2
$\begingroup$

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 $\Phi$?

$\endgroup$
  • 2
    $\begingroup$ What do you mean by "fundamental invariants"? Your notation suggests that you mean the degrees of algebraically independent generators of the Weyl group invariants. If so, a look at the table of degrees shows quickly that Question 1 has a negative answer. In your Question 2, there might well be no reasonable relationship though of course the list of degrees always starts with 2. (Note in any case that the degrees depend just on the underlying Coxeter group and not the root system.) $\endgroup$ – Jim Humphreys Oct 20 '14 at 13:11
  • $\begingroup$ Yes, I do mean the degrees of algebraically independent generators of the Weyl group invariants. Thanks for pointing out the negative answer. I guess a modification of Question 1 is: is there a relationship between fundamental invariants of $\Psi$ and those of $\Phi$? $\endgroup$ – Dr. Evil Oct 21 '14 at 20:24
  • $\begingroup$ No, I don't see any interesting relationship, though it might require some case-by-case study to see if patterns are being missed. Note that your $\Psi$ involves the affine Coxeter graph (or extended Dynkin diagram) from which one or more vertices and attached edges have been removed. This idea goes back to classical work of Borel and de Siebenthal. The adjective "pseudo-Levi" is sometimes applied here, including actual Levi subalgebras of parabolic subalgebras as well as others of the type you mention. $\endgroup$ – Jim Humphreys Oct 21 '14 at 22:55
  • 1
    $\begingroup$ P.S. Looking more closely at the full list of degrees, I don't see any counterexamples to your Question 1 (unlike what I recalled previously). Still, I don't see any intrinsic relationship in terms of the invariant theory. $\endgroup$ – Jim Humphreys Oct 22 '14 at 0:15
  • $\begingroup$ Thanks Jim. As you point out, there is a more appropriate way to phrase my question: Suppose $\mathfrak{l}$ is a (pseudo-)Levi subalgebra of $\mathfrak{g}$. Is there a relationship between polynomial invariants of $\mathfrak{g}$ and $\mathfrak{l}$? $\endgroup$ – Dr. Evil Oct 22 '14 at 18:35
2
$\begingroup$

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(𝔽q) is a subgroup of G(𝔽q), 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.

$\endgroup$
4
$\begingroup$

[EDIT] Maybe it's useful after all this time to give a more complete and uniform answer to both of the questions asked, by referring to Theorem 3.4(i) in Springer's 1974 paper on regular elements of finite reflection groups here. In your situation, a Weyl group or other finite Coxeter group (= finite real reflection group) acts on a real euclidean space, which can be complexified to deal with eigenvalues. Springer denotes an arbitrary complex reflection group by $G$, which includes your groups.

The answer "yes" to Question 1 (or similarly the answer to Question 2) is clear from the classification of finite Coxeter groups (which include the Weyl groups along with non-crystallographic groups of type $ I_2(m), H_3, H_4$ of ranks 2, 3, 4). The numbers you call "fundamental invariants" are usually just called the degrees and were worked out long ago by Coxeter and others. (There is an exposition in Chapter 3 of my 1990 text on reflection groups, along with a table of degrees.) None of this concerns root systems or Lie algebras directly, just the theory of finite Coxeter groups. I've added to your tags accordingly.

But Springer's result provides a quick uniform proof without case-by-case verification. Here you consider a "parabolic subgroup" of $G$ (conjugate to a standard parabolic subgroup generated by some of the chosen simple reflections) or more generally a reflection subgroup corresponding in the Weyl group case to a pseudo-Levi subgroup of an algebraic or Lie group, e.g., $A_2$ inside $G_2$. Call the subgroup $H$ and note that the complexification of its natural reflection representation embeds naturally in that of $G$. So an eigenvalue for an element of $H$ (giving a nonzero eigenspace) is automatically an eigenvalue for the same element in $G$. Any such eigenvalue is a $d$th root of 1 for some $d \geq 1$.

Now Springer shows (applying some fairly elementary algebraic geometry to the hypersurfaces defined by basic invariant polynomials) that a given $d$th root $\zeta$ occurs as an eigenvalue for some element of $G$ if and only if $d$ divides one of the degrees of $G$.

$\endgroup$
  • 1
    $\begingroup$ Thank you for the extensive answer Jim. Another related question one can ask is: suppose $P_1,\cdots, P_\ell$ are algebraically independent generators for invariant polynomials associated to a finite Coxeter group $G$. Now let $H$ be a parabolic subgroup of $G$. Can we somehow start with $P_i$'s and obtain algebraically independent generators for invariant polynomials associated to $H$? From your explanations above, I gather that this question is not settled in the literature (though there are some partial information when there is folding, etc.) $\endgroup$ – Dr. Evil Oct 29 '15 at 23:21
  • 1
    $\begingroup$ Since the literature on polynomial invariants is so diverse and scattered, I'm not absolutely sure what's there. But the question itself is natural, given the divisibility of degrees. Some case study is perhaps needed, starting with type $A_n$ and the elementary symmetric polynomials. $\endgroup$ – Jim Humphreys Oct 30 '15 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.