ON EXPLICIT RECIPROCITY LAW OVER FORMAL GROUPS

PII. S0161171204301249 http://ijmms.hindawi.com

© Hindawi Publishing Corp.

ON EXPLICIT RECIPROCITY LAW OVER FORMAL GROUPS

SHAOWEI ZHANG Received 16 January 2003

LetFbe a one-dimensional Lubin-Tate formal group overZp. Colmez (1998) and Perrin-Riou (1994) proved an explicit reciprocity law for tempered distributions over the formal group Gm. In this paper, the general explicit reciprocity law over the formal groupFis proved.

2000 Mathematics Subject Classification: 11S40.

1. Formal groups. In this section, we review some basic facts on formal groups.

De Shalit [3, Chapter I] is the source for this section. We just give the description on Coleman power series and the associated measures, and then study such measures extensively. These measures are the basic examples for our theory. For admissible dis- tributions, see [8].

Letpbe an odd prime. Forα∈Z^×p, letπ=pα. Letfπ(x)∈Zp[[x]]be a Frobenius corresponding toπ, sofπ(x)≡π x(mod deg 2),fπ(x)≡x^p(modp). LetFbe the one- dimensional Lubin-Tate formal group overZpcorresponding tofπ, let[+]denote the formal addition. LetW_πⁿ:= {x∈Cp|fπ⁽ⁿ⁾(x)=0}, Kn=Qp(W_πⁿ), K_∞= ∪n≥1Kn. Hence, K_∞/Qpis a totally ramified extension with Galois groupZ^×p. We call this tower the Lubin- Tate tower corresponding to the formal groupF. Let R=Zp[[T ]],ᐁ=lim←K_n^×, where the map is with respect to the norm map. Letη:Gm→Ᏺπ be an isomorphism, then η∈Q^pur[[T ]]andη(T )=ΩT+···, such thatΩ^ϕ−1=α, whereϕ=Frobpis a generator of Gal(Q^urp /Qp). We havef◦η=η^ϕ◦[p], where[p]=(1+T )^p−1 is the Frobenius of the formal groupGm. Letωn=η^ϕ−n(ζpⁿ−1). Then it is easy to see that

(i) ωn∈Wπ⁽ⁿ⁾, (ii) fπ(ωn)=ω_n−1,

asfπ(ωn)=fπη^ϕ−n(ζpⁿ−1)=η^ϕ−n+1◦[p](ζpⁿ−1)=η^ϕ−n+1(ζ_pn−1−1)=ω_n−1. LetTπ=lim←Wπ⁽ⁿ⁾ be the Tate module, where the inverse limit is taken with respect tofπ. Letκ: Gal(K_∞/Qp)→Z^×p be the character given by the action ofG_Qp onTπ. If the formal group isGm, this character is justχ, the cyclotomic character. We know that κ=χψ, whereψis an unramified character.

Assumeβ∈ᐁ, then Coleman’s theorem tells us that there is a unique (Coleman) power seriesgβ∈Zp[[T ]]such that

(i) gβ(ωi)=βi,for alli≥1, (ii) g^ϕ_β◦fπ(x)=

w∈Wπ¹gβ(x[+]w).

To get a rough idea for what Coleman power series is, we look at some examples.

Consider the Gm case, let βn = 2 +ω⁻¹(2)(ζpⁿ−1), where ω is the Teichmüller character. Then we havegβ(T )= 2 +ω⁻¹(2)T. If we takeβn=(ζ_p^an−1)/(ζ_p^bn−1), thengβ(T )=((1+T )^a−1)/((1+T )^b−1).

Assumeβ∈ᐁ^{such thatβ}n≡1(modωn). Thengβ(T )≡1 mod(p, T ), wherepis the maximal ideal inZp, hence we can define

logg β(T ):=loggβ(T )−1 p

w∈W_π¹

loggβ

T [+]w

, (1.1)

property (ii) of the Coleman power series implies thatlogg β(T )has integral coefficients.

Define an algebraic distributionµβ∈Ᏸ⁺alg(Zp,Qp^ur)such that

Zp

(1+T )^xµβ(x)=loggβ◦η(T ). (1.2) Proposition1.1. (i)The restrictionµβ|_Z^×_pis a measure and its Amice transformation islogg β◦η(T ).

(ii)The distributionµβcan be extended to a distribution inᏰ1(Qp,Q^urp )^Φ=1and has the following Galois property:

σ

Qp

f (x)µβ =

Qp

f ψ(σ )x

µβ, ∀σ , (1.3)

for allf (x):Qp→Qp.

Proof. It is easy to see that

Z^×p

(1+T )^xµβ=

Zp

(1+T )^xµβ−

pZp

(1+T )^xµβ. (1.4)

By property (ii),

gβ◦fπ(X)=

w∈Wπ¹

gβ

X[+]w

. (1.5)

PutX=η(T ), then gβ◦fπ

η(T )

=

ζ∈µp

gβ

η(T )[+]η(ζ−1)

=

ζ∈µp

gβη

ζ(1+T )−1

. (1.6)

By usingfπ◦η=η^ϕ◦[p], we see gβ◦ηϕ

◦[p]=

ζ

gβη

ζ(1+T )−1

. (1.7)

Taking logarithm and using the definition ofµβ, we have ϕ

Zp

1+[p]Tx

µβ =

ζ

Zp

ζ^x(1+T )^xµβ=p

pZp

(1+T )^xµβ. (1.8)

We write down this useful property as follows.

Proposition1.2. Forµβas above,

ϕ

Zp

(1+T )^pxµβ =p

pZp

(1+T )^xµβ. (1.9)

Now continue the proof ofProposition 1.1.

The integral

Z^×p

(1+T )^xµβ=loggβη(T )−1 pϕ

loggβ◦η^ϕ◦ [p]T

=loggβ◦η−1

ploggβ◦fπ◦η(T )

=logg β◦η(T )

(1.10)

has integral coefficients, henceµβ|Z^×p is a measure. ByProposition 1.2we extendµβto Qpby defining

p⁻ⁿZp

f (x)µβ=pⁿϕ⁻ⁿ

Zp

f p⁻ⁿx

µβ, (1.11)

for anyf (x)with support inp⁻ⁿZp. It is easy to see that this definition does not depend onn.

To prove the second property, since

η(T ):Gm →Ᏺπ, (1.12)

we can show that σ

η(T )

=η

(1+T )^{ψ(σ )}−1

, ∀σ∈G_Qp. (1.13)

To see this, define

hσ(T ):=σ η(T )

−η

(1+T )^{ψ(σ )}−1

. (1.14)

ForTn=ζpⁿ−1,η(Tn)∈Wπ⁽ⁿ⁾, (σ η)

σ Tn

=σ η

Tn

= κ(σ )

πη Tn

=η κ(σ )

Tn

=η

ζ_p^{κ(σ )}n −1

=η 1+σ Tn

ψ(σ )

−1

, (1.15)

hencehσ(σ Tn)=0 for alln≥1. The function hσ(T )has infinitely many zeros by Weierstrass lemma, thenhσ(T )=0.

From this property, we see that σ

Zp

(1+T )^xµβ =σ

loggβ◦η(T )

=loggβ◦σ η(T )

=loggβ◦η

(1+T )^{ψ(σ )}−1

=

Zp

(1+T )^{ψ(σ )x}µβ,

(1.16)

so for generalf, we have

σ

Zp

f (x)µβ =

Zp

f ψ(σ )x

µβ. (1.17)

From the extension definition (1.11) ofµβ, we have, for allf,

σ

Qp

f (x)µβ =

Qp

f

ψ(σ )x

µβ. (1.18)

To show thatµβis 1-admissible, by definition and [8, Proposition 3.3], we only need to show thatp^n(1−j)

a+pⁿZp(x−a)^jµβ isr-bounded forj=0,1. Forj=0, ifa≠0, then sinceµβ|Z^×p is a measure, the integralpⁿ

a+pⁿZpµβis always bounded. If a=0, then pⁿ

pⁿZpµβ=ϕⁿ(

Zpµβ)=ϕⁿloggβ(0)=loggβ(0), hence bounded.

Forj=1, ifa≠0, then

a+pⁿZpxµβis bounded. Ifa=0, then

pⁿZpxµβ=ϕⁿ(

Zpxµβ)= ϕⁿ(Ω·(g_β(0)/gβ(0)))=αⁿΩ(g_β(0)/gβ(0)), hence bounded.

Next, we will restrict to theGmcase.µβis an example which is a distribution onQp

but not a measure. We will show an example of distribution which is not a tempered distribution.µβcan be extended to negative power. Fork >0, define

Zp

x^−kµβ=

1−p^−k−1₋1

Z^×p

x^−kµβ, (1.19)

defineνβ∈Ᏸ⁺alg(Zp, K_∞,cyc)such that

a+pⁿZp

x^kνβ:=p^nkk!

Zp

ε ax

pⁿ µβ

x^k, k≥0. (1.20)

The relation betweenµβandνβis that

νβ⊗e t=Ᏺalg

µβ⊗e

t . (1.21)

Lemma 1.3. The distribution νβ is a distribution over Zp, but it is not a tempered distribution ifloggβ(T )≡logg β(0)(mod(p, T^p−1)).

Proof. The additivity ofνβfollows from the relation

p−1 i=0

ε a+ipⁿ x pⁿ⁺¹

=





 pε

ax

pⁿ⁺¹ , x≡0

0, x≡0

(modp). (1.22)

Assumelogg β(T )=

aiTⁱ, sinceloggβ≡0 mod(p, T^p−1), hence there existsi,1≤i≤ p−1, such thatai≡0(modp). Then

1+pⁿZpνβ=logβnhas denominator with valuation n−1−i/(p−1), soνβis not a measure.

Ifνβisr-admissible, assumer∈N, then for allX,

sup

a∈X

p^[n(r−j)]

a+pⁿZp

(x−a)^jνβ

(1.23)

is r-bounded. Takingj=r, we would have that

a+pⁿZp(x−a)^rνβ is bounded when n→ ∞.

We will calculate the valuation of

a+pⁿZp(x−a)^rνβ. Fora=1,

a+pⁿZp

(x−1)^rνβ= r k=0

r k

·(−1)^r−k

1+pⁿZp

x^kνβ= r k=0

dk. (1.24)

Claim1.4. v(dk) > i/(p−1)−(n−1)=v(d0).

To prove the claim, keeping using the relation

pZp

ε x

p^m µβ

x^k= 1 p^k+1

Zp

ε x

p^m−1 µβ

x^k (1.25)

and the decomposition

Zp=Z^×p∪pZ^×p∪···∪pⁿ⁻¹Z^×p∪pⁿZp, (1.26) we have

1+pⁿZp

x^kνβ

=p^kn·k!

Zp

ε x

pⁿ µβ

x^k

=p^kn·k!

_n−1

i=0

pⁱZ^×p

ε x

pⁿ µβ

x^k+

pⁿZp

ε x

pⁿ µβ

x^k

=p^kn·k!

_n−1

i=0

1 p^i(k+1)

Z^×p

ε x

pⁿ⁻ⁱ µβ

x^k+ 1 p^n(k+1)

Zp

µβ

x^k

=p^kn·k!

n−1 i=0

1 p^i(k+1)

Z^×p

ε x

pⁿ⁻ⁱ µβ

x^k+ 1 p^n(k+1)

1−p^−k−1₋1

Z^×p

µβ

x^k

.

(1.27)

The last two terms ford0equal 1

pⁿ⁻¹

Z^×p

ε x

p µβ+p−1 pⁿ⁻¹

Z^×p

µβ= 1

pⁿ⁻¹loggβ

ζ1−1

+(p−1)loggβ(0)

. (1.28)

Assumeloggβ(T )=a0+a1T+ ···. From the hypothesis we know that there exists a minimaliwith 1≤i < p−1 such thatai≡0 modp, hence the above expression equals

1 pⁿ⁻¹

pa0+a1

ζ1−1

+···+ai

ζ1−1i

+···

, (1.29)

and therefore v(d0)=i/(p−1)−(n−1). For k >0, we easily see that v(dk)≥k− (n−1) > v(d0), hence the claim follows. The sequence

1+pⁿZp(x−1)^rνβcould not be bounded, hence this completes the proof of the lemma.

Ifr∉ R, takingj=[r ], thenp^{[n(r−[r ])]}

1+pⁿZp(x−1)^{[r ]}νβcannot tend to zero.

2. Galois action onBdR. LetFbe a one-dimensional height-one formal group over Zp, letπ=pαbe the uniformizer withα∈Z^×p, and letf be a Frobenius power series corresponding toF, then we havef (x)≡π x(mod deg 2), f (x)≡x^p(modp). Letη(x): Gm→Fwhich is an isomorphism, such thatη(x)=Ωx+···, whereΩis ap-adic unit with Ω^ϕ−1=α. Letωn=η^ϕ−n(ζpⁿ−1),Kn=Qp(ωn), andK_∞= ∪n≥0Kn, letTπ be the Tate module, and letκ: Gal(K_∞/Qp)→Z^×pbe the character given by the action on Tπ, then we know thatκ=χψwithψan unramified character such thatψ(Frobp)=α andσ (ωn)=[κ(σ )]ωn, where fora∈Zp, [a]denotes the unique endomorphism of Fsuch that[a]=aX+ ··· (see [4, Section 20.1]). LetΞbe the completion ofQ^urp . Let tπ=Ωt. Thentπhas the property thatσ (tπ)=κ(σ )tπ, ϕ(tπ)=π tπ.

Ifx∈K_∞andn∈N, we defineTn(x):=(1/p^m)TrK_m/K_n(x)form1. We extend Tn to a map fromK_∞((tπ))toKn((tπ))byTn(

akt_π^k)=

Tn(ak)t_π^k. Note that this definition does not depend on the choice oftπ since the different choice only differs a multiple inQpandTnisQp-linear. We also define Tr/Kn=(1/[Km:Kn])TrKm/Kn(x) form1.

Recall that᏾is the projective limit of the following diagram:

O¯← O¯← ···. (2.1)

Forx∈᏾,x=(xn)_n∈N, x_n^p₊₁=xn. Choosexn∈O_Cpsuch thatxn≡xn(modp).

Lemma2.1. The limitlim_n→∞f^(m)(x_n+m)exists and does not depend on the choice of

xn. Denote it byx^(m)and then we havef (x^(m))=x^(m−1). Proof. We first prove thatfhas the property, fory∈᏾,

f⁽ⁿ⁾(x+py)≡f⁽ⁿ⁾(x)

modpⁿ⁺¹y

. (2.2)

This is true forn=1 from the definition off. Assume it is true forn, then we have f⁽ⁿ⁾(x+py)=f⁽ⁿ⁾(x)+pⁿ⁺¹yzfor somez∈᏾^{, hence}

f⁽ⁿ⁺¹⁾(x+py)=f

f⁽ⁿ⁾(x)+pⁿ⁺¹yz

=f⁽ⁿ⁺¹⁾(x)

modp·pⁿ⁺¹yz

=f⁽ⁿ⁺¹⁾(x)

modpⁿ⁺²y .

(2.3)

From this we see that f⁽ⁿ⁾

x_n+m

−f⁽ⁿ⁺¹⁾ x_n+m+1

=f⁽ⁿ⁾ x_n+m

−f⁽ⁿ⁺¹⁾

x_n+m+py

≡0

modpⁿ⁺¹

, (2.4)

hence the lemma follows.

On the other hand, if(x^(m))satisfies the relationf (x^(m))=x^(m−1), then(¯x^(m))∈᏾^, hence᏾is one-to-one corresponding to

x^(m)

m∈Nf x^(m)

=x^(m−1)

. (2.5)

LetW= {w∈Cp| ∃n,s.t.[π ]ⁿw=0}. Forw∈W, ifn∈Nsuch that[π ]ⁿw≠0, [π ]ⁿ⁺¹w=0, then we say thatwhas ordern. Tow∈W, we can associate an element

¯

w=(w, f⁽⁻¹⁾(w), f⁽⁻²⁾(w), . . . ). The association is not unique.

Lemma2.2. All elements ofC^GpK∞ can be written in the formx=

w∈Waw(x)w with aw(x)∈Qptending to zero when the order ofwtends to infinity.

Proof. SinceK_∞=Q¯^Gp^K∞,W is a base ofOK_∞/ZpandC^Gp^K∞ is a separated comple- tion ofK_∞with respect top-adic topology, hence the proof follows from Tate-Sen-Ax theorem.

Proposition2.3. (i)The fieldK_∞((tπ))is dense in(B_dR⁺ )^GK∞ andTncan be extended to a continuousQp-linear map from(B_dR⁺ )^GK∞ toKn((tπ)).

(ii)IfF∈(B_dR⁺ )^GK∞, thenlim_n→∞pⁿTn(F )=F.

Proof. (i) For allx∈(B_dR⁺ )^GK∞,θ(x)∈C^GpK∞, henceθ(x)=

w∈Waw(x)wfor some aw(x)∈QpfromLemma 2.2. Let

R(x)=t⁻¹_π

x−

w∈W

aw(x)[w]¯

∈ B⁺_dRG_K∞

, (2.6)

so we can repeat the above process and get x=t_π^k+1R^k+1(x)−

k i=0

t_πⁱ

w

aw

R⁽ⁱ⁾(x) [w]¯

, (2.7)

and this shows thatK_∞((tπ))is dense in(B⁺_dR)^GK∞.

(ii) ForF∈K_∞,F∈Km₀for somem0, hence forn≥m0, Tn(F )=(1/p^m)TrK_m/K_n(F )= (1/pⁿ)F. Hence, limn→+∞pⁿTn(F )=F. By the definition ofTnonK_∞((tπ))we have for F∈K_∞((tπ)), lim_n→+∞pⁿTn(F )=F.

ForF∈(B_dR⁺ )^GK∞, (i) shows that we can takeFk∈K_∞((tπ))such that lim_k→+∞Fk=F. Hence

nlim→+∞pⁿTn(F )= lim

n→+∞pⁿTn

k→+∞lim Fk = lim

k→+∞ lim

n→+∞pⁿTn Fk

= lim

k→+∞Fk=F . (2.8) We can change the order of the limit sincepⁿTnis continuous.

Recall that LA denotes the space of locally analytic functions with compact sup- port in Qp taking values inQp. For a p-adic Banach space A, defineᏰcont(Qp, A)= Homcont(LA, A)with respect to Morita topology. Define

Ᏸcont Qp,Ξψ

:=

µ∈Ᏸcont Qp,Ξ

|σ

Qp

f (x)µ

=

Qp

f

ψ(σ )x

µ,∀σ∈G_Qp

. (2.9)

Forµ∈Ᏸcont(Qp,Ξ)^ψwith compact support, we can define an element inBdRas Ᏺcont(µ)=

Qp

ε^x

µ. (2.10)

Colmez called this element as the continuous Fourier transformation ofµand we con- tinue using his notation. Recall from [8, Section 4] thatµ∈Ᏸcont(Qp,Ξ)is said to be of orderr∈R¯⁺if for all open compact setX⊂Qpand allj≥0, the following sequence isr-bounded:

supa∈X

p^E(n(r−j))

a+pⁿZp

(x−a)^jµ

. (2.11)

DefineB⁺_cont:= ∩n≥0ϕⁿ(B⁺_max). An elementF ∈B⁺_cont is said to be of orderr if the sequencep^{[nr ]}ϕ⁻ⁿF isr-bounded. The power seriesA(T )=

anTⁿ∈Ξ[[T ]]is said to be of orderr ifn^−r|an|isr-bounded. DefineΞ[[T ]]^ψ:= {h(T )∈Ξ[[T ]]such that for allσ∈G_Qp, σ (h(T ))=h((1+T )^{ψ(σ )}−1)}.

Proposition2.4. (i)Forµ∈Ᏸcont(Qp,Ξ)^ψ,Ᏺcont(µ)∈(B⁺_cont)^GK∞.

(ii)Forµ∈Ᏸcont(Zp,Ξ)^ψ, the Amice transformationᏭµ(T )is inΞ[[T ]]^ψ,µhas order rif and only ifᏭµhas.

(iii)Forµ∈Ᏸcont(Zp,Ξ)^ψ, ifµhas orderr, thenᏲcont(µ)has orderr.

(iv)For a crystalline representationV, forµ∈Ᏸcont(Zp,Ξ⊗D(V ))^ψ, if µ has order r, thenᏲcont(µ)has orderr+r (V ), wherer (V )=min{k∈Z|ϕ(d)=p^kdfor some d∈D(V )}.

Proof. (i) Forσ∈G_Qpandµ∈Ᏸcont(Qp,Ξ)^ψ,

σ

Ᏺcont(µ)

=σ

Qp

ε^x µ =

Qp

εψ(σ )xχ(σ ) µ=

Qp

ε^{κ(σ )x}

µ. (2.12)

Ifκ(σ )=1, thenσ (Ᏺcont(µ))=Ᏺcont(µ), soᏲcont(µ)∈(B⁺_max)^GK∞. On the other hand, letFn=

Qp[ε_n^α−nx]µ; we have ϕ

Fn

=ϕⁿ

Qp

ε^α_n^−nx µ =

Qp

ε^x

µ (2.13)

and this shows thatᏲcont(µ)∈(B⁺_cont)^GK∞. (ii) The Galois action gives

σ Ꮽµ(T )

=σ

Zp

(1+T )^xµ=

Zp

(1+T )^{ψ(σ )x}µ=Ꮽµ

(1+T )^{ψ(σ )}−1

, (2.14) henceᏭµ(T )∈Ξ[[T ]]^ψ.

The second statement can be found in [1].

(iii) Assumeµhas orderr, henceᏭµ(T )has orderrby (ii).

Case1. Assumer∈N, then there exists a constantC >0 such that|ak| ≤Ck^r, that is,−vp(ak)≤(logC+rlogk)/logp (here the log is the logarithmic function on real variable). The real functionf (x, a, r )=ax−r (logx/logp)has a minimumg(a, r )at

x=r /alogp. Forx∈R>0, we have

vp

ak

+r n+ k

(p−1)pⁿ⁻¹≥r n+ k

(p−1)pⁿ⁻¹−rlogk logp−logC

logp

=f k

pⁿ, p

p−1, r −logC logp

≥g p

p−1, r −logC logp.

(2.15)

Takem≤ −1+g(p/(p−1), r )−logC/logpand recall that

Fn=

Qp

ε^α_n^−nx µ=

ak

ε_n^α−n

−1k

, Fn=sup

k

p^{−(vp(ak)+kvp([εnα−n]−1))}=sup_kp^{−(vp(ak)+k/(p−1)pn−1)};

(2.16)

we get

p^nrFn=sup

k

p^{−(nr+vp(ak)+k/(p−1)pn−1)}≤p^−m, (2.17)

henceᏲcont(µ)isr-bounded.

Case2. Forr∈R¯⁺\R⁺, then we have lim_n→+∞n^−r|an| =0. Hence, we can choose a sequenceck→0 such that|ak|< ckk^r, then the above proof shows that we can take

mk= −1+g p

p−1, r −logck

logp

!

, (2.18)

thenp^{[nr ]}Fn ≤p^−mn hence tends to zero.

(iv) Assumed1, . . . , dkare a base ofD(V )andµ∈Ᏸ(Qp,Ξ⊗D(V ))^ψ, thenᏲcont(µ)= bi⊗di for some bi ∈B⁺cont with order r, p^−nr|ϕ⁻ⁿ(bi) ≤cn for somecn (cn is bounded if r ∈ R, otherwise cn → 0), so p^{−n(r+r (V ))}|ϕ⁻ⁿ(bi⊗di)| ≤ cn|di|, hence Ᏺcont(µ)is(r+r (v))-bounded.

Remark2.5. Properties (iii) and (iv) are even true forµhas support inQp. To prove this we only need to extend (ii) to this case.

Lemma2.6. (i)Form≥n≥1,

σ∈Gal(Km/Kn)

ε κ(σ )x

=





p^m−nε(x), ifx∈p⁻ⁿZp,

0, otherwise. (2.19)

(ii)Form >0,

σ∈Gal(Km/K₀)

ε κ(σ )x

=p^m1_Zp−p^m−11_p−1Zp. (2.20)

Proof. (i) Ifx∈p⁻ⁿZp, then

σ∈G(K_m/K_n)

ε κ(σ )x

=

p^m−n−1 a=0

ε

1+a·pⁿ x

=p^m−nε(x). (2.21)

Otherwise,

σ∈G(K_m/K_n)

ε κ(σ )x

=ε(x)

p^m−n−1 a=0

ε ab

p^k =0, (2.22)

with someb≠0 ap-unit,k≥1.

(ii) Ifx∈Zp, then

σ∈G(Km/K₀)

ε κ(σ )x

=p^m−p^m−1. (2.23)

Ifx∈p⁻¹Zp\Zp, then

σ∈G(Km/K₀)

ε κ(σ )x

=p^m−1

p−1

a=1

ε a

p =p^m−1·(−1)= −p^m−1. (2.24)

And it is easy to see that ifx∉p⁻¹Zp, then

σ∈G(Km/K₀)

ε κ(σ )x

=0. (2.25)

So the sum equals 1_Zp·(p^m−p^m−1)−1_p−1Zp\Zpp^m−1=p^m1_Zp−p^m−11_p−1Zp. Proposition2.7. Ifµ∈Ᏸcont(Qp,Ξ)^ψ, then

Tn

Ᏺcont(µ)

=

+∞

k=0

t^k

p⁻ⁿ

p−nZp

ε(x)x^k k!µ

, ifn≥1,

T0

Ᏺcont(µ)

=

+∞

k=0

t^k

Zp

x^k k!µ−p⁻¹

p⁻¹Zp

x^k k!µ

.

(2.26)

Proof. The Fourier transformation gives

Ᏺcont(µ)=

Qp

ε^x µ=

Qp

ε(x)exp(tx)µ

=

Qp

ε(x) ∞ k=0

(tx)^k k! µ=

∞ k=0

t^k_π

Qp

ε(x) x^k Ω^k·k!µ.

(2.27)

To prove the first identity, we need to show that Tn

Qp

ε(x) x^k Ω^k·k!µ

=p⁻ⁿ

p⁻ⁿZp

ε(x) x^k

Ω^k·k!µ (2.28) (note that the right-hand side is indeed inKn).

Now, assumeµhas compact supportp^−mZpfor somem, then Tn

p−mZp

ε(x)x^k

Ω^kµ = 1 p^m

σ∈G(K_m/K_n)

σ

p−mZp

ε(x)x^k Ω^kµ

= 1 p^m

σ

p^−mZp

ε

κ(σ )xx^k Ω^kµ

=p⁻ⁿ

p⁻ⁿZp

ε(x)x^k Ω^kµ.

(2.29)

The same proof works for the second formula.

The Galois action onBdRhas the following property.

Lemma 2.8. Suppose V is a p-adic representation ofGK_∞, then (BdR⊗V )^GK∞ is a finite-dimensional vector space overB_dR^GK∞ with dimensiondim_QpVand can have a base consisting of elements in(Bmax^ϕ=1⊗V )^GK∞.

Proof. By an argument similar to [2, Corollary B.14], we can see that(B_dR⁺ ⊗V )^GK∞ is a finite-dimensional vector space over(B⁺_dR)^GK∞ and we can have a basev1, . . . , vdsuch thatv1, . . . , vd∈(W (᏾)⊗V )^GK∞. Assumee1, . . . , edare a base ofV /Qp, and(v1, . . . , vd)= (e1, . . . , ed)AwithA∈GLd((B⁺_dR⊗V )^GK∞), thenθ(det(A))≠0.

Fork≥1,1≤i≤d, letvi,k=

m∈Zπ^−kmϕ^m−1(η([ε]−1)^k+1vi), then this is a con- vergence sum and it converges to an element in (B⁺_crys⊗V )^GK∞. Consider that v1,k/ ϕ⁻¹(η([ε]−1))^k+1tends toϕ⁻¹(vi)whenk→ ∞, hence whenk1,vi,k/ϕ⁻¹(η([ε]

−1))^k+1, . . . , vd,k/ϕ⁻¹(η([ε]−1))^k+1 are linearly independent, hence v1,k, . . . , vd,k are linearly independent.

Fromϕ(t_π^−kvi,k)=(t_π^−kvi,k)we see that t_π^−kv1,k, . . . , t^−k_π vd,k∈(Bmax^ϕ=1⊗V )^GK∞ are a base of(BdR⊗V )^GK∞ overB^G_dR^K∞.

Lemma2.9. H¹(K_∞, BdR⊗V )=0andH¹(K_∞,Ᏸr(Z^×p, Bmax^ϕ=1⊗V ))=0.

Proof. Assumeτ→cτ is a cocycle fromGK_∞ →V. From [2] we know thatH¹(K_∞, W (m᏾)⊗T )=0, whereT⊂Vis a Galois invariant lattice. Letω=(ω0, ω1, ω2, . . .)∈᏾^, then[ω]cτ∈H¹(K_∞, W (m᏾)⊗T )=0, so we can find ac∈W (᏾)⊗Vsuch that[ω]cτ= (1−τ)c. From the ramification property, we can show that[ω]^p−1≡0(modp)inAcrys, hencea=

n∈Z(ϕⁿ([ω])/pⁿ)is convergent inA^Gcrys^K∞ andϕ(a)=pa. Let c= 1

pa ϕⁿ

pⁿ

[ω]⁻¹ ϕ[ω]

c

, (2.30)

then it is easy to see that(1−τ)c=cτ. Sincec∈(Bmax^ϕ=1⊗V ), this shows that the inclu- sion maph¹(K_∞, V )→H¹(K_∞,Fil⁻¹(Bmax^ϕ=1⊗V ))is the zero map. By similar arguments as