Covariant derivative of the metric tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; 9.

One interesting and useful theorem is that the covariant derivative of any metric tensor is always zero. We can show this by using the expression for the covariant derivative of a general tensor to say:

\displaystyle \nabla_{j}g_{kl}=\partial_{j}g_{kl}-\Gamma_{\; jk}^{m}g_{ml}-\Gamma_{\; jl}^{m}g_{km} \ \ \ \ \ (1)

We can combine this with the explicit expression for the Christoffel symbols:

\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right) \ \ \ \ \ (2)

Substituting, we get

\displaystyle \nabla_{j}g_{kl}=\partial_{j}g_{kl}-\frac{1}{2}g^{mn}\left(\partial_{k}g_{jn}+\partial_{j}g_{nk}-\partial_{n}g_{kj}\right)g_{ml}-\frac{1}{2}g^{mn}\left(\partial_{l}g_{jln}+\partial_{j}g_{nl}-\partial_{n}g_{lj}\right)g_{km} \ \ \ \ \ (3)

Since {g^{mn}g_{ml}=\delta_{\; l}^{n}}, we get

\displaystyle \nabla_{j}g_{kl}=\partial_{j}g_{kl}-\frac{1}{2}\left(\partial_{k}g_{jl}+\partial_{j}g_{lk}-\partial_{l}g_{kj}\right)-\frac{1}{2}\left(\partial_{l}g_{jk}+\partial_{j}g_{kl}-\partial_{k}g_{lj}\right) \ \ \ \ \ (4)

Now we use the symmetry of the metric tensor: {g_{kl}=g_{lk}}:

\displaystyle \nabla_{j}g_{kl} \displaystyle = \displaystyle \partial_{j}g_{lk}-\frac{1}{2}\left(\partial_{k}g_{jl}+\partial_{j}g_{lk}-\partial_{l}g_{kj}\right)-\frac{1}{2}\left(\partial_{l}g_{kj}+\partial_{j}g_{lk}-\partial_{k}g_{jl}\right)\ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle 0 \ \ \ \ \ (6)

6 thoughts on “Covariant derivative of the metric tensor

  1. Mark Rupright

    Technically, this is a property of a “metric-compatible connection”. The Christoffel symbols could have been defined first from the requirement that the covariant derivative of the metric vanishes. There are more general connections, and covariant derivatives, than the metric-compatible one.

    Reply
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  4. Valery Khersonsky

    Yes, what is discussed above is true only for Levi-Civita connection (metric compatible and torsion free). Can we prove the same without introducing this type of connection?

    Reply
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