- #1
George Keeling
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- TL;DR Summary
- Metric compatibility? Why is it an additional property?
In chapter 3 of Sean Carroll's Introduction to General Relativity he 'makes the demand' of metric compatibility of a connection that ##\nabla_\mu g_{\lambda\nu}=0##. Metric compatibility becomes a phrase that is used frequently. However metric compatibility seems to arise naturally. One only has to require that the covariant derivative is a tensor (!) and that it obeys the Leibniz rule:$$
\nabla\left(T\ \bigotimes\ \ S\right)=\left(\nabla T\right)\ \ \bigotimes\ \ S+T\ \bigotimes\ \ \left(\nabla S\right)
$$If the covariant derivative is a tensor then we could say that ##\ \nabla_\mu V^\nu\equiv\ \left(\nabla V\right)_\mu^{\ \ \ \nu}## or ##\nabla V## is the tensor and ##\mu,\nu## are its indices. In that case we must have$$
g_{\lambda\nu}\nabla_\mu V^\lambda=g_{\lambda\nu}\left(\nabla V\right)_\mu^{\ \ \ \lambda}=\left(\nabla V\right)_{\mu\nu}^{\ \ \ }=\nabla_\mu V_\nu
$$
By the Leibniz rule we have $$
\mathrm{\nabla}_\mu V_\nu=\nabla_\mu\left(g_{\lambda\nu}V^\lambda\right)=V^\lambda\nabla_\mu g_{\lambda\nu}+g_{\lambda\nu}\nabla_\mu V^\lambda
$$Picking bits out of those we get$$
V^\lambda\nabla_\mu g_{\lambda\nu}=0
$$Since that holds for any ##V^\lambda## we must have$$
\nabla_\mu g_{\lambda\nu}=0
$$We can call that metric compatibility but why say it is an additional property of a connection?
This harks back back to an answer StevenDaryl :
\nabla\left(T\ \bigotimes\ \ S\right)=\left(\nabla T\right)\ \ \bigotimes\ \ S+T\ \bigotimes\ \ \left(\nabla S\right)
$$If the covariant derivative is a tensor then we could say that ##\ \nabla_\mu V^\nu\equiv\ \left(\nabla V\right)_\mu^{\ \ \ \nu}## or ##\nabla V## is the tensor and ##\mu,\nu## are its indices. In that case we must have$$
g_{\lambda\nu}\nabla_\mu V^\lambda=g_{\lambda\nu}\left(\nabla V\right)_\mu^{\ \ \ \lambda}=\left(\nabla V\right)_{\mu\nu}^{\ \ \ }=\nabla_\mu V_\nu
$$
By the Leibniz rule we have $$
\mathrm{\nabla}_\mu V_\nu=\nabla_\mu\left(g_{\lambda\nu}V^\lambda\right)=V^\lambda\nabla_\mu g_{\lambda\nu}+g_{\lambda\nu}\nabla_\mu V^\lambda
$$Picking bits out of those we get$$
V^\lambda\nabla_\mu g_{\lambda\nu}=0
$$Since that holds for any ##V^\lambda## we must have$$
\nabla_\mu g_{\lambda\nu}=0
$$We can call that metric compatibility but why say it is an additional property of a connection?
This harks back back to an answer StevenDaryl :
So 'rule 3' of covariant derivatives is that they are tensors.stevendaryl said:Here's the way I understand #3.
##g^{\lambda \nu} (\nabla_\mu T_{\nu \lambda \rho}) = \nabla_\mu (g^{\lambda \nu} T_{\nu \lambda \rho})##