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If $\Phi$ is the root system of a complex semisimple Lie algebra $\mathfrak{g}$, then one way to choose a set $\Phi^+$ of positive roots is by choosing a hyperplane which does not contain any root, which will partition (by looking at the corresponding two half-spaces) the set of roots $\Phi$ into two subsets. Either one of these two subsets can be taken to be $\Phi^+$.

My question is about the converse. Can any set $\Phi^+$ of positive roots be obtained by this construction?

I am mostly interested in $A_n$ but the question makes sense in general.

Malkoun
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    Trusting Wikipedia, the answer is yes. "... any two sets of positive roots differ by the action of the Weyl group." The citation for this statement is Proposition 8.28 in Brian Hall's Lie Groups, Lie Algebras, and Representations: An Elementary Introduction. It would be nice if someone with a copy of that book could type up (a sketch of) the proof as an answer. :) – Mike Pierce Mar 02 '20 at 08:31
  • @MikePierce, it makes sense. Thank you. I guess one can see that this way. There is a one-to-one correspondence between the set of positive roots and the set of simple roots. Associate to each set of simple roots its fundamental Weyl chamber. It is known that the Weyl group acts freely and transitively on the set of Weyl chambers. – Malkoun Mar 02 '20 at 08:55

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Let $V$ be the real vector space spanned by the root system, and let $L: V \rightarrow \mathbb R$ be any linear functional with the property that $L(\Phi^+) \subset \mathbb R_{> 0}$; then $\ker(L)$ is a hyperplane that does the job.

One obvious choice for $L$ is as follows: From your $\Phi^+$, choose a set of simple roots $\alpha_i$, $1 \le i \le n$. It is well known that the $\alpha_i$ are a basis of $V$, and a root $\alpha = \sum_{i=1}^n c_i \alpha_i$ is in $\Phi^+$ if and only if all $c_i \in \mathbb Z_{\ge 0}$. Now set $L(\sum_{i=1}^n c_i \alpha_i) := \sum_{i=1}^n c_i$.

(With respect to the basis $(\alpha_i)_i$, this is the functional $(1,1, ...,1)^{tr}$. Notice that because of the fact that for all roots, the coefficients $c_i$ are either all $\ge 0$ or all $\le 0$, one actually has a lot of "wiggle room": any $(\ell_1, ..., \ell_n)^{tr}$ with all $\ell_i > 0$ gives another suitable hyperplane.)

Torsten Schoeneberg
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