4

Polchinski Vol. 1 (Sec. 2.4): I'm trying to understand the Eq. 2.4.26 where he shows how the stress tensor transforms under a conformal transformation ($z \rightarrow w$):

$$(\partial w)^2 T(w) = T(z) - \frac{c}{12} \{w,z\} \tag{2.4.26}$$

Firstly, what is the reason behind the appearance of a Schwarzian derivative $\{z,w\}$ and why is it sitting with the central charge term or the anomaly term $\frac{c}{12}$?

One observation is under a fractional linear transformation $$z \rightarrow \frac{az+b}{cz+d}$$ the Schwarzian vanishes, thereby making the transformation look "normal", but then why does this transformation play a special role in a CFT?

Jon Snow
  • 575
  • 1
    This seems like two or three questions in one. It is a policy of this site to have only one question per post. Kindly modify your question to reflect that. – Prahar Oct 02 '15 at 06:08
  • However, they are pretty related. I'd rather rewrite instead of asking multiple questions. Hope the edited version is coherent enough to meet the policy requirements. – Jon Snow Oct 02 '15 at 06:29

1 Answers1

8

The significance of the Möbius transformations $\mathrm{PSL}(2,\mathbb{C})$ in 2D conformal field theory is that they are the globally defined conformal transformations on the Riemann sphere.

While the infinitesimal conformal transformations form the infinite-dimensional Witt algebra spanned by the vector fields $$ L_n = -z^{n+1}\partial_z$$ we must be mindful that those vector fields are not globally defined on the Riemann sphere $S^2 = \mathbb{C}\cup\{\infty\}$. Obviously, they are singular at $z = 0$ for $n < -1$. Changing coordinates by $z\mapsto w = \frac{1}{z}$, we get $$ L_n = -w^{1-n}\partial_w$$ which is singular at $w=0$, i.e $z=\infty$, for $n > 1$.

Therefore, the only globally defined conformal generators are $L_{-1},L_0,L_1$. These three generate precisely the group of Möbius transformations $z\mapsto \frac{az+b}{cz+d}$.

Thus, the symmetry group of a conformal field theory on the Riemann sphere is just $\mathrm{PSL}(2,\mathbb{C})$, and we have the requirement that the stress-energy tensor also should be invariant under this symmetry group. No such requirement can be said for the infinitesimal transformation of the Witt algebra. Nevertheless, classically, the stress-energy tensor transforms with its usual conformal weight also under those, since there is no central charge.

In the course of quantization, we incur a central charge for the Witt algebra, turning it into the Virasoro algebra1. Since the energy-momentum tensor is $T(z) = \sum_n L_n z^{n-2}$, the appearance of the central charge means the classical transformation law under the infinitesimal transformations generated by the $L_n$ may change by a quantum correction - this is precisely the Schwarzian derivative term. In the classical case $c = 0$, it vanishes, as a quantum correction (in this case often interpreted as a normal ordering constant) should.

However, if this also changed its behaviour under the global transformations, then the quantum theory would become anomalous, in particular, it would break the conservation of the Noether currents associated to $L_{-1},L_0,L_1$, which are $T(z),zT(z),z^2T(z)$. That is, anomalous transformation under a $\mathrm{PSL}(2,\mathbb{C})$ transformation would break energy-momentum conservation. This is undesirable, and, in fact, does not happen (as you may convince yourself by just chugging through the calculation of the transformation behaviour of $T$).

Now, why does the Schwarzian derivative appear as the quantum correction? If you start from the requirement that the quantum correction must vanish for $c=0$ and for $\mathrm{PSL}(2,\mathbb{C})$ transformations, then it is clear that it must be proportional to $c$. Furthermore, whatever $\{z,w\}$ is, it has to respect the group composition law that two successive transformations $z\mapsto w \mapsto u$ give the same as mapping $z\mapsto u$ directly. This is equivalent to the equation $$ \{u,z\} = \{w,z\} + \left(\frac{\mathrm{d}w}{\mathrm{d}z}\right)^2\{u,w\} \tag{1}$$ since $$ T(u) = \left(\frac{\mathrm{d}w}{\mathrm{d}u}\right)^2 \left(T(w) + \{u,w\}\right)$$ and $$ T(w) = \left(\frac{\mathrm{d}z}{\mathrm{d}w}\right)^2 \left(T(z) + \{w,z\}\right)$$ but also $$ T(u) = \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \left(T(z) + \{u,z\}\right)$$ so we obtain $$ \left(\frac{\mathrm{d}w}{\mathrm{d}u}\right)^2 \left(\left(\frac{\mathrm{d}z}{\mathrm{d}w}\right)^2 \left(T(z) + \{w,z\}\right) + \{u,w\}\right) = \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \left(T(z) + \{u,z\}\right)$$ which gives $$ \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \{w,z\} + \left(\frac{\mathrm{d}w}{\mathrm{d}u}\right)^2\{u,w\} = \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 \{u,z\} $$ after subtracting $\left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 T(z)$ from both sides. Multiplying by $\left(\frac{\mathrm{d}u}{\mathrm{d}z}\right)^2$ now yields eq. (1).

It can be shown that $(1)$ together with the requirement of $\mathrm{PSL}(2,\mathbb{C})$ invariance define the Schwarzian derivative uniquely.


1Shameless self-promotion: See this Q&A of mine for why we get a central charge in the quantum theory.

ACuriousMind
  • 124,833
  • Could you give a hint on how to prove PSL2 invariant + chain rule determine the Schwarzian derivative uniquely? – Yingfei Gu May 05 '17 at 19:56
  • 1
    @YingfeiGu It's a highly nontrivial statement, related to the fact that eq. (1) essentially determines that the Schwarzian derivative is a "1-cocycle with values in the quadratic differentials", and that Möbius-invariant cocycles are highly restricted. I can't quite recall where I heard this, but it seems to be that e.g this paper by Bouarroudj and Ovisenko contains the proof, if read correctly. – ACuriousMind May 06 '17 at 01:09
  • 1
    Thanks. A related question, consider more generally we take cocycle values in the tensor density of degree $n$, and it seems although the 1-cohomology is nontrivial for $n=0,1,2$, however, only in the $n=2$ "quadratic differentials'' we can find a cocycle that is mobius invariant. I am curious on how the "2'' is related to the mobius group. My motivation for this question is: in some detail calculation, e.g., expand cross ratio w.r.t. infinitesimal reparametrization of coordinates, the Schwarzian derivation appears always in second order. I am a bit curious if these facts are related. – Yingfei Gu May 08 '17 at 06:27
  • @YingfeiGu : In case it is still of interest, you can find a path integral derivation of the Schwarzian derivative here: https://physics.stackexchange.com/a/483863/83405. – Wakabaloola Jun 02 '19 at 14:42
  • @ACuriousMind: in case it is still of interest, https://physics.stackexchange.com/a/483863/83405. – Wakabaloola Jun 02 '19 at 14:44
  • 1
    I am not getting how to arrive at expression (1) of your answer. Because from the expression (2.4.26) given in the question we get $(\frac{\partial u}{\partial w})^2 T(u) = T(w) - \frac{c}{12} {u,w} = (\frac{\partial z}{\partial w})^2 (T(z) -\frac{c}{12} {w,z}) -\frac{c}{12} {u,w}$, which implies $T(u)= (\frac{\partial z}{\partial u})^2 (T(z) -\frac{c}{12} {w,z}) - \frac{c}{12} {u,w}$. On the other hand, $T(u)=(\frac{\partial z}{\partial u})^2 (T(z) -\frac{c}{12} {u,z})$. Equating both expressions fot $T(u)$ gives $ {u,z}= {w,z} + (\frac{\partial u}{\partial z})^2 {u,w}$. – Slayer147 Oct 04 '19 at 22:38
  • 1
    @Slayer147 Eq. (1) is correct, but all my derivatives after that were the wrong way around. I've corrected the answer. – ACuriousMind Oct 05 '19 at 09:27