Introduction to Local and Global Euler Characteristic Formulas

0.16 MB·English

by  Wei Lu

INTRODUCTION TO LOCAL AND GLOBAL EULER
CHARACTERISTIC FORMULAS
2
1 WEILU
0
2
Abstract. This is a note of talks I gave at the number theory seminar at

n TsinghuaUniversityinFall2011.
a We willintroduce the local and global Euler characteristic formulas given
J byJohnTate(1962)forGaloiscohomology. Wewillgiveadetailedproofbased

1 ontheideainHida’sbook[1,Ch4.4.4and4.4.5]andMilne’sbook[2,Ch1.5].
1

]
T This note is organized as follows. In preliminary, we review the definition of
N group cohomology and some basic properties. In Section 1, we will give a detailed
proof of the local case. In Section 2, we also prove the global case by using a
.
h powerful theorem given by John Tate. Both of these proofs roughly follow Hida’s
t
a book[1, Ch4.4.4 and 4.4.5] and Milne’s book[2,Ch1.5].
m
Preliminary
[

In the section, we recall some basic facts on Galois cohomology without proof.
5
v Readers can see [4] or [5] for more details.
3
7 Cohomology and Cochains. Let G be a finite group, and M be a G-module.
3 The functor M MG from the category of G-modules to the category of Abelian
7→
4 groups is left-exact. The derived functor is denoted by Hn(G, ). Hn(G,M) is
. −
2 called the cohomology group, and it can be computed by the complex of cochains
1 in the following way.
1 We define Cn(G,M) := Map(Gn,M), an element of Cn(G,M) is a function f
1
of n variables in G with codomain M. The differential maps
:
v
d :Cn(G,M) Cn+1(G,M)
i n →
X
are defined by
r
a n
d (f )(g , ,g ) = g f(g , ,g )+ ( 1)if(g , ,g g , ,g )
n n 1 n 1 2 n 1 i i+1 n+1
··· · ··· − ··· ···
Xi=1
+( 1)n+1f(g , ,g ),
1 n
− ···
and we can check d d =0 directly. The cohomologicalgroups are given by
n+1 n
◦
Hn(G,M)=Hn(C•(G,M))=kerd /Imd .
n n−1
When G is a profinite group, the discrete abelian groups on which G acts con-
tinuously form an abelian category C , which is a full subcategory of the category
G
of all G-modules. For a (discrete) G-module M, we define
Hn(G,M):= lim Hn(G/H,MH),
H−→⊳G
Date:November30,2011and,inrevisedform,December7,2011.
1
2 WEILU
where H runs over the open normal subgroups of G. If we want to compute the
cohomologygroupsHn(G,M),wecanstillusethesamemethodappliedtocompute
the cochains. The only change is that the cochains must be continuous.
cor , res . Let U be a closed subgroup of G, we have restriction map
G/U G/U
res :Hn(G,M) Hn(U,M).
G/U
→
If U is a open subgroup of G of finite index, we have corestriction map
cor :Hn(U,M) Hn(G,M).
G/U
→
Then we have the following proposition.
Proposition 0.1. The following properties hold:
(1) cor res (x)=[G:U]x, where x Hn(G,M);
G/U G/U
◦ ∈
(2) If G is a finite group, M is a finite G-module, and (G, M ) = 1, then
| | | |
Hq(G,M)=0 for all q >0.
Inflation and Restriction Sequences.
Proposition0.2. LetU beaclosednormalsubgroupofG,andsupposeHp(U,M)=
0 for all p=1,2, ,q 1. Then the following sequence is exact:
··· −
0 Hq(G/U,MU) Hq(G,M) H0(G/U,Hq(U,M)) Hq+1(G/U,MU).
→ → → →
Shapiro’s Lemma. Let U be a closed subgroup of G. The induced module is
given by
IndGM =Hom (Z[G],M).
U Z[U]
Proposition 0.3. We have isomorphisms:
Hn(G,IndGM)=Hn(U,M)
U
for all n 0.
≥
Tate Cohomology. LetGbeafinite groupandM beaG-module,thenthe Tate
groups are defined by:
Hn(G,M), n 1
≥
 H0(G,M), n=0
Hn(G,M)=
T  kerNm/IGM, n= 1
−
H (G,M), n< 1
i−1
where N :m gm, I Z[G] is generated by 1. −
m G
7→ ⊂ −
gP∈G
Proposition 0.4. Suppose that G is a finite cyclic group generated by g. Then
Hn(G,M)=Hn+2(G,M).
T ∼ T
Corollary 0.5. All the notations are the same as above, then we have:
H2n(G,M)=H0(G,M)=MG/N (M)
∼ T m
for all n 1, and
≥
H2n−1(G,M)=kerN /I M.
∼ m G
Proposition 0.6. If G is cyclic, and M is finite, then
H0(G,M) = H1(G,M).
| T | | T |
INTRODUCTION TO LOCAL AND GLOBAL EULER CHARACTERISTIC FORMULAS 3
1. Local Euler Characteristic Formula
The main result is the following theorem.
Theorem 1.1 (Local Case). Let K/Q be a finite extension for a prime p, G =
p
Gal(Q /K) and M be a finite (discrete) G-module. We have local Euler character-
p
istic formula:
H0(G,M) H2(G,M) H0(G,M) H0(G,M∗(1))
| |·| | = | |·| | = M ,
H1(G,M) H1(G,M) || ||K
| | | |
where M∗(1)=Hom(M,K×), n =[O :nO ]−1 for a positive integer n.
K K K
| |
The first equation is as a result of Tate duality.
Proposition 1.2. (Tate duality) Let M be a finitely generated discrete Z[G]-
module,
Hr(G,M∗(1))=H2−r(G,M)∗
∼
for all 0 r 2, where M∗(1) = Hom(M,K×), N∗ = Hom(N,Q/Z) is the
≤ ≤
Potryagin dual module of an abelian group N.
In particular, if M is finite, all cohomology groups Hr(G,M) are finite and
Hr(G,M)=0 for r 3.
≥
Proof of the Local Case. We simply write Hn(M) for Hn(G,M). Since M =
M[l∞] for prime l, so we only need to prove the case M = M[l∞] because of
Ll
additions. Now we may assume that M = M[l∞], then Hq(M) is a Z -module of
l
finite length. For any finite Zl-module N, we have N = llengthZl(N) because a
| |
simple non-zero Z -module must be isomorphic to Z /l. Here the length(N) is the
l l
length of the Jordan-Holder sequence of Z-module M.
l
We define the local Euler character by
2
χ(M)=χ(G,M)= ( 1)qlength Hq(M),
− Zl
Xq=0
[K :Q ]length M l =p,
χ′(M)=χ′(G,M)=log ( M )= − p Zp
l || ||K (cid:26) 0 l =p.
6
Note that the left and the right side of this formula, so we only need to prove
[K :Q ]length M l =p,
χ(M)=χ′(M)= − p Zp
(cid:26) 0 l =p.
6
We first check the formula for the trivial case: M = F = Z/lZ (G acts on F
l l
trivially). By Tate Duality, dim H0(G,F ) = dim FG = 1, dim H2(G,F ) =
Fl l Fl l Fl l
dim H0(G,µ )∗ = dim µ∗(K) = dim µ (K), where µ (K) = z K zl = 1 .
Fl l Fl l Fl l l { ∈ | }
On the other hand, by Kummer theory, dim H1(G,F ) = dim H1(G,µ )∗ =
Fl l Fl l
dim H1(G,µ )=dim K×/(K×)l. The reason for the last equation is that
Fl l Fl
× x7→xl ×
1 µ K K 1.
l
−−−−→ −−−−→ −−−−→ −−−−→
Then we have
1 H0(G,µ ) H0(G,K×) H0(G,K×) H1(G,µ ) H1(G,K×)
l l
−−−−→ −−−−→ −−−−→ −−−−→ −−−−→
1 µ ((cid:13)(cid:13)K) K(cid:13)(cid:13)× x7→xl K(cid:13)(cid:13)× H1(G(cid:13)(cid:13),µ ) 1(cid:13)(cid:13),
l l
−−−−→ −−−−→ −−−−→ −−−−→ −−−−→
4 WEILU
where H1(G,K×) = 1 is given by Hilbert 90. So H1(G,µ ) = K×/(K×)l. Since
l ∼
K× ∼= OK× ×Z, O× ∼= OK ×µ ∼= OK ×µl∞(K)× µq∞(K), where OK is the
qQ6=l
integer ring of K. So
Z/lZ µ (K) l =p,
K×/(K×)l = ⊕ l 6
∼(cid:26) Z/pZ OK/pOK µp(K) l =p.
⊕ ⊕
When l=p,
χ(F ) = dim H0(G,F ) dim H1(G,F )+dim H2(G,F )
p Fp p − Fp p Fp p
= 1 dim (F O /pO µ (K))+dim µ (K)
− Fp p⊕ K K ⊕ p Fp p
= dim O /pO = [K :Q ]dim F =χ′(F ).
− Fp K K − p Fp p p
And when l=p,
6
χ(F )=1 dim (F µ (K)) dim µ (K)=0=χ′(F ).
l − Fl l⊕ l − Fl l l
By Tate duality, the formula also holds for M =µ =F∗(1).
l l
We willprovethatif 0 L M N 0 is anexactsequenceof finite Z [G]-
l
→ → → →
modules,thenχ′(M)=χ′(N)+χ′(L),χ(M)=χ(N)+χ(L),andχ′(M)=χ′(Mss),
n
χ(M) = χ(Mss), where Mss = M /M for a Jordan-Holder sequence 0 =
q q−1
qL=1
M M M =M of Z[G]-modules.
0 1 n l
⊂ ⊂···
In fact, χ′(M) = [K : Q ]length M = [K : Q ](length L+length N) =
− p Zp − p Zp Zp
χ′(N)+χ′(L). And 0 H0(L) H0(M) H0(N) H1(L) H1(M)
→ → → → → →
H1(N) H2(L) H2(M) H2(N) 0 is an exact sequence which follows
→ → → →
Proposition 1.2. So we have
2
χ(M) = ( 1)qlengthHq(M)
−
Xq=0
2
= ( 1)q(lengthHq(L)+lengthHq(N))
−
Xq=0
2 2
= ( 1)qlengthHq(L)+ ( 1)qlengthHq(N)
− −
Xq=0 Xq=0
= χ(L)+χ(N).
Thus we get
n
χ′(Mss) = χ′( M /M )
q q−1
Mq=1
n
= χ′(M /M )
q q−1
Xq=1
n
= (χ′(M ) χ′(M ))
q q−1
−
Xq=1
= χ′(M )=χ′(M).
n
And the same reason for χ(M)=χ(Mss).
INTRODUCTION TO LOCAL AND GLOBAL EULER CHARACTERISTIC FORMULAS 5
However, Mss is a F [G]-module because M /M is a F [G]-module. Then we
l q q−1 l
may assume that M itself is a F [G]-module. At this time dim M =length M.
l Fl Zl
Now we recall the notation of Grothendieck groups. Let G be a profinite group
andE beafield. WeconsiderthecategoryRep (G)madeupofthefollowingdata:
E
(1) ObjectsarefinitedimensionalE vectorspaceswithacontinuousactionof
−
G under the discrete topology;
(2) Morphisms are E[G] linear maps.
−
Grothendieck group R (G) of Rep (G) is an Abelian group which is defined by
E E
generators and relations: R (G) is generated by symbols [M] for objects M
E
∈
Rep (G). The only relation is [M]=[N]+[L] if 0 L M N 0 is a short
E
→ → → →
exact sequence.
NowweconsiderthecategoryRep (G)whichismadeofallfiniteF [G]-module.
Fl l
Its Grothendieck group is R (G), we can regard χ and χ′ as functions on the
Fl
GrothendieckgroupR (G)withvalueZ. Weneedtochecktheformulaforasetof
Fl
generators of R (G). As Z is torsion-free, we only check it for a set of generators
Fl
of R (G) Q. And we can find a set of generators by the following proposition:
Fl ⊗Z
Proposition1.3. (see[2,lemma2.10])LetGbeafinitegroupand,foranysubgroup
H of G, let IndG be the homomorphism R (H) Q R (G) Q taking the class
H Fp ⊗ → Fp ⊗
of an H-module to the class of the corresponding induced G-module. R (G) Q
Fp ⊗
is generated by the images of the IndG as H runs over the set of cyclic subgroups
H
of G of order prime to p.
WetakeafiniteGaloisextensionF/K suchthatGal(Q /F)actstriviallyonM,
p
writeG=Gal(F/K). Hence,Weonlyneedtochecktheformulaforasetofgenera-
torsofR (G). However,byProposition1.4,R (G) QisgeneratedbyIndGρfor
Fl Fl ⊗Z H
cyclicsubgroupsH oforderprimetol andcharacterρ:H × forafiniteexten-
→K
sion /F . Thus we can assume M = IndGρ = IndGρ, where H = Gal(Q /FH).
K l H H p
Then by Shapiro’s lemma, Hq(G,IndGρ) = Hq(G,IndGρ) = Hq(H,ρ), thus
H ∼ H ∼
χ(G,IndGρ) = χ(H,ρ). So we only need to check the formula for ρ (or for one-
H
dimensional single module V(ρ) on which H acts via ρ).
Thus we may assume FH = K, then H = G, M = V(ρ) is one-dimensional over
and G = H is a cyclic group, (G,l) = 1. By Proposition 0.1(2), we know
K | |
Hq(G,M)=0 for all q 1. Hence, we have inflation and restriction sequence,
≥
Hq(G/G′,MG′) Hq(G,M) H0(G/G′,Hq(G′,M)) Hq+1(G/G′,MG′)
−−−−→ −−−−→ −−−−→
(cid:13) (cid:13) (cid:13) (cid:13)
0=Hq(cid:13)(G,M) Hq(G(cid:13),M) H0(G,H(cid:13)q(G′,M)) Hq+1(G(cid:13),M)=0,
−−−−→ −−−−→ −−−−→
where G′ = Gal(Q /F). Moreover, we have Hq(G,M) = H0(G,Hq(G′M)) for
p ∼
q =0,1,2, and we note that
q =0;
K
Hq(G′,M)=Hq(G′,K)=Hq(G′,Fl)⊗Fl K= (µ(∗F(F×)/(F×)l)∗⊗Fl K qq ==12.;
l ⊗Fl K

6 WEILU
Then
χ(G,M)=dim G dim (((F×)/(F×)l)∗ )G+dim (µ∗(F) )G.
FlK − Fl ⊗Fl K Fl l ⊗Fl K
Since we have checked the cases M = F and M = µ , we may assume that ρ is
l l
neither trivialnor cyclotomiccharacter. Hence G =(µ∗(F) )G =0 because
K l ⊗FlK
the action of the Galois group on is via nontrivial character ρ on µ is via the
p
K
cyclotomic character. Therefore, χ(G,M)= dim (((F×)/(F×)l)∗ )G.
− Fl ⊗Fl K
When l =p, we need to show that
χ(G,M)= dim (((F×)/(F×)l)∗ )G = [K :Q ]dim M =χ′(G,M).
− Fl ⊗Fl K − p Fp
Since µ is the maximal torsion-subgroupof F×, then we have
1 µ F× F×/µ 1
−−−−→ −−−−→ −−−−→ −−−−→
p p p
  
1 yµ Fy× F×y/µ 1,
−−−−→ −−−−→ −−−−→ −−−−→
where p:x xp. By snake Lemma, we have
7→
1 µ/µp (F×)/(F×)p (F×/µ)/(F×/µ)p 1.
→ → → →
As ”∗” is defined as following: ”∗” = Hom( ,Q/Z) = Hom( ,F ) (only in this
p
− −
situation) is a contravariantand left exact functor, thus
1 ((F×/µ)/(F×/µ)p)∗ ((F×)/(F×)p)∗ (µ/µp)∗ Ext1((F×/µ)/(F×/µ)p,F )=1.
p
→ → → →
We know that is flat, then
K
1 ((F×/µ)/(F×/µ)p)∗ ((F×)/(F×)p)∗ (µ/µp)∗ 1.
→ ⊗Fp K→ ⊗Fp K→ ⊗Fp K→
Hence
1 H0(G,((F×/µ)/(F×/µ)p)∗ ) H0(G,((F×)/(F×)p)∗ ) H0(G,(µ/µp)∗ ).
→ ⊗FpK → ⊗FpK → ⊗FpK
However, µ/µp =µ (F), then
∼ p
H0(G,(µ/µp)∗ )=(µ∗(F) )G =0.
⊗Fp K p ⊗Fp K
Therefore,
H0(G,((F×/µ)/(F×/µ)p)∗⊗Fp K)∼=H0(G,((F×)/(F×)p)∗⊗Fp K).
In other words,
dim (((F×)/(F×)p)∗ )G =dim (((F×/µ)/(F×/µ)p)∗ )G =dim (((F×/µ) F )∗ )G.
Fp ⊗FpK Fp ⊗FpK Fp ⊗Z p ⊗FpK
Writing the additive valuation of F as v : F× Z, then we have an exact
→
sequence:
1 O×/µ F×/µ Z 0.
→ F → → →
Then the exact sequence is torsion-free, and after tensor F , we still have an exact
p
sequence
1 O×/µ F F×/µ F F 0.
→ F ⊗Z p → ⊗Z p → p →
Using functor again, we have:
∗
0 Hom(F ,F )=F (F×/µ F )∗ (O×/µ F )∗ Ext1(F ,F )=0.
→ p p p → ⊗Z p → F ⊗Z p → p p
So we have
0 G ((F×/µ F )∗ )G ((O×/µ F )∗ )G H1(G, )=0.
→K → ⊗Z p ⊗FpK → F ⊗Z p ⊗FpK → K
INTRODUCTION TO LOCAL AND GLOBAL EULER CHARACTERISTIC FORMULAS 7
Therefore,
dim ((F×/µ F )∗ )G =dim ((O×/µ F )∗ )G.
Fp ⊗Z p ⊗Fp K Fp F ⊗Z p ⊗Fp K
Now we want to lift the representation ρ to characteristic 0 representation ρ by
the following proposition.
e
Proposition 1.4. (see [1, corollary 2.7]) Let K be a finite extension of Q with
p
p adic integer ring O. Let E = O/m for the maximal ideal m of O. Suppose
O O
−
p G is not true and that all irreducible representations of G over K are abso-
|| |
lutely irreducible. Then all irreducible representations of G over E are absolutely
irreducible, and the reduction map ρ (ρ mod m ) induces a bijection between
O
7→
isomorphism classes of absolutely irreducible representations of G over K and over
E, preserving dimension.
For that we take the unique unramified extension L of Q of degree dim .
p FpK
Then we have O /(p)= , O× =(1+pO ) ×, where O is the p-adic integer
L ∼K L ∼ L ×K L
ring of L. By the isomorphism, we may think ρ has valuation in O×. We write
L
the character ρ:G O×, which is called the Teichmuller lift of ρ. Since O×/µ is
→ L F
torsionfree and(G,ρ)=1, by Proposition1.4 for the unique Teichmuller lift ρ of
e | |
ρ, we have:
e
dim ((O×/µ F )∗ )G = dim ((O×/µ F )∗ )G
Fp F ⊗Z p ⊗Fp K Fp F ⊗Zp p ⊗Fp K
= dim (Hom(O×/µ F ,F ) )G
Fp F ⊗Zp p p ⊗Fp K
= Rank (Hom(O×/µ Z ,Z ) O )G
Zp F ⊗Zp p p ⊗Zp L
= dim (Hom(O×/µ Q,Q ) L)G.
Qp F ⊗Z p ⊗Qp
By p-adic logarithm, we know that OF×/µ⊗ZQ∼=F as G-module. Hence
dim (Hom((O×/µ) Q,Q ) L)G =dim (Hom(F,Q ) L)G.
Qp F ⊗Z p ⊗Qp Qp p ⊗Qp
By normal base theorem, F ∼=K[G]∼=Qp[G][K:Qp]. Then
dimQp(Hom(F,Qp)⊗Qp L)G = dimQp(Hom(Qp[G][K:Qp],Qp)⊗Qp L)G
= [K :Q ]dim (Hom(Q [G],Q ) L)G.
p Qp p p ⊗Qp
We can easily check that Hom(Q [G],Q ) = Q [G] as G-module by ψ : f
p p ∼ p 7→
a σ, where a =f(σ), then
σ σ
σP∈G
[K :Q ]dim (Hom(Q [G],Q ) L)G
p Qp p p ⊗Qp
= [K :Q ]dim (Q [G] L)G
p Qp p ⊗Qp
= [K :Q ]dim (Q [G] L0)G ( )
p Qp p ⊗Qp ······ ∗
= [K :Q ]dim (Q [G]G L0)
p Qp p ⊗Qp
= [K :Q ]dim (Q L0)
p Qp p⊗Qp
= [K :Q ]dim L0
p Qp
= [K :Q ]dim L
p Qp
= [K :Q ]dim
p FpK
= [K :Q ]dim M,
p Fp
8 WEILU
where Q [G]G =Q and (*) follows from the isomorphism:
p ∼ p
Qp[G]⊗Qp L∼=Qp[G]⊗Qp L0
given by
σ m σ σ−1m,
⊗ 7→ ⊗
where L0 is the trivial G-module with L=L0 as Q vector spaces.
∼ p−
When l=p,we onlyneedto checkχ(G,M)= dim((F×/(F×)l)∗ )G =0.
6 − ⊗FlK
Discuss it again, we know ((F×/(F×)l)∗⊗Fl K)G ∼=((F×/µ⊗ZFl)∗⊗Fl K)G. But
F×/µisaZ -module,andlisinvertibleinZ . Thus,F×/µ F =0,χ(G,M)=0.
p p Z l
⊗
Remark 1.5. : Actually, dim MG¯ = dim (M∗)G¯, because G¯ is a finite cyclic
Fp Fp
group, let σ be the generator of G¯, then we have
dim (M∗)G¯ = dim (Hom (M,F ))G¯
Fp Fp Fp p
= dimFpHomFp[G¯](M,Fp)
= dimFpHomFp[G¯](M/(σ−1)M,Fp)
= dim Hom (M/(σ 1)M,F )
Fp Fp − p
= dim M/(σ 1)M =dim MG¯,
Fp − Fp
where M is the F [G¯]-module. Thus, by using the conclusion, we know that
p
dim ((F×/(F×)p)∗ )G¯ =dim (((F×/(F×)p)∗ )∗)G¯
Fp ⊗Fp K Fp ⊗Fp K
=dim ((F×/(F×)p) ∗)G¯.
Fp ⊗Fp K
Thus, we only need to show that
dim ((F×/(F×)p) ∗)G¯ =[K :Q ]dim M.
Fp ⊗Fp K p Fp
Then we can compute it a little easily.
2. Global Euler Characteristic Formula
Theorem 2.1 (Global Case). Let K/Q be a finite extension, S be a finite set
of places of Q including the archimedean places and Σ be the set of places of K
above S. Ks/K is the maximal algebraic extension unramified outside S. We write
S = Gal(Ks/K). Assume M is a finite S -module such that if l M , then
S S
| | |
l S. Then we have:
∈
H2(S ,M) H0(S ,M) H0(G ,M)
S S v
| |·| | = | |,
H1(S ,M) M
| S | v∈YΣ∞ || ||Kv
where |n|Kv =n[Kv:Q], v is archimedean place, Gv =Gal(Kv/Kv).
Inthisformula,weknowthatM isS -module,wealsoregardM asG -module.
S v
Before proving this formula, we will give a very powerful theorem which is proved
by John Tate. This theorem is the key to prove it. But we do not plan to prove it
here. We only narrate it. If you are interested in it ,you can see the reference.
Proposition 2.2. Let S be a finite set of places of Q including the archimedean
places. Let K be a number field and Σ be the set of places of K above S. Fix a
prime p S, and let M be a discrete finite S -module with p power order. Then
S
∈ −
we have:
INTRODUCTION TO LOCAL AND GLOBAL EULER CHARACTERISTIC FORMULAS 9
(1)
Hr(S ,M)= Hr(G ,M) for all r 3,
S ∼ v ≥
v∈YΣ(R)
where Σ(R) is the set of real places of K.
(2) We have the following long exact sequence:
0 H0(S ,M) H0(G ,M) H0(G ,M) H2(S ,M∗(1))∗
→ S → v × T v → S
vY∈Σ0 v∈YΣ∞
H1(S ,M) H1(G ,M) H1(S ,M∗(1))∗
S v S
→ → →
vY∈Σ
H2(S ,M) H2(G ,M) H0(S ,M∗(1))∗ 0.
S v S
→ → → →
vY∈Σ
Proof of the Global Case. Since M = M[l∞], we may assume M =M[l∞].
l
Now, we prove it for l>2. Let L
ϕ(M)=χ(S ,M) (length H0(G ,M) [K :R]length M),
S − Zl v − v Zl
vX∈Σ∞
and we need to prove that ϕ(M)=0.
By Proposition 2.2(1), we have Hq(S ,M) = Hq(G ,M) for all q 3.
S v
≥
v∈QΣ(R)
However, gcd(G ,M) = 1 because of G = 2 and l > 3. Then by Proposition
v v
| | | |
0.1(2), we have Hq(G ,M)=0 for q 1, so Hq(S ,M)=0 for all q 3.
v S
≥ ≥
When 0 L M N 0 is a short exact sequence of Z [S ]-module, we
l S
→ → → →
have a long exact sequence:
0 H0(S ,L) H0(S ,M) H0(S ,N) H2(S ,N) 0
S S S S
→ → → →···→ →
offiniteZ -module. Hence, χ(M)=χ(L)+χ(N)andϕ(M)=ϕ(L)+ϕ(N),where
l
χ and ϕ factor through the Grothendieck group R (S ) and have values in Z.
Fl S
Thenby Proposition2.2(2),wehaveχ(M)+χ(M∗(1))= χ (G ,M), where
v v
vP∈Σ
2
( 1)qlength Hq(G ,M), v is a non-archimedean place;
χ (G ,M)= qP=0 − Zl v
v v  2 ( 1)qlength Hq(G ,M), v is an archimedean place.
− Zl T v
Then  qP=0
χ (G ,M)= χ (G ,M)+ χ (G ,M),
v v v v v v
vX∈Σ vX∈Σ∞ vX∈Σ0
where Σ is the set of finite places. We know
0
2
χ (G ,M)= ( 1)qlength Hq(G ,M).
v v − Zl T v
vX∈Σ∞ vX∈Σ∞Xq=0
However, G is a cyclic group, by Proposition 0.6,
v
length H2(G ,M)=length H1(G ,M),
Zl T v Zl T v
length H0(G ,M)=length H1(G ,M)=length H1(G ,M).
Zl T v Zl T v Zl v
10 WEILU
Thus,
χ (G ,M)= length H1(G ,M).
v v Zl v
vX∈Σ∞ vX∈Σ∞
On the other hand, since M = 1 for v Σ, by the product formula
|| ||Kv 6∈
M =1 and the local Euler Characteristic formula,
|| ||Kv
Qv
χ (G ,M)=log ( M )= log ( M )= [K :R]length M.
v v l || ||Kv − l || ||Kv − v Zl
vX∈Σ0 vY∈Σ0 v∈YΣ∞ vX∈Σ∞
Therefore,
ϕ(M)+ϕ(M∗(1))=χ(M)+χ(M∗(1))
(length H0(G ,M)+length H0(G ,M∗(1)) 2[K :R]length M)
− Zl v Zl v − v Zl
vX∈Σ∞
= (length H1(G ,M) length H0(G ,M) length H0(G ,M∗(1))+[K :R]length M)
Zl v − Zl v − Zl v v Zl
vX∈Σ∞
=0.
The last equation follows the following proposition:
Proposition2.3. (in[2,theorem2.3(c)])LetK =RorCandletG =Gal(K /K ),
v v v v
|n|Kv =n[Kv:R]. For any finite Gv-module M, Then we have
H0(G ,M)) H0(G ,M∗(1))
v v
| |·| | = M .
H1(G ,M) || ||Kv
v
| |
So we only need to show that ϕ(M)=ϕ(M∗(1)).
We take a finite Galois extension F/K such that S′ = Gal(Ks/F) acts on
S
M and µ trivially. We write G = Gal(F/K). By the same argument of local
l
case, we may assume G is cyclic of degree prime to l and M is a F [G]-module.
l
We know Hq(S′ ,M) = 0 for q 3, since gcd(G,l) = 1, Hq(G,M) = 0 for all
S ≥ | |
q > 0, and by the inflation and restriction sequence again, we get Hq(S ,M) =
S ∼
H0(G,Hq(S′ ,M)).
S
Since weconsiderϕasahomomorphismfromthe GrothendieckgroupR [G]to
Fl
Q. let
χ′ :R (G) Q R (G) Q,
Fl ⊗ → Fl ⊗
2
[M] ( 1)i[Hi(S′ ,M)]
7→ − S
Xi=0
and
θ :R (G) Q Q,
Fl ⊗ →
[M] dim MG,
7→ Fl
then
χ=θ χ′.
◦
We know that [H0(S′ ,µ )]=[µ ], and
S l l
(i): H1(S′ ,µ )=[O× /l]+[Cl (F)[l]],
S l F,S S
(ii): [H2(S′ ,µ )]=[Cl (F)/l] [F ]+[ F ]+[ H0(G ,F )],
S l S − l l T p l
p∈S\LS∞(F) p∈SL∞(F)