$2$ out of $3$ property of the unitary group

Question

I am trying to understand the $2$ out of $3$ property of the unitary group. I have almost got it, but I am not completely sure about the interaction between an inner product and a symplectic form to obtain an almost complex structure.

---

Let $V$ be a real vector space.

An inner product on $V$ is a positive definite symmetric bilinear form $g$. An endomorphism $T \in \operatorname{End}(V)$ preserves $g$ if $g(T(u), T(v)) = g(u, v)$ for all $u, v \in V$; the collection of all such endomorphisms forms a group $O(V, g)$ called the orthogonal group.

An almost complex structure on $V$ is an endomorphism $J \in \operatorname{End}(V)$ such that $J^2 = -\operatorname{id}_V$. An endomorphism $T \in \operatorname{End}(V)$ is complex linear if $T \circ J = J\circ T$; the collection of all such endomorphisms forms a group $GL(V, J)$ called the complex general linear group.

A symplectic form on $V$ is a skew-symmetric non-degenerate bilinear form $\omega$. An endomorphism $T \in \operatorname{End}(V)$ preserves $\omega$ if $\omega(T(u), T(v)) = \omega(u, v)$ for all $u, v \in V$; the collection of all such endomorphisms forms a group $Sp(V, \omega)$ called the symplectic group.

---

Almost Complex Structure & Inner Product

For an inner product $g$ and a compatible almost complex structure $J$ (i.e. $J \in O(V, g)$), we obtain a symplectic form by defining $\omega(u, v) := g(u, J(v))$.

It follows that $O(V, g)\cap GL(V, J) \subseteq Sp(V, \omega)$.

---

Almost Complex Structure & Symplectic Form

For a symplectic form $\omega$ and a compatible almost complex structure $J$ (i.e. $J \in Sp(V, \omega)$) which tames $\omega$ (i.e. $\omega(u, J(u)) > 0$ for all $u \in V\setminus\{0\}$) we obtain an inner product by defining $g(u, v) := \omega(J(u), v)$.

It follows that $Sp(V, \omega)\cap GL(V, J) \subseteq O(V, g)$.

---

Inner Product & Symplectic Form

This is the part I am unsure about.

Denote by $\Phi_g$ the isomorphism $V \to V^*$ induced by $g$; that is $\Phi_g(v) \in V^*$ is defined by $\Phi_g(v)(u) = g(u, v)$. Likewise, denote the isomorphism $V \to V^*$ induced by $\omega$ by $\Phi_{\omega}$; that is $\Phi_{\omega}(v) \in V^*$ is defined by $\Phi_{\omega}(v)(u) = \omega(u, v)$.

1. Is there any compatibility restriction that we must impose on $\Phi_g$ and $\Phi_{\omega}$?

2. Is $J = \Phi_g^{-1}\circ\Phi_{\omega}$ an almost complex structure on $V$?

3. How do we use this to deduce $O(V, g)\cap Sp(V, \omega) \subseteq GL(V, J)$?

For question 3, I can use the previous relationships between the three groups, but I'd like to be able to deduce it from the structures themselves.

Answer 1

For question 1, the compatibility relation we are looking for is

$$ \Phi_g^{-1} \circ \Phi_{\omega} = -\Phi_{\omega}^{-1} \circ \Phi_g.$$

Define $J = \Phi_g^{-1} \circ \Phi_{\omega}$. Then, unrolling the definitions, we get the relation $\Phi_g(Jv)(u) = \omega(u,v)$, i.e., $$g(u,Jv) = \omega(u,v). $$

Since we also have the relation $-J = \Phi_{\omega}^{-1} \circ \Phi_g$, we have $$\omega(u, -Jv) = g(u,v).$$

To see that $J$ defines an almost complex structure, note that, $$g(u, J^2v) = \omega(u, Jv) = -g(u,v)$$ so by the nondegeneracy of $g$, we have $J^2 = -1$. This answers question 2.

For question 3, let $T$ be any automorphism of $V$ preserving both $g$ and $\omega$. Then $$g(Tu, JTv) = \omega(Tu, Tv) = \omega(u,v) = g(u, Jv) = g(Tu, TJv)$$ and hence by nondegeneracy of $g$, we have $JT = TJ$, i.e., $T \in GL(V,J)$.