I found this statement in several paper, but I don't have a clear reasoning why. It's said that the correlation function of the energy-momentum tensor $T(z)$ like $<T(z)O_1O_2O_3>$ vanishes like $z^{-4}$ when $z$ goes to infinity. Could anyone explain why?
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In fact, it is true if $T$ is replaced by any other quasi-primary operator with the scaling being $|z|^{-2\Delta}$, $\Delta=h+\bar h$. Also, the same holds in higher dimensional CFT's. For $T$ in 2d you have $\Delta_T=2$. There are various ways to see this.
One way is that actually the Euclidean correlation functions can be defined on a sphere, which (with north pole removed) is conformal to the plane via the stereographic projection. This projection relates correlators on the sphere $\langle T(z)\ldots\rangle_{S^2}$ and on the plane $\langle T(z)\ldots\rangle_{R^2}$. Taking $z$ to infinity is equivalent to sending $T$ to the north pole of the sphere. On the sphere this is point is no special, and the correlator with $T$ at the north pole is regular and non-zero for a generic configuration of the remaining operators, $$ \langle T(\infty)\ldots\rangle_{S^2} \neq 0,\infty $$ When you write down the relation induced by the stereographic projection, you find something like $$ \left|\langle T(z)\ldots\rangle_{R^2}\right|\simeq |z|^{-2\Delta_T}\left|\langle T(z)\ldots\rangle_{S^2}\right|, $$ which explains the $z^{-4}$ damping. This formula follows from the standard formula for change of the correlator under a Weyl transformation (the correlators are assumed to be normalized as $\langle 1 \rangle_{S^2}=\langle 1 \rangle_{R^2}=1$, otherwize you will have to care about Weyl anomaly).
Another way is through the OPE. You move $T$ to infinity, while the remaining operators are somewhere at fixed positions. At some point you can draw a circle around all the other operators such that it will not contain $T$. It means that you can now use the OPE. In your example, you write $$ O_1(z_1,\bar z_1)O_2(z_2,\bar z_2)O_3(z_3,\bar z_3)=\sum_i (C_i(z_1,\bar z_1,z_2,\bar z_2,z_3,\bar z_3)O_i(z_1,\bar z_1)+\text{desc.}), $$ where the sum is over all quasi-primaries in the theory and "desc." denote the contribution from the $sl_2(\mathbb{C})$ descendants of $O_i$ (i.e. descendants involving $L_{-1}$ and $\bar L_{-1}$ only). Assume the basis of quasi-primaries is chosen to be diagonal, i.e. $\langle O_iO_j\rangle\propto \delta_{ij}$. Then taking the expectation value with $T$ we find $$ \langle T(z)O_1(z_1,\bar z_1)O_2(z_2,\bar z_2)O_3(z_3,\bar z_3)\rangle= \langle T(z) (C_T(z_1,\bar z_1,z_2,\bar z_2,z_3,\bar z_3)T(z_1)+\text{desc.})\rangle. $$ If the four-point function is non-zero at all, the so is $C_T$. Now, the two-point function $\left|\langle T(z) T(z_1) \rangle\right|\propto |z-z_1|^{-2\Delta_T}$. The descendant contributions fall off quicker than that since they are all proportional to the derivatives of $\langle T(z) T(z_1)\rangle$ over $z_1$.
Peter Kravchuk
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The stress energy tensor is a quasi-primary field of dimension 2 (when the central charge vanishes). This means that in a operator product expansion $$ T(z)\phi(w) = \frac{h}{(z-w)^2}\phi(w) + \frac{1}{z-w}\partial\phi(w) + \textrm{regular terms}\ldots $$ with $\phi(w)$ being a conformal field of scaling dimension $h$. This is due to contour integration and residue theorem applied to the left hand side of the above, when calculating all possible variations (that are subject to conformal invariance).
Taking the expectation values of both sides should give back the results. For product of more than two operators just apply the rule more than once, with $\phi(w)$ being the new $T(z=z_0)\Psi(w)$. Another answer along the same lines can be found here.
