0
$\begingroup$

I was wondering : given a full rank lattice $\Lambda$ of $R^n$ (a discrete subgroup spanning $R^n$) the successive minima of $\Lambda$ are for $1\leqslant i \leqslant n$ $\lambda_i= \min\{r>0 \mid \text{exists i linearly independants vectors of $\Lambda$ un the ball centered in 0 of radius r}\}$. Every text talking about theses say that there exists $u_1,...,u_n$ in $\Lambda$ such as for all $1\leqslant i \leqslant n$ $||u_i||=\lambda_i$, without giving a full proof. As I can't manage to proove it myself, maybe someone could help me.. Thank you.

$\endgroup$
1
$\begingroup$

Define the linearly independent vectors $u_1,\dots,u_n\in\Lambda$ recursively as follows. If $u_1,\dots,u_{m-1}$ has already been defined, then let $u_m$ be the shortest lattice vector that is linearly independent of $u_1,\dots,u_{m-1}$. By the definition of $\lambda_m$, it is clear that $|u_m|\leq\lambda_m$. By induction, it is also clear that $|u_1|\leq\dots\leq|u_n|$. Therefore, $|u_m|<\lambda_m$ would contradict the definition of $\lambda_m$, hence in fact $|u_m|=\lambda_m$. QED

$\endgroup$
3
  • $\begingroup$ Why $ |u_m| \leqslant \lambda_i $ ? As $u_1,..,u_m$ are independants and $|u_1|$ is the smallest vector we have $|u_1|=\min |u_i|\leqslant\lambda_m \leqslant \max |u_i|$ but i do not understand why $ |u_m| \leqslant \lambda_i $. Also , $u_1,..u_{m-1}$ are specials and maybe by chosing them a little bigger |u_m| would have been smaller . For me, it is'nt clear, but thank you for this quick review. $\endgroup$
    – Swann
    Apr 15 '18 at 6:52
  • $\begingroup$ $|u_m|\leqslant \lambda_m$ and not $\lambda_i$ I meant. $\endgroup$
    – Swann
    Apr 15 '18 at 10:46
  • $\begingroup$ @Swann: There exist $m$ linearly independent vectors in the ball $|u|\leqslant\lambda_m$. At least one of these is linearly independent of $u_1,\dots,u_{m-1}$, because the latter span an $(m-1)$-dimensional subspace. Hence the shortest vector that is linearly independent of $u_1,\dots,u_{m-1}$ has length at most $\lambda_m$, and this is what we called $u_m$. $\endgroup$
    – GH from MO
    Apr 15 '18 at 13:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.