I think that general point of view about central charge in books is considering OPE $T(z) T(w)$ for different field theories and finding that general expression for the most singular term is about to be $\sim \frac{c/2}{(z-w)^4}$. Then we do operator expansion $$T(z) = \sum \frac{L_n}{z^{n+2}}$$ and see that operators $L_n$ form Virasoro algebra. Or we can write transformation law for $T(z)$ and see that generally it's not a tensor, and here central charge comes into play again.

My question: I wonder if there is any way to start from Virasoro algebra generators and to prove that energy-momentum tensor in arbitrary theory must contain the term $\frac{c/2}{(z-w)^4}$ in its OPE with itself, where $c$ is a central charge of Virasoro algebra?

Another way to show this is to prove that operator expansion of $T(z)$ is always $\sum \frac{L_n}{z^{n+2}}$, where $L_n$ are Virasoro generators.

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    $\begingroup$ I'm not exactly sure what the question here is. By definition of quantization, a central charge appears when we quantize the Witt algebra, cf. this Q&A of mine, and then the central charge has to appear in the expansion of $T(z) = L_n z^{-n-2}$ since it is in the commutation relation of $L_n$ of equal level, and the $-2$ already shows this charge will thus play its role at $(z-w)^{-4}$. Are you asking why the central charge appears, or why the e-m tensor is $\sum_n L_n z^{-n-2}$? $\endgroup$
    – ACuriousMind
    Commented Dec 24, 2015 at 14:12
  • $\begingroup$ Well, I think, both)) $\endgroup$
    – newt
    Commented Dec 24, 2015 at 15:13

2 Answers 2

  1. The central charge itself appears in the commutation relations because the quantum theory allows not only linear, but projective representations of the conformal Witt algebra to be physically admissible. Such projective representations are in bijection to linear representations of the central extensions of the Witt algebra $$[L_m,L_n] = (m-n)L_{m+n}$$ which are the Virasoro algebras $$ [L_m,L_n] = (m-n)L_{m+n} + \frac{c}{12}(m^3-m)\delta_{m,-n} $$ for $c\in\mathbb{R}$. For more on projective representations and central extensions, see this Q&A of mine.

  2. The energy-momentum tensor classically has conformal weight $2$, so its conformal mode expansion is $T(z) = \sum_n T_n z^{-n+2}$. The question now becomes how one shows that $T_n = L_n$. Since the energy-momentum tensor is the conserved current for the translations generated by $L_{-1} = \partial_z$ (since $L_n = z^{-n+1}\partial_z$), its integral has to be the generator $L_{-1}$ itself (the integral of the conserved current is the Noether charge, which is the generator of the symmetry in the Hamiltonian formulations): $$ L_{-1} = \frac{1}{2\pi\mathrm{i}}\int T(z)\mathrm{d}z$$ and inserting the mode expansion we get $L_{-1} = T_{-1}$. More generally, for any conformal transformation $z\mapsto z+\epsilon(z)$, we get a conserved current $\epsilon(z)T(z)$. The $L_n$ generate the transformations $z\mapsto z + \epsilon_n z^{n+1}$. As above, it follows that $$ L_n = \frac{1}{2\pi\mathrm{i}}\int z^{n+1}T(z)\mathrm{d}z$$ and thus $L_n = T_n$. Therefore, the energy-momentum tensor of a conformal field theory is $T(z) = \sum_z L_n z^{-n+2}$.

Here we only considered the holomorphic parts, but the arguments for the anti-holomorphic pieces are exacctly the same

  • $\begingroup$ How do you prove that, if $T(z)=\sum_n L_n z^{-n-2}$ then $T(y)T(z)=c/2(y-z)^{-4}+2T(z)(y-z)^{-2}+\partial T(z)(y-z)^{-1}+O(1)$? I guess the second and third term follow by $L_{-1}T(z)=\partial T(z)$ by def. of the $L_{-1}$ action and $L_0 T(z)=2T(z)$ by the conformal dimension of $T$. Furthermore, the triple pole can be excluded by symmetry reasons. But what about the first term? How do I see that $L_{-2}T=c/2$? $\endgroup$
    – Brightsun
    Commented Feb 11, 2019 at 20:48

$T(z)$ has dimension 2. Thus, the OPE $T(z)T(0)$ has dimension 4. This implies that the most singular piece in this OPE that can possibly appear is $z^{-4}$. The operator that multiplies this singularity has dimension $0$. In a unitary CFT, the only such operator is the identity operator. Thus, the coefficient of $z^{-4}$ can only be a $c$-number. Further since this is appearing in the $TT$ OPE it must be real and positive (again due to unitarity).

Thus the most singular piece must take the form $$ T(z)T(0) \sim \frac{c}{2z^4} + \cdots $$

This doesn't say that this term must exist, since we could easily have $c=0$. However, you can derive the Virasoro algebra with $c=0$ (also called the Witt algebra) and ask what kind of representations can occur in a unitary theory. It is easily shown that with $c=0$, only a identity module can occur in the quantum theory which makes the theory trivial.

Thus, in non-trivial unitary quantum field theories, you must have $c > 0$.


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