# locally-free Lie group action not preserving any measure

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.

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By closed, do you mean "compact"? –  Igor Rivin Dec 7 '11 at 12:45
@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. –  Robert Bryant Dec 7 '11 at 13:30
@Robert: Nice example! Indeed by amenability of $\mathbb{R}$ it {\it will} preserve some Borel probability measure on $S^3$. –  Alain Valette Dec 7 '11 at 14:36
@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 $4$. –  Robert Bryant Dec 7 '11 at 16:21
@Robert: is there any reason why you suggest a covering of PSL(2,R) and not PSL(2,R) itself? –  Łukasz Grabowski Dec 7 '11 at 23:17
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).
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. –  Yves Cornulier Dec 8 '11 at 17:49
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... –  Alejandro Dec 8 '11 at 21:19
@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 $M$. –  Alain Valette Dec 8 '11 at 21:20
@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. –  Alejandro Dec 8 '11 at 21:37
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. –  Alain Valette Dec 9 '11 at 11:04