Consider an irreducible root system associated to a complex simple Lie algebra $\mathfrak{g}$. Let $\rho$ be the half sum of positive roots and let $W$ be the Weyl group. Then what is the lattice $L$ generated by $\{w(\rho) - \rho\,\vert\,w\in W\}$?
It is easy to see that $L$ is a sublattice of the root lattice $Q$, and I have checked that $L$ coincides with $Q$ for $A_1, A_2, A_3$ and $G_2$ root systems. Do $L$ and $Q$ always coincide?
I just figured that this is an easy question. Yes, $L$ is always the same as $Q$. The proof goes as follows :
A choice of positive roots is the same as the choice of a Weyl chamber, and there is Weyl group-worth of such choices. For any root $\alpha\in \Delta$, there are always a pair of choices of positive roots, say $\Delta_+$ and $\Delta_+'$ such that $\Delta_+ \setminus \Delta_+' = \{\alpha\}$ and $\Delta_+' \setminus \Delta_+ = \{-\alpha\}$. Let $w\in W$ be the element of Weyl group that transforms $\Delta_+$ to $\Delta_+'$. Let $\rho$ be the half sum of elements of $\Delta_+$. Then, $\rho - w(\rho) = \alpha$. Hence $L\supseteq Q \supseteq L$, and therefore $L = Q$.