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

  1. $J^2=-id_V$

  2. 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?

  1. For all $u,v\in V$ $\omega(Au,v)=-\omega(u,Av)$
  2. 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?

  • $\begingroup$ If 1. is true, then $g_\omega(Au,v)=\omega(Au,Jv)=-\omega(u,AJv)\neq -g_\omega(u,Av)$, unless $A$ and $J$ commute. Do you have an explicit operator $A$ in mind? $\endgroup$
    – Avitus
    Commented Jul 27, 2014 at 10:00

2 Answers 2


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.


Let's focus on the finite-dimensional case to begin with. Throughout I'll assume that $F$ is $\mathbb{R}$, since this is really what you want in order to talk about a "symmetric, positive definite bilinear form". Let $\dim V = 2n$.

It's a standard fact (which you may well know) that given $V$ (finite dimensional) and $\omega$, $J$ and $g$ as you describe (a "compatible triple"), we can pick a basis for $V$ with respect to which $\omega$, $J$ and $g$ take their standard forms. So $g$ is given by the $2n \times 2n$ identity matrix, $J$ is given by the matrix $J_0$ representing the complex structure on $\mathbb{C}^n \cong \mathbb{R}^{2n}$, and $\omega$ is given by $-J_0$. We may therefore assume that $V$ is just this standard prototypical example.

Now try writing out what the two statements mean in matrix form, and see where that gets you.

  • $\begingroup$ Thanks Jez, this really gives a good overwiew. In this setting, condition 1. means that $ A=\begin{pmatrix}a && b \\ c && -a^T\end{pmatrix}$ where $ a,b,c,d$ are $ n\times n$ matrices, and $ b^T=b$, $ c^T=c$, condition 2. means that $ A=\begin{pmatrix}a && b \\ -b^T && d\end{pmatrix}$, $ a^T=-a$, $ d^T=-d$, while the $ AJ=JA$ condition means, that $ A=\begin{pmatrix}a && b \\ -b && a\end{pmatrix}$. It is easy to check, that any two of these 3 conditions implies the third one. $\endgroup$
    – mma
    Commented Jul 29, 2014 at 5:45

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