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Let $L\subset \Bbb{C}$ be a lattice, assume there is a $\gamma \in \Bbb{C}$ such that $\gamma L = L$, prove the following two facts:

(1) $\gamma$ must be roots of unity

(2) if $\gamma$ is not real, then for any element $\ell \in L$ with minimal length, we can generate the lattice using $\ell, \gamma\ell$.

My attempt:

first I prove $\|\gamma\| = 1$ :assume $m\in L$ has minimal length, then $\|\gamma m\| \ge \|m\|$ which implies $\|\gamma\|\ge 1$ similarily assume $\gamma n$ has minimal length then $\|n\|\ge \|n\gamma\|$ therefore $\|\gamma\| = 1$.

Second , if it's not roots of unity, then assume minimal length is $k$ then the points on the lattice will dense on the circle with radius $k$ (by repeating apply $\gamma$) that's not possible by change of coordinate assume generator are $(a,0)$ and $(p,q)$ ($a\ne 0, q\ne 0$) then there are infinite many point of the form $(na+mp,mq)$ with $m,n\in \Bbb{Z}$ lies on the circle, but the norm of them $(na+mp)^2 +(mq)^2\to \infty$ (since there are infinite many points $m \to \infty$ or $n\to \infty$ , if $m\to \infty$ the second term will goes to infty, if not the first term will goes to infty)

And I don't know how to prove (2)?

yi li
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  • For (1) there are only finitely many elements of minimal length. Otherwise, we have no discrete set. Let $v$ be one of them, then for all $n\in \mathbb N$ the element $\gamma^n v$ is of minimal length like you have already shown. By pigeon's hole's principle we get that there are $m, n\in \mathbb N$ with $m>n$ and $\gamma^mv=\gamma^nv$. Hence, $\gamma^{m-n}=1$. – Jochen Nov 24 '22 at 11:21
  • thank you @Jochen, how to show the lattice is discrete set? – yi li Nov 24 '22 at 11:30
  • Hm, isn't it part of the definition? For me a lattice is a discrete subgroup of $\mathbb R^n$. But maybe you use another one? – Jochen Nov 24 '22 at 11:33
  • In my definition $L= {am+bn\mid a,b\in \Bbb{Z}}$ with $m,n \in \Bbb{C}$ not colinear – yi li Nov 24 '22 at 11:34
  • Ah ok, yes, then these definitions are equivalent. Wait a minute. – Jochen Nov 24 '22 at 11:35
  • Emm, I have proved the norm of infinite sequence on the lattice must goes to infty, therefore there can no have accumulated points. – yi li Nov 24 '22 at 11:36
  • I found someone asked the same question for (2) https://math.stackexchange.com/a/250354/360262 – yi li Nov 24 '22 at 11:46

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