Let $V$ be a vector space over a field $F$ endowed by an $\omega$ symplectic (i.e. nondegenerate, skew-symmetric bilinear) form and by a $J: V\to V$ compatible complex structure, i.e. by such a linear operator, that
$J^2=-id_V$
The map $g_\omega: V^2\to F: (u,v)\mapsto\omega(u,Jv)$ is a nondegenerate symmetric, positive definite bilinear form (i.e. an inner product) on $V$.
Let $A$ be a linear operator on $V$.
Is it true, that the following two statements are equivalent?
- For all $u,v\in V$ $\omega(Au,v)=-\omega(u,Av)$
- For all $u,v\in V$ $g_\omega(Au,v)=-g_\omega(u,Av)$
Is the answer the same for finite dimensional and infinite dimensional vector spaces?
The answer to your first question is in general no; you need linear operators $A$ commuting with $J$. In fact, if 1. is true, then $$g_\omega(Au,v):=\omega(Au, Jv)=(1.)=-\omega(u,AJv); $$
if $JA=AJ$ then $-\omega(u,AJv)=-\omega(u,JAv):=-g_\omega(u, Av)$ and you obtain 2. Viceversa, if 2. is true then
$$g_\omega(Au,v)=-g_\omega(u,Av)\Leftrightarrow \omega(Au,Jv)=-\omega(u,JAv); $$
once again, if $JA=AJ$ then $\omega(Au,Jv)=-\omega(u,JAv)=-\omega(u,AJv),$ i.e.
$$ \omega(Au,w)=-\omega(u,Aw)$$
for all $u\in V$ and $w=Jv$ for $v\in V$.
To obtain condition 1. we are left to prove surjectivity of $J$ on $V$:
$$\forall w\in V~\exists v\in V~:~ w= Jv.$$
choosing $v:=-Jw$ one gets it.