Fundamental form of almost complex manifold is $(1,1)$-form

Question

Let $M$ be an almost complex manifold with almost complex structure $J$, a compatible Riemannian metric $g$ and fundamental form $\omega$. Consider the eigenspaces $T_p^{1,0}M=\{v\in T_pM\otimes\mathbb{C}:J_pv=iv\}$ and $T_p^{0,1}M=\{v\in T_pM\otimes\mathbb{C}:J_pv=-iv\}$ of the tangent spaces to $M$ at some point $p$. I need to show that:

• $\omega_p(u,v)=0$ for $u,v\in T_p^{1,0}M$
• $\omega_p(u,v)=0$ for $u,v\in T_p^{0,1}M$

For this, we consider of course the complex linear extension of the maps involved. I know that $\omega_p(u,v)=g_p(J_pu,v)$ by definition of the fundamental form and that $g_p(J_pu,J_pv)=g_p(u,v)$ by compatibility of the metric. But I do not manage to prove the two results. Can sombebody help me here?

And how can I dedude that the fundamental form is a $(1,1)$-form, ie. $\omega\in\Omega_M^{1,1}$? Thanks for your help!

Answer 1

Note that

$$\omega_p(J_pu, J_pv) = g_p(J_p(J_pu), J_pv) = g_p(J_pu, v) = \omega_p(u, v)$$

for any $u, v \in T_pM\otimes_{\mathbb{R}}\mathbb{C}$. Now consider how to simplify the left-hand side when $u, v \in T^{1,0}_pM$ or $u, v \in T^{0,1}_pM$.