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I'm coming with a very basic question for which I can't find an answer. Please forgive me if I didn't search efficiently enough.

What can the closure of an orbit of an element $X$ of $\mathbb{R}^2$ under the action of $SL(2,\mathbb{Z}) $ look like ?

1) Obviously if the vector space spanned by the coordinates of X has dimension $1$, $SL(2,\mathbb{Z}) \cdot X$ is a lattice.

2) The $SL(2,\mathbb{Z}) $-action is known to be ergodic on $\mathbb{R}^2$, which implies that there is a set of full measure of which every element has a dense orbit. Is this set exactly the remaining cases ? Namely when the coordinates of $X$ are $\mathbb{Q}$-linearly independent ? This reduces to the question :

If $\theta \notin \mathbb{Q}$, is $SL(2,\mathbb{Z}) \cdot (1, \theta) $ dense in $\mathbb{R}^2$ ?

3) Is the question solved in the case of the $SL(n,\mathbb{Z}) $ action on $\mathbb{R}^n$ ?

Any good reference is welcome !

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2 Answers 2

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There are very precise results on the repartition of orbits of $SL(2,\mathbb{Z})$ of irrational points in $\mathbb{R}^2$, such as this one.

In particular these orbits are dense, and this can be seen easily like this.

If $\theta$ is irrational, there are coprime $(a,b)$ such that $a+b\theta$ is less than $\epsilon$ for any $\epsilon>0$. Then there are $(c,d)$ such that $ad-bc=1$, an the various $c+d\theta$ form an $\epsilon$-net in $\mathbb{R}$, as they are $c_0+d_0\theta + k(a+b\theta)$ for $k\in\mathbb{Z}$.

So the closure of the $SL(2,\mathbb{Z})$)-orbit of $(1,\theta)$ contains at least the vertical axis $0\times\mathbb{R}$.

But this closure is invariant under $SL(2,\mathbb{Z})$, hence contains all images of the vertical line, that is all rational lines, hence the closure is the whole plane, and the orbit is dense.

ADDED: As you asked also about the situation for higher $n$, I can direct you to this very recent arXiv preprint.

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Another interpretation of this fact is as follows. Consider the unit tangent bundle $PSL(2,\mathbb{Z})\backslash PSL(2,\mathbb{R})$ of the modular surface $PSL(2,\mathbb{Z}) \backslash \mathbb{H}^2$. The horocyclic flow has two kinds of orbits: periodic orbits turning around the cusp, and dense orbits. This is known since Hedlund's work in the 30's at least.

Moreover, all dense orbits are equidistributed towards the Liouville measure, which is the unique horocyclic-invariant ergodic probability measure of full support. (Results of Dani and Dani-Smillie at the end of the 70's)

And the space $\mathbb{R}^2\setminus\{0\}$ identifies naturally with the space of horocycles, and the action of $PSL(2,\mathbb{Z})$ by isometries on the horocycles becomes the usual linear action of $SL(2,\mathbb{Z})$ on $\mathbb{R}^2$.

Thus, a point in $\mathbb{R}^2$ coming from a horocycle centered at a rational point (or equivalently turning around the cusp) has a $SL(2,\mathbb{Z})$-orbit which is discrete, whereas the other points have dense orbits (equidistributed towards a measure which is not exactly Lebesgue on $\mathbb{R}^2$, but $dr d\theta$ in polar coordinates)

Of course with this point of view, the good group is not $SL(2,\mathbb{Z})$ but $PSL(2,\mathbb{Z})$...

Barbara

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