quantum field theory - Traceless of stress-energy tensor in $d=2$

This is a question regarding Francesco, section 4.3.3. In this section, he considers the two-point function

$$S_{\mu\nu\rho\sigma}(x) = \left< T_{\mu\nu}(x) T_{\rho\sigma}(0)\right>$$ He then goes on to claim that symmetry of the stress-energy tensor implies $$S_{\mu\nu\rho\sigma}(x) = S_{\nu\mu\rho\sigma}(x)~~~(1)$$ Though he doesn't mention this, I presume this is true only when $x \neq 0$ since the EM tensor is symmetric in a correlation as long as the other fields in the correlator are not evaluated at the same point.

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EDIT: Due to some comments, I'll explain why I think so. If a theory is Poincare invariant, it has conserved currents $T^{\mu\nu}$ for translations and

$$j^{\mu\nu\rho} = T^{\mu\nu} x^\rho - T^{\mu\rho} x^\nu$$ for Lorentz transformations. For completeness, we also note that if the theory has scale invariance the dilation current is $$j^\mu_D = T^{\mu\nu} x_\nu$$ In a classical theory, conservation of these currents implies symmetry and tracelessness of the stress-energy tensor. In a quantum theory, we have a Ward Identity, which for each of the currents reads $$\begin{equation} \begin{split} \partial_\mu \left< T^\mu{}_\nu X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \frac{\partial}{\partial x_i^\nu} \left< X \right> \\ \partial_\mu \left< j^{\mu\nu\rho} X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \left( x_i^\rho\frac{\partial}{\partial x_i^\nu} - x_i^\nu\frac{\partial}{\partial x_i^\rho} - i S_i^{\mu\nu} \right) \left< X \right> \\ \partial_\mu \left< j^\mu_D X \right> &= - \sum\limits_{i=1}^n \delta^d(x-x_i) \left( x_i^\alpha \frac{\partial}{\partial x_i^\alpha} + \Delta_i \right) \left< X \right> \end{split} \end{equation}$$ where $X = \Phi_1(x_1) \cdots \Phi_n(x_n)$, $S^{\mu\nu}_i$ is the representation of the Lorentz algebra under which $\Phi_i(x_i)$ transforms and $\Delta_i$ is the scaling dimension of $\Phi_i(x_i)$. Now plugging in the exact forms of the currents $j^{\mu\nu\rho}$ and $j^\mu_D$, we find $$\begin{equation} \begin{split} \partial_\mu \left< T^\mu{}_\nu X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \frac{\partial}{\partial x_i^\nu} \left< X \right> \\ \left< \left( T^{\mu\nu} - T^{\nu\mu} \right) X \right> &= i \sum\limits_{i=1}^n \delta^d(x-x_i) S_i^{\mu\nu} \left< X \right> \\ \left< T^\mu{}_\mu X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \Delta_i \left< X \right> \end{split} \end{equation}$$ Clearly, the EM tensor is not symmetric under correlation functions at the points $x = x_i$.

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Now, using these symmetry properties and certain other properties under parity, he argues that

$$S^\mu{}_\mu{}^\sigma{}_\sigma(x) = \left< T^\mu{}_\mu(x) T^\sigma{}_\sigma(0)\right> = 0$$ Following the above arguments, this should then only be true at $x \neq 0$. However, Francesco claims that this holds everywhere and therefore concludes that $\left< T^\mu{}_\mu(0)^2 \right> = 0$. How does this makes sense?