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Let $\Delta$ be a irreducible root system and $\Delta^+$ be its positive roots.

We say a subset $\Delta^{\prime}\subset \Delta^+$ can generate the Weyl group if reflections of roots in $\Delta^{\prime}$ generate the whole Weyl group.

If we simply consider a partition of $\Delta^+$ into disjoint union $$ \Delta^+=\Delta^+_1\amalg \Delta^+_2, $$ then it is quite possible that neither $\Delta^+_1$ nor $\Delta^+_2$ can generate the Weyl group. For example take $\Delta=B_2$ and $\Delta^+_1=$ the two long positive roots and $\Delta^+_2=$ the two short positive roots. For example $\Delta^+$ itself or the set $\Sigma$ of simple roots generates the Weyl group.

My question is: if we consider a partition of $\Delta^+$ by a hyperplane which does not contain any root. Then it is true that either $\Delta^+_1$ or $\Delta^+_2$ can generate the Weyl group by root reflections in the above sense?

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  • $\begingroup$ I feel like there should be a way to reduce this to considering $2$-dimensional subspaces... $\endgroup$
    – Sam Hopkins
    Commented Oct 2, 2019 at 22:57
  • $\begingroup$ I have the impression this is trivially true. Changing the hyperplane corresponds to a different choice of positive roots. So, your example already covers this. $\endgroup$
    – Christoph Mark
    Commented Oct 6, 2019 at 4:55
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    $\begingroup$ @user66288 This is not trivially true. Actually let $\Phi^+$ and $\Psi^+$ be too positive root systems. My question is equivalent to the question whether one of $\Phi^+\cap \Psi^+$ and $\Phi^+\cap \Psi^-$ generates the whole Weyl group. $\endgroup$
    – Zhaoting Wei
    Commented Oct 6, 2019 at 9:26
  • $\begingroup$ You are right. I got it wrong. $\endgroup$
    – Christoph Mark
    Commented Oct 6, 2019 at 10:55

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