Algebraic (?) proof that Ricci form is closed

Question

Let $(M,\omega, J, g)$ be a Kähler manifold. The Ricci form of $M$ is defined as $\rho(X,Y)={\rm Ric}(JX,Y)$. I wanted to give a possibly coordinate-free proof that ${\rm d}\rho=0$. From the condition $\nabla J=0$ we have that $${\rm d}\rho(X,Y,Z) = (\nabla_X{\rm Ric})(JY,Z) +(\nabla_Y{\rm Ric})(JZ,X)+(\nabla_Z{\rm Ric})(JX,Y).$$I'm guessing that there is a smart way of using the second Bianchi identity to get the result from the above, but I can't see how to deal with terms of the form $\nabla_{JX}$. Also the fact that the second Bianchi identity is true for all connections makes me think that it might not be powerful enough. Help?

Answer 1

Andrei Moroianu gives such an algebraic proof in his "Lectures on Kähler Geometry" on p. 88.

First, he proves the formula $$ 2\operatorname{Ric} (X,Y) = \operatorname{Tr}( R(X,JY) \circ J) $$ using the first Bianchi identity.

Then, start computing $d \rho$ as done above, but the $J$s disappear, because $R$ preserves $J$, i.e. $R(JX, JY) = R(X, Y)$: \begin{align*} 2d \rho &= 2( (\nabla_X \operatorname{Ric})(JY, Z)+ (\nabla_Z \operatorname{Ric})(JX, Y)+ (\nabla_Y \operatorname{Ric})(JZ, X) )\\ &= \operatorname{Tr}(( (\nabla_X R)(Y, Z) + (\nabla_Z R)(X, Y) + (\nabla_Y R)(Z, X))\circ J) =0 \end{align*} where we finally could use the second Bianchi identity.