Set of roots such that sum of them with given positive root is not a root

Question

Let $\Phi$ be a indecomposable root system of rank greater than 1 and fix a positive root $\alpha$. How can one describe the set $ B = \{ \beta \in \Phi \;| \; \alpha + \beta \notin \Phi \}$? It's obvious that all orthogonal roots, at least one simple root and sum of all simple roots lie there, but what else?

Answer 1

It is not the case that all orthogonal roots will live in $B$. A root $\alpha$ being orthogonal to $\beta$ will not always force $\alpha + \beta \notin \Phi$. Instead it merely implies that the $\beta$-root string is symmetric around $\alpha$ i.e. $\alpha - k\beta,\dots,\alpha,\dots,\alpha+k\beta$ are all roots for some $k$ but $\alpha - (k+1)\beta,\alpha+(k+1)\beta$ are not.

As an example look at the root system $B_2$ where there are orthogonal short roots which can be added to form a long root.

Instead the condition you might want is called strongly orthogonal: $\alpha,\beta$ are strongly orthogonal if $\alpha\pm\beta \notin \Phi \cup \{0\}$.

As stated your set will also contain $-\beta$ since $0\notin\Phi$.

I'm not sure that the sum of all simple roots is guaranteed either (again $B_2$ provides a counterexample taking $\beta$ to be the short simple root). The highest root however will be in there.

Edit: If we're allowed to rechoose the ordering so that $\beta$ is the highest root (assuming it is a long root of course) then $B$ is the union of the positive roots, the roots which are strongly orthogonal to $\beta$ and $\{-\beta\}$ (I suspect you may want to exclude this by replacing $\Phi$ by $\Phi \cup \{0\}$ in your definition of $B$). I'm not sure of an easy way to describe it in the case of a short root or if the ordering is fixed.