.46 and 6.53–6.57.

For positivity, additionally assume radial reflection positivity, the adjoint $J_n^\dagger=J_{-n}$, and the normalized neutral vacuum with $J_n|0\rangle=0$ for $n\ge0$. Then

$$\|J_{-1}|0\rangle\|^2 =\langle0|[J_1,J_{-1}]|0\rangle=k\ge0.$$

The contact term can be stated without an unspecified spacetime-current convention. On the unit circle $z=e^{i\theta}$ at radial time zero, write its angle as $\theta\sim\theta+2\pi$, distinct from the boson $\varphi$, and define

$$\begin{aligned} j(\theta)&=\frac{1}{2\pi}\sum_{n\in\mathbb Z}J_ne^{-in\theta},\\ \delta_{2\pi}(x)&=\frac{1}{2\pi}\sum_{n\in\mathbb Z}e^{-inx}. \end{aligned}$$

Thus $j(\theta)=e^{i\theta}J(e^{i\theta})/(2\pi)$ and $\int_0^{2\pi}j(\theta)\,d\theta=J_0$. Interpret the series after smooth periodic smearing, on a common current-mode domain where the smeared sums exist. The mode algebra implies

$$\begin{aligned} [j(\theta),j(\theta')] &=\frac{k}{4\pi^2}\sum_n n e^{-in(\theta-\theta')}\\ &=\frac{ik}{2\pi}\partial_\theta \delta_{2\pi}(\theta-\theta'). \end{aligned}$$

This derivative-of-delta central term is a Schwinger term in the declared circle convention. It vanishes for $k=0$ and contributes nothing to the zero-mode commutator. A spacetime density/current formula would require its own component and continuation dictionary. The current and stress central OPE poles have orders two and four; the highest derivatives in their central contact distributions have orders one and three.

The analogous stress calculation uses the normalized plane identity vacuum already specified and, for the norm, $L_n^\dagger=L_{-n}$. Its stress state is exactly

$$|T\rangle=T(0)|0\rangle=L_{-2}|0\rangle$$

in the state–operator construction. Since $L_2|0\rangle=0$,

$$\langle T|T\rangle =\langle0|[L_2,L_{-2}]|0\rangle =\langle0|\left(4L_0+\frac c2\right)|0\rangle =\frac c2.$$

More generally,

$$\langle0|L_nL_{-n}|0\rangle =\frac c{12}n(n^2-1),\qquad n\ge2.$$

Thus unitarity requires $c\ge0$. This is a necessary test, not a sufficient condition for a complete representation to be unitary; other descendant norms impose further conditions. The general highest-weight norm is derived in Di Francesco, Mathieu and Sénéchal 1997, § 7.2.1, p. 205, Eq. 7.19. Nonunitary theories can have negative central charge.

Free-field checks

The normalization above gives familiar values:

$$\text{real free boson:}\quad c=1, \qquad \text{Majorana fermion:}\quad c={1\over2}, \qquad \text{Dirac fermion:}\quad c=1.$$

The Ising CFT contains a Majorana fermion $\psi(z)$ with

$$\psi(z)\psi(w)\sim {1\over z-w}.$$

The holomorphic stress tensor is

$$T(z)=-{1\over2}:\psi\partial\psi:(z).$$

The fermionic sign can be checked directly. Put $x=z-w$ and order the four fields as $\psi(z),\partial\psi(z),\psi(w),\partial\psi(w)$. Their cross contractions are

$$C_{13}=\frac1x,\quad C_{24}=-\frac2{x^3},\quad C_{14}=\frac1{x^2},\quad C_{23}=-\frac1{x^2}.$$

The two complete cross pairings have opposite Wick signs. Including the two factors $-1/2$ from the stress tensors gives

$$\frac14\left(-C_{13}C_{24}+C_{14}C_{23}\right) =\frac14\frac{2-1}{x^4} =\frac1{4x^4}.$$

For the single-cross terms, Taylor expansion about $w$ and fermionic normal ordering give

$$-\frac{:\psi\partial\psi:(w)}{x^2} -\frac{:\psi\partial^2\psi:(w)}{2x} =\frac{2T(w)}{x^2}+\frac{\partial T(w)}x.$$

Here $:\partial\psi\partial\psi:=0$. Combining the terms gives

$$T(z)T(w) \sim {1/4\over (z-w)^4} +{2T(w)\over (z-w)^2} +{\partial T(w)\over z-w}.$$

Comparing with

$$T(z)T(w) \sim {c/2\over (z-w)^4}+\cdots$$

gives

$$c={1\over2}.$$

The source uses $\psi\psi\sim1/[2\pi g(z-w)]$ and $T=-\pi g:\psi\partial\psi:$; choosing its unit field $\psi_{\rm unit}=\sqrt{2\pi g}\,\psi$ gives exactly our normalization. See Di Francesco, Mathieu and Sénéchal 1997, § 5.3.2, pp. 131–132, Eqs. 5.95–5.100. The bosonic counterpart in Exercise 3 uses the unit field corresponding to $\varphi_{\rm unit}=\sqrt{4\pi g}\,\varphi$ in Di Francesco, Mathieu and Sénéchal 1997, § 5.3.1, pp. 128–129, Eqs. 5.73–5.83.

The critical Ising model has this stress-tensor normalization: one chiral Majorana fermion in each sector, with half the central charge of a chiral complex fermion.

The cylinder vacuum shift

The central charge also appears in the cylinder energy calculated in Lesson 22. Use a flat cylinder of circumference $2\pi$ with the map

$$z=e^w, \qquad w=\tau+i\theta.$$

The stress tensor does not transform as a purely classical quadratic differential. The quantum result is

$$T_{\rm cyl}(w)=z^2T_{\rm plane}(z)-{c\over24}.$$

The constant term is a Casimir-energy shift. For both holomorphic and antiholomorphic sectors, the cylinder Hamiltonian is

$$H_{\rm cyl}=L_0+\bar L_0-{c+\bar c\over24}.$$

For the plane identity vacuum, the energy is $-(c+\bar c)/24$. If another sector has a lowest-weight state with weights $h_0,\bar h_0$, its energy is $h_0+\bar h_0-(c+\bar c)/24$; the stress shift alone does not identify that sector’s ground state. The circumference and plane-vacuum assumptions are explicit in Di Francesco, Mathieu and Sénéchal 1997, § 5.4.2, pp. 138–139, Eqs. 5.137–5.139. The same $c$ controls short-distance stress fluctuations and this finite-size shift. Lesson 24 develops the Schwarzian transformation further; the Virasoro algebra and stress tensor provides the canonical treatment of that connection.

Summary

The stress tensor is the generator of local conformal transformations. Its modes

$$L_n={1\over2\pi i}\oint dz\,z^{n+1}T(z)$$

act on primary fields and generate descendant towers. A primary state obeys

$$L_n|h\rangle=0\quad(n>0), \qquad L_0|h\rangle=h|h\rangle,$$

and formal descendants are obtained by applying $L_{-n}$ with $n>0$. Null relations must be imposed before counting independent physical states.

The stress tensor is special because its OPE with itself contains the identity pole

$$T(z)T(w) \sim {c/2\over (z-w)^4}+\cdots.$$

The coefficient of this pole defines the central charge. It produces the Virasoro algebra

$$[L_m,L_n] =(m-n)L_{m+n} +{c\over12}m(m^2-1)\delta_{m+n,0}.$$

For unitary CFTs, $c$ is nonnegative because it is proportional to the norm of the stress-tensor state. For the critical Ising model, the Majorana fermion gives $c=1/2$.

Common pitfalls

Do not identify $(L_{-2}^{(a)}\mathcal O)(a)$ with $\partial^2\mathcal O(a)$ by definition. The latter is the repeated translation descendant; a relation between the two requires a null-state or other representation-specific identity.

Do not treat $T$ as an ordinary primary field when $c\ne0$. It has primary-like terms in the OPE, but the fourth-order identity pole makes it anomalous under general local conformal maps.

Do not regard $c$ as removable by rescaling $T$. The normalization of $T$ is fixed by the Ward identity; once $T$ generates transformations correctly, $c$ is a physical coefficient of $\langle TT\rangle$.

Do not forget the antiholomorphic sector. A full local CFT has both $L_n$ and $\bar L_n$, and generally two central charges $c$ and $\bar c$. Parity symmetry exchanging the sectors requires $c=\bar c$.

Exercises

Exercise 1: Level of a Virasoro descendant

Let $|h\rangle$ be a primary state satisfying

$$L_0|h\rangle=h|h\rangle, \qquad L_n|h\rangle=0\quad(n>0).$$

Assuming

$$[L_0,L_{-n}]=nL_{-n},$$

show that every nonzero level-$N$ descendant that survives the physical quotient has $L_0$ eigenvalue $h+N$. The algebraic equation should also hold when the descendant is zero.

Solution

A general descendant has the form

$$|\chi\rangle=L_{-n_1}\cdots L_{-n_k}|h\rangle, \qquad n_i>0.$$

Using $[A,BC]=[A,B]C+B[A,C]$ repeatedly,

$$[L_0,L_{-n_1}\cdots L_{-n_k}] =(n_1+\cdots+n_k)L_{-n_1}\cdots L_{-n_k}.$$

Therefore

$$\begin{aligned} L_0|\chi\rangle &=L_{-n_1}\cdots L_{-n_k}L_0|h\rangle +[L_0,L_{-n_1}\cdots L_{-n_k}]|h\rangle \\ &=h|\chi\rangle+(n_1+\cdots+n_k)|\chi\rangle. \end{aligned}$$

Thus

$$L_0|\chi\rangle=(h+N)|\chi\rangle, \qquad N=n_1+\cdots+n_k.$$

Exercise 2: Virasoro algebra from the TT OPE

Use the OPE

$$T(z)T(w) \sim {c/2\over (z-w)^4} +{2T(w)\over (z-w)^2} +{\partial T(w)\over z-w}$$

and the mode definition

$$L_n={1\over2\pi i}\oint dz\,z^{n+1}T(z)$$

with the counterclockwise radial-contour and no-extra-singularity assumptions of the derivation above, to derive

$$[L_m,L_n] =(m-n)L_{m+n} +{c\over12}m(m^2-1)\delta_{m+n,0}.$$

Solution

Compute the commutator by taking the $z$ contour around $w$:

$$[L_m,L_n] ={1\over(2\pi i)^2}\oint_0 dw\,w^{n+1} \oint_w dz\,z^{m+1}T(z)T(w).$$

The double pole gives

$${1\over2\pi i}\oint_w dz\,{2z^{m+1}T(w)\over(z-w)^2} =2(m+1)w^mT(w).$$

The simple pole gives

$${1\over2\pi i}\oint_w dz\,{z^{m+1}\partial T(w)\over z-w} =w^{m+1}\partial T(w).$$

Thus the noncentral part is

$${1\over2\pi i}\oint_0 dw\, \left[2(m+1)w^{m+n+1}T(w)+w^{m+n+2}\partial T(w)\right].$$

Integrating the second term by parts gives

$$-(m+n+2){1\over2\pi i}\oint_0dw\,w^{m+n+1}T(w).$$

Hence the coefficient is

$$2(m+1)-(m+n+2)=m-n,$$

so this part is $(m-n)L_{m+n}$.

For the central term,

$${1\over2\pi i}\oint_w dz\,{z^{m+1}\over(z-w)^4} ={1\over 3!}\partial_w^3w^{m+1} ={m(m^2-1)\over6}w^{m-2}.$$

Multiplying by $c/2$ and then by $w^{n+1}$, the $w$ integral is nonzero only when

$$m+n=0.$$

The coefficient is

$${c\over2}{m(m^2-1)\over6} ={c\over12}m(m^2-1).$$

Therefore

$$[L_m,L_n] =(m-n)L_{m+n} +{c\over12}m(m^2-1)\delta_{m+n,0}.$$

Exercise 3: Central charge of a free boson

Let a chiral free boson have the OPE

$$\varphi(z)\varphi(w)\sim -\log(z-w),$$

so that

$$\partial\varphi(z)\partial\varphi(w)\sim -{1\over (z-w)^2}.$$

With

$$T(z)=-{1\over2}:\partial\varphi\partial\varphi:(z),$$

show that the central charge is $c=1$.

Solution

The fourth-order pole in $T(z)T(w)$ comes from double contractions:

$$T(z)T(w) ={1\over4}:\partial\varphi\partial\varphi:(z):\partial\varphi\partial\varphi:(w).$$

There are two ways to contract the two $\partial\varphi$ fields at $z$ with the two $\partial\varphi$ fields at $w$. Each contraction contributes

$$\left(-{1\over(z-w)^2}\right)^2={1\over(z-w)^4}.$$

Therefore the identity singularity is

$$T(z)T(w)\sim {1\over4}\cdot 2\,{1\over(z-w)^4}+\cdots ={1/2\over(z-w)^4}+\cdots.$$

Comparing with

$$T(z)T(w)\sim {c/2\over(z-w)^4}+\cdots$$

gives

$$c=1.$$

Exercise 4: Norm of the stress-tensor state

Assume a normalized plane identity vacuum, $\langle0|0\rangle=1$, with $T(z)|0\rangle$ regular at the origin, so that

$$L_n|0\rangle=0\quad(n\ge -1),$$

and that $L_n^\dagger=L_{-n}$. Use the Virasoro algebra to show

$$\|L_{-2}|0\rangle\|^2={c\over2}.$$

What does unitarity imply?

Solution

The norm is

$$\|L_{-2}|0\rangle\|^2 =\langle0|L_2L_{-2}|0\rangle.$$

Since $L_2|0\rangle=0$, we may replace $L_2L_{-2}$ by the commutator:

$$\langle0|L_2L_{-2}|0\rangle =\langle0|[L_2,L_{-2}]|0\rangle.$$

The Virasoro algebra gives

$$[L_2,L_{-2}] =4L_0+{c\over12}2(2^2-1) =4L_0+{c\over2}.$$

The vacuum has $L_0|0\rangle=0$, so

$$\|L_{-2}|0\rangle\|^2={c\over2}.$$

A unitary Hilbert space has nonnegative norms. Therefore

$$c\ge0.$$

References

• P. Di Francesco, P. Mathieu, and D. Sénéchal, Conformal Field Theory, Graduate Texts in Contemporary Physics (Springer, New York, 1997). DOI: 10.1007/978-1-4612-2256-9.

Further reading

• A. A. Belavin, A. M. Polyakov, and A. B. Zamolodchikov, “Infinite conformal symmetry in two-dimensional quantum field theory,” Nuclear Physics B 241 (1984) 333–380, for Virasoro representation theory and the conformal bootstrap framework.
• P. Ginsparg, “Applied Conformal Field Theory,” in Fields, Strings and Critical Phenomena, Les Houches Session XLIX, edited by E. Brézin and J. Zinn-Justin (Elsevier, 1989), pp. 1–168, for stress-tensor modes, central charge, and cylinder quantization.
• J. Polchinski, String Theory, Volume 1 (Cambridge University Press, 1998), Chapter 2, for the Virasoro algebra and the cylinder interpretation of the central charge in worldsheet theory.
• A. M. Polyakov, Gauge Fields and Strings (Harwood Academic Publishers, 1987), Chapter 9, for the stress tensor and central charge in the random-surface and string-theory setting.