Let $(M,\omega)$, be pre-symplectic, then can we say, we have a foliation of $M$, with tangent spaces $ker\omega$.What can we say about its trajectories. ?
1 Answer
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If your definition of pre-symplectic involves constant rank and $\ker\omega$ defines a smooth distribution of hyperplanes in the tangent bundle, then an application of the Frobenious theorem shows that you have a foliation tangent to $\ker \omega$. I'm not sure that one can say more than that in general.
answered Jun 30, 2013 at 23:53
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1$\begingroup$ Seems like a full answer to me. Only obstruction to the characteristic distribution (ker($\omega$) being an honest distribution is that its rank jumps. As Igor says, if this rank is const. , then it is involutive as a distribution. $\endgroup$ Commented Jul 1, 2013 at 5:05
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1$\begingroup$ Even if the rank is not constant, still the kernel distribution is integrable. I first read this in the unpublished math.unipd.it/~bottacin/papers/presympred.pdf $\endgroup$ Commented Jul 1, 2013 at 6:49
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1$\begingroup$ @Nicola, thanks. That's very interesting! But in general it is integrable only in a generalized sense: some integral curves may terminate, that is, they are no longer inextensible in the manifold. $\endgroup$ Commented Jul 1, 2013 at 7:53
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1$\begingroup$ Well integrable to me means to me each point is contained in a maximal integral submanifold, which does not imply each "Hamiltonian" flow can be indefinitely extended. This happens also for the symplectic foliation of Poisson manifolds. $\endgroup$ Commented Jul 1, 2013 at 8:28
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1$\begingroup$ Hassan, it's possible you're missing the connection. When the dimension of $\ker \omega$ is larger than 1, a leaf of the foliation is the generalization of the $1$-dimensional notion of a trajectory. $\endgroup$ Commented Jul 1, 2013 at 10:33