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 $\mathrm{\Phi}=\{{\alpha}_{1},\dots ,{\alpha}_{l}\}$, 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?

${\alpha}_{1}+\cdots +{\alpha}_{l}\in \mathrm{\Delta}$

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?

gozaderaradiox5

Beginner2022-07-04Added 19 answers

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 ${\mathrm{\Phi}}_{k}:=\{{\alpha}_{1},\dots ,{\alpha}_{k}\}$ 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

$\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}({\mathrm{\Phi}}_{k})\perp \mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}(\mathrm{\Phi}\mathrm{\setminus}{\mathrm{\Phi}}_{k}),$

so $\mathrm{\Delta}$ is not indecomposable, a contradiction.

This is enough to obtain strict inequality since, by construction, we must have $({\alpha}_{j},{\alpha}_{l})<0$ for some $j<l$.

Suppose you have chosen ${\mathrm{\Phi}}_{k}:=\{{\alpha}_{1},\dots ,{\alpha}_{k}\}$ 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

$\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}({\mathrm{\Phi}}_{k})\perp \mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}(\mathrm{\Phi}\mathrm{\setminus}{\mathrm{\Phi}}_{k}),$

so $\mathrm{\Delta}$ is not indecomposable, a contradiction.

This is enough to obtain strict inequality since, by construction, we must have $({\alpha}_{j},{\alpha}_{l})<0$ for some $j<l$.

Lucian Maddox

Beginner2022-07-05Added 8 answers

Great, but maybe there is another solution?

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