If $D$ is the Riemannian covariant differentiation associated to $g$ and $J$ is an almost complex structure, then $DJ=0 \implies g$ is a Kähler metric

Question

If $D$ is the Riemannian covariant differentiation associated to $g$ and $J$ is an almost complex structure, then $DJ=0 \implies g$ is a Kähler metric.

Since $Dg = 0$ and $\omega(X,Y)=g(JX,Y)$ the vanishing of $DJ$ implies that of $D\omega:$

\begin{align*} D_Z\omega(X,Y)&= Z\omega(X,Y)-\omega(D_ZX,Y) - \omega(X,D_ZY) \\ &= Zg(JX,Y) - g(JD_ZX,Y) - g(JX,D_ZY) \\ &= (D_Zg)(JX,Y) + g(D_Z(JX),Y) + g(JX,D_ZY)- g(JD_ZX,Y) - g(JX,D_ZY) \\ &= 0+g((D_ZJ)X,Y) + g(J(D_ZX),Y) + g(JX,D_ZY) - g(JD_ZX,Y) - g(JX,D_ZY) \\ &= g((D_ZJ)X,Y) = 0 \end{align*}

can anyone explain the equality \begin{align*} &(D_Zg)(JX,Y) + g(D_Z(JX),Y) + g(JX,D_ZY)- g(JD_ZX,Y) - g(JX,D_ZY) \\ &= 0+g((D_ZJ)X,Y) + g(J(D_ZX),Y) + g(JX,D_ZY) - g(JD_ZX,Y) - g(JX,D_ZY) \\ \end{align*} to me? I think there is a clever addition of $0$ since $$(D_Zg)(JX,Y) + g(D_Z(JX),Y) + g(JX,D_ZY)=Zg(JX,Y)$$

but I don't see where $g((D_ZJ)X,Y)$ comes from.

Answer 1

First note that $(D_Zg)(JX, Y) = 0$ because $D_Zg = 0$.

Secondly, as $D_Z(JX) = (D_ZJ)X + J(D_ZX)$, we have $$g(D_Z(JX), Y) = g((D_ZJ)X, Y) + g(J(D_Z X), Y).$$

Combining these two observations yields the equality.