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).

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.

  • $\begingroup$ The condition $(*)$ doesn't seem to appear anywhere in Dani's paper? $\endgroup$
    – Claude
    Mar 28, 2022 at 3:50
  • $\begingroup$ @Fancydressfatima That $A$ is in $\textbf{BA}$ is same as $(g_tu_A\mathbb Z^d)$ being bounded. See Corollary 2. 5 and Proposition 2. 12 in the paper... $\endgroup$
    – No One
    Mar 28, 2022 at 17:37
  • $\begingroup$ Sorry sir, I don't understand where the condition comes in although it doesn't matter since there's an answer now. $\endgroup$
    – Claude
    Mar 29, 2022 at 2:27
  • $\begingroup$ @Fancydressfatima Well that condition is a partial conclusion in the paper that is not explicitly written down, and as of now the answer doesn't show why the Lebesgue measure of BA is zero. It only talks about the Haar measure on that homogeneous space... $\endgroup$
    – No One
    Mar 29, 2022 at 14:26

2 Answers 2


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.


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.

  • $\begingroup$ Sorry but your answer still focuses on the Haar measure on homogeneous space level, but my question is how do you pass it to Lebesgue measure in mn dimensional space $\endgroup$
    – No One
    Mar 29, 2022 at 14:14
  • 1
    $\begingroup$ @question I did write the Haar measure on $U$, what you call ``the Lebesgue measure''... $\endgroup$
    – Asaf
    Mar 29, 2022 at 22:01
  • $\begingroup$ Well, $u$ is a matrix and this identification of Haar measure and Lebesgue measure on R^mn is a little unclear to me. (Please see the last paragraph of my question) $\endgroup$
    – No One
    Mar 30, 2022 at 1:19
  • $\begingroup$ Chief, the group $U$ is isomorphic to $\mathbb{R}^{mn}$ with usual addition. Don't you think the Haar measure on $U$ is related to the $mn$-dimensional Lebesgue measure? He's already mentioned the decomposition of the Haar measure $\mu$ on $G$ into `products' of Haar measures on the various subgroups. $\endgroup$
    – Claude
    Mar 30, 2022 at 2:21
  • $\begingroup$ @Fancydressfatima 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). $\endgroup$
    – No One
    Mar 30, 2022 at 16:18

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