\text{SL}(2,\mathbb Z)$

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Sep 14, 2023 at 21:52	history	edited	user506835	CC BY-SA 4.0	added 163 characters in body
Sep 14, 2023 at 20:59	answer	added	Asaf		timeline score: 0
Sep 14, 2023 at 19:14	comment	added	user506835		@Asaf As I said above, this interpretation would violate your identity $\frac{1}{T}\int_{t=1}^{T}h_{t}.f(x)dt = \frac{1}{T}\sum_{i=0}^{\log T -1}h_{2^{i}}.F(x)$. Since LHS is $\frac{1}{T}(T-1)$ and RHS is $\frac{1}{T}(\log T)$
Sep 14, 2023 at 17:33	comment	added	user506835		But I suppose when you were writing this, you were treating $h_{2^{i}}.F(x)=\int_1^2 h_{t}.f(2^i x)dx$ and when $f\equiv 1$ we have $h_{2^{i}}.F(x)=1$ and thus $\frac{1}{T}\sum_{i=0}^{\log T -1}h_{2^{i}}.F(x)= \frac{1}{T}\sum_{i=0}^{\log T -1} 1 \to 0$ and therefore you wanted to conlude $\frac{1}{T}$ is not the right normalization. But under this interpretation, your identity $\frac{1}{T}\int_{t=1}^{T}h_{t}.f(x)dt = \frac{1}{T}\sum_{i=0}^{\log T -1}h_{2^{i}}.F(x)$ also fails! So your example is not valid.
Sep 14, 2023 at 17:28	comment	added	user506835		@Asaf Thanks for your effort. But in your example (you meant $\log T = \log_2 T$ probably), when you write $\frac{1}{T}\int_{t=1}^{T}h_{t}.f(x)dt = \frac{1}{T}\sum_{i=0}^{\log T -1}h_{2^{i}}.F(x)$, what do you mean by $h_{2^{i}}.F(x)$? In order to make this equality true, I have to interpret this term as $\int_{2^i}^{2^{i+1}}h_{t}.f(x)dt$. If this is the case, then when you take $f\equiv 1$ (well, $f$ has to be compactly supported...) the individual summand is getting bigger and bigger with $i$. Therefore, you cannot conclude $\frac{1}{T}\sum_{i=0}^{\log T -1}h_{2^{i}}.F(x) \to 0$.
Sep 14, 2023 at 0:09	history	edited	user506835	CC BY-SA 4.0	added 53 characters in body
Sep 14, 2023 at 0:04	comment	added	user506835		@Asaf I don't understand what you meant by "doesn't make any sense". Do you mean for different $f$ we have different convergence outcomes? Can you construct any examples showing the almost everywhere equidistribution is too much to hope for? without just saying looking at the discrete analog? I also don't understand that discrete analog you mentioned. Why are there $O(\log N)$ terms?
Sep 13, 2023 at 18:33	history	edited	user506835	CC BY-SA 4.0	deleted 11 characters in body
S Sep 12, 2023 at 21:25	history	edited	user506835	CC BY-SA 4.0	some typos fixed
S Sep 12, 2023 at 21:25	history	suggested	No One	CC BY-SA 4.0	some typos fixed
Sep 12, 2023 at 21:20	review	Suggested edits
S Sep 12, 2023 at 21:25
Sep 11, 2023 at 2:51	review	Close votes
Sep 13, 2023 at 19:32
Sep 11, 2023 at 2:28	comment	added	Asaf		This question doesn’t make any kind of sense, even with the “reasonable” modification of $t\geq 1$. One can move into the analogous discrete average where one can easily see that one sums over $O(\log N)$ elements but normalizes by $N$.
Sep 6, 2023 at 17:12	comment	added	user506835		@Echo Since $f\in C_c(X)$, it is bounded by some $M_f$. Therefore $\frac{1}{T}\int_0^1 f(\gamma(t)\Lambda)dt \le \frac{M_f}{T} \to 0$ as $T\to \infty$. I couldn't get your point for continuity when $t\to 0$. It doesn't matter if the integral starts at $0$ or $1$.
Sep 5, 2023 at 13:54	comment	added	user473423		The convergence problem remains, as the function is of compact support on $X$, which does not imply that the function in $t$ is of compact support. If it were, the theorem would be false. Also, my argument keeps, as the claim is for all $f$.
Sep 4, 2023 at 21:29	history	edited	Michael Hardy	CC BY-SA 4.0	added 19 characters in body
Sep 4, 2023 at 19:04	history	edited	LSpice	CC BY-SA 4.0	`\DeclareMathOperator` + `\eqref`
Sep 4, 2023 at 17:51	comment	added	user506835		@echo The convergence problem shouldn't be an issue since $f$ is compactly supported and thus zero as $t\to 0$. When I perform a change of variable, the integrand and limits of integral are both changed. One can't conclude such a formula no longer holds simply because of this.
Sep 4, 2023 at 17:44	comment	added	user473423		Your integral has a convergence problem at 0. If you integrate from 1 instead, you can perform a simple change of variables to reduce it to the known case but with a different integrand, which shows that such a formula cannot hold.
Sep 4, 2023 at 17:35	history	edited	user506835	CC BY-SA 4.0	edited title
Sep 4, 2023 at 14:43	history	edited	user506835	CC BY-SA 4.0	added 128 characters in body
Sep 4, 2023 at 14:27	history	edited	user506835	CC BY-SA 4.0	added 2 characters in body
Sep 4, 2023 at 14:00	history	asked	user506835	CC BY-SA 4.0

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