I wonder if there are any 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 (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 shrink it along another direction and it should change the Haar measure in some ways---that is my intuition.

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.

I wonder if there are any 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 (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, 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 shrink it along another direction and it should change the Haar measure in some ways---that is my intuition.

I wonder if there are any 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 (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 shrink it along another direction and it should change the Haar measure in some ways---that is my intuition.

I wonder if there are any 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 (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, 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.

I wonder if there are any 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 (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, 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 shrink it along another direction and it should change the Haar measure in some ways---that is my intuition.