The (last step of the) proof that the set of badly approximable matrices has measure zero

Question

An $m \times n$ matrix $A$ is called badly approximable if there exists $c > 0$ such that for all integer vectors $p \in \mathbb Z^m$ and $q \in \mathbb Z^n-\{0\}$ we have

$$ \|A q + p \| \ge c \| q \|^{-\frac{n}{m}}. $$

We denote the collection of badly approximable $m \times n$ matrices by $\textbf{BA}(m,n)$.

Let $d=m+n$ and $X_d:=\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ be the space of unimodular lattices in $\mathbb R^d$, equipped with an $\text{SL}(d,\mathbb R)$-invariant measure $\mu$.

The classical proof that $m(\textbf{BA}(m,n))=0$, where $m$ is the Lebesgue measure on $\mathbb R^{mn}$, goes as follows:

For each $t \in \mathbb R$ and for each matrix $A$, let $g_t:=\begin{bmatrix} e^{t/m}I_m & A \\ 0 & e^{-t/n}I_n \end{bmatrix}$ and $u_A:=\begin{bmatrix} I_m & A \\ 0 & I_n \end{bmatrix}$, where $I_k$ denotes the $k$-dimensional identity matrix.

By Dani's Correspondence (see the paper I linked below), $A$ is badly approximable if and only if the trajectory $(g_tu_A\mathbb Z^d)$ is bounded. Using this and the ergodicity of the action of $(g_t)$, one can conclude that

$$\mu (\{u_A \mathbb Z^d: A\in \textbf{BA}(m,n) \})=0. (*)$$

My question is why $\mu (\{u_A \mathbb Z^d: A\in \textbf{BA}(m,n) \})=0$ implies $m(\textbf{BA}(m,n))=0$.

This last step seems to be omitted in any literature I have seen, including Dani's original paper I believe. They just argue that $(*)$ holds (or with $P^-$ matrix multiple) and take $m(\textbf{BA}(m,n))=0$ for granted from that (See the top of the page 66 from the paper I linked).

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In essence, this question is about the relation between Haar measure and Lebesgue measure. Please note in general, $\mu(\{u_A \mathbb Z^d: A\in S \subset \mathbb R^{mn}\})=0$ does not imply $m(S)=0$ (take $S=\mathbb R^{mn}$ for example). I really wish someone could explain the Haar measure passing to Lebesgue measure part well. I think I am okay with dynamical system part.

Answer 1

You have that $\mu\left\lbrace x \in X_d : \{g_tx\}_{t>0} \text{ is bounded}\right\rbrace = 0\ \ $ $(\star)$.

If you look at a generator $H$ for $g_t$ in the Lie algebra, you see that under the action of $ad_H$, $\operatorname{Lie}(G)$ is diagonable. Decompose the lie algebra into three subspaces, those with positive eigenvalue $\mathfrak{g}^+$, those with negative eigenvalue $\mathfrak{g}^-$ and those with zero eigenvalue $\mathfrak{h}$. Under the exponential map, $\mathfrak{g}^+$ corresponds with the set $\{u_A\}$, $\mathfrak{h}$ corresponds to some matrices which commute with $g_t$, and $\mathfrak{g}^-$ corresponds with the set $\{(u_A)^t\}$. For small values of $(Y^-,T,Y^+) \in \mathfrak{g}^-\times \mathfrak{h}\times \mathfrak{g}^+$, mapping this to $\exp(Y^-)\exp(T)\exp(Y^+)u_A\Gamma$ gives a coordinate system of the manifold $X_d$ near $u_A\Gamma$.

Moreover, the Haar measure restircted to this chart is mutually absolutely continuous with respect to $(d^2-1)$-dimensional Lebesgue measure on this chart. Thus $(\star)$ gives that Lebesgue almost no lattice here has bounded orbit. Now it is easy to check using commutators that if some $\exp(Y^+)u_A\Gamma$ is bounded under $g_t$, then so is $\exp(Y^-)\exp(T)\exp(Y^+)u_A\Gamma$ (For every small $Y^-, T$!). Now Fubini's theorem tells you that $mn$-dimensional Lebesgue almost no $\{u_B :B \text{ near } A\}$ has bounded orbit.

I think this is exactly what Asaf says although he's dealing with a global decomposition of the group (which is probably a more professional way to think about this).

Hope this helps somewhat.

Answer 2

Notice that every $m\in SL_{n}(\mathbb{R})$ can be written as $m=u\cdot q$, where $q$ is an ``opposite horospherical'' (maybe with center), namely it is not getting expanded by the $g_{t}$ action. Therefore $g_{t}.m=g_{t}.u.g_{-t} \cdot g_{t}.q$ with $g_{t}.q$ uniformly bounded for $t\geq 0$. Hence one only worries about what happens to the expanding horospherical subgroup of $g_{t}$ (=the unstable part), regarding divergence. Ergodicity of $g_{t}$ shows that for almost every $x=m\Gamma$, the $g_{t}$-orbit of $x$ diverges. Using Fubini over $G$ (notice that in this decomposition I mentioned, the measures decompose appropriately), for almost every $u\in U$ (this is with respect to the Haar measure on $U$), the orbit $g_{t}.u.\Gamma$ is unbounded, hence the set $BA$ is of measure zero.

Edit: Starting from the lower-upper decomposition of any element $g\in G$, we have $g=qp$ where $q$ is lower triangular namely $q\in P^{t}$, $p\in P$ for $P=AN$. As the $g_{t}$ action is non-expanding over $q$, we may ignore the $q$ part, namely the divergence of $g\in\Gamma$ and of $p\in\Gamma$ is the same. Write $P=AN$, $A$ commutes with $g_{t}$, hence we may ignore the $A$ part as well. Now consider $N$. We may write $N=N^{0}\cdot N^{+}$ where $N^{0}$ is not expanded by the $g_{t}$ action. This can be done for example via an appropriate Lie algebra decomposition, and noticing that commutators are expanded (which is essentially saying that $N^{+}$ is normal inside $N$). So again, you only need to worry about $N^{+}$ for divergence, and this is exactly your subgroup $u_{A}$. All of the measure considerations clearly follow by disintegration and Fubini arguments.