Every positive system of roots contains a unique simple system.

Question

The following question is in effort to understand a proof to a theorem appearing in "Reflection Groups and Coxeter Groups" by Humphreys on page 8.

Let $\Phi$ be a root system in the euclidean space $V$, that is, a set of elements in $V$ such that:

1. $\Phi \cap \mathbb{R}\alpha = \{\pm\alpha\}$ for all $\alpha\in\Phi$.
2. $s_\alpha\Phi = \Phi$ for all $\alpha\in\Phi$ ($s_\alpha$ is the reflection sending $\alpha$ to $-\alpha$ and fixing the hyperplane $H_\alpha$ orthogonal to $\alpha$).

A positive system in $\Phi$ is a subset $\Pi$ of $\Phi$ containing all the positive roots in $\Phi$ relative to some linear order on $V$.

A simple system in $\Phi$ is a subset $\Delta$ of $\Phi$ which is a base to the space spanned by $\Phi$ and every element in $\Phi$ is a linear combination of $\Delta$ in which all the coefficients have the same sign.

Suppose that a simple system $\Delta$ is contained in a positive system $\Pi$ (relative to some linear ordering of $V$).

I would like to prove that $\Delta$ may be characterized as the set of all roots $\alpha$ in $\Pi$ such that $\alpha$ is not expressible as a linear combination with strictly positive coefficients of two or more elements of $\Pi$ (which Humphreys states should follow easily from the definitions).

Any help would be appreciated.

Answer 1

Let $\Phi$ be a root system in $V$, $\Pi$ a positive system and $\Delta\subseteq \Pi$ a simple system. We need to prove that $\alpha\in \Delta \Leftrightarrow \alpha$ is not a linear combination of two or more elements of $\Pi$ with strictly positive coefficients.

(*) Note that for any linear order of $V$, if $\alpha\in\Pi$ then $-\alpha\in -\Pi$. Thus, by the definition of a root system, $\alpha$ is not a scalar multiple of any other element in $\Pi$.

$\Leftarrow$: If $\alpha$ is not a linear combination of two or more elements in $\Pi$, then (together with (*)) it must be in $\Delta$ (since otherwise $\alpha\notin \operatorname{span}(\Delta)$).

$\Rightarrow$: Assume the contrary that $\alpha\in \Delta$ and that $\alpha$ is a linear combination $\sum_{i=1}^{n}b_i\beta_i$ of two or more elements of $\Pi$ with strictly positive coefficients.
Every $\beta_i\in \operatorname{span}(\Delta)$. At least one element in this linear combination, say $\beta_t$, must be in $\Pi\setminus\Delta$, otherwise $\Delta$ is linearly dependent.
By (*) $\beta_t$ is a linear combination of at least two elements in $\Delta$ with strictly positive coefficients.
Thus, the entire linear combination of positive roots $\sum_{i=1}^{n}b_i\beta_i$ can be replaced by a linear combination of simple roots (elements of $\Delta$), with at least two summands with strictly positive coefficients. This implies that $\Delta$ is linearly dependent, contradicting the definition of $\Delta$.

Answer 2

It is actually very easy to show that: $\alpha\in\Delta\Leftrightarrow\alpha$ cannot be decompose into sum of two or more positive roots. Now prove it by contradition

Actually, if assume that $\alpha\in\Delta$ and $\alpha=\alpha_1+\alpha_2$ with $\alpha_1,\alpha_2$ is positive roots.

Then for any $X\in\mathfrak{h}$, where $\mathfrak{h}$ is compact Lie algebra, if $\alpha_1(X)>0$ and $\alpha_2(X)>0$, then we have $\alpha(X)>0$. Now consider the fundamental Weyl Chamber $C_0=\cap_{\alpha\in\Pi}\{X\in\mathfrak{h}\mid \alpha(X)>0\}$. We know that system of simple roots $\Delta$ is the unque minimal set s.t. $C_0=\cap_{\alpha\in\Delta}\{X\in\mathfrak{h}\mid \alpha(X)>0\}$. thus we get the contradiction.