The closure of the orbit of an irrational grid contains the fiber

Question

Let $G=\text{SL}(d,\mathbb R)$ and $\Gamma = \text{SL}(d,\mathbb Z)$. The homogeneous space is identified with the space of unimodular lattices, denoted $X_d$. Let $Y_d$ denote the space of unimodular grids, namely the translation of lattices $Y_d:=\{x+v: x\in X_d, v\in \mathbb R^d \}$. Let $\pi:Y_d \to X_d$ be the natural projection $x+v \mapsto x$.

Let $A$ denote the subgroup of $G$ consisting of all diagonal matrices of positive entries (of course, the determinant has to be one).

Let $x_0\in X_d$. If $y_0\in \pi^{-1}(x_0)$ is irrational, namely $y_0$ is not in the rational span of $x_0$ (span of basis of the lattice $x_0$ with rational coefficients), then it is intuitively correct to me that

$$\overline{Ay_0} \supset \pi^{-1}(x_0).$$

Namely the closure of the $A$-orbit will contain the whole fiber $\pi^{-1}(x_0)$.

The intuition comes from the fact that irrational lines on a torus are dense.

Is this necessarily correct? How do we prove this rigorously? Note that if $y_0 = x_0 + v_0$, then we do not necessarily have $Ay_0 = Ax_0 + Av_0$ (LHS is contained in RHS, though).

Answer 1

First of all, $Y$ is not called the “grid space”. It is sometimes called the affine space and can be identified with a quotient of the affine group $\operatorname{ASL}_{n}$, namely the semi-direct product of $\operatorname{SL}_{n}$ with $\mathbb R^{n}$.

Anyhow, the question you are referring to was addressed by U. Shapira in his well-known thesis where he studied the “big Littlewood conjecture”.

The main new ingredient here (for which one needs to assume $n\geq 3$, compared to the results of Einsiedler–Katok–Lindenstrauss over the Littlewood conjecutre) is the famous $\times 2$, $\times 3$ theorem of Furstenberg, showing that for any irrational number $\alpha$, the set $\left\{2^{n}3^{m}\alpha\right\}$ is dense modulo $1$. This theorem has been generalized by D. Berend, a student of Fursternberg's, to actions of commuting matrices over $\mathbb{T}^d$. Using Berend's theorem (with the observation that the derived $A$-action over the fiber (so one has a higher-rank commutative Anosov action over the torus)), your question follows (you can also consider the PhD thesis of Z. Wang).

Notice that in order to conclude your question, you just need to show that the origin of your fiber is a point contained in the orbit closure. On the other hand, due to Berened's theorem, if the origin is not a point in your orbit closure, the orbit will be stuck in some rational subgroup, so it is essentially if and only if.

Answer 2

As Asaf pointed out, this has more to do with linear actions on tori, and relies on very advanced homogeneous dynamics which use crucially the higher rank assumption ($d\geq 3$).

For $d=2$, the dynamics is much wilder: the action of $A$ on the space of unimodular lattices in this case is the geodesic flow of the modular surface, which is essentially an Anosov flow.

A first issue is that, for many points $x_0$, the $A$ orbit of $x_0$ is not returning to $x_0$ (take for instance a geodesic that accumulates in positive and negative time to two distinct closed geodesics).

To overcome this let us now say that you chose a point $x_0$ whose orbit is closed. Then the return map of $A$ on the fiber $\pi^{-1}(x_0)$ is an Anosov transformation of the torus of dimension $2$, and you are asking whether the orbit of every irrational point is dense.

But Anosov diffeomorphisms have very wild dynamics. In particular, there are many (irrational) points with orbit closure a Cantor set.