Hodge Groups of certain Superelliptic Jacobians II

0.15 MB·English

by  Jiangwei Xue

HODGE GROUPS OF CERTAIN SUPERELLIPTIC JACOBIANS
II
1
1 JIANGWEIXUE
0
2

n
a
1. introduction
J

0 Throughout this paper C is the field of complex numbers, K ⊆ C is a subfield
1 of C, f(x) ∈ K[x] a polynomial without multiple roots and of degree n ≥ 4. Let

p ∈ N be a prime that does not divide n and q = pr ∈ N an integral power of
]
G p. We write C for the superelliptic K-curve yq = f(x), and J(C ) for the
f,q f,q
A Jacobian of Cf,q. By definition, Cf,q is the smooth projective model of the affine
curve yq =f(x). The Jacobian J(C ) is an abelian variety over K of dimension
. f,q
h
(n−1)(q−1)
t
a dimJ(Cf,q)=g(Cf,q)= .
2
m
If q >p, the map
[
C →C , (x,y)7→(x,yp)
f,q f,q/p
1
v inducesbyAlbanesefuctorialityasurjectiveK-mapbetweentheJacobiansJ(Cf,q)→
3 J(C ). We write J(f,q) for the identity component of the kernel. If q = p, we
f,q/p
4 set J(f,p) = J(C ). It is follows easily that J(f,q) is an abelian variety over K of
7 f,p
dimension(n−1)ϕ(q)/2,whereϕdenotesthe Eulerϕ-function. Moreover,J(C )
1 f,q
. is K-isogenous to the product r J(f,pi)(See [15]).
1 i=1
0 Since K ⊆C, we may viewQJ(f,q) as a complex abelian variety. We refer to [5],
1 [10,Sect. 6.6.1and6.6.2]forthedefinitionandbasicpropertiesoftheHodgegroup
1 (aka special Mumford–Tate group). In [9], assuming that n > q and some other
v: conditions on n,q and f(x), the authors showed that the (reductive Q-algebraic
i connected) Hodge group of J(f,q) coincides with the largest Q-algebraic subgroup
X
ofGL(H1(J(f,q),Q))that’s“cutout”bytheinducedpolarizationfromthecanonical
ar principal polarizationof J(Cf,q) and the endomorphism ring of J(f,q). Notice that
whenq =2(i.e.,inthe hyperellipticcase)thisgroupwascompletelydeterminedin
[12](whenf(x)has“large”Galoisgroup). Inthis paper,westudysomeadditional
properties of J(f,q) which will allow us to extend the result to the case n < q as
well. This case is necessary in order to treat the infinite towers of superelliptic
jacobians, which, in turn, are useful for the study of the ranks of Mordell-Weil
groups in infinite towers of function fields (See [6]).
To state our main result, we make explicit the endomorphism ring and the po-
larizationmentionedabove. LetX beanabelianvarietyoverK¯. WewriteEnd(X)
for the ring of all its K¯-endomorphisms and End0(X) for the endomorphism al-
gebra End(X)⊗ Q. In a series of papers [11, 13, 14, 15], Yuri Zarhin discussed
Z
the structure of End0(J(C )), assuming that n ≥5 and the Galois group Gal(f)
f,q
of f(x) over K is, at least, doubly transitive. Here Gal(f) ⊆ S is viewed as a
n
permutation group on the roots of f(x). It is well known that f(x) is irreducible
1
2 JIANGWEIXUE
over K if and only if Gal(f) acts transitively on the roots. For the sake of sim-
plicity let’s assume that K contains a primitive q-th root of unity ζ . The curve
q
C :yq =f(x) admits the obvious periodic automorphism
f,q
δ :C →C , (x,y)7→(x,ζ y).
q f,q f,q q
Byanabuseofnotation,wealsowriteδ fortheinducedautomorphismofJ(C ).
q f,q
The subvariety J(f,q) is δ -invariant and we have an embedding
q
Z[ζ ]֒→End(J(f,q)), ζ 7→δ .
q q q
In particular, the q-th cyclotomic filed E := Q(ζ ) is contained in End0(J(f,q)).
q
Zarhinshowed([11,15,17])thatEnd(J(f,q))is isomorphicto Z[ζ ]ifeither Gal(f)
q
coincides with the full symmetric group S , n ≥ 4 and p ≥ 3, or Gal(f) coincides
n
with the alternating group A (or S ), and n ≥ 5. This result has also been
n n
extended to the case Gal(f)=S or A , n≥5 and p|n in [7].
n n
The first rational homology group H (J(f,q),Q) carries a natural structure of
1
E-vector space of dimension
dim H (J(f,q),Q) 2dimJ(f,q) (n−1)ϕ(q)
dim H (J(f,q),Q)= Q 1 = = =n−1.
E 1 [E :Q] [E :Q] ϕ(q)
Notice that if q >2, then E is a CM field with complex conjugation e7→e¯. Let
E+ ={e∈Q(ζ )|e¯=e}
q
be the maximal totally real subfield of E and let
E− ={e∈Q(ζq)|e¯=−e}.
The canonical principal polarization on J(C ) induces a polarization on J(f,q),
f,q
which gives rise to a nondegenerate E-sesquilinear Hermitian form ([9])
φ :H (J(f,q),Q)×H (J(f,q),Q)→E.
q 1 1
WewriteU(H (J(f,q),Q),φ )fortheunitarygroupofφ oftheQ(ζ )-vectorspace
1 q q q
H (J(f,q),Q), viewed as an Q-algebraic subgroup of GL(H (J(f,q),Q)) (via Weil’s
1 1
restriction of scalars from E+ to Q ([5])). Since the Hodge group respects the
polarization and commutes with endomorphisms of J(f,q),
Hdg(J(f,q))⊂U(H (J(f,q),Q),φ ).
1 q
If End0(J(f,q)) = E, then U(H (J(f,q),Q),φ ) is the largest connected reductive
1 q
Q-algebraicsubgroupofGL(H (J(f,q),Q))thatboth respectsthe polarizationand
1
commutes with endomorphisms of J(f,q).
The following theorem is a natural extension of [9, Theorem 0.1].
Theorem 1. Suppose that n ≥ 4 and p is a prime that does not divide n. Let
f(x) ∈ C[x] be a degree n polynomial without multiple roots. Let r be a positive
integer and q =pr. Suppose that there exists a subfield K of C that contains all the
coefficients of f(x). Let us assume that f(x) is irreducible over K and the Galois
group Gal(f) of f(x) over K is either S or A . Assume additionally that either
n n
n≥5 or n=4 and Gal(f)=S .
4
Suppose that one of the following three conditions holds:
(A) n=q+1;
(B) p is odd and n6≡1 mod q;
(C) p=2, n6≡1 mod q and n6≡q−1 mod 2q.
HODGE GROUPS 3
Then Hdg(J(f,q))=U(H (J(f,q),Q),φ ).
1 q
Corollary 2. Corollary 0.3, Theorem 4.2 and Theorem 4.3 of [9] all hold without
the assumption that n>q.
Remark 3. We assume that n<q throughoutthe restofthe papersince the case
n>q has already been treated in [9].
Remark 4. Since both Hdg(J(f,q)) and U(H (J(f,q),Q),φ ) are connected Q-
1 q
algebraic groups, to prove Theorem 1, it suffices to show that
dimHdg(J(f,q))≥dimU(H (J(f,q),Q),φ ).
1 q
It is known that
dimU(H (J(f,q),Q),φ )=dim E+· dim H (J(f,q),Q) 2.
1 q Q E 1
(cid:0) (cid:1)
Let hdg be the Q-Lie algebra of Hdg(J(f,q)). It is a reductive Q-Lie subalgebra of
End H (J(f,q),Q) , and thus splits into a direct sum
Q 1
(cid:0) (cid:1) hdg=c⊕hdgss,
of its center c and the semisimple part hdgss =[hdg,hdg]. By [8, Theorem 1.3], if
Gal(f) = S and n ≥ 4, or Gal(f) = A and n ≥ 5, the center c coincides with
n n
E−. Notice that
dimQE− =dimQE+ =[E :Q]/2.
Theorem 1 follows if we show that
1
(1) dim hdgss ≥ [E :Q] (dim H (J(f,q),Q))2−1 .
Q E 1
2
(cid:0) (cid:1)
The paper is organized as follows. In section 2 we study the Galois actions on
certainvectorspaces. Insection3werecallsomefactsabouttheHodgeLiealgebra
hdg. TheproofofTheorem1isgivenattheendofsection3exceptakeyarithmetic
lemma, which is proven in Section 4.
2. Galois Actions
Throughout this section, let E be a field that is a finite Galois extension of
Q with Galois group G. Let V be a E-vector space of finite dimension. We
write V for the underlying Q-vector space of V, and V for the C-vector space
Q C
V ⊗ C = V ⊗ C. Let Aut(C) be the group of all automorphisms of C. It
Q Q Q
act semilinearly on V = V ⊗ C through the second factor. More explicitly,
C Q
∀κ ∈ Aut(C),v ⊗z ∈ V ⊗ C, we define κ(v ⊗z) := v ⊗κ(z). It follows that
Q
∀x∈V⊗ Candc∈C,κ(cx)=κ(c)x. Ontheotherhand,E actsonV =V⊗ C
Q C Q
throughits firstfactor. ItfollowsthatV isafreeE⊗ C moduleofrankdim V,
C Q E
and the action of E = E⊗1 ⊆ E⊗ C commutes with that of Aut(C). In other
Q
words,
κ((e⊗1)x)=(e⊗1)κ(x), ∀κ∈Aut(C),e∈E, and x∈V .
C
Let’s fix an embedding E ֒→C. This allowsus to identify eachGalois automor-
phism σ : E → E with the embedding σ : E → E ⊂ C of E into C. It is well
known that
E :=E⊗ C= E⊗ C= C , where C :=E⊗ C.
C Q E,σ σ σ E,σ
σM∈G σM∈G
4 JIANGWEIXUE
So every EC module W splits as a direct sum W =⊕σ∈GWσ, where
W :=C W ={w∈W |(e⊗1)w =σ(e)w,∀e∈E}.
σ σ
Inparticular,VC =⊕σ∈GVσ,andeachVσ isaC-vectorspaceofdimensiondimEV.
For each σ ∈G, let P : V →V be the C-linear projection map from V to the
σ C σ C
summandV . Similarly,foreachpairσ 6=τ,wewriteP =P ⊕P :V →V ⊕V
σ σ,τ σ τ C σ τ
for the projection map onto this pair of summands.
We claim that Aut(C) permutes the set {V | σ ∈ G}, and the action factors
σ
through the canonical restriction
Aut(C)։G, κ7→κ| .
E
Indeed, for all κ∈Aut(C),e∈E and x ∈V ,
σ σ
(e⊗1)κ(x )=κ((e⊗1)x )=κ(σ(e)x )=κ(σ(e))κ(x )=κσ(e)κ(x ).
σ σ σ σ σ
Clearlyκσ(e)=((κ| )σ)(e). Byanabuseofnotation,wewriteκfortherestriction
E
κ| . So it follows that κ(x )∈V , and thus κ(V )=V for all κ∈Aut(C) and
E σ κσ σ κσ
σ ∈G.
Let us define an action of Aut(C) on the set of projection P ={P |σ ∈G} by
σ
κ∗Pσ :=κ◦Pσ◦κ−1.
Then for any element xσ ∈⊕σ∈GVσ =VC and Pτ ∈P,
(κ∗Pτ)( xPσ)=κ◦Pτ κ−1(xσ) =κ(κ−1(xκτ))=xκτ,
X (cid:16)X (cid:17)
where all summations runs through σ ∈ G, and we used the fact that κ−1(x )
σ
belongs to V if and only if σ =κτ. Therefore,
τ
κ∗Pσ =Pκσ.
Clearly Aut(C) acts transitively on P. Since P =P ⊕P , we have similarly an
σ,τ σ τ
action of Aut(C) on the set PP :={P |(σ,τ)∈G2,σ 6=τ} by
σ,τ
κ∗Pσ,τ =κ◦Pσ,τ ◦κ−1 =Pκσ,κτ.
TheAut(C)-orbitO ofeachP ∈PP consistsofallelementsoftheformP
σ,τ σ,τ κσ,κτ
with κ∈G.
Lemma 5. Let W ⊆ V be any Q-subspace of V , and W := W ⊗ C ⊆ V
Q Q Q C Q Q C
be its complexification.
(i) If there exists σ ∈G such that P (W )=V , then P (W )=V for all
0 σ0 C σ0 σ C σ
σ ∈G.
(ii) If there exists a pair (σ ,τ ) ∈ G2 with σ 6= τ such that P (W ) =
0 0 0 0 σ0,τ0 C
V ⊕V , then P (W )=V ⊕V for all P ∈O .
σ0 τ0 σ,τ C σ τ σ,τ σ0,τ0
Proof. Clearly,W is Aut(C)-invariant. Foreachσ ∈G, letus chooseκ∈Aut(C)
C
such that σ =κσ . Then
0
Pσ(WC)=(κ∗Pσ0)(WC)=κ◦Pσ0 ◦κ−1(WC)=κ◦Pσ0(WC)=κ(Vσ0)=Vσ.
This proves part (i). Similarly, suppose that P (W ) = V ⊕ V . For all
σ0,τ0 C σ0 τ0
P ∈O , there exists κ∈Aut(C) suchthat σ =κσ andτ =κτ . So we have
σ,τ σ0,τ0 0 0
Pσ,τ(WC)=(κ∗Pσ0,τ0)(WC)=κ◦Pσ0,τ0 ◦κ−1(WC)=κ◦Pσ0,τ0(WC)
=κ(V ⊕V )=κ(V )⊕κ(V )=V ⊕V ,
σ0 τ0 σ0 τ0 σ τ
and part (ii) follows. (cid:3)
HODGE GROUPS 5
Let R be a commutative ring with unity, and N be a free R-module of finite
rank. We write Tr :End (N)→R for the trace map, and
R R
sl (N):={g ∈End (N)|Tr (g)=0}
R R R
for the R-Lie algebra of traceless endomorphisms of N. It is well-known that
slE(V)⊗QC=slEC(VC)=slEC(⊕σ∈GVσ)= slC(Vσ).
σM∈G
We will denote the projection map sl (V) ⊗ C → sl (V ) again by P , and
E Q C σ σ
similarly for P . Clearly, each sl (V ) has C-dimension (dim V)2−1.
σ,τ C σ E
For the rest of the section, we assume additionally that E is a CM-field. For
any σ ∈ G, let σ¯ : E → E be the complex conjugation of σ. In other words, σ¯
σ
is the composition E −→ E → E, where the second arrow stands for the complex
conjugation map e7→e¯.
Lemma 6. Let k be a semisimple Q-Lie subalgebra of sl (V), and k := k⊗ C
E C Q
be its complexification. Suppose that the following two conditions holds:
(I) there exists σ ∈G such that P (k )=sl (V );
0 σ0 C C σ0
(II) For each pair (σ,τ) ∈G2 with σ 6=τ and σ 6=τ¯, there exists P ∈O
σ0,τ0 σ,τ
such Pσ0,τ0(kC) =slC(Vσ0)⊕slC(Vτ0).
Then
1
dim k≥ [E :Q] (dim V)2−1 .
Q E
2
(cid:0) (cid:1)
Proof. Applying Lemma 5 with k in place of W and sl (V) in place of V, we see
E
that
P (k )=sl (V ), ∀σ ∈G;
σ C C σ
P (k )=sl (V )⊕sl (V ), ∀(σ,τ)∈G2 with σ 6=τ and σ 6=τ¯.
σ,τ C C σ C τ
Let us fix a CM-type Φ of E. By definition, Φ is a maximal subset of G =
Hom(E,C) such that no two elements of Φ are complex conjugate to each other.
Clearly, |Φ|=[E :Q]/2, and
1
dim sl (V ) = [E :Q](dim (V)2−1).
C C σ E
2
(cid:16)σM∈Φ (cid:17)
Let k′C be the projection of kC on ⊕σ∈ΦslC(Vσ). It follows that the projection
k′ → sl (V ) is surjective for all σ ∈ Φ, and k′ also projects surjectively onto
C C σ C
slC(Vσ)⊕slC(Vτ) for all distinct pairs σ,τ ∈Φ. Therefore, k′C = ⊕σ∈ΦslC(Vσ) by
the Lemma on pp.790-791 of [4]. In particular, we get
1
dim k=dim k ≥dim k′ = [E :Q] (dim V)2−1 .
Q C C C C 2 E
(cid:0) (cid:1)
(cid:3)
In the next section, we will show that our semisimple part of Hodge Lie algebra
hdgss = [hdg,hdg] satisfies (I) and (II) of Lemma 6 and thus prove our Main
Theorem.
6 JIANGWEIXUE
3. the hodge lie algebra
Wekeepallnotationandassumptionsoftheprevioussections. Morespecifically,
ζ is a primitive q-th root of unity, E = Q(ζ ) and G = Gal(E/Q) = (Z/qZ)∗,
q q
where each a ∈ (Z/qZ)∗ maps ζ to ζa. In order to simplify the notation, we
q q
write X for the abelian variety J(f,q), and V for its first rational homology group
H (X,Q). In addition, we assume that End0(X)=E.
1
Recall that E = E ⊗ C. Let Lie(X) be the complex tangent space to the
C Q
originofX. Byfunctoriality,E actsonLie(X)andprovidesLie(X)withanatural
structure of E -module. Therefore, Lie(X) splits into a direct sum
C
Lie(X)=⊕a∈GLie(X)a.
where Lie(X) :={x∈Lie(X)|(ζ ⊗1)x=ζax}. Let us put n =dim Lie(X) .
a q q a C a
It is known that n =[na/q] (see [15, 16]), where [x] is the maximal integer that’s
a
less or equal to x, and we take the representative 1≤a≤q−1.
Remark 7. By [9, Proposition2,1,2.2],the assumptions (A)(B)(C) ofTheorem1
guarantee that there exists an integer a such that
1≤a≤q−1, gcd(a,p)=1
and the integers [na/q] and dim V = n−1 are relative prime. We note that the
E
conditions (A)(B)(C) of Theorem 1 are equivalent to the conditions (A)(B)(C) of
[9, Theorem 0.1].
Since V = H (X,Q) carries a natural structure of E-vector space, the first
1
complex homology group V = H (X,C) = H (X,Q)⊗ C carries a structure of
C 1 1 Q
E -module, and therefore splits into a direct sum
C
VC =⊕a∈GVa.
Each V is a C-vector space of dimension dim V =n−1.
a E
There is a canonical Hodge decomposition ([3, chapter 1], [1, pp. 52–53])
V =H (X,C)=H−1,0(X)⊕H0,−1(X)
C 1
whereH−1,0(X)andH0,−1(X)aremutually“complexconjugate”dim(X)-dimensional
complexvectorspaces. This splitting is E-invariant,andH−1,0(X)andLie(X)are
canonically isomorphic as E -modules. In particular,
C
dim H−1,0(X) =dim Lie(X) =n .
C a C a a
Let f0 =f0 :V →V be the C-linear operator such that
H H,Z C C
f (x)=−x/2 ∀ x∈H−1,0(X); f0 (x)=x/2 ∀ x∈H0,−1(X).
H H
Since the Hodge decomposition is E-invariant, f0 commutes with E. Therefore,
H
eachV isf0 -invariant. Itfollowsthatthe linearoperatorf0 :V →V issemisim-
a H H a a
ple and its spectrum lies in the two-element set {−1/2,1/2}. The multiplicity of
eigenvalue −1/2 is n = dim H−1,0(X) , while the multiplicity of eigenvalue 1/2
a C a
is dim V −n . Clearly, the complex conjugate of a ∈ Gal(E/Q) = (Z/qZ)∗ is
E a
a¯=q−a. It is known ([1], [2]) that
(2) n +n =dim V.
a a¯ E
This implies that the multiplicity of the eigenvalue 1/2 is n .
a¯
HODGE GROUPS 7
TheHodgeLiealgebrahdgofX isareductiveQ-LiesubalgebraofEnd (V). Its
Q
naturalrepresentationin V is completely reducible and its centralizer in End (V)
Q
coincides with End0(X)=E. Moreover,its complexification
hdg =hdg⊗ C⊂End (V)⊗ C=End (V )
C Q Q Q C C
contains f0 [8, Sect. 3.4]. Recall that hdg = c⊕hdgss, with c being the center of
H
hdg and hdgss = [hdg,hdg] the semisimple part. Let c := c⊗ C be the com-
C Q
plexification of c and hdgss :=hdgss⊗ C the complexification of hdgss. Clearly,
C Q
hdgss ⊂sl (V), and thus
E
hdgsCs ⊂slEC(VC)=⊕a∈GslC(Va).
We write hdgss for the image of projection P : hdgss → sl (V ). Clearly, each
a a C C a
hdgss is a semisimple complex Lie subalgebra of sl (V ).
a C a
Remark 8. Let us decompose f0 as f +f′ with f′ ∈ c and f ∈ hdgss. By
H C C
[9, Remark 3.2], the natural representation V of hdgss is simple for all a ∈ G.
a a
It follows from Schur’s Lemma that when restricted to each V , f′ coincides with
a
multiplication by scalar c ∈ C. Therefore, hdgss contains an operator (namely,
a C
f) whose restriction on each V is diagonalizable with at most two eigenvalues:
a
−1/2−c of multiplicity n and 1/2−c of multiplicity n =dim V −n .
a a a a¯ E a
Lemma 9. Let the assumptions be the same as in Theorem 1. There exists an
a∈G=(Z/qZ)∗ such that hdgss =P (hdgss) coincides with sl (V ).
a a C C a
Proof. The idea is to combine Remark 7, 8 together with Lemma 3.3 of [9]. This
resultis alreadycontainedin the proofof[9,Theorem3.4],where wenote thatthe
assumption n > q in [9, Theorem 3.4] is not used for this particular step of the
proof. (cid:3)
Notice that this is the place where assumptions (A)(B)(C) in Theorem 1 are
used, since we need to make sure that there exists a∈G such that n and dim V
a E
are relative prime in order to apply Lemma 3.3 of [9].
Let h : (Z/qZ)∗ → R be the function such that for all 1 ≤ a ≤ q −1 with
gcd(a,q)=1,
2 2
dim V n−1 na
E
(3) h(a)= −n = − .
a
(cid:18) 2 (cid:19) (cid:18) 2 (cid:20) q (cid:21)(cid:19)
By (2), n +n = dim V, so h(a) = h(a¯) = h(q−a), which is also easy to check
a a¯ E
directly from (3). The function h is non-increasing on the set of integers
[1,q/2] :={a|1≤a≤q/2,gcd(a,p)=1}.
Z
By Remark 3, we have 4≤n<q. In particular, [n/q]=0. On the other hand, let
t be the maximal element of [1,q/2] . Then t 6= 1 and [nt/q] 6= 0. It follows that
Z
h is not a constant function.
Lemma 10. Let the assumption be the same as Theorem 1. Let (a,b) ∈ G2 be a
pair such that h(a)6=h(b). Then P (hdgss)=sl (V )⊕sl (V ).
a,b C C a C b
Proof. By (3),
h(a)−h(b)=(n −n )(dim V −n −n ).
a b E a b
Soh(a)6=h(b)if andonlyif n 6=n andn 6=dim V −n . Letkss =P (hdgss).
a b a E b a,b C
By Lemma 9 and part (i) of Lemma 5, both projections kss → sl (V ) and kss →
C a
8 JIANGWEIXUE
sl (V ) are surjective. By Remark 8, P (f) is a semisimple element of kss ⊆
C b a,b
End (V )⊕End (V ) such that P (f) acts on V with (at most) 2 eigenvalues
C a C b a,b a
of multiplicities n and dim V −n respectively, and similarly for b. Lemma 10
a E a
followsby setting d=2 in [9, Lemma 3.6]. Last, we point out that the assumption
that the multiplicities a are positive in [9, Lemma 3.6] is not used in its proof, so
i
thelemmaappliestothecasethatn orn iszero,whichmayhappenifn<q. (cid:3)
a b
Proof of Theorem 1. As remarkedat the end of Section 2, Theorem 1 follows if we
showthattheconditions(I)and(II)ofLemma6holdsfork=hdgss. Condition(I)
holds by Lemma 9. To show that Condition (II) holds, by Lemma 10 it is enough
to prove that for each (a,b) ∈ G2 with a 6= b and a 6= ¯b, there exists x ∈ G such
thath(xa)6=h(xb). Supposethatthisisnotthecase,thenthereexistsapair(a,b)
such that h(xa) = h(xb) for all x ∈ G. Without loss of generality, we may and
will assume that b = 1 ∈ (Z/qZ)∗, thus a 6= ±1. It follows that h(xa) = h(x) for
all x ∈ (Z/qZ)∗. Since h is not a constant function, such an a does not exists by
Lemma 11 of next section. Contradiction. (cid:3)
4. Arithmetic Results
Throughout this section, G = (Z/qZ)∗. For each a ∈ G, let θ : G → G be the
a
translationmap: b7→ab. Afunctionh:G→Rissaidtobeeven ifh◦θ−1=h. For
any x≤y ∈R, we write [x,y] for the set of integers {i|x≤i≤y,gcd(i,p)=1}.
Z
Lemma 11. Let h : (Z/qZ)∗ → R be an even function that’s monotonic on
[1,q/2] . If h◦θ = h for some a ∈ (Z/qZ)∗ and a 6= ±1, then h is a constant
Z a
function.
Proof. We prove the Lemma in seven steps.
Step 1. Let h±ai be the subgroup of (Z/qZ)∗ generated by a and −1. Clearly
h◦θ = h for any b ∈ h±ai since h◦θ = h and h is even. In particular, this
b a
holds true for the maximalelement b in the set in h±ai∩[1,q/2] . If b =1,
max Z max
the group h±ai is necessarily {±1}. Therefore, it is enough to prove that h being
nonconstant implies that b =1. So with out lose of generality, we assume that
max
a = b throughout the rest of the proof. Notice that if a 6= 1, then 2a2 > q, for
max
otherwise it contradicts the maximality of a.
Step 2. Lemma 11 holds if p=2.
Every even function on (Z/qZ)∗ is constant if q is 2 or 4 so we assume that
q = 2r ≥ 8. The group (Z/2rZ)∗ is isomorphic to Z/2Z×Z/2r−2Z, where the
factor Z/2Z is generated by −1. Let us assume that h±ai has order 2s. Since
h±ai⊇h±1i,itfollowsthath±ai∼=Z/2Z×Z/2s−1Z. Inparticular,ifh±ai6=h±1i,
then Z/2s−1Z is nontrivial, therefore h±ai contains 3 elements of order two. But
thereareexactly3elementsofordertwoin(Z/qZ)∗ :−1,2r−1−1,2r−1+1. Hence
h±ai contains all the above elements of order 2. So a = 2r−1 −1 since it is the
largest element in [1,q/2] . Therefore,
Z
h(q/2−1)=h(2r−1−1)=h(a)=(h◦θ )(1)=h(1).
a
Since h is monotonic on [1,q/2] , the above equality implies that h is constant on
Z
[1,q/2] and therefore a constant function.
Z
HODGE GROUPS 9
Step 3. Let p be an odd prime. Lemma 11 holds if either a is even, or a is odd
and 3a≥q.
It is enough to prove that if a 6= 1, then h(1) = h((q −1)/2). Since h(1) =
(h◦θ )(1)=h(a), by monotonicity h is constant on [1,a] . Therefore it is enough
a Z
to find b such that h((q−1)/2)=h(b) and b∈[1,a] .
Z
First, let’s assume that a=2b is even. Then
q−1
a· =(q−1)b≡−b mod q.
2
So h((q−1)/2)=h(a(q−1)/2)=h(−b)=h(b). Clearly b=a/2 lies in [1,a] .
Z
Next, assume that a is odd. Then
q−1 qa−a q−a
a· = ≡ (mod q).
2 2 2
So h((q − 1)/2) = h((q − a)/2). Let b = (q − a)/2. When 3a ≥ q, we have
b=(q−a)/2≤a hence b lies in [1,a] as desired.
Z
Step 4. Lemma 11 holds if p=3.
When p is odd, (Z/prZ)∗ is cyclic of order ϕ(pr)=(p−1)pr−1. For p=3,
(Z/3rZ)∗ ∼=Z/(2·3r−1)Z∼=Z/2Z×Z/3r−1Z.
In particular, if q ≥ 9, (Z/qZ)∗ contains a unique subgroup of order 3 which is
generatedby3r−1+1. Ifthe orderofh±aiis coprimeto3,thenh±aiisnecessarily
{±1}, which leads to an contradiction. If the order of h±ai is divisible by 3, then
q ≥9 and h±ai contains 3r−1+1. By assumption on the maximality of a we must
have a≥3r−1+1 and hence 3a>q.
Step 5. Assume that both p and a are odd, p6=3 and 3a<q. Lemma 11 holds if
7a≥q.
Sincep6=3,(q−3)/2liesin[1,q/2] . Itisenoughtoprovethata6=1impliesthat
Z
h(1)=h((q−3)/2). Indeed, itfollowsfromthe proofofStep3 thath((q−1)/2)=
h((q−a)/2). Butifa6=1thena≥3 so(q−a)/2≤(q−3)/2. Ifweprovethath is
constant on [1,(q−3)/2] , then h((q−1)/2)=h((q−a)/2)=h(1) and it follows
Z
that h is a constant function.
By our assumption 3a<q, so (q−3a)/2 lies in [1,q/2] . Notice that
Z
q−3 q−3a
a· ≡ mod q.
2 2
We see that h((q − 3)/2) = h((q − 3a)/2). Since h is constant on [1,a] , the
Z
inequality h(1) 6= h((q −3)/2) would imply that a < (q−3a)/2, or equivalently
5a < q. In particular, 2a < q/2. But 2 ∈ [1,a] since p is odd and a ≥ 3. So
Z
h(2)=h(1), therefore h(2a)=h(1) and h is constant on [1,2a] . But now by our
Z
assumption 7a≥q, or equivalently 2a≥(q−3a)/2, it follows that
q−3 q−3a
h =h =h(1).
(cid:18) 2 (cid:19) (cid:18) 2 (cid:19)
Step 6. Assume that bothp anda areodd, p6=3,5and7a<q. Lemma11 holds.
10 JIANGWEIXUE
Since 7a < q and p 6= 5, (q−5a)/2 lies in [1,q/2] . By similar argument as in
Z
Step 5, h((q−5)/2)=h((q−5a)/2). We claim that now it is enoughto show that
h(1) = h((q −5)/2). Indeed, by the proof of the Step 5, all we need to show is
that h(1)=h((q−3)/2), but since a≥3, then (q−3a)/2<(q−5)/2. So h being
constant on [1,(q−5)/2] implies that h(1)=h((q−3a)/2)=h((q−3)/2).
Z
Let S be the set of all integers
S ={b|b≥1,p∤b,(2b+1)a<q}.
Clearly 1 ∈ S so S is not empty. Let x be the maximal element of S. By Step 1,
2a2 > q so necessarily x < a. Since h is constant on [1,a] , we must have h(1) =
Z
h(x). Notice that xa<q/2 by assumption. So h(ax)=h(x)=h(1) and it follows
that h is constant on [1,ax] . Assume that h(1) 6= h((q−5)/2). It is necessary
Z
thatax<(q−5a)/2,orequivalently,(2x+5)a<q. Butwecanchoosex′ fromthe
two elements set {x+1,x+2} such that x′ is coprime to p. It follows that x′ ∈S.
This contradicts the maximality of x.
Step 7. Lemma 11 holds if p=5.
If the order of h±ai is divisible by 5, then h±ai contains the unique subgroupof
order5in(Z/5rZ)∗. Inparticular,2·5r−1+1∈h±ai. Itfollowsthata>2·5r−1+1
and therefore 3a>5r. The Lemma holds by Step 3.
If the order of h±ai is coprime to 5. Then from the isomorphism
Z/5rZ∼=Z/4Z×Z/5r−1Z,
we see that h±ai is has either order 2 or 4. If h±ai has order 2, then h±ai is
necessarily h±1i and this leads to a contradiction. So we assume that h±ai has
order 4 and a is the unique element such that 1<a<5r/2 and a2 ≡−1 mod 5r.
In particular, a2 + 1 ≥ 5r. If a is even then the Lemma holds by Step 3. In
particular,thisworksforq =p=5sincea=2inthiscase. Weassumethatq ≥25
and a is odd through out the rest of the proof. First we claim that a≥7. Indeed,
If q = 25, then a = 7 by direct calculation; if q > 25, then a > 7 since a2+1 ≥q.
This implies that (q −a)/2 ≤ (q −7)/2. Therefore, it is enough to prove that
h((q−7)/2)= h(1) since it then follows that h((q−1)/2)= h((q−a)/2) = h(1).
By Step 5 we may also assume that 7a < q. It follows that (q−7a)/2 ∈ [1,q/2]
Z
and h((q−7)/2)=h((q−7a)/2).
Let c = [q/a]. Since a2 + 1 ≥ q and a < q/2 we see that 2 ≤ c ≤ a. Let
x = [c/2] if [c/2] is not divisible by 5, and x = [c/2]−1 otherwise. Notice that
a > x ≥ max{1,(c − 3)/2} and xa ≤ q/2 by our choice of x. It follows that
x ∈ [1,a] therefore h(x) = h(1), and therefore h(ax) = h(x) = h(1). So h is
Z
constant on [1,ax] . If h(1) 6= h((q −7)/2), we must have xa < (q −7a)/2, or
Z
equivalently, (2x+7)a<q. Then it follows that
q c−3 q
>2x+7≥2 +7=c+4= +4,
a (cid:18) 2 (cid:19) hai
which is absurd.
Lemma 11 is proved by combining all the above steps. (cid:3)
References
[1] P.Deligne,Hodge cycleson abelian varieties(notes byJ.S.Milne).LectureNotesinMath.,
vol.900(Springer-Verlag,1982),pp.9–100.