Schwarzian Derivative and the Virasoro Algebra

The central pole in the stress-tensor OPE has a finite geometric consequence. In a two-dimensional CFT, the stress tensor is not quite a primary field. Classically it behaves like a field of holomorphic weight $2$, but quantum mechanically its transformation law contains an additive term controlled by the central charge $c$. That additive term is the Schwarzian derivative.

This is one of the most compact places where the whole structure of two-dimensional conformal field theory shows itself. The same coefficient $c$ appears in three equivalent ways:

$$T(z)T(w) \sim {c/2\over (z-w)^4}+{2T(w)\over (z-w)^2}+{\partial T(w)\over z-w},$$ $$T_z(z)=\left({df\over dz}\right)^2T_f(f(z))+{c\over 12}\{f,z\},$$

and

$$[L_n,L_m]=(n-m)L_{n+m}+{c\over 12}n(n^2-1)\delta_{n+m,0}.$$

The first is local short-distance data, the second is the finite coordinate-transformation law, and the third is the operator algebra of infinitesimal conformal transformations. The miracle is that they are not three separate facts; they are the same fact viewed through OPEs, geometry, and radial quantization.

Required background. Lesson 23 supplies the $T T$ OPE, the mode convention, and the central extension used below.

The stress tensor is almost primary

Passive finite maps and radial generators

We work in one holomorphic sector. The antiholomorphic sector has the same formulas with $z,T,L_n,c$ replaced by $\bar z,\bar T,\bar L_n,\bar c$. The mode convention is

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

An infinitesimal conformal transformation is generated by the holomorphic vector field $\epsilon(z)\partial_z$. We use the radial-quantization convention

$$\delta_\epsilon \mathcal O(w)={1\over 2\pi i}\oint_w dz\,\epsilon(z)T(z)\mathcal O(w),$$

so that a primary field of weight $h$ transforms as

$$\delta_\epsilon \mathcal O=\epsilon\partial\mathcal O+h(\partial\epsilon)\mathcal O.$$

For a finite change of coordinate $f=f(z)$, the pulled-back holomorphic stress tensor is written

$$T_z(z)=\left({df\over dz}\right)^2T_f(f(z))+{c\over 12}\{f,z\}.$$

Changing from this passive coordinate convention to an active field-variation convention reverses some signs in intermediate formulas, but not in the Virasoro algebra stated above.

A primary field of holomorphic weight $h$ has OPE

$$T(z)\mathcal O(w) \sim {h\mathcal O(w)\over (z-w)^2} +{\partial\mathcal O(w)\over z-w}.$$

Multiplying by $\epsilon(z)$ and taking the residue at $z=w$ gives

$$\delta_\epsilon\mathcal O(w)=\epsilon(w)\partial\mathcal O(w)+h\partial\epsilon(w)\mathcal O(w).$$

If the stress tensor were an ordinary primary of weight $2$, we would have only

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

But the stress tensor has the OPE with itself

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

The fourth-order pole is a c-number singularity proportional to the identity, rather than to $T$ or one of its descendants. It is the local shadow of the central extension. Applying the contour prescription gives

$$\delta_\epsilon T(w) ={1\over 2\pi i}\oint_w dz\,\epsilon(z)T(z)T(w).$$

The weight-$2$ terms are immediate. For the central term, use

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

Therefore

$$\boxed{ \delta_\epsilon T =\epsilon\partial T+2(\partial\epsilon)T+{c\over 12}\partial^3\epsilon. }$$

The stress tensor is therefore not a Virasoro primary when $c\ne0$. It is quasi-primary under the global Möbius subgroup, for which the inhomogeneous term vanishes, but a general local conformal transformation acts on it as a weight-$2$ field plus the third-derivative term.

The stress tensor would transform as a weight-$2$ primary without the fourth-order pole in $T(z)T(w)$. The central pole contributes the residue $c\partial^3\epsilon/12$, the infinitesimal form of the Schwarzian anomaly.

A useful immediate check is the global conformal subgroup. On the Riemann sphere, the globally defined holomorphic vector fields are

$$\epsilon(z)=a+bz+cz^2.$$

For these transformations,

$$\partial^3\epsilon=0.$$

Thus translations, dilatations/rotations, and special conformal transformations do not see the central term. The anomaly appears only for genuinely local conformal transformations.

The Schwarzian derivative

The finite version of the anomalous stress-tensor transformation is expressed through the Schwarzian derivative

$$\boxed{ \{f,z\} ={f'''(z)\over f'(z)}-{3\over2}\left({f''(z)\over f'(z)}\right)^2. }$$

For a finite holomorphic coordinate change $f=f(z)$, the stress tensor in the $z$ coordinate is

$$\boxed{ T_z(z)=\left(f'(z)\right)^2T_f(f(z))+{c\over12}\{f,z\}. }$$

This formula says that $T$ is a projective connection rather than an ordinary quadratic differential. The first term is the ordinary tensor transformation. The second term is an inhomogeneous quantum correction.

Under a finite holomorphic map $f=f(z)$, the pulled-back stress tensor has a tensor piece $(f')^2T_f$ and an additive Schwarzian piece $c\{f,z\}/12$. This is the finite counterpart of the infinitesimal term $c\partial^3\epsilon/12$.

To check consistency with the infinitesimal result, take

$$f(z)=z+\epsilon(z), \qquad |\epsilon|\ll 1.$$

Then

$$f'(z)=1+\epsilon'(z), \qquad (f')^2T(f)=T+\epsilon\partial T+2\epsilon'T+O(\epsilon^2),$$

and

$$\{z+\epsilon,z\}=\epsilon'''(z)+O(\epsilon^2).$$

Therefore

$$T_z(z)-T_f(z) =\epsilon\partial T+2(\partial\epsilon)T+{c\over12}\partial^3\epsilon+O(\epsilon^2),$$

which is exactly the infinitesimal Ward-identity result.

The Schwarzian has two structural properties that make it almost inevitable.

First, it vanishes for Möbius transformations,

$$f(z)={az+b\over cz+d}, \qquad ad-bc\ne0,$$

so the global conformal group acts without an anomaly:

$$\{f,z\}=0 \qquad \text{for } f\in PSL(2,\mathbb C).$$

Second, it obeys the chain rule

$$\boxed{ \{f\circ g,z\}=(g'(z))^2\{f,g(z)\}+\{g,z\}. }$$

This is exactly what is required for two successive coordinate changes to act associatively on $T$. Without the Schwarzian chain rule, the finite transformation law would fail to define a representation of the conformal transformations.

The central charge as a cocycle

The inhomogeneous term in $T$ is a cocycle. In less compact language, it measures the failure of local conformal transformations to be represented without a central extension after quantization.

A helpful analogy is a projective representation in quantum mechanics. A symmetry may act on states only up to a phase,

$$U(g_1)U(g_2)=e^{i\alpha(g_1,g_2)}U(g_1g_2),$$

where associativity imposes a cocycle condition on $\alpha$. The Virasoro central term is the infinitesimal version of the same idea. The Schwarzian derivative is the geometric finite version.

The Schwarzian obeys a composition law. This is why the anomalous transformation of $T$ is compatible with doing two conformal maps in sequence. The central charge $c$ multiplies this cocycle.

The fact that the coefficient is the same $c$ as in the $T T$ OPE is not an extra assumption. The OPE determines the infinitesimal transformation of $T$, and integrating that infinitesimal law gives the Schwarzian term. Conversely, expanding the finite Schwarzian law near the identity recovers the third-derivative term, hence the fourth-order pole.

Virasoro algebra from the TT OPE

Now derive the mode algebra directly from the OPE. Define

$$L_n={1\over2\pi i}\oint_0 dz\,z^{n+1}T(z).$$

The commutator is computed by radial ordering: one contour surrounds the other, and the difference is the contour that picks up the singularity of $T(z)T(w)$ at $z=w$.

The difference between the two radial orderings of $L_nL_m$ collapses to a small contour around the OPE singularity at $z=w$. The three singular terms in $T(z)T(w)$ produce the two pieces of the Virasoro algebra.

Using only the singular part of the OPE,

$$[L_n,L_m] ={1\over2\pi i}\oint_0 dw\,w^{m+1} \operatorname{Res}_{z=w}\left[z^{n+1}T(z)T(w)\right].$$

The double-pole term gives

$$\operatorname{Res}_{z=w}\left[z^{n+1}{2T(w)\over (z-w)^2}\right] =2(n+1)w^nT(w).$$

The simple-pole term gives

$$\operatorname{Res}_{z=w}\left[z^{n+1}{\partial T(w)\over z-w}\right] =w^{n+1}\partial T(w).$$

After multiplying by $w^{m+1}$ and integrating, the derivative term is integrated by parts:

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

Together these two operator terms give

$$\left[2(n+1)-(m+n+2)\right]L_{m+n}=(n-m)L_{m+n}.$$

The central pole gives

$$\operatorname{Res}_{z=w}\left[z^{n+1}{c/2\over (z-w)^4}\right] ={c\over2}{1\over3!}\partial_w^3w^{n+1} ={c\over12}n(n^2-1)w^{n-2}.$$

Multiplying by $w^{m+1}$ and taking the $w$ residue yields

$${c\over12}n(n^2-1)\delta_{n+m,0}.$$

Therefore

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

This is the Virasoro algebra. If $c=0$, it reduces to the Witt algebra of holomorphic vector fields

$$\ell_n=-z^{n+1}\partial_z, \qquad [\ell_n,\ell_m]=(n-m)\ell_{n+m}.$$

The central term is invisible classically and unavoidable quantum mechanically in most interesting CFTs.

Global conformal transformations and the missing central term

The modes

$$L_{-1},\quad L_0,\quad L_1$$

generate the globally defined holomorphic transformations on the sphere:

$$1,\quad z,\quad z^2.$$

The central term is proportional to

$$n(n^2-1),$$

so it vanishes for $n=-1,0,1$. Hence

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

This is the Lie algebra of the global conformal group. The anomaly begins only when the contour generator corresponds to a vector field with a pole or higher-order zero in a local coordinate patch. Equivalently, the Schwarzian derivative vanishes on Möbius maps but not on a general holomorphic map.

The modes $L_{-1},L_0,L_1$ form the global conformal subalgebra and have no central term. The remaining local modes complete the infinite Virasoro algebra and carry the central extension.

This distinction is conceptually important. Global conformal invariance fixes two- and three-point functions and constrains four-point functions. Local conformal invariance, through the full Virasoro algebra, organizes an infinite tower of descendants and gives much stronger constraints. Minimal models owe their solvability to this enhancement.

A useful consistency check is the action of the global generators on the first descendant of a primary state. If $|h\rangle$ is primary, then

$$[L_1,L_{-1}]|h\rangle=L_1L_{-1}|h\rangle=2h|h\rangle.$$

The same result follows directly from the OPE. Differentiate

$$T(z)\phi(w) \sim {h\phi(w)\over(z-w)^2} +{\partial\phi(w)\over z-w}$$

with respect to $w$ and set $w=0$:

$$T(z)\,\partial\phi(0) \sim {2h\phi(0)\over z^3} +{(h+1)\partial\phi(0)\over z^2} +{\partial^2\phi(0)\over z}.$$

The coefficient of $z^{-3}$ is $L_1L_{-1}\phi(0)=2L_0\phi(0)$. This elementary check is the level-one version of the general contour derivation.

Plane-to-cylinder map and Casimir energy

The Schwarzian term has a famous concrete consequence. Map the complex plane to a cylinder by

$$z=e^w, \qquad w=\tau+i\sigma, \qquad \sigma\sim\sigma+2\pi.$$

Here $w$ is the cylinder coordinate and $z$ is the plane coordinate. The Schwarzian derivative is

$$\{e^w,w\} ={z'''\over z'}-{3\over2}\left({z''\over z'}\right)^2 =1-{3\over2}=-{1\over2}.$$

Therefore the cylinder stress tensor is

$$\boxed{ T_{\mathrm{cyl}}(w)=z^2T_{\mathrm{pl}}(z)-{c\over24}. }$$

Even if the plane vacuum has

$$\langle0|T_{\mathrm{pl}}(z)|0\rangle=0,$$

the cylinder vacuum has

$$\boxed{ \langle T_{\mathrm{cyl}}\rangle=-{c\over24}. }$$

The map $z=e^w$ turns radial quantization on the plane into time evolution on the cylinder. The Schwarzian derivative is $-1/2$, so the cylinder stress tensor is shifted by $-c/24$ in each holomorphic sector.

This is the conformal Casimir energy. In a nonchiral theory with $c=\bar c$ on a spatial circle of circumference $L$, the ground-state energy is

$$E_0=-{\pi c\over 6L}.$$

The finite-size spectrum is correspondingly

$$E={2\pi\over L}\left(h+\bar h-{c\over12}\right), \qquad P={2\pi\over L}(h-\bar h),$$

where $h$ and $\bar h$ are the eigenvalues of $L_0$ and $\bar L_0$; descendants are included by using their shifted eigenvalues. Thus the same central charge that appears in a local fourth-order pole also shifts the vacuum energy on a compact spatial circle. This is why $c$ is not just algebraic decoration; it is measurable in finite-size physics.

Reflection positivity and adjoints

The Euclidean path integral of a unitary Lorentzian theory obeys reflection positivity. In flat Euclidean coordinates, the reflection is $t\mapsto -t$. For an operator $A$ supported at positive Euclidean time, reflection positivity says schematically

$$\langle \Theta A\,A\rangle\ge0,$$

where $\Theta$ reflects the operator support and complex conjugates coefficients. For a scalar operator,

$$\Theta\phi(\mathbf x,t)\Theta^{-1}=\phi(\mathbf x,-t).$$

In radial quantization, Euclidean time is $\tau=\log|z|$. Reflection through the quantization surface $|z|=1$ sends

$$z\mapsto {1\over \bar z}.$$

For the holomorphic stress tensor this gives the adjoint relation

$$T(z)^\dagger=\bar z^{-4}T\left({1\over \bar z}\right), \qquad L_n^\dagger=L_{-n}.$$

Reflection positivity in radial quantization reflects the insertion point through the unit circle. This gives $L_n^\dagger=L_{-n}$ and turns Virasoro commutators into norm constraints.

A quick consequence is the positivity of the central charge in unitary theories. Since the vacuum is invariant under the global conformal group,

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

we get for $n\ge2$

$$\|L_{-n}|0\rangle\|^2 =\langle0|L_nL_{-n}|0\rangle =\langle0|[L_n,L_{-n}]|0\rangle ={c\over12}n(n^2-1).$$

Reflection positivity therefore requires

$$c\ge0.$$

This argument is only a necessary condition for unitarity, not a classification. But it is a useful sanity check: the central term is not merely a formal anomaly; it is visible as the norm of stress-tensor descendants.

Summary

The central charge can be read in several mutually reinforcing ways.

The local OPE statement is

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

The infinitesimal transformation statement is

$$\delta_\epsilon T=\epsilon\partial T+2(\partial\epsilon)T+{c\over12}\partial^3\epsilon.$$

The finite geometric statement is

$$T_z=(f')^2T_f+{c\over12}\{f,z\}.$$

The mode-algebra statement is

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

The physical finite-size statement is that the cylinder vacuum energy is shifted by the Schwarzian term. All these are one phenomenon: the quantum stress tensor represents local conformal transformations projectively, and $c$ measures the central extension.

Common pitfalls

The stress tensor is not a primary field under arbitrary local conformal maps. It is primary-like under the global Möbius group because the Schwarzian vanishes there, but under general holomorphic maps it has an additive term.

The sign of the Schwarzian term depends on whether one writes the transformation as a pullback from the $f$ coordinate to the $z$ coordinate or as an active variation of fields at fixed coordinate. The invariant content is the pairing of $c/2$ in the $T T$ OPE with $c/12$ in the infinitesimal third-derivative term.

The central term in the Virasoro algebra is not optional once the $T T$ OPE has a fourth-order pole. Removing it would destroy the Ward identities and the plane–cylinder Casimir shift.

The global conformal algebra is not the whole story in two dimensions. The modes $L_{-1},L_0,L_1$ form an $sl(2)$ subalgebra, but the local stress-tensor algebra contains all $L_n$ and is much more restrictive.

Exercises

Exercise 1: Infinitesimal stress-tensor anomaly

Use the OPE

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

to derive

$$\delta_\epsilon T=\epsilon\partial T+2(\partial\epsilon)T+{c\over12}\partial^3\epsilon.$$

Solution

By the Ward-identity contour prescription,

$$\delta_\epsilon T(w) ={1\over2\pi i}\oint_w dz\,\epsilon(z)T(z)T(w).$$

Only singular terms contribute. The simple pole gives

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

The double pole gives

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

The fourth-order pole gives

$${c\over2}{1\over2\pi i}\oint_w dz\,{\epsilon(z)\over (z-w)^4} ={c\over2}{1\over3!}\partial^3\epsilon(w) ={c\over12}\partial^3\epsilon(w).$$

Adding the three pieces gives the result.

Exercise 2: Möbius maps have vanishing Schwarzian

Show that the Schwarzian derivative vanishes for

$$f(z)={az+b\over cz+d}, \qquad ad-bc\ne0.$$

Solution

Let $D=ad-bc$. Then

$$f'(z)={D\over (cz+d)^2}, \qquad f''(z)=-{2cD\over (cz+d)^3}, \qquad f'''(z)={6c^2D\over (cz+d)^4}.$$

Hence

$${f'''\over f'}={6c^2\over (cz+d)^2}, \qquad \left({f''\over f'}\right)^2={4c^2\over (cz+d)^2}.$$

Therefore

$$\{f,z\} ={6c^2\over (cz+d)^2} -{3\over2}{4c^2\over (cz+d)^2}=0.$$

This is why the stress tensor transforms as an ordinary weight-$2$ field under global conformal transformations.

Exercise 3: Central term from the fourth-order pole

Starting from the $T T$ OPE and the mode definition

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

derive the central term in $[L_n,L_m]$.

Solution

The central term comes only from

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

In the contour computation,

$$[L_n,L_m]_{\mathrm{cent}} ={1\over2\pi i}\oint_0 dw\,w^{m+1} \operatorname{Res}_{z=w}\left(z^{n+1}{c/2\over (z-w)^4}\right).$$

The residue is

$${c\over2}{1\over3!}\partial_w^3w^{n+1} ={c\over12}n(n^2-1)w^{n-2}.$$

Thus

$$[L_n,L_m]_{\mathrm{cent}} ={c\over12}n(n^2-1){1\over2\pi i}\oint_0dw\,w^{m+n-1} ={c\over12}n(n^2-1)\delta_{m+n,0}.$$

Exercise 4: Plane-to-cylinder Casimir shift

Compute the Schwarzian derivative of $z=e^w$ and use it to find the vacuum expectation value of the holomorphic stress tensor on the cylinder, assuming $\langle T_{\mathrm{pl}}(z)\rangle=0$ on the plane.

Solution

For $z=e^w$,

$$z'=z, \qquad z''=z, \qquad z'''=z.$$

Therefore

$$\{e^w,w\} ={z'''\over z'}-{3\over2}\left({z''\over z'}\right)^2 =1-{3\over2}=-{1\over2}.$$

The finite transformation law gives

$$T_{\mathrm{cyl}}(w)=z^2T_{\mathrm{pl}}(z)+{c\over12}\{e^w,w\} =z^2T_{\mathrm{pl}}(z)-{c\over24}.$$

Taking the plane vacuum expectation value gives

$$\langle T_{\mathrm{cyl}}\rangle=-{c\over24}.$$

The antiholomorphic sector gives the analogous shift $-\bar c/24$.

Exercise 5: Positivity of the stress-tensor state

Use the Virasoro algebra to compute $[L_2,L_{-2}]$. Then, assuming the vacuum obeys $L_n|0\rangle=0$ for $n\ge -1$, compute the norm of $L_{-2}|0\rangle$.

Solution

The Virasoro algebra gives

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

The norm is

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

because $L_2|0\rangle=0$ and $\langle0|L_{-2}=0$ by conjugation. Since $L_0|0\rangle=0$,

$$\langle0|L_2L_{-2}|0\rangle={c\over2}.$$

In a unitary theory this norm is nonnegative, so this calculation gives the simple necessary condition $c\ge0$.

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.
• P. Di Francesco, P. Mathieu, and D. Sénéchal, Conformal Field Theory (Springer, 1997), Chapters 4–6.
• 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.
• P. G. O. Freund, Introduction to Supersymmetry (Cambridge University Press, 1986), appendix material on the Virasoro algebra.
• J. Polchinski, String Theory, Volume 1 (Cambridge University Press, 1998), Sections 2.4–2.6, for the plane–cylinder map, stress-tensor anomaly, and Virasoro generators in worldsheet CFT.