Lattice generators - MathOverflow

Lattice generators

2011-05-17T13:46:44Z

I hope this question isn't trivial. Let $L$ be a lattice in $\mathbb{C}$ generated by two complex numbers $w_1,w_2$ which are linearly independent over $\mathbb{R}$. Let $\gamma\in\mathbb{C}$ be a root of unity such that $\gamma\notin\mathbb{R}$ and $\gamma L=L$, and let $l\in L\backslash{0}$ be an element of shortest length. Is it true that the set ${l,\gamma l}$ generates $L$? It seems almost trivial, but I have been unable to prove it. Thanks!

Answer by Charles Matthews for Lattice generators

2011-05-17T14:00:36Z

Since L actually must be the Gaussian integers or Eisenstein integers, up to a scalar factor, this is true and easy to see. There is a little algebra involved, behind scenes, in that statement. I guess you need something equivalent to class number one to be sure (a non-principal ideal in the Gaussian integers would give a different situation).

Answer by Vladimir Dotsenko for Lattice generators

2011-05-17T14:04:31Z

Since $\gamma L=L$, we have $\gamma w_1=aw_1+bw_2$, $\gamma w_2=cw_1+d w_2$ with integer $a,b,c,d$, which immediately implies that $\gamma$ is a root of a polynomial of degree $2$ with integer coefficients, so the root of unity in question can be of degree $3,4,6$ only (otherwise the cyclotomic polynomial is irreducible over $\mathbb{Z}$ and is of degree greater [or smaller] than $2$). In each of these cases the lattice generated by $l$ and $\gamma l$ satisfies the following property: for each point $A$ inside the fundamental parallelogram of that lattice there is a vertex $X$ of that parallelogram such that the distance between $A$ and $X$ is less than the length of $l$. Now, if our lattice is not $L$, take a point not in $L$ and bring it into the fundamental parallelogram by subtracting an integral combination of $l$ and $\gamma l$, thus finding a shorter element of $L$.