Timeline for Direct proof that $g_t=\text{diag}(e^{t/m}I_m,e^{-t/n}I_n)$'s action on $\operatorname{SL}(d,\mathbb R)/{\operatorname{SL}(d,\mathbb Z)}$ is ergodic
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| Oct 6, 2022 at 2:34 | comment | added | Asaf | @NoOne, I think we are going in circles. I explained how one can deduce that $g_{1}$ acting ergodically based on the Hopf argument (and the fact that the horospherical groups generate $SL_{n}$ which is a simple group). | |
| Oct 4, 2022 at 18:06 | comment | added | No One | @Asaf sorry for my typo. Z should be R. Yes, the big group acts ergodically doesn't mean small ones act ergodically. And yes that is what Howe Moore theorem is giving. But I am thinking about maybe just directly proving that $g_1$ acts ergodically. | |
| Oct 3, 2022 at 15:19 | comment | added | Asaf | @NoOne , clearly $g_{n}$ is not a subgroup of $SL_{d}(\mathbb{Z})$. The fact that $g_{1}$ acts ergodically does not in general imply that $g_{n}$ is (say think of exchanging two points). It is true in the case of mixing (or in more generality - if your spectrum does not contain rational eigenvalues). I am still puzzled with one thing - how would you know that $g_{1}/g_{n}$ acts ergodically to begin with? P.S. In general, for general $G$-actions, having an ergodic action does not mean that a subgroup acts ergodically (this is exactly the situation the Mautner phenomenon is trying to solve). | |
| Oct 2, 2022 at 18:22 | comment | added | No One | @Asaf By unbounded I mean "non-compact". Subgroup of $\text{SL}(d,\mathbb Z)$. | |
| Oct 1, 2022 at 9:58 | comment | added | Calamardo | @NoOne Pfft, how will you become the next big thing if you don't know $C^*$-algebras, Type I groups, structure theorems for semisimple groups. :p | |
| Sep 30, 2022 at 3:02 | comment | added | Asaf | P.S. Even in the usual proofs of decay of matrix coefficients (say Cowling-Howe-Haagerup), one studies decay of the matrix coefficient under $g_{t}$, then extends by KAK decomposition. I think the estimates of Harish Chandra are also following the decay along the Cartan group (as it allows HC to get the differential equations). | |
| Sep 30, 2022 at 3:00 | comment | added | Asaf | I don't understand the statement ''unbounded subgroup'', unbounded subgroup of what? It is correct that every unbounded subgroups act ergodically (and even mixingly), and that follows from say decay of matrix coefficients (which is rather ''direct proof'' but very technical, as it quantifies Howe-Moore). All the other proofs I can think of will definitely use some sort of Mautner phenomenon and fallback on ergodicity/mixing of $g_{t}$. | |
| Sep 30, 2022 at 1:05 | comment | added | No One | @Asaf Maybe I am saying something stupid but the unbounded subgroup $(g_n)_{\mathbb Z}$ acts ergodically and I guess this should imply the generator $g_1$ is ergodic just from the definition? | |
| Sep 30, 2022 at 1:00 | history | edited | LSpice | CC BY-SA 4.0 |
Mild tidying, while this is on the front page
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| Sep 29, 2022 at 19:01 | comment | added | Asaf | @NoOne, my comments above, in view of the fact that $SL_{n}$ is a simple group, give a full proof of ergodicity in the lines of the Hopf argument. It's an interesting question whether you can upgrade ergodicity to mixing, I will need to think about it. What you are proposing is not possible, there is a plethora of invariant measures (and sets) for the diagonal action (say in $SL_{2}$, the action is Bernoulli), including fractal ones of essentially any dimension between $1$ and full dimension. | |
| Sep 29, 2022 at 17:30 | history | edited | No One | CC BY-SA 4.0 |
edited tags
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| Sep 29, 2022 at 17:24 | history | edited | No One | CC BY-SA 4.0 |
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| Sep 29, 2022 at 17:23 | comment | added | No One | @Latimer Actually I did try to read Moore's paper myself and it is actually not quite self-contained, by citing many results of Borel and Mackey etc. Also I am just asking about the ergodicity, not mixing! I feel like for such specific setting there must be a more explicit and self-contained proof. Note that for $\operatorname{SL}(d,\mathbb R)/\operatorname{SL}(d,\mathbb Z)$, there are relatively more explicit descriptions for the Haar measure (Siegel domain etc.). I am thinking maybe we can solve this by studying what kind of sets in this space have invariant measure under $g_1$ | |
| Sep 29, 2022 at 17:03 | history | edited | No One | CC BY-SA 4.0 |
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| S Sep 19, 2022 at 18:32 | history | suggested | Calamardo | CC BY-SA 4.0 |
The dimension of the matrix is m+n. I define d:=m+n
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| Sep 19, 2022 at 15:56 | comment | added | Asaf | Actually the shorter proof would always work for $SL_n$, no matter about your (m,n) as $SL_n$ is a SIMPLE group, and the Lie algebra generated by $Lie(G±)$ is an ideal. In the general semisimple case, one needs to be careful with the roots, etc... Probably the smart thing to do is simply to enumerate all the root spaces of the maximal horospherical subgroups and construct by hand the required spaces by the Lie brackets... | |
| Sep 19, 2022 at 13:36 | review | Suggested edits | |||
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| Sep 18, 2022 at 22:44 | comment | added | Asaf | Probably a slightly easier argument that would work in many cases would go like this - one knows that any $g_{t}$-inv function is $G^{\pm}$ inv. One can consider maximal horosphericals (the abelianizations), of-course the function is invariant under them as well. Using commutators (from opposite root spaces), one may show that the function is invariant under the whole $A$ - the split Cartan. And actually in many many cases, the horosphericals will generate (i.e. in your case for m=2,n=1). The main trick is to consider commutators of $[n_{a+b},n_{-a}]$, etc... | |
| Sep 17, 2022 at 14:05 | comment | added | Calamardo | @NoOne Why don't you read Moore's 'Ergodicity of flows on homogeneous spaces'? :) | |
| Sep 16, 2022 at 21:10 | comment | added | Asaf | As I said, you conclude that every ergodic component which is $g_{t}$-invariant is invariant under the subgroup generated by $G_{a}^{-}AMG_{a}^{+}$, which is everything, hence by uniqueness of the Haar measure, every ergodic component is Haar. The Hopf argument gives invariance of the general (a-priori not ergodic) measure under the strong stable/unstable foliations. Obviously this also applies to each ergodic component... | |
| Sep 16, 2022 at 20:07 | comment | added | No One | @Asaf My question is about how to prove directly "$g_t=\text{diag}(e^{t/m}I_m,e^{-t/n}I_n)$'s action on $\operatorname{SL}(n,\mathbb R)/\operatorname{SL}(n,\mathbb Z)$ is ergodic". What is the aim of your comment? Invariance of what under stable/unstable horospherical groups? | |
| Sep 16, 2022 at 19:52 | review | Close votes | |||
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| Sep 16, 2022 at 19:30 | comment | added | Asaf | That's not a research level question. Anyhow the usual Hopf argument gives you invariant under stable/unstable horospherical groups. This does not require ergodicity. You just need to show extra invariance under commuting central directions. Pick an ergodic component, use the ergodic theorem for $g_{t}$, pick your favorite generic point and show that it is preserved under any element from the center direction by commutation. Hence this ergodic component is preserved by $g_{t}$, horospherics $G_{a}^{\pm}$ and $AM$. Now just notice by Lie alg, that $G_{a}^{-}AMG_{a}^{+}$ is generating everything | |
| Sep 16, 2022 at 18:40 | history | edited | No One | CC BY-SA 4.0 |
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| Sep 16, 2022 at 18:31 | history | edited | No One | CC BY-SA 4.0 |
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| Sep 16, 2022 at 18:20 | history | asked | No One | CC BY-SA 4.0 |