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In maths that is

## Question

Verify the consequences of metric compatibility: If\begin{align}

{\mathrm{\nabla }}_{\sigma}g_{\mu \nu }=0 & \phantom {10000}(1) \\

\end{align}then (a)\begin{align}

{\mathrm{\nabla }}_{\sigma}g^{\mu \nu }=0 & \phantom {10000}(2) \\

\end{align}and (b)\begin{align}

{\mathrm{\nabla }}_{\lambda}{\varepsilon }_{\mu \nu \sigma \rho }=0 & \phantom {10000}(3) \\

\end{align} I am not sure if we are assuming ##{\mathrm{\Gamma }}^{\tau }_{\lambda \mu }={\mathrm{\Gamma }}^{\tau }_{ \mu \lambda }## or not.

## Answer

Part (a) was quite simple but I struggled with the part (b) until 23 March and had to give up. Along the way I had lots of practice at index manipulation, I reacquainted myself with Cramer's rule for solving simultaneous equations, proved (b) on the surface of a sphere, found the 'dynamite' version of Carroll's streamlined matrix determinant equation (2.66) and added some equation shortcut keys to my keyboard. The time was not wasted.We make frequent use here of the fact that ##g_{\mu \nu }g^{\mu \rho }\mathrm{=}{\delta}^{\rho }_{\nu }## and the indexing effect of the Kronecker delta: ##{\delta}^{\lambda }_{\beta }\mathrm{\Gamma }^{\mu }_{\sigma \lambda }={\mathrm{\Gamma }}^{\mu }_{\sigma \beta }## because we are summing over ##\lambda ## and the only non-zero term is when ##\beta =\lambda ##. In this case ##\mathrm{\Gamma }## can be replaced by any symbol or tensor of any rank.

Here is the full effort Ex 3.01 Consequences of metric compatibility.pdf (7 pages of which 4 might be worth looking at).

Check out this link for some really clever solutions to the porblem from Carroll's book. The Part B is so easier than you realize.

ReplyDeletehttps://indexguy.wordpress.com/2007/07/13/some-identities-regarding-metric-compatibility/

The proof for part a is great. But I don't follow the proof for part b. It is true that the determinant will only depend on the metric entries, but why would that mean that the covariant derivative will vanish?

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