I'm stuck in this question and I would like some help.
Question: In a reduced root system let $\alpha= n_1 \alpha_1 + …+ n_k \alpha_k$ be a root such that each $\alpha_i$ is a simple root. Show that
$$n_i \frac{\langle \alpha_i , \alpha_i \rangle}{\langle \alpha , \alpha \rangle} $$
is an integer.
I tried things like writing
$$n_i \frac{\langle \alpha_i , \alpha_i \rangle}{\langle \alpha , \alpha \rangle} = n_i \frac{2\frac{\langle \alpha , \alpha_i \rangle}{\langle \alpha , \alpha \rangle}}{ 2\frac{\langle \alpha , \alpha_i \rangle}{\langle \alpha_i , \alpha_i \rangle}}, $$
and somehow try to conclude that $\left.2\frac{\langle \alpha , \alpha_i \rangle}{\langle \alpha_i , \alpha_i \rangle}\right| n_i\cdot 2\frac{\langle \alpha , \alpha_i \rangle}{\langle \alpha , \alpha \rangle}$. But this led me to anywhere.
Best Answer
We can suppose that the root system is irreducible. The only nontrivial case is when $\alpha_i$ is a small root and $\alpha$ is a long root.
When there is a double edge in the Dynkin diagram, one observes that the long roots belong to the additive subgroup generated by long simple roots and twice short simple roots. This can be seen, since long roots are inductively created either as sum of two longs roots, or one long root plus twice a small root. The same holds in type $G_2$, with 2 replaced by 3.
So, in either case, one has $\|\alpha_i\|^2/\|\alpha\|^2=1/k$, where $k\in\{2,3\}$, and $k$ divides $n_i$, so we're done.