Complex structure is determined in a compatible triple in a Kähler space

Question

Let $V$ be a real (finite dimensional) vector space, $\Omega$ be a symplectic form on $V$, and $g$ be a pseudo-Euclidean scalar product. Using $g$ to obtain an isomorphism $\sharp\colon V^* \to V$, we can set $J(v) = \Omega(v,\cdot)^\sharp \in V$. Since $\Omega$ is bilinear and $\sharp$ is linear, the map $J$ thus defined is also linear. It also satisfies $g(Jv,w) = \Omega(v,w)$ for all $v,w \in V$.

Can we prove that $J^2 =-{\rm Id}_V$?

My guess would be "no". If this is the case, then what I want to confirm is: when people are talking about compatible triples in a Kähler vector space and they say that either two of the structures actually determines the remaining one, is $J$ assumed to be a complex structure (or even exist) to begin with?

Answer 1

Nevermind, I got it. The following are equivalent:

• $J^2 = -{\rm Id}_V$
• $J \in {\rm Sp}(V,\Omega)$;
• $J \in {\rm O}(V,g)$.