Equidistribution of the orbit $\{\text{diag}(t^a,t^{-a})\Lambda \}_{t>0}$ for a.e. $\Lambda\in \text{SL}(2,\mathbb R)/\text{SL}(2,\mathbb Z)$

Question

$\DeclareMathOperator\diag{diag}\DeclareMathOperator\SL{SL}$It is well-known that geodesic flow $g_t=\{\diag(e^t,e^{-t}) \}_{t>0}$ acts ergodically (actually mixing) on $\SL(2,\mathbb R)$ (Howe–Moore theorem) and thus for almost all lattice $\Lambda\in X:=\SL(2,\mathbb R)/{\SL(2,\mathbb Z)}$ and any $f\in C_c(X)$, we have

$$\lim_{T\to \infty} \frac{1}{T}\int_0^Tf(g_t.\Lambda)\,dt =\int_X f(x) \, d\mu(x),$$ where $\mu$ is the $\SL(2,\mathbb R)$ invariant probability Haar measure on $\SL(2,\mathbb R)/{\SL(2,\mathbb Z)}$.

Now what happens if we replace $(g_t)$ with $(h_t):=\{\diag(t^a,t^{-a})\}_{t>0}$ (assume $a>0$)? Do we still have for almost all $\Lambda \in X$ and compactly supported continuous function $f$ (in particular $f(h_t.\Lambda)$ is bounded on $[0,1]$ and thus the limit for the normalization by $T$ of this integral and the integral starting from $t=1$ are the same) that

$$\lim_{T\to \infty} \frac{1}{T}\int_0^Tf(h_t.\Lambda) \, dt=\int_X f(x) \, d\mu(x)? \tag{*}\label{star}$$

Or the equidistribution depends on the choice of $a>0$ or there is no equidistribution at all?

The difference between $g_t$ and $h_t$ is that $g_{s+t}=g_s g_t$ but $h_{st}=h_t h_s$ and thus not a one-parameter subgroup. Ideally, I wish someone had studied an analog of Weyl's criterion for the equidistribution of $\diag\{f(t),1/f(t)\}_{t>0} \cdot \Lambda$ on $X$.

By the way, I know one can perform a change of variable on $\frac{1}{T} \int_0^Tf(g_t.\Lambda) \, dt$ to reparametrize it to become $h_t$ action. But that is not what I meant. My question is about \eqref{star}, literally.

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I hope those who claim that this is a poorly stated question could provide some solid evidence showing for example when $a=1$ the a.e. equidistribution is impossible.

Answer 1

This thing cannot hold no matter what. As Echo rightfully commented, the expression doesn't even compile when $t=0$. It is true one might temper the integral away from $0$, but that's not what you asked. This asymmetry between $[1,T]$ and $[0,1]$ suggests something fishy is going on in general.

Anyhow, I will demonstrate the answer is STRONGLY NEGATIVE even with the "reasonable" adjustment (say taking $t\in [1,T]$).

What follows holds for any lattice $\Gamma\leq SL_{2}(\mathbb{R})$.

Define $A_{T}f(x\Gamma) = \frac{1}{T}\int_{1}^{T}g_{t}.f(x\Gamma)dt$. [I will call the modified family $g_{t}$ as well, the usage of $h_{t}$ here is rather confusing as this is usually reserved for the horocyclic flow].

Assume now that $f\in L^{2}_{0}$. A computation a-la Furstenberg shows $$ \left\lVert A_{T}f\right\rVert^{2} = \frac{1}{T^2}\int_{1\leq t,t' \leq T } \left\langle g_{t}.f,g_{t'}.f\right\rangle dtdt' = \frac{1}{T^2}\int_{1\leq t,t' \leq T } \left\langle g_{1/t'}g_{t}.f.f\right\rangle dtdt'.$$ The last expression shows something is fishy (as this is not a unitary family of operators, otherwise it should have been $-t'$). So we get $$ \left\lVert A_{T}f\right\rVert^{2} = \frac{1}{T^2}\int_{1\leq t,t' \leq T } \left\langle g_{t/t'}.f.f\right\rangle dtdt'.$$ So we just need to analyze this "convolution" integral over the matrix coefficients. Assume now that $f$ is nice (in general, one can say more here with a proper assumption over the automorphic support of $f$).

By HC's bound, we have that $\left\lvert \left\langle g_{r}.f,f\right\rangle\right\rvert \ll_{f} e^{-s\cdot a \cdot \log(r^{\pm})},$ for some spectral gap $s=s(\Gamma)>0$.

We split the set of parameters into two subsets, one is when $\left\lvert t/t'\right\rvert^{\pm} -1 \gg R$ and the complement. The matrix coefficients over the first set have some decay. The second set is the culprit, for that set, one only have a negligible decay if any. Now the area estimate for the second set is $O(T^{2}\cdot R)$. So if one picks some $R$ say, there's a set of positive density (essentially $O(R)$) such that the $L^{2}$ norm is greater than $O(R)\cdot \min_{\left\lvert \ell\right\rvert \leq R}\left\lvert \langle g_{\ell}.f,f \rangle \right\rvert$, the rest is trivially bounded by $O(R^{-a\cdot s})$. As such, we see that $\left\lVert A_{T}f\right\rVert_{L^{2}}^2 \gg R^{1-a\cdot s}$, which clearly does not decay with $T$. Hence there is not even convergence in mean.

This is in sharp contrast with the regular parameterization, where one have $\left\lvert t’-t\right\rvert \leq R$ is of size $O(R\cdot T)$, hence the integral along that diagonal becomes negligible as $T\to\infty$.