Q:

Lorentz symmetry and Noether's theorem

I'm trying to overcome some misunderstanding that I have in Noether's theorem. 
There is formula in David Gross's Lectures on QFT for Noether's theorem: 
$$J^\mu_\alpha=\mathcal{L}X^\mu_\alpha+\Pi^\mu_i\left[\Psi_{i\alpha}-\partial_\nu\phi_i X^i_\alpha\right]$$
$J^\mu_\alpha$ - conserved vector current, $\Pi^\mu_i$-canonical momentum, $X^\mu_\alpha$ and $\Psi_{i\alpha}$ are generators of symmetry which are defined by $$x'^\mu=x^\mu+X^\mu_\alpha \omega^\alpha,$$
$$\phi'(x'^\mu)=\phi(x^\mu)+\Psi_{i\alpha} \omega^\alpha,$$
Particularly I'm working out Lorentz symmetry:
$$x'^\mu=\Lambda^\mu_\nu x^\nu$$
Infinitesimal version of this transformation is: $\Lambda^\mu_\nu=\delta^\mu_\nu+\omega^\mu_\nu$, where $\omega^\mu_\nu=-\omega^\nu_\mu$ is antisymmetric. 
So, my problem that I dont understand why (they write this expression in many sources, not only in Gross)
$$X^{\mu,\alpha}_\beta=\delta^\mu_\alpha x^\beta-\delta^\mu_\beta x^\alpha$$.
If I'm calculating $X^{\mu,\alpha}_\beta\omega^\beta_\alpha$, I get $2\omega^\mu_\nu x^\nu$, not the $\omega^\mu_\nu x^\nu$. Also I've done this calculation:
$$f(x'^\mu)=f(x^\mu+\omega^\mu_\alpha x^\alpha)\approx f(x^\mu)+\omega_{\alpha \mu}x^\alpha \partial^\mu f = f(x^\mu) + \frac 1 2 \omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)f $$
If $f=x^\mu$ (coordinate function), then $$X^{\mu,\alpha}_\beta=\frac 12\left(\delta^\mu_\alpha x^\beta-\delta^\mu_\beta x^\alpha\right).$$
What exactly am I doing wrong?

A:

I think this calculation is the answer for your question: 
$$\sum_{\alpha,\mu}\frac 1 2 \omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)=\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)+\frac 1 2 \sum_{\mu<\alpha}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)=\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)+\frac 1 2 \sum_{\alpha<\mu}\omega_{\mu\alpha } \left(x^\alpha  \partial^\mu-x^\mu \partial^\alpha  \right)=\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)+\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)=\sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)$$
Where I 

Splited sum in "$\alpha<\mu$" and "$\mu<\alpha$" parts.
Changed $\alpha \leftrightarrow \mu$
Used antisymmetric property of $\omega_{\alpha \mu}$ and $\left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)$

