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Let $\Lambda$ be a lattice in $\mathbb{R}^n$ (i.e. a discrete subgroup spanning the whole space). Given a basis $e_1,\dots,e_n$ of the lattice, we can consider the fundamental parallelogram $P$ induced by these vectors.

The following result holds:

Let $P'$ be a non-degenerate parallelogram induced by some $f_1,\dots,f_n\in\Lambda$. If $P$ contains no points of $\Lambda$ excepts its vertices, then $f_1,\dots,f_n$ is a basis for $\Lambda$.

We can write $(f_1,\dots,f_n)^t=M(e_1,\dots,e_n)^t$ with $M\in \text{GL}_n(\mathbb{R})\cap M_n(\mathbb{Z})$. To conclude, we would need to show that $M\in\text{GL}_n(\mathbb{Z})$.

I have found a nice geometric argument (although maybe missing some details), but I was wondering if one could give a purely algebraic proof of this result ? I haven't been able to find it in the books I consulted.

Klaus
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1 Answers1

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Points inside $P'$ are linear combinations with all coefficients in $[0,1]$, but not all in $\{0,1\}$. Assume $f$ is not a basis. Then there exists a point $x\in\lambda$ that is not in $\sum f_i\mathbb Z$, hence (as $f$ is at least a basis of $\mathbb R^n$) $$ x=\sum_{i=1}^n c_if_i$$ with not all $c_i\in\mathbb Z$. Let $r_i=c_i-\lfloor c_i\rfloor$. Then $$ y:=\sum_{i=1}^n r_i f_i$$ is in $\Lambda$ and also in $P'$ because all $r_i$ are in $[0,1]$. Since at least one $c_i$ is $\notin\mathbb Z$, the corresponding $r_i$ is $\in(0,1)$, i.e. $y$ is not a vertex of $P'$.

Thus by contraposition, if $P'$ contains no lattice points apart from the vertices, $f$ is a basis of $\Lambda$.

Hagen von Eitzen
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