1
$\begingroup$

For Ratner's orbit closure theorem, one may refer to the following Wikipedia page.

Let $\{u_t\mathrel: t\in \mathbb{R} \}\subset G$ be a unipotent one-parameter subgroup of a connected Lie group. Let $\Gamma$ be a lattice of $G$. From the theorem, we know that the closure of every full orbit $\{x\cdot u_t\mathrel: t\in \mathbb{R}\}$ is dense in the orbit $xS\subset \Gamma\backslash G$ of a certain subgroup $S$ of $G$.

Do we also have the density of the semigroup orbit $\{x\cdot u_t\mathrel: t\geq 0\}$ in certain orbits of subgroups? Is $\{x\cdot u_t\mathrel: t\geq 0\}$ just dense in $xS$ as well?

Improve this question
$\endgroup$
2
  • $\begingroup$ I guess, one should look at the proof to see whether it uses negative times. $\endgroup$
    – ThiKu
    Commented Jul 15, 2020 at 6:30
  • $\begingroup$ An easier special case (with the full Borel group rather than just U) is discussed in Section 2.2 of perso.ens-lyon.fr/ghys/articles/dynamiqueflots.pdf . In the cocompact case, density is reduced to density of $\Gamma$ in $G/B$, so one may look at $G/B^+$ and wonder whether density of $\Gamma$ still holds. $\endgroup$
    – ThiKu
    Commented Jul 15, 2020 at 6:45

1 Answer 1

Reset to default
2
$\begingroup$

$\DeclareMathOperator\supp{supp}$The theorem holds for semigroups as well (well, in the finite volume setting! in the infinite volume setting there are subtleties between two-sided and one-sided averages).

Assume $\mu$ is an $S$-invariant and ergodic probability measure, where $S$ is a semigroup inside a one-parameter unipotent subgroup $U$. We will show that $\supp(\mu)$ contains a full $U=\langle S\rangle$ orbit, that's enough. Pick some $x\in \supp(\mu)$. Consider $\overline{S\cdot x}=P$. For a generic point $x$, $P=\supp(\mu)$ by the ergodic theorem. Notice that $S\cdot P\subset P$, moreover $S^{2}\cdot P=S\cdot P$, so this is an $S$-invariant subset. By ergodicity $\mu(S\cdot P)=1$, but we also have that $\mu(P)=1$ as well, so $P=S\cdot P$ up to a measure zero set, or in other words $S^{-1}\cdot P=P$ as well, hence $P$ is invariant under $S\cup S^{-1}$, now you may apply the orbit closure theorem.

P.S. you might want to consider the paper by Nimish Shah - "Invariant measures and orbit closures on homogeneous spaces for actions of subgroups generated by unipotent elements" (MSN) in Lie groups and ergodic theory, where he discusses the move from Ratner's theorem to discrete subgroup actions!

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ I did some light proofreading, which I hope is OK. One thing that I didn't change is that you refer to a semigroup $S$ inside a unipotent group $U$, and later say $U = \langle S\rangle$. I think you meant to replace $U$ by the group generated by $S$, but I wasn't sure if this was instead an additional hypothesis. $\endgroup$
    – LSpice
    Commented Aug 7, 2020 at 22:45
  • $\begingroup$ @LSpice, thanks, I meant $S$ is the positive semigroup inside $U$, which is a one-parameter unipotent group. $\endgroup$
    – Asaf
    Commented Aug 8, 2020 at 0:30
  • $\begingroup$ @LSpice, he meant the full-positive semigroup. Not a discrete one. Dealing with a discrete subgroup/group situation is a bit more complicated (and needs a suspension construction, as in Shah's article I have mentioned). $\endgroup$
    – Asaf
    Commented Aug 8, 2020 at 0:33
  • $\begingroup$ I don't think I mentioned discreteness, and I didn't have it in mind, so I'm not sure to what that is responding (although certainly I believe you that there is more complication involved). Your first comment is what I meant. $\endgroup$
    – LSpice
    Commented Aug 8, 2020 at 1:17
  • $\begingroup$ @LSpice, it is clear that for $\mathbb{R}$, the group generated by the positives (or say any non-discrete semigroup) is the whole group. I suspected that you might have considered a discrete semigroup like $\mathbb{N}$. $\endgroup$
    – Asaf
    Commented Aug 8, 2020 at 2:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .