How to make derivative of antisymmetric tensor？

Question

When I read the Conformal Field Theory of Di Francesco et al, I found I will get two different answers from the same expression. To get the (2.131) $$ \dfrac{\delta x^{\mu}}{\delta\omega_{\rho\nu}}=\dfrac{1}{2}(\eta^{\rho\mu}x^{\nu}-\eta^{\nu\mu}x^{\rho}),\tag{2.131} $$ I compare the (2.125) $$ x^{'\mu}=x^{\mu}+\omega_{\alpha\beta}\dfrac{\delta x^{\mu}}{\delta \omega_{\alpha\beta}}\tag{2.125} $$ and (2.130) $$ x^{'\mu}=x^{\mu}+\omega_{\rho\nu}\eta^{\rho\mu}x^{\nu},\tag{2.130} $$ where $\omega_{\alpha\beta}$ is an antisymmetry infinitesimal parameter. Then I get the right answer by divide the $\omega$ parameter, $$ \omega_{\alpha\beta}\frac{\delta x^{\mu}}{\delta \omega_{\alpha\beta}}=\dfrac{1}{2}\omega_{\alpha\beta}(\eta^{\alpha\mu}x^{\beta}-\eta^{\beta\mu}x^{\alpha}) $$ But if I think $\delta x^{\mu}=x^{'\mu}-x^{\mu}$, and differentiate(2.130), then I get $$ \dfrac{\delta x^{\mu}}{\delta\omega_{\alpha\beta}}=\dfrac{\delta\omega_{\rho\nu}}{\delta\omega_{\alpha\beta}}\eta^{\rho\mu}x^{\nu}=(\delta_{\rho}^{\alpha}\delta_{\nu}^{\beta}-\delta_{\rho}^{\beta}\delta_{\nu}^{\alpha})\eta^{\rho\mu}x^{\nu}=\eta^{\alpha\mu}x^{\beta}-\eta^{\beta\mu}x^{\alpha}, $$ where the factor $\frac{1}{2}$ is missing. I can't find the mistake I made, could someone point out where I'm wrong? I'd appreciate some help.

Answer 1

Since the matrix $\omega_{\mu\nu}=-\omega_{\nu\mu}$ is antisymmetric, we must demand that $$\begin{align} \delta \omega_{\mu\nu}~=~&-\delta \omega_{\nu\mu}\cr ~=~&\frac{1}{2}\left(\delta \omega_{\mu\nu}-\delta \omega_{\nu\mu}\right)\cr ~=~&\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} - \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta \omega_{\alpha\beta},\end{align}\tag{1}$$ and therefore the differentiation rule contains a half: $$ \frac{\delta \omega_{\mu\nu}}{\delta \omega_{\alpha\beta}} ~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} - \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right).\tag{2} $$ See also this and this related Phys.SE posts.