EDIT: This question has been edited thanks to a comment. One of my definitions was wrong, so I have rewritten the whole question.

I was reading this paper about $T \bar{T}$ deformations of $2d$-QFTs in the plane. All is fine until the beginning of section 3. There they start talking about the limit of a large number of degrees of freedom. This means large central charge $c$ in a CFT and something similar for general QFTs. They say something like:

In the large $c$ limit correlation functions of $T_{ij}$ factorize ($T_{ij}$ is the energy-momentum tensor in Euclidean coordinates).

The connected contribution to an $n$-point function of energy-momentum tensors is proportional to $c$ at large $c$, so that when we compute a general correlation function and look at the contribution to it which is a product of $k$ connected components, then this will scale as $c^k$.

I will assume connected correlation functions are defined in a similar way as connected parts of the $S$-matrix in Weinberg's QFT I. For instance, for a 6 point function of the energy-mometum tensor:

$$ \langle T_1 T_2 T_3 T_4 T_5 T_6 \rangle = \langle T_1 T_2 T_3 T_4 T_5 T_6 \rangle_c \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad + \langle T_1 T_2 T_3 \rangle_c \langle T_4 T_5 T_6 \rangle_c + \text{permutations} \\ \qquad \qquad \qquad \qquad \qquad \qquad + \langle T_1 T_2 \rangle_c \langle T_3 T_4 \rangle_c \langle T_5 T_6 \rangle_c + \text{permutations} $$

There is no term like $\langle T_1\rangle_c \langle T_2 \rangle_c \langle T_3 \rangle_c \langle T_4 \rangle_c \langle T_5 \rangle_c \langle T_6 \rangle_c$ because $ \langle T_{ij} \rangle_c = \langle T_{ij} \rangle = 0$ ( in the plane you can set this to zero). Note that this implies $\langle T_{ij} T_{kl} \rangle_c = \langle T_{ij} T_{kl} \rangle$.

QUESTION: Why do connected $n$-point functions scale as $c$ at large $c$? I mean, why is the following true for any $n$?

$$ \langle T_{i_1 j_1} T_{i_2 j_2} ... T_{i_n j_n} \rangle_c \underset{c \rightarrow \infty}{\propto} c .$$

The only thing that comes to my mind that relates the energy-momentum tensor correlation functions and $c$ is the OPE

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

This works inside correlation functions, and I don't know how would it act inside connected correlation functions. Moreover, if this was the correct approach, it would give a factor of $c$ for every pair of $T$s inside the correlator. For instance

$$ \langle T(z_1) T(z_2) T(z_3) T(z_4) \rangle \sim \frac{1/4}{(z_1-z_2)^4(z_3-z_4)^4} c^2. $$

  • $\begingroup$ I think you have already given the answers? take a $2n$-point correlator of $T$. Then the formulas you wrote down say $\langle T_1 \cdots T_{2n} \rangle \sim \langle T_1 T_2 \rangle \langle T_3 T_4 \rangle \cdots \langle T_{2n-1} T_{2n} \rangle$. $\endgroup$ Jul 18, 2019 at 15:27
  • $\begingroup$ @LorenzMayer but then the correlator $\langle T_1 ... T_{2n} \rangle$ would scale as $c^n$, not as $c$ (which is what they say). $\endgroup$
    – MBolin
    Jul 18, 2019 at 22:55
  • $\begingroup$ where do they say that? on page 7 at the bottom they say that connected $n$-point functions scale like $c^{n-1}$. i mean this doesn't help with your problems 2) and 3), but 1) should be clarified by this? $\endgroup$ Jul 21, 2019 at 16:01
  • $\begingroup$ I think they say $n$-point functions scale as $t^{n-2} c^{n-1}$. But the large $c$-limit in this paper is a 't Hooft-like limit, keeping $tc$ constant. So actually $t^{n-2} c^{n-1} \propto c$, as I say. They say this explicitly at the bottom of page 5. $\endgroup$
    – MBolin
    Jul 21, 2019 at 16:16
  • 1
    $\begingroup$ Then maybe i misunderstood what you were asking for in question 1). Sorry about that. I thought you were talking about correlators, not connected correlators. If you are unsure what is the definition of a connected correlator and why $n$-point functions should be decomposable into $k$-connected components: in Weinbergs QFT book, section 4.3, it is shown how to decompose the S-matrix (which is essentially the same) into its connected parts. Maybe this helps. I haven't checked, but the algorithm he gives there together with the OPE should give the correct scaling. $\endgroup$ Jul 21, 2019 at 18:56

1 Answer 1


I think i can give a hint for the connected $4$-point function which maybe helps to understand why it should be true for the general case. For this pick four point $z_1,z_2,z_3,z_4$, let $z_{ij} = z_i - z_j$ and $T_i = T(z_i)$. Then the four point function, in the limit $z_1 \rightarrow z_2$ and $z_3 \rightarrow z_4$:

$$ \langle T_1 T_2 T_3 T_4 \rangle \sim \langle ( \langle T_1 T_2 \rangle + 2 z_{12}^{-2} T_2)(\langle T_3 T_4 \rangle + 2z_{34}^{-2}T_4) \rangle + \text{less singular} = \\ = \langle T_1 T_2 \rangle \langle T_3 T_4 \rangle + 4 z_{12}^{-2} z_{34}^{-2} \langle T_2 T_4 \rangle + \text{less singular} \ .$$

It follows for the connected correlation function

$$ \langle T_1 T_2 T_3 T_4 \rangle_c \sim 4 z_{12}^{-2} z_{34}^{-2} \langle T_2 T_4 \rangle + \text{less singular} $$

which is of order $c$.

As the $n$-th connected correlation function is defined by induction, this might be the easiest way to prove it for general connected correlation functions.

Here is also a reference for these operator algebra calculations, in particular chapter 6.

  • $\begingroup$ That could explain it. I don't see however why can you treat each $T_{ij}$ (in Euclidean coordinates) as a $T_{zz}$ (in holomorphic coordinates). I mean you are using the OPE $T_{zz} T_{zz}$, but you could equally take $T_{zz} T_{\bar{z} \bar{z}} \sim 0$, and you would get no $c$ at all. $\endgroup$
    – MBolin
    Jul 22, 2019 at 15:47

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