How do I derive this series for this unitary operator? I want to derive eq. (2.4.3) in S. Weinberg, The Quantum Theory of fields, Vol. 1. The derivations start from expanding inhomogenous Lorentz transforms near identity 
$$\Lambda^{\mu}_{\nu} ~=~ \delta^{\mu}_{\gamma}+\omega^{\mu}_{\nu}, \qquad a^{\mu}~=~\epsilon^{\mu}.$$ 
 $\Lambda^{\mu}_{\nu}= \delta^{\mu}_{\gamma}$ and $a^{\mu}=0$ at identity.
Then the Unitary operator is expanded as follows:
$$U(1+\omega,\epsilon) ~=~ 1 + \frac{1}{2} i \omega_{\rho \sigma} J^{\rho \sigma}- i \epsilon_{\rho}P^{\rho}\ldots $$
I was wondering how this equation was derived. I know that near the identity, the Unitary operator can be expanded as 
$$ U ~=~ 1+i \epsilon t.$$
Not able to see how to extend this to above equation. 
 A: Here is one possible derivation.
I) The unitary operator $U=U(\Lambda,a)$ depends on a Lorentz transformation $\Lambda$ and a translation $a$.
II) It is assumed that 
$$U(\Lambda={\bf 1},a=0)~=~{\bf 1}.$$
III) Define
$$ \tag{2.4.1}
 \Lambda^{\mu}{}_{\nu}~=~\delta^{\mu}_{\nu}+\omega^{\mu}{}_{\nu}. $$
IV) Lower the indices with the Minkowski metric $\eta_{\mu\nu}$,
$$ \omega_{\mu\nu}~=~\sum_{\lambda}\eta_{\mu\lambda} \omega^{\lambda}{}_{\nu} , \qquad a_{\mu}~=~ \sum_{\nu}\eta_{\mu\nu} a^{\nu}.  $$
V) Prove that 
$$\tag{2.4.2}\omega_{\mu\nu}~=~-\omega_{\nu\mu}$$
is an antisymmetric matrix if $\omega^{\mu}{}_{\nu}$ is infinitesimal. 
VI) Assume e.g. that the entries $\omega_{\mu\nu}$ above the diagonal $\mu<\nu$ are the independent d.o.f. of the $\omega$ matrix. (The entries $\omega_{\mu\nu}$ below the diagonal $\mu>\nu$ are then fully determined as the opposite values.)
VII) For $\mu<\nu$, define angular momentum
$$J^{\mu\nu}~=~ -i \left.\frac{\partial U(\Lambda={\bf 1}+\omega,a=0)}{\partial\omega_{\mu\nu}}\right|_{\omega=0}.$$
Extend $J^{\mu\nu}$ to an antisymmetric matrix $J^{\mu\nu}=-J^{\nu\mu}$.
VIII) Similarly, define $4$-momentum
$$P^{\mu}~=~ i \left.\frac{\partial U(\Lambda={\bf 1},a)}{\partial a_{\mu}}\right|_{a=0}.$$
IX) Taylor expand to first order
$$
U({\bf 1}+\omega,a) ~=~ {\bf 1} + i\sum_{\mu<\nu} \omega_{\mu\nu} J^{\mu\nu}
- i \sum_{\mu}a_{\mu}P^{\mu}+\ldots $$
$$\tag{2.4.3} ~=~ {\bf 1} + \frac{i}{2}\sum_{\mu\nu} \omega_{\mu\nu} J^{\mu\nu}
- i \sum_{\mu}a_{\mu}P^{\mu}+\ldots.$$
