Generators of the lorentz group

Question

I have the following minus sign problem:

Consider an infinitesimal Lorentz transformation for which $\Lambda^{\mu}_{\nu}=\delta^{\mu}_{\nu}+\lambda^{\mu}_{\nu}$, where $\lambda^{\mu}_{\nu}$ is infinitesimal small. Define the vector fields $M_{\mu\nu}=x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}$. Show that acting on $x^{\mu}$, we have

$\frac{1}{2}\lambda^{\rho\sigma}M_{\rho\sigma}(x^{\mu})=\lambda^{\mu}_{\nu}x^{\nu}$

If i make the derivations:

$\frac{1}{2}\lambda^{\rho\sigma}M_{\rho\sigma}(x^{\mu})=\frac{1}{2}\lambda^{\rho\sigma}(x_{\rho}\partial_{\sigma}-x_{\sigma}\partial_{\rho})x^{\mu}= \lambda^{\rho\sigma}x_{\rho}\partial_{\sigma}x^{\mu}=\lambda^{\rho\sigma}x_{\rho}\delta^{\mu}_{\sigma}=\lambda^{\rho\mu}x_{\rho}=-\lambda^{\mu\rho}x_{\rho}=-{\lambda^\mu}_{\rho}x^{\rho}$

I can't see how lose the minus sign.. Probably trivial, but it keeps me busy.

Correction: orignal question had in the last step $\lambda^\mu_\rho$ which should be ${\lambda^\mu}_\rho$

Answer 1

You get a sign ambiguity because of your notation, as you simplify both ${\lambda^\mu}_\nu$ and ${\lambda_\nu}^\mu$ (which differ by a sign) to the same symbol $\lambda_\nu^\mu$.