CHAPTER 26

G₂ Examples: Construction in Characteristic Two

We treat the case of characteristic two separately because it is somewhat simpler than the case of odd characteristic. Recall from the first paragraph of the previous chapter that for k a finite field of characteristic 2, and any character χ of k^×, the Tate-twisted Kloosterman sheaf of rank seven

F(χ, k) := Kl(, , , χ, χ, χ, χ)(3)

has G_geom = G_arith = G₂. Our first task is to express its stalk at a fixed point a ∈ k^× as the finite field Mellin transform of the desired object N(a, k).

We abbreviate

Kl₂ := Kl(, ), Kl₃ = Kl(, , ).

Lemma 26.1. For a finite extension k/F₂, and for a ∈ k^×, consider the lisse perverse sheaf on G_m × G_m/k, with coordinates (z, t), given by

M = M(z, t) := Kl₃(a/(z²t)) ⊗ Kl₂(z) ⊗ Kl₂(zt)(3)[2].

Then we have the following results.

(1) The object N(a, k) := R(pr₂)_!M on G_m/k is perverse, of the form N_a[1] for a sheaf N_a.

(2) Denote by π : (G_m)³ → G_m the multiplication map (x, y, z) xyz. Given a character χ of k^×, consider the lisse perverse sheaf on (G_m)³ with coordinates x, y, z given by

Kχ := Kl₃(x) ⊗ Kl₂(y) ⊗L_χ(y) ⊗ Kl₂(z) ⊗L_χ(z)(3)[3].

We have an isomorphism of perverse sheaves on G_m/k,

F(χ, k)[1] ≅ Rπ_!K_χ.

(3) For any character χ of k^×, we have H⁰_c(G_m/k, N(a, k)⊗L_χ) ≅ F(χ, k)_a as Frob_k-module, and Hⁱ_c(G_m/k, N(a, k) ⊗L_χ) = 0 for i ≠ 0.

(4) The object N(a, k) on G_m/k is pure of weight zero, lies in P_arith, has no bad characters, and is totally wild at both 0 and ∞.

Proof. (1) The Verdier dual of M is

M:= Kl₃(–a/(z²t)) ⊗ Kl₂(z) ⊗ Kl₂(zt)(3)[2].

But in characteristic 2, this is just M. So the Verdier dual of N(a, k) is R(pr₂)_?M. Because pr₂ is an affine morphism, and M is perverse, R(pr₂)_?M is semiperverse. So to show that N(a, k) is perverse, it suffices to show that it is semiperverse, i.e., that for all but at most finitely many values of t₀ ∈ k^×, we have

H²_c(G_m/k, Kl₃(a/(z²t₀)) ⊗ Kl₂(z) ⊗ Kl₂(zt₀)) = 0.

In fact, this H²_c vanishes for all t_o, because for each t₀ ≠ 0, the coefficient sheaf is totally wildly ramified at z = 0 (from the Kl₃(a/(z²t₀) factor, the other two factors being tame at z = 0). Thus N(a, k)[–1] is a single sheaf, which we name N_a.

(2) This is simply the expression of the Kloosterman sheaf F(χ, k) as an iterated ! multiplicative convolution [Ka-GKM, 5.4, 5.5], taking into account the isomorphisms Kl₂(χ, χ) ≅ Kl₂ ⊗ L_χ and Kl₂(χ, χ) ≅ Kl₂ ⊗L_χ [Ka-ESDE, 8.2.5].

(3) If we make the substitutions x = w/(z²t), y = zt, z = z, then, using the identity χ(zt)χ(z) = χ(t), K_χ becomes

Kχ = Kl₃(w/(z²t)) ⊗ Kl₂(y) ⊗L_χ(t) ⊗ Kl₂(z)(3)[3].

The restriction of K_χ[–1] to the locus w = a, which we view as being G_m × G_m with coordinates z, t, is just M ⊗ pr^?₂(L_χ). So from the isomorphism F(χ, k)[1] ≅ Rπ_!K_χ of part (2), and proper base change, we get

F(χ, k)_a = RΓ_c(G²_m/k, M ⊗ pr₂(L_χ)).

This last term is just RΓ_c(G_m/k, N(a, k) ⊗L_χ), by Leray and the projection formula.

(4) For E/k a finite field extension, and t₀ ∈ E^×, the stalk of N_a at t₀ is (by proper base change)

H¹_c(G_m/k, Kl₃(a/(z²t₀)) ⊗ Kl₂(z) ⊗ Kl₂(zt₀)(3)).

As already noted, the coefficient sheaf is totally wildly ramified at z = 0.

We first claim that for t₀ ≠ 1, the coefficient sheaf is also totally wildly ramified at z = ∞. As the first factor Kl₃(a/(z²t₀)) is tame, in fact unipotent, at z = ∞, we must show that Kl₂(z)⊗Kl₂(zt₀) is totally wild at z = ∞, so long as t₀ ≠ 1. As I(∞)-representations, Kl₂(z) and its translate Kl₂(zt₀) are both irreducible, so their tensor product is I(∞)-semisimple. In addition, both are I(∞)-self-dual and of trivial determinant. We argue by contradiction. If Kl₂(z) ⊗ Kl₂(zt₀) is not totally wild, then it contains some tame character L_χ as a summand, and hence we have an I(∞)-isomorphism

Kl₂(zt₀) ≅ Kl₂(z) ⊗ L_χ.

Taking determinants, we infer that χ² = . As p = 2, this forces χ = . But the isomorphism class of Kl₂(z) as I(∞)-representation detects nontrivial translations, hence t₀ = 1, contradiction.

For any t₀ ≠ 0, Kl₃(a/(z²t₀)) has Swan₀ = 1 (because p = 2). Hence the coefficient sheaf Kl₃(a/(z²t₀))⊗Kl₂(z)⊗Kl₂(zt₀)has Swan₀ = 4.

We next show that the Swan conductor at ∞ of Kl₂(z) ⊗ Kl₂(zt₀), for t₀ ≠ 1, is 2. To see this, we argue as follows. Both Kl₂(z) and its translate Kl₂(zt₀) have both ∞-slopes 1/2, so the four ∞-slopes of Kl₂(z)⊗Kl₂(zt₀) are each ≤ 1/2, and each > 0 (by the total wildness). Taking into account the Hasse-Arf theorem, that the multiplicity of any slope is a multiple of its denominator, the only possible slopes are 1/2 or 1/3 or 1/4. No slope can be 1/3, for then there would be 3 slopes 1/3, and the remaining slope would be an integer. So either we have all slopes 1/2, or we have all slopes 1/4. We argue by contradiction. If we had all slopes 1/4, then Kl₂(z) ⊗ Kl₂(zt₀) would be a lisse sheaf on G_m which is tame at 0, and totally wild at ∞ with Swan_∞ = 1. Any such sheaf is a multiplicative translate of a Kloosterman sheaf, cf. [Ka-GKM, 8.7.1]. The local monodromy at 0 is Unip(2) ⊗ Unip(2) ≅ Unip(3) ⊕ Unip(1), but the only Kloosterman sheaves which are unipotent at 0 have their local monodromy at 0 a single Jordan block.

Hence the coefficient sheaf Kl₃(a/(z²t₀)) ⊗ Kl₂(z) ⊗ Kl₂(zt₀) has Swan_∞ = 6, for t₀ ≠ 1. Thus N_a|G_m{1} is lisse of rank 10 (by Deligne’s semicontinuity theorem [Lau-SCCS, 2.1.2]) and pure of weight –1.

We next claim that N_a on G_m is a middle extension, i.e., that for j₁ : G_m {1} ⊂ G_m the inclusion, we have N_a ≅ j_1?j^*₁N_a. To see this, recall that N(a, k) = N_a[1] is perverse, so the adjunction map is injective: N_a ⊂ j₁j?₁N_a as sheaves. So we have a short exact sequence of sheaves with a punctual third term supported at 1, say

0 → N_a → j_1? j^?₁N_a → δ₁ ⊗ V → 0,

with V a Gal(k/k)-module. Because N_a|G_m {1} is lisse and pure of weight –1 (in fact, we only “need” that it is lisse and mixed of weight ≤ –1), V is mixed of weight ≤ –1, cf. [De-Weil II, 1.8.1]. In the long exact cohomology sequence with compact supports, the group H⁰_c(G_m/k, j₁j_?1N_a) = 0 (as the sheaf j₁j^?₁N_a has no nonzero punctual sections), so we get an injection V ⊂ H¹_c(G_m/k, N_a). By part (3), this last group is the stalk F(, k)_a, which is pure of weight 0. Therefore V = 0.

Therefore N(a, k) = N_a[1] is the middle extension from G_m {1} of a (lisse) perverse sheaf which is pure of weight 0, hence is itself pure of weight 0. The isomorphism of part (3), applied also to all extensions of scalars N(a, E), shows that for any finite extension field E/k, and for any character χ of E^×, the group H¹_c(G_m/k, N_a ⊗ L_χ) ≅ F(χ, E)_a is pure of weight 0, and the H²_c vanishes. Therefore N(a, k) has, geometrically, no quotient which is a Kummer sheaf L_χ[1]. But as N(a, k) is pure, it is geometrically semisimple, so it has, geometrically, no Kummer subsheaf either. Hence N(a, k) lies in P_arith. By Theorem 4.1, the purity of H¹_c(G_m/k, N_a ⊗ L_χ) for every χ implies that N(a, k) has no bad characters, which, as already noted, is equivalent to the fact that N_a is totally wild at both 0 and ∞.

It remains to show that the object N(a, k) we have constructed in characteristic 2 is orthogonally self-dual, of “dimension” seven, and geometrically Lie-irreducible. That it is of “dimension” seven is obvious from part (3) of the previous lemma, applied with any one choice of χ. We next show that N(a, k) is self-dual as an object of P_arith.

Lemma 26.2. We have the following results.

(1) The object N(a, k) is isomorphic to its pullback by multiplicative inversion t 1/t on G_m.

(2) The object N(a, k) is isomorphic to its Verdier dual D(N(a, k)).

(3) As an object in the Tannakian category P_arith, N(a, k) is self-dual.

Proof. Assertion (3) is immediate from (1) and (2), since N(a, k)^∨ ≅ [t 1/t]^? D(N(a, k)).

Assertion (1) is obvious from the definition of N(a, k) given in part (1) of the previous lemma, by the change of variable z z/t, t t in the sheaf M := Kl₃(a/(z²t)) ⊗ Kl₂(z) ⊗ Kl₂(zt)(3)[2] on G_m × G_m. It remains to show (2), that N(a, k) is its own Verdier dual. It suffices to show that this holds over the open set G_m {1}, since middle extension of perverse sheaves from G_m{1} to G_m commutes with Verdier duality. On G_m × G_m, the coefficient sheaf M is self-dual (remember we are in characteristic 2, so Kl₃(1) is self-dual), so the Verdier dual of N(a, k) := R(pr₂)_!M is R(pr₂)_?M. It suffices to show that the canonical “forget supports” map

R(pr₂)_!M → R(pr₂)_?M

is an isomorphism over G_m {1}. As we have seen in the proof of the previous lemma, R(pr₂)_!M is lisse over G_m {1}. Therefore its Verdier dual, R(pr₂)_?M, is also lisse over G_m {1}, and its formation is compatible with arbitrary change of base on G_m{1}. Sotoshowthat the “forget supports” map is an isomorphism over G_m {1}, it suffices to check point by point in G_m{1}. We have already observed that over any point t₀ ∈ G_m {1}, the restriction of M(–3)[–2] to that fibre, namely the lisse sheaf Kl₃(a/(z²t₀))⊗Kl₂(z)⊗Kl₂(zt₀), is totally wild at both 0 and ∞, and hence we have the desired isomorphism H_c ≅ H of its cohomology groups.

Although the “dimension” of N(a, k) is odd, namely seven, we do not yet know that N(a, k) is irreducible, much less that it is geometrically Lie-irreducible. So we cannot assert that the autoduality we have shown N(a, k) to admit is necessarily orthogonal (or indeed that it has a sign at all). To clarify this question, we need information on the local monodromy of the sheaf N_a = N(a, k)[–1] at the point 1.

Lemma 26.3. We have the following results.

(1) The stalk

(N_a(–3))₁ = H¹_c(G_m/k, Kl₃(a/z²) ⊗ Kl₂(z) ⊗ Kl₂(z)) .

has rank seven. It has six Frobenius eigenvalues of weight 5, and one Frobenius eigenvalue of weight 2.

(2) As I(1)-representation, N_a is Unip(1)⁶ ⊕ Unip(4).

(3) The sheaf N_a is totally wild at both 0 and ∞, with Swan₀ = Swan_∞ = 2.

Proof. (1) Because we are in characteristic 2, we have the Carlitz isomorphism

Kl₂(z) ⊗ Kl₂(z) ≅ Q_ℓ(–1) ⊕ Kl₃(z),

cf. [Ka-ClausCar, 3.1]. So our coefficient sheaf is

Kl₃(a/z²)(–1) M Kl₃(a/z²) ⊗ Kl₃(z).

Because we are in characteristic two, the map z → a/z² is radicial and bijective, so we have

H¹_c(G_m/k, Kl₃(a/z²)(–1)) ≅ H¹_c(G_m/k, Kl₃(z)(–1)) = Q_ℓ(–1),

the last equality by Mellin inversion, cf. [Ka-GKM, 4.0]. This is the eigenvalue of weight 2. The sheaf Kl₃(a/z²) ⊗ Kl₃(z) is lisse on G_m, pure of weight 4, totally wild at both 0 and ∞ with all slopes 1/3 (remember we are in characteristic 2, so Kl₃(a/z²) still has its I(0)-representation totally wild with Swan conductor 1, rather than the 2 it would be in odd characteristic). So Swan₀ = Swan_∞ = 3, and the cohomology group H¹_c(G_m/k, Kl₃(a/z²) ⊗ KL₃(z)) has dimension six and is pure of weight 5.

(2) Since N_a(–3) is the middle extension across 1 of a lisse sheaf on G_m {1} which is pure of weight 5, the weights of the Frobenius eigenvalues on the space of I(1)-invariants tell us that the unipotent part of the I(1)-representation is precisely Unip(1)⁶ ⊕ Unip(4), cf. [De-Weil II, 1.8.4 and 1.7.14.2-3] or [Ka-GKM, 7.0.7]. As this lisse sheaf N_a|G_m {1} has rank 10, there is room for nothing more in its I(1)-representation.

(3) From the fact that N(a, k) has no bad chararcters, we know that N_a is totally wild at both 0 and ∞. From its invariance under t 1/t, we know that Swan₀ = Swan_∞. Because N_a is lisse outside 1 and unipotent (hence tame) there, we have the equation

Swan₀(N_a)+ Swan_∞(N_a)+ drop₁(N_a) = “dim”(N(a, k)) = 7.

But drop₁ = 3, so Swan₀ + Swan_∞ = 4. As Swan₀ = Swan_∞, we are done.

Lemma 26.4. The object N(a, k) is geometrically Lie-irreducible, and orthogonally self-dual.

Proof. It suffices to prove the first statement, since N(a, k) is self-dual and of “dimension” seven. It suffices to show that N(a, k) is geometrically irreducible, since it has a unique singularity, namely 1, in G_m, so cannot be geometrically isomorphic to any nontrivial multiplicative translate of itself.

We argue by contradiction. Suppose we have, geometrically, a direct sum decomposition

N(a, k) ≅ A[1] ⊕B[1].

Then both A and B must be totally wild at both 0 and ∞, so each of the four Swan conductors (of A and B at 0 and ∞) is ≥ 1. As their direct sum A⊕B = N_a has Swan₀ = Swan_∞ = 2, we must have

Swan₀(A) = Swan_∞(A) = 1, Swan₀(B) = Swan_∞(B) = 1.

Furthermore, both A and B are lisse on G_m {1}, and middle extensions across 1. So both A[1] an B[1] are geometrically irreducible. Their local monodromies at 0 are both totally wild of Swan conductor 1, so detect nontrivial multiplicative translations. So each of A[1] and B[1] is geometrically Lie-irreducible. The local monodromy at 1 of N_a is Unip(1)⁶ ⊕ Unip(4), so one of our summands is lisse at 1, say A at 1 is Unip(1)^a, and the other is B = Unip(1)^b ⊕ Unip(4). Thus A[1] has “dimension” 2, and B[1] has “dimension” 5. As they have different “dimensions” and are geometrically irreducible, any autoduality of N(a, k) must make each of the summands self-dual. For B[1], the auto-duality must be orthogonal, since its “dimension” is odd. For A[1], the autoduality must be symplectic, for otherwise its G_geom lies in O(2), which contradicts its Lie-irreducibility.

Since G_geom is a normal subgroup of G_arith, G_arith must permute the G_geom-irreducible constituents of !(N(a, k)). As these two constituents have different dimensions, each must be G_arith-stable. Thus N(a, k) = A[1] ⊕B[1] is an arithmetic direct sum decomposition into irreducibles of “dimensions” 2 and 5. As N(a, k) admits an arithmetic autoduality, this autoduality must make each summand self-dual, in a way compatible with its geometric autoduality. Thus A[1] is symplectically self-dual, and B[1] is orthogonally self-dual. So G_arith,A[1] ⊂ SL(2), and G_arith,B[1] ⊂ O(5). In fact, G_arith,B[1] ⊂ SO(5), because “det”(N(a, k)) is trivial. Indeed, by Lemma 26.1, part (3), every Frobenius Frob_E,χ of N(a, k) lies inside a G₂ inside an SO(7). But “det”(N(a, k)) ≅ “det”(A[1])^? _mid “det”(B[1]) ≅ “det”(B[1]) (this last because “det”(A[1]) is trivial, as G_arith,A[1] ⊂ SL(2)).

As A[1] is geometrically Lie-irreducible with its G_arith,A[1] ⊂ SL(2), G⁰geom, A[1] is an irreducible connected subgroup of SL(2), so must be SL(2), so we have

G_geom,A[1] = G_arith,A[1] = SL(2).

As B[1] is geometrically Lie-irreducible with its G_arith,B[1] ⊂ SO(5), G⁰geom, B[1] is an irreducible connected subgroup of SO(5). By Gabber’s theorem on prime-dimensional representations, either G⁰_geom,B[1] is SO(5) or it is the image SL(2)/ ± 1 of SL(2) in SO(5) via Sym⁴(std₂). Both of these groups are their own normalizers in the ambient SO(5), so

G_geom,B[1] = G_arith,B[1] = SO(5) or G_geom,B[1] = G_arith,B[1] = SL(2)/ ± 1.

Now we apply Goursat’s lemma to conclude that we have one of three possibilities. Either

G_{geom, N(a, k)} = G_{arith,N(a, k)} = SL(2) × SO(5),

or

G_{geom, N(a, k)} = G_{arith,N(a, k)} = SL(2) × SL(2)/ ± 1,

or

G_{geom, N(a, k)} = G_{arith,N (a, k)} = SL(2).

In this last case, SL(2) acts by the representation std₂ ⊕ Sym⁴(std₂), i.e., this SL(2) is the subgroup of SL(2) × SL(2)/ ± 1 which is the graph of the projection of the first factor onto the second.

Even for this smallest group, its compact form contains elements Diag(e^iθ, e^–iθ) with, e.g., θ = 3π/4, whose trace in std₂ ⊕ Sym⁴(std₂), namely

2cos(θ)+ 1+ 2cos(2θ)+ 2cos(4θ),

is –1 – √2 = –2.414… < –2.4. So in the compact form K of any of the three possible groups G_arith,N (a, k), the set “Trace < –2.4” is a nonempty open set in the compact space K^#. So by equidistribution we will find Frobenius elements Frob_E,χ for N(a, k) whose traces are < –2.4. But each such Frobenius is, by Lemma 26.1, part (3), an element of UG₂, and so its trace lies in [–2, 7]. This contradiction completes the proof that N(a, k) is in fact geometrically irreducible.

This concludes the construction of the objects N(a, k) in characteristic 2.