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?

  • $\begingroup$ I'm not an expert in the area at all, but wanted to point out the paper "Hausdorff dimension of singular vectors" by Cheung and Chevallier, which is, at least, seems to closely related to your question. $\endgroup$ Jan 26 at 4:34
  • $\begingroup$ @AnthonyQuas Thanks, I have seen a lot of works of Cheung et al. But what I was asking is actually something they believe "obvious" in their (and others') papers, but is very hard to me... $\endgroup$
    – No One
    Jan 26 at 18:51

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 $\underline{h}$.


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