Let $L$ be an discrete subgroup (integral lattice, if needed) of finite-dimensional euclidian vector space, say that vector $x$ is indecomposable if it can not be represented as the sum of two shorter vectors of $L$. Define $I(L)$ - set of indecomposable vectors of $L$. Then, the following relation on vectors of $I(L)$: $(x,y) \neq 0$ is an equivalence (where $( \cdot, \cdot)$ is positive definite bilinear form). Can someone explain why is it transitive?
Many thanks!
I don't believe this is true. Consider $\Bbb Z^3$. The indecomposable vectors will be $0$ and the $\pm e_i$ where $e_i$ are the unit vectors. Perturb the lattice slightly by replacing the basis vector $e_3$ by $e_3'=a e_1+b e_2+e_3$ where $a$ and $b$ are very small but nonzero. Then $e_1$, $e_2$ and $e_3'$ will be indecomposable in the new lattice, but $(e_1,e_3')\ne0$, $(e_3',e_2)\ne0$ and $(e_1,e_2)=0$.