9
$\begingroup$

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.

$\endgroup$

1 Answer 1

7
$\begingroup$

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., $T \in GL(V,J)$.

$\endgroup$
2
  • $\begingroup$ I should add that the way I found this compatibility relation was to simply do calculations like the above until I got stuck, and then ask myself what relations would have to hold in order for things to work out. $\endgroup$
    – Ted
    Commented Jun 19, 2013 at 7:22
  • 2
    $\begingroup$ It makes sense though. There are two ways of obtaining an endomorphism of $V$ from $\Phi_g$ and $\Phi_{\omega}$, and the two endomorphisms are inverses of one another. Therefore, if one of them defines an almost complex structure, the inverse must be its negative (and hence is an almost complex structure itself). $\endgroup$ Commented Jun 19, 2013 at 7:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .