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)?


1 Answer 1


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
    Commented Feb 25, 2016 at 15:11

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