Lie algebra of lorentz group I'm stuck in following calcualtion from sredniki's QFT book.(Its actually in the solution manual)
How can i get from
$$\delta\omega_{\rho\sigma}(g^{\sigma\mu}M^{\rho\nu} - g^{\rho\nu}M^{\mu\sigma})
$$
to
$$
\frac{1}{2}\delta\omega_{\rho\sigma}(g^{\sigma\mu}M^{\rho\nu} - g^{\rho\nu}M^{\mu\sigma} -g^{\rho\mu}M^{\sigma\nu} + g^{\sigma\nu}M^{\mu\rho})
$$
 A: Since $\delta \omega_{\rho\sigma}$ is antisymmetric, the part of what's inside the parentheses symmetrized over $\rho$ and $\sigma$ will vanish because if you contract antisymmetric indices with symmetric ones, you get zero.  So you can anti-symmetrize over that pair of indices inside the parentheses without changing anything, which is precisely what's inside the parentheses in the second line:
\begin{align}
\delta \omega_{\rho \sigma} (g^{\sigma \mu} M^{\rho \nu} - g^{\rho \mu} M^{\sigma \nu})
&=
\delta \omega_{\rho \sigma} \bigg[ \frac{1}{2} (g^{\sigma\mu}M^{\rho\nu} - g^{\rho\nu}M^{\mu\sigma} -g^{\rho\mu}M^{\sigma\nu} + g^{\sigma\nu}M^{\mu\rho})
\\
& \qquad \qquad +\frac{1}{2} (g^{\sigma\mu}M^{\rho\nu} - g^{\rho\nu}M^{\mu\sigma} + g^{\rho\mu}M^{\sigma\nu} - g^{\sigma\nu}M^{\mu\rho}) \bigg]
\end{align}
Here, I've just split the $gM$ factor into parts antisymmetric and symmetric in $\rho \sigma$.  (You can check that the terms inside the brackets cancel directly to give the same thing as the left-hand side.)  It's an easy exercise to check that for a general symmetric tensor $T^{\rho\sigma}$, you have $\delta \omega_{\rho \sigma}T^{\rho\sigma}=0$.  Adding more indices to $T$ doesn't change that, so the second term inside the brackets just drops out, and you get your result.
