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Nondegenerate Poisson bracket and evendimensional manifold 
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#1
Oct1704, 03:01 PM

P: 1,235

From this reference:
http://www.amazon.com/exec/obidos/tg...glance&s=books titled From Classical to Quantum Mechanics, I quote the following: ( [tex]\xi^i [/tex] are coordinate functions) Let M be a manifold of dimension n. If we consider a nondegenerate Poisson bracket, i.e. such that [tex]\{\xi^i,\xi^j\} \equiv \omega^i^j[/tex] is an inversible matrix, we may define the inverse [tex]\omega_i_j[/tex] by requiring [tex]\omega_i_j \omega^j^k = \delta_i^k[/tex] We define a tensorial quantity [tex]\omega \equiv \frac{1}{2}\;\omega_i_j \; d\xi^i \wedge d\xi^j[/tex] which turns out to be a nondegenerate 2form. This implies that the dimension of the manifold M is necessarily even. My questions are the following: I dont understand the two statement that I have put in red above. What is a nondegenerate 2form? Why does this one above 'turns out' to be nondegenerate? Why does that imply that M is even? Additional comments would be welcome. Like concerning the meaning of [tex]\omega [/tex] above. In addition, I guess the point here by a shorter way: I think that all odddimensional antisymmetric matrices are singular. Is there a link with the language used above? Warm thanks in advance, Michel 


#2
Oct2104, 09:50 AM

Sci Advisor
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P: 9,396

A nondegenerate two form is essentially a nondegenerate symplectic form on the tangent bundle at all points. This implies that the tangent bundle has even dimension. Which is what the things in red are saying.
given any twoform, it is not necessarily nondegenerate, just as any symplectic form is not necessarily nondegenerate. 'it turns out' means 'in this case with these hypotheses we can prove it is' 


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