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

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?

• Not sure what kind of description you are looking for here that is simpler than the given one. I assume you have inspected this in the case of rank $2$, where one already sees that neither this subset nor its complement are root subsystems, and there is just a little symmetry within. It seems like one succinct way to describe its complement would be as some $\alpha$-strings through certain roots, excluding the end of the string. Commented May 9, 2022 at 20:13

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.

• Thank you for your answer! It really helped to specify the problem. The real problem i want to solve is to determine whether i can express all roots of Φ as γ such that γ+β∈Φ for some β∈B or as −β (you are right i don't want −β to be part of the set) for some β∈B. Or if it is true for some roots ( i can consider α a simple root for example). Oh, and im considering an irreducible root system. Maybe you know in which direction i should go? Hints and literature advices would be appreciated. Commented May 9, 2022 at 14:11
• Sorry I have just realised that I swapped the roles of $\alpha$ and $\beta$ in my answer. Hopefully that wasn't too confusing. If I understand your comment it is true for any root except perhaps $\alpha$ itself (assuming again that $\alpha$ is long). We can without loss of generality take $\alpha$ to be the highest root. Then $B$ contains the positive roots and every root $\gamma$ has the property that $\gamma + \zeta$ is a root for some positive root $\zeta$ except for the highest root $\alpha$ itself. Then if as in your original statement $-\alpha \in B$ you can include $\alpha$. Commented May 11, 2022 at 12:24