Let $M$ be a semi-Riemannian manifold and $G\subset Iso(M)$ a closed connected Lie subgroup which acts properly on $M$. It is known that every orbit of the action is a (closed) submanifold of $M$. My question is that is every orbit of the action a **semi-Riemannian** submanifold of $M$? precisely, for any orbit $O=G(x)$, $x \in M$, is it true that all subspaces $T\_yO\subset T\_yM$, $y\in O$, have same casual character? ($T\_yO$ and $T\_yM$ are tangent space to $O$ resp. $M$ at the point $y$)  

Thank you.



Are all orbits semi-Riemannian submanifolds?
No!


Consider the adjoint action of $SL(2,\mathbb R)$ on its Lie algebra with the Killing form,
which is $\mathbb R^{1,2}$. The orbits are:


* Non-closed: future light cone, past light cone. 0 is in the closure of both.
* Closed: 1-sheeted hyperboloids (signature $1,1$)
* Closed: 1 sheet of the two-sheeted hyperboloids (Riemannian)


This action is not proper, so go over to $SO(2)\subset SL(2,\mathbb R)$. This keeps invariant also a Euclidean metric, so the orbits are intersections of the above orbits with spheres.
Those in the lightcone are null-lines.


The following paper explains this in many details:


* Hilgert, Joachim; Hofmann, Karl H. Old and new on Sl(2). Manuscripta Math. 54 (1985), no. 1-2


