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)≡xp(modp). LetFbe the one-dimensional Lubin-Tate formal group overZpcorresponding tofπ, let[+]denote the formal addition. LetWn
π:= {x∈Cp|fπ(n)(x)=0}, Kn=Qp(Wπn), 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←Kn×, where the map is with respect to the norm map. Letη:Gm→Ᏺπ be an isomorphism, then η∈Qur
p [[T ]]andη(T )=ΩT+···, such thatΩϕ−1=α, whereϕ=Frobpis a generator of Gal(Qur
p /Qp). We havef◦η=ηϕ◦[p], where[p]=(1+T )p−1 is the Frobenius of the formal groupGm. Letωn=ηϕ
−n
(ζpn−1). Then it is easy to see that (i) ωn∈Wπ(n),
(ii) fπ(ωn)=ωn−1,
asfπ(ωn)=fπηϕ −n
(ζpn−1)=ηϕ−n+1◦[p](ζpn−1)=ηϕ−n+1(ζpn−1−1)=ωn−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 ofGQp 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) gϕβ◦fπ(x)=
w∈Wπ1gβ(x[+]w).
Assumeβ∈ᐁsuch thatβn≡1(modωn). Thengβ(T )≡1 mod(p, T ), wherepis the maximal ideal inZp, hence we can define
loggβ(T ):=loggβ(T )−1 p
w∈Wπ1
loggβT [+]w, (1.1)
property (ii) of the Coleman power series implies thatloggβ(T )has integral coefficients. Define an algebraic distributionµβ∈Ᏸ+alg(Zp,Qpur)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
isloggβ◦η(T ).
(ii)The distributionµβcan be extended to a distribution inᏰ1(Qp,Qurp )Φ=1and has
the following Galois property:
σ
Qp
f (x)µβ =
Qp
fψ(σ )xµβ, ∀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◦fπη(T )=
ζ∈µp
gβη(T )[+]η(ζ−1)=
ζ∈µp
gβηζ(1+T )−1. (1.6)
By usingfπ◦η=ηϕ◦[p], we see
gβ◦η
ϕ
◦[p]=
ζ
gβη
ζ(1+T )−1. (1.7)
Taking logarithm and using the definition ofµβ, we have
ϕ
Zp
1+[p]Txµβ =
ζ
Zp
ζx(1+T )xµβ=p
pZp
(1+T )xµβ. (1.8)
Proposition1.2. Forµβas above,
ϕ
Zp
(1+T )pxµ β =p
pZp
(1+T )xµ
β. (1.9)
Now continue the proof ofProposition 1.1. The integral
Z×p
(1+T )xµβ=loggβη(T )−1 pϕ
loggβ◦ηϕ◦[p]T
=loggβ◦η− 1
ploggβ◦fπ◦η(T ) =loggβ◦η(T )
(1.10)
has integral coefficients, henceµβ|Z×
p is a measure. ByProposition 1.2we extendµβto
Qpby defining
p−nZpf (x)µβ=p nϕ−n
Zp
fp−nxµ
β, (1.11)
for anyf (x)with support inp−nZ
p. 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
ση(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=σηTn=κ(σ )πη
Tn=ηκ(σ )Tn
=ηζpκ(σ )n −1
=η1+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 =σloggβ◦η(T )=loggβ◦ση(T )
=loggβ◦η(1+T )ψ(σ )−1=
Zp
(1+T )ψ(σ )xµβ,
so for generalf, we have
σ
Zp
f (x)µβ =
Zp
fψ(σ )xµβ. (1.17)
From the extension definition (1.11) ofµβ, we have, for allf,
σ
Qp
f (x)µβ =
Qp
fψ(σ )xµβ. (1.18)
To show thatµβis 1-admissible, by definition and [8, Proposition 3.3], we only need to show thatpn(1−j)
a+pnZp(x−a)jµβ isr-bounded forj=0,1. Forj=0, ifa≠0, then sinceµβ|Z×
p is a measure, the integralp n
a+pnZpµβis always bounded. If a=0, then pn
pnZpµβ=ϕn(
Zpµβ)=ϕ nlogg
β(0)=loggβ(0), hence bounded. Forj=1, ifa≠0, thena+pnZpxµβis bounded. Ifa=0, then
pnZpxµβ=ϕn(
Zpxµβ)=
ϕn(Ω·(g
β(0)/gβ(0)))=αnΩ(gβ(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+pnZpx kν
β:=pnkk!
Zp
ε ax
pn µβ
xk, 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 )≡loggβ(0)(mod(p, Tp−1)).
Proof. The additivity ofνβfollows from the relation
p−1
i=0
ε a+ip nx pn+1
=
pε ax
pn+1 , x≡0
0, x≡0
(modp). (1.22)
Assumeloggβ(T )=
aiTi, sinceloggβ≡0 mod(p, Tp−1), hence there existsi,1≤i≤ p−1, such thatai≡0(modp). Then
Ifνβisr-admissible, assumer∈N, then for allX,
sup a∈X
p[n(r−j)]
a+pnZp(x−a) jν
β
(1.23)
is r-bounded. Takingj=r, we would have thata+pnZp(x−a)rνβ is bounded when n→ ∞.
We will calculate the valuation ofa+pnZp(x−a)rνβ. Fora=1,
a+pnZp(x−1) rν
β= r
k=0
r k
·(−1)r−k
1+pnZpx kν
β= 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
pm µβ xk=
1
pk+1
Zp
ε x
pm−1
µβ
xk (1.25)
and the decomposition
Zp=Z×p∪pZ×p∪···∪pn−1Z×p∪pnZp, (1.26)
we have
1+pnZpx kν
β
=pkn·k!
Zp ε x pn µβ xk
=pkn·k!
n−1
i=0
piZ×pε
x
pn µβ xk+
pnZpε
x
pn µβ xk
=pkn·k!
n−1
i=0 1
pi(k+1)
Z× p
ε x
pn−i µβ xk+
1
pn(k+1)
Zp
µβ xk
=pkn·k!
n−1
i=0 1
pi(k+1)
Z×p
ε x
pn−i µβ xk+
1
pn(k+1)
1−p−k−1−1
Z×p
µβ xk
.
(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
ζ1−1
i
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)≡xp(modp). Letη(x):
Gm→Fwhich is an isomorphism, such thatη(x)=Ωx+···, whereΩis ap-adic unit with Ωϕ−1=α. Letω
n=ηϕ− 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 ofQur
p . Let tπ=Ωt. Thentπhas the property thatσ (tπ)=κ(σ )tπ, ϕ(tπ)=π tπ.
Ifx∈K∞andn∈N, we defineTn(x):=(1/pm)TrKm/Kn(x)form1. We extend
Tn to a map fromK∞((tπ))toKn((tπ))byTn(aktπk)=
Tn(ak)tπk. 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 thatis the projective limit of the following diagram:
¯
O← O¯← ···. (2.1)
Forx∈,x=(xn)n∈N, xnp+1=xn. Choosexn∈OCpsuch thatxn≡xn(modp).
Lemma2.1. The limitlimn→∞f(m)(xn+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)=ff(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
≡0modpn+1, (2.4)
On the other hand, if(x(m))satisfies the relationf (x(m))=x(m−1), then(x¯(m))∈, henceis 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 ofCGK∞
p can be written in the formx= w∈W
aw(x)w with
aw(x)∈Qptending to zero when the order ofwtends to infinity.
Proof. SinceK∞=Q¯GK∞
p ,W is a base ofOK∞/ZpandC GK∞
p 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(BdR+ )GK∞ andTncan be extended
to a continuousQp-linear map from(BdR+ )GK∞ toKn((tπ)). (ii)IfF∈(BdR+ )GK∞, thenlimn→∞pnTn(F )=F.
Proof. (i) For allx∈(BdR+ )GK∞,θ(x)∈CGK∞
p , 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+dRGK∞, (2.6)
so we can repeat the above process and get
x=tk+1
π Rk+1(x)− k
i=0
ti π
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/pm)TrKm/Kn(F )=
(1/pn)F. Hence, lim
n→+∞pnTn(F )=F. By the definition ofTnonK∞((tπ))we have for F∈K∞((tπ)), limn→+∞pnTn(F )=F.
ForF∈(BdR+ )GK∞, (i) shows that we can takeFk∈K∞((tπ))such that limk→+∞Fk=F. Hence
lim n→+∞p
nT
n(F )= lim n→+∞p
nT n
lim
k→+∞Fk =klim→+∞nlim→+∞p
nT
nFk= lim
k→+∞Fk=F . (2.8)
We can change the order of the limit sincepnT
nis 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
ᏰcontQp,Ξψ:=
µ∈ᏰcontQp,Ξ|σ
Qp
f (x)µ
=
Qp
fψ(σ )xµ,∀σ∈GQp
.
Forµ∈Ᏸcont(Qp,Ξ)ψwith compact support, we can define an element inBdRas
Ᏺcont(µ)=
Qp
εxµ. (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:
sup a∈X
pE(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 )=a
nTn∈Ξ[[T ]]is said to be of orderr ifn−r|a
n|isr-bounded. DefineΞ[[T ]]ψ:= {h(T )∈Ξ[[T ]]such that for allσ∈GQp, 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)=pkdfor some
d∈D(V )}.
Proof. (i) Forσ∈GQpandµ∈Ᏸcont(Qp,Ξ)ψ,
σᏲcont(µ)
=σ
Qp
εxµ =
Qp
εψ(σ )xχ(σ )µ=
Qp
εκ(σ )xµ. (2.12)
Ifκ(σ )=1, thenσ (Ᏺcont(µ))=Ᏺcont(µ), soᏲcont(µ)∈(B+max)GK∞. On the other hand, letFn=
Qp[ε α−nx
n ]µ; we have
ϕFn
=ϕn
Qp
εαn−nx
µ =
Qp
εxµ (2.13)
and this shows thatᏲcont(µ)∈(B+cont)GK∞. (ii) The Galois action gives
σᏭµ(T )
=σ
Zp
(1+T )xµ=
Zp
(1+T )ψ(σ )xµ=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).
x=r /alogp. Forx∈R>0, we have
vp
ak
+r n+(p−1k)pn−1≥r n+
k
(p−1)pn−1−r logk
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
εα−nx
n
µ=ak
εα−n
n
−1k, Fn=sup
k
p−(vp(ak)+kvp([εα− n
n ]−1))=sup
kp−(vp(ak)+k/(p−1)p n−1)
;
(2.16)
we get
pnrFn=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 limn→+∞n−r|a
n| =0. Hence, we can choose a sequenceck→0 such that|ak|< ckkr, then the above proof shows that we can take
mk= −1+g
p
p−1, r − logck
logp !
, (2.18)
thenp[nr ]F
n ≤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)
εκ(σ )x=
pm−nε(x), ifx∈p−nZ p,
0, otherwise. (2.19)
(ii)Form >0,
σ∈Gal(Km/K0)
εκ(σ )x=pm1 Zp−p
m−11
Proof. (i) Ifx∈p−nZ p, then
σ∈G(Km/Kn)
εκ(σ )x=
pm−n−1
a=0
ε1+a·pnx=pm−nε(x). (2.21)
Otherwise,
σ∈G(Km/Kn)
εκ(σ )x=ε(x) pm−n−1
a=0
ε ab
pk =0, (2.22)
with someb≠0 ap-unit,k≥1. (ii) Ifx∈Zp, then
σ∈G(Km/K0)
εκ(σ )x=pm−pm−1. (2.23)
Ifx∈p−1Z
p\Zp, then
σ∈G(Km/K0)
εκ(σ )x=pm−1 p−1
a=1
ε a
p =p
m−1·(−1)= −pm−1. (2.24)
And it is easy to see that ifx∉p−1Z p, then
σ∈G(Km/K0)
εκ(σ )x=0. (2.25)
So the sum equals 1Zp·(pm−pm−1)−1p−1Zp\Zppm−1=pm1Zp−pm−11p−1Zp.
Proposition2.7. Ifµ∈Ᏸcont(Qp,Ξ)ψ, then
Tn
Ᏺcont(µ)
=
+∞
k=0
tk
p−n
p−nZpε(x)
xk k!µ
, ifn≥1,
T0
Ᏺcont(µ)
=
+∞
k=0
tk
Zp
xk k!µ−p
−1
p−1Zp
xk k!µ
.
(2.26)
Proof. The Fourier transformation gives
Ᏺcont(µ)=
Qp
εxµ=
Qp
ε(x)exp(tx)µ
=
Qp
ε(x) ∞
k=0
(tx)k k! µ=
∞
k=0
tk π
Qp
ε(x) x k
Ωk·k!µ.
To prove the first identity, we need to show that
Tn
Qp
ε(x) x k
Ωk·k!µ
=p−n
p−nZpε(x)
xk
Ωk·k!µ (2.28)
(note that the right-hand side is indeed inKn). Now, assumeµhas compact supportp−mZ
pfor somem, then
Tn
p−mZpε(x)
xk
Ωkµ = 1
pm
σ∈G(Km/Kn)
σ
p−mZpε(x)
xk
Ωkµ
= 1 pm
σ
p−mZpε
κ(σ )xx k
Ωkµ
=p−n
p−nZpε(x)
xk
Ωkµ.
(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 overBdRGK∞ with dimensiondimQpVand can have a base
consisting of elements in(Bmaxϕ=1⊗V )GK∞.
Proof. By an argument similar to [2, Corollary B.14], we can see that(BdR+ ⊗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((BdR+ ⊗V )GK∞), thenθ(det(A))≠0.
Fork≥1,1≤i≤d, letvi,k=m∈Zπ−kmϕm−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(v
i)whenk→ ∞, hence whenk1,vi,k/ϕ−1(η([ε]
−1))k+1, . . . , v
d,k/ϕ−1(η([ε]−1))k+1 are linearly independent, hence v1,k, . . . , vd,k are linearly independent.
Fromϕ(t−k
π vi,k)=(tπ−kvi,k)we see that tπ−kv1,k, . . . , t−πkvd,k∈(B ϕ=1
max ⊗V )GK∞ are a base of(BdR⊗V )GK∞ overB
GK∞ dR .
Lemma2.9. H1(K
∞, BdR⊗V )=0andH1(K∞,Ᏸr(Z×p, B ϕ=1
max⊗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 inAGcrysK∞ andϕ(a)=pa. Let
c= 1 pa
ϕn pn
[ω]−1ϕ[ω]c, (2.30)
then it is easy to see that(1−τ)c=cτ. Sincec∈(Bmaxϕ=1⊗V ), this shows that the inclu-sion maph1(K
in [2, Lemmas B.18 and B.19] we conclude thatH1(K
∞, BdR⊗V )=0, H1(K∞,Ᏸr(Z×p, B ϕ=1 max⊗
V ))=0.
Lemma2.10. For ap-adic representationV ofGQp, as aᏰ0(Z×p,Qp)-module the
fol-lowing sequence is exact:
0 →H1Γ,Ᏸ 0
Z×
p, V
GK∞ →H1Q p,Ᏸ0
Z×
p, V
→H1K
∞,Ᏸ0
Z×
p, V
Γ →0.
(2.31)
Proof. The proof can be found in [2].
3. Perrin-Riou exponential map. ForI ⊂ Z, we have defined LPI, ᏰI
alg(Qp, A), Φ, andᏲalgin [8, Section 4]. For a Galois moduleA, in this section, the Galois action on
ᏰI
alg(Qp, A)is defined by
f (x)σ (µ)=σ
fκ(σ )xµ , (3.1)
whereµ∈ᏰI
alg(Qp, A)andσ∈GQp. Recall that we have the following formulas:
Ᏺalgxk1a+pnZp(y)=p−n k!
(−ty)kε(ay)1p−nZp(y), (3.2)
Ᏺalg
xk1
Z× p
=(−tx)1 k·1Zp−p−1 1
(−tx)k1p−1Zp. (3.3)
Proposition3.1. For a de Rham representationV, ifµ∈Ᏸ(−∞,h−1]
alg (Qp,Ξ⊗D(V ))ψ,
thenᏲ(h) alg(µ)∈Ᏸ
[1−h,+∞)
alg (Qp, BdR⊗V )is fixed byGQp.
Proof. Forσ∈GQp,
a+pnZpx kσᏲ(h)
alg(µ)
=σ
κ(σ )−1a+pnZp
κ(σ )xkᏲalg(h)(µ)
=p−n(k+h−1)!κ(σ )k
×
σ
p−nZpε
κ(σ )−1ax(−tx)−kµ
=p−n(k+h−1)!
p−nZpε(ax)(−tx)
−kµ
=
a+pnZpx kᏲ(h)
alg(µ),
(3.4)
hence we haveσ (Ᏺ(h)alg(µ))=Ᏺ(h) alg(µ).
Let exp denote the connecting map of the following exact sequence:
0 →Ᏸ+
alg
Z×
p, V
→Ᏸ+
alg
Z×
p, B ϕ=1
max⊗V →Ᏸ+alg
Z×
p, BdR/B+dR⊗V
Proposition3.2. The following diagram is commutative:
Ᏸ+
alg
Z×
p, BdR/B+dR⊗V
GQp exp
1+pnZpxkµ
H1Q p,Ᏸ+alg
Z×
p, V
1+pnZpxkµ
DKn
Vκk exp H1K, Vκk.
(3.6)
Proof. For µ ∈ Ᏸ+
alg(Z×p, BdR/BdR+ ⊗V )
GQp, it is easy to see that
1+pnZpxkµ ∈
DKn(V (κ
k)) and for ξ∈H1(Q
p,Ᏸ+alg(Z×p, V )),
1+pnZpxkξ∈H1(Kn, V (κk)), then the proposition follows from the following commutative diagram:
0 V Bmaxϕ=1⊗V BdR/BdR+ ⊗V 0
0 Ᏸ+alg
Z×
p, V
1+pnZpxk
Ᏸ+
alg
Z×
p, B ϕ=1 max⊗V
1+pnZpxk
Ᏸ+
alg
Z×
p, BdR/B+dR⊗V
1+pnZpxk
0.
(3.7)
Now we define the Perrin-Riou exponential mapE=Eh,V as the composition of the following ones:
Ᏸ−
alg
Qp,Ξ⊗D(V )
Φ=1,ψ Ᏺalg
→Ᏸ+
alg
Qp, BdR⊗V
GQp
Fil<0
→Ᏸ+
alg
Z×
p, BdR/BdR+ ⊗V
GQp
exp
→H1Qp,Ᏸ+alg
Z×
p, V
.
(3.8)
The invariance ofΦµ=µgives the following identity:
p−nZpf (x) 1
(−tx)kµ=ϕ
−n
Zp
f x
pn µ
(−tx)k (3.9)
for alln∈Z, k≥0, andf a local constant function with compact supportp−nZ p.
Theorem3.3. The integrals of the exponential map give
Z× p
xkE
h,V(µ)=(k+h−1)! expV (κk)
(1−ϕ)−11−p−1ϕ−1 Z×
p
µ (−tx)k ,
a+pnZpx kE
h,V(µ)=(k+h−1)! expV (κk)
ϕ−n
pn
Zp
ε ax
pn · µ (−tx)k
.
Proof. By formulas (3.2) and (3.3), we have
Z×p
xkᏲ(h)alg(µ)=
Qp
Ᏺ(h) alg
xk1Z×
p
µ
=
Qp
(k+h−1)!
1
(−tx)k1Zp−p−
1 1
(−tx)k1p−1Zp µ
=(k+h−1)!(1−ϕ)−11−p−1ϕ−1 Z×p
µ (−tx)k ,
a+pnZpx kᏲ(h)
alg(µ)=
Qp
Ᏺ(h) alg
xk1
a+pnZpµ
=
Qp
(k+h−1)!p−n·(−tx)1 kε(ax)1pnZp
=(k+h−1)!ϕ
−n
pn
Zp
ε ax
pn µ (−tx)k .
(3.11)
Hence, byProposition 1.2, we have
Z× p
xkE
h,V(µ)=expV (κk)
Qp
Ᏺalgxk·1Z×pµ
=expV (κk)
(k+h−1)!(1−ϕ)−11−p−1ϕ−1 Z×
p
µ (−tx)k ,
a+pnZpx kE
h,V(µ)=expV (κk)
Qp
Ᏺalg
xk1a+pnZpµ
=expV (κk)
(k+h−1)!ϕ
−n
pn
Zp
ε ax
pn µ (−tx)k
.
(3.12)
Lemma 3.4. For a de Rham representation V such that Fil−1D(V ) = D(V ), µ ∈ Ᏸ(−∞,0]
alg (Qp,Ξ⊗D(V ))Φ=1,ψ,a∈Z×p, n≥1, andj≥0,
a+pnZp(x−a)jEh,V(µ)restricted
toH1(K
∞,Ᏸ[alg0,∞)(Z×p, V ))is represented by the cocycle
τ →(τ−1)EulB βa,n,j(µ), (3.13)
where
βa,n,j(µ)=j!pnj(−t)j ϕ−n
pn
Z×p
[ε]ax µ
xj. (3.14)
Proof. We first consider
a+pnZpx kE
1,V(µ)=expV (κk)
k!ϕ
−n
pn
Zp
ε ax
pn µ
Put
γa,n,k(µ)=k! ϕ−n
pn Zp ε ax pn µ (−tx)k,
γa,n,k(µ)=k! ϕ−n
pn Zp εax k
i=0
(−atx)i i!
µ (−tx)k.
(3.16)
By the formulaε(1/pn)=[ε
n]exp(−t/pn), we have
ε x
pn ≡
εxn
k
i=0 1
i!
−tx
pn i
=ϕ−n
εx
k
i=0
(−tx)i i!
modFilk+1BdR
. (3.17)
Let
δa,n,k(x)=ε
ax
pn −ϕ
−n εax k
i=0
(−atx)i i!
∈Filk+1BdR
=ϕ−n
ε ax
pn −
εax
k
i=0
(−atx)i i! , (3.18) then
γa,n,k(µ)−γa,n,k(µ)=k! ϕ−n
pn
Zp
ϕnδa,n,k
µ (−tx)k
∈BdR1 ⊗D(V )⊂BdR1 ⊗B−dR1⊗V⊂B+dR⊗V .
(3.19)
Hence, γa,n,k(µ)∈ Bmax⊗V is a lifting of γa,n,k under the projection map Bmax →
BdR/BdR+ . By [2, Lemma 0.5.1], we see that the above cohomology class expV (κk)(γa,n,k(µ)) is represented by the cocycleτ→(1−τ)EulB((1−ϕ)γa,n,k(µ)). Moreover,
(1−ϕ)γa,n,k(µ)=(1−ϕ)k! ϕ−n
pn Zp εax k
i=0
(−atx)i i!
µ (−tx)k
=k!ϕ− n pn Z× p εax k
i=0
(−atx)i i!
µ (−tx)k
(3.20) since ϕ Zp εax k
i=0
(−atx)i i!
µ (−tx)i
= Zp εax k
i=0
(−atx)i i!
1
(−tx)kΦµ, (3.21)
andΦ(µ·1Zp)=1pZp·µ. Hence, fora∈Z×
p, n≥1,
a+pnZpxkE1,V(µ)is represented by
τ →(τ−1)EulB
where
γ∗a,n,k(µ)=k! ϕ−n
pn
Z×p
εax
k
i=0
(−atx)i i!
ν (−tx)k
. (3.23)
Therefore,
a+pnZp(x−a) jE
1,V(µ)= j
k=0
a+pnZp
j k
(−a)j−kxkE 1,V(µ)
=
j
k=0
j k
(−a)j−k
a+pnZpx kE
1,V(µ)
(3.24)
is represented by
τ →(τ−1) j
k=0
j k
(−a)j−kEul B
γ∗a,n,k(µ)
=(τ−1)EulB
j
k=0
j k
(−a)j−kk!ϕ−n pn Z× p εax k
i=0
(−atx)i i!
µ (−tx)k
=(τ−1)EulB
ϕ−n
pn
Z× p
eatx·
j
k=0
k! j k
(atx)j−k
k
k=0
(−atx)i i!
× µ
(−tx)j
=(τ−1)EulB
ϕ−n
pn
Z×p
j!εax µ (−tx)j
,
(3.25)
anda+pnZp(x−a)jE1,V(µ)is represented by
τ →(τ−1)EulB
j!pnj(−t)−jϕ− n
pn
Z×p
εaxµ
xj , (3.26)
and the lemma follows.
Lemma3.5. The following diagram is commutative:
Ᏸ(−∞,h] alg
Qp,Ξ⊗D
V (κ)Φ=1,ψ µ→(−tx)µ
Eh+1,V (κ)
Ᏸ(−∞,h−1] alg
Qp,Ξ⊗D(V )
Φ=1,ψ
Eh,V
H1K,Ᏸ[−h,∞) alg
Z×
p, V (κ)
ξ→x−1ξκ−1
H1K,Ᏸ[1−h,∞) alg
Z×
p, V
.
Proof. First note that all the maps are well defined. For µ ∈ Ᏸ(−∞,h]
alg (Qp,Ξ⊗ D(V (κ)))ψ,(−tx)µ∈Ᏸ(−∞,h−1]
alg (Qp,Ξ⊗D(V ))ψ,
Z× p
xkE h,V
(−tx)µ
=(k+h−1)! expV (κk)
(1−ϕ)−11−p−1ϕ−1 Z×
p
µ (−tx)k−1
,
Z×p
xk·x−1Eh+1,V (κ)(µ)
=(k+h−1)! expV (κk)
(1−ϕ)−11−p−1ϕ−1 Z×
p
µ (−tx)k−1
,
(3.28)
hence the lemma follows.
Theorem3.6. AssumeVis a crystalline representation andh∈Zsuch thatFil−hD(V ) =D(V ). Forµ∈Ᏸtemp(Qp,Ξ⊗D(V ))Φ=1,ψ, under the map
Ᏸ(−∞,h−1] alg
Qp,Ξ⊗D(V )Φ=1,ψ →H1
Qp,Ᏸ[alg1−h,∞)
Z×
p, V
res
→H1K
∞,Ᏸ[alg1−h,∞)
Z×
p, V
Γ ,
(3.29)
the image is inH1(K
∞,Ᏸtemp(Z×p, V ))ΓwithΓ=Gal(K∞/Qp).
Proof. Sincex1−hE
1,V (κ1−h)((−tx)h−1µ)=Eh,V(µ), byLemma 3.5, replacingV by V (1−h)andµby(−tx)h−1µ, we can assumeh=1. FromLemma 3.5fora∈Z×
p, n≥1, andj≥0,a+pnZp(x−a)jE1,V(µ)is represented by the cocycle
τ →(τ−1)EulB
j!pnj(−t)−jϕ−n pn
Z×p
εaxµ
xj
. (3.30)
From [8, Lemma 14], we know that ifµis of orderr, thenµ/xjis also of orderr. From
Proposition 2.4, we know thatZ× p[ε
ax](µ/xj)has orderr+r (V ), that is,
p[n(r+r (V )+1)]ϕ−n
Z× p
εaxµ
xj (3.31)
isr-bounded. Hence,p[n(r+r (V )+1)]
a+pnZp(x−a)jE1,V(µ)represented by
τ →(τ−1)EulB
j!(−t)−j
p[n(r+r (V )+1)]ϕ−n
Z×p
εaxµ
xj
(3.32)
is(r+r (V ))-bounded. Hence, the image is inH1(K
∞,Ᏸr+r (V )+1(Z×p, V ))Γ.
4. The logarithmic map. The proof ofTheorem 3.6suggests thatᏲcont(µ)has some deep arithmetic meaning for the distributionµ. In this section, we construct a logarith-mic map on the cohomology side, which is related toᏲcont(µ).
SupposeV is a de Rham representation. The Qp-space ∪n∈N(Bmaxϕ=1⊗V (κ−i))GKn ⊂
subspace ofBϕmax=1⊗V. LetWibe one supplement. Supposeh∈Z, h≥1, µ∈H1(Qp,
Ᏸh−(Z×p, V )), andτ→µτ is a 1-cocycle representation ofµ. Supposeµ has the prop-erty 1+pnZpx−iµ ∈He1(Kn, V (κ−i))=ker{H1(Kn, V (κ−i))→H1(Kn, B
ϕ=1
max ⊗V (κ−i))}.
For such aµ, there is acn,i∈Bmaxϕ=1⊗V (κ−i)such that
1−κ(σ )−iσc n,i=
1+pnZpx
−iµ
σ. (4.1)
The choice ofWimakes it possible to choosecn,i∈Wiuniquely forn1.
Theorem4.1. SupposeVis a de Rham representation ofGQp,h≥1is an integer, and µ∈H1(Q
p,Ᏸh−(Z×p, V ))such that
1+pnZpx−iµ∈He1(Kn, V (κ−i))forn≥1,0≤i≤h−1.
Choosecn,i∈Wisuch that
1−κ(σ )−iσc n,i=
1+pnZpx
−iµ, (4.2)
then the sequence (1/(h−1)!)pnh−1 i=0
h−1 i
cn,i converges to an element in (Bϕmax=1⊗
V )GK∞ which is denoted byLh,V(µ).
Proof. ByLemma 2.9, H1(K
∞,Ᏸh−(Z×p, B ϕ=1
max ⊗V ))=0, the inflation-restriction se-quence gives the following isomorphism:
H1Q p,Ᏸh−
Z×
p, B ϕ=1 max ⊗V
∼→H1Γ,Ᏸ h−
Z×
p,
Bmaxϕ=1⊗V
GK∞. (4.3)
Letµ∈H1(Γ,Ᏸh−(Z×
p, (B ϕ=1
max ⊗V )GK∞))correspond to the image ofµ under the map
H1(Q
p,Ᏸh−(Z×p, V ))→H1(Qp,Ᏸh−(Z×p, B ϕ=1
max⊗V )). Thenµ−µis a coboundary inH1(Qp,
Ᏸh−(Z×p, B ϕ=1
max ⊗V )), that is, there existsν∈Ᏸh−(Z×p, B ϕ=1
max⊗V )such thatµσ−µσ =(1−
σ )ν. Letun,i=cn,i−
1+pnZpx−iν, thenun,i∈(Bmaxϕ=1⊗V (κ−i))GK∞and(1−κ(σ )−iσ )un,i
=1+pnZpx−iµσ forσ∈GKn, and we have
lim n→∞
pn
h−1
i=0
(−1)i
h−1
i
cn,i−pn h−1
i=0
(−1)i
h−1
i
un,i
=lim n→∞p
n
1+pnZp
1−x−1h−1
ν=0.
(4.4)
Here, again, we use [8, Lemma 14]. Note that there exists aksuch thatun,i∈(t−kBmax+ )ϕ=1
⊗V (κ−i)(which does not depend onn, i) [2]. The theorem will follow from the following lemma.
Lemma 4.2. Suppose µ ∈H1(Γ,Ᏸ
h−(Z×p, (B ϕ=1
max ⊗V )GK∞)) and k∈N such that for
n≥1and0≤i≤h−1, the image1+pnZpx−iµ is zero inH1(Γn, (Fil−kBmaxϕ=1⊗V )GK∞).
Suppose
Fil−kBϕmax=1⊗VGK∞ = ∪+∞n=0
Fil−kBmaxϕ=1⊗Vκ−i GKn
is a direct summand decomposition (since the first component is closed). Supposeσ → µσ is a cocycle representation of µ, take un,i ∈ Wi such that
1+pnZpx−iξσ = (1− κ−i(σ )σ )u
n,ifor allσ∈GKn, then the sequencevn=pnhi=−01(−1)i
h−1
i
un,iconverges
to an element in(Fil−kBϕmax=1⊗V )GK∞.
Proof. The cocycles satisfyµσ−1τ=µσ−1+σ−1µτ=µτ+τµσ−1, and this means that
σ−1µ
τ=µτ+(τ−1)µσ−1,
1+pnZpx
−iσ−1µ τ=σ−1
κ(σ )+pnZpκ
i(σ )x−iµ τ .
(4.6)
This gives
κ(σ )+pnZpx
−iµ
τ=κ(σ )−iσ
1+pnZpx
−i·σ−1µ τ
=tπi·σ
1+pnZp
tπx
−i
·σ−1µτ
=ti
π(1−τ)σ
t−i
π un,i−
1+pnZp
tπx
−i µσ−1
(4.7)
by
tπ−i(1−τ)un−1,i=t−πi·
σ∈Γn−1/Γn
κ(σ )+pnZpx
−iµ σ
=(1−τ) σ∈Γn−1/Γn
σt−i
πun,i
−σ
1+pnZp
tπx
−i
µσ−1 .
(4.8)
From the choice ofWi, we can cancel(1−τ)and get
vn−vn−1=pn−1 h−1
i=0
(−1)i
h−1
i
tπi
σ∈Γn−1/Γn
(σ−1)t−πiun,i
−σ
1+pnZp
tπx
−i
µσ−1 .
(4.9)
The sum
pn h−1
i=0
(−1)i
h−1
i
ti πσ
1+pnZp
tπx
−i µσ−1
=pnσ 1+pnZ
p
1−κ(σ )−1xh−1µ
σ−1 →0
(4.10)
asκ(σ )∈1+pn−1Z
p, µσ−1 →µ1 which is of order 1−. Hence, to show that vn is a Cauchy sequence, we only need to show that forσn∈Γn−1the sequence
pn h−1
i=0
(−1)i
h−1
i
κσniσn−1
un,i (4.11)
Case1(h=1). This is equivalent to showing that
lim n→+∞p
nσ
n−1un,i=0. (4.12)
PutTσ=
p−1
j=0σj, then
Tσn
σn−1
un,0=
σnp−1
un,0= −
1+pnZpµσ p
n. (4.13)
Sinceµσp
n has order 1
−, we have
lim n→∞p
n
1+pnZpµσ p
n=0, (4.14)
hence
lim n→∞p
nT σn
σn−1
un,0
=0. (4.15)
The theorem follows from the following proposition.
Proposition 4.3. There exists a constantC >0such that, for alln∈N, σ∈Γn, F∈(Fil−kBmax⊗V )GKn,
Tσ(F )≥CF, (4.16)
where·is the norm onBmax.
In the following we will develop several lemmas to prove the above proposition.
Lemma4.4. There exist aζ∈Z×
pand aBmax-morphismg:Bmax⊗V→Bmaxd such that
g(Bmaxϕ=1⊗V )GK∞ is contained in((Bmaxϕ=ζ)GK∞)d, whered=dimQpV.
Proof. Supposee1, . . . , ed∈(Bmaxϕ=1⊗V )GK∞ such that they are a base ofBdR⊗Vover BdR byLemma 2.8. Assumev1, . . . , vd are a base ofV /Qpand assumeei=
d
j=1aijvj, thenaij∈Bϕmax=1, theσ action gives det(σ )·det(σ (aij))=det(aij). Consider the one-dimensional representation detQp(V )which has a basee=v1∧···∧vd, then the above formula givese1∧···∧ed=(det(aij))v1∧···∧vd, henceσ (e)=(det(aij)/det(σ (aij)))e. But detQp(V )is a one-dimensional representation, so it must be of the formφ0φκk, where φ0 is a finite-order N of character, φ is an unramified character such that
φ(Frobp)=ufor some u∈Z×p. Take∈Qpur such that Frobp()=u, then the above identity means that det(aij)/σ (det(aij))=φ0(σ )·φ(σ )·κk(σ ). Assumeσ∈GK∞ and raiseNth power, we get(det(aij)/σ (det(aij)))N=φN(σ ). Take∆=(det(aij))N, then this givesσ (∆)=∆for allσ∈GK∞. On the other hand,ϕ(∆)=ϕ(det(aij)·)N= det(aij)N·N·uN =uN·∆. Now, takingζ=uN, g:x→∆x, theng(ei)=∆ei and ϕ(∆ei)=uNδei=ζ∆ei, and the lemma follows.
Corollary4.5. Fork∈N, there existαk∈Zpandgk:t−πkBmax+ ⊗V→(B+max)d such
Proof. Use thegin Lemma 4.4, assumets
π·g(Bmax+ ⊗V )(Bmax+ )d, and letgk= tk+s
π g, αk=πk+sζ.
Lemma 4.6. For k ∈N, >0, there exists n∈ N such that for σ ∈Γn and F ∈ (Fil−kBϕmax=1⊗V )GK∞,
gk
σ (F )−σgk(F )≤F. (4.17)
Proof. Letγ1 be a topology generator ofΓ1and γ1∈GQp a lifting ofγ1; we can chooseγ1in thep-Sylow subgroup ofGQp.γ1induces a continuous map fromΓ1→GQp by sendingxtoγ"1logγ1x. So, forσ∈Γ1, we can getσ∈GQpunder such a map. The map
gk◦σ−σ◦gk:Bmax⊗V→(Bmax)dand the mapgk◦σ−σ◦gk:(Fil−k(Bmaxϕ=1⊗V ))GK∞ →
(Bd
max)are the same if we restrict to(B ϕ=1
max⊗V )GK∞, and
gk
σtπ−kvi
−σ g k
tπ−kvi
=gk
σ−1tπ−kvi
+1−σgk
tπ−kvi
(4.18)
sincet−k
π v1, . . . , tπ−kvdare a base oftπ−kBmax+ ⊗V overBmax+ , and whenσ →1 the above goes to zero, hence the lemma follows.
Lemma4.7(Coleman-Colmez exact sequence). Forα∈Zp, α≠0, r=vp(α), (i) ifα∉pN, then(Bmax+ )ϕ=α,GK∞
∼
→Ᏸr−(Z×p,Ξ⊗Qp)ψ, (ii) ifα=pr, then the following sequence is exact:
0→Qptr →
Bmax+
ϕ=pr,G K∞ →Ᏸ
r−
Z×
p,Ξ⊗Qp
ψ
→Qp →0. (4.19)
Proof. See [2, Appendix A].
Lemma4.8. Forr∈R+, there existsCr>0such that forµ∈Ᏸr(Zp,Ξ⊗D(V ))ψand a∈Zp,
p−1
i=0
δia∗µ
≥Crµ. (4.20)
Proof. This is the same as [2, Lemma III.2.6]. Note the∗above is induced by the addition inZp.
Corollary4.9. Forr∈R+, there existsCr>0such that forµ∈Ᏸr(Zp,Ξ⊗D(V ))ψ
anda∈1+pZp,
p−1
i=0
δai∗µ
≥Crµ. (4.21)
Here the∗is induced by the multiplication inZ×
p.
Corollary4.10. There existC >0and ann0∈Nsuch that forF∈(Bmax+ )ϕ=α,GK∞
andσ∈Γn0,
Proof. UsingCorollary 4.9,F corresponds toµ,F = µ, andTσ(F )corresponds toδai∗µ, hence
Tσ(F )=
δai∗µ≥Cµ =CF. (4.23)
Corollary4.11. There exist aC >0and ann∈Nsuch thatTσ(gk(F ))≥Cgk(F )
for allF∈(Fil−kBmaxϕ=1⊗V )GK∞.
Proof. The corollary follows sincegk(F )∈(Bmax+ )ϕ=α,GK∞.
NowProposition 4.3follows sincegkis just a multiple bypk+s∆.
Case2(generalh). In this case we need to show that
pn
h−1
i=0
(−1)i
h−1
i
κσn−iσn−1
un,i
→0 (4.24)
asn→ ∞.
Lemma4.12. Forl∈Nandσ∈Γ, putTl,σ=pj=−01κ(σ )−ljσj, then the sequence
pn k
l=0
Tl,σn
h−1
i=0
h−1
i
κσn
−i σn−1
un,i
(4.25)
tends to zero asn→ ∞.
Proof. Forl≥0 andl≤k
Tl,σn
κσn−iσn−1
un,i=
κσnp−i·σnp−1
un,i= −
1+pnZpx
−iµ
σnp, (4.26)
then (4.25) equals
−pn h−1
i=0
(−1)i
h−1
i h−1
l=0,l≠i Tl,σn
1+pnZpx
−iµ
σnp= −p n
h−1
l=0
Rl·Yl (4.27)
with Rl∈pn(h−1−l)Zp[[T ]]andpnRl·Yl∈pn(h−l)Zp[[T ]]
1+pnZp(x−1)lµσp
n →0. By [2, Lemma III.3.3], the proof of the lemma follows.
Lemma4.13. There exist aC >0and ann∈Nsuch that
h−1
i=0
Ti,σ(F )
≥CFh (4.28)
