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.

  • $\begingroup$ 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? $\endgroup$ Sep 29, 2019 at 15:38
  • $\begingroup$ @SamHopkins Actually one of $\text{Span}_{\mathbb{R}}(\Phi_{[\lambda]})$ and $\text{Span}_{\mathbb{R}}(\Phi_{[\mu]})$ must be the whole $\text{Span}_{\mathbb{R}}(\Phi)$. $\endgroup$ Sep 29, 2019 at 18:52
  • 1
    $\begingroup$ 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? $\endgroup$ Sep 29, 2019 at 18:54
  • $\begingroup$ @SamHopkins I cannot find one either. Maybe you are right, they both span the whole space. $\endgroup$ Sep 29, 2019 at 19:34

1 Answer 1


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 $(\Phi_{[\lambda]})^{\vee}$.

The corresponding answer for closed sub-root systems of a irreducible root system is answered for this MO question.


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