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I'm reading some notes on CFT, and there's a strange topic that I find quite confusing. We define the Witt algebra to be the generators of conformal transformations on the complex plane.

$l_n = -z^{n+1}\partial_z$

$[l_n,l_m] = (m-n)l_{m+n}$

But then we perform something called the central extension which changes our algebra to the Virasoro algebra.

I'm okay with that. But then we go on to define the generators of conformal symmetry as contour integrals of the stress-energy tensor, and these satisfy not the Witt algebra but the Virasoro algebra. Is there a good reason for this?

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  • $\begingroup$ which notes are you reading?, I've never came across Witt algebra in my CFT readings $\endgroup$
    – CGH
    May 20, 2016 at 19:16
  • $\begingroup$ Actually, it's Blumenhagen's text, Intro to CFT with applications to string theory. They're also mentioned in Ginsparg's notes, though I don't know if he calls it the Witt algebra. $\endgroup$
    – Aurey
    May 20, 2016 at 19:18

1 Answer 1

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Basically the reason is that the classical conformal symmetry no longer holds at the quantum level due to the presence of the trace anomaly. More precisely, the tracelessness of the quantum stress-energy tensor is incompatible with the normal ordering needed to define it. By cohomological reasons, the trace of the stress-energy tensor, although non-vanishing, must be a central element of the "quantized Witt algebra", i.e. it must commute with all its generators. This implies that the "quantized Witt algebra" must be a non-trivial central extension of the Witt algebra, which in this particular case must be unique up to isomorphism (thanks to user106422 for recalling this point below) - that is, a Virasoro algebra.

An exposition I particularly like on such matters is the little book of Martin Schottenloher, "A Mathematical Introduction to Conformal Field Theory" (2nd. edition), Lecture Notes in Physics 759 (Springer--Verlag, 2008).

Edit (June 15th 2022): this comes more than six years late, but only now have I managed to get time to think about Peter Kravchuk's two (correct) points in his comments. I apologize to all.

  1. Indeed the trace anomaly only appears in curved space-time. More precisely, if your reference state has a short-distance behavior similar to that of the Minkowski vacuum state (in the case of Lagrangian free fields, this reduces to the Hadamard property for the two-point function in that state), the expectation value of the trace of the normal-ordered stress-energy tensor (w.r.t. the aforementioned reference state by e.g. point splitting and exploiting the operator product expansion) is state-independent and proportional to both the central charge and the scalar curvature (even if e.g. in the case of Lagrangian free fields you conformally couple the field to the curvature so as to keep the conformal symmetry of the model at the classical level). In other words, the trace anomaly is always there; we just do not see it where the scalar curvature vanishes. The role of the latter is to introduce a scale into the model through the coupling of the field dynamics with the background metric. Curvature, however, is not the only way to do this; another one is to impose boundary conditions on the conformal field theory - the Casimir energy is then also seen to be proportional to the central charge. As a rule, the central charge manifests itself physically as a way the conformal field model reacts to the introduction of an external macroscopic scale by means of a partial (anomalous, not spontaneous) breakdown of that symmetry. A more detailed discussion on this may be found e.g. in pp. 138-146 of the book by P. Di Francesco, P. Mathieu and D. Sénéchal, Conformal Field Theory (Springer-Verlag, 1997).
  2. In four dimensions the conformal Lie algebra is (semi)simple and therefore has no nontrivial central extensions, but both the trace anomaly in the presence of curvature and the Casimir energy in the presence of boundary conditions persist, so what changes? Well, recall that in four dimensions conformal transformations (in Lorentz signature) are composites of Poincaré transformations, (rigid) scale transformations and the so-called special conformal transformations, whereas the group of conformal transformations in two dimensions is much larger than that (in fact, infinite dimensional!). In Euclidean signature, (one of the two chiral components of) the conformal subgroup generated in the same way as in higher dimensions consists of Möbius transformations on the complex plane and therefore equals the semisimple Lie group $SL(2,\mathbb{R})$. The corresponding "rigid" (Möbius) part of the Witt Lie algebra (generated by $l_0$, $l_1$ and $l_{-1}$), which generates all Möbius transformations, sees no change in the commutation relations among their generators when we perform the central extension to obtain the Virasoro algebra, since the central part of $[l_n,l_m]$ is then given by $\frac{c}{12}(n^3-n)\delta_{m+n,0}$, where $c$ is the central charge. In other words, the Möbius Lie subalgebra does not "see" $c$, just like in higher dimensions. Nonetheless, the appearance of the scalar curvature in the trace anomaly formula shows that this anomaly really only affects "local" (i.e. non-"rigid") conformal transformations. In the case of boundary conditions, they affect the field algebra itself (e.g. in the case of free fields the causal commutator Green function must respect the boundary conditions) and hence also the normal ordering of the stress-energy tensor through shifting by central elements of the field algebra. Needless to say, usually global (resp. "rigid" = Möbius) conformal transformations (resp. in two dimensions) do not survive either the introduction of curvature or boundary conditions, so the role of the ("rigid") conformal Lie (sub)algebra in the present matter becomes moot since they no longer generate symmetries of the field theory anyway.
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  • $\begingroup$ Just to add to the answer you wrote, the central extension for the Witt algebra is unique, although it doesn't have to be necessarily unique for other Lie algebras. $\endgroup$
    – user106422
    May 20, 2016 at 20:29
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    $\begingroup$ I am a little confused by your answer. Is there a direct relationship between the trace anomaly (which is non-zero only in curved backgrounds and I am not sure that it is related to normal-ordering as directly as you seem to imply) and the central extension of the algebra? In 4D CFTs the conformal anomaly is still present, while the conformal algebra is simple and free of central extensions. On the other hand the central extensions arise due to physical possibility of projective representations. (continued) $\endgroup$ May 21, 2016 at 6:13
  • $\begingroup$ (continued) The shortest relation between the $c$ in Virasoro algebra and $c$ in trace anomaly I know is that Virasoro determines correlation functions of $T$ and these in turn (not quite trivially) determine the trace anomaly in curved background. Is there a more direct relationship? $\endgroup$ May 21, 2016 at 6:15
  • $\begingroup$ Hmm... I'm not recalling right now the details of the relation and I'm currently away from my office. I'll answer your point when I get there. $\endgroup$ May 21, 2016 at 13:02
  • $\begingroup$ Dear @PedroLauridsenRibeiro please do answer the questions as I would be grateful for your enlightenment on the topic. :)))))) $\endgroup$
    – Yalom
    Aug 9, 2020 at 15:31

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