$V^{1,0}$ as a Lagrangian subspace of $V^c$

Question

Let $V$ be a real vector space, $V^c = V\otimes_\mathbb{R}\mathbb{C}$ its complexification and $J:V\to V$ an almost complex structure.

Recall that the complexified map $J^c:V^c\to V^c$ defines an eigenspace decomposition $V^c = V^{1,0}\oplus V^{0,1}$. Also, an inner product $g$ on $V$ compatible with $J$ defines a symplectic structure on $V$ by $\omega(v,w) = g(Jv,w)$.

In a review by Blau, at page 31 we read the following. My question is: what does this mean?

If $\omega$ is compatible with $J$, then $V^{1,0}$ and $V^{0,1}$ are Lagrangian subspaces of $V^c$.

1. This phrase implies we're extending $\omega$ (a real 2-form on $V$) to $\omega^c$, a (complex?) 2-form on $V^c$. How is this done? My naive intuition asks for $\mathbb{C}$-bilinearity; for $\lambda,\mu\in\mathbb{C}$, we set

$$ \omega^c(\lambda \cdot v,\mu \cdot w)=\lambda\mu\cdot\omega(v,w) $$

It's easy to see that $J$ is compatible with $\omega\implies J^c$ is compatible with $\omega^c$. Furthermore, we indeed find out that $V^{1,0}$ is a Lagrangian subspace, as it has half the dimension of $V^c$ and for all $v,w\in V^{1,0}$ we have \begin{align*} \omega^c(v,\cdot w) &= \omega^c(J(v), \cdot J(w)) \\ &= \omega^c(iv,iw) \\ &= i^2\cdot\omega(v,w) \\ &=-\omega^c(v, w)\\ \implies \omega^c(v,w) = 0 \end{align*}

Good!

2. Is this definition of $\omega^c$ correct, or perhabs usually phrased differently in the literature? It seemingly implies that the Kähler form of a Kähler manifold is complex-bilinear, which sounds untrue.

3. One problem I found is that with respect to the natural extension $g^c$ of $g$ to $V^c$ we don't find $g^c(v,w) = \omega^c(v,J^c(w))$, instead $$ g^c(v,\lambda\cdot w) = \bar\lambda\cdot g(v,w) = \bar\lambda\omega(v,Jw) = \omega^c(v,J(\bar\lambda\cdot w)) = g^c(v,\bar\lambda\cdot w) $$

Seemingly this calls for sesquilinearity of $\omega^c$, but this would imply $\omega^c\equiv 0$.

Answer 1

Following Arctic's comment, we replace Hermitian by bilinear.

We're extending the inner product $g$ on $V$ - a $\mathbb R$-bilinear symmetric and positive definite quadratic form. What we find is a $\mathbb C$-bilinear symmetric quadratic form $g^c$ on $V^c$, which can't possibly be positive definite: $$ g^c(iv,iv) = i^2g(v,v) = - g(v,v)<0 $$

However, this time we do have a pseudo-Kähler structure on $V^c$ given by $(\omega^c,J^c,g^c)$, i.e. $\omega^c(X,Y) = g^c(J^cX,Y)$.