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Let $g$ be a positive definite symmetric form on a $2n$-dimensional vector space (a metric), $\mathbb{J}$ the symplectic unit and $c$ a real number. Define the symmetric form

$h = g + c[g,\mathbb{J}],$

where $[\cdot,\cdot]$ is the commutator. Have you ever encountered this or similar objects? If so, what is the geometrical meaning of requiring it to be positive definite,

$h>0$

?

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  • $\begingroup$ how do you define the commutator between $g$ and $\mathbb{J}$? As operators that doesn't make sense, so you'd be doing this as matrices w.r.t. a fixed basis? If so the result would be basis dependent. Or do you mean $[g,\mathbb{J}]$ = g\mathbb{J} - \mathbb{J}^Tg$? $\endgroup$
    – Willie Wong
    Commented Apr 23, 2011 at 12:06
  • $\begingroup$ (Also, can you give the context in which you encountered this object?) $\endgroup$
    – Willie Wong
    Commented Apr 23, 2011 at 12:07
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    $\begingroup$ What I am trying to get you to realise is that the fact $\mathbb{J}^2 = -\mathbb{I}$ tells you that $\mathbb{J}:V\to V$, whereas $g$, a symmetric bilinear form, maps $V\otimes V \to \mathbb{R}$, or equivalently $g: V\to V^*$. Hence $\mathbb{J}g$ is ill defined as a composition of maps. What you really want is probably the induced action $\mathbb{J}^T: V^*\to V^*$ instead in that direction. If $g$ is positive definite and symmetric on a finite dimensional vector space, there always exists (Graham-Schmidt) an orthonormal basis. In this basis, $g = \mathbb{I}$, and the matrix commutator of $\endgroup$
    – Willie Wong
    Commented Apr 23, 2011 at 21:52
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    $\begingroup$ any matrix $\Omega$, completely independent of whatever property $\Omega$ has, against $\mathbb{I}$, is $[\mathbb{I},\Omega] = 0$. If your statement is completely dependent on the choice of vector basis, it can have no natural geometric interpretation! On the other hand, if what you want is actually $\mathbb{J}^Tg - g \mathbb{J}$, this object is geometric, and in the simplest case where $g = \mathbb{I}$ and $\mathbb{J}$ is the standard $[ 0,-\mathbb{I};\mathbb{I},0]$ matrix, you get something nontrivial: the commutator is now $\pm 2\mathbb{J}$ (as a matrix) or in fact the antisymmetric form $\endgroup$
    – Willie Wong
    Commented Apr 23, 2011 at 21:56
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    $\begingroup$ induced by $2g\mathbb{J}$. In this sense the computation is coordinate invariant and may admit a nice geometrical interpretation. $\endgroup$
    – Willie Wong
    Commented Apr 23, 2011 at 21:57

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I have met the combination of complex structures, symplectic forms, scalar products on the same real vector space in the following context.

Let $V$ be a real vector space.
The complex vector space structures compatible with the assigned real vector space structure of $V$ are in correspondence one-to-one with the complex operators on $V$, i.e. the linear operators $J$ on $V$ such that $\mathbb{J}^2=-id_V\equiv \mathbb{I}$; this correspondence is realized through the relation $(a+ib).v=av+b\mathbb{J}v$, for any $a,b\in\mathbb{R}$, and $v\in V$.
Obviously such structures $(V,\mathbb{J})$ exist if and only if $\dim{V}$ is pair.

Definition. Let $\Omega$ be a symplectic form and $\mathbb{J}$ a complex operator on $V$. The complex operator $\mathbb{J}$ is said to be adapted to $\Omega$ when there exists a pseudo-hermitian form $\eta$ on the complex vector space $(V,\mathbb{J})$ such that $\Im\eta=\Omega$.

Theorem $\mathbb{J}$ is adapted to $\Omega$ if and only if $\mathbb{J}$ is an isomorphism of $(V,\Omega)$, i.e. $\mathbb{J}^T\circ\Omega^\flat\circ\mathbb{J}=\mathbb{J}\circ\Omega\in L(V,V^*)$; and in the affirmative case there is a unique hermitian form $\eta$ on $(V,\Omega)$ with $\Im\eta=\Omega$, it is given by $\eta=(\Omega^\flat\circ\mathbb{J})^\sharp+i\Omega$.

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  • $\begingroup$ This seems to go in my direction. I'll try to think about it. $\endgroup$
    – tomate
    Commented Apr 24, 2011 at 10:40

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