Conserved charge operator CFT In David Simmons-Duffin's TASI lectures on conformal bootstrap, there is a discussion about conserved charges and operators, which is as follows:

Given $\epsilon = \epsilon^{\mu}(x) \partial_{\mu}$, we can define an operator
$$ Q_{\epsilon}(\Sigma) = - \int_{\Sigma}dS_{\mu}\epsilon_{\nu}(x)T^{\mu\nu}(x)$$
which is conserved if
$$\frac{1}{2}\left(\partial_{\mu}\epsilon_{\nu}+\partial_{\nu}\epsilon_{\mu}\right)T^{\mu\nu} = 0.$$

I am trying to prove that, given the charges $Q_{p_{\mu}}$ and $Q_{m_{\mu\nu}}$ are conserved charges, where
$$p_{\mu} = \partial_{\mu}$$
$$m_{\mu\nu} = x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}$$
For the first operator, I showed that
$$\partial_{\mu} Q_{p_{\nu}} = 0 \Leftrightarrow \partial_{\mu}\partial_{\nu}T^{\mu\nu} = 0$$
which is clearly true, since (exchanging derivatives) $\partial_{\mu}T^{\mu\nu} = 0$.
Regarding the second operator: given that it has two open indexes, how should I contract them in order to obtain the charge expression?
I tried something of the sort
$$Q_{m_{\mu\nu}} = - \int_{\Sigma}dS_{\mu}^{\alpha}\epsilon_{\alpha\nu}(x) T^{\mu\nu}(x)$$
but I am not quite sure this is accurate/correct at all. What am I doing wrong?
 A: The fact that you are writing the conservation law as $\partial_\mu Q_{p_\nu} 
= 0$ makes me question whether you did things correctly. If $\Sigma$ were a flat codimension 1 surface (as it is in more introductory sources), the conservation law would be $n^\mu \partial_\mu Q_{p_\nu}$ where $n$ is normal to $\Sigma$. To make things as simple as possible, we can let $\Sigma$ be a constant time slice. In this special case, the momentum is annihilated by $\partial_0$ as it should be.
Conservation in this context means the ability to smoothly deform $\Sigma$. This could be a rigid motion of $\Sigma$ to a new value of time or a more complicated deformation. To show that this follows from $\partial_\mu (\epsilon_\nu T^{\mu\nu}) = 0$, consider the diagram

where the red line is $\Sigma$, the blue line is a deformation of it $\Sigma^\prime$ and the black lines are at infinity. Assuming our currents vanish at infinity, we can take
$$
Q(\Sigma) - Q(\Sigma^\prime) = \int_{\Sigma^\prime - \Sigma} dS_\mu \epsilon_\nu T^{\mu\nu}(x)
$$
and add the black lines to the domain of integration for free. The above then becomes the surface integral of something over a closed surface.
\begin{align}
Q(\Sigma) - Q(\Sigma^\prime) &= \oint dS_\mu \epsilon_\nu T^{\mu\nu} \\
&= \int dx \partial_\mu ( \epsilon_\nu T^{\mu\nu} ) \\
&= 0
\end{align}
by Stokes' theorem. This works for both translations and rotations since they solve the Killing equation. For translations, $\epsilon_\mu = a_\mu$ where $a$ is a constant vector. This means all the indices in the definition of $Q(\Sigma)$ are contracted. You can set all but one component of $a$ to be zero if you want to talk about momentum in a specific direction. Similarly for rotations, $\epsilon_\mu = \omega_{\mu\nu} x^\nu$ where $\omega$ is anti-symmetric. Again, everything in $Q(\Sigma)$ is contracted and you can set components of $\omega$ to be zero to focus on angular momentum in a specific plane.
