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HAL Id: hal-00440358

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Preprint submitted on 10 Dec 2009

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Isotypic Decomposition of the Cohomology and Factorization of the Zeta Functions of Dwork

Hypersurfaces

Philippe Goutet

To cite this version:

Philippe Goutet. Isotypic Decomposition of the Cohomology and Factorization of the Zeta Functions

of Dwork Hypersurfaces. 2009. �hal-00440358�

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Isotypic Decomposition of the Cohomology and Factorization of the Zeta Functions of Dwork Hypersurfaces

Philippe Goutet December 10, 2009

Abstract

The aim of this article is to illustrate, on the Dwork hypersurfacesxn1+· · ·+xnn−nψx1. . . xn= 0 (withn an integer ≥3 andψ ∈Fq a parameter satisfyingψn 6= 1), how the study of the representation of a finite group of automorphisms of a hypersurface in its etale cohomology allows to factor its zeta function.

1 Introduction

Letnbe an integer≥3 andFq a finite field of characteristicp6= 2 not dividing n; to simplify the results, we will assume that q ≡1 modn. We consider the projective hypersurface Xψ ⊂ PFqn−1 given by

xn1 +· · ·+xnn−nψx1. . . xn= 0,

whereψ is a non zero parameter belonging toFq. The zeta function ofXψ is defined as ZXψ/Fq(t) = exp

+∞

X

r=1

#Xψ(Fqr)tr r

.

We assume thatψn 6= 1, so thatXψ=XψFqFqis nonsingular. AsXψis a non-singular hypersurface ofPn−1, we know that the dimension of the etaleℓ-adic cohomology spacesHeti (Xψ,Q) is zero for i >2n−4 ori <0 and that, for 0≤i≤2n−4,

dimHeti(Xψ,Q) =

i ifi6=n−2, δi+(n−1)n+(−1)n n(n−1) ifi=n−2,

where δi = 0 ifiis odd and δi = 1 ifi is even (see §2.2). As we will recall in Remark 2.3 page 3, the zeta function ofXψ is related to how the Frobenius acts onHetn−2(Xψ,Q).

We set

A={(ζ1, . . . , ζn)∈µn(Fq)n1. . . ζn= 1}/{(ζ, . . . , ζ)}; Aˆ={(a1, . . . , an)∈(Z/nZ)n |a1+· · ·+an= 0}/{(a, . . . , a)},

and denote by [ζ1, . . . , ζn] the class of (ζ1, . . . , ζn) in A and [a1, . . . , an] that of (a1, . . . , an) in ˆA.

We will identify the group ˆA with the group of characters ofA taking values inFq. The groupA

2000 Mathematical Subject Classification:Primary 14G10; Secondary 11G25, 14G15, 20C05.

Keywords: Zeta function factorisation, Dwork hypersurfaces, isotypic decomposition.

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acts onXψ by coordinatewise multiplication; the symmetric groupSn acts on the right on Xψ by permutation of the coordinates

[x1:. . .:xn]σ= [xσ(1):. . .:xσ(n)], and on the left onAand ˆA by

σ1, . . . , ζn] = [ζσ1(1), . . . , ζσ1(n)];

σ[a1, . . . , an] = [aσ1(1), . . . , aσ1(n)].

The semidirect productG=A⋊Sn acts on the right onXψ, and hence on the left onHetn−2(Xψ, Q) as the functorg7→gis contravariant.

The aim of this article is to describe the structure ofHetn−2(Xψ,Q) as aQ[G]-module in order to deduce a factorization of the zeta function ofXψ. More precisely, we will show that the primitive part ofHetn−2(Xψ,Q) (as defined in§2.2) admits an isotypic decomposition

M

a,ω

Wa,ωDaVa,ω,

where a describes (Sn ×(Z/nZ)×)\A,ˆ ω belongs to a certain set of roots of unity (see Corol- lary 5.12 page 19),Wa,ωis a simpleQ[G]-module which is independent ofℓ,Da is the division ring EndQ[G](Wa,ω)opp, andVa,ωis a free module overDaQQwhose rank is independent ofℓ. Because the Frobenius stabilizes these isotypic spaces, its characteristic polynomial splits in as many factors (the idea to use this method is inspired by an argument given in [Hulek et al., 2006,§6.2]).

The first step is to decompose the Q[G]-module Hetn−2(Xψ,Q); we follow the same method Br¨unjes used for the case ψ = 0 (Fermat hypersurface), but, thanks to a more powerful trace formula, we avoid the tedious induction of [Br¨unjes, 2004, Proposition 11.5]. Our methods can be generalized to other families of hypersurfaces, allow us to obtain factorizations slightly finer than those of Kloosterman [2007] (who uses thep-adic Monsky-Washnitzer cohomology), and also allow us to express each factor as the norm of a polynomial with coefficients in a certain finite extension ofQ, hence explaining a numerical observation of Candelas, de la Ossa and Rodriguez-Villegas in the case n= 5 where this extension is Q(√

5) (see [Candelas et al., 2003, Table 12.1 page 133]1).

Let us also mention that, in a recent article, Katz [2009] studies the action ofA(but not ofA⋊Sn) on the cohomology ofXψand establishes a motivic link betweenXψ and objects of hypergeometric type.

The article is organized as follows. After preliminaries (§2), we describe the structure ofHetn−2(Xψ, Q) as aQ[A]-module (§3) and then as aQ[G]-module (§4). We then deduce the structure of the Q[G]-moduleHetn−2(Xψ,Q) (§5) and explain the link between this structure and the existence of a factorisation of the zeta function ofXψ (§6). An index of all notations introduced in the article is given in§A and a table of the main formulas appears in§B.

2 Preliminaries

We begin by recalling a Lefschetz-type trace formula by Deligne and Lusztig which allows to express the alternating sum of the traces of an automorphism on theℓ-adic cohomology spaces as the Euler–

Poincar´e characteristic of the fixed-point scheme of this automorphism. We then recall the value of this Euler–Poincar´e characteristic in the cases we will encounter in what follows (smooth projective hypersurfaces). Finally, we link the trace of an element ofGto the Euler–Poincar´e characteristic of a subscheme of fixed points.

1They make this observation only in the case ψ = 0, but their numerical data in §13.3 suggests the same phenomenon happens whenψ6= 0 andq1 mod 5.

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2.1 Lefschetz trace formula

Let us recall that the Euler–Poincar´e characteristic of a proper scheme overFp is given by χ(X) =

2 dimX

X

i=0

(−1)idimHeti (X,Q), whereℓ is a prime number6=p. It is an integer independent ofℓ.

Theorem 2.1. Let X be a proper scheme over Fp. If f is an automorphism ofX of finite order prime top, and if Xf denotes the fixed-point subscheme of f of the schemeX, then

2 dimX

X

i=0

(−1)itr(f|Heti (X,Q)) =χ(Xf).

Proof. See [Deligne and Lusztig, 1976, Theorem 3.2, page 119].

2.2 Euler–Poincar´ e characteristic of a non-singular hypersurface

In this§2.2, exceptionally, we do not assume thatn≥3.

Theorem 2.2(Hirzebruch formula). Letnbe an integer≥1andf ∈Fp[x1, . . . , xn]a homogeneous polynomial of degree d such that f, ∂x∂f

1, . . . , ∂x∂fn have no common zero in Fpn except (0, . . . ,0).

Then the hypersurface X ⊂PFpn−1 defined by f = 0 is non-singular (and irreducible if n≥3) and its Euler–Poincar´e characteristic is

χ(X) = (n−1) +(1−d)n+ (d−1)

d .

Proof. Ifn≥3, we use Corollary 7.5.(iii) of [SGA5, expos´e VII]: indeed, the subschemeX ofPFn−1 is smooth, connected and of dimensionn−2; its Euler–Poincar´e characteristic is hence p

χ(X) =d

n−2

X

i=0

(−1)n−i n

i

dn−2−i= 1 d

n−2

X

i=0

(−1)n−i n

i

dn−i

=(1−d)n+nd−1

d ,

which is the announced formula. If n= 2, the hypersurfaceX of PFp1 consists of d distinct points and soχ(X) =d, which shows the result as (2−1) +1d[(1−d)2+ (d−1)] =d. Finally, ifn= 1, X=∅and soχ(X) = 0, which also shows the result in this case.

Remark 2.3. Whenn≥3, Theorem 2.2 can be refined as follows. We keep the same notations and denote byj the canonical injectionX →PFn−1

p . By the Weak Lefschetz Theorem, (see for example [Freitag and Kiehl, 1988, Corollary 9.4, page 106]), fori < n−2 (respectivelyi=n−2), the linear mapj:Heti(PFn−1p ,Q)→Heti (X,Q) is bijective (respectively injective). If we set δi = 0 if i odd andδi= 1 ifi is even, we thus have dimHeti(X,Q) =δi fori < n−2, and this result stays valid forn−2< i≤2(n−2) by Poincar´e duality. Fori=n−2, the image of the mapj:Hetn−2(PFpn−1, Q)→ Hetn−2(X,Q) has dimension δi. We will denote it by Hetn−2(X,Q)inprim and set Hetn−2(X, Q)prim = Hetn−2(X,Q)/Hetn−2(X,Q)inprim. Because the Frobenius acts as the multiplication by q(n−2)/2 onHetn−2(Xψ,Q)inprimand by multiplication by qi on eachHet2i(Xψ,Q), we have

ZXψ/Fq(t) =det(1−tFrob|Hetn−2(Xψ,Q)prim)(−1)n−1 (1−t)(1−qt). . .(1−qn−2t) .

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2.3 Character of G acting on H

etn−2

(X

ψ

, Q

)

prim

The isomorphism class of aQ[G]-module is completely determined by its character. In this§2.3, we will express in terms of Euler–Poincar´e characteristics the values of the character of theQ[G]- moduleHetn−2(Xψ,Q)primfor the elementsg∈Gwhich are of order prime to p.

Lemma 2.4. Each g ∈G acts as the identity on Hetn−2(Xψ,Q)inprim and on Heti(Xψ,Q) when i6=n−2.

Proof. As g is the restriction of an automorphism of PFn−1

q , it results from Remark 2.3 and the following lemma.

Lemma 2.5. If his an automorphism of PFqn−1, thenh acts as the identity on Heti(PFqn−1,Q) for alli.

Proof. The groupP GLn(Fq) acts on the right onHeti (PFn−1q ,Q) byu7→u; asHeti(PFqn−1,Q) is of dimension 0 or 1, this action is by homothety, and thus factors by an abelian quotient ofP GLn(Fq).

Since Fq is algebraically closed, P GLn(Fq) is equal to its commutator subgroup and thus has no nonzero abelian quotient. Hence, for allu∈P GLn(Fq),u= Id.

Theorem 2.6. If g∈Gis of order prime top, then tr(g|Hetn−2(Xψ,Q)prim) = (−1)n−1

(n−1)−χ(Xgψ)

. (2.1)

Proof. Using the trace formula of Theorem 2.1, we can write

2 dimX

X

i=0

(−1)itr(g|Heti(Xψ,Q)) =χ(Xgψ).

By Lemma 2.4, we have (with, as previously,δi= 0 ifiis odd andδi= 1 ifiis even) tr(g|Heti(Xψ,Q)) =

i ifi6=n−2, δi+ tr(g|Heti(Xψ,Q)prim) ifi=n−2, and thus

χ(Xgψ) = (n−1) + (−1)n−2tr(g|Hetn−2(Xψ,Q)prim), which is exactly the announced formula.

3 Action of A on H

etn2

(X

ψ

, Q

)

prim

The irreducible representations overQof the finite abelian groupAare its characters (of degree 1).

Finding the structure of theQ[A]-moduleHetn−2(Xψ,Q) hence amounts to figuring out the mul- tiplicity of each character ofA in the representation g7→g of A in Hetn−2(Xψ,Q); it is the aim of this §3. The choice, in §3.1, of an isomorphism between µn(Fq) and µn(Q) allows to identify Aˆ to the group of characters of A taking values in Q. After determining the character of the Q[A]-module Hetn−2(Xψ,Q)prim in §3.2, we will prove in §3.3 that the multiplicity of a ∈ Aˆ is ma= #(Z/nZ\ {a1, . . . , an}).

3.1 Characters of A with values in Q

As we only consider the caseq≡1 mod n, the groupµn(Fq) consisting of thenth roots of unity of Fq is isomorphic to the group ofnth roots of unity ofQ. We calltan isomorphism of µn(Fq) onto µn(Q) and use it to identify the group ˆA with the group of characters of A taking values in Q thanks to the isomorphism [a1, . . . , an]7→([ζ1, . . . , ζn]7→t(ζ1)a1· · ·t(ζn)an).

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3.2 Character values of the Q

[A]-module H

etn−2

(X

ψ

, Q

)

prim

As pis prime to n by assumption, the elements ofA have an order prime to p; we may thus use Formula (2.1) to obtain the values taken by the characters of theQ[A]-moduleHetn−2(Xψ,Q)prim. Theorem 3.1. Consider1, . . . , ζn)∈µn(Fq)n such thatζ1. . . ζn= 1and let g be the correspond- ing element1, . . . , ζn]of A. For allζ∈µn(Fq), denote by k(ζ) the number ofi∈[[1;n]]such that ζi=ζ. We have

tr(g|Hetn−2(Xψ,Q)prim) = (−1)n n

X

ζ∈µn(Fq)

(1−n)k(ζ). (3.1)

Proof. A point ofXψ with homogeneous coordinates [x1:. . .:xn] is a fixed point ofgif and only if (x1, . . . , xn) is proportional to (ζ1x1, . . . , ζnxn). The proportionality coefficient is necessarily a root of unityζ∈µn(Fq), and we must havexi = 0 ifζi 6=ζ. Hence, the subscheme of fixed points of g ofXψ is the disjoint union overζ∈µn(Fq) of the subvarieties

Yζ ={x∈Xψ|xi= 0 ifζi 6=ζ}.

Ifk(ζ) =n, we haveYζ =Xψ. If 2≤k(ζ)≤n−1,Yζ is isomorphic to the hypersurface ofPk(ζ)−1 defined byy1n+yn2 +· · ·+yk(ζ)n = 0. Finally, ifk(ζ) = 0 or 1,Yζ is empty. In each of these cases, we can apply Theorem 2.2 and obtain

χ(Yζ) =k(ζ)−1 +(1−n)k(ζ)+n−1

n =k(ζ)−1

n+(1−n)k(ζ)

n .

Consequently, sinceP

ζ∈µn(Fq)k(ζ) =nandP

ζ∈µn(Fq)1 n = 1, χ(Xgψ) = X

ζ∈µn(Fq)

χ(Yζ) =n−1 + X

ζ∈µn(Fq)

(1−n)k(ζ)

n .

Using trace formula (2.1) page 4, we deduce the announced result.

Remark 3.2. A recent preprint proves, in a more general setting, formulas of the type given in Theorem 3.1 and Theorem 4.12 page 10; see [Chˆenevert, 2009, Corollary 2.5].

3.3 Decomposition in irreducible representations

The following theorem gives a simple expression for the multiplicityma of a charactera∈Aˆin the Q[A]-moduleHetn−2(Xψ,Q)prim.

Theorem 3.3. The multiplicity of the irreducible character a = [a1, . . . , an] of A in the Q[A]- moduleHetn−2(Xψ,Q)primis

ma = #(Z/nZ\ {a1, . . . , an}) =n−(number of distinctai).

Proof. Consider (ζ1, . . . , ζn) ∈ µn(Fq)n such that ζ1. . . ζn = 1 and let g be the corresponding element [ζ1, . . . , ζn] ofA. From the definition, we have

tr(g|Hetn−2(Xψ,Q)prim) =X

a∈Aˆ

maζ1a1. . . ζnan

= 1 n

X

(a1,...,an)∈(Z/nZ)n a1+···+an=0

maζ1a1. . . ζnan.

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We will show that if we replacema by the number of elements ofZ/nZ\ {a1, . . . , an} in the right hand side, we recover Formula (3.1) above, which will show the announced result. We write

1 n

X

(a1,...,an)∈(Z/nZ)n a1+···+an=0

#(Z/nZ\ {a1, . . . , an})ζ1a1. . . ζnan

= 1 n

X

(a1,...,an)∈(Z/nZ)n a1+···+an=0

X

k∈Z/nZ

∀i,ai6=k

1

ζ1a1. . . ζnan

= 1 n

X

k∈Z/nZ

X

(a1,...,an)∈(Z/nZ)n a1+···+an=0

∀i,ai6=k

ζ1a1. . . ζnan

= 1 n

X

k∈Z/nZ

X

(a1,...,an)∈(Z/nZ)n a1+···+an=0

∀i,ai6=0

ζ1a1. . . ζnan

= X

(a1,...,an)∈(Z/nZ)n a1+···+an=0

∀i,ai6=0

ζ1a1. . . ζnan

= X

a1,...,an∈(Z/nZ)\{0}

a1+···+an=0

ζ1a1. . . ζnan.

We now conclude by using the following lemma.

Lemma 3.4. Let r be an integer≥1 and ζ1, . . . , ζr elements of µn(Fq). If k(ζ) =k1,...,ζr)(ζ) denotes the number ofi∈[[1;r]]such thatζi=ζ, then

X

a1,...,ar∈(Z/nZ)\{0}

a1+···+ar=0

ζ1a1. . . ζrar =(−1)r n

X

ζ∈µn(Fq)

(1−n)k(ζ).

Proof. We proceed by induction onr. Forr= 1, the equality is the relation 0 =−1

n

(1−n)1+ (n−1)(1−n)0 .

We now assume thatr≥2 and that the result is known forr−1. We write X

a1,...,ar∈(Z/nZ)\{0}

a1+···+ar=0

ζ1a1. . . ζrar = X

a1,...,ar−1∈(Z/nZ)\{0}

a1+···+ar−16=0

ζ1a1. . . ζr−1ar−1ζr−a1−···−ar−1

= X

a1,...,ar−1∈(Z/nZ)\{0}

ζ1

ζr

a1

. . . ζr−1

ζr

ar−1

− X

a1,...,ar−1∈(Z/nZ)\{0}

a1+···+ar−1=0

ζ1a1. . . ζr−1ar−1.

Givenζ∈µn(Fq), we have

X

a∈(Z/nZ)\{0}

ζa =

(−1 ifζ6= 1, n−1 ifζ= 1.

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This little remark allows to compute the first sum:

X

a1,...,ar−1∈(Z/nZ)\{0}

ζ1

ζr

a1

. . . ζr−1

ζr

ar−1

= (−1)r−k(ζr)(n−1)k(ζr)−1,

wherek(ζ) =k1,...,ζr)(ζ). To compute the second sum, we use the induction assumption:

X

a1,...,ar−1∈(Z/nZ)\{0}

a1+···+ar−1=0

ζ1a1. . . ζr−1ar−1 =(−1)r−1 n

X

ζ6=ζr

(1−n)k(ζ)+ (1−n)k(ζr)−1

.

We conclude by noting that

(−1)r−k(ζr)(n−1)k(ζr)−1−(−1)r−1

n (1−n)k(ζr)−1

=−(−1)rn(1−n)k(ζr)−1

n +(−1)r

n (1−n)k(ζr)−1= (−1)r

n (1−n)k(ζr).

Remark 3.5. As a consequence of Theorem 3.3, the multiplicity ma of the character a ∈ Aˆ is nonzero unlessabelongs to the orbit of [0,1,2, . . . , n−1] underSn (which imposesnodd, or else 1 + 2 +· · ·+ (n−1) is not divisible byn).

4 Action of G on H

etn2

(X

ψ

, Q

)

prim

4.1 A decomposition of the Q

[G]-module H

etn−2

(X

ψ

, Q

)

prim

For everyabelonging to ˆAidentified to the group of characters ofAtaking values inQ, we denote byHathe isotypic component relatively toaof theQ[A]-moduleHetn−2(Xψ,Q)prim(see [Bourbaki, 1958,§3.4]). It is aQ-vector space of dimensionma, wherema is the multiplicity computed in§3.3, and we have

Hetn−2(Xψ,Q)prim=M

a∈Aˆ

Ha.

The groupGacts on the left onAby inner automorphisms, and thus acts on the left on ˆA: if g∈Aσ, withσ∈Sn, and ifa= [a1, . . . , an], we havega=σa= [aσ1(1), . . . , aσ1(n)].

Considera∈A. Denote byˆ haithe orbit ofaunderSn. The stabilizerGa ofain Gis equal to A⋊Sa, whereSa ={σ∈Sn | σa=a}. We have gHa =Hga for all g ∈G and the space Ha is stable byGa. The subspaceL

a∈haiHa ofHetn−2(Xψ,Q)primis stable byG; it is aQ[G]-module canonically isomorphic to IndGGaHa. We thus deduce the following result.

Theorem 4.1. Denote by R⊂Aˆ a set of representatives ofSn\A. Theˆ Q[G]-module Hetn−2(Xψ, Q)prim is isomorphic to

M

a∈R

IndGGaHa.

The aim of the rest of this§4 is to determine how the groupSa acts onHa. The strategy is the following: after showing thatSais a semi-direct productSa⋊Σa(§4.2), we compute tr(σ|Hetn−2(Xψ, Q)prim) for σa generator ofSa and compare it to the trace of the identity (§4.4) to deduce that Sa acts asǫ(σ) IdHaonHa (see§4.5). We then show, using a method similar to§3, that Σa acts as a multiple of the regular representation (§4.6–4.8).

The approach we use to study the action ofSa is the same that Br¨unjes used in [Br¨unjes, 2004, Proposition 11.5, page 197] for the case ψ = 0, the only difference being that our trace formula allows us to avoid a tedious proof by induction.

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4.2 Structure of S

a

Considera= [a1, . . . , an]∈A, where (aˆ 1, . . . , an) is an element of (Z/nZ)nsuch thata1+· · ·+an= 0.

The set ofj∈Z/nZsuch that (a1+j, . . . , an+j) is a permutation of (a1, . . . , an) is a subgroup of Z/nZ; it can be written asnaZ/nZfor some integerna≥1 dividingn; letda=n/na be the order of this group. These two integers only depend onaand not on the choice ofa1, . . . ,an.

Remark 4.2. For all b ∈Z/nZ, denote by I(b) the set ofi∈ {1, . . . , n} such thatai =b. The set naZ/nZis the set of j∈Z/nZsuch that I(b+j) has the same number of elements asI(b) for all b∈Z/nZ.

Lemma 4.3. There is a permutation σ∈Sn such that a) if 1≤i≤n, we have aσ(i)=ai+na ;

b) σ is the product ofna disjoint cycles of lengthda.

Proof. Let us note that the condition 4.3.ais equivalent to the fact thatσ(I(b)) =I(b+na). For all b∈Z/nZsuch that I(b)6=∅, choose a numbering i1(b), . . . ,i#I(b)(b) of the elements ofI(b) and denote byσthe element ofSn which sendsil(b) toil(b+na) for allb∈Z/nZand 1≤l≤#I(b).

From the definition, we have aσ(i) =ai+na and, inspecting the orbits of each of theai under b7→b+na, we see thatσis a product ofna disjoint cycles of lengthda.

Denote bySa the fixator of (a1, . . . , an)∈(Z/nZ)n in Sn; it is a group which can be identified with Q

b∈Z/nZSI(b) (it is hence generated by transpositions) and we set γa = [Sn :Sa]. Consider σ∈Sn satisfying the conditions of the preceding lemma and let Σa =hσibe the cyclic subgroup of orderda ofSn generated byσ.

Proposition 4.4. The fixatorSa ofa= [a1, . . . , an]∈Aˆ can be written as the semi-direct product Sa=Sa ⋊Σa.

Proof. If s∈Sa, there exists a uniquej ∈naZ/nZsuch that s(a1, . . . , an) = (a1+j, . . . , an+j).

This element only depends on a, not on the choice ofa1, . . . ,an; we denote it byja(s). The map ja:Sa →naZ/nZ thus defined is a group homomorphism. This homomorphism is surjective and its kernel is the fixatorSa of (a1, . . . , an)∈(Z/nZ)n in Sn.

Moreover, asaσ(i)=ai+na and thusaσ1(i)=ai−na, we haveja(σ) =−na by construction, henceja induces an isomorphism of Σa=hσionto the imagenaZ/nZofja, which shows that

Sa=Sa ⋊Σa.

Remarks 4.5. a) In particular, the groupSa is a normal subgroup ofSa and the quotient group Sa/Sa is isomorphic tonaZ/nZand hence of orderda.

b) Let us insist on the fact that na, da, Sa and ja only depend on a and not on the choice of the representative (a1, . . . , an) ∈ Z/nZ. The group Σa also only depends on a, but its construction is not canonical as it depends on an arbitrary choice of numbering.

c) Let us also note that ifk∈(Z/nZ)×, thendka=da,nka=na,Ska =Sa andSka=Sa, but jka=kja.

4.3 Character values on a transposition τ

Theorem 4.6. For any transposition τ∈Sn, we have tr(τ|Hetn−2(Xψ,Q)prim) = (−1)n

(1−n)n−1+ (n−1)

n −δn

, (4.1)

where, as previously,δn= 0if nis odd and δn = 1if nis even.

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Proof. We may assume that τ = (1,2). We look for the fixed points of τ, i.e. the set of points [x1:. . .:xn] such that [x1:x2:x3:. . .:xn] = [x2:x1:x3:. . .:xn] andxn1+· · ·+xnn−nψx1. . . xn = 0.

For such a point, we havex21=x22, so that we are in one of the following two cases.

a) We have x1 = x2 and 2xn2 +xn3 +· · ·+xnn−nψx22x3. . . xn = 0. The hypersurface of Pn−2 defined by this equation is smooth becauseψn 6= 1 and its Euler–Poincar´e characteristic is (n−2) +n1[(1−n)n−1+ (n−1)] (Theorem 2.2).

b) We have x1 =−x2 6= 0, in which casex3 =· · ·=xn = 0 and xn1 +xn2 = 0. This can only happen if nis odd and [x1:. . .:xn] = [1 :−1 : 0 :. . .: 0].

The Euler–Poincar´e characteristic of the fixed-point subvariety ofτ ofXψ is thus χ(Xψτ) = (n−2) +(1−n)n−1+ (n−1)

n + 1−δn

= (n−1) +(1−n)n−1+ (n−1)

n −δn,

and consequently, asτ is of order 2 andFq is of characteristic6= 2, Theorem 2.6 applies:

tr(τ|Hetn−2(Xψ,Q)prim) = (−1)n−1

(n−1)−χ(Xτψ)

= (−1)n

(1−n)n−1+ (n−1)

n −δn

.

4.4 Sum of the dimensions of the spaces H

a

for a ∈ A ˆ

τ

Proposition 4.7. Letτ ∈Sn be a transposition. Denote byτ the set of elements offixed byτ.

We have

X

a∈Aˆτ

ma= (−1)n−1

(1−n)n−1+ (n−1)

n −δn

,

where, as previously,δn= 0if nis odd and δn = 1if nis even.

Proof. We may assume thatτ= (1,2). Denote byBthe set of elements (b1, . . . , bn)∈(Z/nZ\{0})n such thatb1=b2andb1+· · ·+bn= 0. The map (b1, . . . , bn)7→[b1, . . . , bn] fromBto ˆAτ is surjective and each elementa∈Aˆτ has exactlyma elements in its preimage. We thus haveP

a∈Aˆτma= #B and conclude thanks to the following lemma.

Lemma 4.8. Letrbe an integer≥2. The number ofr-uples(b1, . . . , br)belonging to(Z/nZ\{0})r such thatb1=b2 andb1+· · ·+br= 0 is

(−1)r−1

(1−n)r−1+ (n−1)

n −δn

.

Proof. Denote byurthe number we want to compute. We haveu2nandur+ur+1is the number of (r+ 1)-uples (b1, . . . , br, br+1)∈(Z/nZ\ {0})r×Z/nZsuch thatb1=b2andb1+· · ·+br+1= 0, that is,ur+ur+1= (n−1)r−1. We deduce the announced result by induction onr.

4.5 Action of S

a

on H

a

We start with a general result on automorphisms of finite order with trace equal to the dimension of the space.

Lemma 4.9. Let k be a field of characteristic zero, V a vector space of finite dimension over k anduan automorphism ofV of finite order. If tru= dimV, thenu= IdV.

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Proof. LetM be the matrix ofuin a certain basis ofV overk. The subfield k ofk generated by the coefficients ofM embeds itself inC; we can thus restrict ourselves to the casek=C.

Let λ1, . . . ,λm (where m = dimV) be the (complex) eigenvalues of M, each repeated with multiplicity. They are all roots of unity. As we have, according to the assumptions of the lemma,

1+· · ·+λm|=|tru|=m=|λ1|+· · ·+|λm|,

theλi’s are positively proportional, hence equal. As their sum is m, they are all equal to 1. The endomorphismuofV is thus unipotent; as it is of finite order, it is equal to IdV.

Remark 4.10. Let k be a field having characteristic zero, and (Vi)i∈I a finite sequence of vector space of finite dimensions overk. For each i∈I, let ui be an automorphism of Vi of finite order.

IfP

i∈Itrui is equal toP

i∈IdimVi (respectively to −P

i∈IdimVi), then ui = IdVi (respectively ui = −IdVi) for all i ∈ I. This results from Lemma 4.9 applied to the automorphism uof V = L

i∈IVi which is equal toui (respectively to−ui) overVi for alli∈I.

Letτ ∈Snbe a transposition. AsHetn−2(Xψ,Q)prim=L

a∈AˆHa and asτsendsHa intoHτa, we have

tr(τ|Hetn−2(Xψ,Q)prim) = X

a∈Aˆτ

tr(τ|Ha).

By Theorem 4.6 and Proposition 4.7, we also have

tr(τ|Hetn−2(Xψ,Q)prim) =− X

a∈Aˆτ

dimHa.

We thus deduce from Remark 4.10 that, for eacha∈Aˆτ, τ acts onHa by−IdHa.

Theorem 4.11. Considera∈Aˆandσ∈Sa. If we denote byǫ(σ)the signature of σ, we have σ|Ha=ǫ(σ) IdHa.

Proof. The subgroupSaofSn is generated by the transpositionsτ satisfyingτa=a(see§4.2) and we have just seen thatτ|Ha=−IdHa =ǫ(τ) IdHa.

4.6 Character values on Aσ where σ is a product of n

disjoint cycles of length d

Letn anddbe integers≥1 such thatnd=nand letσ∈Sn be a product ofn disjoint cycles of lengthd. Letζ1, . . . ,ζn be elements of µn(Fq) such thatζ1. . . ζn= 1 and denote byg the element [ζ1, . . . , ζn]σ of G = A⋊Sn. Let O1, . . . , On be the n orbits of σ in {1, . . . , n} and, for each ζ ∈µn(Fq), denote byk(ζ) the number ofj ∈ {1, . . . , n} such that Q

i∈Ojζi =ζ. The following theorem generalizes Theorem 3.1 (which is recovered by takingd= 1 and n=ni.e.σ= Id).

Theorem 4.12. Under the preceding assumptions, tr(g|Hetn−2(Xψ,Q)prim) = (−1)n

n

X

ζ∈µn(Fq)

(1−n)k(ζ).

Proof. We may assume that σ is the product of ((j−1)d+ 1, . . . , jd) for 1 ≤ j ≤ n and that Oj ={(j−1)d+ 1, . . . , jd}. The fixed points of g inXψ(Fq) are the points [x1:. . .:xn] ofXψ(Fq) such that

σ1(1)xσ1(1):. . .:ζσ1(n)xσ1(n)] = [x1:. . .:xn] i.e.

1x1:. . .:ζnxn] = [xσ(1):. . .:xσ(n)].

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The subscheme Xgψ of these fixed points is thus the disjoint union, over λ ∈ Fq, of the closed subschemesYλ ofXψ defined by

(Yλ)

(xn1+· · ·+xnn−nψx1. . . xn= 0, xσ(i)=λζixi for 1≤i≤n.

Let j ∈ {1, . . . , n}. If Q

i∈Ojζi 6= λ−d, the second relation shows that xi = 0 for alli ∈ Oj. If Q

i∈Ojζi−d, we haveλ∈µnd(Fq) and the second relation shows that X

i∈Oj

xni =xnjd d

X

i=1

n)i

=

(dxnjd ifλ∈µn(Fq), 0 ifλ /∈µn(Fq).

Considerλ∈Fq and letζ=λ−d (asn=nd, we haveζn = 1 ⇐⇒ λn = 1). Denote byJ the set ofj ∈ {1, . . . , n} such thatQ

i∈Ojζi =ζ and let yj =xjd for each j ∈J. Ifζ /∈µn(Fq), J is empty and henceYλ is empty. Assume now that ζ∈µn(Fq). The number of elements ofJ is k(ζ).

We consider two cases.

a) First case:ζ∈µn(Fq). According to what we have just done, the schemeYλis isomorphic to the hypersurface ofPFk(ζ)−1

q defined by d

X

j∈J

ynj

= 0 ifJ 6={1, . . . , n},

d(y1n+· · ·+ynn)−nψy1d. . . ynd = 0 ifJ ={1, . . . , n},

whereψ is the product ofψby an element ofµn(Fq). This hypersurface is smooth (because, in the second case, we have (ψ)nn 6= 1 and thus (ψ)n 6= 1), hence, by Theorem 2.2 page 3, we have

χ(Yλ) =k(ζ)−1 + (1−n)k(ζ)+ (n−1)

n =k(ζ) +(1−n)k(ζ)−1

n .

b) Second case: ζ ∈ µn(Fq)\µn(Fq). This time, the schemeYλ is isomorphic toPFk(ζ)−1q if J 6={1, . . . , n}and to the hypersurface ofPFqn−1defined by (y1. . . yn)d= 0 ifJ={1, . . . , n}. In the first case, we haveχ(Yλ) =k(ζ). In the second case, we necessarily haven ≥2 and the Euler–Poincar´e characteristic ofYλ is equal to that of Yλred, which is the union inPFqn−1 of the hyperplanes defined byyj = 0, hence

χ(Yλ) = X

L⊂{1,...,n} L6=∅

(−1)#L−1(n−#L) =

n

X

l=1

(−1)l−1 n

l

(n−l)

=n

n−1

X

l=1

(−1)l−1 n−1

l

=n(1−(1 + (−1))n−1) =n=k(ζ).

For eachζ∈µn(Fq), there exists exactlydvalues ofλsuch thatλ−d=ζ. Thus χ(Xgψ) = X

λ∈Fq

χ(Yλ) =d X

ζ∈µn(Fq)

k(ζ) +d X

ζ∈µn(Fq)

(1−n)k(ζ)−1 n

=dn+ X

ζ∈µn(Fq)

(1−n)k(ζ)−1

n =n−1 + X

ζ∈µn(Fq)

(1−n)k(ζ) n .

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The order ofg dividesndand hence is prime toq; thus, by Theorem 2.6, tr(g|Hetn−2(Xψ,Q)prim) = (−1)n−1

(n−1)−χ(Xgψ)

= (−1)n n

X

ζ∈µn(Fq)

(1−n)k(ζ).

4.7 Trace of a product σ of n

disjoint cycles of length d acting on H

a

when a ∈ A ˆ

σ

We keep the notations of§4.6.

Lemma 4.13. If σ∈Sn is a product of n disjoint cycles of length d, X

a∈Aˆsuch thatσ∈Sa¯

a(ζ1, . . . , ζn)ma= (−1)n n

X

ζ∈µn(Fq)

(1−n)k(ζ).

Proof. Denote by B the set of (b1, . . . , bn) ∈ ((Z/nZ)\ {0})n such that b1+· · ·+bn = 0 and

σ(b1, . . . , bn) = (b1, . . . , bn). The image of the map B→A, (bˆ 1, . . . , bn)7→[b1, . . . , bn] is the set of a∈Aˆ such thatσ∈Sa; such an elementa has exactlyma elements in its preimage. The sum we must compute can hence be rewritten as

X

(b1,...,bn)∈B

ζ1b1. . . ζnbn.

If (b1, . . . , bn) ∈ B, all the bi, for i belonging to an orbit Oj of σ, are equal to a common cj ∈(Z/nZ)\ {0} and we haved(c1+. . . cn) = 0 inZ/nZi.e.c1+· · ·+cn ∈nZ/nZ. Our sum can thus be rewritten as

X

c1,...,cn∈(Z/nZ)\{0}

c1+···+cn∈nZ/nZ

µc11. . . µcnn,

where µj = Q

i∈Ojζi. We conclude by using the following generalization of Lemma 3.4 (which is recovered by takingd= 1 andn=ni.e.σ= Id).

Lemma 4.14. Let r be an integer≥1 andµ1, . . . ,µr elements of µn(Fq). For each ζ∈µn(Fq), we denote byk(ζ)the number of j∈ {1, . . . , r} such that µj=ζ. We have

X

c1,...,cr∈(Z/nZ)\{0}

c1+···+cr∈nZ/nZ

µc11. . . µcrr =(−1)r n

X

ζ∈µn(Fq)

(1−n)k(ζ).

Proof. We prove the result by induction onr. Forr= 1, we have X

c1∈nZ/nZ\{0}

µc11=

(d−1 = −1n((1−n)1+ (n−1)(1−n)0) ifµ1∈µn(Fq),

−1 = −1n(n(1−n)0) ifµ1∈/ µn(Fq),

hence the result in that case. Assume now thatr≥2 and that the result is proved forr−1. We

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