regarding metric and symplectic forms

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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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$? – Willie Wong Apr 23 '11 at 12:06 (Also, can you give the context in which you encountered this object?) – Willie Wong Apr 23 '11 at 12:07 What I am trying to get you to realise is that the fact$\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}$, and the matrix commutator of – Willie Wong Apr 23 '11 at 21:52 any matrix$\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 – Willie Wong Apr 23 '11 at 21:56 induced by$2g\mathbb{J}$. In this sense the computation is coordinate invariant and may admit a nice geometrical interpretation. – Willie Wong Apr 23 '11 at 21:57 1 Answer 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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This seems to go in my direction. I'll try to think about it. –  tomate Apr 24 '11 at 10:40