In an earlier paper [Ka5], one of us, inspired by a paper [Gr] of Gross, defined, for each n ≥ 1, a local system on A 1 /F 2 of rank 2 n (2 2n+1 -1), whose geometric monodromy group we conjectured to be the Suzuki group Sz(q) := 2 B 2 (q), q = 2 2n+1 , in one of its two lowest dimensional nontrivial irreducible representations. These representations, complex conjugates of each other, are of dimension 2 n (2 2n+1 -1) and have traces in Z[i], the Gaussian integers.
The definition involved p-Witt vectors of length 2 for p = 2, with values in F 2 -algebras. We identify W 2 (F 2 ) with Z/4Z, by the map [a, b] → a 2 + 2b, with the usual convention that we first lift a, b to Z and then reduce a 2 + 2b modulo 4. We take the additive character of Z/4Z given by n → i n , and view it as the additive character ψ 2 : W 2 (F 2 ) → µ 4 (Z[i]) given by ψ 2 ([a, b]) := i a 2 +2b .
Attached to a Witt vector of length 2 with coefficients in F 2 [x], say [a(x), b(x)], is the Artin-Schreier-Witt sheaf L ψ 2 ([a(x),b(x)]) on A 1 /F 2 . It is lisse of rank one, and its trace function is given as follows. For k/F 2 a finite extension, and x ∈ k,
If instead we take a(x), b(x) ∈ k 0 [x] for some finite extension k 0 /F 2 , then the Artin-Schreier-Witt sheaf L ψ 2 ([a(x),b(x)]) is lisse of rank one on A 1 /k 0 , and for k/k 0 a finite extension and x ∈ k, the trace of Frob x,k is given by the same formula (which only makes sense when k is an extension of k 0 ).
The unique nontrivial additive character ψ of F 2 is related to ψ 2 by the formula
this is simply the identity (-1) b = i 2b . Under Witt vector addition, Quite generally, for an Artin-Schreier-Witt sheaf L := L ψ 2 ([a(x),b(x)]) on A 1 /k 0 with Swan conductor Swan ∞ = n ≥ 2, its Fourier transform FT ψ (L) on A 1 /k 0 is an Airy sheaf in the sense of Šuch [Such]. It is lisse of rank n -1, and all its ∞-slopes are n n-1 . It is pure of weight one. Its trace function is given as follows: for k/k 0 a finite extension, and t ∈ k, ([a(x), b(x) + tx]) .
Some key facts about Airy sheaves and their monodromy groups are due to Šuch [Such], and are fundamental for the investigations reported on here.
We now turn to the Suzuki "candidates" of [Ka5]. Here n ≥ 1 , q 0 := 2 n , q := 2q 2 0 , t(q) := q + 1 -2q 0 .
We take the Witt vector q) , n i=1
x (1+2 i )t(q) ],
form the Artin-Schreier-Witt sheaf L := L ψ 2 ([x t(q) , n i=1 x (1+2 i )t(q) ]) , form its FT ψ (L), and twist by the constant field twist ( 11-(-1) n i ) deg , to arrive at the local system F q on A 1 /F 2 , whose trace function is given as follows. For k/F 2 a finite extension, and t ∈ k, q) , n i=1
x (1+2 i )t(q) + tx]) .
A key fact about F q is that the input Witt vector [x t(q) , n i=1 x (1+2 i )t (q) ] is a function of x t (q) . So in the trace formula for F q , if we restrict to t = 0 and make the substitution x → x/t, the formula becomes Trace(Frob t,k |F q ) = -1 q) , n i=1
x (1+2 i )t(q) t (1+2 i )t(q) + x]) ,
which is a function of t t (q) . Thus F q |G m has a descent to a lisse sheaf G q on G m /F 2 whose trace function is given by Trace(Frob t,k |F q ) = -1
x (1+2 i )t(q) t (1+2 i ) + x]) ,
and such that under the t(q) Kummer pullback, [t(q)] G q = F q |G m .
For the case n = 1, i.e. for 2 B 2 (8), we prove in Theorem 6.1 that F 8 has the predicted geometric monodromy group G geom,F 8 = 2 B 2 (8) and arithmetic monodromy group G arith,F 8 ,F 2 = Aut( 2 B 2 (8)) over F 2 , and thus establish [Ka5,Conjecture 2.2] in full in this case. For each n ≥ 1, we show that G geom,Fq is either 2 B 2 (q) or SL D for D = rank(F q ) = q 0 (q -1). A (huge) calculation of fourth moment for F 8 shows that G geom,F 8 cannot be SL 14 . It remains an open problem to prove (or disprove) that each F q has G geom,Fq = 2 B 2 (q) when q ≥ 32.
In fact, we show in Theorem 8.4 that we have such a dichotomy of possible geometric monodromy groups G geom , either 2 B 2 (q) or SL D , D = q 0 (q -1), for a general class of local systems of "the same shape" as F q (see Remark 8.6 and Theorem 9.18 for examples of such local systems with G geom = SL D ). Key to our investigation is the fact that these sheaves all satisfy the condition (S+) of [KT1,Definition 1.2]. Somewhat to our surprise, the most difficult part of establishing condition (S+) was to show the sheaves in question are geometrically primitive, i.e. that the representation of their G geom is not induced. Our proof utilizes the existence of primitive prime divisors of the integer t(q), cf. Theorem 3.4. We give a second proof which will be useful in future studies of geometric monodromy groups of quite general Airy sheaves. Condition (S+) implies that for each of these sheaves, either G geom is a finite, almost quasisimple group, or has G • geom acting irreducibly. This initial dichotomy, plus a substantial group-theoretic analysis, in which Theorem 4.5 plays a key role, is what leads to the 2 B 2 (q)/SL D dichotomy. Pursuing the study of primitive prime divisors, we prove in Theorems 3.4, 3.7, and Corollary 3.5 the existence of primitive prime divisors in the orders of maximal tori of the Suzuki-Ree groups 2 B 2 (q), 2 G 2 (q), and 2 F 4 (q). We also extend in §7 the classification of low-dimensional representations of classical groups in characteristic p ≥ 0, beyond the bounds in [KlL] and [Lu]. These results will be useful in other situations as well. Finally, the structure of arithmetic monodromy groups, assuming finiteness, is determined in Theorem 9.14.
For each odd power q = 2 2n+1 of 2, starting with q = 8, the finite simple group 2 B 2 (q) has two (complex conjugate) lowest dimensional nontrivial irreducible representations, of dimension d(q) := q 0 (q -1), with q 0 := 2 n . In each case we have a factorization 1 + d(q) = (q 0 + 1)t(q) with t(q) := q + 1 -2q 0 .
In [Ka5], for each such q there is proposed an Airy sheaf in the sense of Šuch [Such], call it F q on A 1 /F 2 which is lisse of rank d(q) and with Swan ∞ (F q ) = 1 + d(q). Its I(∞)-representation is irreducible of dimension d(q), with all ∞-slopes 1+d (q) d(q) . Moreover, F q is given [ Ka5,Section 4] with an explicit descent to a lisse sheaf G q on G m /F 2 whose Kummer pullback by t(q) th power is (the restriction to G m /F 2 of) F q :
[t(q)] G q ∼ = F q |G m . Thus G q is lisse on G m /F 2 , tame at 0 with I(0)-representation a direct sum of Kummer characters of order dividing t(q), and whose I(∞)-representation is irreducible of dimension d(q), with all ∞-slopes q 0 +1 d(q) = q 0 +1 q 0 (q-1) . Next we give a slight improvement of [KRLT2,Theorem 6.5].
Theorem 1.1. Let H 0 be a lisse sheaf on G m /F q which is tame at 0 and pure of weight zero. Let a, b be nonnegative integers, and consider the moment M a,b . Denote by Then we have the following estimate.
Proof. For any lisse sheaf F on G m , the Lefschetz trace formula gives
If F is pure of weight zero, then H 2 c is pure of weight 2, and H 1 c is mixed of weight ≤ 1, indeed H 1 c = H 1 c (wt = 1) ⊕ H 1 c (wt ≤ 0). Thus for F pure of weight zero,
When F is tame at 0, the Euler-Poincaré formula gives
To compute the dimension of H 1 c (wt = 1), we use the fact that for the inclusion j : G m ⊂ P 1 , the group H 1 (P 1 /F q , j F ) is pure of weight one. We exploit this by looking at the short exact sequence of sheaves on P 1 /F q given by 0 → j ! F → j F → (F I(0) ) 0 ⊕ (F I(∞) ) ∞ → 0, where the last two summands are skyscraper sheaves at 0 and ∞. The group H 2 (P 1 /F q , j F ) = H 2 c (G m /F q , F ) is the Tate-twisted group of π geom
• -1, multiplicity 52429, • 0, multiplicity 112347, • 1, multiplicity 60495, • 2i, multiplicity 16512, • 14, multiplicity 9. [We have not been able to find any conceptual explanation for the multiplicities of these traces, let alone for the fact that all traces are algebraic integers. Had we known that the traces of all Frobenii over all finite extensions of F 2 are algebraic integers, we would have been able to conclude that G 8 has finite geometric monodromy group G geom , and the proof of the main result Theorem 6.1 would have been much simpler.] This computation was carried out using Magma 2.26-6 [BCP] at the University of Auckland. It exploited a new feature of Magma which supports running tasks in parallel on multiple processors: each task performed the necessary computation for a given t ∈ k × . Using 55 3.0GHz processors, the computation completed in 5.75 days, taking about 7500 hours of CPU time.
The empirical M 2,2 for G 8 over F × 2 18 is thus approximately 3.99963378766551080898593515753. Applying Theorem 1.1, we find Corollary 1.2. For the lisse sheaf G 8 on G m /F 2 , M 2,2 > 2.
Proof. The sheaf F 8 is a geometrically irreducible Airy sheaf, lisse on A 1 /F 2 of rank 14, whose I(∞) representation is irreducible, with all slopes 15/14. Its descent G 8 is thus lisse and geometrically irreducible on G m , its I(0) representation is the direct sum with multiplicities of the characters of order dividing 5, and its I(∞) representation is irreducible, with all slopes 3/14.
Let us denote by K := (G 8 ) ⊗2 ⊗ (G ∨ 8 ) ⊗2 . In the proof of Theorem 1.1, the estimate The equality h 1 c (wt = 1) = Swan ∞ (K) + 2M 2,2 -dim K I(0) -dim K I(∞) can be weakened to h 1 c (wt = 1) ≤ Swan ∞ (K) + M 2,2 -dim K I(0) , simply because in the exact sequence, the space H 0 of global invariants of K injects into the space K I(∞) of I(∞)-invariants.
Thus K is lisse on G m , its I(0) representation is the direct sum with multiplicities of the characters of order dividing 5, and all its I(∞) slopes are ≤ 3/14. So we have the crude estimate Swan ∞ (K) ≤ rank(K)(biggest slope) ≤ 14 4 (3/14) = 8232.
The I(0) representation of K is End(End(the I(0)-representation of G 8 )).
We now turn to the I(0) representation of G 8 . The action of I(0) is through µ 5 . So in terms of a fixed character χ of order 5, it is a direct sum a1 + bχ + cχ 2 + dχ 3 + eχ 4 , with non-negative integers a, b, c, d, e which sum to 14. We also know from Deligne's "independence of ℓ" result [De2,Theorem 9.8] or from Theorem 2(ii)], that the character of this representation of I(0) has values in the field Q(i) (because G 8 is part of a compatible system over Q(i)). On the other hand, this character has values in Q(ζ 5 ). But the intersection Q(i) ∩ Q(ζ 5 ) is just Q, so the trace has values in Q. In other words, the quantity a + bζ 5 + cζ 2 5 + dζ 3 5 + eζ 4 5 lies in Q. This in turn forces b = c = d = e, and so a + bζ 5 + bζ 2 5 + bζ 3 5 + bζ 4 5 = a -b. But a + 4b = 14, so there are only four possibilities for the character, namely 21 + 3χ + 3χ 2 + 3χ 3 + 3χ 4 , , 61 + 2χ + 2χ 2 + 2χ 3 + 2χ 4 , 101 + 1χ + 1χ 2 + 1χ 3 + 1χ 4 , 141.
More intrinsically, let us denote by Reg the regular representation of µ 5 :
Then the I(0) representation of G 8 is one of -1 + 3Reg, 41 + 2Reg, 91 + Reg, 141.
Because the character takes real (in fact integer) values, dim K I(0) is the coefficient of 1 in the fourth power of the character. For each of our four candidates, this is easily computed by hand, because Reg 2 = 5Reg, Reg 3 = 5 2 Reg, Reg 4 = 5 3 Reg. The least of the possible values of dim K I(0) is 7684, the value attained by the candidate -1 + 3Reg. So the weakened estimate becomes
Dividing by 2 18 -1, we get
Here 2 18 -1 = 262143, and 2 9 /(2 18 -1) ≤ (2 9 + 1)/(2 18 -1) = 1/(2 9 -1) = 1/511.
Thus if M 2,2 were 2, we would get 3.999 ≤ (1 + (1/262143))2 + 550/511 + 8332/262143, which is nonsense. 2. Background results on determinants, rationality, and slopes Theorem 2.1. Let p be a prime, k/F p a finite extension, ν ≥ 1 an integer, and F a lisse Q ℓ sheaf on a smooth, geometrically connected scheme U/k which is pure of weight zero and part of a compatible system of lisse sheaves on U whose trace functions take values in Q(ζ p ν ). Denote by G geom ≤ G arith the geometric and arithmetic monodromy groups of F . Then we have the following results. (i) det(G arith ) has finite order dividing 2p ν if p is odd, dividing 2 ν if p = 2. (ii) det(G geom ) has finite order dividing 2p ν if p is odd, dividing 2 ν if p = 2. (iii) Suppose that U is a dense open set of P 1 , and that at each point x ∈ P 1 (k)\U (k), all I(x)-slopes of F are < 1. Then det(G geom ) has order dividing 2, and is trivial if p = 2. Proof. (i) For any finite extension L/k and any point t ∈ U (L), det(Frob t,L |F ) lies in Q(ζ p ν ). It has absolute value 1 at every archimedean place of Q(ζ p ν ) (purity of weight zero), and is a λ-adic unit at every finite place λ of residue characteristic = p (being part of a compatible system). Because Q(ζ p ν ) has a unique place above p, the product formula tells us that this determinant is a unit at all finite places. Thus it is a root of unity in Q(ζ p ν ), so of order dividing 2p ν for p odd, and of order dividing 2 ν if p = 2.
As G geom is a subgroup of G arith , we trivially obtain (ii).
To prove (iii), the slope hypotheses imply that det as a character of G geom has slope < 1, hence 0, at each missing point, and thus is everywhere tame, and hence (being on a dense open set of P 1 ) has order prime to p.
Theorem 2.2. Let p be a prime, k/F p a finite extension, ν ≥ 1 an integer, and F a lisse Q ℓ sheaf on a smooth, geometrically connected scheme U/k which is pure of weight zero and part of a compatible system of lisse sheaves on U whose trace functions take values in a number field E. Suppose further that there exists a proper smooth curve with geometrically connected fibres π : C → U, a finite group Γ of automorphisms of C/U , and a linear character
Then we have the following results.
(i) For any finite extension L/k and any point t ∈ U (L), the action of Frob t,L |F is semisimple.
(ii) If U is a curve, with complete nonsingular model X, then for each point x ∈ X(k) \ U (k), the character of the action of the inertia group I(x) acting on F has values in K.
Proof. The semisimplicity of Frobenii on H 1 of curves goes back to Weil. By Serre-Tate [Se-Ta, Theorem 2(ii)], the character of I(x) on R 1 π (Q ℓ ) has values in Z. When we project the I(x) action onto the χ component, the character of the resulting I(x) action has values in Q(χ), a subfield of K.
Remark 2.3. Theorem 2.2 applies to the Airy sheaves of Šuch [Such] and any of their descents, where the family of curves in question is a family of Artin-Schreier-Witt coverings of A 1 (compactified by adding its one point at ∞).
Next we record a general result on G geom of lisse sheaves on open sets of A 1 :
Theorem 2.4. Let U be a dense open set of A 1 /F p , and F a lisse Q ℓ -sheaf on U , with ℓ = p. Suppose all the ∞-slopes of F are at most σ, for some 0 < σ < 1. Suppose the geometric monodromy group G = G geom of F admits a representation Φ : G → GL d (F) over some algebraically closed field F of characteristic = p, of dimension d < 1/σ. Then Φ is tame at ∞.
The hypothesis that all ∞-slopes of F are ≤ σ is that for all y > σ, the upper numbering subgroup I(∞) y acts trivially on F , i.e. dies in G geom , and hence dies under Φ. Thus all ∞-slopes of F are ≤ σ, and hence Swan ∞ (Φ) ≤ dσ < 1. As Swan conductors are non-negative integers, Swan ∞ (Φ) = 0. This means P (∞) acts trivially in Φ, i.e. Φ is tame at ∞. or an e-dimensional projective representation
which is nontrivial over S, over some algebraically closed field
Proof. Note that the given Ψ is faithful. Indeed, Ker(Ψ) ¡ R does not contain S, so it intersects S trivially by simplicity of S. Because both S and Ker(Ψ) are normal in R, the commutator
Let Ṽ denote the Ḡ-module Ind Ḡ R (V ) for Ḡ := G/Z(G). Note that Ḡ acts faithfully on Ṽ . Indeed, let K ¡ Ḡ denote the kernel of the action of Ḡ on Ṽ . By the construction of V as the induced representation, the R-module Ṽ contains V as a submodule. But S acts faithfully on V , hence S ∩ K = 1. As S ¡ Ḡ, it follows that [S, K] = 1, and so
Now view Ṽ as a representation of G, of dimension < 1/σ. By Theorem 2.4, this representation is tame at ∞. Thus the image Q in G of P (∞) acts trivially on Ṽ . But G/Z(G) acts faithfully on Ṽ . Therefore Q lands in Z(G). Recall that I(∞) has finite image J in G, and J/Q is cyclic. As Q ≤ Z(J), it follows that J is abelian. Thus all simple J-summands in H are one-dimensional, and at least one of them is wild, as H is not tame at ∞. Each one-dimensional wild component has Swan a strictly positive integer, which is also its slope, contradicting the hypothesis that all ∞-slopes of H are < 1.
The order of a finite group of Lie type G(F q ) over a field F q is usually a product of a power of q = p f (p the defining characteristic) and the values at q of cyclotomic polynomial Φ m (q) for various m. In a number of problems on G(F q ), the existence of primitive prime divisors ppd(q, m) or ppd(p, mf ) for certain m was helpful. Recall [Zs] that for a, m ∈ Z ≥2 , a primitive prime divisor ℓ = ppd(a, m) is a prime divisor of a m -1 that does not divide m-1 i=1 (a i -1); such a prime divisor always exists unless (a, m) = (2, 6) or m = 2 and a + 1 is a 2-power. For the Suzuki-Ree groups 2 B 2 (q) with q = 2 n , 2 n ≥ 3, 2 G 2 (q) with q = 3 n , 2 n ≥ 3, and 2 F 4 (q) with q = 2 n , 2 n ≥ 3, some factor Φ m (q) of |G(F q )| decomposes further into values at √ q of polynomials over
Φ 4 (q) = q 2 + 1 = (q -2q + 1)(q + 2q + 1) for 2 B 2 (q), Φ 6 (q) = q 2 -q + 1 = (q -3q + 1)(q + 3q + 1) for 2 G 2 (q), and Φ 12 (q) = q 4 -q 2 + 1 = (q 2 -q 2q + q -2q + 1)(q 2 + q 2q + q + 2q + 1) for 2 F 4 (q). In applications, it is desirable to prove that these factors also possess primitive prime divisors ppd(p, 4n), respectively ppd(p, 6n), ppd(p, 12n), whose existence does not follow from [Zs].
The main results of this section establish the existence of such prime divisors for Suzuki-Ree groups.
3A. Almost equidistribution of coprime integers in congruence classes. For n ∈ Z ≥1 , let φ(n) denote the Euler function of n, let µ(n) denote the Möbius function of n, and let ω(n) denote the number of distinct prime divisors of n (not counting multiplicities). First we prove the following Proposition 3.1. Let m, n ∈ Z ≥1 be coprime integers. For any integer 0 ≤ a ≤ m -1, the number
If d|n, then gcd(d, m) = 1, so we can find 1 ≤ e ≤ m -1 such that de ≡ 1 (mod m). Now write every 1 ≤ k ≤ n with d|k as k = dl with 1 ≤ l ≤ n/d. Then the condition that k ≡ a (mod m) is equivalent to l ≡ ea (mod m), in which case we can write l = s + mi with 0 ≤ r ≤ m -1, ea ≡ r (mod m), and i ∈ Z. To count the number N (a, d) of i occurring, write n/d = qm + r with 0 ≤ r ≤ m -1 and q ∈ Z ≥0 . Certainly, every 0 ≤ i ≤ q -1 works, but neither i = -1 nor i = q + 1 can occur. It follows that
We also note by [MNZ,(4.1)] that φ(n) = d|n µ(d)n/d and that d|n |µ(d)| is the number of square-free divisors of n and hence equals to 2 ω(n) . Combining with (3.1.1) and (3.1.2), this yields
We will also need the following analogue of Proposition 3.1:
Proposition 3.2. Let n ∈ Z ≥1 be an odd integer divisible by 3. For any integer 0 ≤ a ≤ 11 coprime to 3, the number N a of integers 1 ≤ k ≤ n such that gcd(k, n) = 1 and k ≡ a (mod 12) satisfies
Proof. As in the proof of Proposition 3.2, we have Next, µ(d) = 0 if 9|d, and µ(d) = -µ(d/3) if 3|d. It follows from [MNZ,(4.1)] that
If d|n and 3 d, then gcd(d, 12) = 1, so we can find 1 ≤ e ≤ 11 such that de ≡ 1 (mod 12). The proof of (3.1.2) repeated verbatim shows that
Combining with (3.2.1) and (3.2.2), this yields
3B. Primitive prime divisor for Suzuki groups. We make the choice √ 2 > 0. For odd n ∈ Z ≥1 and a = 1, 3, set
Proposition 3.3. If 2 n ≥ 2603 and a = 1, 3, then P 2,a (n) > 2n.
Proof. (i) First we note that
when 2 n and n = 1, 3,9,15,21,33,45,75,105,165,195. Indeed, suppose s := ω(n) ≥ 1 and write n = s i=1 p a i i for some prime divisors 2 < p 1 < p 2 < . . . < p s of n. If p 1 ≥ 5, then
and so φ(n) > n 6/7 . If p 1 ≥ 7, then
If p 1 = 5 and s ≥ 2, then
If p 1 = 5 and s = 1, but n = 5 then φ(n) ≥ 20 > 2 4s .
In the rest of the proof of (3.3.1), we may assume that p 1 = 3. First suppose that s ≥ 4. As (11 6 /10 7 ) s-3 < (2
If p 3 < 11 then (p 1 , p 2 , p 3 ) = (3, 5, 7); thus if n = 105 then n > 3 • 105 and so
Finally, assume that s = 1, and n = 3 a with a ≥ 3. Then n/φ(n) = 3/2 < 3 a/7 = n 1/7 , and φ(n) ≥ 18 > 2 4s , completing the proof of (3.3.1).
(ii) Now assume that n ≥ 2602. By (3.3.1) m := φ(n) ≥ n 6/7 > 846, so m ≥ 847, and 2 ω(n) ≤ m 5/11 . Fix a ∈ {1, 3} and let S j := {(j-1)n ≤ k < jn | gcd(k, 8n) = 1, k ≡ a (mod 8)} for 0 ≤ j ≤ 7. For each j, observe that k ∈ S j if and only if 0 ≤ k := k-jn < n is coprime to n and k ≡ a-jn (mod 8). By Proposition 3.1, n) . By the above,
Now g (t) = log(A)/4 -log(B)/(1.1t 6/11 ) -(7/6t) is increasing, so g (t) ≥ g (847) > 0.13 when t ≥ 847. It follows that g(m) ≥ g(847) > 0.75, and so f (m) = exp(g(m)) > 2.11. Thus, for m = φ(n) ≥ 847 P 2,a (n) ≥ f (m)m 7/6 > 2m 7/6 > 2n, as desired.
As we will see in part (iii) of the proof of the following theorem, Proposition 3.3 actually holds for all odd n ≥ 7.
Theorem 3.4. Let q = 2 n with 2 n. If n ≥ 7 then t(q) = t -(q) := q -√ 2q + 1 is divisible by a primitive prime divisor ppd(2, 4n) of q 2 + 1. In all cases, t + (q) := q + √ 2q + 1 is divisible by a primitive prime divisor ppd(2, 4n) of q 2 + 1.
Proof. (i) The second statement is obvious for n ≤ 5; also note that t(2) = 1, t(8) = 5. Henceforth we may assume n ≥ 7. Consider the sets
for j = 1, 3, 5, 7. Then the map k → 8n -k yields bijections
and the map k → k + 4n yields bijections
for each j = 1, 3, 5, 7.
(ii) Make the choice of ζ := ζ 8n = exp(πi/4n) and consider the cyclotomic polynomial
In particular,
(iii) To show that P 2,1 (n) and P 2,3 (n) are integers, we use gcd(8, n) = 1 to write 1 = ns + 8t for some s, t ∈ Z with gcd(s, 8) = gcd(t, n) = 1, and set ζ = ζ 8n = αβ with α := ζ ns , a 8 th root of unity, and β := ζ 8t , an n th root of unity. When k runs over A 1 B 1 , ζ k = α k β k = αβ k , and k (mod n) runs over units in Z/nZ, each at most once. Using (3.4.2), we see that each unit is met exactly once. Repeating the same argument for k ∈ A 7 B 7 , we get Φ 8n,1 (X) :=
Note that α + α -1 = √ 2 for some = ±1. It follows from (3.4.5) that
hence it is an integer. The same arguments show that P 2,3 (n) is the norm Norm Q(ζn)/Q of the algebraic integer 2 + 2 ζ n + ζ 2 n , hence it is an integer. Using this norm interpretation for P 2,1 (n) and P 2,3 (n), a calculation with Magma shows that P 2,a (n) > 2n for odd integers 7 ≤ n ≤ 2601. Together with Proposition 3.3, this shows that (3.4.7) P 2,a (n) > 2n
for a = 1, 3 and odd n ≥ 7.
(iv) We also set
Certainly, any root of
Similarly, if k ≡ ±3 (mod 8), then
Comparing to (3.4.4) and (3.4.5), we see that
) is an integer, and P 2,1 (n) is an integer by (iii). Hence P 2,1 (n) divides t -(q). Similarly, P 2,3 (n) divides t + (q).
(v) By (3.4.7), P 2,1 (n) > n. Consider any prime divisor ℓ of P 2,1 (n), which then divides t -(q) by the result of (iv), and divides Φ 4n (2) by (3.4.6). Suppose that ℓ is not a primitive prime divisor of 2 4n -1. By [Lun, Satz 1] (cf. [Ro,Proposition 2]) ℓ|n, and moreover ℓ 2 Φ 4n (2). It follows that the ℓ-part of P 2,1 (n) is ℓ. Hence, if t -(q) is not divisible by any primitive prime divisor of 2 4n -1, then P 2,1 (n) divides n, a contradiction.
The proof for t + (q) is entirely similar.
Corollary 3.5. Let q = 2 n with 2 n. If n ≥ 3 then Φ 24 := q 2 -q √ 2q + q -√ 2q + 1 is divisible by a primitive prime divisor ppd(2, 12n) of q 4 -q 2 + 1. In all cases, Φ 24 := q 2 + q √ 2q + q + √ 2q + 1 is divisible by a primitive prime divisor ppd(2, 12n) of q 4 -q 2 + 1.
Proof. Note that if n = 1 then Φ 24 = 1 and Φ 24 = 13. Assume that n ≥ 3. By Theorem 3.4, t -(q 3 ) = q 3 -q 2q + 1 = (q + 2q + 1)(q 2 -q 2q + q -2q + 1) = t + (q)Φ 24 is divisible by a primitive prime divisor ℓ 1 = ppd(2, 12n). Since t + (q)|(q 2 + 1) and ℓ 1 |(q 4 -q 2 + 1), we see that ℓ 1 t + (q), and so ℓ 1 |Φ 24 . The argument for Φ 24 is similar, using t + (q 3 ) = q 3 + q 2q + 1 = (q -2q + 1)(q 2 + q 2q + q + 2q + 1) = t -(q)Φ 24 .
3C. Primitive prime divisor for Ree groups. The results in this subsection are not needed for the rest of the paper, however they will be used elsewhere [KT2]. We make the choice √ 3 > 0. For odd n ∈ Z ≥1 and a = 1, 5, set
Proposition 3.6. If 2 n ≥ 3 and a = 1, 5, then P 3,a (n) > 2n.
Proof. A computation with Mathematica shows that P 3,a (n) > 2n when 3 ≤ n ≤ 353. Now assume that n ≥ 354. By (3.3.1) m := φ(n) ≥ n 6/7 > 153, so m ≥ 154, and 2 ω(n) ≤ m 5/11 . Fix a ∈ {1, 5} and let R j := {(j -1)n ≤ k < jn | gcd(k, 12n) = 1, k ≡ a (mod 12)} for 0 ≤ j ≤ 11. For each j, observe that k ∈ R j if and only if 0 ≤ k := k -jn < n is coprime to n and k ≡ a -jn (mod 12). If 3 n, then according to Proposition 3.1, n) . On the other hand, if 3|n then 3 (a -jn), so by Proposition 3.2,
Setting b := 12 when 3 n and b := 8 when 3|n, by the above consideration now
with A := 364(4 -2 √ 3) and B := 364/(4 -2 √ 3). Setting f (t) := A t/6 B -2t 5/11 t -7/6 , g(t) := log f (t) = (t/6) log(A) -2t 5/11 log(B) -(7/6) log(t). Now g (t) = log(A)/6-log(B)/(1.1t 6/11 )-7/6t is increasing, so g (t) ≥ g (154) > 0.48 when t ≥ 154. It follows that g(m) ≥ g(154) > 0.74, and so f (m) = exp(g(m)) > 2.09. Thus, for m = φ(n) ≥ 154
Theorem 3.7. Let q = 3 n with 2 n. If n ≥ 3 then t(q) = t -(q) := q -√ 3q + 1 is divisible by a primitive prime divisor ppd(3, 6n) of q 2 -q + 1. In all cases, t + (q) := q + √ 3q + 1 is divisible by a primitive prime divisor ppd(3, 6n) of q 2 -q + 1.
Proof. (i) Note that t(3) = 1. Henceforth we will assume n ≥ 3. Consider the sets
and the map k → k + 6n yields bijections
for each j = 1, 5, 7, 11.
(ii) Make the choice of ζ := ζ 12n = exp(πi/6n) and consider the cyclotomic polynomial
In particular,
(iii) Here we show that P 3,1 (n) and P 3,5 (n) are integers. Clearly, they are algebraic integers in Q(ζ). Hence it suffices to show that each of them is fixed by any Galois automorphism σ : ζ → ζ l , gcd(l, 12n) = 1. First consider the case j ≡ ±1 (mod 12). Then σ fixes
12 , and fixes each of the sets C := j=1,11 (A j ∪ B j ) and D := j=1,11 (A j ∪ B j ) modulo 12n. Since
it follows that σ fixes each of P 3,1 (n) and P 3,5 (n). Now assume that j ≡ ±5 (mod 12). Then σ sends √ 3 to -√ 3 and ζ k to ζ kl , and thus
Note that modulo 12n, when k runs over C, 6n + kl runs over C, covering each element of C exactly once. Also, |C| = φ(3n) by (3.7.2), and so |C| is even. It follows that σ sends P 3,1 (n) to (-1) |C| P 3,1 (n) = P 3,1 (n), and similarly σ fixes P 3,5 (n).
(iv) We also set
Similarly, if k ≡ ±5 (mod 12), then
Comparing to (3.7.4) and (3.7.5), we see that
) is an integer, and P 3,1 (n) is an integer by (iii). Hence P 3,1 (n) divides t -(q). Similarly, P 3,5 (n) divides t + (q). (v) By Proposition 3.6, P 3,1 (n) > n. Consider any prime divisor ℓ of P 3,1 (n), which then divides t -(q) by the result of (iv), and divides Φ 6n (3) by (3.7.6). Since ℓ|(q 2 -q + 1), ℓ = 2, 3. Suppose that ℓ is not a primitive prime divisor of 3 6n -1. Again by [Lun, Satz 1] ℓ|3n, whence ℓ|n as ℓ ≥ 5, and moreover ℓ 2 Φ 6n (3). It follows that the ℓ-part of P 3,1 (n) is ℓ. Hence, if t -(q) is not divisible by any primitive prime divisor of 3 6n -1, then P 3,1 (n) divides n, a contradiction.
The proof for t + (q) is entirely similar.
3D. Primitive prime divisors for Suzuki-Ree groups: another approach. For δ ∈ Z + , α ∈ (Z/δ) × , and x ∈ R + with x ≥ 1, let
Note that for such δ, α, and x, f
.
Proof. The claim is evident for x = 1, and we will show that
and it suffices to show each factor is increasing in a) , and rad(a) := p|a p.
Lemma 3.9.
(1). This follows by repeatedly applying the following. If p|n is such that k := v p (δn) ≥ 2, then, writing δn =:
Therefore without changing f (α mod δ) n
(1) we may assume without loss of generality that n is squarefree and such that gcd(n, δ) = 1. Now if p|n, writing δn =: p • n 0 and letting p , n 0 ∈ Z be such that pp + n 0 n 0 = 1,
(1)
.
Since the lemma is evident for n = 1, by induction on the number of prime factors of n we find that
, and we are done.
Corollary 3.10. Let n ∈ Z + with n ≥ 3 be odd. Then
Proof. If there is a p|n with p ≡ ±1 (mod 8), then pair S -{p} with S ∪ {p} in the conclusion of Lemma 3.9 to see that f
(1 mod 8) n
(1) = f
(3 mod 8) n
(1) = 1. Otherwise every p|n is such that p ≡ ±3 (mod 8), so that by Lemma 3.9
By the same reasoning
We are done by Lemma 3.8.
Proof. If there is a p|m := n gcd(n,6 ∞ ) with p ≡ ±1 (mod 12), then pair S -{p} with S ∪ {p} in the conclusion of Lemma 3.9 to see that f
(1) = 1. Otherwise every p|m is such that p ≡ ±5 (mod 12), so that by Lemma 3.9 m) .
By the same reasoning
We are done by Lemma 3.8.
We begin this section with a group theoretic lemma which will be used in the proof of primitivity given in Theorem 4.5.
Lemma 4.1. Suppose A is a cyclic group of prime order p that acts faithfully on a finite q-group G, where p = q are primes. Let n be the order of q modulo p. Let χ ∈ Irr(G) be A-invariant and faithful, and write χ(1) = q a . Then n ≤ 2a. [MW,Example 2.7], say, |G/N | = q n . Let θ ∈ Irr(N ) be A-invariant under χ; such exists by [Is3,Theorem 13.27]. Now, since G/N is an abelian chief factor of the semidirect product GA and χ is GA-invariant, by the Isaacs "going down" theorem [Is3,Theorem 6.18], either χ N = θ, or χ N = eθ with e 2 = |G/N |, or θ G = χ.
In the third case, χ(1) = q n θ(1) = q n+b = q a , for some b ≥ 0. Thus n ≤ n + b = a ≤ 2a. In the second case, q n is a square, and that χ(1)/θ(1) = q n/2 . Then χ(1) = q n 2 +c = q a , for some c ≥ 0. Then a = n 2 + c, and again n ≤ 2a. In the first case, χ N = θ ∈ Irr(N ). If A does not act trivially on N , then A acts faithfully on N . Since θ is A-invariant and faithful, then we are done by the inductive hypothesis. Otherwise, A acts trivially on N , and therefore
Since χ is faithful, A acts trivially on G. But this cannot happen.
We will now introduce a general class of Airy sheaves F (q, f ) that includes the sheaves F q . Recall that q = 2 2n+1 = 2q 2 0 and t(q) := q -2q 0 + 1. Let k 0 /F 2 be a finite extension, and f (x) ∈ k 0 [x] a polynomial of degree (1 + q 0 )t(q). Form the Artin-Schreier-Witt lisse sheaf
Assume in addition that
and whose Kummer pullback by r th power is (the restriction to
We next give a key lemma of Šuch, which is proved in [Such,Proposition 11.1] but not stated there explicitly. Lemma 4.2. ( Šuch) Let F be an Airy sheaf of rank n ≥ 2 on A 1 /F p , i.e. F is the Fourier transform FT(L) of a lisse, rank one sheaf L on A 1 /F p with Swan ∞ (L) = n + 1. If F is induced, then it is induced from an Artin-Schreier covering of A 1 /F p . In particular, if F is induced, then it is induced from a normal subgroup of index p of its G geom .
Proof. Let us recall the argument, which is contained in the proof of [Such,Proposition 11
and hence End 0 (F ) ⊇ g 1/1. But g 1/1 has rank < n, and all its slopes are ≤ (n + 1)/n = 1 + 1/n. But each slope of g 1/1 has denominator in lowest terms at most the rank of g 1/1, which is < n. Therefore each slope, being at most 1 + 1/n, is in fact ≤ 1. Then by [Such,Corollary 3.3
The rest of the argument is given in the first eight lines of the second paragraph of the proof of [Such,Proposition 11.1].
We will need the following general result, a slight generalization of [KT1,Theorem 4.6].
Theorem 4.3. Let F be a semisimple lisse sheaf on G m /F p which is tame at 0. Denote by J the image of I(∞) in G := G geom,F . Then G is the Zariski closure of the normal subgroup of G generated by all G-conjugates of J.
Proof. Denote by G ∞ this Zariski closure. Then G is reductive, and hence its quotient G/G ∞ is reductive. It suffices to show that every irreducible representation of G/G ∞ is trivial. But a d-dimensional irreducible representation is a lisse sheaf of rank d on G m /F p which is tame at 0 (because P (0) dies in G) and lisse at ∞. By multiplicative inversion, this is an irreducible local system on A 1 /F p which is tame at ∞, so a representation of π tame 1 (A 1 /F p ), which is the trivial group.
We will also need a very special case (about subgroups of index 2) of (the second part of) the following proposition. Since we cannot find a reference for it, we give a proof. Proof. (a) We give a proof of statement (i), which is due to Jason DeVito, posted on MathStack-Exchange.
(a1) First we show that if G is connected, then G is generated by its divisible subgroups. Indeed, there is an open set U ⊆ G containing the identity such that U ⊆ exp(L) for the Lie algebra L of G. For any u in U , if u = exp(X), then u lies in the divisible subgroup {exp(tX) | t ∈ R} of G. Since G is generated by U , the claim follows.
(a2) Next we show that if G is connected and H ≤ G is of finite index, then H = G. Indeed, by considering the action via left translation of G on the finite set G/H of left H-cosets, we see that H contains a normal subgroup K ¡ G of finite index. By (i), G, and so G/K, is generated by its divisible subgroups. But G/K is finite, so K = G.
(a3) In the general case, consider the map ι :
and so H is a union of a finite number of G • , each of which is closed. Hence H is closed.
(b) For statement (ii), the argument in (a3) shows that it suffices to prove that if G is connected reductive and H has finite index in G, then H = G. The argument in (a2) allows us to further assume that H ¡ G. Recall [C, p. 16 [Tits, Main Theorem] implies that H ∩ G i is either equal to G i or contained in Z(G i ). The finite index assumption now ensures that H ≥ G i for all i, and thus
In particular, H contains t m for all t ∈ T . But the torus T is (k × ) r for some r, so k = k implies that any element in T is an m th power, and hence H ≥ T , completing the proof. [The referee kindly pointed out a much simpler argument as follows.
But G, being connected reductive, is generated by its maximal tori, and the map g → g N is surjective on each maximal torus. Hence H = G.]
As pointed out by the referee, in both of the cases of Proposition 4.4, once H is closed, it is also open (indeed, G H is a disjoint union of a finite number of cosets gH, each being closed, and so it is closed).
Theorem 4.5. Let q = 2 2n+1 with n ∈ Z ≥1 and n = 2. Under the assumption (4.1.3), the group G geom,F (q,f ) of the Airy sheaf F (q, f ) has no subgroups of index 2. As a consequence, the Airy sheaf F (q, f ) is not geometrically induced, and hence none of the sheaves G(q, f, r) is geometrically induced.
Proof. Assume to the contrary that the Airy sheaf F (q, f ) is induced. Then, by Lemma 4.2, the underlying representation V of the geometric monodromy group
(i) For each divisor r > 1 of t(q), consider the descent G(q, f, r). Because F (q, f ) is lisse at 0, the image of I(0) in the geometric monodromy group H of G(q, f, r) is the cyclic group µ r (F 2 ) of order r. Because the ∞-slopes of F (q, f ) are
We next show that for r > 1, H has no subgroup of index 2. We argue by contradiction. Suppose that H has a subgroup H 1 of index 2. Since every H-conjugate of the image of I(0) has odd order r, all such conjugates are contained in H 1 , and thus
and hence H = H 1 , a contradiction. We have shown that H has no subgroup of index 2.
(ii) We also know that G is a normal subgroup of H of index dividing r, simply because by the r th power map, G m /F 2 becomes a finite étale Galois covering of itself with cyclic group of order r, and G and H are respectively the Zariski closure of the images of π 1 (G m /F 2 ) and of a normal subgroup of cyclic index r of that group. Now if H = G, then the existence of G 1 leads to a contradiction by (i). If we now take r to be a prime dividing t(q), then H/G ∼ = C r .
Let J and Q denote the images of I(∞) and of P (∞) in H. By Theorem 4.3, H is equal to the Zariski closure H Zar ∞ in H of the normal closure H ∞ in H of J. Suppose for the moment that r |J|. Then every H-conjugate of J has order coprime to r, and so they are all contained in G, and thus
We have shown that r divides |J|. Recall that J = Q C, with C a cyclic 2 -group that permutes cyclically and transitively the q -1 simple Q-submodules V i in V , each of dimension q 0 = 2 n . As Q is a 2-group, C is of order divisible by r.
(iii) Since n = 2, by Theorem 3.4 the integer t(q) = q -√ 2q + 1 admits a prime divisor (4.5.1) r = ppd(2, 4(2n + 1)).
At this point, we take for r a ppd(2, 4(2n + 1)) which divides t(q). Fix c ∈ C of order r. Since r|t(q), r (q -1), and so c stabilizes each of the subspace V i , and certainly normalizes Q. For each i, let Φ i denote the representation of Q, c on V i . We claim that Φ i (c) centralizes Φ i (Q). Otherwise, Φ i (c) acts faithfully on Φ i (Q), of prime order r, and Φ i (Q) is faithful and irreducible of degree 2 n . Hence by Lemma 4.1, the order of 2 modulo r is at most 2n, which contradicts (4.5.1). Therefore, for each i,
But the cyclic group C permutes the V i 's transitively and c ∈ C, so α 1 = . . . = α q-1 =: α, i.e. c acts on V as α • Id V . As |c| = r, α = 1 is a primitive r th root of unity. Thus we may assume that c ∈ J has trace dim(V ) • ζ r . On the other hand, by Theorem 2.2 and (4.1.2), the trace of every element in J belongs to Q(i). Thus ζ r ∈ Q(i), a contradiction since r is an odd prime.
(iii) Since the irreducible representation V of G = G geom of F (q, f ) is not induced, for any r|t(q), the representation of G geom of G(q, f, r) on V , which contains G, is not induced.
In this section, we continue to consider Airy sheaves F (q, f ) of the same general shape (4.1.1), but we consider them only geometrically, i.e. as lisse sheaves on A 1 /F 2 . The key insight on which the results of this section are based is due to Šuch, cf. [Such,Proposition 11.1].
Lemma 5.1. ( Šuch) Let F be an Airy sheaf of rank D ≥ 2 on A 1 /F p . Let H be a direct factor of End(F ) of rank r < D. Then H is a direct sum of L ψ(ax) for various a ∈ F p .
Proof. The ∞-slopes of F are all 1 + 1/D. All slopes of End(F ) are therefore ≤ 1 + 1/D. Thus all slopes of H are ≤ 1 + 1/D. But each slope of H, written in lowest terms, has denominator ≤ r. Therefore H has all slopes ≤ 1. Now take an irreducible constituent K of H. Its Fourier transform, i.e. t → H c (A 1 /F p , K ⊗ L ψ(tx) ), is perverse irreducible on A 1 /F p , so is either a single delta function δ a or is the extension by direct image of a lisse sheaf on a dense open set. But on a dense open set of the t-line, K ⊗ L ψ(tx) has all ∞-slopes 1, so H 2 c (A 1 /F p , K ⊗ L ψ(tx) ) = 0 and by the Euler-Poincaré formula χ c (A 1 /F p , K ⊗ L ψ(tx) ) = 0, so FT(K) is punctual, hence a single δ a . This means in turn that each irreducible constituent of H is an L ψ(ax) . Because F is irreducible, End(F ) is completely reducible, hence H is completely reducible, and so it is the sum of its irreducible constituents.
Theorem 5.2. Let F be an Airy sheaf of rank at least 2 on A 1 /F p . Then the following conditions are equivalent.
(i) F is geometrically induced.
(ii) End(F ) contains a summand L ψ(ax) for some a = 0 in F p .
(iii) There exists a geometric isomorphism F ∼ = F ⊗ L ψ(-ax) for some a = 0 in F p .
Proof. Let us denote by n the rank of
and hence End 0 (F ) ⊇ g 1/1. But g 1/1 has rank less than n. So by Lemma 5.1, g 1/1 is the direct sum of various L ψ(ax) . As End 0 (F ) does not contain the trivial sheaf (by irreducibility of F ), we find that End 0 (F ), and hence End(F ), contains some L ψ(ax) for some nonzero a ∈ F 2 .
Conversely, suppose End(F ) contains a summand L ψ(ax) for some a = 0 in F p . Thus there is a nonzero geometric homomorphism from F to F ⊗ L ψ(-ax) . But source and target are geometrically isomorphic, so every nonzero homomorphism is an isomorphism. Thus we have a geometric isomorphism
That F is geometrically induced is the special case, where
, N is p, and E is Q ℓ , of the following general statement in characteristic zero representation theory.
Theorem 5.3. Over an algebraically closed field E of characteristic zero, let V be a finite dimensional irreducible representation of dimension d ≥ 2 of a group G, and L a one-dimensional representation of G which, viewed as a linear character χ of G, has finite order
Proof. We first reduce to the case when N is prime.
for every integer n. Let r be a prime dividing N . Replacing L by L ⊗(N/r) and Ker(χ) by the overgroup K := Ker(χ (N/r) ), we are reduced to the case when
for some e ∈ Z ≥1 . Note that Ind G K (E) ∼ = ⊕ r i=1 L ⊗i , so by Frobenius reciprocity, cf. [CR2,(10.20)],
Again by Frobenius reciprocity [CR2,(10.8)],
The following general result is stated for ease of later reference.
Theorem 5.4. Let F be a lisse irreducible Q ℓ -sheaf of rank d ≥ 2 on a smooth, geometrically connected X/F p . Suppose that End 0 (F ) contains a rank one summand L. Then F is induced.
Proof. Because L occurs in End 0 (F ), it cannot be trivial (as by irreducibility of F , End(F ) contains 1 exactly once). Exactly as in the proof of the implication (ii) =⇒ (iii) in Theorem 5.2, we infer that F ∼ = F ⊗ L -1 . Taking determinants of this isomorphism, we find that L ⊗d is trivial. Thus L has finite order N > 1 as a character of π 1 (X). Now apply Theorem 5.3.
Theorem 5.5. Let k 0 /F 2 be a finite extension, and f (x) ∈ k 0 [x] a polynomial of degree (1 + q 0 )t(q). Then the Airy sheaf F (q, f ) introduced in (4.1.1) is geometrically primitive, i.e. is not geometrically induced.
Proof. (i) In view of Theorem 5.2, it suffices to show that there is no geometric isomorphism from F (q, f ) to F (q, f ) ⊗ L ψ(ax) for any a = 0 in F p . We argue by contradiction. Because F (q, f ) was geometrically a Fourier transform, so is F (q, f ) ⊗ L ψ(ax) ; the effect of tensoring with L ψ(ax) after FT is the same as translating additively by -a before FT:
Since FT is invertible, an isomorphism of F (q, f ) with F (q, f ) ⊗ L ψ(ax) gives an isomorphism of L(q, f ) with its additive translate by -a:
We will show no such isomorphism exists.
(ii) Suppose that n ≥ 2. In this case, we argue as follows. If two lisse rank one sheaves are isomorphic, then so are their tensor squares.
Quite generally, addition of Witt vector of length 2 over an F 2 -algebra is given by
) . So we would have a geometric isomorphism
or equivalently a geometric isomorphism
Now we use the explicit shape of t(q) = q + 1 -2q 0 = 2q 2 0 -2q 0 + 1 = (q 0 -1)(2q 0 ) + 1. The key point is that t(q) = 1 + dQ with d ≥ 2 prime to p = 2 (here d = q 0 -1) and Q a strictly positive power of p = 2 (here
Thus up to Artin-Schreier equivalence,
Thus L ψ((x-a) 1+dQ -x 1+dQ ) has Swan = 1 + (d -1)Q > 0, so is not geometrically trivial.
(iii) We now turn to the case n = 1, which requires a more delicate analysis. What makes the n ≥ 2 case argument work is that q 0 -1 = 2 n -1 has n ≥ 2 binary digits. In the n = 1 case, t(8) = 5, so the above argument would involve examining (x -a) 5 -x 5 = (x -a)(x 4 -a 4 ) -x 5 = ax 4 -a 4 x + a 5 , but this is Artin-Schreier equivalent to (a 1/4 -a 4 )x, which for a ∈ µ 15 is Artin-Schreier trivial.
Thus we must look instead at
The first factor has Swan ∞ ≤ 2 × 4, while the second factor, which is L ψ(a polynomial of degree 13) , has Swan ∞ = 13, and hence their tensor product has Swan ∞ = 13, so is not geometrically trivial.
Recall that a lisse sheaf H is Lie-irreducible if, in the underlying representation of its G geom , the identity component G • geom acts irreducibly. It is Lie-self-dual if the given representation of G geom , when restricted to G • geom , is self-dual. Theorem 5.6. Suppose that a lisse sheaf F (q, f ) as in (4.1.1) is Lie-irreducible and Lie-self-dual. Then F (q, f ) is self-dual.
Proof. The two sheaves F (q, f ) and its dual F (q, f ) ∨ are irreducible representations of G geom whose restrictions to the identity component G • geom are isomorphic. So the two, as representations of G geom , differ by a linear character of
for some lisse, rank one L on A 1 . Both F (q, f ) ∨ and F (q, f ) have all ∞-slopes 1+1/rank(F (q, f )) < 2. Therefore Swan ∞ (L) ≤ 1 (otherwise F (q, f ) ∨ would have all ∞-slopes ≥ 2).
If Swan ∞ (L) = 0, then L is lisse on A 1 and tame at ∞, so geometrically trivial, and F (q, f ) is geometrically self-dual. If Swan ∞ (L) = 1, then L is L ψ(ax) for some nonzero a ∈ F p . We will show that this case cannot arise.
Recall that F (q, f ) := FT ψ (L ψ 2 ([x t(q) ,f (x)]) ). In general, the interaction of FT with duality is a geometric isomorphism
In characteristic 2, where ψ takes values ±1, we have ψ = ψ. Thus
while
By Fourier inversion, this is equivalent to a geometric isomorphism
We first treat the case n ≥ 2. Already the tensor squares of these two lisse rank one sheaves are not geometrically isomorphic, by the identical argument used to treat the case n ≥ 2 in the proof of Theorem 5.5.
In the case n = 1, we must show that the lisse rank one sheaf
is not geometrically trivial. The second tensor factor is
So we must show that
is not geometrically trivial. Exactly as in the proof of the n = 1 case of Theorem 5.5, the first factor has Swan ∞ ≤ 2 × 4 = 8, while the second factor has Swan ∞ = 13.
Theorem 5.7. No sheaf F (q, f ) introduced in (4.1.1) is geometrically self-dual.
Proof. Recall that F (q, f ) := FT ψ L ψ 2 ([x t(q) ,f (x)]) . In general, the interaction of the Fourier transform FT with the duality functor D(•) is a geometric isomorphism
In characteristic 2, where ψ takes values ±1, we have
) . We have the Witt vector addition law [x, y] = [x, 0] + [0, y], and, for Witt vectors over
) , and the asserted isomorphism is
This holds if and only if there is an isomorphism
or equivalently if L ψ((x t(q) ) 2 ) is geometrically trivial. By the Artin-Schreier reduction we have L ψ((x t(q) ) 2 ) ∼ = L ψ(x t(q) ) . The latter sheaf is not geometrically trivial, because x t(q) is a polynomial of odd degree t(q) and so the sheaf has Swan ∞ (L ψ(x t(q) ) ) = t(q) = 0.
Corollary 5.8. Suppose that the sheaf F (q, f ) introduced in (4.1.1) is Lie-irreducible. Then it is not Lie-self-dual.
Proof. Combine Theorem 5.6 and Theorem 5.7.
Theorem 5.9. The sheaf F (q, f ) introduced in (4.1.1) is tensor indecomposable.
Proof. We will show that F (q, f ) is tensor indecomposable as a representation of π 1 := π 1 (A 1 /F p ).
Because π 1 has cohomological dimension ≤ 1 (this is true for the π 1 of any smooth, connected affine curve over an algebraically closed field), the argument of [KRLT1,Corollary 10.4] shows that if F (q, f ) is tensor decomposable, then it is linearly tensor decomposable, i.e. we have an isomorphism of local systems on A 1 /F p ,
with both rank(A), rank(B) ≥ 2. To fix ideas, suppose rank(A) ≤ rank(B). Then
Thus End 0 (F (q, f )) contains End 0 (A) as a direct factor. Now End 0 (A) has rank less than (rank(A)) 2 ≤ rank(A)rank(B) = rank(F (q, f )).
So by Lemma 5.1, End 0 (F (q, f )) contains some L ψ(ax) with a = 0 (a = 0 because End 0 (F (q, f )) only contains 1 once, by irreducibility of F ). The proof of Theorem 5.5 shows this is impossible.
Lemma 5.10. For every integer n ≥ 2, the integer 2 n -1 is never a perfect power x m with x ∈ Z and m ≥ 2.
Proof. We argue by contradiction. If 2 n -1 = x m , then x is odd and x m = 2 n -1 ≡ 3 (mod 4). Thus m is odd, and hence 2 n = x m + 1 is divisible by x + 1. The quotient x m +1 x+1 > 1 is the alternating sum of m powers of the odd integer x, so is itself odd. Thus x m +1
x+1 is an odd divisor of 2 n , the desired contradiction.
Corollary 5.11. For n ≥ 1, no lisse sheaf of rank 2 n (2 2n+1 -1) can be tensor induced. In particular, F (q, f ) is not tensor induced.
Theorem 5.12. The local systems F (q, f ) on A 1 /F 2 introduced in (4.1.1) all satisfy the condition (S+) of [KT1, Definition 1.2].
Proof. First, by [Such,Proposition 7.4] the underlying representation V of G geom of F (q, f ) is irreducible. That F (q, f ) is primitive, tensor indecomposable, and not tensor induced is the content of Theorem 5.5, Theorem 5.9 and Corollary 5.11. That det(F (q, f )) has finite order results from the fact that F (q, f ) began life over a finite subfield of F p (in fact any subfield containing the coefficients of f ).
Theorem 5.13. For the local system F (q, f ) on A 1 /F 2 introduced in (4.1.1), every irreducible constituent of End 0 (F (q, f )) has dimension ≥ rank(F (q, f )). In particular, if G geom,F (q,f ) is not finite, then G • geom,F (q,f ) is a simple algebraic group of dimension ≥ rank(F (q, f )). Proof. By Lemma 5.1, any irreducible constituent of dimension < D := rank(F (q, f )) is a single L ψ(ax) , while Theorem 5.5 shows that End 0 (F (q, f )) contains no L ψ(ax) . Because F (q, f ) satisfies condition (S+) by Theorem 5.12, if G geom,F (q,f )) is not finite, then its identity component G • geom,F (q,f )) is a simple algebraic group. In that case, Lie(G • geom,F (q,f )) ) is an irreducible constituent of End 0 (F (q, f )), so has dimension ≥ D.
Theorem 5.14. Consider the sheaf F (q, f ) in (4.1.1) subject to the condition (4.1.3). Then, for each r|t(q), the descent G(q, f, r) satisfies the condition (S+) of [KT1, Definition 1.2].
Proof. Because G geom,F (q,f ) is a subgroup of G geom,G(q,f,r) of finite index, the fact that F (q, f ) satisfies condition (S+), see Theorem 5.12, implies that G(q, f, r) does as well. [Condition (S+) holds for (G, V ) if it holds for (H, V ) with H a subgroup of G of finite index.]
We also record the following Theorem 5.15. Under the condition (4.1.3), if the sheaf F (q, f ) in (4.1.1) is Lie-irreducible, then it is not Lie-self-dual.
Proof. This is just a restatement of Corollary 5.8, under the more restrictive hypotheses of (4.1.3). Given Theorem 5.7, it suffices to show that if F (q, f ) is Lie-irreducible and Lie-self-dual, then it is self-dual.
However, there is a simpler proof of this last fact, using the descent G(q, f ) := G(q, f, t(q)) in Theorem 5.14. Exactly as in the proof of Theorem 5.6, we find an isomorphism
for some lisse L on G m of rank one. Because G(q, f ) ∨ and G(q, f ) are both tame at 0 and with all ∞-slopes < 1, L is tame on G m , hence a Kummer sheaf L χ . Because G(q, f ) ∨ and G(q, f ) both have I(0) representations which are sums of characters of order dividing t(q), χ is a ratio of characters of order dividing t(q), so χ has order dividing t(q). Pulling back this isomorphism by t(q) th power, we get an isomorphism F (q, f ) ∨ ∼ = F (q, f ). 6. A local system for the Suzuki group 2 B 2 (8) Now we can prove the first main result of the paper, which establishes [Ka5, Conjecture 2.2] in the case q = 8: Theorem 6.1. Let q = 8. Both the local systems F q and G q have geometric monodromy group G geom ∼ = 2 B 2 (8) in one of its irreducible representation of degree 14. Over F 2 , the local systems F q and G q have arithmetic monodromy group G arith ∼ = Aut( 2 B 2 (8)).
Proof. (a) Let G, respectively H, denote the geometric monodromy group of F q , respectively of G q . Similarly, let G arith , respectively H arith , denote the arithmetic monodromy group of F q , respectively of G q , over F 2 . We will use the fact that f := Frob 1,F 2 has order 15 and trivial determinant. Indeed, a Magma calculation shows that Trace(Frob 1,F 2 15 |G q ) = Trace(Frob 1,F 2 15 |F q ) = 14. By Theorem 2.2, Frob 1,F 2 15 |G q is semisimple, hence (being of weight zero in a 14-dimensional representation) is the identity. Therefore det(Frob 1,F 2 15 |G q )) is a root of unity of order dividing both 4 (by Theorem 2.1) and 15, so this determinant is trivial. But H arith = f, H , and H has trivial determinant (by Theorem 2.1), and hence H arith also has trivial determinant. As G ¡ H and G arith ≤ H arith are subgroups, both G and G arith have trivial determinants.
Recall that G is normal in H of index dividing t(q) = 5. By Theorem 5.12, H satisfies (S+). Since the rank of the sheaves is 14, not a prime power, by [KT1,Lemma 1.1] this implies that either the identity component H • of H is a simple algebraic group that acts irreducibly on the underlying representation V of H, or H is a finite almost quasisimple group with L := H (∞) also acting irreducibly on V .
(b) Consider the former case. Note that if H • is classical of rank r, then either it is of type A and r ≤ 13, or r ≤ 7 by [KlL,Proposition 5.4.11]. Using the tables of [Lu], we see that H • is of type SL 2 , SL 14 , Sp 4 , Sp 6 , Sp 14 , SO 14 , or
Suppose first that H • = SL 14 (C). In this case, V is just the natural module for H • , hence M 2,2 (V ) takes the smallest possible value 2 for all H • , H, G • , and G. But this contradicts Corollary 1.2.
Next suppose that G • = H • is of type SL 2 , Sp 4 , Sp 6 , Sp 14 , SO 14 , or G 2 . In all these cases, the G • -module V is self-dual, and this contradicts Corollary 5.8.
(c) We have shown that H is finite, and we are in the almost quasisimple case. In this case, H arith is also finite, and
Let ϕ denote the character of H arith in the underlying representation V . Then the formula for the trace function of G q and the existence of an element with trace 2ζ 4 show that (6.1.1)
). Since any element z of C H arith (L) = Z(H arith ) acts as a root of unity γ on V , this implies that γ 4 = 1. Since H arith has trivial determinant, γ 14 = 1 and thus γ = ±1. It follows that (6.1.2)
Now, using (6.1.1) and the fact that L acts irreducibly on V = C 14 , the classification results of [HM] show that L = PSL 2 (13), SL 2 (13),
In all but the last case, ϕ| L is real-valued; furthermore, |Out(L)| ≤ 2. As shown in the proof of Theorem 4.5, G has no subgroups of index 2, so G can induce only inner automorphisms of L. But G arith = G, f , implying G arith /G ↩→ C 15 ; so the same conclusion holds for G arith , and hence G arith = C G arith (L)L = Z(G arith )L, with |Z(G arith | ≤ 2 by (6.1.2). On the other hand, as we saw in the computation for the proof of Corollary 1.2, some g ∈ G arith has trace 2ζ 4 on V . Write g = zh with z ∈ Z(G arith ) and h ∈ L. It follows that 2ζ 4 = ϕ(g) = ±ϕ(h), a contradiction since ϕ(h) ∈ R.
We have therefore shown that L = 2 B 2 (8). As |Out(L)| = 3, (6.1.2) implies that |H/L| divides 6. On the other hand, H is the normal closure of the image of order 5 of I(0), so we conclude that H = L. Since L ¡ G ¡ H, we also get G = L.
Recalling again that G arith /G ↩→ C 15 , we now see from (6.1.2) that Z(G arith ) = 1 and thus G arith ↩→ Aut(L) ∼ = L C 3 . A Magma computation shows the existence of a Frobenius element with trace ζ 4 , and this implies that G arith > L, whence G arith = Aut(L). Since H arith is generated over H by f = Frob 1,F 2 , an element of order 15, Z(H arith ) = 1 by (6.1.2). Thus H arith ↩→ Aut(L), and so H arith = Aut(L) as H arith ≥ G arith .
In this section, we will extend the classification results obtained in [KlL,Proposition 5.4.11] and [Lu,Theorem 5.1]. Even though the intended applications in the paper only need the complex case of these results, we establish them in the modular case, which is interesting in its own right.
Let F be an algebraically closed field of characteristic p ≥ 0 and let G be a simple, simply connected, classical algebraic group of rank r over F. Fixing a maximal torus in G, we consider the set of simple roots {α 1 , . . . , α r } and the corresponding set of fundamental weights { 1 , . . . , r } (in the ordering of [OV]). Then the set
of dominant weights admits the partial ordering where λ µ precisely when λ -µ = r i=1 k i α i for some non-negative integers k i . As usual, W denotes the Weyl group. If λ ∈ Λ + , let L(λ) denote the irreducible FG-module with highest weight λ.
We will rely on the following two results.
Theorem 7.1. [Pr] Let G be a simple, simply connected algebraic group in characteristic p > 0. If the root system of G has different root lengths, then we assume that p = 2, and if G is of type G 2 , then we also assume that p = 3. Let λ be a restricted dominant weight. Then the set of weights Π(λ) of the irreducible G-module L(λ) is the union of the W -orbits of dominant weights µ with λ µ. P Lemma 7.2. [H,Lemma 10.3B] Let λ = r i=1 a i i be a dominant weight. Then the stabilizer of λ in the Weyl group is the Young subgroup generated by the reflections ρ i along the simple roots α i for which a i = 0. P Our first result treats groups of type A and includes a strengthening for SL 22 :
Theorem 7.3. Let G = SL n (F) with n = r+1 ≥ 8, and let L(λ) be an irreducible FG-representation, which is restricted if p = Char(F) > 0. Suppose that dim L(λ) ≤ M , where M := n 4 if n = 22 and M := 8176 if n = 22. Then λ = 0, a 1 or a r with 1 ≤ a ≤ 3, 1 + r , 2 or r-1 , 3 or r-2 , 4 or r-3 , 1 + 2 or r-1 + r , 2 1 + r or 1 + 2 r , and 2 + r or 1 + r-1 .
Proof. Write the highest weight λ as r i=1 a i i with a i ∈ Z ≥0 . (a) First suppose that there is some weight µ = r i=1 b i i ∈ Λ + with λ µ and b j = 0 for some s + 1 ≤ j ≤ r -s, where s := 3 if n = 22 and s := 4 if n = 22. By Theorem 7.1, Π(λ) contains the W -orbit O(µ) of µ. By Lemma 7.2, Stab W (µ) ≤ ρ 1 , . . . , ρ j-1 , ρ j+1 , . . . , ρ r = S j × S n-j (where
showing µ is the unique dominant weight of L(λ), whence µ = λ. This also forces Stab W (λ) = S j × S n-j , and so λ = a j j . Now if a ≥ 2, then λ λ -α j = j-1 + (a j -2) j + j+1 , and the latter is another dominant weight of L(λ), a contradiction. So λ = 4 or r-3 in such a case.
We may therefore assume that (7.3.1) a j = 0 for every s + 1 ≤ j ≤ r -s.
Next suppose that s i=1 ia i ≥ s + 1. By [GLT,Lemma 2.6] (applied with m = s), there is some weight µ = r i=1 b i i ∈ Λ + with λ µ and b s+1 > a s+1 . This situation is already considered by the preceding analysis. Using the symmetry under the graph automorphism τ of G,
(b) Suppose a 4 > 0 or a r-3 > 0. By symmetry, we may assume a 4 > 0. By (7.3.2) n = 22, a 4 = 1, and a 1 = a 2 = a 3 = 0. Now if λ = 4 , then a j ≥ 1 for some r -3 ≤ j ≤ r by ( 7
a contradiction. Hence from now on we may assume that a 4 = a r-3 = 0. (c) Suppose that a 3 > 0 or a r-2 > 0. By symmetry, we may assume a 3 > 0, and so a 3 = 1 by (7.3.2). Suppose a j ≥ 1 for some r -3 ≤ j ≤ r. By Lemma 7.2,
with Y 2 a proper Young subgroup of S n-3 and hence
(as n ≥ 8), a contradiction. Now, if n = 22, then using (7.3.1) and (7.3.2) we see that
is also a weight of L(λ) by Theorem 7.1, and since
We may now assume that a i = 0 for 3 ≤ i ≤ r -2. Suppose that a 2 > 0 or a r-1 > 0. By symmetry, we may assume a 2 > 0. If a 2 ≥ 2, then by (7.3.2) n = 22, (a 2 , a 1 ) = (2, 0). Note that λ λ -α 2 = 1 + 3 + a r-1 r-1 + a r r . It follows from Theorem 7.1 that Π(λ) contains 1 + 3 + a r-1 r-1 + a r r and 4 + a r-1 r-1 + a r r . The lengths of W -orbits of these two weights are at least 11395 by (7.3.4), and so dim L(λ) > M .
If a 2 = 1 but a 1 ≥ 2, then by (7.3.2) n = 22, (a 2 , a 1 ) = (1, 2). Note that
The preceding arguments show that Π(λ) contains the weights 1 + 3 + a r-1 r-1 + a r r and 4 + a r-1 r-1 + a r r , and again dim L(λ) > M .
Hence a 2 = 1 and a 1 ≤ 1. If a r = a r-1 = 0, then λ = 2 or 1 + 2 . So assume that a r + a r-1 > 0.
Suppose that a r-1 > 0. Then by Lemma 7.2
Suppose next that a r ≥ 2. Then
Thus ν 1 ∈ Π(λ) ∩ Λ + , and (7.3.5) applied to
We may therefore assume that a r = 1. If
(e) We may now assume that a i = 0 for 2 ≤ i ≤ r -1, i.e. λ = a 1 + b r with s ≥ a ≥ b ≥ 0 (by symmetry). Suppose a ≥ 4, whence n = 22 and a = 4 by (7.3.2). Then λ λ -2α 1 = 2 2 + b r , and so dim L(λ) > M by the first paragraph of (d).
Suppose a = 3 but b > 0. Then λ ν and (7.3.3)
by using (7.3.5) for ν 4 . So a + b ≤ 3, and thus λ = 2 1 + r , 2 1 , 1 + r , 1 , or 0.
To handle the other classical groups, we first consider a special case. Again, we use the weight labeling as in [OV].
Proposition 7.4. Let G be a simply connected simple algebraic group over F of type B r , C r , or D r , with r ≥ 7. In what follows we may assume p = 3. We realize the roots and the weights of G using an orthonormal basis (e i | 1 ≤ i ≤ r) of R r (with scalar product (•, •)); in particular, 1 = e 1 , 2 = e 1 + e 2 , 3 = e 1 + e 2 + e 3 . Consider the simple (Weyl) module V (λ) of the corresponding algebraic group over C. Then ν ∈ Λ + is a weight of V (λ) precisely when λ ν. Writing i in terms of simple roots (see [OV,Table 2]), it is straightforward to check that this is equivalent to
Next we use Freudenthal's formula [H, p. 122] to find the multiplicity m V (λ) (µ) of µ as a weight of V (λ). Then we must find all multiples lα of positive roots α, with l ∈ Z ≥1 , such that µ + lα is a weight of V (λ), i.e. W -conjugate to one of the weights listed in (7.4.1). Again using [OV,Table 2], we can check that this happens precisely when l = 1 and α = e 1 -e 2 , e 1 -e 3 , e 2 -e 3 . In all these cases, µ + lα is W -equivalent to λ, and we readily obtain m V (λ) (µ) = 2.
Suppose now that
This can happen only when µ is a weight of some composition factor L(ν 0 ) of a reduction modulo p of V (λ), where ν 0 is listed in (7.4.1). Note that if ν is such a weight and ν = µ, then ν µ. It follows that ν 0 = µ, i.e. L(µ) is a composition factor of a reduction modulo p of V (λ). By the linkage principle, see [Jan], this implies that w • λ = µ for some w ∈ W p . Here, the affine Weyl group W p is generated by the map ρ α,l • λ = ρ α • λ + lpα, where α is a simple root, l ∈ Z and ρ α denotes the reflection corresponding to α; furthermore,
Observe that |α| 2 ∈ Z and
It follows that |ρ α,l • λ + δ| 2 ≡ |λ + δ| 2 (mod gcd(2, p)). Thus we have shown (Note that a slightly weaker result than (7.4.3), namely only modulo p, was obtained in [T, Lemma 2.1].) In our case, |λ + δ| 2 -|µ + δ| 2 = 6. Applying (7.4.3), we conclude that (7.4.2) can happen only when p = 3.
Theorem 7.5. Let G be a simply connected simple algebraic group over F of type B r , C r , or D r , with r ≥ 12. Let L(λ) be an irreducible FG-representation, which is restricted if p := Char(F) > 0.
Suppose that dim L(λ) ≤ M , where M := 2(r + 1) 3 if r ≥ 14, and M := r 3 if r = 12, 13. Then λ = a 1 with 0 ≤ a ≤ 3, 2 , 3 , or 1 + 2 . Moreover, if λ = 1 + 2 and r ≥ 16, then p = 3.
Proof. Write the highest weight λ as r i=1 a i i with a i ∈ Z ≥0 (note that L( 1 ) is the natural module for G).
(a) First suppose that a i > 0 for some r -2 ≤ i ≤ r. In this case, by Lemma 7.2, the length of the
Next suppose that a i > 0 for some 4
Applying this argument to µ = r i=1 b i i ∈ Π(λ) ∩ Λ + , we deduce that (7.5.1)
(since r ≥ 9), contradicting the bound on dim L(λ). So λ = a 3 3 . Now, if a 3 ≥ 2 (and so p = 2 as λ is restricted), then by Theorem 7.1 violating (7.5.1). Hence λ = 3 in this case.
We have shown that a 3 = 0. Suppose a 2 > 0. Now, if a 2 ≥ 2 (so again p = 2), then
leading to a contradiction by applying (7.5.2) to λ -α 2 . If a 2 = 1 but a 1 ≥ 2, then p = 2 and
again yielding a contradiction by applying (7.5.2) to λ -
We are left with the case λ = a 1 . If a ≥ 4, then again p = 2 and
leading to a contradiction by applying (7.5.2) to λ -2α 1 -α 2 . So 0 ≤ a ≤ 3 as stated.
(c) We make some more comments about the cases λ ∈ {3 1 , 3 , 1 + 2 }. Note that
So, assuming p = 2 when λ = 1 + 2 , we may assume 3 ∈ Π(λ) in all these three cases. It now follows from Lemma 7.2 that in these cases dim L(λ) ≥ |O( 3 )| = 4r(r -1)(r -2)/3 > r 3 .
Proposition 7.4 shows that if p = 2, then dim L( 1 + 2 ) > 8 r 3 > r 3 (as r ≥ 12), and if r ≥ 16 and p = 3, then dim L( 1 + 2 ) > 4r(r -1)(2r -1)/3 > 2(r + 1) 3 . Thus,
This statement (7.5.3) was recorded in [Lu,Theorem 5.1], but we note that the treatment of the weights a 1 1 + a 2 2 therein is incorrect. Also note that dim L( 1 + 2 ) may be smaller than 2(r + 1) 3 when p = 3, see [Lu] for examples.
We will need some preliminary facts:
Lemma 8.1. Let n ∈ Z ≥1 and let D := 2 n (2 2n+1 -1). Then none of the following equations
has a solution in the positive integers.
Proof. (i) Suppose x 2 -1 = D for some x ∈ Z ≥1 . Checking the cases 1 ≤ n ≤ 7 directly, we may assume n ≥ 8. Now x > 1 is odd, and gcd(x -1, x + 1) = 2, but 2 n |(x 2 -1). It follows that there is some = ±1 such that 2 n-1 |(x -). Write x -= 2 n-1 y for some y ∈ Z ≥1 . Then
and so y + 1 = 2 n-2 (2 n+3 -y 2 ). This implies y > 1, and y + is divisible by 2 n-2 . Hence y + = 2 n-2 z for some z ∈ Z ≥1 . In this case, y ≥ 2 n-2 -1, x ≥ 2 2n-3 -2 n-1 -1, and so x 2 -1 > 2 4n-7 ≥ 2 3n+1 > D (as n ≥ 8), a contradiction. (ii) Suppose x(x -1)/2 = D for some x ∈ Z ≥1 . Then gcd(x -1, x) = 1, but 2 n+1 |x(x -1). It follows that there is some ∈ {0, 1} such that 2 n+1 |(x -). Write x -= 2 n+1 y for some y ∈ Z ≥1 . If = 0, then 2 3n+2 -2 n+1 = 2D = x(x -1) = 2 2n+2 y 2 -2 n+1 y, and so y -1 = 2 n+1 (y 2 -2 n ). This implies y > 1, and y -1 is divisible by 2 n+1 . Hence y -1 ≥ 2 n+1 , x > 2 2n+2 , and so
and so y + 1 = 2 n+1 (2 n -y 2 ). This implies 1 ≤ y < 2 n/2 , and y + 1 is divisible by 2 n+1 , which is impossible.
(iii) Suppose n ≥ 2 and
and so 3y -1 = 2 n+1 (y 2 -2 n ). This implies that 3y -1 ≥ 2 is divisible by 2 n+1 . Hence y ≥ (2 n+1 + 1)/3 > 2 n-1 , x + 1 > 2 2n , and so (x + 1)(x -2) > 2 2n (2 2n -3) > 2 3n+1 > D (as n ≥ 2), a contradiction. If = 2, then 2 3n+2 -2 n+1 = 2D = (x + 1)(x -2) = 2 2n+2 y 2 + 3 • 2 n+1 y, and so 3y + 1 = 2 n+1 (2 n -y 2 ). This implies 1 ≤ y < 2 n/2 ≤ 2 n-1 , and 3y + 1 is divisible by 2 n+1 , which is impossible when n ≥ 2.
Note that if (n, D) = (1, 14), then D = 6 2 -1 = 6 3 -6 = (3 3 + 1)/2. Theorem 8.2. Suppose the sheaf F (q, f ) in (4.1.1), of rank D = 2 n (2 2n+1 -1), has infinite geometric monodromy group G = G geom . Then G • = SL D . Under the more restrictive condition (4.1.3) G = SL D ; moreover, the sheaf G(q, f, t(q)) has geometric monodromy group equal to G.
Proof. By Theorem 5.12, G satisfies (S+). Thus G
(a) Suppose n ≥ 3, so that D ≥ 1016. Then G • must be a classical group of rank say r, where r(2r
In the case of A r , we can apply Theorem 7.3 and, using the dimension formula for L(λ) given in [OV,Table 5] and the bound (8.2.1), we see that the highest weight λ of the G • -module V is, up to duality, a 1 with a = 1, 2, 2 , or 1 + r . If λ = 1 , then D = r + 1, G • = SL D , and hence G = SL D as stated. In the other cases, D = r+1 2 , r 2 , or (r + 1) 2 -1; all are impossible by Lemma 8.1.
In the case of types B r , C r , and D r , we can apply Theorem 7.5 (more precisely, (7.5.3)), and, using the dimension formula for L(λ) given in [OV,Table 5] and the bound (8.2.2), we see that the highest weight λ of the G • -module V is a 1 with a = 1, 2, or 2 . If λ = 1 , then G • = Sp(V ) or SO(V ), whence F (q, f ) is Lie-self-dual, contrary to Corollary 5.8. In the other cases, D = m 2 or m 2 -1 for some integer m ≥ 2, and this is impossible by Lemma 8.1. (b) Suppose n = 2, so that D = 124. Then G • is of type A r with r ≥ 11, B r , C r , or D r with r ≥ 8, E 7 , or E 8 . Neither E 7 nor E 8 has irreducible representations of degree 124, see [Lu], so G • is classical of rank r. If G • is of type B r , C r , or D r with 8 ≤ r ≤ 11, then using [Lu] we check that G • has no irreducible representation of degree 124. So r ≥ 12, and we apply Theorem 7.3, respectively (7.5.3), as above to conclude that G = SL D .
Finally, assume that n = 1, so that D = 14. The arguments in part (b) of the proof of Theorem 6.1 repeated verbatim show that either
In the former case G = SL D as in (a), and the latter case is ruled out by Corollary 5.8.
(c) Now assume (4.1.3). Then we can consider the descent G(q, f, t(q)), for which the trace function takes values in Q(ζ 4 ) and all slopes are less than 1. Since p = 2, Theorem 2.1(iii) implies that G(q, f, t(q)) has trivial determinant, and thus G(q, f, t(q)) has geometric monodromy group
Proof. (a) Let V denote the underlying representation. By Theorem 5.12, G satisfies (S+) on V . But the dimension D = dim(V ) = q 0 (q -1), with q 0 := 2 n , is not a prime power. Hence G is almost quasisimple by [KT1, Lemma 1.1]: S ¡ G/Z(G) ≤ Aut(S) for some finite, non-abelian simple group S. Then the quasisimple subgroup L := G (∞) acts irreducibly on V by [KT1,Lemma 1.4].
The condition (4.1.3) allows us to consider the descent G(q, f, t(q)) on G m , with geometric monodromy group H. Then G ¡ H of finite index, whence H is finite and satisfies (S+), and L = H (∞) , as H/G ↩→ C t(q) . The representation of H on V has ∞-slopes σ := (q 0 + 1)/D < 1, and is not tame at ∞. Hence Theorem 2.5 applies to G and H. We collect some further facts about (G, V ) and the character ϕ of H on V that we will use in the proof:
Indeed, by Theorem 2.1(i), the arithmetic monodromy group H arith,k 0 of G(q, f, t(q)) over k 0 has finite determinant. But it normalizes the finite irreducible subgroup H, so finite determinant implies that H arith,k 0 is finite. Now, by Chebotarev density, the finiteness of H arith,k 0 implies that all elements of it are Frobenii, and all Frobenii have traces in
Since each element of Z(H) acts as a root of unity on V , and the only roots of unity in
has no nontrivial finite p -quotient. Since F (q, f ) lives on A 1 /F 2 , G has no nontrivial quotient of odd order. On the other hand, the proof of Theorem 4.5 shows that G has no quotient of order 2 when n = 2. (iii) The image J = QC of I(∞) has Q = O 2 (J) and C = g ∞ , where the central order ō(g ∞ ) is divisible by q -1. Indeed, since the ∞-slope of F (q, f ) is 1 + 1/D, this implies, by [Ka1,Proposition 1.14], that I(∞) acts irreducibly on V , of dimension D = q 0 (q -1). Since the image J of I(∞) is cyclic of p -order modulo the image Q of P (∞), it follows that g ∞ permutes the pairwise non-isomorphic q -1 simple Q-summands on V , each of dimension q 0 , transitively.
(b) Consider the case n ≥ 3, so that D ≥ 1016, and G = L by (ii).
(b1) First suppose that S = A m for some m ≥ 3. As G/Z(G) = A m ↩→ GL m-1 (C), by Theorem 2.5, m -1 ≥ 1/σ. Note that 1/σ = q -2q 0 + 1, so m ≥ q -2q 0 + 2 ≥ 114. It follows that D = q 0 (q -1) < (m 2 -5m + 2)/2. In this case, by [GT,Lemma 6.1] m = D + 1, G = A D+1 , and V = C D is the heart of the natural permutation module. But then V is self-dual, contrary to (i).
(b2) Next suppose that S is a sporadic simple group. By (iii), the maximum order meo(S) of elements in S is at least q -1 ≥ 127. On the other hand, meo(S) ≤ 119, as one can check using the [Atlas] (see also [KT1,Table 2]), a contradiction.
(b3) Consider the case S is a simple group of Lie type in characteristic r = 2. By Theorem 2.5, the degree e of every nontrivial projective representation over F r of S satisfies e 2 -1 ≥ 1/σ, in particular, e ≥ 11. Similarly, the degree d of every faithful linear representation over F r of S satisfies d ≥ 1/σ, in particular, d ≥ 114. This rules out all classical groups of types A m or 2 A m with m ≤ 9, B m with m ≤ 56, and C m or D m with m ≤ 7 (as PSp 2m (r a ) and PΩ ± 2m (r a ) have faithful representations of degree m(2m -1) over F r ). This also rules out exceptional groups of types G 2 , 2 G 2 , and 3 D 4 . For the remaining exceptional groups of type F 4 (s), E 6 (s), 2 E 6 (s), E 7 (s), and E 8 (s), with s = r a , by [KS, Table A7] we have the following upper bounds s(s + 1)(s 2 + 1), s(s 6 -1)/((s -1) gcd(3, s -1)), (s + 1)(s 2 + 1)(s 3 -1)/ gcd(3, s + 1), (s + 1)(s 2 + 1)(s 4 + 1)/2, and (s + 1)(s 2 + s + 1)(s 5 -1) for meo(S), respectively. On the other hand, q 0 (q -1) = D is at least the smallest degree d(S) of nontrivial projective complex representations of S, which in turn is at least s 6 (s 2 -1), s 9 (s 2 -1), s 9 (s 2 -1), s 15 (s 2 -1), s 27 (s 2 -1), respectively, see e.g. [TZ, Table I], and we arrive at a contradiction in all five cases, as q -1 ≤ meo(S) by (iii) and D < q 3 /2.
If S = PSL m (s) or PSU m (s) with m ≥ 11, then m 2 -1 ≥ 1/σ, and so m 2 ≥ q -2q 0 + 2, by Theorem 2.5. Since m ≥ 11, by [TZ,Theorem 1
a contradiction. For S of type BCD m with m ≥ 7, we have min(m(2m -1), 2m + 1) ≥ 1/σ, and so 2m 2 ≥ q -2q 0 + 1, by Theorem 2.5. Since m ≥ 7, by [TZ,Theorem 1.1]
again a contradiction.
(b4) We may now assume that S is a simple group of Lie type in characteristic 2, defined over a field F s with s = 2 a . Since n ≥ 3, by Theorem 3.4, t(q) = q -2q 0 + 1 admits a divisor ℓ = ppd(2, 4(2n + 1)). Next, we use the fact that the image of I(0) in H has order q -2q 0 + 1, which implies that H has an element of prime order ℓ that normalizes G = L. But C H (G) = Z(H) has order dividing 4 by (i), so
First suppose that S = Sp 2m (s) with m ≥ 2 or PΩ ± 2m (s) with m ≥ 3. Then, (8.3.1) implies that ℓ ≥ 4(2n + 1) + 1 ≥ 29 divides a|S|. If moreover ℓ a, then ℓ divides m i=1 (s 2i -1), which implies 2ma ≥ 4(2n + 1) by primitivity of ℓ. In either case, (8.3.2)
Now, applying [TZ,Theorem 1.1] for m ≥ 3 we obtain
a contradiction. If S = PSp 4 (s), then s ≥ q ≥ 2 7 by (8.3.2), and so
Next suppose that S = PSL m (s) or PSU m (s) with m ≥ 2. The same arguments as above show that (8.3.2) still holds; in fact, s m ≥ q 4 for S = PSL m (s). Assume that S = PSL m (s) with m ≥ 2; in particular, s m-1 ≥ s m/2 ≥ q 2 . Then using [TZ,Theorem 1.1] we obtain that
a contradiction. Next, assume that S = PSU m (s) with m ≥ 5; in particular, q 3/2 ≤ s 3m/4 ≤ s m-5/4 . Then using [TZ, Theorem 1.1] we obtain that
again a contradiction. When S = PSU 4 (s), ℓ|a or ℓ divides (s + 1)(s 2 -1)(s 3 + 1)(s 4 -1), so instead of (8.3.2) now s 3 ≥ q 2 . Again using [TZ, Theorem 1.1] we obtain
a contradiction. Finally, if S = PSU 3 (s), then s 3 ≥ q 2 ≥ 2 14 from (8.3.2). But every irreducible character of SU 3 (s) of even degree has degree divisible by s ≥ q 2/3 , and hence cannot be equal to D = q 0 (q -1).
Let S be one of the exceptional groups of type G 2 (s) with s > 2, 3 D 4 (s), 2 F 4 (s) with s > 2, F 4 (s) with s > 2, E 6 (s), 2 E 6 (s), E 7 (s), and E 8 (s). That ℓ divides |Aut(S)| implies q 4 ≤ s c , where 6, 12, 12, 12, 12, 18, 18, 30, respectively. It follows that D < q 3/2 ≤ s 3c/8 , with 3c/8 = 9/4, 9/2, 9/2, 9/2, 9/2, 27/4, 27/4, 45/4, respectively. But this contradicts the lower bounds D ≥ d(S) ≥ s(s 2 -1), s 3 (s 2 -1), s 4 (s -1) s/2, s 6 (s 2 -1), s 9 (s 2 -1), s 9 (s 2 -1), s 15 (s 2 -1), s 27 (s 2 -1), respectively, see e.g. [TZ, Table I], and we arrive at a contradiction in all eight cases. The cases 2 F 4 (2) and F 4 (2) are ruled out because |Aut(S)| is not divisible by the prime ℓ ≥ 29.
The only remaining case is that S = 2 B 2 (s). That ℓ divides |Aut(S)| implies q 4 ≤ s 4 , i.e. q ≤ s. But D = q 0 (q -1) is the degree of some irreducible character of G, so q = s, i.e. S = 2 B 2 (q). In this case also G = L = 2 B 2 (q), as stated.
(c) Now we consider the case n = 2, i.e. D = 124.
(c1) As the quasisimple subgroup L = G (∞) acts irreducibly on V = C D , by [HM] we have the following possibilities for S:
Here, a generator g 0 of the image of I(0) in H has order t(q) = 25 and normalizes G and L. Since C H (L) = Z(H) ≤ C 4 by (i), 25 divides |Aut(S)|. This rules out the first three cases.
In the two cases S = G 2 (5) and A 125 , L = S and V | L is self-dual. Since C H (L) ≤ C 4 and |Out(S)| ≤ 2, it follows that H/L is a 2-group. On the other hand, H has no quotient of order 2, as shown in part (i) of the proof of Theorem 4.5 (which also works for n = 2). Hence H = L = G, and so F (q, f ) is self-dual. But this contradicts Theorem 5.7.
The only remaining case is S = 2 B 2 (32), in which case L = S.
(c2) The rest of this paragraph applies to all n ≥ 1, for which we know L = S = 2 B 2 (q). Since Out(S) ↩→ C t(q) and Z(H) ≤ C 4 . In particular, Z(H)G/G ∼ = Z(H)/Z(G) has order dividing both 4 and t(q), whence Z(H) = Z(G). Theorem 8.4. For the geometric monodromy group G = G geom of the sheaf F (q, f ) in (4.1.3), of rank D = 2 n (2 2n+1 -1), either G = SL D or G = 2 B 2 (q) with q = 2 2n+1 . Furthermore, for any r|t(q), the geometric monodromy group of the descent G(q, f, r) is also equal to G.
Proof. Suppose G is infinite. Then the statements follow from Theorem 8.2.
From now on, assume that G is finite. Then G = S = 2 B 2 (q) by Theorem 8.3. Since for any r|t(q), the geometric monodromy group of G(q, f, r) contains G and is contained in the geometric monodromy group H of F (q, f, t(q)), it suffices to show that H = S.
First, note that the ∞-slope of G(q, f, t(q)) is σ = (q 0 + 1)/D and D + 1 = (q 0 + 1)t(q), so gcd(D, q 0 + 1) = 1. It follows from [Ka1,Proposition 1.14] that I(∞) acts irreducibly on V , of dimension D = q 0 (q -1). Since the image J of I(∞) is cyclic of p -order modulo the image Q of P (∞), Q = O 2 (J) and J = Q, g ∞ , where the p -element g ∞ transitively permutes the pairwise non-isomorphic q -1 simple Q-summands on V , each of dimension q 0 . Since q = 2 2n+1 , using [Zs] we can find a primitive prime divisor ℓ = ppd(2, 2n + 1), and fix a power h of g ∞ that has order ℓ. Clearly, the prime ℓ is at least 2n + 3, so it is coprime to 2n + 1 = |Out(S)|. On the other hand, S ≤ H ≤ Aut(S) by (8.3.3). Hence h ∈ S and Q < S.
We can write h = x (q-1)/ℓ , where on the natural module U = F 4 q for S = 2 B 2 (q) < Sp(U ), the spectrum of x ∈ S consists of 4 eigenvalues ξ 2 n , ξ -2 n , ξ 1-2 n , ξ 2 n -1 where ξ ∈ F 2 × has order q -1 see [Bur], [Suz]. We may write Aut(S) = S, θ , where θ acts as the Galois automorphism λ → λ 2 of F 2 . Suppose that for some 1 ≤ a ≤ 2n and for some y ∈ S, the element yθ a centralizes h. Note that θ a sends x to x 2 a and h = x (q-1)/ℓ to x 2 a (q-1)/ℓ . It follows that
x 2 a (q-1)/ℓ = θ a hθ -a = y -1 (yθ a )h(yθ a ) -1 y = y -1 hy is S-conjugate to h = x (q-1)/ℓ . On the other hand, it is known [Bur] that if b, c ∈ Z then x b and x c are S-conjugate if and only if c ≡ ±b (mod (q -1)). It follows that ℓ divides 2 a ± 1; in particular, ℓ divides 2 2a -1. The primitivity of ℓ then implies that 2n + 1 divides 2a, a contradiction. We have shown that C Aut(S) (h) ≤ S. As g ∞ centralizes h, it follows that g ∞ ∈ S. Hence J = Q g ∞ < S. Now S ¡ H and H is finite, so H = S by Theorem 4.3. Theorems 6.1 and 8.4 imply the following.
Corollary 8.5. The sheaves F q and G q in §1, of rank D = 2 n (2 2n+1 -1), have the same geometric monodromy group G = G geom . Furthermore, either G = 2 B 2 (q) with q = 2 2n+1 , or n ≥ 2 and G = SL D .
Remark 8.6. It is plausible that for each q = 2 2n+1 ≥ 32, we can find a polynomial f 1 ∈ F 2 [x] of degree 2 n + 1 such that the sheaf F (q, f ) with f (x) = f 1 (x t(q) ) as in (4.1.3) has infinite geometric monodromy group, which then is SL D by Theorem 8.4. Indeed, a Magma calculation shows for each 2 ≤ n ≤ 25 that Frob 1,F 2 k(n) has non-integral trace on F (q, x (1+2 n )t(q) ) for some choice of k(n) ∈ Z ≥1 . For instance, Frob 1,F 2 k(n) has trace (2i -7)/2 for (n, (k(n)) = (2, 7), (5i + 3)/2 for (n, k(n)) = (3, 5), (7 -3i)/4 for (n, k(n)) = (4, 7), and 5/2 for (n, k(n)) = (5, 7). In fact, for infinitely many integers n ≥ 2, we offer in Theorem 9.18 a construction of a sheaf F (q, f ) with G geom = SL D . 9. Arithmetic vs. geometric monodromy groups 9A. Glauberman and Dade correspondences. Our next results depend on the Glauberman correspondence. Recall that if A is a solvable finite group acting by automorphisms on another finite group S with (|A|, |S|) = 1, then there exists a canonical bijection * : Irr A (S) → Irr (C), where C = C S (A) is the fixed-points subgroup and Irr A (S) is the set of A-invariant irreducible characters of S. (See [Is3,Chapter 13].) Since the map is canonical, it is not difficult to see that
where n is odd. Then it is well-known that S admits a field automorphism a of order n. Assume further that (m, |S|) = 1 for some divisor m > 1 of n.
has 5 irreducible characters; two of its four linear characters are rational, the others have field of values Q(i). It also has a rational irreducible character of degree 4. In particular, it follows that Irr(S) has exactly 5 irreducible A-invariant characters, and exactly two of them have field of values Q(i). On the other hand, if m < n, then C = C S (A) = 2 B 2 (2 n/m ) has exactly 2 irreducible A-invariant characters with field of values Q(i), namely the ones of degree (r -1) r/2 with r := 2 n/m ; see [Bur]. This proves the following.
Lemma 9.1. Suppose that S = 2 B 2 (2 n ), where n is odd. Assume further that (m, |S|) = 1 for a divisor m > 1 of n, and let A be the subgroup of field automorphisms of S of order m. Let C = C S (A). If θ ∈ Irr A (S) has field of values Q(i), then θ * has degree (r -1) r/2 with r := 2 n/m . Lemma 9.2. Suppose that A is a cyclic group of order m acting faithfully and coprimely on S. Proof. Let ψ ∈ Irr(G) be the canonical extension of θ to G. (This is the unique extension ψ such that the determinantal order is coprime with |A|, see [Is3,Corollary 6.28]). Since every extension of θ to G is a multiple of ψ by a linear character λ of G/S, we may assume that ψ = λψ 0 , where ψ 0 is the canonical extension.
Observe that m divides the order of x. Since each of the two irreducible characters of S := 2 B 2 (q) of degree (q -1) q/2 has field of values Q(i) and is Aut(S)-invariant, Lemmas 9.1 and 9.2 imply the following.
Corollary 9.3. Let q = 2 n with 2 n and let θ be either of the two irreducible characters of S := 2 B 2 (q) of degree (q -1) q/2. Then θ extends to G = Aut(S) ∼ = S C n . If furthermore n is coprime to |S|, then |ψ(x)| = 1 for any extension ψ of θ to G and for any x ∈ G S with G = x, S .
We remark that the character θ in Corollary 9.3 has a canonical extension to G. Indeed, by [N1, Theorem A], there exists a unique ψ ∈ Irr(G) such that the field of values of ψ is Q(i). In particular, notice that the restriction ψ to C n is rational-valued.
In some non-coprime situations, we can use the following statement. This also follows from results in [Da,§9], but in the situation under consideration, our approach is more straightforward.
Lemma 9.4. Suppose that G = SA, where A = a , S G and A ∩ S = 1. Let C = C S (A), and assume that χ ∈ Irr A (S) has an extension ψ ∈ Irr(G) such that ψ A is rational-valued. Suppose that for every x ∈ G S, there exists some g ∈ G such that x g = cb for some c ∈ C and b ∈ A. Then there exist a character θ ∈ Irr(C) and a sign = ± such that ψ(cb) = θ(c) for every c ∈ C and every generator b of A.
Proof. By the proof of [Is3,Lemma 13.5], we can write
where ψ β is a rational-valued character of S. Now, define θ(c) = ψ(ca) for c ∈ C. Then
Thus θ is a virtual character of C. We claim that [θ, θ] = 1. Let T be a set of representatives for the right cosets of C in S. In order to use the proof of [Is3,Theorem 13.6], we claim that
and the claim follows. The rest of the proof follows as in [Is3,Theorem 13.6].
Finally, to address the general situation in the Suzuki case, we must go much deeper into [Da,§9].
Corollary 9.5. Let q = 2 n with 2 n and let θ be either of the two irreducible characters of S := 2 B 2 (q) of degree (q -1) q/2. Then θ extends to G = Aut(S) ∼ = S C n and |ψ(x)| = 1 for any extension ψ of θ to G and for any x ∈ G S with G = x, S .
Proof. We adapt the notation of [Da,§9] to ours. We know that C S (A) = B R, where B = β has order 5, and R is a cyclic group of order 4. Now, let G = N G (B) = S A, where S = N S (B). Let G 0 be the set of x ∈ G such that xS = G/S, and let G 0 be the set of x ∈ G such that xS = G /S . By [Da,Lemma 9.3], S = C R, where C is a cyclic group described there. By [Da,Proposition 9.7], G 0 is a trivial intersection subset of G with normalizer G . In Dade's language, it satisfies (6.4) of [Da]. In particular, by [Da,Lemma 6.5],
is a disjoint union, where G = τ ∈T G τ is a disjoint union. (In that lemma right cosets are used.) In particular, if x ∈ G 0 and η is a class function of G , then
By [Da,Theorem 9.8], θ naturally corresponds to some irreducible character η ∈ Irr A (S ), which therefore has field of values Q(i). Now, S only has two A-invariant irreducible characters with field of values Q(i), so using the inverse of Dade's natural correspondence from that theorem, necessarily η is one of the two linear characters of S /C = R.
By hypothesis x, S = G, and so x ∈ G 0 . Since
we may assume that x ∈ G 0 . By [Is2,Lemma G], ψ(x) = γ(x), where γ ∈ Irr(S A) is an extension of η, and is a root of unity. In particular, γ is linear. We conclude that |ψ(x)| = 1, as desired.
Lemma 9.6. Let S ¡ G < GL(V ) ∼ = GL d (C), where G is finite, S = 2 B 2 (q), D = q 0 (q -1) with q 0 = 2 n , q = 2 2n+1 , and V = C D is irreducible over S. Suppose that G contains two elements g 0 , g 1 such that Trace(g 0 ) = ±i, Trace(g 1 ) = ±1, and g 1 ∈ g 0 S. Suppose in addition that the trace of every element in Z(G) belongs to Q(i). Then g 0 and g 1 both induce non-inner automorphisms of S.
Proof. Assume the contrary. As g 1 ∈ g 0 S, g 0 induces an inner automorphism of S. Hence we can write g 0 = zs for some s ∈ S and some z
±i, but this is impossible for any element in S, see [Suz]. Hence, ζ = ±i. Recalling that g 1 ∈ g 0 S, we can write g 1 = g 0 t = zst for some t ∈ S. In such a case, Trace(st) = ζ -1 Trace(g 1 ) = ±i, and we again arrive at a contradiction since st ∈ S.
9B. Traces of Frobenii and arithmetic monodromy groups. We now give a lemma on traces for those local systems F (q, f ) in which the polynomial f (x) of degree (q 0 + 1)t(q) lies in F 2 [x] and has f (0) = 0. Recall that q 0 = 2 n , n ≥ 1, q = 2q 2 0 , t(q) = q + 1 -2q 0 ; for k/F 2 a finite extension and i = ζ 4 , the trace function of F (q, f ) is q) , f (x) + tx])).
Lemma 9.7. For F (q, f ) as above, i.e. with f (x) ∈ F 2 [x] of degree (q 0 + 1)t(q) and f (0) = 0, define A := the number of nonzero monomials in f (x), so that A ≡ f (1) (mod 2). Then the traces at points of F 2 are given as follows, with i = ζ 4 :
Furthermore, for any finite extension k/F 2 and any t
If (-1) A = -(-1) n , these traces are respectively -1 and -(-1) n i. If (-1) A = (-1) n , these traces are respectively -(-1) n i and -1. For the second statement, note that Trace(Frob t,k |F (q, f )) is a sum of #k terms, each of absolute value 1, divided by a clearing factor of absolute value √ #k, hence it has absolute value ≤ √ #k.
Theorem 9.8. Let n ∈ Z ≥1 and q = 2 2n+1 . Suppose the Airy sheaf F (q, f ) defined in (4.1.3)
, and f (0) = 0, has finite geometric monodromy group S. Then the following statements hold for S and the arithmetic monodromy group
if Trace(Frob 0,F q 2 ) = -q 0 or Trace(Frob 0,Fq ) = ±q 0 i, and Z(G) = 1 if Frob 1,F 2 has odd order.
Proof. (i) By Theorem 8.4, the geometric monodromy group of F (q, f ) is S = 2 B 2 (q) in its irreducible representation of degree D = q 0 (q -1). By Lemma 9.7, for the images g 0 of Frob 0,F 2 and g 1 of Frob 1,F 2 in G, one has trace ±i and the other has trace ±1. Note that g 1 ∈ g 0 S, and since G is finite, the traces of all elements in G belong to Q(i) by Chebotarev density. By Lemma 9.6, g 0 induces a non-inner automorphism of S. (ii) Now assume that 2n + 1 is coprime to |S|, but m < 2n + 1. Let ψ be the character of R = S C m acting on the sheaf, which extends the character θ of S. Let η denote the Glauberman correspondent of θ as in Lemma 9.2, in particular, it has degree (r -1) r/2 for r := 2 (2n+1)/m ≥ 8. We can write g 0 = zh 0 and g 1 = zh 1 for some h 0 , h 1 ∈ R and z ∈ Z(G) (recall g 0 S = g 1 S). Now z acts as a root of unity β with β 4 = 1, and ψ(h j ) = γ j η(c j ) with c j ∈ C = 2 B 2 (r) and γ 2m j = 1 for j = 0, 1 by Lemma 9.2. Since η(c j ), β, ψ(g j ) ∈ Q(i), we see that γ j ∈ Q(i). But γ 2m j = 1 and 2 m, so γ j = ±1.
Then for any subfield k of of F q 2 (9.9.1) Trace(Frob 0,k |F q ) = Trace(Frob 0,k |H q ), precisely because the map x → x t(q) is bijective on k.
Let us rewrite the input Witt vector for H q as (9.9.2) x,
With this rewriting, we apply the idea of van der Geer-van der Vlugt, cf. [vG-vV, §5], as follows.
Let us define (9.9.3)
In Witt vector addition in F 2 -algebras, using the fact that R(x) is an additive polynomial, we get
= [0, x, y ] for (9.9.4)
x, y := xy + xR(y) + yR(x).
The key point is that x, y on k × k is a symmetric F 2 -bilinear map to k, and Trace k/F 2 ( x, y ) is a symmetric F 2 -bilinear form on k × k as F 2 vector space. Then |Trace(Frob 0,k |H q )| 2 = (1/#k)
x,y∈k
(by the shearing transformation (x, y) → (x + y, y))
The second summand vanishes unless the given x ∈ k has Trace k/F 2 ( x, y ) = 0 for all y ∈ k, in which case it is 1.
But x ∈ k has Trace k/F 2 ( x, y ) = 0 for all y ∈ k if and only if x ∈ Ker(k). To see this, note that for x, y ∈ k, x, y = xy + xR(y) + yR(x) = xy + n i=1 xy 2 i + n i=1 yx 2 i has the same Trace k/F 2 as x + n i=1 x 1/2 i + n i=1 x 2 i y. So by nondegeneracy of the trace, x ∈ k has Trace k/F 2 ( x, y ) = 0 for all y ∈ k if and only if x + n i=1 x 1/2 i + n i=1 x 2 i = 0, i.e. if and only if 2n i=0
x 2 i = 0. the value of Trace Fq/F 2 (V n (x)) is constant, and this constant value is then repeated p times as we sum over this orbit. So we have a congruence modulo pZ[i]:
(remembering that F q /F 2 has odd degree p, and both V n (0), V n (1) are already F 2 -rational)
So when 2n + 1 = p, we have a congruence modulo pZ[i]:
(1 -(-1) (p-1)/2 i) p Trace(Frob 0,Fq |F q ) ≡ -1 -i.
Multiplying both sides by (1 + (-1) (p-1)/2 i) p , we get a congruence modulo pZ[i]:
Recalling that 2 p ≡ 2 (mod p), and that (p -1)/2 = n, we get a congruence modulo pZ[i]:
Trace(Frob 0,Fq |F q ) ≡ -i2 n i (p-1)/2 , if p ≡ 1 (mod 4), 2 n i (p-1)/2 , if p ≡ 3 (mod 4).
(9.10.1) Trace(Frob 0,Fq |F q ) ≡p 2 n i, where p is given by i (p-1)/2 = (-1) (p-1)/4 = (-1) (p 2 -1)/8 when p ≡ 1 (mod 4), and by i (p+1)/2 = (-1) (p+1)/4 = (-1) (p 2 -1)/8 when p ≡ 3 (mod 4). Thus in both cases p is the Legendre symbol 2 p = (-1) (p 2 -1)/8 . In view of Proposition 9.9, Trace(Frob 0,Fq |F q ) is either 0 or an element of Z[i] of absolute value 2 n . It cannot be 0 because of (9.10.1). So it must be one of ±2 n or ±2 n i. Of these four possibilities, onlyp 2 n i is congruent modulo pZ[i] top 2 n i (this is just the statement that for an odd prime p, the four powers of i are distinct modulo pZ[i]).
(iii) We now turn to the general case, where we proceed by induction on the total number (counting multiplicity) of primes dividing 2n + 1. Thus we write 2n + 1 = ps, s = 2a + 1, p = 2b + 1, with a, b ≥ 1 and p prime.
We will need to deal with both F 2 ps and F 2 s . To simplify notation, let us write Q := 2 ps , q := 2 s .
ps = s(2b + 1) = 2bs + s = 2(sb + a) + 1.
The idea is that F Q F q is the disjoint union of F s -orbits, each of length p, and on each of these orbits, the value of Trace F Q /F 2 (V sb+a (x)) is constant, and this constant value is then repeated p times as we sum over this orbit. Thus we get a congruence modulo pZ[i]:
But for x ∈ F q , the trace from F Q down to F q is just multiplication by p, so
Suppose first that b is even, i.e. that p ≡ 1 (mod 4). Recall 4[x, y] = 0. So for x ∈ F q ,
where the last equality holds because b is even. So in this even b case,
Suppose now that b is odd, i.e. that p ≡ -1 (mod 4). Then for x ∈ F q ,
where the last equality holds because Trace F q /F 2 (x 2 ) = Trace F q /F 2 (x) = (Trace F q /F 2 (x)) 2 . So in this odd b case as well, also
We have shown that (9.10.
By the induction hypothesis, (9.10.3) Trace(Frob 0,F q |F q ) =s 2 a i.
The clearing factors are invertible modulo pZ[i]. We next show that the clearing factors are equal modulo pZ
If b is odd, then (p + 1)/2 is even, and this ratio is
(1 + (-1) a i) ps+s 2 s = (2(-1) a i) s(p+1)/2 2 s = 2 s(p-1)/2 ((-1) a i) s(p+1)/2 = 2 s(p-1)/2 i s(p+1)/2 .
If b = (p -1)/2 is even, this ratio is
(1 -(-1) a i) (p-1)s = (-2(-1) a i) s(p-1)/2 = 2 s(p-1)/2 i s(p-1)/2 .
Let χ p,quad be the quadratic character of F × p , so 2 (p-1)/2 ≡ χ p,quad (2) (mod pZ[i]). As s is odd, the ratio of clearing factors modulo pZ[i] is χ p,quad (2)i s(p-1)/2 when p ≡ 1 (mod 4), and χ p,quad (2)i s(p+1)/2 when p ≡ 3 (mod 4). Next, when p ≡ 1 (mod 4), i (p-1)/2 = χ p,quad (2).
When p ≡ 3 (mod 4), i (p+1)/2 = χ p,quad (2). As s is odd, we find that in all cases the ratio of clearing factors is 1 (mod pZ[i]), as stated. Hence (9.10.2) and (9.10.3) imply the congruence Trace(Frob 0,F Q |F Q ) ≡ Trace(Frob 0,F q |F q ) =s 2 a i (mod pZ[i]).
Note that 2 sb = (2 (p-1)/2 ) s ≡ χ p,quad (2) s = χ p,quad (2) = p (mod p). Hence ps 2 sb+a ≡ ps p 2 a = s 2 a (mod p) by multiplicativity of the Jacobi symbol. It follows that Trace(Frob 0,F Q |F Q ) ≡ps 2 sb+a i (mod pZ [i]).
This congruence shows that Trace(Frob 0,F Q |F Q ) is nonzero, so by Proposition 9.9 it is one of ±2 sb+a or ±2 sb+a i. Again, of these four possibilities, onlyps 2 n i is congruent modulo pZ[i] tops 2 n i, and the induction step is complete. Proposition 9.11. For q = 2 2n+1 , consider the Airy sheaf F q . Then for any subfield k of F q |Trace(Frob 1,k |F q )| 2 = 1.
Proof. Let k be a subfield of F q 2 . Note that N := (q + 2q 0 + 1)q 2 /2 divides (q 2 + 1)q 2 and so is coprime to q 2 -1. Hence the map x → x N is a bijection on k, and it sends x t(q) to x (q 2 +1)q 2 /2 = x for any x ∈ k. As in the proof of Proposition 9.10, let us denote by F the absolute Frobenius x → x 2 . For each integer j ≥ 0, we define R j (x) := We now take k to be a subfield of F q . We examine the function x → x N on k. This function depends only on N (mod (q -1)). Then N ≡ (1 + 1 + 2q 0 )q/2 = (2 + 2q 0 )q/2 = q + qq 0 ≡ 1 + q 0 = 1 + 2 n (mod (q -1)). At this point, we repeat the van der Geer-van der Vlugt argument of Theorem 9.9. We find that if k is a subfield of F q , then The key observation is that for k a subfield of F q , Ker (k) = {0}. Indeed, since 0 ∈ Ker (k) ⊆ Ker (F q ), it suffices to show that Ker (F q ) = {0}. But for x ∈ Ker (F q ), Trace Fq/F 2 (x) = 2n-2 i=0 F i (x) +F 2n-1 (x) + F 2n (x) = F 2n-1 (x) + F 2n (x).
Thus for x ∈ Ker (F q ), Trace Fq/F 2 (x) = F 2n-1 (x + F (x)). As Trace Fq/F 2 (x) ∈ F 2 , this gives Trace Fq/F 2 (x) = x + x 2 .
If Trace Fq/F 2 (x) = 0, then x + x 2 = 0, i.e. x ∈ F 2 , so x is 0 or 1. Of these, only x = 0 has Trace Fq/F 2 (x) = 0. If Trace Fq/F 2 (x) = 1, then x + x 2 = 1 and so F 2 (x) = F 4 . But F 4 is not a subfield of F q , which has odd degree 2n + 1 over F 2 . So in this second case, there are no possible Proposition 9.12. For q = 2 2n+1 and the Airy sheaf F q , Trace(Frob 1,Fq |F q ) = -1.
Proof. (i) The quantity Trace(Frob 1,k |F q ) lies in Q(i) and is integral outside the unique place over 2, so by Proposition 9.11, Trace(Frob 1,Fq |F q ) is one of {1, -1, i, -i}. For any odd prime p, these four elements are distinct modulo pZ [i]. We proceed by induction on the total number (counting multiplicity) of primes in the factorization of 2n + 1. For the induction base, suppose that k = F q = F 2 p . Then F q F 2 is the disjoint union of F -orbits of length divisible by p. On each such F -orbit, the value of Trace Fq/F 2 (V (x)) is constant, and this constant value is then repeated a multiple of p times as we sum over this orbit. So we have a congruence modulo pZ[i]:
RawTrace(Frob 1,Fq |F q ) ≡ - (1 -(-1) (p-1)/2 i) p Trace(Frob 1,Fq |F q ) = RawTrace(Frob 1,Fq |F q ) ≡ -1 + i.
Multiplying both sides by (1 + (-1) (p-1)/2 i) p , we get a congruence modulo pZ[i]:
2 p Trace(Frob 0,Fq |F q ) ≡ -(1 -i)(1 + (-1) (p-1)/2 i) p .
If p ≡ 3 (mod 4), the right side is -(1 -i) p+1 = -(2i) (p+1)/2 = -2 (p+1)/2 (-1) (p+1)/4 ≡ -2(-1) (p 2 -1)/8+(p+1)/4 = -2 (mod p).
If p ≡ 1 (mod 4), the right side is -(1 -i)(1 + i) p , which is -2(1 + i) s-1 = -2(2i) (p-1)/2 = -2 (p+1)/2 (-1) (p-1)/4 ≡ -2(-1) (p 2 -1)/8+(p-1)/4 = -2 (mod p).
Thus in both cases qTrace(Frob 1,Fq |F q ) ≡ -2 (mod pZ[i]). Using Proposition 9.11, we conclude that Trace(Frob 1,Fq |F q ) = -1 in this case.
(ii) Suppose now that 2n + 1 = ps with an odd prime p and an odd integer s. Then just as in (iii) of the proof of Proposition 9.10 we write s = 2a + 1, p = 2b + 1, ps = 2(sb + a) + 1.
We will need to deal with both F 2 ps and F 2 s . To simplify notation, let us write Q := 2 ps , q := 2 s .
Then, just as in the proof of Proposition 9.11,
The elements of F Q F q fall into F -orbits of length p, and the elements from each of these orbits have the same Trace F Q /F 2 ([x, R sb+a-1 (x)]). So we get a congruence modulo pZ[i]
Notice that for x ∈ F q , R sb+a-1 (x) = R a-1 (x) + bTrace F q /F 2 (x).
Suppose first that p ≡ 1 (mod 4). Then b is even,
and hence RawTrace(Frob 1,F Q |F Q ) ≡ RawTrace(Frob 1,F q |F q ) (mod pZ[i])
when p ≡ 1 (mod 4). Suppose next that p ≡ -1 (mod 4). Since b is odd, p[x, R sb+a-1 (x)] = -[x, R sb+a-1 (x)] = [x, x 2 + R sb+a-1 (x)] = [x, x 2 + xTrace F q /F 2 (x) + xR a-1 (x)] = [x, xR a-1 (x)] + [0, x 2 + xTrace F q /F 2 (x)].
that G arith,Fq,F 2 = Aut( 2 B 2 (q)) is false when q = 2 m with m ≡ ±1 (mod 8); it should be corrected in that case by replacing F q by Fq as in (iv)(β) of Theorem 9.14.
9C. Local systems with infinite monodromy groups. Theorem 9.8 allows us to prove the following criterion for infinite monodromy.
Proposition 9.16. Let n ∈ Z ≥1 , q = 2 2n+1 , and consider the Airy sheaf F (q, f ) defined in (4.1.3) with q = 2 2n+1 , f (x) ∈ F 2 [x], and f (0) = 0. Suppose that (i) 2n + 1 is coprime to | 2 B 2 (q)|, and (ii) there is some odd integer m coprime to 2n + 1 such that |Trace Frob 0,F 2 m |F (q, f ) | = 1.
Then F (q, f ) has infinite geometric monodromy group.
Proof. Assume the contrary: F (q, f ) has finite geometric monodromy group S. By Theorem 9.8, S = 2 B 2 (q) and the arithmetic monodromy group of F (q, f ) over F 2 is G = Z(G) × Aut(S), where Z(G) ≤ C 4 . In particular, |G/S| divides 4(2n + 1), and so S equals the arithmetic monodromy group of F (q, f ) over F q 4 . Moreover, G = g, S for the image g of Frob 0,F 2 in G, and we can write g = zs with z ∈ Z(G), s ∈ Aut(S), and Aut(S) = s, S . Applying Corollary 9.5, we see that |Trace(g)| = |Trace(s)| = 1, a contradiction.
Proposition 9.17. Let n ∈ Z ≥2 , q := 2 2n+1 , r := [(n -1)/2], and define
x 1+2 n-2i .
Consider the sheaf F (q, f ) with f (x) := f 1 (x t(q) ) as in (4.1.3). Then for m := 2[n/2] + 1 we have |Trace Frob 0,F 2 m |F (q, f ) | = 1.
Proof. As in the proof of Proposition 9.9, the starting point is that gcd(t(q), 2 m -1) = 1. To see this, note that t(q) = q + 1 -2q 0 divides q 4 -1, and gcd(q 4 -1, 2 m -1) = 2 gcd(4(2n+1),m) -1 = 1. Thus for the field k := F 2 m , the map x → x t(q) is bijective on k.
The sheaf F (q, f ) was built out of the Witt vector x t(q) , f 1 (x t(q) . Let us denote by H(q, f 1 ) the sheaf built by the same recipe, with same clearing factor, out of the Witt vector [x, f 1 (x)]. Then we have Trace(Frob 0,k |F (q, f )) = Trace(Frob 0,k |H(q, f 1 )), precisely because the map x → x t(q) is bijective on k.
Next, as in (9.9.2), we write the input vector [x, f 1 (x)] as [x, xR(x)], with R(x) := r i=0
x 2 n-2i = F n (x) + F n-2 (x) + . . . + F n-2r (x), and F denotes the absolute Frobenius. Now we can repeat the arguments in the proof of Proposition 9.9, and compute the form Trace k/F 2 ( x, y ) on k × k with x, y := xy + xR(y) + yR(x) as in (9.9.4). So in both cases, the symmetric bilinear form Trace k/F 2 ( x, y ) on k × k has kernel consisting of the elements x ∈ k with Trace k/F 2 (x) = 0, that is, of exactly 2 m-1 elements. Then the proof of Proposition 9.9 shows that |Trace Frob 0,F 2 m |F (q, f ) | is either 0 or 2 (m-1)/2 > 1, and hence can never be equal to 1 (since by hypothesis n ≥ 2, and hence m ≥ 3).
Theorem 9.18. Let n ∈ Z ≥2 , q := 2 2n+1 , and consider the sheaf F (q, f ) of rank D := 2 n (2 2n+1 -1), with f (x) = f 1 (x t(q) ) and f 1 (x) as defined in Proposition 9.17. Assume in addition that 2n + 1 is coprime to | 2 B 2 (q)|; for instance, take 2n + 1 = ℓ a for any odd prime ℓ = 5 and any a ∈ Z ≥1 . Then the geometric monodromy group of F (q, f ) is SL D .
Proof. We first apply Proposition 9.9, and note that m is coprime to 2(2n + 1). It then follows from Proposition 9.16 that F (q, f ) has infinite geometric monodromy group G geom . By Theorem 8.4, G geom = SL D . Suppose that 2n + 1 = ℓ a for a prime ℓ, but ℓ divides | 2 B 2 (q)|. Then ℓ divides q 4 -1 = 2 4ℓ a -1. Since ℓ|(2 ℓ-1 -1), ℓ divides gcd(2 4ℓ a -1, 2 ℓ-1 -1) = 2 gcd(4ℓ a ,ℓ-1) -1, and so ℓ divides 2 4 -1 = 15. But one knows that 3 | 2 B 2 (q)|, so r = 5.
More generally, to ensure that 2n + 1 is coprime to | 2 B 2 (q)| for q = 2 2n+1 , we can take any n such that 2n + 1 = p a 1 1 p a 2 2 . . . p at t , where p 1 < p 2 < . . . < p t are primes, p i = 5, a i ∈ Z ≥1 , and p i (p j -1) whenever i < j. Indeed, suppose p j divides | 2 B 2 (q)| for some j. Then p j divides both 2 p j -1 -1 and q 4 -1 = 2 4(2n+1) -1. Since gcd(p j -1, 4(2n + 1)) = gcd(p j -1, 4 t i=1 p a i i ) divides 4, it follows that p j divides 2 4 -1 = 15. But p j = 5 by assumption, and p j = 3 since 3 | 2 B 2 (q)|, a contradiction.