# Can we have a nontrivial division of a irreducible root system as the union of two closed sub-root systems?

The question is related to this MO question. Let $$(\Phi, E)$$ be a irreducible crystallographic root system where $$\Phi$$ is the set of all roots and $$E$$ is the $$\mathbb{R}$$-span of $$\Phi$$. As in the standard terminology, we call a sub-root system $$\Phi^{\prime}\subset \Phi$$ closed if for any $$\alpha$$, $$\beta\in \Phi^{\prime}$$, $$\alpha+\beta\in \Phi$$ implies $$\alpha+\beta\in \Phi^{\prime}$$.

The Borel-de Siebenthal theorem classifies all closed sub-root systems of irreducible crystallographic root systems. See also Chapter 12 of Kane's book Reflection Groups and Invariant Theory.

My question is: for a irreducible crystallographic root system $$(\Phi, E)$$, can we find two closed sub-root systems $$\Phi_1$$ and $$\Phi_2$$ such that $$\Phi_i \neq \Phi$$ for $$i=1,2$$, and $$\Phi_1\cup \Phi_2=\Phi$$?

I believe the answer is negative and since we have the classification, we may get a proof through a exhausting all maximal closed sub-root systems. I wonder if we can prove it more theoretically.

• I originally said that the answer was positive, because we could write any non-simply laced root system as the union of its systems of long and short roots (specifically thinking of $\mathsf B_2$ as the union of two non-orthogonal $\mathsf A_1 + \mathsf A_1$'s); but that fails the 'closed' condition. Oct 1, 2019 at 19:30
• In your setup, one can show that $\Phi_1 \setminus \Phi_2$ is orthogonal to $\Phi_2 \setminus \Phi_1$, so the hard part will presumably be handling the case where $\Phi_1 \cap \Phi_2$ is large. Oct 1, 2019 at 19:32

$$\def\abs#1{\lvert#1\rvert}\DeclareMathOperator\Span{Span}$$I think I get a proof inspired by the comment of @LSpice.
First we can prove that $$\Phi_1\setminus \Phi_2$$ is orthogonal to $$\Phi_2\setminus \Phi_1$$. Pick $$\alpha\in \Phi_1\setminus \Phi_2$$ and $$\beta\in \Phi_2\setminus \Phi_1$$. Without loss of generality we can assume that $$\abs\alpha\geq \abs\beta$$. It is clear that $$s_{\alpha}\beta=\beta-2(\alpha,\beta)/(\alpha,\alpha)\alpha \in \Phi_2\setminus \Phi_1$$. On the other hand since $$\abs\alpha\geq \abs\beta$$ we have $$2(\alpha,\beta)/(\alpha,\alpha)=0$$ or $$\pm 1$$. If $$(\alpha,\beta)\neq 0$$ then $$s_{\alpha}\beta=\beta\pm \alpha \in \Phi_2\setminus \Phi_1$$. We know $$\alpha=\pm \beta\pm s_{\alpha}\beta$$ is a root in $$\Phi$$ so by the closedness of $$\Phi_2$$, $$\alpha\in \Phi_2$$. Contradiction.
Moreover for any $$\alpha\in \Phi_1$$ and $$\beta\in \Phi_2\setminus \Phi_1$$ we have $$s_{\alpha}\beta \in \Phi_2\setminus \Phi_1$$. Hence $$s_{\alpha}$$ preserves $$\Span_{\mathbb{R}}(\Phi_2\setminus \Phi_1)$$. So either $$\alpha\in \Span_{\mathbb{R}}(\Phi_2\setminus \Phi_1)$$ or $$\alpha\in (\Span_{\mathbb{R}}(\Phi_2\backslash \Phi_1))^\perp$$. Similarly for any $$\beta\in \Phi_2$$ either $$\beta\in \Span_{\mathbb{R}}(\Phi_1\setminus \Phi_2)$$ or $$\beta\in (\Span_{\mathbb{R}}(\Phi_1\setminus \Phi_2))^\perp$$.
As a result we can decompose the root system $$\Phi$$ into three disjoint parts: $$\Phi'_1=\Phi\cap (\Span_{\mathbb{R}}(\Phi_1\setminus \Phi_2))\\ \Phi'_2=\Phi\cap (\Span_{\mathbb{R}}(\Phi_2\setminus \Phi_1))\\ \Phi'_0=\Phi\cap (\Span_{\mathbb{R}}(\Phi_1\setminus \Phi_2)\oplus \Span_{\mathbb{R}}(\Phi_2\backslash \Phi_1))^\perp.$$ It is clear that $$\Phi=\bigsqcup_{i=0}^2\Phi'_i$$ and $$\Phi'_i$$, $$i=0,1,2$$ are sub-root systems. So it is contradictory to the fact that $$\Phi$$ is irreducible.
• Actually $\Phi_1 \setminus \Phi_2$ and $\Phi_2 \setminus \Phi_1$ are mutually strongly orthogonal: if $\alpha_1$ is in the former and $\alpha_2$ in the latter, and if $\alpha_1 + \alpha_2$ were a root, then it would lie in $\Phi_1$, and hence $\alpha_2 = (\alpha_1 + \alpha_2) - \alpha_1$ would also lie in $\Phi_1$; or it would lie in $\Phi_2$, and hence $\alpha_1 = (\alpha_1 + \alpha_2) - \alpha_2$ would also lie in $\Phi_2$. (Standard references show that strong orthogonality implies orthogonality, since a non-$0$ inner product tells you which of $\alpha_1 \pm \alpha_2$ is a root.) Oct 2, 2019 at 15:00
• (Our arguments for orthogonality are basically the same, but I like getting to avoid distinctions based on root lengths.) This is a nice argument! I had defined $\tilde\Phi_i = \Phi \cap (\operatorname{Span}_{\mathbb R}(\Phi_i \setminus \Phi_{\ne i}))$, roughly in the vein of your $\Phi'_i$, but couldn't figure out the analogue of your $\Phi'_0$. Oct 2, 2019 at 15:08