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


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.

  • $\begingroup$ Why necessary torsion free? $\endgroup$
    – jan
    Apr 11 '13 at 10:47
  • 1
    $\begingroup$ 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 '13 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.