Pink-type results for general subgroup of $\operatorname{SL}_2(\mathbb{Z}_\ell)^n$

0.39 MB·English

by  Davide Lombardo

Pink-type results for general subgroups of SL (Z )n
2 ℓ
5
1 Davide Lombardo
0
2

g Abstract
Au WestudyopensubgroupsGofSL2(Zℓ)n intermsofsomeassociated Liealgebraswithout
assuming that G is a pro-ℓ group, thereby extending a theorem of Pink. The result has
9 applications tothe studyof families of Galois representations.

]
R 1 Motivation and statement of the result
G
. The ultimate goal of this work is the study of the images of certain Galois representations with
h values in GL (Z )n, such as those affordedby the Tate modules of elliptic curves,or those arising
t 2 ℓ
a from modular forms. In view of this aim, one would wish to provide a manageable description
m
of these images; however, it turns out that it is beneficial, and in a sense simpler, to consider
[ arbitrary subgroups G of GL (Z )n without making any reference to their origin, and in the
2 ℓ
present work Galois representations will play virtually no role. In most applications to the study
2
v of Galois representations, the main object of interest is actually the intersection G SL2(Zℓ)n,
∩
8 and furthermore it is an easy matter to pass from results on subgroups of SL (Z )n to results on
2 ℓ
3 subgroups of GL (Z )n, so we shall actually mostly work with subgroups G of SL (Z )n. Any
2 ℓ 2 ℓ
9 such G is the extension of a ‘finite’ part, the image G(ℓ) of the reduction G SL (F )n, by a
2 ℓ
3 →
‘Lie’ part, the kernel of this reduction.
0
. When G is closed and G(ℓ) is trivial (or more generally when G is pro-ℓ), and ℓ is odd, a
1
construction due to Pink [Pin93] gives a very concrete and handy description of G in terms of a
0
certain Z -Lie algebra (G) (together with some additional data which is not very important to
5 ℓ
L
1 our present discussion). Furthermore, if G is the image of a representation of Gal K/K (K a
: number field), the condition that G(ℓ) be trivial can always be met by replacing K by a finite
v (cid:0) (cid:1)
extension,sothatPink’stheoremapplies. Notehoweverthatthedegreeofthisextensiondepends
i
X on ℓ: while this is often perfectly fine when considering a single Galois representation, it may
r become a major drawback when dealing with infinite families G indexed by the rational primes
a ℓ
(as is the case, for example, with the action of Gal K/K on the various Tate modules of an
abelian variety). Furthermore, Pink’s theorem does not apply to ℓ = 2, which might again be
(cid:0) (cid:1)
quite a hindrance when trying to study the whole system G at once.
ℓ
WhilewecannothopetogiveacompletedescriptionofGintermsofPink’sLiealgebraswhen
G is not pro-ℓ, we could try and settle for less, namely a result of the form ‘when (G) contains
L
a largeneighbourhoodof the identity (givenexplicitly), we canexplicitly find a neighbourhoodof
theidentityofSL (Z )n thatisincludedinG’. NotethatwhendealingwithGaloisrepresentations
2 ℓ
weareofteninterestedin‘largeimage’results,forwhichthisweakerformofPink’stheoremwould
still be adequate. Unfortunately, even this is not possible (cf. for example [Lom14, 4.5]), and
§
the best we can hope for is for such a statement to hold not quite for G, but for a subgroupH of
G such that the index [G : H] is bounded by a function of n alone. In order to give a concrete
statement we shall need some preliminary definitions:
Definition 1.1. For a prime ℓ and a positive integer s we let (s) be the open subgroup of
ℓ
B
SL (Z ) given by
2 ℓ
x SL (Z ) x Id (mod ℓs) .
2 ℓ
∈ ≡
Wealsoset ℓ(0)=SL2(Zℓ)(cid:8),andfornon-(cid:12)negativeintegersk1(cid:9),...,kn wedenoteby ℓ(k1,...,kn)
the open subgroBup n (k ) of SL (Z )n. (cid:12) B
j=1Bℓ j 2 ℓ
Q
1
Definition 1.2. (cf. [Pin93]) Let ℓ be a prime, n a positive integer and G be an open subgroup of
GL (Z )n. If ℓ=2, assume further that the reduction modulo 4 of G is trivial. Writing elements
2 ℓ
of G as n-tuples (g ,...,g ) of elements of GL (Z ), we define a map Θ by the formula
1 n 2 ℓ n
Θ : G n sl (Z )
n → i=1 2 ℓ
(g ,...,g ) g 1tr(g ),...,g 1tr(g ) ,
1 n 7→ 1− 2 L1 n− 2 n
and we let (G) sl (Z )n be the Z -span o(cid:0)f Θ (G). We call (G) the Lie(cid:1)algebra of G.
2 ℓ ℓ n
L ⊆ L
Theorem 1.3. Let ℓ be an odd prime, n be an integer, and G be an open subgroup of SL (Z )n.
2 ℓ
There exists an open subgroup H of G, of index at most 24n48n(n−1), with the following property:
if (H) contains n ℓksl (Z ) for a certain integer k > 0, then H contains B (p,...,p) for
L i=1 2 ℓ ℓ
p=80(max n,2 1)k.
{ }−L
Similarly, let n be a positive integer and G be an open subgroup of SL (Z )n. There exists an
2 2
open subgroup H of G that satisfies [G : H] 96n, is trivial modulo 4 (so that (H) is defined),
and has the following property: if (H) contains n 2ksl (Z ) for a certain inLteger k >0, then
L (cid:12) i=1 2 2
H contains B (p,...,p) for p=607(max n,2(cid:12) 1)k.
2
{ }−L
While it is certainly true that both this theorem and its proof are quite technical, it should
be remarked that this statement does enable us to show exactly the kind of ‘large image’ results
we alluded to: the case n =1 has been used in [Lom14] to show an explicit open image theorem
for elliptic curves (without complex multiplication), and in [Lom15] we apply the case n = 2 to
extend this result to arbitrary products of non-CM elliptic curves.
A few more words on the proof of theorem 1.3: as it will be clear from section 5, the crucial
cases are n = 1 and n = 2. While the former has essentially been proven in [Lom14], the latter
forms the core of the present paper, and we shall actually prove it in a slightly more precise form
thanstrictlynecessarytoestablishtheorem1.3. Thiswillbedoneinsections3and4below,where
we also give analogous statements for GL (Z )2.
2 ℓ
Notation. Throughoutthepaperℓisafixedprimenumber. IfGisaclosedsubgroupofGL (Z )n
2 ℓ
and k is a positive integer,we denote by G(ℓk) the image of the projectionG GL (Z/ℓkZ)n. If
2
→
H isasubgroupofSL (Z )n (resp.ofSL (F )n),wedenotebyN(H)the largestnormalsubgroup
2 ℓ 2 ℓ
of H which is pro-ℓ (resp. an ℓ-group); this object is well-defined by lemma 2.3 below. If x is an
element ofGL (Z ) (resp. ofZ ), we write [x] for its image in GL (F ) (resp.in F ). Since special
2 ℓ ℓ 2 ℓ ℓ
careis neededtotreatthe caseℓ=2,towriteuniformstatementswesetv either0or1according
to whether ℓ is odd or ℓ = 2. Finally, it will also be useful to introduce some standard elements
of SL (Z ): for every a Z we set
2 ℓ ℓ
∈
1 0 1+a 0 1 a
L(a)= , D(a)= , R(a)= .
a 1 0 (1+a)−1 0 1
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
Notice that L(a),R(a) belong to SL (Z ) for any a Z , while D(a) is in SL (Z ) if and only if
2 ℓ ℓ 2 ℓ
∈
a 1 (mod ℓ).
6≡−
2 Preliminary lemmas
Inthissectionweprovesomegenerallemmaswhichwillbeusedrepeatedlythroughoutthepaper.
2.1 Logarithms and exponentials
We recall the following fundamental properties of logarithms and exponentials, both for ℓ-adic
integers and for elements of M (Z ), the set of 2 2 matrices with coefficients in Z . The proofs
2 ℓ ℓ
×
of these statements are immediate upon direct inspection of the involved power series, and will
not be included.
2
tn
Lemma 2.1. The power series log(1+t)= ( 1)n−1 converges for t ℓZ (respectively for
ℓ
− n ∈
n≥1
X
t ℓM (Z )),andestablishesabijectionbetween1+ℓ1+vZ andℓ1+vZ (resp.between (1+v)and
2 ℓ ℓ ℓ ℓ
∈ tn B
ℓ1+vM (Z )). Theinversefunctionisgivenbythepowerseriesexpt= ,whichconvergesfor
2 ℓ
n!
n≥0
X
t ℓ1+vZ (resp.t ℓ1+vM (Z )). Furthermore,foreveryt ℓ1+vZ wehavev log(1+t)=v (t),
ℓ 2 ℓ ℓ ℓ ℓ
∈ ∈ ∈
and for every t ℓ1+vZ we have v (exp(t) 1)=v (t).
ℓ ℓ ℓ
∈ −
Lemma 2.2. Let A,B M (Z ) and n 2 be an integer. Suppose that A 0 (mod 4) and
2 2
∈ ≥ ≡
B 0 (mod 2n): then exp(A+B) exp(A) (mod 2n) and log(A+B) log(A) (mod 2n).
≡ ≡ ≡
2.2 Subgroups of SL2(Zℓ)n and Zℓ-Lie algebras
In this section we consider various properties of closed subgroupsof SL (Z )n, including their Lie
2 ℓ
algebras, generating sets, and derived subgroups.
Lemma 2.3. Let n be a positive integer and G be a closed subgroup of SL (Z )n. The collection
2 ℓ
of all normal pro-ℓ subgroups of G has a unique maximal element, which we denote by N(G).
Likewise, if G is a subgroup of SL (F )n, the collection of all normal subgroups of G whose order
2 ℓ
is a power of ℓ admits a unique maximal element, which we denote again by N(G).
Proof. Denote by π : G SL (F )n the canonical projection and let N be a normal, pro-ℓ
2 ℓ
→
subgroup of G. Clearly π(N) is an ℓ-group that is normal in G(ℓ), so it suffices to show that
N(G(ℓ)) is well-defined, for then N(G) = π−1(N(G(ℓ))) is the maximal normal pro-ℓ subgroup
of G. To treat the finite case consider first n = 1. Then G is a subgroup of SL (F ), and it easy
2 ℓ
to see that the collection of its maximal normal subgroups of order a power of ℓ has a maximal
element: it follows from the Dickson classification that this is given by the unique ℓ-Sylow if G
is of Borel type, and by the trivial group otherwise. Finally, if G is a subgroup of SL (F )n with
2 ℓ
n > 1, we denote by G the projection of G on the i-th factor SL (F ); it is then immediate to
i 2 ℓ
check that N(G)= (g ,...,g ) G g N(G ) .
1 n i i
∈ ∈
Lemma 2.4. Let t(cid:8)be a non-negativ(cid:12)e integer. L(cid:9)et W sl (Z ) be a Lie subalgebra that does
(cid:12) ⊆ 2 ℓ
not reduce to zero modulo ℓt+1. Suppose that W is stable under conjugation by (s) for some
ℓ
B
non-negative integer s, where s 2 if ℓ = 2 and s 1 if ℓ = 3 or 5. Then W contains the open
set ℓt+4s+4vsl (Z ). ≥ ≥
2 ℓ
Proof. Fix an element w of W that does not vanish modulo ℓt+1 and write w =µ x+µ y+µ h
x y h
for some µ ,µ ,µ Z , where
x y h ℓ
∈
1 0 0 1 0 0
h= , x= , y =
0 1 0 0 1 0
(cid:18) − (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
and min v (µ ),v (µ ),v (µ ) t. Set α := 1 + ℓs and consider the matrices C = L(ℓs),
ℓ x ℓ y ℓ h ℓ
{ } ≤
C = R(ℓs), and C = D(ℓs). For l,d,r define as the linear operator of W into itself
r d •
• ∈ { } C
given by
(x)=C −1xC
• • •
C
and set := Id. As W is stable under conjugation by (s), it is in particular also stable
• • ℓ
D C − B
under and . One can easily check the following identities:
• •
C D
α4( α2Id) (w)=(α4 1)(α2 1)µ x W
d d x
C − ◦D − − ∈
(α2 Id) (w)=(α4 1)(α2 1)µ y W,
d d y
C − ◦D − − ∈
and by considering the decomposition of w we deduce that also (α4 1)(α2 1)µ h belongs to
h
− −
W. By the assumptions on s we have v ((α4 1)(α2 1)) = 2s+3v, and therefore W contains
ℓ
− −
at least one of
M =ℓt+2s+3vh, M =ℓt+2s+3vx, M =ℓt+2s+3vy.
1 2 3
3
In the first case W contains ℓt+3s+4vsl (Z ) because of the following identities:
2 ℓ
(M )=2ℓt+3s+3vx, (M )= 2ℓt+3s+3vy.
r 1 l 1
D D −
In the second and third case W contains ℓt+4s+4vsl (Z ) because of the following identities:
2 ℓ
(M )= 2ℓt+4s+3vy, (M ) 2 (M )= 2ℓt+3s+3vh
l l 2 l l 2 l 2
D ◦D − D ◦D − D −
(M )= 2ℓt+4s+3vx, (M ) 2 (M )=2ℓt+3s+3vh.
r r 3 r r 3 r 3
D ◦D − D ◦D − D
Lemma 2.5. Let n and m be positive integers. Let g End(Zm) and write p for the character-
∈ ℓ g
istic polynomial of g. Let furthermore λ Z ,w Zm be such that gw λw (mod ℓn). Suppose
∈ ℓ ∈ ℓ ≡
that w 0 (mod ℓα+1) holds for some 0 α<n. Then we have p (λ) 0 (mod ℓn−α).
g
6≡ ≤ ≡
Proof. Denote by (g λId)∗ the adjugate matrix of (g λId), that is the unique operator such
− −
that (g λId)∗(g λId)=det(g λId) Id holds. Multiplying (g λId)w 0 (mod ℓn) on the
− − − · − ≡
left by (g λId)∗ we obtain det(g λId) w 0 (mod ℓn), and by considering the coordinate of
− − · ≡
w of smallest valuation we obtain p (λ)=det(g λId) 0 (mod ℓn−α) as claimed.
g
− ≡
Definition2.6. Ifn 2andifg ,...,g areelementsofagroupG,wewriteComm (g ,...,g )
1 n n−1 1 n
≥
for the (n 1) times iterated commutator, which is an element of G that can be defined via the
−
following recursion:
Comm (g ,g )=[g ,g ]=g g g−1g−1,
1 1 2 1 2 1 2 1 2
Comm (g ,...,g )=[Comm (g ,...,g ),g ].
n−1 1 n n−2 1 n−1 n
Furthermore,ifG ,...,G aresubgroupsofatopologicalgroupG,wewriteComm (G ,...,G )
1 n n−1 1 n
for the subgroup of G topologically generated by
Comm (g ,...,g ) g G for i=1,...,n .
n−1 1 n i i
∈
(cid:8) (cid:12) (cid:9)
We also write [G1,G2] for Comm1(G1,G2). (cid:12)
Lemma 2.7. Let s be a non-negative integer (with s 2 if ℓ = 2 and s 1 if ℓ = 3). Let
≥ ≥
a,b,c Z all have exact valuation s. Suppose that c 1 (mod ℓ), so that D(c) belongs to
ℓ
∈ 6≡ −
SL (Z ). Let G be a closed subgroup of SL (Z ).
2 ℓ 2 ℓ
1. Suppose G contains L(a) (resp. R(b)): then for all d Z such that v (d) s the group G
ℓ ℓ
∈ ≥
contains L(d) (resp. R(d)).
2. Suppose s 1 and G contains D(c): then for all d Z such that v (d) s the group G
ℓ ℓ
≥ ∈ ≥
contains D(d).
3. Suppose ℓ 5 or s 1. If G contains L(a),R(b),D(c), then it contains all of (s).
ℓ
≥ ≥ B
4. Suppose ℓ 5 and G contains L(a),R(b): then G contains all of (2s).
ℓ
≥ B
Proof. 1. Let u= d Z . It is easy to check that for every integer n we have L(a)n =L(an).
a ∈ ℓ
Fixasequencen ofrationalintegersthatconvergetouintheℓ-adictopology: thenwehave
k
limk→∞L(a)nk = limk→∞L(ank) = L(limk→∞ank) = L(au) = L(d). Now observe that G
containsevery term ofthe sequence L(a)nk, so – since it is closedin SL2(Zℓ) by assumption
– it also contains their limit L(d). The same proof also applies to the case of R(b).
2. The assumptions ons imply that log(1+c) is defined, andthat v log(1+c)=s (cf. lemma
ℓ
log(1+d)
2.1). Let u = , which is a well-defined element of Z since v log(1 +d) s.
ℓ ℓ
log(1+c) ≥
4
Choose as above a sequence n that converges to u in the ℓ-adic topology: then G contains
k
the limit
lim D(c)nk = lim (1+c)nk 0
k→∞ k→∞ 0 (1+c)−nk
(cid:18) (cid:19)
exp(n log(1+c)) 0
= lim k
k→∞ 0 exp( nklog(1+c))
(cid:18) − (cid:19)
exp(ulog(1+c)) 0
= =D(d).
0 exp( ulog(1+c))
(cid:18) − (cid:19)
3. Suppose first s = 0: then it is easy to see that G(ℓ) = SL (F ). Since ℓ 5, by [Ser97,
2 ℓ
≥
IV-23, Lemma 3] this implies G = SL (Z ). Suppose on the other hand that s 1: then
2 ℓ
≥
parts (1) and (2) imply that G contains L(d), R(d), D(d) for all d Z of valuation atleast
ℓ
∈
s, and by [Lom14, Lemma 3.1] these elements generate (s).
ℓ
B
4. If s=0 we notice – as in part (3) – that G(ℓ)=SL (F ), hence G=SL (Z ). Otherwise, it
2 ℓ 2 ℓ
sufficesto applypart(3)to (a2,b2,ab): indeedpart(1)implies thatGcontainsL(a2),R(b2)
a b
and D(ab)=L R( b)L( a)R .
1+ab − − 1+ab
(cid:18) (cid:19) (cid:18) (cid:19)
Lemma 2.8. Let s ,...,s be non-negative integers (where for every j =1,...,n we require that
1 n
s 2ifℓ=2ands 1ifℓ=3). TheiteratedcommutatorComm ( (s ), (s ),..., (s ))
j j n−1 ℓ 1 ℓ 2 ℓ n
≥ ≥ B B B
contains B (s + +s +(n 1)v).
ℓ 1 n
··· −
Proof. Consider first the case n = 2. Let s be a non-negative integer (with s 2 if ℓ = 2 and
≥
s 1 if ℓ = 3). By lemma 2.7 (3), if a,b,c are any three elements of Z of valuation s and such
ℓ
≥
that c 1 (mod ℓ), the group (s) is topologically generated by L(a),R(b),D(c). It is easy to
ℓ
6≡− B
check the following identities:
ℓs2 +2
[L(ℓs1),D(ℓs2)]=L ℓs1+s2
(ℓs2 +1)2 !
[R(ℓs1),D(ℓs2)]=R (2+ℓs2)ℓs1+s2 ,
−
(cid:0) ℓs2 +2 (cid:1)
where (by the assumptions on s ,s ) we havev =v (2+ℓs2)=v. To conclude the
1 2 ℓ (ℓs2 +1)2! ℓ
proof for n=2 it thus suffices to show that [ (s ), (s )] contains an element of the form D(c)
ℓ 1 ℓ 2
B B
with v (c) s +s +v. This is easily achieved thanks to the following identity:
ℓ 1 2
≤
ℓs1+2s2 ℓ2s1+s2
L [R(ℓs1),L(ℓs2)]R =D ℓs1+s2 +ℓ2(s1+s2) ,
−1+ℓs1+s2 +ℓ2s1+2s2 1+ℓs1+s2 +ℓ2s1+2s2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:16) (cid:17)
ℓs1+2s2 ℓ2s1+s2
where we know that L and R both belong to
−1+ℓs1+s2 +ℓ2s1+2s2 1+ℓs1+s2 +ℓ2s1+2s2
(cid:18) (cid:19) (cid:18) (cid:19)
[ (s ), (s )] by what we already proved and by lemma 2.7 (1). The case of an arbitrary n
ℓ 1 ℓ 2
B B
follows by induction from the case n=2.
The following lemma can be considered as an integral analogue of [Rib76, Lemma on p. 790].
Lemma 2.9. Let n 2 be an integer and G be a closed subgroup of n SL (Z ). Write π
≥ i=1 2 ℓ i
for the projection on the i-th factor, and suppose that for every i = j there is some non-negative
6 Q
integer s (with s 2 if ℓ = 2 and s 1 if ℓ = 3) such that the group (π π )(G) contains
ij ij ij i j
≥ ≥ ×
(s ,s ). Then G contains n s +(n 2)v .
Bℓ ij ij i=1Bℓ j6=i ij −
(cid:16) (cid:17)
Q P
5
Proof. Without loss of generality, by the symmetry of the problem it is enough to prove that G
contains Id Id s +(n 2)v . By lemma 2.8 we know that the group
{ }×···×{ }×Bℓ j6=n nj −
(cid:16) (cid:17)
s +(n 2)v is generPatedby Comm (g ,...,g ) g (s ) , so it suffices
Bℓ j6=n nj − n−2 1 n−1 j ∈Bℓ nj
mtoe(cid:16)nshtPo(wIdt,h..a.t,Ifodr,Ceovmermy c(cid:17)ho(igce,.o.f.(,gg1,..).),.g(cid:8)nB−1y)h∈ypotjh6=ensiBsℓ(wsenjc)anth(cid:12)(cid:12)efingdroxup,.G..,cxo(cid:9)ntains tGhesuelceh-
n−2 1 n−1 1 n−1
Q ∈
that π (x ) = Id and π (x ) = g for all i between 1 and n 1. Consider now the iterated
i i n i i
−
commutator g˜ = Comm (x ,...,x ) G . For i n 1, the i-th component of g˜ is triv-
n−2 1 n−1
∈ ≤ −
ial because we have π (g˜) = Comm π (x ),...,π (x ),...,π (x ) = Id; moreover, our
i n−2 i 1 i i i n−1
 
Id
choice of x ,...,x ensures that π (g˜) = Comm (g ,...,g ) holds. We have thus shown
1 n n n−2 1 n−1
| {z }
that (Id,...,Id,Comm (g ,...,g )) is an element of G.
n−2 1 n−1
Lemma 2.10. Let ℓ>2 and G be a closed subgroup of SL (Z )2. Suppose G contains an element
2 ℓ
g =(x,y)with [x]=[ Id]and[y] nontrivial andoforder primetoℓ. Then Gcontainsan element
±
of the form ( Id,z), where the order of [z] is the same as the order of [y].
±
Proof. Such an element is given by any limit point of the sequence gℓn.
Lemma 2.11. Let n 1 and G ,G be two closed subgroups of GL (Z )n (with G (4),G (4)
1 2 2 ℓ 1 2
trivialifℓ=2). Suppos≥ethatthetwogroups λg λ Z×,g G and µg µ Z×,g G
1 ∈ ℓ 1 ∈ 1 2 ∈ ℓ 2 ∈ 2
coincide: then G ,G have the same derived subgroup and the same Lie algebra.
1 2 (cid:8) (cid:12) (cid:9) (cid:8) (cid:12) (cid:9)
(cid:12) (cid:12)
Proof. The hypothesis is symmetric in G ,G , so it suffices to show the inclusions G′ G′ and
(G ) (G ). The hypothesis implies i1n pa2rticular that for all g G there exists1ν⊆(g )2 Z×
L 1 ⊆L 2 1 ∈ 1 1 ∈ ℓ
suchthatν(g )g G . Nowletx,ybeelementsofG : then[x,y]=[ν(x)x,ν(y)y] G′,so–since
1 1 ∈ 2 1 ∈ 2
G′ isgeneratedbyelementsoftheform[x,y]forx,y varyinginG –wehaveG′ G′. Asforthe
1 1 1 ⊆ 2
Lie algebra,simply notice that for all g G we have Θ (g )= 1 Θ (ν(g )g ) (G ).
1 ∈ 1 n 1 ν(g1) n 1 1 ∈L 2
Lemma 2.12. Let n be a positive integer and g SL (Z ) a matrix satisfying g Id (mod ℓ1+v)
2 ℓ
∈ ≡
and g Id (mod ℓn). Then Θ (g) 0 (mod ℓn).
1
6≡ 6≡
a b a−d b
Proof. Write g = , so that Θ (g)= 2 . Suppose by contradiction that we had
c d 1 c d−a
(cid:18) (cid:19) (cid:18) 2 (cid:19)
Θ (g) 0 (mod ℓn), that is c d a−d 0 (mod ℓn). It then follows a d (mod ℓn+v),
1 ≡ ≡ ≡ 2 ≡ ≡
and since 1 = det(g) = ad bc we find a2 d2 1 (mod ℓn+v). Since a 1 (mod ℓ1+v) by
− ≡ ≡ ≡
hypothesis, this implies a d 1 (mod ℓn), that is, g Id (mod ℓn), contradiction.
≡ ≡ ≡
2.3 Teichmu¨ller lifts
We now recall the definition of Teichmu¨ller lifts and some of their basic properties.
Definition 2.13. Let F be a finite extension of Q with residue field F = F . For an element
ℓ ℓk
[f] F we denote by ω([f]) the Teichmu¨ller lift of [f], that is to say the only element g that
F
∈ ∈O
reduces to [f] in F and satisfies gℓk =g.
Lemma 2.14. With the notation of the previous definition, the sequence fℓkn converges to ω([f])
when n tends to infinity.
Proof. Writef =ω([f]) u,whereureducesto1intheresiduefieldF. Itisclearthatthesequence
·
uℓkn converges to 1 in F, so fℓkn =ω([f])ℓkn uℓkn =ω([f]) uℓkn converges to ω([f]).
· ·
Lemma 2.15. Let g be an element of SL (Z ) such that [g] has order prime to ℓ and strictly
2 ℓ
greater than 2: then the sequence gℓ2n for n N converges to a certain g that satisfies gℓ2 =g .
∈ ∞ ∞ ∞
Moreover, if [g] is diagonalizable over F , the limit g even satisfies gℓ =g .
ℓ ∞ ∞ ∞
6
Proof. Theassumptionimplies that[g]hasdistincteigenvalues,hence thereexistsanextensionF
ofQ ,ofdegreeatmost2,overwhichg canbe writtenasg =P−1DP,whereD =diag(λ,λ−1)is
ℓ
diagonal and P is a base-change matrix. By the previous lemma, the sequence Dℓ2n converges to
diag(ω([λ]),ω([λ−1])),sothesequencegℓ2n =PDℓ2nP−1convergestoP−1diag(ω([λ]),ω([λ−1]))P,
whichsatisfiestheconclusionsinceω([λ])ℓ2 =ω([λ]). ForthesecondstatementwecantakeF =Q
ℓ
and use the fact that the Teichmu¨ller lifts satisfy ω([λ])ℓ =ω([λ]).
Proof. Let[λ±1]=[λ±1]betheeigenvaluesof[g ]=[g ]. Asintheproofofthepreviouslemmawe
1 2 1 2
seethatthesequence(g ,g )ℓ2n convergestoalimit(h ,h )suchthatthematricesh ,h havethe
1 2 1 2 1 2
same eigenvalues ω([λ ]),ω([λ−1]). The existence of Q is then clear if [h ]=[h ] is diagonalizable
i i 1 2
over F , and follows from [Lom14, Lemma 4.7] otherwise. The claim on the orders of [h ],[h ] is
ℓ 1 2
clear.
3 Odd ℓ, n = 2
In this section we establish the case n=2 of the main theorem when ℓ is an odd prime, together
with a refined version of it that applies to subgroups of GL (Z )2:
2 ℓ
Theorem 3.1. Let ℓ>2 be a prime number and G an open subgroup of GL (Z )2. Let G ,G be
2 ℓ 1 2
theprojectionsofGonthetwofactorsGL (Z ),andfori=1,2letn beapositiveintegersuchthat
2 ℓ i
G contains (n ). Suppose furthermore that for every (g ,g ) G we have det(g ) = det(g ).
i ℓ i 1 2 1 2
B ∈
At least one of the following holds:
1. G contains (20max n ,n ,20max n ,n )
ℓ 1 2 1 2
B { } { }
2. there exists a subgroup T of G, of index dividing 2 482, with the following properties:
·
(a) if (T) contains ℓksl (Z ) ℓksl (Z ) for a certain integer k 0, then T contains
2 ℓ 2 ℓ
L ⊕ ≥
(p,p), where p=2k+max 2k,8n ,8n .
ℓ 1 2
B { }
(b) for any (t ,t ) in T, if both [t ] and [t ] are multiples of the identity, then they are
1 2 1 2
equal.
We will derive this theorem from the corresponding statement for SL (Z )2:
2 2
Theorem 3.2. Let ℓ > 2 be a prime number and G an open subgroup of SL (Z )2. For i = 1,2
2 ℓ
let n be a positive integer such that G contains (n ). At least one of the following holds:
i i ℓ i
B
1. G′ contains (20max n ,n ,20max n ,n )
ℓ 1 2 1 2
B { } { }
2. there exists a subgroup T of G, of index dividing 482, with the following properties:
(a) if (T) contains ℓksl (Z ) ℓksl (Z ) for a certain integer k 0, then T′ contains
2 ℓ 2 ℓ
L ⊕ ≥
(p,p), where p=2k+max 2k,8n ,8n .
ℓ 1 2
B { }
(b) for any (t ,t ) in T, if both [t ] and [t ] are multiples of the identity, then they are
1 2 1 2
equal.
3.1 Strategy of proof
Given the amount of technical details required to establish theorem 3.2, before plunging into the
actual proof we try to give a general overview of the argument. Several ingredients are needed.
To begin with, when an element of G exists whose two projections g ,g on G ,G are sensibly
1 2 1 2
different (for example, g has trivial reduction modulo ℓ but g does not), then this element can
1 2
be used to show that G is stable under certain ‘projections’ from G to G Id and Id G :
1 2
×{ } { }×
inthis case,the assumptionthatG ,G containopentopologicalballs is enoughto conclude that
1 2
the same is true for G (in a quantitative manner). This line of argument is what leads to lemma
3.7. Thus, inprovingtheorem3.2one canassume –amongother things –thatG(ℓ) lookslike the
7
graph of an isomorphism: the reduction to this case is carried out in proposition 3.10. Moreover,
theorem 3.2 is essentially known (and due to Pink) when G is pro-ℓ, cf. theorem 3.4 below. We
wouldthereforeliketoreducetothecaseofpro-ℓgroups,whichwedobyconsideringthemaximal
normalpro-ℓ subgroupof G (denoted by N(G) in all that follows). Most of the remaining part of
the proof(sections3.3to3.6)isconcernedwithshowingthatwhenthe LiealgebraofT (a certain
subgroup of G of bounded index) is large, the same is true for the Lie algebra of N(T), to which
we can subsequently apply Pink’s theorem. The key observation in carrying out this last step is
the following: since N(T) is (by definition) normal in T, its Lie algebra is acted upon by all of
T, including those elements that do not belong to N(T). A careful study of the action of these
elements, whose details depend on the specific shape of T(ℓ), shows that the stability of (N(T))
L
by the conjugation action of T forces it to be not very different from (T), and an application of
L
Pink’s theorem (in the form of lemma 3.8) then concludes the argument.
Onefinaltechnicalingredientisourabilitytochangebasestosimplifythecomputations. More
precisely,letg beanyelementofGL (Z )2: thenthegroupGsatisfiestheassumptionsoftheorem
2 ℓ
3.2 (for certainvalues of n ,n ) if and only if gGg−1 does, because the topologicalballs (s) are
1 2 ℓ
B
invariantunderanyintegralchangeofbasis. Likewise,theconclusionofthetheoremonlyinvolves
topological balls, multiples of the identity, and open sets of the form ℓksl (Z ) ℓksl (Z ): all
2 ℓ 2 ℓ
⊕
of these objects are invariant under change of basis – notice that Θ(gtg−1) = gΘ(t)g−1 for any
t GL (Z )2 – and therefore the claim of theorem 3.2 is true for G if and only if it is true for
2 ℓ
∈
gGg−1. In other words, as it should be expected, theorem 3.2 depends on G only through its
conjugacy class, and therefore we are free to change basis every time doing so simplifies some of
our calculations.
Remark 3.3. The proof of theorem 3.2 is constructive,in the sense that the groupT – when we
are in case (2) – is described explicitly. Property (2b) will be clear from the construction, and we
will not comment further on it.
3.2 Preliminary reductions
One of the key ingredients of the proof of theorem 3.2 is the following result:
Theorem 3.4. Let ℓ > 2 be a prime number and k be a non-negative integer. Suppose G is
a closed pro-ℓ subgroup of SL (Z ) such that (G) contains ℓksl (Z ): then G′ contains (2k).
2 ℓ 2 ℓ ℓ
Similarly, if G is a closed pro-ℓ subgroup of SLL (Z )2 and (G) contains ℓksl (Z ) ℓksBl (Z ),
2 ℓ 2 ℓ 2 ℓ
L ⊕
then G′ contains (2k,2k).
ℓ
B
Proof. The proof is identical in the two cases, so let us only consider subgroups of SL (Z )2 (the
2 ℓ
case of SL (Z ) is also treated in [Lom14, proof of theorem 4.2]). Consider the basis
2 ℓ
0 1 0 0 1 0
x= , y = , h=
0 0 1 0 0 1
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) − (cid:19)
of sl (Z ). Since [h,x] = 2x, [h,y] = 2y and [x,y] = h, the derived subalgebra of ℓksl (Z )
2 ℓ 2 ℓ
is ℓ2ksl (Z ). By assumption (G) co−ntains ℓksl (Z ) ℓksl (Z ), so the derived subalgebra
2 ℓ 2 ℓ 2 ℓ
[ (G), (G)] contains ℓ2ksl (Z )L ℓ2ksl (Z ). Furtherm⊕ore, in the notation of [Pin93] we have
2 ℓ 2 ℓ
L L ⊕
that C = tr( (G) (G)) contains both tr (ℓkh,0) (ℓkh,0) = (2ℓ2k,0) and (0,2ℓ2k). Since C
L ·L ·
is by definition a topologically closed additive subgroup of Z2, this shows in particular that C
(cid:0) (cid:1)ℓ
contains ℓ2kZ ℓ2kZ . It then follows from [Pin93, Theorem 2.7] that
ℓ ℓ
⊕
G′ = g SL (Z )2 Θ (g) [ (G), (G)], tr(g) 2 C (2k,2k).
2 ℓ 2 ℓ
∈ ∈ L L − ∈ ⊇B
(cid:8) (cid:12) (cid:9)
(cid:12)
Proof. (of theorem 3.1 assuming theorem 3.2) Write det∗ for the map G det Z2 π1 Z given
−−→ ℓ −→ ℓ
by the composition of the usual determinant with the projection on the first coordinate of Z2.
ℓ
By assumption, an element (g ,g ) of G satisfies det(g ,g ) = (1,1) if and only if it satisfies
1 2 1 2
det∗(g ,g )=1.
1 2
8
Assume first that ℓ 5. Denote by T˜ the inverse image in G of an ℓ-Sylow of G(ℓ), and set
≤
T :=ker T˜ det∗ Z× Z×ℓ .
−−−→ ℓ → Z×2
(cid:18) ℓ (cid:19)
Using the fact detg = detg for every (g ,g ) G, and that the index of an ℓ-Sylow of G(ℓ)
1 2 1 2
∈
is prime to ℓ, it is easily checked that the index of T in G divides 2 1 |GL2(Fℓ)| 2 2 482.
· ℓ−1 ℓ ·
NLeomwmseat3T.116=(2)λ]goneλs∈eeZs×ℓe,agsi∈lyTtha∩t STL2a(nZdℓ)T2.1Uhsaivnegtlheme msaam2e.1d1erainvdedthsue(cid:16)bagrrgouumpeannt(cid:17)doft(cid:12)(cid:12)h[Leosma1m4e,
(cid:8) (cid:12) (cid:9)
Lie algebra, and m(cid:12)oreover it is clear by construction that T1 is a pro-ℓ subgroup of SL (Z )2.
2 ℓ
Furthermore, every element (t ,t ) of T reduces to ([Id],[Id]) modulo ℓ, so T satisfies (2b). Now
1 2
if (T)containsℓksl (Z ) ℓksl (Z ), thenthe same is truefor T1 , hence (T1)′ =T′ contains
2 ℓ 2 ℓ
L ⊕ L
(2k,2k) by theorem 3.4, and T has properties (2a) and (2b) as required.
ℓ
B (cid:0) (cid:1)
Next consider the case ℓ > 5. Let U be the subgroup of G, of index at most 2, given by
1
ker G det∗ Z× Z×/Z×2 . Let U = λg λ Z×,g U SL (Z )2, and notice as above
−−−→ ℓ → ℓ ℓ 2 ∈ ℓ ∈ 1 ∩ 2 ℓ
that(cid:16)U1 andU2 havethe sam(cid:17)ederivedsub(cid:8)grou(cid:12)p. Also notice th(cid:9)atU2 is openinSL2(Zℓ)2: indeed,
as G is open in GL (Z )2 there is an r > 0(cid:12)such that (r,r) G, and the definition of U
2 ℓ ℓ 2
B ⊆
shows that (r,r) is also contained in U . Moreover, since elements in (n ), (n ) have unit
ℓ 2 ℓ 1 ℓ 2
B B B
determinant,thetwoprojectionsofU onthefactorsGL (Z )contain (n ), (n )respectively.
2 2 ℓ ℓ 1 ℓ 2
B B
We can assume that U′ =U′ does not contain (20max n ,n ,20max n ,n ), for otherwise
2 1 Bℓ { 1 2} { 1 2}
the same holds for G and we are done. Apply then theorem 3.2 to (U ,n ,n ) to find a subgroup
2 1 2
T of U (of index dividing 482) that has properties (2a) and (2b) of that statement. Notice
2 2
that we have a well-defined morphism U ψ U /( (Id,Id)) sending g to the class of g/ det∗(g),
1 2
→ ±
where det∗g exists in Z× by construction of U . Let T be the image of T in the quotient
ℓ 1 2 2 p
U /( (Id,Id)). Now ψ is surjective by definition of U , so if we define T to be the inverse image
2 ±p 2
of T through ψ, then the index [U : T] divides 482 and [G : T] divides 2 482. Furthermore,
2 1
·
the Lie algebra of T and that of T agree, as do their derived subgroups (lemma 2.11). Suppose
2
now that (T) contains ℓksl (Z ) ℓksl (Z ): then the same is true for (T ), and therefore by
2 ℓ 2 ℓ 2
L ⊕ L
property(2a)oftheorem3.2weseethatT′ =T′ contains (p,p). Finally,let(t ,t )be inT and
suppose that [t ],[t ] are multiples of the2identity. By conBsℓtruction, there exists1a2scalar λ Z×
1 2 ∈ ℓ
and an element (w ,w ) T such that (t ,t )=λ(w ,w ); in particular, [w ],[w ] are multiples
1 2 2 1 2 1 2 1 2
∈
of the identity, so the properties of T force [t ]=[λ][w ]=[λ][w ]=[t ].
2 1 2 1 2
The following proposition allows us to assume that ℓ>5:
Proposition 3.5. Theorem 3.2 is true for ℓ 5.
≤
Proof. Take T to be the inverse image in G of a ℓ-Sylow of G(ℓ). It is clear that T satisfies (2b).
Furthermore, T is pro-ℓ, so an application of theorem 3.4 shows that T satisfies (2a); finally, it is
immediate to check that [G:T] divides |SL2(Fℓ)| 2 482.
ℓ
Assumption 3.6. From now on, we wo(cid:16)rk under(cid:17)the(cid:12)(cid:12)additional assumption that ℓ>5.
Our final objective is to compute, in terms of k,n and n , an integer p such that T contains
1 2
(p,p). This would be immediate if T were a pro-ℓ group, for then we would simply apply the-
ℓ
B
orem 3.4 as it is. In general, however, one needs to take into account the structure of T(ℓ), and
many different possibilities arise, according to the type of T (ℓ),T (ℓ) in the Dickson classifica-
1 2
tion. In some situations which we now discuss, the two projections G and G behave essentially
1 2
independently of one another; in this case the problem is greatly simplified, and it is possible to
exhibit an integer p as above without examining too closely the structure of G(ℓ):
Lemma 3.7. Let ℓ>5 and G, n ,n be as in theorem 3.2. Suppose that G contains an element
1 2
(a,b) such that [a] = [ Id] and the prime-to-ℓ part of the order of [b] is at least 3. Then G′
±
contains (4n +16n ,8n ) (20max n ,n ,20max n ,n ). The same conclusion is true
ℓ 1 2 2 ℓ 1 2 1 2
B ⊃B { } { }
if [b]=[ Id] and the prime-to-ℓ part of the order of [a] is at least 3.
±
9
Proof. The statement is clearly symmetric in a and b, so let us assume [a]=[ Id]. Since ℓ2 does
±
not divide the order of SL (F ), the element [b]ℓ has order prime to ℓ; replacing (a,b) with (a,b)ℓ
2 ℓ
allows us to assume that the order of [b] is prime to ℓ and not less than 3. By lemma 2.10, G
contains an element of the form ( Id,b′), where the order of [b′] is the same as the order of [b].
±
We can therefore assume a = Id. By hypothesis, for any g in (n ) there exists a g G
2 ℓ 2 1 1
± B ∈
such that (g ,g ) belongs to G: it follows that for any g (n ) the element
1 2 2 ℓ 2
∈B
( Id,b′)−1(g ,g )( Id,b′)(g ,g )−1 =(Id,(b′)−1g b′g−1)=(Id,[(b′)−1,g ]) (1)
± 1 2 ± 1 2 2 2 2
d 0
belongstoG. Uptoachoiceofbasis,wecanassumethateitherb′ = foracertainunitd,
0 1/d
(cid:18) (cid:19)
c dε 0 1
orb′ = forcertainunitsc,dandacertainεsuchthat[ε]isnotasquare,orb′ = −
d c 1 0
(cid:18) (cid:19) (cid:18) (cid:19)
(for the lasttwocasescf. [Lom14, Lemma 4.7]). In the firstcase,setting g2 =R(ℓn2)in (1)shows
that Id,R d−2 1 ℓn2 belongs to G. Given that d is not congruent to 1 modulo ℓ, for
− ±
otherwise the order of [b] would be 1 or 2, we see that the ℓ-adic valuation of (d−2 1)ℓn2 is
(cid:0) (cid:0)(cid:0) (cid:1) (cid:1)(cid:1) −
exactly n2. Similarly, G contains Id,L (d2 1)ℓn2 , and by lemma 2.7 (4) this implies that G
−
contains {1}×Bℓ(2n2). In the sec(cid:0)ond ca(cid:0)se we notice(cid:1)(cid:1)that R −2cℓd2(n12+(2ℓ+n2ℓ)n42)2 can be written as
(cid:16) (cid:17)
1+ℓn2 cℓn2(2+ℓn2) 1 cℓn2(2+ℓn2)
(b′)−1, − d(1+ℓn2) (b′)−1, 1+ℓn2 d(1+ℓn2) ,
" 0 1+1ℓn2 !#·" 0 1+ℓn2 !#
hence Id,R 2cℓ2n2(2+ℓn2)2 belongs to G; by lemma 2.7 (1) we then see that G contains
− d(1+ℓn2)4
Id,R (cid:16)ℓ2n2 (cid:16), and analogous(cid:17)id(cid:17)entities prove that G also contains Id,L ℓ2n2 . Lemma 2.7 (4)
then implies that G contains Id (4n ). Finally, in the last case we use the identities
ℓ 2
(cid:0) (cid:0) (cid:1)(cid:1) { }×B (cid:0) (cid:0) (cid:1)(cid:1)
(b′)−1,D(ℓn2) −1 =D(ℓn2(2+ℓn2))
(cid:2) (cid:3) 2+ℓn2
[D(ℓn2),R(ℓn2)]=R ℓ2n2(2+ℓn2) , [D(ℓn2),L(ℓn2)]=L ℓ2n2
− (1+ℓn2)2
(cid:18) (cid:19)
(cid:0) (cid:1)
which prove that G contains Id (2n ).
ℓ 2
{ }×B
Considernowanelementh=(h1,h2)ofGwhosefirstcoordinateish1 =R ℓ21−1ℓn1 ;suchan
elementexistsbyhypothesis. Theℓ(ℓ2 1)-thpowerofhisoftheformh′ = R(cid:16)ℓn1+1 ,h(cid:17)′ ,where
− 2
[h′2] = [h2]ℓ(ℓ2−1) = [h2]|SL2(Fℓ)| = [Id]. The ℓ4n2−1-th power of h′ (recall th(cid:0)at(cid:0)n2 > 0(cid:1)), th(cid:1)erefore,
is a certainh′′ = R ℓn1+4n2 ,h′2′ , where h′2′ ∈Bℓ(4n2). By what we alreadyproved, G contains
(Id,h′′), so it also contains h′′ (Id,h′′)−1 = R ℓn1+4n2 ,Id . The same argument shows that
2 (cid:0) (cid:0) (cid:1) · (cid:1) 2
L ℓn1+4n2 ,Id belongstoG,andwefinally(cid:0)ded(cid:0)uce that(cid:1)Gco(cid:1)ntainsBℓ(2n1+8n2)×{Id},hence
– since we also have G Id (4n ) – that G contains (2n +8n ,4n ). Taking derived
ℓ 2 ℓ 1 2 2
(cid:0) (cid:0) (cid:1) (cid:1) ⊇ { }×B B
subgroups and using lemma 2.8 then shows that G′ contains (4n +16n ,8n ).
ℓ 1 2 2
B
Lemma 3.8. Let ℓ > 5 be a prime, G a closed subgroup of SL (Z )2 and for i = 1,2 let n be
2 ℓ i
a non-negative integer such that G contains (n ). Let t be a non-negative integer and u be an
i ℓ i
elementofsl (Z )suchthatu 0 (mod ℓt+1).BSupposethat (N(G)) sl (Z ) sl (Z )contains
2 ℓ 2 ℓ 2 ℓ
(0,u): then G′ contains Id 6≡ (2t+8n ). Likewise, if u ,Lu sl (Z⊆) are su⊕ch that u ,u 0
ℓ 2 1 2 2 ℓ 1 2
{ }×B ∈ 6≡
(mod ℓt+1) and (N(G)) contains (u ,0) and (0,u ), then G′ contains (2t+8n ,2t+8n ).
1 2 ℓ 1 2
L B
Proof. Note firstthat –by the sameargumentas[Lom14, Lemma4.5]– the Lie algebra (N(G))
L
is stable under conjugation by G. The smallest Lie subalgebra of (N(G)) that contains (0,u)
L
and is stable under conjugation by G is 0 S(u), where S(u) is the smallest Lie subalgebra of
sl (Z ) that contains u and is stable unde⊕r conjugation by G . By virtue of lemma 2.4, and
2 ℓ 2
given that G2 contains ℓ(n2), the algebra S(u) contains ℓt+4n2sl2(Zℓ). It follows that (N(G))
contains0 ℓt+4n2sl2(ZℓB),andapplyingtheorem3.4wededucethatN(G)′(hencealsoG′)Lcontains
⊕
Id (2t+8n ). The second statement is now immediate.
ℓ 2
{ }×B
10