Let $(M, g)$ be a Riemannian manifold and $X, Y, Z$ smooth vector fields on $M$. Let $\theta_X$ be the $1$-form defined as $\theta_X(Y) = g(X,Y)$ and let $d\theta_X$ be its exterior derivative. Let $L_X$ denote the Lie derivative. Let $\nabla$ be the Levi-Civita connection, i.e. the unique torsion free connection on $(M,g)$ which is also compatible with the metric. I want to prove that: $$ 2 g(\nabla_Y X, Z) = (L_Xg)(Y,Z) + (d \theta_X)(Y,Z). $$ Is there a quick way to see that?

EDIT: I'm stuck with the $2$-form $d\theta_X$. I don't know how to deal with it. Can I express it just using the metric and the Lie derivative (without working in local coordinates)?


Using the invariant formula for the exterior derivative, we have that

$$\textrm d \theta_X (Y, Z) = Y \cdot \theta_X (Z) - Z \cdot \theta_X (Y) - \theta_X ([Y, Z]) = Y \cdot g(X, Z) - Z \cdot g(X, Y) - g(X, [Y, Z]) .$$

Similarly, using the properties of the Lie derivative, we have

$$(L_X g) (Y, Z) = X \cdot g(Y, Z) - g([X, Y], Z) - g(Y, [X, Z]) .$$

Putting all these together we obtain

$$(L_X g) (Y, Z) + \textrm d \theta_X (Y, Z) = X \cdot g(Y, Z) + Y \cdot g(Z, X) - Z \cdot g(X, Y) - g([X, Y], Z) - g([Y, Z], X) - g([X, Z], Y) .$$

This last expression is known to be equal to $2 g (\nabla_Y X, Z)$ by Koszul's formula (warning: the link to Wikipedia points to formulae that obtain $\nabla_X Y$, not $\nabla_Y X$, so a sign will differ!).

  • $\begingroup$ Thank you! Sorry, but differential forms are a very new thing to me. ;) $\endgroup$ – Onil90 Feb 25 '16 at 15:11

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.