Let $\Phi$ be an irreducible root system of rank $\ell$. The fundamental invariants of $\Phi$ is a set of $\ell$ integers $d\_1, \cdots, d\_\ell$ canonically attached to $\Phi$. 


Now suppose $\Psi$ is a closed subsystem of $\Phi$ and suppose $\Psi$ is generated by a subset of simple roots of $\Phi$. (In the language of Lie algebras, $\Psi$ defines a Levi subalgebra of the simple Lie algebra corresponding to $\Phi$.) 


Question 1: Is it true that every fundamental invariant of $\Psi$ is a divisor of a fundamental invariant of $\Phi$? 


Now suppose I drop the condition that $\Psi$ is generated by a subset of simple roots of $\Phi$. For example, $G\_2$ contains $A\_2$ as a closed subsystem, but this subsystem is not generated by simple roots of $G\_2$. (In the language of Lie algebra, this is saying that $\Psi$ defines a generalized Levi subalgebra. These arise in the study of parahoric subgroups of the loop group.)


Question 2: What is the relationship between fundamental invariants of $\Psi$ and $\Phi$? 



Fundamental invariants for root subsystems
I will give the following positive answer:


Let H be a split reductive subgroup of the split reductive group G of the same rank. Let WH and WG be their Weyl groups. Then every fundamental invariant of WH divides a fundamental invariant of WG.


Unsatisfactionally, this proof will not give a bijection between fundamental invariants satisfying a divisibility condition.


This includes all the usual examples of Levi subgroups of reductive groups, as well as some funkier examples such as Spin(9)⊂F4.


Let PG(t) be the Poincaire polynomial
$$P\_G(t)=\sum\_{w\in W\_G} t^{\ell(w)}$$
and similarly for PH(t).


Since H(q) is a subgroup of G(q), Lagrange's theorem, together with the formula for the order of a reductive group over a finite field, gives us the divisibility relation
$$P\_H(q)\mid q^NP\_G(q)$$
for some positive integer N. (I've already cancelled the common factor of (q-1)r since G and H both have rank r).


And as PH has constant term 1, we can strengthen this to PH(q) dividing PG(q).


Once we have this divisibility relation for infinitely many q, this implies that PH(t) divides PG(t) in ℤ[t]. This may not be immediately obvious so here is the proof: Since PH is monic, you can use the division algorithm to write PG=qPH+r with the degree of r less than the degree of PH. For infinitely many prime powers q, |r(q)|<|PH(q)|. Thus the divisibility relation implies that for such q, r(q)=0. r now has infinitely many roots, so is the zero polynomial.


Now let d be a fundamental invariant of WH and let ζ be a primitive d-th root of 1.Then PH(ζ)=0. The divisibility result we've just established implies that PG(ζ)=0 as well. Since PG(t) divides the product of te-1 as e runs over all fundamental invariants of WG, it must be that d divides one of these fundamental invariants. QED.


