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It is not quite intuitive , at least from its origin. Could any one can give me an intuitive explanation?Thank you!

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4 Answers

Any two of (a) symplectic structure, (b) almost-complex structure, and (c) Riemannian structure determine the third. This reflects the fact that $$ U(n) = O(2n) \cap GL(n,\mathbb{C}) \cap Sp(2n),$$ but the intersection of any two is already $U(n)$. This gives rise to the theory of Kähler manifolds (cf. wikipedia), which are central to mirror symmetry (and the mirror symmetry conjecture).

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I suppose a question to ask here is : what kinds of applications?

One of the areas where symplectic geometry is used is Hamiltonian dynamics, where the geometric framework allows one to generalize away from cotangent bundles, but this direction is directly related to the origins of the subject.

Another direction comes from algebraic geometry, where the fact that complex and Riemannian structure determine a symplectic one (see Aaron's answer) makes all complex projective (and affine) algebraic varieties symplectic. Viewing them as such allows one to use differential-geometric methods, while the symplectic form provides some control (in the theory of holomorphic curves, symplectic form gives the crucial energy bounds without which things fall apart; this is why there is no Gromow-Witten theory of almost-complex manifolds). Add a bit of mirror symmetry to the mix and you have interaction with many other areas of math.

There is also the fact that the wedge product is anti-symmetric, which leads the space of connections on a principal bundle over 2-d Riemannian manifold to be an (infinite dimensional) symplectic manifold (the gauge group is Hamiltonian, and curvature is the moment map, leading to flat connections being the reduction!). This leads, for G=SO(3) and the surface coming from a Heegard splitting, to Atiyah-Floer conjecture and some related stuff leads to Heegard-Floer theory. Thus connections to low-dimensional topology. The "naive" connection here is the anti-symmetry underlying symplectic form and wedge product, but probably there is a deeper reason known to wise people...

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Symplectic geometry is applicable in lot of areas such as Classical mechanics, Hamiltonian mechanics. While browsing through the web i have found a reason as to why SG is good for classical mechanics. You may read it here: http://research.microsoft.com/en-us/um/people/cohn/thoughts/symplectic.html

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I think the question is more about why symplectic geometry has found applications beyond classical mechanics. –  Qiaochu Yuan Sep 18 '10 at 5:34
    
@Qiaochu Yuan: I answered the question, w.r.t my thinking. –  anonymous Sep 18 '10 at 7:21
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There is a simple "philosophical" answer: Symplectic forms allow you to only measure two-dimensional quantities, not one-dimensional ones (you can measure area infinitesimally). That's the basic difference with Riemannian geometry, where you can measure length infinitesimally.

Furthermore symplectic geometry pops up in connection with the cotangent bundle. For instance if you have a (scalar) first order differential operator on a manifold, then the principal symbol transforms like a covector (when changing coordinates). Note that this does not require a metric.

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