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This question is dealing with Mahler's criterion which tells us when a subset of unimodular lattices has compact closure when identify set of all unimodular lattices, denoted by $\mathcal{L}$, with ${\rm SL}(n,\mathbb{R})/{\rm SL}(n,\mathbb{Z})$. Chabauty was a first mathematician who noticed $\mathcal{L}$ enjoys a topology which is homeomorphism as a topological space with ${\rm SL}(n,\mathbb{R})/{\rm SL}(n,\mathbb{Z})$. There is another way to look at lattices which is try to define a metric on $\mathcal{L}$. This metric can be defined:

Let $\Gamma, \Lambda\in \mathcal{L}$ and assume $$\mu:={\rm Inf}_{\sigma\Lambda=\Gamma}|\sigma-I| $$

and

$$\\eta:={\rm Inf}_{\tau\Gamma=\Lambda}|\tau-I|$$

Where here $\tau,\sigma$ are linear transformations and $||$ is usual distance on linear transformation. Then define $$d(\Lambda,\Gamma)=\max[\log(1+\eta),\log(1+\mu)]$$

Now I have two question:

1) I have tried to show $d(\Lambda,\Gamma)=0$ implies $\Lambda=\Gamma$ but I could not. Is it clear?

2) Is it clear this topology is locally compact?

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  • $\begingroup$ Is "Is it clear?" your question? i.e., is it clear to you? $\endgroup$
    – Wadim Zudilin
    Jan 7, 2011 at 6:37
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    $\begingroup$ (1) is easy, since $\sigma$ s.t. $\sigma\Lambda=\Gamma$ is a coset of $GL(n, \mathbb{Z})$ in $GL(n, \mathbb{R})$, and so in particular it is a discrete subset of $n\times n$ matrices. $\endgroup$
    – Daniel Litt
    Jan 7, 2011 at 7:51
  • $\begingroup$ I was worried that an argument for (2) would be too long to fit in the margins of MO last night, but I don't think that's the case. It shouldn't be hard to show that a sequence of lattices $(\Lambda_i)$ converges to $\Lambda$ iff one can pick bases $(e^i_j)$ for the $\Lambda_i$ and $(e_j)$ for $\Lambda$, such that for each $j$ $e^i_j\to e_j$ (incidentally, this observation identifies the metric topology on $\mathcal{L}$ with the usual topology on $SL(n, \mathbb{R})/SL(n, \mathbb{Z})$). But then any closed $\delta$-ball is compact, by e.g. choosing convergent subsequences of short bases. $\endgroup$
    – Daniel Litt
    Jan 7, 2011 at 18:29

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