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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


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 ${\lambda^\mu}_\rho$

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1 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 $\lambda_\nu^\mu$.

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Thank you for your reaction and sorry for the small mistake in typing but a mistake which is essential to my question: The correction, i made the derivation: $\frac{1}{2}\lambda^{\rho\sigma}M_{\rho\sigma}(x^{\mu})=\frac{1}{2}\lambda^{\rho‌​\sigma}(x_{\rho}\partial_{\sigma}-x_{\sigma}\partial_{\rho})x^{\mu}= \lambda^{\rho\sigma}x_{\rho}\partial_{\sigma}x^{\mu}=\lambda^{\rho\sigma}x_{\rho‌​}\delta^{\mu}_{\sigma}=\lambda^{\rho\mu}x_{\rho}=-\lambda^{\mu\rho}x_{\rho}=-{\la‌​mbda^\mu}_{\rho}x^{\rho}$ I assume the mistake is in the last step which i can't figure out. –  BB73 Mar 26 '12 at 18:55
This looks correct to me; the last step can be verified by inserting explicitly the metric. Maybe your source has a typo, and the leftmost $\lambda^{\rho\sigma}$ should have its superscript interchanged? –  Arnold Neumaier Mar 26 '12 at 19:20
@BB73, I'd suggest that you submit an edit to the question which fixes your mistake. –  David Z Mar 26 '12 at 20:07

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