If G is a Lie group acting on a manifold M through $\Psi$, one can argue that orbit of any $p\in M$ is an integral submanifold of the generators of group action. Roughly the proof is :

1) Fixing p first one proves that the orbit is a submanifold through following commuting diagram:

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$G_p$ being the isotropy group of p. $\pi$ is a sumbersion and $\widetilde{\Psi^p}$ a immersion. Thus the image of $\mathfrak{g}$ under $d\Psi^p$ covers the tangent space of $O_p$.

2)Also \begin{equation} d\Psi^p(X_g)=d\Psi^p\circ dL_g(X_e)=d(\Psi^p\circ L_g)(X_e)= d\Psi^{g.p}(X_e)= \frac{d}{d\epsilon}\Big|_0\Psi(exp(\epsilon\textbf{v}),x) \end{equation}

The curve $\Psi(exp(\epsilon\textbf{v}),x)$ is in the orbit, and hence $d\Psi^p(X_g)$ is in the tangent space of the orbit.

The above two facts establish that orbit is integral manifold of generators of group action.

My question is when can one say that the orbits are maximal integral manifolds.

I have a feeling that oribits will be maximal integral manifolds if G is connected else the integral manifold through p will be a subset of the orbit of p.

  • $\begingroup$ What is your definition of " integral manifolds." $\endgroup$ – Thomas Feb 24 '16 at 17:28
  • $\begingroup$ The generators give a involutive distribution and any submanifold whose tangent space at each point is same as the subspace spanned by the Distribution at that point is being called the integral manifold. $\endgroup$ – Himanshu Feb 25 '16 at 15:55

Your assertion comes the fact that $M$ is the disjoint union of maximal integral manifolds and of orbits. A maxiam integral manifold $N$ satisfies the following : if $p\in N$ its orbit is also in $N$, as an open sub-manifold, (open for the manifold topology). So $N$ is the disjoint union of orbits, these orbits are open. By connectedness $N$ is an orbit.

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  • $\begingroup$ "A maxiam integral manifold N satisfies the following : if p∈N its orbit is also in N" - why should this be true ? (the definition of orbit i am taking is Op = G.p) $\endgroup$ – Himanshu Feb 26 '16 at 17:23
  • $\begingroup$ There another notion of orbit i see here projecteuclid.org/download/pdf_1/euclid.bams/1183534326, and your assertion holds true for this definition. Infact the theorem 2 of same link is the exact statement. But I was curious about what happens if we take the definition from my previous comment. $\endgroup$ – Himanshu Feb 26 '16 at 17:27
  • $\begingroup$ an orbit is tangent to the to the distibution, so included in integral manifolds. $\endgroup$ – Thomas Feb 26 '16 at 18:02
  • $\begingroup$ yes, but integral manifold of a point is the maximal connected set. what reason do we have to beleive that orbit of that point is also connected ? $\endgroup$ – Himanshu Feb 28 '16 at 13:12
  • $\begingroup$ If you assume the Lie group is connected, as the image of a connected space by the map $g\to g.p$ $\endgroup$ – Thomas Feb 28 '16 at 16:26

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