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$\DeclareMathOperator\SL{SL}$Let $G\mathrel{:=}\SL(n,\mathbb R)$ and $\Gamma\mathrel{:=}\SL(n,\mathbb Z)$. Consider the action of $G$ on $(G/\Gamma,\mu)$ by left translation, where $\mu$ is the Borel probability measure on the homogeneous space $G/\Gamma$ that is left-invariant w.r.t $G$. I wonder where I can find the proof that the action of $G$ is ergodic?

If this is true in more general settings say when $G$ is a simple Lie group as in the $G=\SL(n,\mathbb R)$ case, please let me know.

Many comments below mentioned the Howe-Moore theorem. But I am aware of the fact (as a corollary of Howe-Moore) that every unbounded subgroup of $G$ also acts ergodically/mixingly after we proved that $G$ itself acts ergodically.

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    $\begingroup$ 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. $\endgroup$
    – Asaf
    Commented Dec 20, 2020 at 2:22
  • $\begingroup$ Don't they usually also take a quotient by a compact (central) subgroup? $\endgroup$
    – markvs
    Commented Dec 20, 2020 at 2:55
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    $\begingroup$ The $G$-action is transitive, so,... $\endgroup$
    – Moishe Kohan
    Commented Dec 20, 2020 at 3:20
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    $\begingroup$ 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 $G/H$. $\endgroup$
    – YCor
    Commented Dec 20, 2020 at 12:44
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    $\begingroup$ @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. $\endgroup$
    – YCor
    Commented Dec 20, 2020 at 16:19

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The question is why a measurable function which is invariant under the action of all $g \in G$ must be a.e. constant. But since the action of $G$ is transitive on $G/\Gamma$ it is easy to see that this is the case: take a non constant function $\varphi$ on $G/\Gamma$, there are points $x,y$ such that in small neighborhood $U$ and $V$ of each where the function takes different values in at least 99% of the measure of each. Now, taking some $g$ that maps $x$ to $y$ you can see that $\varphi$ cannot be $g$-invariant.

Once this is seen it is that one can apply Howe-Moore to deduce that every unbounded subgroup also acts ergodically (in fact, mixing). Improvements exist as Asaf points out.

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    $\begingroup$ non-constant should be "not generically constant"? Also the assumption is not that it's invariant under each $g$, but for each $g$, $f-gf$ has support of measure zero. $\endgroup$
    – YCor
    Commented Dec 20, 2020 at 16:17
  • $\begingroup$ sorry, I had not seen your comment to the question that essentially points the same. I just wanted to state that Howe Moore is not really the answer to the question (though related of course, and probably every reference that makes HM) proves this too. Your comments are correct, I was not explicit enough in saying that non constant means in the sense of functions being "the same" if coincide in full measure. $\endgroup$
    – rpotrie
    Commented Dec 20, 2020 at 16:44
  • $\begingroup$ Note that $\phi$ is measurable but not continuous. So the behavior of $\phi$ in the neighborhoods $U,V$ can be very weird... how do you make sure that " the function takes different values in at least 99% of the measure of each"? $\endgroup$
    – No One
    Commented Dec 24, 2020 at 16:11
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    $\begingroup$ Luisin's theorem provides a (compact) set of almost full measure where the function is continuous. If the function is non constant you can also assume the restriction isn't. Then, pick density points where the function takes different values to get that. $\endgroup$
    – rpotrie
    Commented Dec 25, 2020 at 17:25
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Posting YCor's comment to hopefully end some confusion -

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.

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  • $\begingroup$ 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! $\endgroup$
    – No One
    Commented Mar 20, 2022 at 1:48
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    $\begingroup$ 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/… $\endgroup$
    – Calamardo
    Commented Mar 20, 2022 at 3:36
  • $\begingroup$ What do you mean by "nonvanishing top forms"? $\endgroup$
    – No One
    Commented Mar 20, 2022 at 3:50
  • $\begingroup$ 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. $\endgroup$
    – Calamardo
    Commented Mar 20, 2022 at 4:13
  • $\begingroup$ 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! $\endgroup$
    – No One
    Commented Mar 20, 2022 at 17:29
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This is equivalent to for an irrational element $g, g \in G / \Gamma$, irrational mean the equivalent class $[g]\cap \Gamma=\emptyset$, $$ \lim _{N \rightarrow+\infty} \sum_{n=1}^{N} \delta_{g^ n} \longrightarrow \text { Haar Messure } \quad (*) $$

And $(*)$ can be proved by calculate the paring $\sum_{n=1}^{N} \delta_{g^ n}$ with the periodic function on $\mathrm{SL}(n, \mathbb{R})$ induced by the lattice $\mathrm{SL}(n, \mathbb{Z})$ in it.

I do not know what is the explicit expression of the periodic function, say them are $\phi$.

But I believe its behavior is the same as $e(nx)=e^{2\pi inx}$ in the case of $\mathbb{T}$, and in particular use the Gauss summation method on $\phi$ will give us the desirable bound for indicating Weyl criterion is true.

I will give the detail in the above argument later, and this argument is stronger than Howe-Moore Theorem because it can give the speed of convergence to mixing.

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    $\begingroup$ This comment is totally wrong. First ergodicity for a (non-amenable) group action is different than for a one-parameter subgroup (for example, you element might be in a compact subgroup). Second, quantitative estimates for matrix coefficients have been known (largely due to the work of Harish-Chandra) and explicit Kazhdan constants have been computed (see for example the work of Oh). One may extract mixing estimates from those (for say $K$-finite functions). $\endgroup$
    – Asaf
    Commented Dec 20, 2020 at 6:43
  • $\begingroup$ Asaf: Thanks a lot, I already understand this above argument is totally wrong, not only I only consider the one-parameter subgroup, but also the method I describe above seems only can treat the function in $L^2(C(\mathrm{SL}(n, \mathbb{R}) / \mathrm{SL}(n, \mathbb{Z})))$ because only this function can be approximated by a linear combination of function comes from irreducible unitary representation. And a more careful looking at the $\mathrm{SL}(n, \mathbb{R}) / \mathrm{SL}(n, \mathbb{Z})$ is necessary to gain the speed of convergence to mixing, still in keeping correct. $\endgroup$
    – katago
    Commented Dec 20, 2020 at 9:18

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