The Hausdorff codimension of singular matrices vs. the Hausdorff codimension of points with divergent trajectories

Question

Let $G:=SL(m+n,\mathbb R)$ and $\Gamma :=SL(m+n,\mathbb Z)$ and $X:=G/\Gamma$.

(1) Let $M$ denote the set of all $m \times n$ matrices with real entries. A matrix $A \in M$ is called $\textit{singular}$ if for all $\epsilon > 0$, there exists $Q_\epsilon$ such that for all $Q \ge Q_{\epsilon}$, there exist integer vectors $p \in \mathbb Z^m$ and $q \in \mathbb Z^n$ such that

\begin{equation} \|Aq+p\|\le \epsilon Q^{-n/m} ~\text{and}~ 0<\| q \| \le Q. \end{equation}

We denote the set of singular $m\times n$ matrices by $\textbf{Sing}_{m,n}$

By Dani's correspondence principle (1985), this is equivalent to saying that $(g_t u_A \mathbb Z^n)$ is divergent in the space of unimodular lattices where $g_t:=\begin{bmatrix} e^{t/m}I_m & 0 \\ 0 & e^{-t/n}I_n \end{bmatrix}$ and $u_A:=\begin{bmatrix} I_m & A \\ 0 & I_n \end{bmatrix}$.

(2) Let $F^+:=\{g_t:t\ge 0\}$ and let $D(F^+, X)$ be the set of points $X$ such that the trajectory $F^+ x$ is divergent ("leaving any compact set").

I wonder how to show the following equation of Hausdorff dimensions:

$$\dim(X)-\dim(D(F^+, X))=mn- \dim(\textbf{Sing}_{m,n})$$

Intuitively this is very true through Dani's correspondence ("codimension"="codimension"!). But to prove it rigorously, are there any key theorems about the Hausdorff dimensions involved?

Answer 1

It is evident that the singular vectors are defined as the ``$u_{A}$-part which is $g_{t}$ divergent in the future'', this gives $m\cdot n$ ($=\dim \left(u_{A}\right)$) minus the dimension of the singular vectors.

The question here is what's going on with the ''extra dimensions''. Notice that $u_{A}$ is a unstable horospherical subgroup of $g_{t}$.

By definition you may consider its Lie algebra as the part of the $Lie(G)$ for which $Ad(g_{1})$ is expanding (notice that $Ad(g)$ is indeed diagonalizable).

For the ''other directions'', as $Ad(g_{t})$ is not expanding, assume you as given something like $\exp(\underline{h}).\mathbb{Z}^{m+n}$ then you may write $g_{t}.\exp(\underline{h}).\mathbb{Z}^{m+n} = \exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n}$. Notice that $g_{t}.\mathbb{Z}^{m+n}$ is diverging, and $$ dist(\exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n},g_{t}.\mathbb{Z}^{m+n}) \ll_{h} 1$$ due to the fact that $Ad(g_{t})$ is not expanding $\underline{h}$.