Timeline for Ergodicity of the action of $\operatorname{SL}(n,\mathbb R)$ on $\operatorname{SL}(n,\mathbb R)/\operatorname{SL}(n,\mathbb Z)$
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25 events
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| Mar 20, 2022 at 10:42 | comment | added | YCor | @question in such a case $G/\Gamma$, $G$ second countable, $\Gamma$ discrete, indeed a subset of $G$ is measure-generic iff its inverse image is measure-generic (this is easy using a measurable fundamental domain, on which the quotient map is measure-preserving) | |
| Mar 20, 2022 at 1:52 | comment | added | No One | @YCor How do you prove that $Ux$ has full measure in $X$? (I guess that is what you mean by measure generic...) Do we need any theorems of Haar measure or it is straightforward with basic measure theory and topology? | |
| Mar 16, 2022 at 16:04 | answer | added | Calamardo | timeline score: 1 | |
| Mar 16, 2022 at 15:31 | comment | added | YCor | @Fancydressfatima of course details are needed. But it seems that Fubini applied to $|f(gx)-f(x)|$ on $G\times X$ implies that for a.e. all $x$ we have $f(gx)=f(x)$ for a.e. all $g$. Since for given $x$ and every measure-generic subset $U$ of $G$ we have $Ux$ measure-generic in $X$ (for $G$ second-countable and $X$ being $G$ mod discrete this seems quite clear), it follows that $f$ is a.e. constant. | |
| Mar 16, 2022 at 14:54 | comment | added | Calamardo | @YCor I'm scared that it will be wrong, which is why I asked you first. Can you please look at my first comment and see if it looks ok? If so I will post it as an answer. | |
| Mar 16, 2022 at 14:42 | comment | added | YCor | @Fancydressfatima I didn't use the word technicality, and actually I don't guess what you're asking. If you have a proof in mind using Fubini, it would certainly be appreciated that you post it as an answer. | |
| Mar 16, 2022 at 14:13 | comment | added | Calamardo | @YCor Hi, does Fubini as in the above comment account for the technicality mentioned in your first comment? | |
| Mar 15, 2022 at 13:49 | comment | added | Calamardo | Wait, is the assumption that there exists some $f\in L^2(X)$ such that for every $g$, we have $f(gx)=f(x)$ for all $x \in F_g$, some full measure set in X? But then wouldn't Fubini give the existence of a horizontal strip $\{(g,x_0) \in G\times X\}$ such that for almost every $g$, we have $f(gx_0)=f(x_0)$? | |
| Feb 2, 2021 at 20:02 | comment | added | No One | @JHM do you the page number for this proof? Thanks a lot! | |
| Jan 26, 2021 at 20:43 | comment | added | JHM | I think Mostow's book Strong Rigidity of Locally Symmetric Spaces. (AM-78) contains a proof. | |
| Jan 26, 2021 at 19:35 | history | edited | No One | CC BY-SA 4.0 |
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| Dec 20, 2020 at 16:19 | comment | added | YCor | @Asaf I think of Howe-Moore as the assertion that every unitary rep without nonzero invariant vectors is $C^0$. The point here being that it indeed has no nonzero invariant vectors. When the Howe-Moore theorem was proved in the 70s this was obviously considered as obvious old stuff. | |
| Dec 20, 2020 at 16:11 | comment | added | Asaf | @YCor, when I think of Howe-Moore I think about vanishing of matrix coefficients (formed out of say $L^{2}_{0}$ vectors). | |
| Dec 20, 2020 at 15:40 | answer | added | rpotrie | timeline score: 4 | |
| Dec 20, 2020 at 12:44 | comment | added | YCor | The "trivial proof" consists in using transitivity to infer that the only invariant subsets are $\emptyset$ and the whole set. But the acting group is uncountable, so one has to prove a little more (we have to consider subsets that are invariant up to measure zero). Howe-Moore can't be used because it takes ergodicity (absence of nonzero invariant vectors) as an assumption and indeed deduces mixing. I'd guess ergodicity is true for an arbitrary locally compact group $G$ and closed finite covolume subgroup $H$ for $G$ acting on $G/H$. | |
| Dec 20, 2020 at 4:22 | answer | added | katago | timeline score: -4 | |
| Dec 20, 2020 at 4:02 | comment | added | katago | why it should not be for the irrational element $g$, $g\in G / \Gamma$, the action of $g$ on $G / \Gamma$ is ergodic? | |
| Dec 20, 2020 at 3:40 | review | Close votes | |||
| Dec 25, 2020 at 4:21 | |||||
| Dec 20, 2020 at 3:34 | comment | added | markvs | Transitivity is not the same as ergodicity. I am not sure that if the quotient by the center is not taken, the action is ergodic. | |
| Dec 20, 2020 at 3:20 | comment | added | Moishe Kohan | The $G$-action is transitive, so,... | |
| Dec 20, 2020 at 2:55 | comment | added | markvs | Don't they usually also take a quotient by a compact (central) subgroup? | |
| Dec 20, 2020 at 2:39 | history | edited | LSpice | CC BY-SA 4.0 |
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| Dec 20, 2020 at 2:22 | comment | added | Asaf | This follows easily from the Howe-Moore theorem (it shows the action is actually mixing), see a proof in the new Einsiedler-Ward book. Another source is Bekka-Mayer. | |
| Dec 20, 2020 at 1:33 | history | edited | No One |
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| Dec 20, 2020 at 0:52 | history | asked | No One | CC BY-SA 4.0 |