TITLE: Positive roots expressible as a combination of simple roots whose reflections pairwise commute?
QUESTION [1 upvotes]: Suppose you have a subset $\alpha_1,\dots,\alpha_r$ of simple roots in a reflection group whose corresponding reflections $\{s_1,\dots,s_r\}$ pairwise commute. 
Why are the only positive roots expressible as a linear combination of these $\alpha_i$ the $\alpha_i$ themselves?
This fact comes up when counting the number of hyperplanes that intersect a particular plane in computing the order of a Coxeter element.

REPLY [3 votes]: Let $\alpha$ be a positive root. Then there exists $1\leq i\leq r$ and $w\in W$ such that $\alpha=w(\alpha_i)$ where $W$ is the group generated by $s_1,\ldots,s_r$. (This is pretty clear, either by definition of positive roots, or by other considerations...) Write $w=s_{i_1}\ldots s_{i_l}$ for some $1\leq i_1,\ldots,i_l\leq r$, e.g. by taking a reduced expression of $w$.
Now, $s_i$ and $s_j$ commute if and only if $i=j$ or either $\alpha_i$ and $\alpha_j$ are orthogonal. (You can check this directly by expanding the refelctions.) 
This means that either $i_j=i$ or $s_{i_j}(\alpha_i)=\alpha_i$ for all $1\leq j\leq l$. In the first case, you get $s_{i_j}(\alpha_i)=s_i(\alpha_i)=-\alpha_i$. Consequently, you have $\alpha=w(\alpha_i)=\pm\alpha_i$ and since $\alpha$ was supposed to be positive it follows that $\alpha=\alpha_i$. QED.
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