Given almost complex structure of $V$, there is a symplectic form

Question

Let $V$ be a vector space, I want to show if there is an almost complex structure $I$, then $V$ has symplectic form $\omega$ such that $I$ is compatible with $\omega$. Basically I want to find inner product G s.t. $I^*=-I$, then just choose $\omega (u,v) = G(Iu,v)$. But how can I make sure that I can find such inner product?

Answer 1

Assume that $(W,J)$ is another complex vector space and we have an isomorphism $T \colon V \rightarrow W$ such that $TI = JT$. Let's say we managed to solve the problem for $(W,J)$ by finding an inner product $g$ and a symplectic form $\omega$ such that $\omega(w_1,w_2) = g(Jw_1,w_2)$ for all $w_1,w_2 \in W$. By defining

$$ \alpha(v_1,v_2) := \omega(Tv_1,Tv_2), h(v_1,v_2) := g(Tv_1,Tv_2) $$

we obtain a symplectic form $\alpha$ and an inner-product $h$ on $V$ and

$$ \alpha(v_1,v_2) = \omega(Tv_1,Tv_2) = g(JTv_1, Tv_2) = g(TIv_1, Tv_2) = h(Iv_1, v_2) $$

and so we have managed to solve the problem for $(V,I)$. Now, choose any isomorphism $T \colon (V,I) \rightarrow (\mathbb{C}^n,J)$ where $J$ is the standard complex structure on $\mathbb{C}^n$, take $g$ to be the standard inner product and $\alpha$ to be the standard symplectic form.