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I wonder if there are any direct proof that $g_t=\operatorname{diag}(e^{t/m}I_m,e^{-t/n}I_n)$'s action on $\operatorname{SL}(d,\mathbb R)/{\operatorname{SL}(d,\mathbb Z)}$ is ergodic (or even stronger, mixing). Here $I_n$ denote the $n \times n$ identity matrix.

The classical proof of this involves the Howe–Moore theorem, which uses Mautner's lemma and can be applied to much more general Lie groups. But I wonder in this concrete setting whether we could have a more simplified/elementary proof. Although I stated $(g_t)$ explicitly as a diagonal flow, I should mention that perhaps only the unboundedness of $(g_t)$ will be used, although $g_t$ being an unbounded diagonal flow might make the proof easier somehow.

Update: this should be true for any $t\ne 0$. So we can actually just fix $t=1$ and just prove that $g_1$ acts ergodically, which is equivalent to saying that the subgroup $(g_n)_{n\in \mathbb Z}$ acts ergodically.

Note that for $\operatorname{SL}(d,\mathbb R)/{\operatorname{SL}(d,\mathbb Z)}$, there are relatively more explicit descriptions for the Haar measure (using Siegel domain/Iwasawa decomposition etc.). I am thinking maybe we can solve this by studying what kind of sets in this space have invariant measure under $g_1$, or any diagonal subgroup of $\operatorname{SL}(d,\mathbb R)$ that is not all $\pm 1$'s. If an open set in the basis has nontrivial measure, then $g_1$'s acts like stretching the set along one direction and shrinking it along another direction and it should change the Haar measure in some ways—that is my intuition.

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    $\begingroup$ 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 $\endgroup$
    – Asaf
    Commented Sep 16, 2022 at 19:30
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    $\begingroup$ @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? $\endgroup$
    – No One
    Commented Sep 16, 2022 at 20:07
  • $\begingroup$ 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... $\endgroup$
    – Asaf
    Commented Sep 16, 2022 at 21:10
  • $\begingroup$ @NoOne Why don't you read Moore's 'Ergodicity of flows on homogeneous spaces'? :) $\endgroup$
    – Calamardo
    Commented Sep 17, 2022 at 14:05
  • $\begingroup$ 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... $\endgroup$
    – Asaf
    Commented Sep 18, 2022 at 22:44

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