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One can define the convergence of a sequence $(\Lambda_k)_k$ of full rank lattices as folow : $(\Lambda_k)\underset{k\rightarrow +\infty}{\longrightarrow} \Lambda \iff \forall k\in \mathbb{N} ,\exists B_k\in (\mathbb{R}^n)^n \text{ a $\mathbb{Z}$-basis of $\Lambda_k$}, \exists B\in (\mathbb{R}^n)^n \text{ a $\mathbb{Z}$-basis of $\Lambda$}, \underset{k\rightarrow + \infty}{\lim}B_k = B $. Also, we can identify the set of all full rank lattices with the quotient $GL_n(\mathbb{R})/GL_n(\mathbb{Z})$ because if $B_1,B_2$ are $\mathbb{Z}$-basis of a lattice $\Lambda$, if $A_1,A_2$ are the $(n\times n )$ matrix with vectors of $B_i$ as columns ($i\in \{1,2\}$), it exists a matrix $P\in GL_n(\mathbb{Z})$ such that $A_1 = PA_2$.

Thus I'm looking for a distance $d$ (if it exists) for $GL_n(\mathbb{R})/GL_n(\mathbb{Z})$ such that $(\Lambda_k)\underset{k\rightarrow +\infty}{\longrightarrow} \Lambda \iff d(\Lambda_k,\Lambda)\underset{k\rightarrow +\infty}{\longrightarrow} 0$.

A guess could be $d(X,Y) = \max(\underset{(A,B)\in X\times Y , P\in GL_n(\mathbb{Z})}{\sup} \Vert AB^{-1} - P\Vert,\underset{(A,B)\in X\times Y , P\in GL_n(\mathbb{Z})}{\sup} \Vert BA^{-1} - P\Vert)$ for $ (X,Y)\in (GL_n(\mathbb{R})/GL_n(\mathbb{Z}))^2$ because $X,Y$ should be close close if $X \approx GL_n(\mathbb{Z})\cdot Y = Y$. But the "$\sup$" is not even well defined...

Thank you.

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    $\begingroup$ You have a connected smooth manifold and hence plenty of distances defining the topology. $\endgroup$
    – YCor
    Commented May 3, 2018 at 22:30

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Fixing any right invariant metric $D$ on $G=\text{GL}_n(\mathbb{R})$ you get a metric $d$ on $G/\Gamma$ ($\Gamma=\text{GL}_n(\mathbb{Z})$) be letting $d(g\Gamma,h\Gamma)=D(g,h\Gamma)$. If $D$ generates the standard topology on $G$ then (as you can check) $d$ will generate the standard topology on $G/\Gamma$. A way to get a right invariant metric on $G$ is by fixing an inner product on its Lie algebra (eg the Hilbert-Schmidt/Frobenius/$L^2$ norm on $M_n(\mathbb{R})$) and right-shifting it around, thus getting a right invariant Riemannian metric.

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  • $\begingroup$ When you say "right-shifting it around" what do you mean exacty, is it $D(g,h)=\int_{\Gamma}\Vert pg-ph\Vert d\mu$ where $d\mu$ is the Haar measure of $GL_n(\Z)$ and $\Vert \ \Vert $ the euclidean norm ? $\endgroup$
    – Swann
    Commented May 4, 2018 at 9:21
  • $\begingroup$ I identify the Lie algebra of $G$ with its tangent space at $e\in G$, $T_eG$, and consider the corresponding inner product $<.,.>_e$ on this space. I define an inner product on $T_gG$ for any $g\in G$ by setting $<X,Y>_g:=<d(R_g)_g X,d(R_g)_g Y>_e$, where $X,Y\in T_gG$, $R_g:G\to G$ is the map $x\mapsto xg^{-1}$ and $d(R_g)_g:T_gG\to T_eG$ is its differential at $g$. $\endgroup$
    – Uri Bader
    Commented May 4, 2018 at 9:44

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