CHAPTER 24

An O(2n) Example

In this chapter, we work over a finite field k of odd characteristic. Fix an even integer 2n ≥ 4 and a monic polynomial f(x) ∈ k[x] of degree 2n, f(x) = ∑²ⁿ_i=0 A_ixⁱ, A_2n = 1. We make the following three assumptions about f.

(1) f has 2n distinct roots in k, and A₀ = –1.

(2) gcd{i|A_i ≠ 0} = 1.

(3) f is antipalindromic, i.e., for f^pal(x) := x²ⁿ f(1/x), we have f^pal(x) = –f(x).

Theorem 24.1. For χ₂ the quadratic character of k^×, form the object N := L_χ2(f)(1/2)[1] ∈ P_arith. We have the following results.

(1) G_geom,N = O(2n).

(2) Choose a nontrivial additive character à of k, and define

the normalized Gauss sum. Then for the constant-field twisted object N_α := α^deg ⊗ N we have

G_geom,Nα = G_arith,Nα = O(2n).

(3) If –1 is a square in k, then G_geom,N = G_arith,N = O(2n).

Proof. (3) is a special case of (2), simply because α = ±1 when –1 is a square in k. And (1) results from (2), since N and N_α are geometrically isomorphic. To prove (2), we argue as follows. The complex conjugate of α is χ₂(–1)α. This, together with the fact that f is antipalindromic, shows that N_α is arithmetically self-dual. As N_α is irreducible, the duality is either symplectic or orthogonal. So either we have

G_geom,Nα ⊂ G_arith,Nα ⊂ O(2n)

or we have

G_geom,Nα ⊂ G_arith,Nα ⊂ Sp(2n).

The second case is impossible, because by Lemma 23.1, the determinant of N (and so also the determinant of N_α) is geometrically δ_–1, so we cannot have G_geom,Nα ⊂ Sp(2n). Therefore the duality is orthogonal. Just as in the proof of Theorem 23.2, hypothesis (2) on f implies that N_α is geometrically Lie-irreducible. The Frobenius torus argument of Theorem 23.2 then shows that G⁰_geom,Nα must be SO(2n). But the determinant of N_α is geometrically nontrivial, so from the inclusions

SO(2n) = G⁰_geom,Nα ⊂ G_geom,Nα ⊂ G_arith,Nα ⊂ O(2n)

we infer (2).

Remark 24.2. A monic antipalindromic polynomial of degree 2n over a field k of odd characteristic can be written uniquely as xⁿ(P(x) – P(1/x)) with P(x) a monic polynomial of degree n with vanishing constant term. In this way, the space MonicAntipal_2n of monic antipalindromic polynomials of degree 2n becomes the affine space Aⁿ–1. In MonicAntipal_2n, the condition of having 2n distinct roots, i.e., of having an invertible discriminant, defines an open dense set. Indeed, the set is obviously open, so it suffices to observe that it is nonempty. If 2n – 1 is invertible in k, then the polynomial x²n + x^2n–1 – x – 1 has all distinct roots (as it is (x + 1)(x^2n–1 – 1)). If If 2n is invertible in k, then x²ⁿ – 1 has all distinct roots.