Consider a Hermitian symmetric pair of complex Lie algebras $(\mathfrak{g},\mathfrak{k})$ and split the set of roots into compact roots (i.e. roots of $\mathfrak{k}$) and noncomapt roots $\Delta = \Delta_c \cup \Delta_n$.

Pick a weight $\lambda$ and define the set of $\lambda$-singular roots $\Psi_\lambda$ as the subset of roots orthogonal to $\lambda+\rho$. Now consider a subgroup $W_\lambda$ of the Weyl group of $\mathfrak{g}$ generated by the reflections $s_\beta$ for $\beta \in M_\lambda$, where $M_\lambda$ is the subset of noncompact roots that satisfy the following three conditions

- $\beta$ is orthogonal to $\Psi_\lambda$
- scalar product of $\beta$ with $\lambda+\rho$ is a natural number
- if there is a long root in $\Psi_\lambda$, then $\beta$ is short

It is known (by a result of Dyer) that $W_\lambda$ is in fact Weyl group of a root subsystem $\Delta_\lambda$ of $\Delta$.

The article Resolutions and Hilbert series of the unitary highest weight modules of the exceptional groups provides an algorithm for computation of simple roots of $\Delta_\lambda$ in section 5.2. It uses a partial ordering (see section 4.6) on positive noncompat roots such that $\beta$ covers $\alpha$ iff $\beta = \alpha +\alpha_i$ for some simple root $\alpha_i$. The claim is that the set of simple roots is given by differences of successive elements of $M_\lambda$.

Q1:How to prove this?

Q2:Is this true also for the classical cases?Q3: Was this order on positive roots already studied?

Note that the weight is not arbitrary but it is such that the irreducible highest weight module $L_\lambda$ is in fact unitarizable.

Math. Annalen(in fact it was published inAnnals of Math.159 (2004), no. 1, 337–375). – Jim Humphreys Mar 22 '13 at 19:36