L-functions of symmetric powers of the generalized Airy family of exponential sums

L-functions of symmetric powers of the generalized Airy family of exponential sums C. Douglas Haessig∗Antonio Rojas-Le´on† December 7, 2010 Abstract This paper looks at the L-function of the k-th symmetric power of the Q`-sheaf Aifover the affine line A1 Fqassociated to the generalized Airy family of exponential sums. Using `-adic techniques, we compute the degree of this rational function as well as the local factors at infinity. Using p-adic techniques, we study the q-adic Newton polygon of these L-function. 1 Introduction In this paper we study the L-function attached to the k-th symmetric power of the Q`-sheaf Aifassociated to the generalized Airy family of exponential sums. Symmetric powers appear in the proofs of many arithmetic problems. For instance, Deligne’s proof [6] of the Ramanujan-Petersson conjecture relies on the construction of a Galois module coming from the k-th symmetric power of a certain `-adic sheaf. The Sato-Tate conjecture [5] [16] [27] relies on the analytic continuation of the L-function attached to the k-th symmetric power of an `-adic representation coming from an elliptic curve. Another equidistribution result concerning Kloosterman angles was proven by Adolphson [4] using results of Robba’s [23] on the L-function of the k-th symmetric power of the `-adic Kloosterman sheaf Kl2. Symmetric powers also arise in the proof of Dwork’s conjecture [29] [30] [31]. To begin, let us recall the general setup of an L-function of an `-adic representation. Let Fqbe the finite field of qelements and characteristic p. Let Ybe a smooth, geometrically connected, open variety defined over Fq; for instance, take Yto be affine s-space As Fqor the torus Gs m. Denote its function field by K, and its corresponding absolute Galois group by GK:= Gal(Ksep/K). Let Vbe a finite dimensional vector space over a finite extension field of Q`, where `6=p. Let ρ:GK→GL(V) be a continuous `-adic representation unramified on Y, and let Fbe the corresponding lisse sheaf on Y. Define the L-function of ρon Yby L(Y, ρ, T ) := Y x∈|Y| 1 det(1 −ρ(Frobx)Tdeg(x)).(1) By the Lefschetz trace formula, this is a rational function whose zeros and poles may be described using ´etale cohomology with compact support: L(Y, ρ, T ) = 2dim(Y) Y i=0 det(1 −FrobqT|Hi c(Y⊗Fq,F))(−1)i+1 Given such a representation, we may construct new L-functions via operations such as tensor, symmetric, or exterior products. Natural questions about these new L-functions concern the determination of their degrees (Euler characteristic) and describing various properties about their zeros and poles. In this paper, we will focus on the symmetric powers of a particular family of exponential sums called the generalized Airy family. Other families whose symmetric powers have been investigated are the Legendre family of elliptic curves [3] [10] and the hyperKloosterman family [12] [13] [23]. We note that the former seems to have been motivated by Dwork’s p-adic interest in the Ramanujan-Petersson conjecture. The generalized Airy family is defined as follows. Let fbe a polynomial over Fqof degree dwith p-d. Let ψbe a nontrivial additive character on Fq. For each ¯ t∈Fqdefine its degree by deg(¯ t) := [Fq(¯ t) : Fq]. It is well-known that the associated L-function of the sequence of exponential sums Sm(¯ t) := X x∈Fqmdeg(¯ t) ψ◦TrFqmdeg(¯ t)/Fq(f(x) + ¯ tx) for m= 1,2,3, . . . ∗Partially supported by NSF grant DMS-0901542 †Partially supported by PO8-FQM-03894 (Junta de Andaluc´ıa), MTM2007-66929 and FEDER 1 is a polynomial of degree d−1: L(f, A1,¯ t;T) := exp ∞ X m=1 Sm(¯ t)Tm m!= (1 −π1(¯ t)T)· · · (1 −πd−1(¯ t)T). As we will describe later, the relative cohomology of this family may be represented `-adically as a lisse sheaf of rank d−1 over A1via Fourier transform. Let us denote this sheaf by Aif. The L-function of the k-th symmetric power of Aiftakes the form: Mk(f, T):=L(A1,Symk(Aif), T) := Y t∈|A1|Y a1+···+ad−1=k (1 −π1(t)a1· · · πd−1(t)ad−1Tdeg(t))−1, where |A1|denotes the set of closed points on A1. By the Lefschetz trace formula, Mk(f, T) is a rational function. The `-adic sheaf Aifwas extensively studied by N. Katz in [18], where its monodromy group is determined and, as a consequence, an equidistribution result is obtained for the exponential sums in the family ([18, Corollary 20]). From these results it follows that, for p > 2d−1, Mk(f, T ) is in fact a polynomial. For d= 3, a study of the monodromy group may be avoided using Adolphson’s method [4]. Our first main result is the computation of the degree of Mk(f, T) for p>d. The degree of the rational function Mk(f, T) equals the k-th coefficient of a generating series which is explicitly given in Corollary 3.4. Simplified formulas are given in section 5 for some particularly nice values of fand p. As an example of this theorem, consider the family generated by f(x) = xd. Then the degree of Mk(xd, T) may be described as follows. Let ζbe a primitive (d−1)-th root of unity in Fq. Denote by Nd−1,k the number of (d−1)- tuples (a0, a1, . . . , ad−2) of nonnegative integers such that a0+a1+· · ·+ad−2=kand a0+a1ζ+· · ·+ad−2ζd−2= 0 in Fq. Theorem 1.1. With the notation defined above, we have deg Mk(xd, T) = 1 d−1k+d−2 d−2−dNd−1,k. It was conjectured in [15] that Mk(x3, T) is a polynomial for all p > 3 since it was shown, in that paper, that Mk(x3, T) is a polynomial for every odd integer k, and also for every keven with k < 2p. Surprisingly, for p= 5, Mk(x3, T) is not a polynomial for infinitely many k. This was communicated to the first author by N. Katz and is a consequence of the geometric monodromy group of Aix3being finite. Theorem 1.2. Suppose p > 2d−1. Then Mk(f, T)is a polynomial which may be factored into a product Qk(f, T)Pk(f, T), where Pk(f, T)satisfies the functional equation Pk(f, T) = cTdeg(Pk)Pk(f, 1/qk+1T)with |c|=qdeg(Pk)(k+1)/2 and Qk(f, T)has reciprocal roots of weight ≤k. Furthermore, writing f(x) = Pd i=0 cixi, if we assume Fqcontains the 2(d−1)-th roots of −dcdthen an explicit description of Qk(f, T )may be given; see Corollary 4.3. Lastly, we wish to describe the p-adic behavior of the reciprocal roots of Mk(f, T). Motivation for such a study comes from Wan’s reciprocity theorem [28] of the Gouvˆea-Mazur conjecture [14] on the slopes of modular forms; see [3] for the connection between symmetric powers of the Legendre crystal with Hecke polynomials. Now, while we study in Section 6 the q-adic Newton polygon of the L-function for general fand k, our most precise results occur in the cubic case f(x) = x3: Theorem 1.3. Assume p≥7,kis odd, and k < p. Write Mk(x3, T) = 1 + c1T+· · · +crTr. Then ordq(cm)≥1 3(m2+m+km)for m= 0,1,2, . . . , r. (2) Furthermore, as a consequence of the functional equation, the endpoints of the q-adic Newton polygon of Mk(x3, T) coincide with the lower bound (2). If p= 5, then the numerator of Mk(x3, T )satisfies (2). We are hopeful that the restrictions kodd and k < p may be removed from the theorem (see [15] for details). There is also reason to believe that the lower bound (2) may be optimal in the sense that the q-adic Newton polygon will in fact equal this lower bound under certain conditions on pand k. As supporting evidence we note two facts. First, as mentioned in the above theorem, the endpoints of the q-adic Newton polygon of Mk(x3, T) 2 and the lower bound coincide. Secondly, the lower bound has the following symmetric property. Consider the points Pm∈R2defined by the lower bound: Pm:= (m, 1 3(m2+m+km)). The slope of the line segment joining Pmand Pm+1 is given by sm:= 1 3(2m+ 2 + k). If we set m0:= k−1 2−m, then we have the symmetry (k+ 1) −sm=sm0. In other words, for every slope smthere is a corresponding slope sm0. This is precisely a consequence of the functional equation for Mk(x3, T). That is, if αis a reciprocal root of Mk(x3, T) of slope s, then pk+1/α is another reciprocal root whose slope is (k+ 1) −s. Acknowledgments. We would like to thank Nicholas Katz and Steven Sperber for their very helpful comments. 2 Cohomological interpretation of Mk(f, T ) In this section we will study the generalized Airy family of exponential sums from the point of view of `-adic cohomology. We will do so by studying the sheaf Aifthat represents this family on the affine line A1over the given finite field Fq. We begin by observing that the map Fq→Cgiven by t7→ Px∈Fqψ(f(x) + tx) is the Fourier transform with respect to ψ, in the classical sense, of the map t7→ ψ(f(t)). This will translate, in the cohomological sense, to the fact that Aifis the Fourier transform, in the sheaf-theoretical sense, of the Q`-sheaf that represents the latter map, which is just the pull-back of the Artin-Schreier sheaf associated to ψvia the map given by f. Let us be more precise. The polynomial fnaturally defines a morphism, also denoted by f:A1 Fq→A1 Fq. Let Lψbe the Artin-Schreier sheaf on A1 Fqassociated to ψ(cf. [7, 1.7]). For every finite extension Fqmof Fq, every t∈A1(Fqm) = Fqmand every geometric point ¯ tover t, we have Trace(Frobt|Lψ,¯ t) = ψ(TraceFqm/Fq(t)), where Frobtdenotes a geometric Frobenius element at t. Consider the pullback Lψ(f):= f?Lψ. By [18, Theorem 17], for d≥2 the Fourier transform with respect to ψof Lψ(f)(which, in principle, is an element of the derived category Db c(A1,Q`)) is in fact a (shifted) lisse sheaf on A1, of rank d−1 and with d/(d−1) as its single slope at infinity. Its Swan conductor is therefore d. Let us denote this sheaf by Aif= R1πt!Lψ(f(x)+tx), where πt:A2→A1is the projection (x, t)7→ t. For every finite extension Fqmof Fq, every t∈Fqmand every geometric point ¯ tover twe have, denoting ψm=ψ◦TraceFqm/Fq: Trace(Frobt|(Aif)¯ t) = −X x∈Fqm ψm(f(x) + tx). The characteristic polynomial of the action of a geometric Frobenius element Frobtat ton the stalk of Aifat a geometric point over thas the form L(Aif, t, T) = (1 −π1(t)T)· · · (1 −πd−1(t)T) where πi(t) is a Weil algebraic number of weight 1 (i.e. all its complex conjugates have absolute value q1/2) and Px∈Fqmψm(f(x) + tx) = −Piπi(t)mfor all m≥1. Its k-th “symmetric power” is given by L(k; Aif, t, T) := Y a1+···+ad−1=k (1 −π1(t)a1· · · πd−1(t)ad−1T). These are the local factors of the L-function of the k-th symmetric power of Aif, which is given by the infinite product Mk(f, T) := Y t∈|A1| L(k; Aif, t, Tdeg(t))−1 The Lefschetz trace formula demonstrates that the zeros and poles of Mk(f, T) may be described in terms of cohomology: Mk(f, T) = 2 Y i=0 det(1 −Frob T|Hi c(A1 Fq,SymkAif))(−1)i+1 . 3 Since SymkAifis a lisse sheaf on the affine line, we have H0 c(A1 Fq,SymkAif) = 0, and the previous formula simplifies to Mk(f, T) = det(1 −Frob T|H1 c(A1 Fq,SymkAif)) det(1 −Frob T|H2 c(A1 Fq,SymkAif)). On the other hand, H2 c(A1 Fq,SymkAif) is just the space of co-invariants of the sheaf SymkAif, regarded as a representation of the fundamental group π1(A1 Fq), which is the k-th symmetric power of Aifregarded as a representation of the same group. This is the same as the space of co-invariants for its monodromy group, which is defined to be the Zariski closure of its image in the group of automorphisms of the generic stalk of Aif, isomorphic to GL(d−1) := GL(d−1,Q`). By [18, Theorem 19], for p > 2d−1 the geometric monodromy group of Aifis either SL(d−1) for deven, or Sp(d−1) for dodd if cd−1= 0 and µp·SL(d−1) for deven or µp·Sp(d−1) for dodd if cd−16= 0 (where f(x) = Pd i=0 cixi). In either case, its k-th symmetric power is still an irreducible representation of rank d+k−2 d−2of the monodromy group (because it is an irreducible representation of its subgroup SL(d−1) or Sp(d−1)), and in particular the space of co-invariants vanishes. More generally, it was proven by O. ˇ Such ([26, Proposition 1.6]) that, for p > 2, either Aifhas finite monodromy or its monodromy group contains SL(d−1) or Sp(d−1). In order to rule out the finite monodromy case for p≤2d−1 one may use for instance [20, Proposition 8.14.3], which implies that Aifhas finite monodromy if and only if for every element t∈Fqthe Newton polygon of the L-function associated to the exponential sum Pψ(f(x) + tx) has a single slope. Consequently, we have the following: Theorem 2.1. If Aifdoes not have finite monodromy (e.g. if p > 2d−1), the L-function of the k-th symmetric power of Aifis a polynomial: Mk(f, T) = det(1 −Frob T|H1 c(A1 Fq,SymkAif)) While it is tempting to believe that Mk(f, T) is always a polynomial, this is not true, as mentioned in the introduction. In fact, the monodromy group can be finite in certain cases; for instance when p= 5 and f(x) = x3, as proven in [21]. In such cases, H2 c(A1 Fq,SymkAif) will be non-trivial for infinitely many values of k, and consequently Mk(f, T) will have a denominator. Remark 2.2.Arithmetic difficulties often arise when the characteristic pis small compared to d, as demonstrated above by the link between the finiteness of the monodromy group when p≤2d−1 and the Newton polygons of the fibres of the family. By the functional equation, if we denote by NP1(t) the slope of the first line segment of the Newton polygon of the fibre tthen NP1(t)≤1/2 with equality if and only if the Newton polygon is a single line segment. If p≡1 modulo d, and in particular when p=d+ 1, then by [25, Theorem 3.11] the Newton polygon of every fibre equals the q-adic Newton polygon of the polynomial Qd−1 i=1 (1 −qi/dT). Thus, NP1(t) = 1/d and so the monodromy group is infinite when p=d+ 1 >3. Let [f(x)]xNdenote the coefficient of xNin f(x). Suppose d 2+ 1 < p ≤2d−1 and fhas coefficients over Fp. By [24, Theorem 2], if [(f(x) + tx)dp−1 de]xp−16≡ 0 modulo pfor some 0 ≤t≤p−1, then NP1(t)≤p−1 d/(p−1) for those t. By the assumption on dand p, notice that p−1 d/(p−1) equals either 1/(p−1) or 2/(p−1). Hence, the mondromy group is infinite when such a texists and p≥7. Their argument may be extended as follows. Let d>p−1. For a polynomial h(x), define (h(x))s:= h(x)(h(x)−1) · · · (h(x)−s+ 1). Define the linear operator U:Fp[x]→Fpby linearly extending the map which sends monomials xnto 0 if (p−1) -nand 1 otherwise. Let cs:= U((f(x) + tx)s)∈Fp. Suppose c1≡ · · · ≡ ck−1≡0 modulo pand ck6≡ 0 mod pfor some t, then NP1(t)≤k p−1. Hence, if this happens for some k < (p−1)/2 then the mondromy group is infinite. For example, for d>p−1 and f(x) = xd+xp−1then c1= 1 and hence the monodromy group is infinite for p≥5. Lastly, we mention the case when d= 4, p= 7 and f∈Fq[x] is not of the form (x+a)4+bx +c. Then by [17, Theorem 4.6] the monodromy of Aifis infinite. 3 Computation of the degree of the L-function We will now study the degree of Mk(f, T) when p > d. From the formula above we have deg(Mk(f, T)) = dim(H1 c(A1 Fq,SymkAif)) −dim(H2 c(A1 Fq,SymkAif)) = −χc(A1 Fq,SymkAif), 4 where χcdenotes the Euler characteristic with compact supports. Using the Grothendieck-N´eron-Ogg-Shafarevic formula, we have then deg(Mk(f, T)) = Swan∞(SymkAif)−rank(SymkAif) = Swan∞(SymkAif)−k+d−2 d−2.(3) In order to compute the Swan conductor of SymkAifwe have to study the sheaf Aifas a representation of the inertia group I∞of A1 Fqat infinity. Since Lψ(f)is lisse on A1, as a representation of the decomposition group at infinity we have Aif∼ =F∞,∞(Lψ(f)), where F∞,∞is the local Fourier transform as defined in [22]. Recently, Fu [11] and, independently, Abbes and Saito [1] have given an explicit description of the different local Fourier transforms for a wide class of `-adic sheaves. We will mainly be using the description given in [1], which works over an arbitrary (not necessarily algebraically closed) perfect base field, and therefore gives an explicit formula for Aifas a representation of the decomposition group D∞. If S(∞)is the henselization of the local ring of P1 Fqat infinity with uniformizer 1/t, the triple (Lψ(f(t)), t, −f0(t)) is a Legendre triple in the sense of [1, Definition 2.16]. Therefore by [1, Theorem 3.9] we conclude that, as a representation of D∞, Aifis isomorphic to (−f0)?(Lψ(f(t)) ⊗ Lψ(−tf0(t)) ⊗ Lρ(1 2f00(t)) ⊗ Q) = (−f0)?(Lψ(f(t)−tf0(t)) ⊗ Lρ(1 2f00(t)) ⊗ Q) where ρis the unique character I∞→Q? `of order 2, Lρthe corresponding Kummer sheaf and Qis the pull-back of the character Gal(Fq/Fq)→Q`mapping the geometric Frobenius to the quadratic Gauss sum g(ψ, ρ) := −Pt∈F? qψ(t)ρ(t). Write f(t) = Pd i=0 citi. For simplicity, from now on we will assume that Fqcontains the 2(d−1)-th roots of −dcd(which can always be achieved by a finite extension of the base field). Following [11, Proposition 3.1] we can find an invertible power series Pi≥0rit−i∈Fq[[t−1]] with rd−1 0=−dcdsuch that u(t) := tPi≥0rit−iis a solution to f0(t) + u(t)d−1= 0 (the other solutions being ζu(t) for every (d−1)-th root of unity ζ). The map φ: 1/t 7→ 1/u(t) defines an automorphism S(∞)→S(∞), and by construction −f0= [d−1] ◦φ, where [d−1] is the (d−1)-th power map. So Aifis isomorphic to [d−1]?φ?(Lψ(f(t)−tf0(t)) ⊗ Lρ(1 2f00(t)) ⊗ Q) = [d−1]?(φ−1)?(Lψ(f(t)−tf0(t)) ⊗ Lρ(1 2f00(t)) ⊗ Q) = [d−1]?(Lψ(f(v(t))+v(t)td−1)⊗ Lρ(1 2f00(v(t))) ⊗ Q) = [d−1]?(Lψ(f(v(t))+v(t)td−1)⊗ Lρ(1 2f00(v(t))))⊗ Q since [d−1]?Q=Q, where v(t) := φ−1(t) = tPi≥0sit−i. Let g(t) be the polynomial of degree dobtained from f(v(t)) + v(t)td−1by removing the terms with negative powers of t. It is important to notice that the coefficients of gare polynomials in the coefficients of f. More precisely, if we write g(t) = Pbiti, the coefficient biis a polynomial in the coeficients ai, ai+1, . . . , ad of f. Since Lψ(h(t)) is trivial as a representation of D∞for any h(t)∈t−1Fq[[t−1]], we have an isomorphism Lψ(f(v(t))+v(t)td−1)∼ =Lψ(g(t)) as representations of D∞. On the other hand, from f0(v(t)) + td−1= 0 we get f00(v(t))v0(t) + (d−1)td−2= 0, so Lρ(1 2f00(v(t))) = Lρ(−d−1 2v0(t)td−2). Since v0(t) = Pi≥0(1−i)sit−i=s0(1+Pi≥2(1−i)si s0t−i) and 1+Pi≥2(1−i)si s0t−iis a square in Fq[[t−1]], we have Lρ(−d−1 2v0(t)td−2)=Lρ(−d−1 2s0td−2)=Lρ(d(d−1) 2cd(s0t)d−2)(since sd−1 0=−1/dcd). So we finally get Aif∼ =[d−1]?(Lψ(g(t)) ⊗ Lρd(s0t))⊗ Lρ(d(d−1)cd/2) ⊗ Q.(4) We can now easily compute the Swan conductor at infinity of its symmetric powers. By [19, 1.13.1], Swan∞SymkAif=1 d−1Swan∞[d−1]?SymkAif=1 d−1Swan∞Symk[d−1]?Aif Lemma 3.1. Let ζbe a primitive (d−1)-th root of unity if Fq,Id−1 ∞the unique closed subgroup of I∞of index d−1. As a representation of Id−1 ∞, the restriction [d−1]?Aifof Aifis isomorphic to the direct sum d−2 M i=0 Lψ(g(ζit)) ⊗ Lρd(s0ζit)∼ = d−2 M i=0 Lψ(g(ζit)) ⊗ Lρd(t) . 5 Proof. Since (ζi)?Lψ(g)=Lψ(g(ζit)), (ζi)?Lρd(s0t)=Lρd(s0ζit)and [d−1] ◦ζi= [d−1] for every i, we have [d−1]?(Lψ(g(ζit)) ⊗Lρd(s0ζit)) = [d−1]?(Lψ(g(t)) ⊗Lρd(s0t)), and therefore by Frobenius reciprocity HomId−1 ∞([d− 1]?Aif,Lψ(g(ζit)) ⊗ Lρd(s0ζit)) = HomI∞(Aif,[d−1]?(Lψ(g(ζit)) ⊗ Lρd(s0ζit))) = HomI∞(Aif,Aif)∼ =Q`since the latter is an irreducible representation of I∞. So for every i,Lψ(g(ζit)) ⊗ Lρd(s0ζit)is a subrepresentation of [d−1]?Aif. Now Lψ(g(ζit)) ⊗Lρd(s0ζit)and Lψ(g(ζjt)) ⊗ Lρd(s0ζjt)are isomorphic if and only if Lψ(g(ζit)) and Lψ(g(ζjt)) are, if and only if g(ζit)−g(ζjt) = hp−hfor some h∈Fq[t]. Since p>d, this can only happen if g(ζit) = g(ζjt). Comparing the highest degree coefficients we conclude that ζiand ζjmust be equal. Therefore the direct sum of the Lψ(g(ζit)) ⊗ Lρd(s0ζit)for i= 0, . . . , d −2 injects into [d−1]?Aifand we conclude that it must be isomorphic to it, since they have the same rank. Consequently, we have an isomorphism of Q`[I∞]-modules Symk[d−1]?Aif∼ =M a0+a1+···+ad−2=k Lψ(Pd−2 i=0 aig(ζit)) ⊗ Lρdk(t). For every finite subset I⊂Zand every integer k≥0 define Sd−1(k, I) := {(a0, . . . , ad−2)∈Zd−1 ≥0|a0+a1+· · · +ad−2=k, a0+a1ζi+· · · +ad−2ζi(d−2) = 0 for every i∈I} It is clear from the definition that Sd−1(k, I) = Sd−1(k, I0) if φ(I) = φ(I0), where φ:Z→Z/(d−1)Zis reduction modulo d−1. Also, Sd−1(k, I) = ∅if pdoes not divide kand I∩(d−1)Z6=∅. The number of elements in Sd−1(k, I) can be conveniently expressed in terms of a generating function: Lemma 3.2. Let Fd−1(I;T) := P∞ k=0 #Sd−1(k, I)Tk. Then Fd−1(I;T) = 1 q#IX γ∈(Fq)I d−2 Y j=0 (1 −ψ(X i∈I γiζji)T)−1 where ψis any non-trivial additive character of Fq. Proof. From the definition, Fd−1(I;T) = X (a0,...,ad−2)∈Zd−1 ≥0Y i∈I δ(a0+a1ζi+· · · +ad−2ζi(d−2))Ta0+a1+···+ad−2 where δ(a) = 1 if a= 0, 0 otherwise. Equivalently, δ(a) = 1 qPγ∈Fqψ(γa). So we get Fd−1(I;T) = X (a0,...,ad−2)∈Zd−1 ≥0Y i∈I 1 qX γi∈Fq ψ(γi(a0+a1ζi+· · · +ad−2ζi(d−2)))Ta0+a1+···+ad−2 =X (a0,...,ad−2)∈Zd−1 ≥0X γ∈(Fq)I 1 q#I Y i∈I ψ(γia0)!Ta0 Y i∈I ψ(γia1ζi)!Ta1· · · Y i∈I ψ(γiad−2ζ(d−2)i)!Tad−2 =1 q#IX γ∈(Fq)IX (a0,...,ad−2)∈Zd−1 ≥0 ψ(X i∈I γi)a0Ta0ψ(X i∈I γiζi)a1Ta1· · · ψ(X i∈I γiζ(d−2)i)ad−2Tad−2 =1 q#IX γ∈(Fq)I X a0∈Z≥0 ψ(X i∈I γi)a0Ta0  X a1∈Z≥0 ψ(X i∈I γiζi)a1Ta1 · · ·  X ad−2∈Z≥0 ψ(X i∈I γiζ(d−2)i)ad−2Tad−2  =1 q#IX γ∈(Fq)I d−2 Y j=0 (1 −ψ(X i∈I γiζji)T)−1. Write g(t) = Pd j=0 bjtj, and let J={1≤j≤d|bj6= 0}and J≥j:= J∩ {j, j + 1, . . . , d}for every j∈ {1, . . . , d, d + 1}. We have Swan∞Symk[d−1]?Aif=X a0+a1+···+ad−2=k Swan∞Lψ(Pd−2 i=0 aig(ζit)) ⊗ Lρdk(t) =X a0+a1+···+ad−2=k deg( d−2 X i=0 aig(ζit)) 6 and d−2 X i=0 aig(ζit) = d−2 X i=0 ai d X j=0 bjζijtj= d X j=0 (bj d−2 X i=0 ζij)tj so its degree is the greatest jsuch that bjPd−2 i=0 ζij 6= 0. Therefore we get (d−1)Swan∞SymkAif= Swan∞Symk[d−1]?Aif =X j∈J j·(#Sd−1(k, J≥j+1)−#Sd−1(k, J≥j)) =dk+d−2 d−2−X j∈J h(j)·#Sd−1(k, J≥j) where h(j) := j−sup(J−J≥j) is the “gap” between the tjterm and the next lower degree term in g(t). Taking the corresponding generating function we get the formula Corollary 3.3. Let G(f;T) := P∞ k=0(Swan∞SymkAif)Tk, then G(f;T) = d (d−1)(1 −T)d−1−1 d−1X j∈J h(j)·Fd−1(J≥j;T) Using the previous formula for the degree, we deduce Corollary 3.4. The degree of Mk(f;T)is the k-th coefficient of the power series expansion of 1 (d−1)(1 −T)d−1−1 d−1X j∈J h(j)·Fd−1(J≥j;T). Corollary 3.5. For every J⊂ {1, . . . , d −1}, let Pd(J)be the subspace of the affine space Pdof polynomials of degree dover ksuch that bj= 0 if and only if j∈J. The sets {Pd(J)|J⊆ {1, . . . , d −1}} define a stratification of Pdsuch that the degree of Mk(f;T)is constant in each stratum. 4 The trivial factor Suppose p > d and the monodromy of Aifis not finite. We will now study the weights of the (reciprocal) roots of the polynomial Mk(f, T). Let us first consider the easier case where dis even, and therefore Aifis isomorphic to [d−1]?Lψ(g(t)) ⊗ Lρ(d(d−1)cd/2) ⊗ Q as a representation of D∞. Let Dd−1 ∞= Gal(Fq((1/t))/Fq((1/t1/(d−1)))), denote by α:Dd−1 ∞→Q? `the character corresponding to the sheaf Lψ(g), and let b∈I∞be a generator of the cyclic group D∞/Dd−1 ∞∼ =I∞/Id−1 ∞. By the explicit description of induced representations, there is a basis {v0, . . . , vd−2}of the underlying vector space Vsuch that a·v0=α(a)v0for every a∈Id−1 ∞and b·vi=vi+1 for i= 0, . . . , d −3. Then b·vd−2=bd−1·v0=α(bd−1)v0. Replacing bby a−1b, where a∈Id−1 ∞is an element such that α(a)d−1=α(bd−1) (which is always possible since the values of αare the p-th roots of unity and d−1 is prime to psince p > d) we may assume without loss of generality that α(bd−1) = 1. Furthermore, for any a∈Id−1 ∞we have a·vi= (abi)·v0= (bib−iabi)·v0=bi·α(b−iabi)v0=α(b−iabi)vi. So the restriction of Aifto Dd−1 ∞is the direct sum of the characters a7→ αi(a) := α(b−iabi). But we already know that it is the direct sum of the characters associated to the sheaves Lψ(g(ζit)) ⊗ Lρ(d(d−1)cd/2) ⊗ Q, so these two sets of characters are identical. Replacing bby a suitable power of itself we may assume that αiis the character associated to Lψ(g(ζit)) ⊗ Lρ(d(d−1)cd/2) ⊗ Q. In particular, Qd−2 i=0 αai iis geometricaly trivial (that is, trivial on Id−1 ∞) if and only if Paig(ζit) is a constant in Fq[t], that is, if and only if Paiζij = 0 for every j∈J. We turn now to the case dodd. Let χbe a multiplicative character of Fqof order 2(d−1) (which exists, since we are assuming that Fqcontains the 2(d−1)-th roots of unity). Then by the projection formula Aifis isomorphic to [d−1]?(Lψ(g(t)) ⊗ Lρ(s0t))⊗ Lρ(d(d−1)cd/2) ⊗ Q ∼ =([d−1]?Lψ(g(t)))⊗ Lχ(s0t)⊗ Lρ(d(d−1)cd/2) ⊗ Q. Let αi:Dd−1 ∞→Q? `(respectively β:D∞→Q? `) be the character corresponding to the sheaf Lψ(g(ζit)) (resp. Lχ(s0t)). Proceeding as in the deven case, we find a generator b∈I∞of D∞/Dd−1 ∞and a basis {v0, . . . , vd−2}of Vsuch that a·vi=αi(a)β(a)vifor a∈Dd−1 ∞and b·vi=β(b)vi+1 for i= 0, . . . , d −3, b·vd−2=β(b)v0. In this case, Qd−2 i=0 αai iβaiis trivial on Id−1 ∞if and only if Paig(ζit) is a constant in Fq[t] and Paiis even (since αihas order pand βrestricted to Id−1 ∞has order 2). 7 We can now compute the dimension of the invariant subspace of the action of I∞on SymkAif, in very much the same way it is done for the Kloosterman sheaf in [12, Lemma 2.1]. Its underlying vector space is SymkV. An element wis given by a linear combination w=X a0+···+ad−2=k ca0···ad−2va0 0· · · vad−2 d−2. In the deven case we have a·X a0+···+ad−2=k ca0···ad−2va0 0· · · vad−2 d−2=X a0+···+ad−2=k ca0···ad−2(αa0 0· · · αad−2 d−2)(a)va0 0· · · vad−2 d−2 for a∈Id−1 ∞and b·X a0+···+ad−2=k ca0···ad−2va0 0· · · vad−2 d−2=X a0+···+ad−2=k ca0···ad−2va0 1va1 2· · · vad−2 0. So wis fixed by I∞if and only if the character αa0 0· · · αad−2 d−2is trivial whenever ca0···ad−26= 0 and ca0···ad−2= cad−2a0···ad−3for all a0, . . . , ad−2. A basis for the invariant subspace is thus given by all distinct sums of the form (setting vd−1+l:= vlfor all l≥0): d−2 X j=0 va0 jva1 j+1 · · · vad−2 j+d−2 for all a0, . . . , ad−2such that αa0 0· · · αad−2 d−2is trivial, that is, such that Paiζij = 0 in Fqfor every j∈J. In the dodd case we get g·X a0+···+ad−2=k ca0···ad−2va0 0· · · vad−2 d−2=X a0+···+ad−2=k ca0···ad−2(αa0 0· · · αad−2 d−2)(g)βk(g)va0 0· · · vad−2 d−2 for g∈Id−1 ∞and h·X a0+···+ad−2=k ca0···ad−2va0 0· · · vad−2 d−2=X a0+···+ad−2=k ca0···ad−2β(h)kva0 1va1 2· · · vad−2 0. So wis fixed by I∞if and only if the character αa0 0· · · αad−2 d−2βkof Id−1 ∞is trivial whenever ca0···ad−26= 0 and ca0···ad−2=cad−2a0···ad−3β(h)kfor all a0, . . . , ad−2. Since all αi’s have order pand the restriction of βto Id−1 ∞has order 2, αa0 0· · · αad−2 d−2βkis trivial if and only if both αa0 0· · · αad−2 d−2and βkare trivial as characters of Id−1 ∞, that is, if and only if Paiζij = 0 in Fqfor every j∈Jand kis even. In particular, there are no non-zero invariants for I∞if kis odd. If kis even, a generating set for the invariant subspace is given by all distinct sums of the form d−2 X j=0 β(h)jkva0 jva1 j+1 · · · vad−2 j+d−2 for all a0, . . . , ad−2such that Paiζij = 0 in Fqfor every j∈J. Let rbe the size of the orbit of (a0, . . . , ad−2) under the action of Z/(d−1)Zby cyclic permutations. If r6=d−1, we can write d−2 X j=0 β(h)jkva0 jva1 j+1 · · · vad−2 j+d−2= r−1 X j=0 β(h)jk(1 + β(h)rk +· · · +β(h)(d−1 r−1)rk)va0 jva1 j+1 · · · vad−2 j+d−2. Notice that kmust be a multiple of d−1 r, since k=Pd−2 i=0 ai=d−1 rPr−1 i=0 ai. If rk d−1is odd we have 1 + β(h)rk +· · · +β(h)(d−1 r−1)rk =1−β(h)(d−1)k 1−β(h)rk = 0, so the above sum vanishes. On the other hand, if rk d−1is even it is clear that the element d−2 X j=0 β(h)jkva0 jva1 j+1 · · · vad−2 j+d−2=d−1 r r−1 X j=0 β(h)jkva0 jva1 j+1 · · · vad−2 j+d−2 is non-zero, and to different orbits correspond different elements. To summarize, we have 8 Proposition 4.1. Let Td−1(k, J)be the set of orbits of the action of Z/(d−1)Zon the set Sd−1(k, J)by cyclic permutations, and let Ud−1(k, J)be the subset of orbits such that rk d−1is even, where ris their cardinality. If dis even, the invariant subspace of the representation SymkAifof I∞has dimension #Td−1(k, J). If dis odd and k is even, it has dimension #Ud−1(k, J). If dand kare odd, the representation has no non-zero invariants. The sequences #Td−1(k, J) and #Ud−1(k, J) can also be described by means of generating functions. By Burnside’s lemma, the dimension of the invariant subspace for deven is given by #Td−1(k, J) = 1 d−1 d−1 X r=1 #{(a0, a1, . . . , ad−2)|ai=ai+rmod d−1}=1 d−1X r|d−1 φ(d−1 r)#Sr(kr d−1, J) where Sr(k, J) = ∅if kis not an integer and φis Euler’s totient function. So the generating function for the sequence {#Td−1(k, J)|k≥0}is Gd−1(J;T) : = ∞ X k=0 #Td−1(k, J)Tk = ∞ X k=0 1 d−1TkX r|d−1 φ(d−1 r)#Sr(kr d−1, J) =1 d−1X r|d−1 φ(d−1 r)X d−1 r|k #Sr(kr d−1, J)Tk =1 d−1X r|d−1 φ(d−1 r) ∞ X s=0 #Sr(s, J)Td−1 rs =1 d−1X r|d−1 φ(d−1 r)Fr(J;Td−1 r) Next, suppose that dis odd, and let (a0, . . . , ad−2)∈Sd−1(k, J). Let rbe the number of elements in its orbit. Then Pr−1 i=0 ai=kr d−1. We want to count the number of orbits such that this value is even. Since k=kr d−1·d−1 r, if the largest power of 2 that divides d−1 is smaller than the largest power of 2 dividing k,kr d−1must always be even. Suppose that the largest power of 2 that divides k, 2α(k), divides d−1. Then kr d−1is odd if and only if 2α(k) divides d−1 r, if and only if rdivides d−1 2α(k). Therefore #Ud−1(k, J)=#Td−1(k, J) if 2α(k)does not divide d−1 and #Td−1(k, J)−#Td−1 2α(k)(k 2α(k), J) if it does. The generating function is then ∞ X k=0 #Ud−1(k, J)Tk= ∞ X k=0 #Td−1(k, J)Tk−X j≥1;2j|d−1X lodd #Td−1 2j(l, J)T2jl =Gd−1(J;T)−X j≥1;2j|d−1 Hd−1 2j(J;T2j) where Hr(J;T) := 1 2(Gr(J;T)−Gr(J;−T)). Let F∈Dd−1 ∞⊂D∞be a geometric Frobenius element, and w=Pd−2 j=0 va0 jva1 j+1 · · · vad−2 j+d−2(resp. w= Pd−2 j=0 β(h)jkva0 jva1 j+1 · · · vad−2 j+d−2) a generator of the I∞-invariant subspace of SymkV.Facts on va0 jva1 j+1 · · · vad−2 j+d−2 via the character corresponding to Lψ(Paig(ζj+it)) ⊗L⊗k ρ(d(d−1)cd/2) ⊗Q⊗k(resp. Lψ(Paig(ζj+it)) ⊗Lρ(Q(s0ζj+it)ai)⊗ L⊗k ρ(d(d−1)cd/2) ⊗ Q⊗k). Since Paig(ζj+it) must be a constant polynomial, we have Lψ(Paig(ζj+it)) ∼ =Lψ(kb0). Additionally, if dis odd and keven, Lρ(Q(s0t)ai)=Lρ(s0t)kis trivial. We conclude: Proposition 4.2. A Frobenius geometric element at infinity acts on the I∞-invariant subspace of SymkAifby multiplication by ψ(kb0)ρ(d(d−1)cd/2)kg(ψ, ρ)k. As an immediate consequence we get Corollary 4.3. The local L-function of SymkAifat infinity det(1 −Frob T|(SymkAif)I∞)is given by (1 − ψ(kb0)ρ(d(d−1)cd/2)kg(ψ, ρ)kT)#Td−1(k,J)if dis even, (1 −ψ(kb0)ρ(d(d−1)cd/2)kg(ψ, ρ)kT)#Ud−1(k,J)if dis odd and kis even, and 1if dand kare odd. 9 The main result in Section 6.4 below will be to reduce the hypothesis (7) to a similar hypothesis which we believe is attainable. It is expected that this similar hypothesis will hold under rather general conditions on the prime p, the degree d, and the symmetric power k. However, it is also expected that this similar hypothesis will fail just as often, yet (7) will still hold. The conditions under which the weaker hypothesis is valid is currently under investigation. The rest of this section is devoted to the proof of Theorem 6.1, whose argument closely follows that of Dwork’s [9, §7] and Adolphson-Sperber’s [2]. The proof rests on relating the Newton polygon of ¯ βto another operator, ¯ β1, whose Newton polygon is much easier to estimate due to the Dwork decomposition of Ngiven in (7). The reason is that Dwork decomposition allows us to work on the chain level, where the operator β1acts in an easily understood way. Once estimates on β1are found on the chain level Dwork decomposition provides estimates in homology of ¯ β1. For b0≤band b≤p/(p−1), define α1:K(b0, b)→ K(b0/p, b) by α1: = τ−1◦ψx◦F(t, x) =1 G(tp, x)◦τ−1◦ψx◦G(t, x). Notice that α1◦D(t) = pD(tp)◦α1, and so α1induces a map ¯α1(t) : H1,t(b0, b)→ H1,tp(b0/p, b). On K(b0, b), since α(t) = ψa x◦Fa(t, x) =ψa x◦Fτa−1(tpa−1, xpa−1)· · · Fτ(tp, xp)F(t, x) =τ−1◦ψx◦F(tpa−1, x)◦ · · · ◦ τ−1◦ψx◦F(tp, x)◦τ−1◦ψx◦F(t, x) =α1(tpa−1)◦ · · · ◦ α1(tp)◦α1(t), it follows that ¯α(t) = ¯α1(tpa−1)◦ · · · ◦ ¯α1(tp)◦¯α1(t).(8) We also have the property that ψt◦¯α1(tp) = ¯α1(t)◦ψt.(9) Consequently, with β1:= ψt◦Symk(¯α1(t)) : H(k) 1,t (b0, b)→ H(k) 1,t (b0, b), we have the relation βa 1=ψa t◦Symk¯α1(tpa−1)◦ · · · ◦ ¯α1(tp)◦¯α1(t) =ψa t◦Symk(¯α(t)) =β where we have used (9) for the first equality and (8) for the second. Since detQq(π)(1 −¯ βT |H1,k)∈Qp(π), we have that detQq(π)(1 −¯ βT |H1,k)a=NormQq(π)/Qp(π)detQq(π)(1 −¯ βT |H1,k) =detQp(π)(1 −¯ βT |H1,k). Thus, detQq(π)(1 −¯ βTa|H1,k)a=detQp(π)(1 −¯ βTa|H1,k) =detQp(π)(1 −¯ βa 1Ta|H1,k) =Y ζa=1 detQp(π)(1 −ζ¯ β1T|H1,k).(10) Counting multiplicities, let midenote the number of reciprocal roots of detQp(π)(1 −¯ β1T|H1,k) which have slope si; note, we say λhas slope siif ordp(λ) = si. Then, from (10), detQq(π)(1 −¯ βTa|H1,k)ahas ami reciprocal roots of slope si, and so detQq(π)(1 −¯ βTa|H1,k) has mireciprocal roots with slope si. We conclude that detQq(π)(1 −¯ βT |H1,k) has mi/a reciprocal roots of slope asi. Next, for the q-adic valuation ordq(·) := 1 aordp(·), we will say a root λhas q-adic slope siif ordq(λ) = si. Observe that the above paragraph has demonstrated that detQq(π)(1 −¯ βT |H1,k) has mireciprocal roots with 16 q-adic slope siif and only if detQp(π)(1 −¯ β1T|H1,k) has amireciprocal roots with p-adic slope si. In terms of Newton polygons, this means the vertices of the q-adic Newton polygon of detQq(π)(1 −¯ βT |H1,k) are (0,0) and n X i=1 mi, n X i=1 misi!n= 1,2, . . . , dimQq(π)(H1,k) if and only if the vertices of the p-adic Newton polygon of detQp(π)(1 −¯ β1T|H1,k) are (0,0) and n X i=1 ami, n X i=1 amisi!n= 1,2, . . . , dimQp(π)(H1,k). Using this relation, any lower bound for the p-adic Newton polygon of the latter may be transformed to a lower bound of the q-adic Newton polygon of the former by dividing the coordinates of the vertices by a. Let us now concentrate on a lower bound for the p-adic Newton polygon of detQp(π)(1 −¯ β1T|H1,k). Define K(b0, b)•:= xK(b0, b), and let Vbe the L(b0)-span of the set {x, x2, . . . , xd−1}in K(b0, b)•. Define K(b0, b;ρ)•:= K(b0, b)•∩ K(b0, b;ρ) and V(b0, b;ρ) := V ∩ K(b0, b;ρ). By our hypothesis on the prime p, we will prove in the following section the Dwork decomposition K(b0, b; 0)•⊂ V(b0, b; 0) ⊕D(t)K(b0, b;e),(11) where e:= b−1 p−1. Consequently, K(b0, b)•=V ⊕D(t)K(b0, b), so we may identify H1with V, a free L(b0)-module of rank d−1 with basis {x, x2, . . . , xd−1}. Consequently, H(k) 1is a free L(b0)-module with basis {ei1 1· · · eid−1 d−1:= ei}, where i= (i1, . . . , id−1), each ijis a nonnegative integer satisfying i1+· · · +id−1=k, and ej:= xj. Set b=p p−1. For i= 1, . . . , d −1, xi∈ K(b p,b p;−b pw0(i))•and so F(t, x)xi∈ K(b p,b p;−b pw0(i))•. Thus, α1(xi)∈ K(b p, b;−b pw0(i))•. By (11) we may write this as α1(xi) = Ai,1x+· · · +Ai,d−1xd−1mod(D(tp)K(b p, b)) (12) where Ai,j ∈L(b p;b p(pw0(j)−w0(i))). Let S(i1, . . . , id−1) denote the set of nonnegative integers (l(r) s)1≤s,r≤d−1that satisfy the system l(1) 1+· · · +l(1) d−1=i1 . . .. . . l(d−1) 1+· · · +l(d−1) d−1=id−1. and T(j1, . . . , jd−1) denote the set of nonnegative integers (l(r) s)1≤s,r≤d−1that satisfy the system l(1) 1+· · · +l(d−1) 1=j1 . . .. . . l(1) d−1+· · · +l(d−1) d−1=jd−1. The k-th symmetric power of ¯α1acts on the basis {ei}as follows: Symk(¯α1(t))ei1 1· · · eid−1 d−1 = (¯α1(t)e1)i1· · · (¯α1(t)ed−1)id−1 =  r X j=1 A1,jej  i1 · · ·   r X j=1 Ad−1,jej  id−1 =X (l(r) s)∈S(i1,...,id−1) Z>0Al(1) 1 1,1· · · Al(1) d−1 1,d−1· · · Al(d−1) 1 d−1,1· · · Al(d−1) d−1 d−1,d−1el(1) 1+···+l(d−1) 1 1· · · el(1) d−1+···+l(d−1) d−1 d−1 =X j:=(j1,...,jd−1)∈Zd−1 ≥0 j1+···+jd−1=k B(i,j)ej1 1· · · ejd−1 d−1, 17 where B(i;j) := X (l(r) s)∈S(i1,...,id−1)∩T(j1,...,jd−1) (Z>0)Al(1) 1 1,1· · · Al(1) d−1 1,d−1· · · Al(d−1) 1 d−1,1· · · Al(d−1) d−1 d−1,d−1 and “Z>0” is some determinable nonzero positive integer. It follows that B(i;j)∈L(b p;b p(pw0(j)−w0(i))), and so Symk(¯α1(t)) : H(k) 1(b p,b p; 0) → H(k) 1(b p, b; 0).(13) Recall, M:= H(k) 1,t and N:= tH(k) 1,t . We are supposing that there exists a free Zq[π]-submodule Vof Nwith basis Γ := {tnei|(n;i)∈A}such that N(b, b; 0) ⊂ V(b, b; 0) ⊕∂M(b, b;) (14) for some ∈R. Now, Γ represents a basis of H1,k over Qq(π), but we need to understand the Fredholm determinant of ¯ β1on H1,k viewed as a vector space over Qp(π). To do this recall that Qq(π) is an unramified extension field of Qp(π). We have denoted by Zq[π] the ring of integers of Qq(π) with uniformizer πand residue field Fq, and Zp[π] the ring of integers of Qp(π) with uniformizer πand residue field Fp. Let {¯η1, . . . , ¯ηa}be a basis of Fqover Fp, and let {η1, . . . , ηa}be a lifting of this basis to an integral basis of Qq(π) over Qp(π). Lemma 6.3 (Dwork).The basis {ηi}has the property of p-adic directness; that is, for any g∈Qq(π), writing g=h1η1+· · · +haηawith hi∈Qp(π), then ordp(g) = min i=1,...,a{ordp(hi)}. Proof. Without loss of generality, we may assume that ordp(g) = 0. Set −c:= (p−1) min{ordp(hi)} ∈ Z. Suppose c > 0. For any ξ∈Zq[π], denote by ¯ ξits image in the residue field Fq. Using this notation, we see that 0 = (πcg) = (πch1)¯η1+· · · + (πcha)¯ηamod(π). Since {¯ηi}is a basis of Fq, we must have πchi= 0 in Fqfor every i. Hence, πchi∈πZq[π], and so hi∈π1−cZq[π] for every i. Thus, for each iwe have ordp(hi)≥1−c p−1=1 p−1+ min j=1,...,a{ordp(hj)}. However, since this is not possible we must have cnonnegative. Thus, −c≥0 which means min{ordp(hi)} ≥ 0 = ordp(g). Since we easily have ordp(g)≥min{ordp(hi)}, we must have equality, proving the lemma. Since tnei∈ M(b p,b p;−b pW(n, i)), we have by (13) that Symk(α1(t))(tnei)∈ N(b p, b;−b pW(n, i)) and so β1(tnei)∈ N(b, b;−b pW(n, i)). By Dwork decomposition (14), this means β1(tnei) = X (m,j)∈Γ C(n, i;m, j)tmejmod(∂M) with ordp(C(n, i;m, j)) ≥b p(pW(m, j)−W(n, i)). From the lemma above, if B∈Zq[π] satisfies ordp(B)≥ρ, then writing B=B1η1+· · · +Baηathe coefficients satisfy ordp(Bi)≥ρ. Thus, we may write C(n, i;m, j) = a X r=1 C(n, i;m, j)rηr with C(n, i;m, j)r∈Zp[π] and ordp(C(n, i;m, j)r)≥b p(pW(m, j)−W(n, i)). Now, a basis of H1,k over the field Qp(π) is given by Γ0:= {ηjtnei|j= 1, . . . , a, (n, i)∈A}. Thus, for ηjtnei∈Γ0, we have β1(ηjtnei) = τ−1(ηj)X (m,j)∈Γ C(n, i;m, j)tmejmod(∂M) =τ−1(ηj)X (m,j)∈Γ a X r=1 C(n, i;m, j)rηrej(15) 18 Writing τ−1(ηj)ηr= a X s=1 bs,j,rηswith bs,j,r ∈Zp[π], then (15) becomes ¯ β1(ηjtnei) = X (m,j)∈Γ a X s=1 D(j, n, i;s, m, j)ηstmej where D(j, n, i;s, m, j) := a X r=1 C(n, i;m, j)rbs,j,r. It follows that ordp(D(j, n, i;s, m, j)) ≥b p(pW(m, j)−W(n, i)). Writing detQp(π)(1 −¯ β1T|H1,k) = P∞ m=0 cmTmthen cm= (−1)mXX σ∈Sm sgn(σ) m Y l=1 D(jl, nl,i(l);jσ(l), nσ(l),i(σ(l))), where Smis the permutation group on {1, . . . , m}and the outer summation runs over all sets consisting of m distinct elements of the form (j, n, i) where ηjtnei∈Γ0. It follows that ordp(cm)≥min (m X l=1 W(nl,i(l))) where the minimum runs over all sets consisting of mdistinct elements of the form (j, n, i) where ηjtnei∈Γ0. Let rN:= #{tnei∈Γ|W(n, i) = N/d}. Then there are arNnumber of elements ηjtnei∈Γ0with weight W(n, i) = N/d. Thus, if m= R X N=0 arN then ordp(cm)≥ R X N=0 arNN d. In other words, the p-adic Newton polygon of detQp(π)(1 −¯ β1T|H1,k) lies on or above the lower convex hull of the points R X N=0 arN, R X N=0 arN N d!R= 0,1, . . . , dimQp(π)H1,k Thus, the q-adic Newton polygon of detQq(π)(1 −βT |H1,k) lies on or above the lower convex hull of the points R X N=0 rN, R X N=0 rN N d!R= 0,1, . . . , dimQq(π)H1,k. This finishes the proof of Theorem 6.1. 6.3 Relative Dwork homology Integral to the proof of Theorem 6.1 was the Dwork decomposition of relative homology given on (11). The main result of this section is to provide a proof of this result. This proof will closely follow arguments of Dwork’s [9, §7] and Adolphson-Sperber’s [2]. We begin by recalling that V(b0, b;ρ) := L(b0)x⊕ · · · ⊕ L(b0)xd−1∩ K(b0, b;ρ) V(b0, b) := [ ρ∈R V(b0, b;ρ). Theorem 6.4. Suppose (p, d)=1. Let band b0be real numbers satisfying b0≤band 1 p−1≤b≤p p−1. Set e:= b−1 p−1. Then 19 1. K(b0, b)•=V(b0, b)⊕D(t)K(b0, b), 2. K(b0, b; 0)•⊂ V(b0, b; 0) ⊕D(t)K(b0, b;e), 3. D(t)is injective if b > 1 p−1, 4. if g∈ K(b0, b)•is divisible by tnand we write g=ξ+D(t)ζwith ξ∈ V(b0, b)and ζ∈ K(b0, b), then tn divides ξand ζ. The proof of this theorem will consist of a series of lemmas which will comprise the rest of this section. Lemma 6.5. Suppose b0≤b. Then K(b0, b; 0)•⊂ V(b0, b; 0) + πˆ fxK(b0, b;e). Furthermore, if ξ∈ K(b0, b)•is divisible by tnthen when we write ξ=ζ+(πˆ fx)νwith ζ∈ V(b0, b)and ν∈ K(b0, b), then tndivides both ζand ν. Proof. Write ˆ f(t, x) = xd+Pd−1 j=0 ˆajxj+tx. Now, with πˆ fx(t, x) = π(dxd+ d−1 X j=1 jˆajxj+tx) we may write, for m≥d, xm=πdxd1 πdxm−d+π  d−1 X j=1 jˆajxj+tx 1 πdxm−d−1 πdxm−d =πˆ fx(t, x)1 πdxm−d− d−1 X j=1 j dˆajxm+j−d−1 dtxm+1−d. Notice that the last right-hand sum consists of terms in xof degree strictly smaller than xm. This is our reduction formula for xm, reducing all monomials xmto some linear combination of {x, x2, . . . , xd−1}. Next, consider Bnmtnxm∈ K(b0, b; 0) with Bnm ∈Zq[π]. The reduction formula takes the form Bnmtnxm=πˆ fx(t, x)Bnm πd tnxm−d− d−1 X j=1 jBnm dˆajtnxm+j−d−Bnm dtn+1xm+1−d.(16) Observe that, since m≥d, it is immediate that Bnm πd tnxm−d∈ K(b0, b;e) while the other terms jBnm dˆajtnxm+j−d and Bnm dtn+1xm+1−dlie in K(b0, b; 0) since b≥b0. Iterating the recursive equation (16), we obtain Bnmtnxm∈ V(b0, b; 0) + πˆ fx(t, x)K(b0, b;e).(17) Next, let ξ=Pn,m≥0Bnmtnxm∈ K(b0, b; 0)•. For each N∈Z≥0we may write ξ=ζ(N)+Pn≥0η(n,N)where η(n,N):= N X m=0 Bnmtnxmand ζ(N):= X n≥0X m≥N+1 Bnmtnxm. Observe that xN+1 |ζ(N)for every N, and tn|η(n,N). By (17), we may write η(n,N)=ν(n,N) 1+ (πˆ fx)ν(n,N) 2 with ν(n,N) 1∈V(b0, b; 0) and ν(n,N) 2∈ K(b0, b;e), both with the property that they are divisible by tn. Hence, Pn≥0ν(n,N) 1and Pn≥0ν(n,N) 2produce well-defined elements of V(b0, b; 0) and K(b0, b;e), respectively. Let us denote these elements by ν(N) 1and ν(N) 2. We have thus constructed sequences of elements {ν(N) 1}N≥1in V(b0, b; 0) and {ν(N) 2}N≥1in K(b0, b;e) which satisfy ξ=ζ(N)+ν(N) 1+ (πˆ fx)ν(N) 2 Now, in the topology of coefficient-wise convergence (i.e. the (π, t, x)-adic topology), Zq[π][[t, x]] is compact. Thus, K(b0, b;ρ) is compact for each ρin the induced topology. Hence, we may restrict ourselves to convergent subsequences of {ν(N) 1}N≥1and {ν(N) 2}N≥1with limits ν1∈V(b0, b; 0) and ν2∈ K(b0, b;e), respectively. Thus, ξ= lim N→∞ ζ(N)+ν(N) 1+ (πˆ fx)ν(N) 2=ν1+ (πˆ fx)ν2, where limN→∞ ζ(N)= 0 since xN+1 |ζ(N)for each N. This proves the lemma. 20 Lemma 6.6. Let band b0be real numbers. Then V(b0, b)∩πˆ fxK(b0, b) = {0}. Proof. Suppose η:= c1x+· · ·+cd−1xd−1∈ V(b0, b)∩πˆ fxK(b0, b). Let ζ:= P∞ j=0 Bjxj∈ K(b0, b) satisfy η=πˆ fxζ. Now, πˆ fx ∞ X j=0 Bjxj=c1x+c2x2+· · · +cd−1xd−1. Writing πˆ fx(t, x) = π(dxd+Pd−1 j=1 jˆajxj+tx), we have ∞ X j=0 πdBjxd+j+ ∞ X r=0 d−1 X j=1 πBrˆajxj+r+ ∞ X j=0 πBjtxj=1 =c1x+· · · +cd−1xd−1. For j≥0, since the coefficient of xd+jin this equation must vanish, we have πdBj+ d+j−2 X r=j+1 πBrˆad+j−r+πBd+j−1t= 0 and so Bj=−1 d d+j−2 X r=j+1 Brˆad+j−r−1 dBd+j−1t. (18) Using (18) recursively, we see that Bj→0 (p, t)-adically. Hence, Bj= 0 for all j≥0 as desired. Lemma 6.7. Let ξ∈ K(b0, b)and suppose πˆ fxξ∈ K(b0, b;ρ). Then ξ∈ K(b0, b;ρ+e). Proof. Let ξ=P∞ j=0 Bjxj∈ K(b0, b) and πˆ fxξ=P∞ j=0 Cjxj∈ K(b0, b;ρ). Writing πˆ fx(t, x) = π(dxd+ Pd−1 j=1 jˆaj(t)xj), where ˆa1(t) := t, we have ∞ X j=0 πdBjxd+j+ ∞ X r=0 d−1 X j=1 πBrˆajxj+r= ∞ X j=0 Cjxj. From this, the coefficient of xd+jsatisfies πdBj+ d+j−1 X r=j+1 πBrˆad+j−r=Cd+j for all j≥0. Rewriting this, we have Bj=1 πdCd+j−1 d d+j−1 X r=j+1 Brˆad+j−r. Iterating this n-times produces Bj=ζ(j) n+ξ(j) 1+ξ(j) 2+· · · +ξ(j) n where ζ(j) n:= ± d+j−1 X r1=j+1 d+r1−1 X r2=r1+1 · · · d+rn−1−1 X rn=rn−1+1 1 dnBrnˆad+j−r1ˆad+r1−r2· · · ˆad+rn−1−rn and ξ(j) 1:= 1 πdCd+j ξ(j) 2:= 1 d2π d+j−1 X r1=j+1 Cd+r1ˆad+j−r1 ξ(j) 3:= 1 d3π d+j−1 X r1=j+1 d+r1−1 X r2=r1+1 Cd+r2ˆad+j−r1ˆad+r1−r2 . . . ξ(j) n:= 1 dnπ d+j−1 X r1=j+1 d+r1−1 X r2=r1+1 · · · d+rn−2−1 X rn−1=rn−2+1 Cd+rn−1ˆad+j−r1ˆad+r1−r2· · · ˆad+rn−2−rn−1. 21 Since Brn∈L(b0;bw0(rn)), ζ(j) n→0 as ntends to infinity. Thus, to complete the lemma, let us show P∞ n=1 ξ(j) n∈ L(b0;bw0(j) + e). We know Cd+rn−1∈L(b0;bw0(d+rn−1) + ρ). We wish to show 1 dn−1πCd+rn−1ˆad+j−r1ˆad+r1−r2· · · ˆad+rn−2−rn−1∈L(b0;bw0(j) + e+ρ).(19) Notice that (19) typically has many ˆaterms equal to 1. These ˆawill not affect the L(b0;σ) space that (19) lies in. It is only when the ˆaequals tthat things change. The worse case is when all ˆaequal t. In this case, ri=id +j−i for i= 1,2, . . . , n −1 making (19) take the form 1 dn−1πCd+(n−1)d+j−(n−1)tn−1, which may easily be shown to lie in L(b0;bw0(j) + e+ρ). The general case is similar. This concludes the proof of the lemma. Lemma 6.8. Let b0≤band 1 p−1≤b≤p p−1. With e:= b−1 p−1we have K(b0, b; 0)•⊂ V(b0, b; 0) + D(t)K(b0, b;e). Furthermore, if ξ∈ K(b0, b)•is divisible by tn, then when we write ξ=η+D(t)ζwith η∈ V(b0, b)and ζ∈ K(b0, b), then ηand ζare also divisible by tn. Proof. Recall, D(t) = x∂ ∂x +H(t, x) with H(t, x) := P∞ j=0 πjpjˆ fτj x(tpj, xpj). Now, observe that ˆ fτj x(tpj, xpj) = ˆ fx(t, x)pj+phj(t, x) for some polynomial hjwith coefficients in Zq[π]. Write H(t, x) = πˆ fx(t, x)Q1(t, x) + K1(t, x) where Q1(t, x) := ∞ X j=0 πjπ−1pjˆ fx(t, x)pj−1 K1(t, x) := ∞ X j=1 πjpj+1hj(t, x). We claim that Q1,1 Q1, K1∈ K(p p−1,p p−1; 0). To see this, note that since ˆ f∈ K(p p−1,p p−1;−p p−1), we have ˆ fpj−1 x∈ K(p p−1,p p−1;−p p−1(pj−1)). Thus, πjπ−1pjˆ fpj−1 x∈ K(p p−1,p p−1; 0) proving the result for Q1and 1/Q1. Next, since hjconsists of terms coming from the expansion of ˆ fpj x, we see that hj∈ K(p p−1,p p−1;−p p−1pj). Thus πjpj+1hj∈ K(p p−1,p p−1; 0) proving the result for K1. We will first suppose b > 1 p−1. Let ξ∈ K(b0, b; 0)•. By Lemma 6.5, there exists η1∈ V(b0, b; 0) and ζ1∈ K(b0, b;e) such that ξ=η1+ (πˆ fx)ζ1. Thus, ξ=η1+ (H−K1)Q−1 1ζ1 =η1+ (Q−1 1K1ζ1−x∂ ∂xQ−1 1ζ1 | {z } =:ν1 ) + D(t)(Q−1 1ζ1 |{z} =:ζ0 1 ). Notice that ν1∈ K(b0, b;e)•and ζ0 1∈ K(b0, b;e). Continuing this same process, but now with ν1instead of ξ, we are lead to ξ= (η1+· · · +ηN) + νN+D(t)(ζ0 1+· · · +ζ0 N) 22 where ηi∈ V(b0, b; (i−1)e), νN∈ K(b0, b;Ne)•, and ζ0 i∈ K(b0, b;ie). Thus, η:= P∞ i=1 ηi∈ V(b0, b; 0) and ζ0:= P∞ i=1 ζ0 i∈ K(b0, b;e). Upon taking the limit in the coefficientwise convergence topology we see that ξ=η+D(t)ζ0 as desired. We now consider the case when b=1 p−1. Let ξ∈ K(b0, b; 0)•. For each N∈Z≥0, we may write ξ= N X n=1 Bnxn+X n≥N+1 Bnxn. Let  > 0. For 1 ≤n≤N, since bw0(n)≥(b+ w0(N))w0(n)−we see that N X n=1 Bnxn∈ K(b0, b + w0(N);−)•. Since b+ w0(N)>1 p−1, there exists η(,N)∈ V(b0, b + w0(N);−) and ζ(,N)∈ K(b0, b + w0(N);−+ w0(N)) such that N X n=1 Bnxn=η(,N)+D(t)ζ(,N). We have just constructed sequences {η(,N)}∞ N=1 ⊂ V(b0, b;−) and {ζ(,N)}∞ N=1 ⊂ K(b0, b;−). Since K(b0, b;ρ) and V(b0, b;ρ) are compact in the coefficientwise convergence topology for each ρ, we may restrict ourselves to convergent subsequences with limits η()∈ V(b0, b;−) and ζ()∈ K(b0, b;−) which satisfy ξ=η()+D(t)ζ(). In the coefficientwise convergence topology, letting →0+, there exists η∈ V(b0, b; 0) and ζ∈ K(b0, b; 0) such that ξ= lim →0+η()+D(t)ζ()=η+D(t)ζ. This proves the first part of the lemma. The second part follows from the divisibility result in Lemma 6.5 and running through the above argument. Lemma 6.9. Let b0≤band 1 p−1≤b≤p p−1. Then V(b0, b)∩D(t)K(b0, b). Proof. Let us first assume b > 1 p−1. Let η∈V(b0, b)∩D(t)K(b0, b) be non-zero, and let ξ∈ K(b0, b) be such that D(t)ξ=η. Find a real number csuch that ξ∈ K(b0, b;c) but ξ6∈ K(b0, b;c+e). We will prove that no such c exists, contradicting the existence of η. Since D(t) = x∂ ∂x + (πˆ fx)Q1+K1, we have η=D(t)ξ= (πˆ fx)Q1ξ+x∂ ∂xξ+K1ξ. Since x∂ ∂x ξ+K1ξ∈ K(b0, b;c)•, by Lemma 6.8 there exists ν1∈ V(b0, b;c) and ζ1∈ K(b0, b;c+e) such that x∂ ∂xξ+K1ξ=ν1+D(t)ζ1. Hence, η= (πˆ fx)Q1ξ+ν1+D(t)ζ1 = (πˆ fx)Q1(ξ+ζ1) + ν1+K1ζ1. Since K1ζ1∈ K(b0, b;c+e), applying Lemma 6.8 again produces ν2∈ V(b0, b;c+e) and ζ2∈ K(b0, b;c+ 2e) such that K1ζ1=ν2+D(t)ζ2. Thus, η= (πˆ fx)Q1(ξ+ζ1+ζ2)+(ν1+ν2) + K1ζ2. 23 Iterating this via induction, and taking the limit in the coefficient-wise convergence topology, we obtain η= (πˆ fx)Q1(ξ+ ∞ X i=1 ζi) + ∞ X i=1 νi. This means (πˆ fx)Q1(ξ+P∞ i=1 ζi)∈ V(b0, b), and so by Lemma 6.6, ξ=−P∞ i=1 ζi∈ K(b0, b;c) which is impossible. Suppose now that b=1 p−1and η∈ V(b0, b)∩D(t)K(b0, b). With this choice of b,α1:= τ−1◦ψx◦F(t, x) is a map from K(b0, b) to K(b0, pb). Therefore, since α1◦D(t) = pD(tp)◦α1, we see that α1(η)∈D(tp)K(b0, pb). Thus, the reduction of α1(η) equals zero in H1,tp(b0, pb). Now, we may also view α1as an endomorphism of K(b0, pb) and so, abusing notation, we obtain a map on homology ¯α1:H1,t(b0, pb)→ H1,tp(b0, pb). By Lemma 6.10 below, this map ¯α1is invertible. Therefore, since the reduction of α1(η) is zero in H1,tp(b0, pb), we must have the reduction of ηin H1,t(b0, pb) equal to zero as well. Hence, η∈D(t)K(b0, pb). Since D(t)K(b0, pb)⊂ K(b0, pb), we see that η∈V(b0, pb)∩D(t)K(b0, pb). However, this intersection equals {0}by the argument above since bp > 1/(p−1), proving η= 0 as desired. Lemma 6.10. Let b0≤1/(p−1). Then ¯α1:H1,t(b0, p/(p−1)) → H1,tp(b0, p/(p−1)) is an isomorphism. Proof. Since b0≤1/(p−1), it follows from definition that α1is a map from K(b0,1/(p−1)) to K(b0, p/(p−1)). Now, define a map α0 1:K(b0, p/(p−1)) → K(b0,1/(p−1)) by α0 1:= F(t, x)−1◦Φx◦τ where Φxis the map x7→ xp. Clearly, α1◦α0 1=id, the identity map on K(b0, p/(p−1)). Hence, we have K(b0, p/(p−1)) = α1α0 1K(b0, p/(p−1)) ⊂α1K(b0,1/(p−1)) ⊂ K(b0, p/(p−1)). Hence, α1maps K(b0,1/(p−1)) isomorphically onto K(b0, p/(p−1)). A similar argument shows α1maps K(b0,1/(p−1))•isomorphically onto K(b0, p/(p−1))•. By Lemma 6.8, we know K(b0,1/(p−1))•⊂ V(b0,1/(p−1)) + D(t)K(b0,1/(p−1)). Applying α1to this we obtain K(b0, p/(p−1))•=α1K(b0,1/(p−1)) ⊂α1V(b0,1/(p−1)) + D(tp)K(b0, p/(p−1)). Since V(b0, p/(p−1)) ⊂ K(b0, p/(p−1))•, we have V(b0, p/(p−1)) ⊂α1V(b0,1/(p−1)) + D(tp)K(b0, p/(p−1)). Now, it follows from the definition that V(b0, b1) = V(b0, b2) for any positive real numbers b1and b2. Thus, V(b0, p/(p−1)) ⊂α1V(b0, p/(p−1)) + D(tp)K(b0, p/(p−1)). Viewing α1as an endomorphism of K(b0, p/(p−1))•, this shows ¯α1:H1,t(b0, p/(p−1)) → H1,tp(b0, p/(p−1)) is surjective. Since both of these spaces are free L(b0)-modules of finite rank, ¯α1must also be injective. This finishes the lemma. Lemma 6.11. Suppose b > 1 p−1. If ξ∈ K(b0, b)and D(t)ξ∈ K(b0, b;ρ), then ξ∈ K(b0, b;ρ+e). Proof. Suppose ξ6= 0. Choose c∈Rsuch that ξ∈ K(b0, b;c) but ξ6∈ K(b0, b;c+e). Then (πˆ fx)Q1ξ=D(t)ξ−x∂ ∂xξ−K1ξ∈ K(b0, b; min{ρ, c}). Thus, (πˆ fx)ξ∈ K(b0, b; min{ρ, c}) which, by Lemma 6.7, implies ξ∈ K(b0, b; min{ρ, c}+e). By our choice of cthe lemma follows. Corollary 6.12. Suppose b > 1 p−1. Then D(t)is injective. Proof. Suppose there exists nonzero ξ∈ K(b0, b) such that D(t)ξ= 0. Then, by Lemma 6.11, since 0 ∈ K(b0, b;ρ) for every ρ, we have ξ∈ K(b0, b;ρ+e) for every ρ. Hence, ξmust be zero. 24 6.4 Dwork decomposition for the symmetric powers of relative homology Theorem 6.1 demonstrated one consequence of Dwork decomposition with the operator ∂. In this section, we wish to take the hypothesis of Dwork decomposition in that theorem and reduce it to a similar hypothesis which replaces the differential operator ∂with an easier operator LΦdescribed below. We expect that this similar hypothesis may be demonstrated for a large class of f(t, x). However, we also expect that it will fail just as often. New ideas will be required to handle the latter case. It is easiest to see the main obstructions to the theory if we generalize a bit. Let dbe a positive integer. We call a function w:Zs ≥0→1 dZ≥0aweight function if it satisfies the following three properties: 1. w(0) = 0, 2. w(cu) = cw(u) for every c∈Q≥0, and 3. w(u+v)≤w(u) + w(v) for all u, v ∈Zs ≥0. Let w1:Zs ≥0→1 d1Z≥0be a weight function. Let band b0be positive real numbers. For each ρ∈Rdefine L(b0;ρ) :=   X j∈Zs ≥0 Ajtj|Aj∈Zq[π], ordp(Aj)≥b0w1(j) + ρ  . This is a p-adic Banach space with norm given by kPAjtjk:= minjordp(Aj). Define L(b0) := [ ρ∈R L(b0;ρ). Let Mbe a free L(b0)-module with basis {e1, . . . , er}. We place a weight on each basis element eias follows: fix a positive integer d0and let w0:{1, . . . , r} → 1 d0Z≥0be any function. Note, w0is not a weight function since the set {1, . . . , r}is finite. Define M(b0, b;ρ) := (r X i=1 Biei|Bi∈L(b0;bw0(i) + ρ))(20) and M(b0, b) := M=[ ρ∈R M(b0, b;ρ). For any subset Uof M, we may define U(b0, b;ρ) := U∩ M(b0, b;ρ) and U(b0, b) := U∩ M(b0, b). Denote by M(k):= SymkMthe k-th symmetric power of Mover L(b0). Similar to (20) define M(k)(b0, b;ρ) :=       X i:=(i1,...,ir)∈Zr ≥0 i1+···+ir=k Biei|Bi∈L(b0;bw0(i) + ρ)       where ei:= ei1 1· · · eir rand w0(i) := w0(i1, . . . , ir) := r X j=1 ijw0(j). Let Nbe a free, finite rank L(b0)-module, and denote by N(k)the k-th symmetric power of Nover L(b0). Let Φ : M→N be an L(b0)-module morphism. Define the Leibnitz operator of Φ to be the operator LΦ:M(k)→ N(k)defined by LΦ(ei1 1· · · eir r) := r X m=1 imei1 1· · · eim−1 m· · · eir rΦ(em). Next, we mention a short technical lemma which will be useful. Lemma 6.13. Suppose Φ : M(b0, b; 0) → N (b0, b;ρ). Then LΦ:M(k)(b0, b; 0) → N (k)(b0, b;ρ). 25