Are all orbits semi-Riemannian submanifolds?

Question

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.

Answer 1

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

Answer 2

It is true for very special classes of actions and a dense subset of orbits. For instance, in the linear case, for a semisimple semi-Riemannian symmetric space $G/H$, consider the linear isotropy action of the isotropy group $H$ at the base point $x_0=1H$ on the tangent space $T_{x_0}(G/H)$. Then every closed $H$-orbit is semi-Riemannian; in this case the union of closed orbits is an open and dense subset of the ambient space, but it may consist of several components on each of which the signature is different. The case of the adjoint representation of $SL(2,\mathbb R)$ on its Lie algebra is included in this class (http://arxiv.org/pdf/0801.0574.pdf).