Existence of at least one compact orbit on the sphere

Question

Good morning,

I've came across this question during my researches. It seems apparently very simple, however I Googled a bit and I couldn't find the answer I was looking for.

The question is as follows: Is it true that any connected subgroup of $GL(d,\mathbb{R})$ admits at least one compact orbit on the sphere $\mathbb{S}^{d-1}$ (or, if you prefer, on the projective space $\mathbb{P}(\mathbb{R}^d))$?

Well, of course this is true if the subgroup is of the form $e^{tA},t\in\mathbb{R}$, but for more complicated subgroups I don't know. I guess that the main point that prevents the use of standard methods is that the subgroup is not, in general, (Zariski) closed. Thus many powerful results are not available. Nonetheless, I'm still guessing that the answer should be yes, but I cannot prove it.

If you can provide me with references, or showing me counter-examples, that would be great!

Thanks in advance for your kindness.

Regards

EDIT: in response to the comments below, I would like to narrow a bit the question.

In fact, what I am really interested in is the following setting: take any Lie subalgebra $\mathfrak{a}$ of $\mathfrak{gl}(d,\mathbb{R})$, and consider the subgroup $G$ of $GL(d,\mathbb{R})$ generated by $\mathfrak{a}$, so to avoid nasty possibilities. Nonetheless, $G$ need not be algebraic.

Also, for any $g\in G$, the action on the sphere is intended to be $x\mapsto \frac{gx}{\|gx\|}$, for any $x\in\mathbb{S}^{d-1}$.

Answer 1

Yes, it is true that for every continuous real linear finite dimensional representation of a connected Lie group there exists a closed orbit for the associated projective representation.

To see this it is enough to assume the group is solvable (as every connected Lie group has a cocompact solvable subgroup) and the representation is irreducible and faithful. It follows that the unipotent radical of the Zariski closure is trivial, thus our group is diagonalizable and the representation is either 1 or 2-dimensional. The 1-dim case is trivial and in the 2-dim case we get that the action on the projective line is transitive.

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Here are some more details.

Let $A$ be the connected, simply connected Lie group associated with the Lie algebra $\mathfrak{a}$. The inclusion map $\mathfrak{a}\to \mathfrak{gl}_d(\mathbb{R})$ gives rise to a continuous homomorphism $A\to \text{GL}_d(\mathbb{R})$. The connected group alluded to in the (edited) question is the image of $A$ under this continuous homomorphism. Thus we need to show that for every linear representation of a connected Lie group, the associated projective action has a closed orbit.

It is enough to show that a cocompact subgroup of $A$ has a closed orbit, as a cocompact subgroup (the stabilizer of a point in this orbit) of a cocompact subgroup is cocompact. Now, use the fact that every connected Lie group has a cocompact solvable subgroup. This is a standard fact (by factoring the solvable radical it is enough to observe this fact for adjoint semi-simple Lie groups, which are in fact real algebraic groups, then it is reduced to Borel's fixed point theorem, which is nothing but the "Zariski closed version of this question" which you seem to be aware of). We thus need to prove the claim: for every linear representation of a connected solvable group, the associated projective action has a closed orbit. Obviously, it is enough to assume the representation is irreducible.

We thus need to prove the claim: for every continuous irreducible finite dimensional real linear representation of a connected solvable group $A$, the associated projective action has a closed orbit. Irreducible representations of solvable groups are either one or two dimensional. This follows for example from Lie-Kolchin (for an invariant line $\mathbb{C}\cdot v$ in the complexification, take the space generated by $v+\bar{v}$ and $v-\bar{v}$). In the 1-dim case the corresponding projective space is a singleton, and there is nothing to prove. in the 2-dim case the projective space is a topologically a circle. We are done by noting that the action of a connected group on a circle is either transitive or have a fixed point. Indeed, its orbits are homogeneous connected subsets, thus points or open intervals, and in the latter case any end point of the interval will be fixed.