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I've always been confused by this very VERY basic and important fact about two-dimensional CFTs. I hope I can get a satisfactory explanation here. In a classical CFT, the generators of the conformal transformation satisfy the Witt algebra $$[ \ell_m, \ell_n ] = (m-n)\ell_{m+n}.$$ In the quantum theory, the same generators satisfy a different algebra $$[ \hat{\ell}_m, \hat{\ell}_n ] = (m-n) \hat{\ell}_{m+n} + \frac{\hbar c}{12} (m^3-m)\delta_{n+m,0}.$$ Why is this?

How come we don't see similar things for other algebras? For example, why isn't the Poincare algebra modified when going from a classical to quantum theory?

Please, try to be as descriptive as possible in your answer.

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    $\begingroup$ Poincare algebra doesn't have a central extension. This i think is proven in Weinberg Vol 1. Those symmetry algebras which admit nontrivial central extension are prone to develop anomaly upon quantization. I am not sure why quantization results into only central extension type deformation of the algebra but i think the reason is that definition of quantum operators involves normal ordering (or more generally renormalization) which makes the definition ambiguous by some constant term e.g. in case of Virasoro algebra quantum definition of L0 is ambiguous by addition of a constant. $\endgroup$
    – user10001
    Commented Sep 6, 2013 at 1:46
  • $\begingroup$ I think I partially agree with you. How is the normal ordering ambiguity somehow equivalent to anomalous breaking of symmetry? $\endgroup$
    – Prahar
    Commented Sep 6, 2013 at 1:56

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The central charge term as an example of a quantum anomaly; a symmetry that is modified in the quantized version of a classical theory. The central charge is, in fact, often referred to as the conformal anomaly. As di-Francesco et. al. put it at the start of section 5.4.2:

The appearance of the central charge $c$, also known as the conformal anomaly, is related to the "soft" breaking of conformal symmetry by the introduction of a macroscopic scale into the system.

They then go on to show that if, for example, you consider a generic conformal field theory on $\mathbb C$, and if you map the theory onto a cylinder of circumference $L$ with coordinate $w$, then \begin{align} \langle T_\mathrm{cylinder}(w)\rangle = -\frac{c\pi^2}{6L^2} \end{align}

They also, in appendix $5A$, go on to show that when a conformal field theory is defined on a curved two-manifold, then the central charge is related to the so-called trace anomaly; \begin{align} \langle T^\mu_{\phantom\mu\mu}(x)\rangle = \frac{c}{24\pi} R(x) \end{align} where $R$ is the Ricci scalar. The central charge can be seen to arise naturally in radial quantization in the operator formalism of CFT: see di-Francesco et. al chapter 6.

Anomalies arizing from quantization aren't restricted to conformal symmetry. See, for example, the chiral anomaly or the gauge anomaly.

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  • $\begingroup$ While on the plane, there is no length-scale, does it then mean that there is no central charge term when the theory is quantized on the plane? $\endgroup$
    – Prahar
    Commented Sep 5, 2013 at 20:53
  • $\begingroup$ @Prahar Short answer: I don't know. Long answer: my understanding is that radial quantization is a kind of mathematical trick for defining CFT on the plane. I've never actually seen any other method of quantization of a 2d CFT in detail, so I'm not sure how you would quantize on $\mathbb C$ otherwise. Perhaps one of us should ask another physics.SE question about if/why radial quantization is necessary, whether there are means of quantization on the plane that don't need something like radial quantization, and what the central charge is in that case... $\endgroup$
    – joshphysics
    Commented Sep 5, 2013 at 21:23
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    $\begingroup$ @Prahar I actually just found this physics.SE post that was illuminating in this regard: physics.stackexchange.com/questions/57970/… $\endgroup$
    – joshphysics
    Commented Sep 5, 2013 at 23:43
  • $\begingroup$ Here's a more relevant physics.SE post I think - physics.stackexchange.com/questions/33195/… $\endgroup$
    – Prahar
    Commented Sep 6, 2013 at 2:00
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    $\begingroup$ If I map conformally the plane into the sphere of radius $a$, the Ricci scalar go from $R=0$ to $R=2/a^2$. Now, if $c\neq 0$ this means that the trace of the energy-momentum tensor is non-zero after the comformal transformation? $\endgroup$
    – Nogueira
    Commented Oct 30, 2016 at 6:29
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I am not sure if this is the way you want to think about it, but I think it is worth pointing out that not having the central charge leads to a trivial quantum theory. The precise statement would be that a positive/unitary theory with c=0 has only one state, the vacuum. The details are demonstrated in

J.F. Gomes. The triviality of representations of the Virasoro algebra with vanishing central element and L0 positive. Phys. Lett. B 171, 75-76, 1986. http://www.sciencedirect.com/science/article/pii/0370269386910014#

Basically, you do the usual tricks. Create some descendant whose norm you can make negative unless the primary has h=0. So you are left with the m=2, c=0, h=0 minimal model, the trivial representation.

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  • $\begingroup$ +1: Very cool. I was completely unaware of this paper, and the proof is so awesomely short! $\endgroup$
    – joshphysics
    Commented Sep 5, 2013 at 21:33

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