Timeline for Ergodicity of the action of $\operatorname{SL}(n,\mathbb R)$ on $\operatorname{SL}(n,\mathbb R)/\operatorname{SL}(n,\mathbb Z)$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 30, 2022 at 3:43 | comment | added | LSpice | @YCor's comment. | |
| Mar 20, 2022 at 17:29 | comment | added | No One | I believe I have found a proof based on the quotient integral formula of Haar measure (nothing about smooth manifold is used!), which should at least answer the situation of OP for special linear groups. It is posted here math.stackexchange.com/a/4408649/967709 Do you think you could take a look to see if it is correct? Thanks! | |
| Mar 20, 2022 at 4:13 | comment | added | Calamardo | I mean that if you pull back the form to any chart, you get something that looks like $f(x_1,\dots,x_n)dx_1\wedge\dots \wedge dx_n$ where $f$ is nonvanishing. Theorem $8.21$ in Knapp's Lie groups 2ed book has an explanation for why the Haar measure is given by such a form. | |
| Mar 20, 2022 at 3:50 | comment | added | No One | What do you mean by "nonvanishing top forms"? | |
| Mar 20, 2022 at 3:36 | comment | added | Calamardo | I'm thinking of the local diffeomorphism $g\mapsto gx$. The Borel measures on $G$ and $G/\Gamma$ are induced by nonvanishing top forms. So the restrictions of these measures to any chart will be absolutely continuous wrt the Lebesgue measure. And then I imagine that null sets are sent to null sets along the lines of this argument - math.stackexchange.com/questions/59105/… | |
| Mar 20, 2022 at 1:48 | comment | added | No One | Could you add more details explaining why $Ux$ has full measure in $X$? ("measure generic" in your language) Sorry I don't see why this is obvious. Thanks! | |
| S Mar 16, 2022 at 16:04 | history | answered | Calamardo | CC BY-SA 4.0 | |
| S Mar 16, 2022 at 16:04 | history | made wiki | Post Made Community Wiki by Calamardo |