Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi to all,

Let $G$ be a Lie group of linear isometries of $\mathbb{R}^n_{\nu}$ ($\mathbb{R}^n_{\nu}$ is the semi-Euclidean space) and $G_1$ ,$G_2$ two Lie subgroups of $G$. Let $G_1 \times G_2$ as a Lie subgroup of $G$ by identifying it with the inner product of $G_1$ and $G_2$ (suppose $G_1$ , $G_2$ have the required properties to form inner direct product) and let $G_1 \times G_2$ acts on $\mathbb{R}^n_{\nu}$ by the induced action from $G$ . What can we say about the relation between the orbits of $G_1 \times G_2$ and the orbits of $G_1$ and $G_2$? Is a $(G_1 \times G_2)$-orbit isometric (or at least diffeomorphic) to the product of the $G_1$-orbit and $G_2$-orbit?

Best for you.

share|improve this question
Just to be clear: Are you assuming that $G_1$ lies in the subgroup of $G$ that consists of the elements that commute with $G_2$ and, vice versa, that $G_2$ lies in the subgroup of $G$ that consists of the elements that commute with $G_1$? Are you also assuming that $G_1\cap G_2$ is the identity subgroup of $G$? –  Robert Bryant Jan 8 '13 at 18:15
Yes, these hypotheses are assumed. –  purelymath Jan 8 '13 at 21:57

1 Answer 1

up vote 1 down vote accepted

If all actions are proper, orbits are closed embedded submanifolds. If not, then they are only initial submanifolds (unique injectively immersed submanifolds with a possibly finer topology - think of a 1-param. subgroup of the torus with irrational slope; see thm 6.4 in (here))

Now the $(G\times G_1)$-orbit through $x$ is diffeomorphic to $(G_1\times G_2)/(G_1\times G_2)_x$ (mod out the isotropy group). If $(G_1\times G_2)_x = (G_1)_x\times (G_2)_x$ then the orbit is a product. We always have $(G_1\times G_2)_x \supset (G_1)_x\times (G_2)_x$ so the $(G\times G_1)$-orbit through $x$ is always a submersive quotient of the product of the two orbits.

The source I cited has more information.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.