I have just started reading through Srednicki's QFT in preparation for a few courses I am about to take. I have taken courses that have covered the basics of special relativity including contravariance and covariance, basic tensors, etc but not in enormous depth. For those who have a copy of the text, my question is about how he obtained equation 2.13 on page 32. The purpose of this discussion in the text seems to be constructing the lie algebra commutators of the Lorentz group. However I am stuck with some working in the middle.
Srednicki defines a Lorentz transformation as a linear change of coordinates that satisfies $$\ g_{\alpha\beta} \Lambda_\gamma^\alpha \Lambda_\sigma^\beta = g_{\gamma \sigma}\tag{2.6} $$
and proceeds to write an infinitesimal Lorentz transformation in the form: $$\ \Lambda_\beta^\alpha = \delta_\beta^\alpha + \delta w_\beta^\alpha \tag{2.7}$$
Then, if $\ U $ is a unitary operator representing a quantum symmetry where: $\ U(x x')= U(x)U(x') $ he writes:
$$\ U(1+\delta w) = I+({i/2h})\delta w_{\alpha\beta}M^{\alpha\beta}\tag{2.12} $$
Where $\ M^{\alpha\beta}$ is a generator of the Lorentz group.
Then he takes $$\ U(\Lambda)^{-1} U(\Lambda')U(\Lambda)=U(\Lambda^{-1}\Lambda'\Lambda) ,$$ sets $$\Lambda'=1+\delta w' $$ and expands both sides in linear order in $\ \delta w$ to obtain:
$$\ \delta w_{\alpha\beta}U(\Lambda)^{-1}M^{\alpha\beta}U(\Lambda) = \delta w_{\alpha\beta} \Lambda_\gamma^\alpha \Lambda_\sigma^\beta M^{\gamma\sigma} \tag{2.13}$$
My question is how this last expression was obtained? How did he expand to linear terms in $\ \delta w$?
Apologies if that explanation is difficult to follow. To be honest, I was able to follow it only by blindly accepting a few of his statements. I am hoping that for people familiar with the content what I have written is enough to follow Srednicki's discussion, as he didn't give much more information.
He moved faster than I expected in this chapter and I was struggling to keep up. I think I am lacking background knowledge he was whizzing over.
