Infinitesimal Poincare transformations , Taylor expansion Let $(\Lambda,a)\in\text{ ISO}_o(3,1)$ be a finite (proper) Poincare transformation and Let $U(\Lambda,b)$ be the corresponding unitary operator implementing this transformation on the Hilbert space of (free) one particle states. 
Consider now an infinitesimal Poincare transformation, $(\Lambda,a)=(1+\omega,\varepsilon)$ where 1 is the identity operator and $\varepsilon^a$, $\omega_{ab}=-\omega_{ba}$ are small parameters.
In most textbooks I have found, the authors simply claim that the corresponding unitary operator for such an infinitesimal transformation is (see Weinberg V1 pg 59 or Srednicki pg 17 for example)
$$U(1+\omega,\varepsilon)=1+\frac{1}{2}i\omega_{ab}\mathcal{J}^{ab}-i\varepsilon_a\mathcal{P}^a+\cdots$$
where $\mathcal{J}^{ab}$ and $\mathcal{P}^a$ are the generators of Lorentz transformations and translations respectively. I assume that this is some sort of taylor expansion in the infinitesimal parameters $\omega$ and $\varepsilon$, where we only keep linear terms in the infinitesimal parameters. 
There are a couple of facets of this expansion which I am unsure about.


*

*Where does the factor of $\frac{1}{2}$ and $i$ come from?

*It is my impression that the signs in front of the second and third terms are based on your metric convention. If this is true, why is it so?


Regarding (1), I get the feeling that we have some freedom of choice. We choose the $1/2$ because it makes the exponential expression for a finite transformation look nicer. The factor of $i$ ensures that $U$ is hermitian. If this is true, $\textit{why}$ do we have this freedom of choice?
 A: For (1), the idea is to expand in all independent quantities, which are $\omega_{01}, \omega_{02},\omega_{03}, \omega_{12},\omega_{13}, \omega_{23}$, and $\epsilon_{0}$, $\epsilon_{1}$,$\epsilon_{2}$,$\epsilon_{3}$. These are just numbers however, and hence need matrix "coefficients". So when you do the expansion you would get something like this 
$$
U=1 + \omega_{01}\tilde{J}^{01} + \omega_{02}\tilde{J}^{02}+\omega_{03}\tilde{J}^{03}+\omega_{12}\tilde{J}^{12}+\omega_{13}\tilde{J}^{13}+\omega_{23}\tilde{J}^{23} + \epsilon_{0}\tilde{P}^0 +\epsilon_{1}\tilde{P}^1+\epsilon_{2}\tilde{P}^2+\epsilon_{3}\tilde{P}^3+\mathcal{O}(\omega^2,\epsilon^2)$$
Now $U$ is supposed to be unitary, $U^\dagger=U^{-1}$. Note that we have $U^{-1}(1+\omega,\epsilon)=U(1-\omega,-\epsilon)$. When you expand both sides, this means you must have, 
$$ \tilde{J}^{12\dagger}=-\tilde{J}^{12}$$ and the same for all the other generators, which means that they are anti-hermitian. It is then a common convention in physics to pull out a factor of $i$, to make them Hermitian, i.e you would define generators $J^{12}$ such that 
$$\tilde{J}^{12}=iJ^{12}$$. Then 
$$\tilde{J}^{12\dagger}=i^\ast J^{12\dagger}=-iJ^{12\dagger}$$ 
but we want $\tilde{J}^{12\dagger}=-\tilde{J}^{12}=-i J^{12}$
and hence the new generators $J$ must be hermitian, i.e. 
$$J^{12\dagger}=J^{12}$$
Hermitian operators in quantum mechanics are associated with observables, so this is just a convenient convention to make them manifest. This answers your second question. 
As for the first part, note that what you wrote is really what I wrote. 
$$\omega_{\mu\nu}J^{\mu\nu}= \omega_{00}J^{00}+\omega_{01}J^{01}+\omega_{01}J^{10}+\dots$$
The terms with $\omega_{\mu\mu}$ (no sum on $\mu$) vanish by the antisymmetry of $\omega$, and similarly by the antisymmetry, you get $\omega_{10}J^{10}=(-1)^2\omega_{01}J^{01}=\omega_{01}J^{01}$ and hence you actually get $2\omega_{01}J^{01}$, instead of $\omega_{01}J^{01}$ as I had in the expansion at the top. The factor of $1/2$ in your expansion is there to compensate for this. 
The minus sign before the momentum generators is basically just convention. I vaguely recall a reason why its convenient so I'll add it here when I get more time later.
Update: I didn't state clearly but because the $\omega$ are antisymmetric it follows that the $J$ must be too because 
$$\omega_{\mu\nu}J^{\mu\nu}=\omega_{\nu\mu}J^{\nu\mu}=-\omega_{\mu\nu}J^{\nu\mu}$$ where for the first equality I rename $\mu\rightarrow \nu$ and vice-versa and then in the second equality use the antisymmetry of $\omega$. Equating the first and last expression then tells you that $J^{\mu\nu}=-J^{\nu\mu}$.
