Is this “irrational torus rotation” dense

analytic-number-theoryergodic-theoryirrational-numbers

I am trying to understand the behavior of certain generalizations of the irrational rotation, and this one has stumped me. Act on the $2$-torus by
$$T(x, y) = (x + \alpha, x + y)$$
where $\alpha$ is of course an irrational rotation. The orbit of $T$ looks like
$$T^n(x, y)= \left(x + n\alpha, nx + y + \frac{n(n-1)}{2}\alpha\right)$$
and I would like to show that the orbit is dense (possibly just for almost every $(x,y)$). This implies, among other things, that the action of $T$ is ergodic.

I intuitively know that since $x – \alpha$ is almost surely irrational, $x \neq 0$ almost surely, and the action of $T$ is just linear in the first variable and quadratic in the second variable (and has a lower-order term that depends linearly on $x$), that the actions should "decouple", and so $T$ should smear out all the points in the second variable, if that makes sense.

One could thus hope to carry out the following argument: Let $(x, y)$, $(x',y')$, $\varepsilon$ be given. There is a strictly increasing sequence of $n_k$ such that $|T^{n_k}x – x'| < \varepsilon$, since the irrational rotation is recurrent. Since these $n_k$ have "density $O(n^{-1})$ in $\mathbb N$", while the set $M$ of $m$ such that
$$|mx + y + \frac{m(m-1)}{2} \alpha – y'| < \varepsilon$$
has "density $O(n^{-2})$ in $\mathbb N$", we may almost surely pass to a subsequence $n_{k_\ell}$ such that every $n_{k_\ell} \in M$. But I don't know how to formalize this (hence the tag for analytic number theory). How do I go about that?

Best Answer

Let $f$ be a $T$-invariant function in $L^2(\mathbb{T}\times\mathbb{T})$. We may write $f = \sum_{k_1,k_2} c_{k_1,k_2} e^{2\pi i (k_1x+k_2y)}$. Then, $\sum_{k_1,k_2} c_{k_1,k_2} e^{2\pi i (k_1x+k_2y)} = f(x,y) = f(T(x,y)) = f(x+\alpha,x+y) = \sum_{k_1,k_2} c_{k_1,k_2}e^{2\pi i (k_1(x+\alpha)+k_2(x+y))} = \sum_{k_1,k_2} e^{2\pi i \alpha k_1}c_{k_1,k_2}e^{2\pi i ((k_1+k_2)x+k_2y)} = \sum_{k_1,k_2} e^{2\pi i \alpha(k_1-k_2)}c_{k_1-k_2,k_2}e^{2\pi i (k_1x+k_2y)}$, so we must have $c_{k_1,k_2} = e^{2\pi i \alpha(k_1-k_2)}c_{k_1-k_2,k_2}$ for each $k_1,k_2$, i.e., $c_{k_1+k_2,k_2} = e^{2\pi i \alpha k_1} c_{k_1,k_2}$ for each $k_1,k_2$. If $c_{k_1,k_2} \not = 0$ and $k_2 \not = 0$, then by iterating, we get infinitely many terms (namely $c_{k_1+mk_2,k_2}$) that have the same, nonzero magnitude, contradicting $f \in L^2$. Therefore, $c_{k_1,k_2} = 0$ if $k_2 \not = 0$. And for $k_2 = 0$, we have $c_{k_1,0} = e^{2\pi i \alpha k_1} c_{k_1,0}$, meaning $c_{k_1,0} = 0$ if $k_1 \not = 0$. Therefore $f$ is (a.e.) constant.

We conclude $T$ is ergodic. It follows that the orbit of almost every point is dense.

.

I think your intuition is not that good, or maybe is just not fleshed out enough. A density $O(n^{-2})$ set need not lie within an $O(n^{-1})$ set, so I'm not sure why you think you should be able to just pass to a subsequence.

Related Question