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Let $\Lambda\subset\mathbb{R}^n$ a lattice, i.e., a discrete subgroup that spans $\mathbb{R}^n$. Now we can look at the torus $T=\mathbb{R}^n/\Lambda$ which naturally carries the metric $d_T$ induced by the euclidean metric $d$ on $\mathbb{R}^n$: $$d_T(x+\Lambda,y+\Lambda):=\min_{a,b\in\Lambda}(d(x+a,y+b)).$$ Now my question is the following: What information about $\Lambda$ can be recovered from the metric space $T$? Can we completely recover $\Lambda$ (up to some orthogonal equivalence)?

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Yes, we can. Consider the universal cover $U$ of your torus $T$. One can easily show that $U=R^n$ equipped with a Euclidean metric. So we have $f:R^n\to T$,

The $f$-preimage of a point is your lattice, up to the shift of the origin (to one point of this preimage) and an orthogonal transformation (an isometry of the Euclidean metric).

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    $\begingroup$ A missing detail: $T$ is considered as metric space. So $U$ is a topological space and hence "locally" a metric space. One then has to endow $U$ with corresponding length distance, i.e., the largest distance which is locally bounded above by the original bounded distance. This is indeed isometric to the original Euclidean distance. $\endgroup$
    – YCor
    Commented Jan 21, 2023 at 8:58

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