Application of a Poincaré group element to a scalar function Let $f(x)$ be a scalar function and let's say that we want to know how it transforms when it's subjected to a translation (by a vector $a^{\mu}$), rotation and a Lorentz boost. Thus we can write an infinitesimal transformation as:
\begin{equation}
\bar{x}^{\mu} = x^{\mu} + a^{\mu} + \omega^{\mu}_{\nu}x^{\nu} = a^{\mu} + \Lambda^{\mu}_{\nu}x^{\nu}
\end{equation}
where $\omega^{\mu}_{\nu}$ is:
\begin{pmatrix}
0 & v^{1} & v^{2} & v^{3} \\
v^{1} & 0 & \theta^{3} & -\theta^{2} \\
v^{2} & -\theta^{3} & 0 & \theta^{1} \\
v^{3} & \theta^{2} & -\theta^{1} & 0 \\
\end{pmatrix}
therefore a function $f(x)$ transforms to:
\begin{align}
f(\bar{x}) & = f(x+a+\omega x) \\
 & = f(x) + a^{\mu}\partial_{\mu}f(x) + \omega^{\mu}_{\nu}x^{\nu}\partial_{\mu}f(x).
\end{align}
So far so good, the real pain (for me) starts when it is required to use the antisymmetry of $\omega_{\mu\nu}$ (obtained by lowering the index using $g_{\mu\lambda}\omega^{\lambda\nu}$ with $g^{\alpha\beta}=g_{\alpha\beta}=diag(1,-1,-1,-1)$) to write:
\begin{equation}
f(\bar{x}) = \left[ 1 + a^{\mu}\partial_{\mu} -\frac{1}{2}\omega_{\mu\nu}(x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu}) \right]f(x).
\end{equation}
I've tried many different roads, but i feel that I'm stuck because, maybe, I didn't really get what does it mean to deal with an antisymmetric tensor. As far as I know I can generally write a tensor $T_{\alpha\beta}$ as a sum of a symmetric and antisymmetric part namely $S_{\alpha \beta} + A_{\alpha \beta}$ yielding $A_{\alpha \beta} = \frac{1}{2}\left( T_{\alpha \beta} - T_{\beta \alpha} \right)$. But when I try to write something like $\omega_{\mu\nu}=\frac{1}{2}(\omega_{\mu\nu} - \omega_{\nu\mu})$, (cause $\omega_{\mu\nu}=-\omega_{\nu\mu}$) leads me to a dead end. I found also a solution to this problem but it's not clear at all to me and it's at this link if you are interested. So if you can clarify these steps I would be infinitely grateful to you.  
 A: There are two ways, basically equivalent, of arriving at this. Let $\omega_{ab}$ be an antisymmetric tensor and $T_{ab}$ any tensor. Then, it can be shown that $\omega_{ab} T_{ab} = - \omega_{ab} T_{ba}$, as follows:
$$\begin{align}
\omega_{ab} T_{ab} &= \omega_{ba} T_{ba} \quad \text{(relabeling indices)} \\
&= -\omega_{ab} T_{ba} \quad \text{(antisymmetry of } \omega \text{)}
\end{align}$$
Therefore, 
$$\omega_{ab} (T_{ab} - T_{ba}) = \omega_{ab} T_{ab} - \omega_{ab} - T_{ba} = \omega_{ab} T_{ab} + \omega_{ab} T_{ab} = 2\omega_{ab} T_{ab},$$
and you get the identity you want by reading right to left, with $T_{ab} = x_a \partial_b$.
An alternative path is to write $T_{ab} = A_{ab} + S_{ab}$, with $A$ and $S$ the symmetric and antisymmetric parts of $T$, and notice that $\omega_{ab} S_{ab} = 0$. This is because  every term in the sum appears twice but with opposite signs. And then you just use $\omega_{ab} T_{ab} = \omega_{ab} A_{ab}$: the trick was to take the antisymmetric part of $T$, not $\omega$.
