I'd like to know if there exists a connected Lie group $G$ and a closed manifold $M$ such that there is a locally-free smooth action $G\times M\to M$ (i.e. the stabilizer of any point of $M$ is a discrete subgroup of $G$) with no invariant (Borel) probability measure.
$\begingroup$
$\endgroup$
8
-
$\begingroup$ By closed, do you mean "compact"? $\endgroup$– Igor RivinDec 7, 2011 at 12:45
-
1$\begingroup$ @Alejandro: What about the flow of a nowhere vanishing vector field on the $3$-sphere that preserves two circles, say $C_1$ and $C_2$, and has the property that every flow line $\alpha$-limits to $C_1$ and $\omega$-limits to $C_2$. Then $G=\mathbb{R}$ and the stabilizer of a point is either trivial (for points off the two circles) or a discrete subgroup of $G$ (for points on the two circles). This won't preserve any continuous volume form on the $3$-sphere, but I don't know about rougher Borel measures. $\endgroup$– Robert BryantDec 7, 2011 at 13:30
-
$\begingroup$ @Robert: Nice example! Indeed by amenability of $\mathbb{R}$ it {\it will} preserve some Borel probability measure on $S^3$. $\endgroup$– Alain ValetteDec 7, 2011 at 14:36
-
$\begingroup$ @Alain: Thanks, but, as you point out, attempts with $G=\mathbb{R}$ are doomed to fail. (Besides, I realized while I was running this morning that there's an even easier $2$-dimensional example of such an action without a smooth invariant volume form based on using two parallel circles on the torus.) I guess the simplest example one could hope for would be to have $G$ be some covering of $PSL(2,\mathbb{R})$ and to have $M$ have dimension $4$. $\endgroup$– Robert BryantDec 7, 2011 at 16:21
-
$\begingroup$ @Robert: is there any reason why you suggest a covering of PSL(2,R) and not PSL(2,R) itself? $\endgroup$– Łukasz GrabowskiDec 7, 2011 at 23:17
|
Show 3 more comments
1 Answer
$\begingroup$
$\endgroup$
7
Let $\Gamma$ be a co-compact lattice in $G=SL_2(\mathbb{R})$, acting linearly on the real projective line $P^1(\mathbb{R})$. Take $M=G\times_\Gamma P^1(\mathbb{R})$, a 4-dimensional closed manifold which is a circle bundle over $G/\Gamma$ (notation $\times_\Gamma$ means we divide out by the diagonal action of $\Gamma$). The left $G$-action is locally free, as stabilizers are conjugate into $\Gamma$. Moreover, since the groupoid $G\ltimes M$ is equivalent to the groupoid $\Gamma\ltimes P^1(\mathbb{R})$ (the former being induced up from the latter), it is enough to check that there is no $\Gamma$-invariant Borel measure on $P^1(\mathbb{R})$, which is classical (see e.g. Cor. 3.2.2 in Zimmer's `Èrgodic theory and semisimple groups'', Birkhauser, 1984).
-
2$\begingroup$ I can't figure out how does $G$ act on $M$? On the other hand, the bare product $(G/\Gamma)\times\mathbb{P}^1(\mathbf{R})$ carries an action of $G$ and works. $\endgroup$– YCorDec 8, 2011 at 17:49
-
$\begingroup$ For me too, it isn't clear how $G$ acts on $M$, but as Yves mentioned above, Alain's argument seems to work for the diagonal left $G$-action on $(G/\Gamma)\times\mathbb{P}^1(\mathbb{R})$. So I'll consider my question as answered... $\endgroup$ Dec 8, 2011 at 21:19
-
2$\begingroup$ @Yves: OK, maybe I wasn't specific enough: $G$ is acting by left multiplications on the first factor, trivially on the second factor of $G\times P^1(\mathbb{R})$, so that the action descends to $M$. $\endgroup$ Dec 8, 2011 at 21:20
-
$\begingroup$ @Alain: if you consider the trivial action on the second factor, I cannot figure it out how the non-existence of $\Gamma$-invariant measures (for the projective action) appears in your argument. $\endgroup$ Dec 8, 2011 at 21:37
-
1$\begingroup$ It may be interesting to observe that, as a consequence of Prop. 4.2.22 in Zimmer's book, Yves'example and mine are measurably $G$-isomorphic. $\endgroup$ Dec 9, 2011 at 11:04