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r ni where c1 = i=1 1+ j=1 bij − 1. To understand the p-adic absolute values of the exponential sums, we study p-adic slopes Q P of the reciprocal roots and poles of the associated L-function ∞ T k (1.1) L(~a, T ) = exp Sk(~a) . k ! Xk=1 When n =4, ni =2 and bij =1, our previous paper [CL20] showed that 6
L(~a, T )=(1 − T )(1 − qT ) (1 − αiT ), i=1 Y where the p-adic norms of the reciprocal roots are given by 1 if i =1. −1 q if i =2, 3. |αi|p = −2 q if i =4, 5. q−3 if i =6. In this paper, we obtain q-adic slopes for the above general class of exponential sums. Note that our main result is consistent with Katz’s complex estimate.
Theorem 1.1. Let D = lcm{bij|1 ≤ i ≤ r, 1 ≤ j ≤ ni}. (1) The associated L-function is a polynomial given by d−r (−1)n r−1 L(~a, T ) = (1 − T ) (1 − αiT ), i=1 Y r ni where d = i=1 1+ j=1 bij . (2) If p ≡ 1(modD), for each m = 0, 1, . . . , nD, the number of reciprocal zeros of nQ P L(~a, T )(−1) with q-adic slope m/D is the coefficient of xm+D in the following generating function ni 1 r (1+ j=1 b )D n−r 1 − x ij G(x)= 1 − xD P , ni D/bij i=1 j=1 1 − x Y and for any rational number m∈ / {0, 1Q, . . . , nD}, there is no reciprocal zero of L(~a, T )(−1)n with q-adic slope m/D.
Corollary 1.2. Assume bij =1 (1 ≤ i ≤ r, 1 ≤ j ≤ ni). (−1)n r (1) The polynomial L(~a, T ) has degree i=1 (1 + ni) − 1. r (2) Let d = (1 + ni). If ni is even for 1 ≤ i ≤ n, we have i=1 Q r2+3r+2 d− 2 Q 2 r (−1)n r−1 r −r L(~a, T ) = (1 − T ) (1 − γ0T )(1 − qT ) 2 (1 − γiT ) (1 − αjT ), i=1 j=1 Y Y L-FUNCTIONS OF CERTAIN EXPONENTIAL SUMS OVER FINITE FIELDS II 3
n−1 where ordq γ0 =0, ordq γi = 1(1 ≤ i ≤ r), |γi| = q 2 (0 ≤ i ≤ r), ordq αj > 1 n−1 and |αj| ≤ q 2 . n (3) For each m = 0, 1, 2,...,n, the number of reciprocal zeros of L(~a, T )(−1) with q-adic slope m is the coefficient of xm+1 in the following generating function r ni G1(x)= (1 + x + ··· + x ) , i=1 Y and for any rational number m∈ / {0, 1,...,n}, there is no reciprocal zero of L(~a, T )(−1)n with q-adic slope m. Our approach is to reduce this theorem to the L-function of toric exponential sums and i then apply the systematic results available for such toric L-functions. Let Ni = l=1 nl for 1 ≤ i ≤ r and N0 =0. Let xij = xNi−1+j and bij = bNi−1+j (1 ≤ i ≤ r, 1 ≤ j ≤ ni). ±1 ±1 P Consider a Laurent polynomial f ∈ Fq[x1 ,...,xn+1] defined by n r ai f(x1, x2,...,xn+1)= xi + xn+1 − 1 , Ni bl i=1 i=1 l=N 1+1 xl ! X X i− ∗ where ai ∈ Fq and bl ∈ Z>0. For any positive integer k,Q let
∗ Sk(f)= ψ(Trk(f)) ∗ xi∈F Xqk be the associated exponential sum. Its generating L-function is defined to be
∞ k ∗ ∗ T L (f, T ) = exp Sk(f) . k ! Xk=1 ∗ The following equation describes the relationship between Sk(~a) and Sk(f), (−1)n 1 S (~a)= + S∗(f). k qk qk k Based on the relationship, it’s easy to check that
n 1 n (1.2) L(~a, T )(−1) = L∗ (f,T/q)(−1) . 1 − T/q So it suffices to evaluate the reciprocal roots or poles of L∗(f, T ). By Adolphson and Sper- ∗ (−1)n ber’s theorem [AS89], L (f, T ) is a polynomial if p ∤ lcm{bij}. Under this assump- n tion, we compute the q-adic Newton polygon of L∗(f, T )(−1) , which indicates the q-adic slope information of the L-function. Furthermore, using Wan’s decomposition theorem, we show that the L-function has the following form
d−r ∗ (−1)n r−1 L (f, T ) = (1 − T )(1 − qT ) (1 − βiT ), i=1 Y r ni where d = i=1 1+ j=1 bij . This leads to our main theorem. This paper is organized as follows. In section 2, we review some technical methods and Q P theorems including Adolphson-Sperber’s theorems and Wan’s decomposition theorems. In section 3, we prove the main results. 4 XIN LIN AND CHAO CHEN
2. PRELIMINARIES
±1 ±1 2.1. Rationality of the generating L-function. Let f ∈ Fq x1 ,...,xn be a Laurent polynomial and its associated exponential sum is defined to be ∗ (2.1) Sk(f)= ψ(Trk(f)), ∗ xi∈F Xqk ∗ where Trk : Fqk → Fp is the trace map and ψ : Fp → C is a fixed nontrivial additive ∗ character. It’s a classical problem to give a good estimate for the valuations of Sk (f) in analytic number theory. In order to compute the absolute values of the exponential sums, ∗ we usually study the generating L-function of Sk(f) given by
∞ k ∗ ∗ T L (f, T ) = exp Sk(f) ∈ Q(ζp)[[T ]]. k ! Xk=1 By a theorem of Dwork-Bombieri-Grothendieck [Dwo62,Gro66], the generating L-function is a rational function, d1 (1 − α T ) L∗(f, T )= i=1 i , d2 Qj=1(1 − βjT ) where all the reciprocal zeros and poles areQ non-zero algebraic integers. After taking loga- rithmic derivatives, we have the formula
d2 d1 ∗ k k (2.2) Sk(f)= βj − αi , k ∈ Z≥1, j=1 i=1 X X which implies that the zeros and poles of the generating L-function contain critical infor- mation about the exponential sums. From Deligne’s theorem on Riemann hypothesis [Del80], the complex absolute values of reciprocal zeros and poles are bounded as follows
ui/2 vj /2 |αi| = q , |βj| = q ,ui ∈ Z ∩ [0, 2n], vj ∈ Z ∩ [0, 2n].
For non-archimedean absolute values, Deligne [Del80] proved that |αi|ℓ = |βj|ℓ =1 when ℓ is a prime and ℓ =6 p. Depending on Deligne’s integrality theorem, we have the following estimates for p-adic absolute values
−ri −sj |αi|p = q , |βj|p = q ,ri ∈ Q ∩ [0, n],sj ∈ Q ∩ [0, n].
The integer ui (resp. vj) is called the weight of αi (resp. βj) and the rational number ri (resp. sj) is called the slope of αi (resp. βj). In the past few decades, it has been tremendous interest in determining the weights and slopes of the generating L-functions. Without any further condition on the Laurent polynomial f or prime p, it’s even hard to determine the number of reciprocal roots and poles. Adolphson and Sperber [AS89] proved that under a suitable smoothness condition of f in n variables, the associated L-function 1 L∗(f, T )(−1)n− is a polynomial and one can determine the slopes of the reciprocal roots n−1 using Newton polygons. Adolphson and Sperber [AS89] also proved that if L∗(f, T )(−1) is a polynomial, its Newton polygon has a lower bound called Hodge polygon. The basic definitions of the Newton polygon and the Hodge polygon will be discussed in the next subsection. L-FUNCTIONS OF CERTAIN EXPONENTIAL SUMS OVER FINITE FIELDS II 5
2.2. Newton polygon and Hodge polygon. Let J Vj f(x1,...xn)= ajx j=1 X ∗ n be a Laurent polynomial with aj ∈ Fq and Vj = (v1j,...,vnj) ∈ Z (1 ≤ j ≤ J). The Newton polyhedron of f, ∆(f), is defined to be the convex closure in Rn generated by the origin and the lattice points Vj (1 ≤ j ≤ J). For δ ⊂ ∆(f), let the Laurent polynomial
δ Vj f = ajx ∈ VXj δ be the restriction of f to δ. Definition 2.1. A Laurent polynomial f is called non-degenerate if for each closed face δ of ∆(f) of arbitrary dimension which doesn’t contain the origin, the n-th partial derivatives ∂f δ ∂f δ ,..., ∂x ∂x 1 n have no common zeros with x1 ...xn =06 over the algebraic closure of Fq. ±1 ±1 Theorem 2.2 (Adolphson and Sperber [AS89]). For any non-degenerate f ∈ Fq[x1 ,...,xn ], n−1 the associated L-function L∗(f, T )(−1) is of the following form, n! Vol(∆(f)) ∗ (−1)n−1 L (f, T ) = (1 − αiT ), i=1 Y ωi/2 where |αi| = q , ωi ∈ Z ∩ [0, n] and i =1, 2,...,n! Vol(∆). Deligne’s integrality theorem implies that the p-adic absolute values of reciprocal roots −ri are given by |αi|p = q where ri ∈ Q ∩ [0, n]. For simplicity, we normalize p-adic −1 absolute value to be |q|p = q . To determine the q-adic slopes of its reciprocal roots, we n−1 compute the q-adic Newton polygon of L∗(f, T )(−1) .
n i Definition 2.3 (Newton polygon). Let L(T )= i=0 aiT ∈ 1+ T Qp[T ], where Qp is the algebraic closure of Qp. The q-adic Newton polygon of L(T ) is defined to be the lower P 2 convex closure of the set of points {(k, ordq(ak)) |k =0, 1,...,n} in R .
Here ordq denotes the standard q-adic ordinal on Qp where the valuation is normalized by assuming ordq(q)=1. The following lemma [Kob84] relates the q-adic valuation of reciprocal roots to the shape of the corresponding q-adic Newton polygon.
Lemma 2.4. In the above notation, let L(T )=(1−α1T ) ... (1−αnT ) be the factorization of L(T ) in terms of reciprocal roots αi ∈ Qp. Let λi = ordq αi. If λ is the slope of a q-adic Newton polygon with horizontal length l, then precisely l of the λi are equal to λ. Assume f is a non-degenerate Laurent polynomial in n variables and consequently 1 1 L∗(f, T )(−1)n− is a polynomial. The q-adic Newton polygon of L∗(f, T )(−1)n− is de- noted by NP(f), which is hard to compute directly. Adolphson and Sperber proved that NP(f) has a topological lower bound called the Hodge polygon, that is easier to calculate. So generally, we first compute its lower bound Hodge polygon and then determine when the Newton polygon coincides with its lower bound. Let ∆ be an n-dimensional integral polytope containing the origin in Rn. Define C(∆) to be the cone generated by ∆ in Rn. For any point u ∈ Rn, the weight function w(u) is 6 XIN LIN AND CHAO CHEN the smallest non-negative real number c such that u ∈ c∆. Let w(u)= ∞ if such c doesn’t exist. Assume δ is a co-dimension 1 face of ∆ not containing the origin. Let D(δ) be the least common multiple of the denominators of the coefficients in the implicit equation of δ, normalized to have constant term 1. We define the denominator of ∆ to be the least common multiple of all such D(δ) given by:
D = D(∆) = lcmδD(δ) where δ runs over all the co-dimension 1 faces of ∆ that don’t contain the origin. It’s easy to check 1 w(Zn) ⊆ Z ∪{+∞}. D(∆) ≥0 For a non-negative integer k, let k (2.3) W (k) = # u ∈ Zn|w(u)= ∆ D be the number of lattice points in Zn with weight k/D. Definition 2.5 (Hodge number). Let ∆ be an n-dimensional integral polytope containing the origin in Rn. For a non-negative integer k, the k-th Hodge number of ∆ is defined to be n n (2.4) H (k)= (−1)i W (k − iD). ∆ i ∆ i=0 X It’s easy to check that H∆(k)=0, if k > nD.
Adolphson and Sperber [AS89] proved that H∆(k) coincides with the usual Hodge number in the toric hypersurface case that D = 1. Based on the Hodge numbers, we define the Hodge polygon of a given polyhedron ∆ ∈ Rn as follows. Definition 2.6 (Hodge polygon). The Hodge polygon HP(∆) of ∆ is the lower convex polygon in R2 with vertices (0,0) and k 1 k Q = H (m), mH (m) , k =0, 1, . . . , nD, k ∆ D ∆ m=0 m=0 ! X X where H∆(k) is the k-th Hodge number of ∆, k =0, 1, . . . , nD. That is, HP(∆) is a polygon startingfromorigin (0,0) with a slope k/D side of horizontal length H∆(k) for k =0, 1, . . . , nD. The vertex Qk is called a break point if H∆(k +1) =06 where k =1, 2, . . . , nD − 1.
Here the horizontal length H∆(k) represents the number of lattice points of weight k/D in a certain fundamental domain corresponding to a basis of the p-adic cohomology space used to compute the L-function. Adolphson and Sperber constructed the Hodge polygon and proved that it’s a lower bound of the corresponding Newton polygon. Theorem 2.7 (Adolphson and Sperber [AS89]). For every prime p and non-degenerate Laurent polynomial f with ∆(f) = ∆ ⊂ Rn, we have NP(f) ≥ HP(∆), n−1 where NP(f) is the q-adic Newton polygon of L∗(f, T )(−1) . Furthermore, the endpoints of NP(f) and NP(∆) coincide. Definition 2.8. A Laurent polynomial f is called ordinary if NP(f) = HP(∆). L-FUNCTIONS OF CERTAIN EXPONENTIAL SUMS OVER FINITE FIELDS II 7
Apparently, the ordinary property of a Laurent polynomial depends on its Newton poly- hedron ∆. In order to study the ordinary property, we will apply Wan’s decomposition theorem [Wan93], decomposing the polyhedron ∆ into small pieces that are much easier to deal with.
2.3. Wan’s decomposition theorems.
2.3.1. Facial decomposition theorem. In this paper, we use facial decomposition theorem to cut the polyhedron into small simplices. For each simplex, we can apply some criteria to determine the non-degenerate and ordinary property. Theorem 2.9 (Facial decomposition theorem [Wan93]). Let f be a non-degenerate Laurent polynomial over Fq. Assume ∆ = ∆(f) is n-dimensional and δ1,...,δh are all the co- dimension 1 faces of ∆ which don’t contain the origin. Let f δi denote the restriction of f δi to δi. Then f is ordinary if and only if f is ordinary for 1 ≤ i ≤ h. In order to describe the boundary decomposition, we first express the L-function in terms of the Fredholm determinant of an infinite Frobenius matrix.
2.3.2. Dwork’s trace formula. Let p be a prime and q = pa for some positive integer a. Let Qp denote the field of p-adic numbers and Ω be the completion of Qp. Pick a fixed primitive p-th root of unity in Ω denoted by ζp. In Qp(ζp), choose a fixed element π satisfying
∞ m πp 1 =0 and ord π = . pm p p − 1 m=0 X By Krasner’s lemma, it’s easy to check Qp(π) = Qp(ζp). Let K be the unramified exten- sion of Qp of degree a. Let Ωa be the compositum of Qp(ζp) and K.
Ωa
K Qp(π)
a p − 1 Qp
Define the Frobenius automorphism τ ∈ Gal(Ωa/Qp(π)) by lifting the Frobenius automor- p phism x 7→ x of Gal(Fq/Fp) to a generator τ of Gal(K/Qp) and extending it to Ωa with p τ(π)= π. For the primitive (q − 1)-th root of unity ζq−1 in Ωa, we have τ(ζq−1)= ζq−1. Let Ep(t) be the Artin-Hasse exponential series,
∞ m ∞ tp E (t) = exp = λ tm ∈ Z [[x]]. p pm m p m=0 ! m=0 X X In Dwork’s terminology, a splitting function θ(t) is defined to be ∞ m m θ(t)= Ep(πt)= λmπ t . m=0 X When t =1, θ(1) can be identified with ζp in Ω. 8 XIN LIN AND CHAO CHEN
±1 ±1 Consider a Laurent polynomial f ∈ Fq[x1 ,...,xn ] given by J Vj f = a¯jx , j=1 X n ∗ q where Vj ∈ Z and a¯j ∈ Fq. Let aj be the Teichmüller lifting of a¯j in Ω satisfying aj = aj. Let J Vj r F (f, x)= θ(ajx )= Fr(f)x ∈ Ωa[[x]]. j=1 ∈ n Y rXZ The coefficients of F (f, x) are given by
J uj u1+···+uJ n Fr(f)= ( λuj aj )π , r ∈ Z , u j=1 X Y where the sum is over all the solutions of the following linear system
J
ujVj = r with uj ∈ Z≥0, j=1 X and λm is m-th coefficient of the Artin-Hasse exponential series Ep(t). Assume ∆ = ∆(f). Let L(∆) = Zn ∩ C(∆) be the set of lattice points in the closed cone generated by origin and ∆. For a given point r ∈ Rn, define the weight function to be J J
w(r) := inf uj| ujVj = r, uj ∈ R≥0 . ~u ( j=1 j=1 ) X X In Dwork’s terminology, the infinite semilinear Frobenius matrix A1(f) is a matrix whose rows and columns are indexed by the lattice points in L(∆) with respect to the weights w(r)−w(s) A1(f)=(ar,s(f))=(Fps−r(f)π ), w(r) where r, s ∈ L(∆). Based on the fact that ordp Fr(f) ≥ p−1 , we have the following estimate w(ps − r)+ w(r) − w(s) ord (a (f)) ≥ ≥ w(s). p r,s p − 1 D p−1 Let ξ be an element in Ω satisfying ξ = π . Then A1(f) can be written in a block form, i A00 ξA01 ··· ξ A0i ··· i A10 ξA11 ··· ξ A1i ··· . . .. . A1(f)= . . . . , i Ai0 ξAi1 ··· ξ Aii ··· . . . . . . .. . where the block Aii is a p-adic integral W∆(i) × W∆(i) matrix and W∆(i) = #{u ∈ n i Z |w(u)= D }. The infinite linear Frobenius matrix Aa(f) is defined to be τ τ a−1 Aa(f)= A1(f)A1(f) ··· A1 (f).
The q-adic Newton polygon of det(I − T Aa(f)) has a natural lower bound which can be identified with the chain level version of the Hodge polygon. L-FUNCTIONS OF CERTAIN EXPONENTIAL SUMS OVER FINITE FIELDS II 9
Definition 2.10. Let P (∆) be the polygon in R2 with vertices (0, 0) and k 1 k P = W (m), mW (m) , k =0, 1, 2,... k ∆ D ∆ m=0 m=0 ! X X Proposition 2.11 ( [Wan04]). The q-adic Newton polygon of det(I − T Aa(f)) lies above P (∆). By Dwork’s trace formula, we can identify the associated L-function with a product of some powers of the Fredholm determinant [Wan04], n ∗ (−1)n−1 i (−1)i(n) (2.5) L (f, T ) = det(I − T q Aa(f)) i . i=0 Y Equivalently, we have,
∞ n+i−1 ∗ i (−1)n−1 ( i ) (2.6) det(I − T Aa(f)) = L (f, q T ) . i=0 Y Proposition 2.12 ( [Wan04]). Notations as above. Assume f is non-degenerate with ∆ = ∆(f). Then NP(f) = HP(∆) if and only if the q-adic Newton polygon of det(I −T Aa(f)) coincides with its lower bound P (∆).
±1 ±1 2.3.3. Boundary decomposition. Let f ∈ Fq[x1 ,...,xn ] with ∆ = ∆(f), where ∆ is an n-dimensional integral convex polyhedron in Rn containing the origin. Let C(∆) be the cone generated by ∆ in Rn. Definition 2.13. The boundary decomposition B(∆) = { the interior of a closed face in C(∆) containing the origin} is the unique interior decomposition of C(∆) into a disjoint union of relatively open cones. If the origin is a vertex of ∆, then it is the unique 0-dimensional open cone in B(∆). Recall that A1(f)=(ar,s(f)) is the infinite semilinear Frobenius matrix whose rows and columns are indexed by the lattice points in L(∆). For Σ ∈ B(∆), we define A1(Σ, f) to Σ be the submatrix of A1(f) with r, s ∈ Σ. Let f be the restriction of f to the closure of Σ. Σ Σ Then A1(Σ, f ) denotes the submatrix of A1(f ) with r, s ∈ Σ. Let B(∆) = {Σ0,..., Σh} such that dim(Σi) ≤ dim(Σi+1), i = 0,...,h − 1. Define Bij = (ar,s(f)) with r ∈ Σi and s ∈ Σj (0 ≤ i, j ≤ h). After permutation, the infinite semilinear Frobenius matrix can be written as
B00 B01 ··· B0h B10 B11 ··· B1h (2.7) A1(f)= . . . . , . . .. . B B ··· B h0 h1 hh h where Bij =0 for i > j. Then det(I − T A1(f)) = i=0 det(I − T Bii) and we have the boundary decomposition theorem. Q ±1 ±1 Theorem 2.14 (Boundary decomposition [Wan93]). Let f ∈ Fq[x1 ,...,xn ] with ∆ = ∆(f). Then we have the following factorization
Σ det(I − T A1(f)) = det I − T A1(Σ, f ) . Σ∈YB(∆) 10 XIN LIN AND CHAO CHEN
±1 ±1 Corollary 2.15. Let f ∈ Fq[x1 ,...,xn ] be a non-degenerate Laurent polynomial with an n-dimensional Newton polyhedron ∆. If the origin is a vertex of ∆, then the associated L-function n! Vol(∆(f))−1 ∗ (−1)n−1 L (f, T ) = (1 − ψ(Tr(c))T ) (1 − αiT ), i=1 Y n/2 where c is the constant term of f, Tr : Fq → Fp is the trace map and |αi| ≤ q .
2.4. Diagonal local theory. In this subsection, we give some non-degenerate and ordinary criteria for the diagonal Laurent polynomials whose Newton polyhedrons are simplices.
±1 ±1 Definition 2.16. A Laurent polynomial f ∈ Fq[x1 ,...,xn ] is called diagonal if f has exactly n non-constant terms and ∆(f) is an n-dimensional simplex in Rn.
Let f be a Laurent polynomial over Fq, n Vj f(x1, x2,...xn)= ajx , j=1 X ∗ n where aj ∈ Fq and Vj =(v1j,...,vnj) ∈ Z , j =1, 2, . . . , n. Let ∆= ∆(f). Then the vertex matrix of ∆ is defined to be
M(∆) = (V1,...,Vn), where the i-th column is the i-th exponent of f. If f is diagonal, M(∆) is invertible.
±1 ±1 Proposition 2.17. Suppose f ∈ Fq[x1 ,...,xn ] is diagonal with ∆ = ∆(f). Then f is non-degenerate if and only if p is relatively prime to det(M(∆)). Let S(∆) be the solution set of the following linear system
r1 r2 M(∆) . ≡ 0(mod1), ri ∈ Q ∩ [0, 1). . r n It’s easy to prove that S(∆) is an abelian group and its order is given by (2.8) |det M(∆)| = n! Vol(∆). By the fundamental structure theorem of finite abelian group, we decompose S(∆) into a direct product of invariant factors, n
S(∆) = Z/diZ, i=0 M where di|di+1 for i = 1, 2,...,n − 1. By the Stickelberger theorem for Gauss sums, we have the following ordinary criterion for a non-degenerate Laurent polynomial [Wan04].
±1 ±1 Proposition 2.18. Suppose f ∈ Fq[x1 ,...,xn ] is a non-degenerate diagonal Laurent polynomial with ∆ = ∆(f). Let dn be the largest invariant factor of S(∆). If p ≡ 1(moddn), then f is ordinary at p. L-FUNCTIONS OF CERTAIN EXPONENTIAL SUMS OVER FINITE FIELDS II 11
3. PROOF OF THE MAIN THEOREM r We prove the main theorem in this section. Let n = i=1 ni be a partition of a positive integer n. Recall that for bij ∈ Z>0 and p ∤ bij (1 ≤ i ≤ r, 1 ≤ j ≤ ni), the exponential P sum Sk(~a) has the expression,
r ni
Sk(~a)= ψ Trk xij . a r i =1 i=1 j=1 !! i=1 n bij Xi x X X P Qj=1 ij ∗ ∗ ∗ where xij ∈ Fqk , ai ∈ Fq, ψ : Fp → C is a nontrivial additive character and Trk : Fqk → Fp is the trace map. The L-function associated to Sk(~a) is defined by ∞ T k L(~a, T ) = exp Sk(~a) . k ! Xk=1 i For the sake of brevity, let Ni = l=1 nl for 1 ≤ i ≤ r and N0 =0. Denote xij = xNi−1+j ±1 ±1 k and bij = bNi−1+j (1 ≤ i ≤ r, 1 ≤ j ≤ ni). Let f ∈ Fq [x1 ,...,xn+1] be the Laurent polynomial defined by P n r ai f(x1, ··· , xn+1)= xi + xn+1 − 1 . Ni bl i=1 i=1 l=Ni−1+1 xl ! X X We have Q 1 (−1)n 1 (3.1) Sk(~a)= k ψ (Trk (f)) = k + k ψ (Trk (f)) . q ∗ q q ∗ x1,··· ,xn∈F xi∈F X qk Xqk xn+1∈Fqk Define ∞ k ∗ ∗ ∗ T Sk(f)= ψ(Trk(f)) and L (f, T ) = exp Sk(f) . ∗ k xi∈F k=1 ! Xqk X It follows from formula (3.1) that
n 1 n (3.2) L(~a, T )(−1) = L∗(f,T/q)(−1) . 1 − T/q The aim of this paper is to determine the slopes of L(~a, T )(−1)n . By formula (3.2), it suf- n n fices to consider the slopes of L∗(f, T )(−1) instead. We will prove it later that L(~a, T )(−1) is a polynomial under a restriction on p. Let ∆ = ∆(f) denote the Newton polyhedron corresponding to the Laurent polynomial f. Vertices of ∆ are as follows.
V0 = (0, ··· , 0)
V1 = (1, 0, ··· , 0), . .
Vn+1 = (0, ··· , 0, 1),
Vn+2 =(−b1, ··· , −bn1 , 0, ··· , 0, 1),
n1 . . | {z } 12 XIN LIN AND CHAO CHEN
Vn+i+1 = (0, ··· , 0, −bNi−1+1, ··· , −bNi , 0, ··· , 0, 1),
ni . . | {z }
Vn+r+1 = (0, ··· , 0, −bNr−1+1, ··· , −bNr , 1).
nr Claim that and has vertices. Let be a co-dimension 1 face dim ∆ = n +1 ∆ |n + r +2{z } δi
V5
V7 V6
V1 V4 V2 V3
V0
FIGURE 1. ∆ for n =4, r =2 of ∆ not containing the origin. Here is a criterion for δi. n+1 Proposition 3.1. Let hi(x) = j=1 ejxj − 1 be the equation of δi, where ej is uniquely determined rational numbers not all zero. For any vertex V of ∆, one has h (V ) ≤ 0. P i This deduces the following proposition.
Proposition 3.2. Every co-dimension 1 face δi contains Vn+1 as a vertex.
Then we can describe some detailed properties for δi.
Proposition 3.3. Let ∆i be the polytope generated by the origin and δi. Denote V = {V1,V2, ··· ,Vn}. Every ∆i has n +2 vertices:
V0,Vn+1, V−{Vi1 , ··· ,Vik } and {Vn+1+j1 ,...,Vn+1+jk }, n−k vertices k vertices where 0 ≤ k ≤ r and N +1 ≤ i ≤ N , i.e., there is a one-to-one correspondence |jl−1 {z l } jl | {z } between vertices Vil and Vn+1+jl for 1 ≤ l ≤ k. Note that for l1 =6 l2, we have jl1 =6 jl2 .
Proposition 3.4. The equation of δi is given by n+1
cixi =1, i=1 X where 1 if i =6 il, Nj c = 1 l i − b − b if i = i . b m il l il m=N +1 Xjl−1 L-FUNCTIONS OF CERTAIN EXPONENTIAL SUMS OVER FINITE FIELDS II 13
Note that bil is the il-th coordinate of Vn+1+jl . r Proposition 3.5. There are i=1 (1 + ni) co-dimension 1 faces of ∆ not containing the origin, i.e., Q r
#{δi | δi face of ∆, 0 ∈/ δi} = (1 + ni) . i=1 Y Proof. Recall the selection of vertices in proposition 3.3. For any fixed k0 (1 ≤ k0 ≤ r), we have face of #{δi | δi ∆, 0 ∈/ δi,k = k0} = ni1 ··· nik0 . i1=6 i2=6 ···6=i X k0
Note that ii, ··· , ik lie in k distinct intervals [Njl−1 +1, Njl ]. Then we have r r
#{δi | δi face of ∆, 0 ∈/ δi} =1+ ni + ninj + ... + n1n2 ··· nr = (1 + ni) . i=1 6 i=1 X Xi=j Y Another way to prove proposition 3.5 leads to the following result in combinatorics. Corollary 3.6. Let n and r be positive integers. Consider a partition
n = n1 + n2 + ··· + nr of n as the sum of r integers ni, we have n + r r n + r − n − 1 r − i = (1 + n ) . n n − n − 1 i i=1 i i=1 X Y Proof. To count the number of δi, it suffices to consider the combination of n +1 linearly independent vertices. Note that all the linearly dependent combinations of Vj(1 ≤ j ≤ n + r + 1) contain vertices
VNi−1+1,VNi−1+2, ··· ,VNi ,Vn+1,Vn+i+1,
ni Since Vn+1 is a fixed vertex| of all δi,{z we have } n + r r n + r − n − 1 #{δ | δ face of ∆, 0 ∈/ δ } = − i . i i i n n − n − 1 i=1 i X As a consequence of proposition 3.4, we have the following result.
Proposition 3.7. (i). The denominator D = lcm{bij | 1 ≤ i ≤ r, 1 ≤ j ≤ ni}. (ii). f is non-degenerate. 1 r ni (iii). Vol(∆) = 1+ b . (n + 1)! ij i=1 j=1 ! Y X Proof. The denominator can be deduced immediately from the equation of δi. For each δi, the restriction of f to δi is defined by
δi Vj f = ajx . ∈ VXj δi 14 XIN LIN AND CHAO CHEN
Note that each f δi is diagonal. According to proposition 2.17, we get the condition when f δi is non-degenerate. Using Wan’s facial decomposition theorem, we obtain (ii). r Now we prove (iii). For 1 ≤ k ≤ i=1 (1 + ni), let ∆k be the polytope generated by δk and the origin. It can be deduced from the facial decomposition of ∆ [Wan93] that Q r i=1(1+ni) Q Vol(∆) = Vol(∆k). Xk=1 By proposition 3.4 and formula (2.8), we obtain Vol(∆) in this proposition. Proposition 3.8. Let D be the denominator of ∆. The polynomial f is ordinary if p ≡ 1(modD). Proof. This proposition follows from proposition 2.18 and Wan’s facial decomposition theorem. Now we are ready to consider the Hodge number of ∆.
Theorem 3.9. Let bij ∈ Z>0 for 1 ≤ i ≤ r and 1 ≤ j ≤ ni. The number of H∆(k) equals the coefficient of xk in G(x) as follow.
n 1+ i 1 D r j=1 bij n−r 1 − x (3.3) G(x)= 1 − xD P . ni D/bij i=1 j=1 1 − x Y Proof. By formula (2.3) and (2.4), to obtain theQ number of H∆(k), it suffices to consider the weight function w(u), where u is a lattice point in Zn+1. Recall that n+r+1 n+r+1
(3.4) w(u) = inf cl clVl = u,cl ∈ R≥0 . ~c ( l=1 l=1 ) X X By the relationship that b = b , non-negative integral constant b (1 ≤ i ≤ r, 1 ≤ ij Ni−1+j ij j ≤ ni) can be relabeled as bk (1 ≤ k ≤ n). For our example, it can be deduced that Z≥0 c ∈ for 1 ≤ l ≤ n and c ∈ Z≥ for n +1 ≤ l ≤ n + r. Since w(u) is determined by l bl l 0 n+r+1 its associated coefficients, we need to discuss when l=1 cl reaches its minimum. For 1 ≤ i ≤ r, divide the first n coordinates of u into r disjoint parts in order, where the P i-th part has ni elements.
u =(u1, ··· ,un1 , ··· ,uNi−1+1, ··· ,uNi , ··· ,uNr−1+1, ··· ,un,un+1)
n1 ni nr
Let Mi = {VNi−|1+1,{z··· ,V}Ni }∪{|Vn+1+i{z} and Si}= {cN|i−1+1, ···{z ,cNi }∪{} cn+1+i}. Ele- ments in Si are coefficients of vertices in Mi. Note that |Si| = ni +1 and Si1 ∩ Si2 = ∅ for 1 ≤ i1 ≤ i2 ≤ r. For our example, formula (3.4) implies that each part of coordinates of u, i.e., {uNi−1+1, ··· ,uNi }, are uniquely determined by the vertices in Mi and thus the coefficients in Si. n+r+1 We claim that l=1 cl reaches its minimum if and only if at least one of elements in Si is 0 for all 1 ≤ i ≤ r. First we prove the necessary condition. If we focus on ni P coordinates of u, we consider the corresponding ni coordinates of each vertex in Mi. For a vertex V =(v1, v2, ··· , vn+1), let
(ni) V =(vNi−1+1, ··· , vNi )
ni | {z } L-FUNCTIONS OF CERTAIN EXPONENTIAL SUMS OVER FINITE FIELDS II 15 denote the i-th part of coordinates of V . Let
(ni) (ni) (ni) (ni) Mi = {VNi−1+1, ··· ,VNi }∪{Vn+1+i}.
(ni) ni (ni) If we take V as a vector in Z , Mi is linearly dependent, while any proper subset of M (ni)is linearly independent. Therefore, if at least one of elements in S vanishes, c i i cl∈Si l is unique and thus minimum. The sum cl reaches minimum for all 1 ≤ i ≤ r cl∈Si P implies n+r+1 c reaches minimum. The sufficient condition holds as well and can be l=1 l P proved by contradiction. Suppose none of the elements in Si equal 0 for some 1 ≤ i ≤ r. P n+r+1 Let c = l=1 cl. By the selection of vertices in proposition 3.3, u doesn’t lie on any co-dimension 1 face δ of c∆, which implies that n+r+1 c can be reduced. P l=1 l It follows that W∆(k) equals the number of solutions to P n+r+1
cl = k/D, Xl=1 satisfying that c = 0 for 1 ≤ i ≤ r. That is to say, W (k) equals the number of cl∈Si l ∆ non-negative integer solutions to Q D D (3.5) x1 + ··· + xn + Dxn+1 + ··· + Dxn+r+1 = k, b1 bn satisfying that xNi−1+1 ··· xNi xn+1+i =0 for 1 ≤ i ≤ r. Now we determine the generating function for W∆(k) with the help of formula (3.5).
r Ni D/bj D 1 j=Ni−1+1 x x g(x)= n r+1 − n r+1 D/bk D D/bk D k=1 (1 − x )(1 − x ) i=1 k=1Q(1 − x ) (1 − x ) N NX Q r i1 xD/bj xD i2 Q xD/bj xD j=Ni1−1+1 j=Ni2−1+1 + n r+1 −··· (1 − xD/bk )(1 − xD) i1,i2=1 Q k=1 Q iX1 n 1+ i 1 D r j=1 bij D n−r 1 − x P = 1 − x n . i 1 − xD/bij i=1 j=1 Y k The number H∆(k) equals the coefficients of x Qin G( x). 16 XIN LIN AND CHAO CHEN As a corollary, we obtain the number of H∆(k) when bij =1. Corollary 3.10. If bij =1 for 1 ≤ i ≤ r and 1 ≤ j ≤ ni, the number of H∆(k) equals the k coefficients of x in G1(x) as follow. r ni (3.6) G1(x)= (1 + x + ··· + x ) . i=1 Y In this case, we have H∆(0) = 1, H∆(1) = r and H∆(k)= H∆(n − k) for 0 ≤ k ≤ n. Then we obtain the slopes of αi. (−1)n Theorem 3.11. Let hk be the number of reciprocal roots of L(~a, T ) with q-adic slope k/D, i.e. hk = #{αi| ordq αi = k/D}. For bij ∈ Z>0 with 1 ≤ i ≤ r and 1 ≤ j ≤ ni, assume p ≡ 1(modD). The number k+D of hk equals the coefficients of x in G(x) for k = 0, 1, ··· , nD and hk = 0 for any rational number k∈{ / 0, 1, ··· , nD}. In particular, for bij = 1 with 1 ≤ i ≤ r and 1 ≤ j ≤ ni, the number of hk equals the k+1 coefficients of x in G1(x) for k < n, and hk =0 for k ≥ n. Proof. When p ≡ 1( mod D), Laurent polynomial f is ordinary. In this case, hk = H∆(k+ D) by lemma 2.4 and the fact that ordq(αi) = ordq(βi) − 1. Combining corollary 3.10 and formula (2.4), we obtain hk in this theorem. Using Wan’s boundary theorem, we can factor L-function as follow. Theorem 3.12. We have d−r (−1)n r−1 L(~a, T ) = (1 − T ) (1 − αiT ), i=1 Y d−r ∗ (−1)n r−1 L (f, T ) = (1 − T )(1 − qT ) (1 − βiT ), i=1 Y r ni where d = i=1 1+ j=1 bij and αi = βi/q. i i Proof. DenoteQ D(q T )P = det (I − q T Aa(g)). Recall the boundary decomposition B(∆) defined in definition 2.13. Let N(i) denote the number of i-dimensional face Σi,j of C(∆), where 0 ≤ i ≤ dim∆ and 1 ≤ j ≤ N(i). For Newton polyhedron ∆ = ∆(f), we have N(0) = 1 and N(1) = n + r. For simplicity, we abbreviate Σi,j to Σi since the following proof with respect to face Σi,j is independent of the choice of j. Note that Σi is an open cone and Σi ∈ B(∆). Let Σi be the closure of Σi. Denote ′ Σi Σi Di(T ) = det I − T Aa(Σi, f ) and Di(T ) = det I − T Aa(Σi, f ) . ′ It is obvious that the unique 0-dimensional cone Σ0 is the origin and D0(T ) = D0(T ) = 1 − T . When i =1, each f Σ1 can be normalized to x by variable substitution. Explicitly, ∞ T k L∗(f Σ1 , T ) = exp − =1 − T. k ! Xk=1 L-FUNCTIONS OF CERTAIN EXPONENTIAL SUMS OVER FINITE FIELDS II 17 By formula (2.6), we have ∞ i (i) ∗ Σ1 i 2 D1(T )= L f , q T = (1 − T ) (1 − qT ) 1 − q T ··· . i=0 Y ′ ′ Since the only boundary of Σ1 is Σ0, we get D1(T ) after eliminating D0(T ), i.e., ∞ D (T ) D′ (T )= 1 = (1 − qT ) 1 − q2T 1 − qiT , 1 D′ (T ) 0 i=3 Y Using Wan’s boundary decomposition theorem, n+1 N(i) ′ n+r D(T )= Di(T )=(1 − T )(1 − qT ) ··· , i=1 j=1 Y Y By theorem 2.2, L∗(f, T )(−1)n is a polynomial. From formula (2.5), we obtain n+1 d−r 2 ( 2 ) ∗ (−1)n D(T )D(q T ) ··· r−1 L (f, T ) = n+1 = (1 − T )(1 − qT ) (1 − βiT ), n+1 3 ( 3 ) D(qT ) D(q T ) ··· i=1 Y r ni where d = i=1 1+ j=1 bij . Together with formula (3.2), we get the factorization for (−1)n L(~a, T ) Q. P Corollary 3.13. Assume bi =1 and ni is even for 1 ≤ i ≤ n. We have r2+r d− 2 2 (−1)n r−1 r −r L(~a, T ) = (1 − T ) (1 − qT ) 2 (1 − αiT ), i=1 Y r2+r d− 2 2 ∗ (−1)n r−1 2 r −r L (f, T ) = (1 − T )(1 − qT ) (1 − q T ) 2 (1 − βiT ), i=1 Y r 2 where d = i=1 (1 + ni) and αi = βi/q. Note that ordq βi ≥ 1 and βi =6 q or q for 2 1 ≤ i ≤ d − r +r . Q2 Proof. Proof of this corollary follows from the proof of theorem 3.12. Similarly, when ∗ Σ2 1 i =2, we have L (f , T )= 1−T . Since the boundary of Σ2 consists of the origin and two sides, we obtain ∞ ′ 2 i i−1 D2(T )= 1 − q T 1 − q T . i=3 Y By proposition 2.12, we have ∞ (3.7) D(T )= (1 − γiT ) , i=1 Y where #{γi | ordqγi = k} = W∆(k) for k ∈ Z≥0. Note that W∆(1) = n + r +1. Using Wan’s boundary decomposition theorem and Dwork’s trace formula, we have n+r n+r 2 ( 2 )+n+r D(T )=(1 − T )(1 − qT ) (1 − q T ) (1 − γ1T )h1(T ), 2 n r −r 1 − γ1T ∗ (−1) r−1 2 2 L (f, T ) = (1 − T )(1 − qT ) (1 − q T ) n+1 h2(T ), (1 − γ1qT ) −1 where |γ1|q = q and the slopes of reciprocal roots of h2(T ) are greater than 1. 18 XIN LIN AND CHAO CHEN 2 Now we claim that γ1 is non-real and (1 − q T ) ∤ h1(T ). Let βi be a reciprocal root of ∗ (−1)n L (f, T ) . Under the assumption that bi = 1 and ni is even for 1 ≤ i ≤ n, f is an ∗ (−1)n odd function. Then the conjugate βi is also a reciprocal root of L (f, T ) . Theorem 2.2 implies that if ordq βi > (n + 1)/2, then βi is non-real and (3.8) ordq βi + ordq βi = ωi ∈ Z ∩ [0, n + 1]. By theorem 3.10, the Hodge number H∆(0) = H∆(n)=1, H∆(1) = H∆(n − 1) = r r2−r and H∆(2) = r + 2 since ni is even. Restricted by formula (3.8), γ1 is non-real and 2 (1 − q T ) ∤ h1(T ). 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MR 3171761 DEPARTMENT OF MATHEMATICS, SHANGHAI MARITIME UNIVERSITY, SHANGHAI 201306, PR CHINA Email address: [email protected] DEPARTMENT OF MATHEMATICS,UNIVERSITY OF CALIFORNIA,IRVINE,IRVINE, CA 92697-3875, USA Email address: [email protected]