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I have the following question. Let $M$ be a Riemannian manifold with metric $g$ and $\nabla$ the Levi-Civita connection. Let furthermore $\alpha \in \Omega^{k}(M)$ be a $k$-form such that $\nabla \alpha = 0$. Why is then $d \alpha = 0$?

thanks, jan

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1 Answer 1

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I'm not sure what sort of approach you're looking for, but in coordinates, it is easy to see that (ignoring possible combinatoric factors) the antisymmetrization of a covariant derivative kills off everything except the partial derivative which corresponds to the exterior derivative in coordinates. This works for any torsion free connection.

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  • $\begingroup$ Why necessary torsion free? $\endgroup$
    – jan
    Apr 11, 2013 at 10:47
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    $\begingroup$ The torsion free property ensures the Christoffel symbol is symmetric in its lower two indices; try writing down the coordinate expression for a covariant derivative and seeing what happens when you antisymmetrize. $\endgroup$ Apr 11, 2013 at 11:27

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