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

Question

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.

Answer 1

$\def\abs#1{\lvert#1\rvert}\DeclareMathOperator\Span{Span}$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.