I have a question related to geometry of numbers, which although seems quite basic, I was rather confused by it so I decided to ask here. Let $\Lambda$ be a lattice in $\mathbb{R}^n$. Let $R_1, ..., R_n$ be the successive minima of $\Lambda$ and $\mathbf{x}^{(1)}, ..., \mathbf{x}^{(n)}$ be the minimal points of $\Lambda$.

In "Analytic Methods for Diophantine Equations and Diophantine Inequalities", by Harold Davenport, in Lemma 12.2, he proves that if $d(\Lambda)=1$, then $$ 1 \leq R_1...R_n \leq 2^nJ_n, $$ where $J_n$ is the volume of a sphere of radius $1$ in $n$ dimensions. The very first step of the proof is to "rotate the $n$-dimensional space about $O$ until" the matrix $[\mathbf{x}^{(1)}, ..., \mathbf{x}^{(n)}]$ is upper triangular.

What I was wondering was that in stead of rotation, could we instead: change coordinates via elementary row operations so that, in the new coordinate system, the minimal points have the given shape? (Since elementary row operations have a non-zero determinant, we can divide the resulting transformation by their determinants to obtain a transformation which has determinant $1$.)

I felt perhaps there may be some issues with this, but I was not quite sure. I would appreciate any clarification. Thank you!


I think this step is fine: you can rotate the lattice $\Lambda$, or alternately the standard basis in $\mathbb{R}^n$, so that the matrix $[\mathbf{x}^{(1)}, ..., \mathbf{x}^{(n)}]$ becomes upper triangular. The reason is the Iwasawa decomposition which tells us that any invertible matrix is the product of an upper triangular matrix and an orthogonal matrix. See, for example, Proposition 1.2.6 in Goldfeld: Automorphic forms and L-functions for the group GL(n,R).

Note that it is important to use rotations in this argument so that minimal points remain minimal points. This is because rotations do not alter distances and linear independence.

  • $\begingroup$ Thank you for this helpful answer! I agree with you that the step "rotating the lattice $\Lambda$" is fine as well. Perhaps I was not clear enough, but the main part of my question was the third paragraph. Can we replace rotation with elementary row operations? $\endgroup$ – SJY Jul 10 '14 at 16:47
  • $\begingroup$ I am guessing from your answer however that the answer to my question is no though... $\endgroup$ – SJY Jul 10 '14 at 16:48
  • $\begingroup$ @SJY: Elementary row operations will not do, because the columns of $[\mathbf{x}^{(1)}, ..., \mathbf{x}^{(n)}]$ will no longer be minimal points. $\endgroup$ – GH from MO Jul 10 '14 at 18:23

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