Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For defining the symplectic space $(V, \omega)$ where $V$ is a vector space, it doesn't necessary to add the condition $d\omega=0$. But, when we work on a manifold instead of vector space, then we need $\omega$ be closed. So why?

share|improve this question
1  
I guess that "symplectic vector space" and "symplectic manifold" are two different notions. In particular not every symplectic vector space is a symplectic manifold. –  user10001 Jun 28 '13 at 15:55
    
Thanks @user10001 , can you give an example for more effectiveness of your comment. In fact every vector space is a manifold, so your example would be good? –  hassan joulani Jun 28 '13 at 16:09
1  
Now when you are asking for an example I think my statement is not correct;) It should be other way around - every symplectic vector space is also a symplect manifold. Doesn't the condition $d\omega=0$ follow from bilinearity? Since matrix elements of $\omega$ will have no dependence on coordinates. –  user10001 Jun 28 '13 at 16:31
2  
Yes, of course, thinking of $V$ as a manifold, the derivative of a constant $2$-form is $0$. Note that the Darboux Theorem says that on any symplectic manifold there are local coordinates in which the symplectic form becomes the standard symplectic form $\sum_{i=1}^n dx_i\wedge dy_i$ on $\mathbb R^{2n}$. –  Ted Shifrin Jun 28 '13 at 16:45
    
We have also a theorem which says :Every symplectic manifold $M$ is locally isomorphic to a symplectic vector space $V$ –  hassan joulani Jun 28 '13 at 16:58

2 Answers 2

up vote 2 down vote accepted

Suppose $\omega$ be a bilinear form on a vector space $V$. Then consider a basis $e_1,...,e_n$ of $V$ wrt to which coordinates of points are $x_1,...,x_n$. Wrt these coordinates we can define one forms $dx_i,i=1,...,n$ and vector fields $\partial_i,i=1,...,n$ in the usual way.

Now

$\omega(a_1\partial_1+\dots a_n\partial_n,b_1\partial_1+\dots+b_n\partial_n)=\displaystyle\sum_{i,j}a_ib_j\omega_{ij}\tag 1$

where $\omega_{ij}=\omega(\partial_i,\partial_j)$ are constants independent of a's and b's.

From (1)

$\omega=\displaystyle\sum_{i,j}\omega_{ij}dx_i\otimes dx_j$

So $d\omega=0$

share|improve this answer

When you have an alternating tensor on a vector space, there's nothing to differentiate. The point of closedness of the symplectic $2$-form on a manifold is that it therefore represents a cohomology class and induces a generator of the top cohomology (in the compact case). [My view, as a complex geometer, is that the symplectic form is quite analogous to the Kähler form in Kähler geometry.] Most of the interplay with the Poisson bracket and Lie derivatives depend on closedness, as well.

share|improve this answer
    
Ted Shifrin@ what do you mean of top cohomology? –  hassan joulani Jun 28 '13 at 16:02
2  
If $\dim M = 2n$, then I mean $H^{2n}(M,\mathbb R)$. –  Ted Shifrin Jun 28 '13 at 16:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.