Timeline for regarding metric and symplectic forms
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2011 at 10:38 | comment | added | tomate | I start to get the point. What if I write $h(v,w) = g(v,w) + c[g(v,Jw) + g(Jv,w)]$. Would that make a little bit more sense? | |
| Apr 23, 2011 at 21:57 | comment | added | Willie Wong | induced by $2g\mathbb{J}$. In this sense the computation is coordinate invariant and may admit a nice geometrical interpretation. | |
| Apr 23, 2011 at 21:56 | comment | added | Willie Wong | 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 | |
| Apr 23, 2011 at 21:52 | comment | added | Willie Wong | 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 | |
| Apr 23, 2011 at 19:21 | answer | added | agt | timeline score: 1 | |
| Apr 23, 2011 at 12:39 | comment | added | tomate | It is true that I'm working in a preferred basis, where the commutator is just te matrix commutator $[g,\mathbb{J}] = g \mathbb{J} - \mathbb{J} g$, and $\mathb{J}^2 = -\mathbb{I}^T$, but for sake of greater generality I don't see any obstruction to having $g$ a symmetric form, say $\Omega$ an antisymmetric form, and define $h=g - c[g,\Omega]$, requiring it to be positive. It is hard to go into the detail of how I got there and it's not really relevant. I tried to put the question into the most generic form possible, in the hope to find similar objects and some geometrical insight. Thanks! | |
| Apr 23, 2011 at 12:07 | comment | added | Willie Wong | (Also, can you give the context in which you encountered this object?) | |
| Apr 23, 2011 at 12:06 | comment | added | Willie Wong | 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$? | |
| Apr 23, 2011 at 9:11 | history | asked | tomate | CC BY-SA 3.0 |