# Can we have a nontrivial division of a irreducible root system as $\Phi=\Phi_{[\lambda]}\cup \Phi_{[\mu]}$?

Let $$(\mathfrak{g},\mathfrak{h},\Phi)$$ be a root system of a complex simple Lie algebra, where $$\Phi$$ is the set of all roots. For each $$\alpha\in \Phi$$, let $$\alpha^{\vee}=2\alpha/(\alpha,\alpha)$$ be the coroot. Let $$\Lambda_r$$ be the root lattice and $$W$$ be the Weyl group. Here the root system is irreducible.

Now for each $$\lambda\in \mathfrak{h}^*$$ we define $$\Phi_{[\lambda]}:=\{\alpha\in \Phi|(\alpha^{\vee},\lambda)\in \mathbb{Z}\}$$ and $$W_{[\lambda]}:=\{w\in W|w\lambda-\lambda\in \Lambda_r\}.$$ Jantzen has prove that $$\Phi_{[\lambda]}$$ is a root system in its $$\mathbb{R}$$-span and $$W_{[\lambda]}$$ is the Weyl group of $$\Phi_{[\lambda]}$$.

Now consider $$\lambda$$, $$\mu\in \mathfrak{h}^*$$. Then we get $$\Phi_{[\lambda]}$$, $$W_{[\lambda]}$$, $$\Phi_{[\mu]}$$, and $$W_{[\mu]}$$.

My question is: if $$\Phi_{[\lambda]}\cup \Phi_{[\mu]}=\Phi$$, then is it true that one of them must be the whole $$\Phi$$?

For example we consider the root system $$B_2$$. Let $$\alpha$$ be the short simple root so $$\alpha^{\vee}=\alpha$$. Consider $$\lambda=\alpha/2$$ and we can show that $$\Phi_{[\lambda]}=\{\text{the four short roots}\}$$. Hence to make sure that$$\Phi_{[\lambda]}\cup \Phi_{[\mu]}=\Phi$$, we must choose $$\mu$$ such that $$\Phi_{[\mu]}$$ contains the four long roots. But we can show that a $$\Phi_{[\mu]}$$ that contains the four long roots must also contain the four short roots.

Of course it is not true if we do not require that the root system is irreducible.

• So it seems that we at least need $\mathrm{Span}_{\mathbb{R}}(\Phi_{[\lambda]}) = \mathrm{Span}_{\mathbb{R}}(\Phi_{[\mu]})=\mathrm{Span}_{\mathbb{R}}(\Phi)$, is that right? Sep 29 '19 at 15:38
• @SamHopkins Actually one of $\text{Span}_{\mathbb{R}}(\Phi_{[\lambda]})$ and $\text{Span}_{\mathbb{R}}(\Phi_{[\mu]})$ must be the whole $\text{Span}_{\mathbb{R}}(\Phi)$. Sep 29 '19 at 18:52
• Hmm, I'm trying to think of an example of an irreducible crystallographic root system $\Phi$ and a non-trivial sub-root system $\Phi' \subseteq \Phi$ for which the set of "missing" roots $\Phi\setminus\Phi'$ does not span the whole space. But I can't think of one. Do you know of such an example? Sep 29 '19 at 18:54
• @SamHopkins I cannot find one either. Maybe you are right, they both span the whole space. Sep 29 '19 at 19:34

I think the answer is yes because $$(\Phi_{[\lambda]})^{\vee}$$ and $$(\Phi_{[\mu]})^{\vee}$$ are closed sub-root systems of the dual root system $$\Phi^{\vee}$$. Closed means if $$\alpha$$ and $$\beta$$ are roots in $$(\Phi_{[\lambda]})^{\vee}$$ and $$\alpha+\beta$$ is a root in $$\Phi^{\vee}$$, then $$\alpha+\beta$$ is also a root in $$(\Phi_{[\lambda]})^{\vee}$$.