# simple roots of a reflection subgroup

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

1. $$\beta$$ is orthogonal to $$\Psi_\lambda$$
2. scalar product of $$\beta$$ with $$\lambda+\rho$$ is a natural number
3. 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?

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