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How can one tell whether a given (finite-dimensional) symplectic manifold is homogeneous (that is, admits a transitive group of symplectomorphisms)?

Note that it is of little help to observe that a homogeneous symplectic manifold must be a covering space of a coadjoint orbit of some Lie group, since this provides no effective test applicable to a given example. Note also that in the Riemannian case, there is a well-known local obstruction to the existence of a transitive group action by sutomorphisms: the covariant derivative of the curvature must vanish. But since all symplectic manifolds of dimension n are locally symplectomorphic, there can be no such obstruction in this case, and any analogous constraint must be simultaneously sensitive to the symplectic form and the topology of the manifold.

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  • $\begingroup$ Do you have any reason to believe that there is such a test? $\endgroup$ – Igor Rivin Dec 7 '11 at 18:38
  • $\begingroup$ As Shlomo Sternberg once pointed out to me, Bott's Morse-theoretic investigations of the topology of semi-simple Lie groups imply strong constraints on the homology of their coadjoint orbits. When these constraints are violated, one can at least rule out the existence of a transitive action by a semisimple symplectomorphism group. (Oh, and of course I meant to ask about finite-dimensional transitive symplectomorphism groups; please excuse the oversight.) $\endgroup$ – Laurens Gunnarsen Dec 7 '11 at 21:04
  • $\begingroup$ Perhaps, if you precisely define what you are going to accept as defining a 'given' symplectic structure and what you will accept as an 'effective' test, someone will be able to help. Obviously, there are necessary conditions (for example, if the dimension is two, I think we can pretty completely tell you whether it's homogeneous or not, provided you can effectively determine from the 'givens' things such as whether it's compact or not, what it's fundamental group is, etc.) $\endgroup$ – Robert Bryant Dec 8 '11 at 4:46
  • $\begingroup$ In the long run, I'm hoping for an elegant intrinsic characterization of the covering spaces of coadjoint orbits. In the short run, I'd be happy with any non-trivial theorem whose conclusion is "then M is symplectomorphic to a covering space of a coadjoint orbit." In particular, in my present state of ignorance, I don't mind assuming that M is compact, that its fundamental group is nice, that it has vanishing odd-dimensional homology, and whatever else is needed to evade any already-evident obstructions. My impression is that no result of the form I've indicated is now known. $\endgroup$ – Laurens Gunnarsen Dec 8 '11 at 6:36
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    $\begingroup$ @Laurens: Oh, by the way, it is not true that the condition for a Riemannian metric to be homogeneous is that the covariant derivative of the curvature vanish; that's the condition for the Riemannian metric to be (locally) symmetric, which is a much stronger condition than homogeneity. For example, any left-invariant metric on a $3$-dimensional Lie group is homogeneous, but the vast majority of them are not symmetric. $\endgroup$ – Robert Bryant Dec 8 '11 at 12:55
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See

How transitive are the actions of symplectomorphism groups ?

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  • $\begingroup$ This link points to a discussion of infinite-dimensional symplectomorphism groups; my interest is in the finite-dimensional case. $\endgroup$ – Laurens Gunnarsen Dec 8 '11 at 3:53

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