Caleb Proctor

2022-07-03

Let $\mathrm{\Delta }$ be an indecomposable root system in a real inner product space $E$, and suppose that $\mathrm{\Phi }$ is a simple system of roots in $\mathrm{\Delta }$, with respect to an ordering of $E$. If $\mathrm{\Phi }=\left\{{\alpha }_{1},\dots ,{\alpha }_{l}\right\}$, prove that
${\alpha }_{1}+\cdots +{\alpha }_{l}\in \mathrm{\Delta }$
I know that any positive root $\gamma$ may be written as a sum of simple roots, and furthermore that every partial sum is itself a root, but I am unsure if that will help me or not. Any hints to get me started?

First, I would let $\mathrm{\Phi }$ be a base for an indecomposable root system, and choose labels for the elements of $\mathrm{\Phi }$ according to the following algorithm:
Suppose you have chosen ${\mathrm{\Phi }}_{k}:=\left\{{\alpha }_{1},\dots ,{\alpha }_{k}\right\}$ such that the corresponding root system ${\mathrm{\Delta }}_{k}$ is indecomposable (when $k=1$ this is automatic). Then, there exists ${\alpha }_{k+1}\in \mathrm{\Phi }$ such that ${\mathrm{\Phi }}_{k+1}$ is a base for an indecomposable root system ${\mathrm{\Delta }}_{k+1}$. Indeed, otherwise
$\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left({\mathrm{\Phi }}_{k}\right)\perp \mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left(\mathrm{\Phi }\mathrm{\setminus }{\mathrm{\Phi }}_{k}\right),$
so $\mathrm{\Delta }$ is not indecomposable, a contradiction.
This is enough to obtain strict inequality since, by construction, we must have $\left({\alpha }_{j},{\alpha }_{l}\right)<0$ for some $j.