How far are p-adic Lie groups from algebraic groups?
aa r X i v : . [ m a t h . G R ] J a n HOW FAR ARE P-ADIC LIE GROUPS FROMALGEBRAIC GROUPS?
YVES BENOIST AND JEAN-FRANC¸ OIS QUINT
Abstract.
We show that, in a weakly regular p -adic Lie group G , the subgroup G u spanned by the one-parameter subgroups of G admits a Levi decomposition. As a consequence, there exists aregular open subgroup of G which contains G u Contents
1. Introduction 21.1. Motivations 21.2. Main results 21.3. Plan 32. Preliminary results 32.1. One-parameter subgroups 42.2. Weakly regular and regular p -adic Lie groups 43. Algebraic unipotent p -adic Lie group 63.1. Definition and closedness 63.2. Lifting one-parameter morphisms 83.3. Unipotent subgroups tangent to a nilpotent Lie algebra 103.4. Largest normal algebraic unipotent subgroup 114. Derivatives of one-parameter morphisms 124.1. Construction of one-parameter subgroups 124.2. The group G nc and its Lie algebra g nc p -adic Lie groups 175.2. The Levi decomposition of G u p -adic Lie groups 23References 24 Key words p-adic Lie group, algebraic group, unipotent subgroup, Levi decom-position, central extension Introduction
Motivations.
When studying the dynamics of the subgroups ofa p -adic Lie group G on its homogeneous spaces, various assumptionscan be made on G . For instance, one can ask G to be algebraic asin [10] i.e. to be a subgroup of a linear group defined by polynomialequations.Another possible assumption on G is regularity. One asks G to satisfyproperties that are well-known to hold for Zariski connected algebraic p -adic Lie groups: a uniform bound on the cardinality of the finitesubgroups and a characterization of the center as the kernel of theadjoint representation (see Definition 2.5). Ratner’s theorems in [13]are written under this regularity assumption.A more natural and weaker assumption on G in this context is theweak regularity of G i.e. the fact that the one-parameter morphismsof G are uniquely determined by their derivative (see Definition 2.3).Our paper [1] is written under this weak-regularity assumption.The aim of this text is to clarify the relationships between these threeassumptions.A key ingredient in the proof is Proposition 5.1. It claims the finiteness of the center of the universal topologicalcentral extension of non-discrete simple p -adic Lie groups. This is a fact which is due to Prasad and Raghunathan in [12].1.2.
Main results.
We will first prove (Proposition 5.5):
In a weakly regular p -adic Lie group G , the subgroup G u spanned by the one-parameter subgroups of G is closedand admits a Levi decomposition ,i.e. G u is a semidirect product G u = S u ⋉ R u of a group S u by a normalsubgroup R u where S u is a finite cover of a finite index subgroup of analgebraic semisimple Lie group and where R u is algebraic unipotent.As a consequence, we will prove (Theorem 5.12): In a weakly regular p -adic Lie group G there always existsan open subgroup G Ω which is regular and contains G u . This Theorem 5.12 is useful since it extends the level of generality ofRatner’s theorems in [13] (see [1, Th 5.15]). More precisely, Ratner’stheorems for products of real and p -adic Lie groups in [13] are provenunder the assumption that these p -adic Lie groups are regular . Ratner’stheorems can be extended under the weaker assumption that these p -adic Lie groups are weakly regular thanks to our Theorem 5.12. -ADIC LIE GROUPS 3 This Theorem 5.12 has been announced in [1, Prop. 5.8] and hasbeen used in the same paper.The strategy consists in proving first various statements for weaklyregular p -adic Lie groups which were proven in [13] under the regularityassumption. To clarify the discussion we will reprove also the resultsfrom [13] that we need. But we will take for granted classical resultson the structure of simple p -adic algebraic groups that can be found in[3], [6], [7] or [11].1.3. Plan.
In the preliminary Chapter 2 we recall a few definitions andexamples.Our main task in Chapters 3 and 4 is to describe, for a weakly regular p -adic Lie group G , the subset g G ⊂ g of derivatives of one-parametermorphisms of G .In Chapter 3, the results are mainly due to Ratner. We first studythe nilpotent p -adic Lie subgroups N of G spanned by one-parametersubgroups. We will see that they satisfy n ⊂ g G (Proposition 3.7).Those p -adic Lie groups N are called algebraic unipotent. We willsimultaneously compare this set g G with the analogous set g ′ G ′ for aquotient group G ′ = G/N of G when N is a normal algebraic unipotentsubgroup (Lemma 3.6). This will allow us to prove that G contains alargest normal algebraic unipotent subgroup (Proposition 3.8).In Chapter 4, we will then be able to describe precisely the set g G using a Levi decomposition of g (Proposition 4.4). A key ingredient isa technic, borrowed from [1], for constructing one-parameter subgroupsin a p -adic Lie group G (Lemma 4.1).In the last Chapter 5, we will prove the main results we have juststated: Proposition 5.5 and Theorem 5.12, using Prasad–Raghunathanfiniteness theorem in [12] (see Proposition 5.1). We will end this textby an example showing that, when a p -adic Lie group G with G = G u isnot assumed to be weakly regular, it does not always contain a regularopen subgroup H for which H = H u (Example 5.14).2. Preliminary results
We recall here a few definitions and results from [13].Let p be a prime number and Q p be the field of p -adic numbers.When G is a p -adic Lie group (see [5]), we will always denote by g theLie algebra of G . It is a Q p -vector space. We will denote by Ad g orAd the adjoint action of G on g and by ad g or ad the adjoint action of g on g . Any closed subgroup H of G is a p -adic Lie subgroup and itsLie algebra h is a Q p -vector subspace of g (see [13, Prop. 1.5]).We choose a ultrametric norm k . k on g with values in p Z . YVES BENOIST AND JEAN-FRANC¸ OIS QUINT
One-parameter subgroups.Definition 2.1.
A one-parameter morphism ϕ of a p -adic Lie group G is a continuous morphism ϕ : Q p → G ; t ϕ ( t ) . A one-parametersubgroup is the image ϕ ( Q p ) of an injective one-parameter morphism. The derivative of a one-parameter morphism is an element X of g forwhich ad X is nilpotent. This follows from the following Lemma from[13, Corollary 1.2]. Lemma 2.2.
Let ϕ : Q p → GL( d, Q p ) be a one-parameter morphism.Then there exists a nilpotent matrix X in gl ( d, Q p ) such that ϕ ( t ) =exp( tX ) for all t in Q p . Here exp is the exponential of matrices : exp( tX ) := P n ≥ t n X n n ! Proof.
First, we claim that, if K is a finite extension of Q p , any con-tinuous one-parameter morphism ψ : Q p → K ∗ is constant. Indeed,since the modulus of ψ is a continuous morphism form Q p to a discretemultiplicative subgroup of (0 , ∞ ), the kernel of | ψ | contains p k Z p , forsome integer k . Since Q p /p k Z p is a torsion group and (0 , ∞ ) has notorsion, ψ has constant modulus, that is, ψ takes values in O ∗ , where O is the integer ring of K . Now, on one hand, O ∗ is a profinite group,that is, it is a compact totally discontinuous group and hence it ad-mits a basis of neighborhoods of the identity which are finite indexsubgroups (namely, for example, the subgroups 1 + p k O , k ≥ Q p is of the form p k Z p , forsome integer k , and hence, has infinite index in Q p and therefore anycontinuous morphism from Q p to a finite group is trivial. Thus, ψ isconstant, which should be proved.Let now ϕ be as in the lemma and let X ∈ gl ( d, Q p ) be the derivativeof ϕ . After having simultaneously reduced the commutative familyof matrices ϕ ( t ) t ∈ Q p , the joint eigenvalues give continuous morphisms Q p → K ∗ where K is a finite extension of Q p . By the remark above,these morphisms are constant, that is, there exists g in GL( d, Q p ),such that, for any t in Q p , the matrix gϕ ( t ) g − is unipotent and uppertriangular. We may assume g = 1. Then X is nilpotent and uppertriangular and it remains to check that the map θ : t ϕ ( t )exp( − tX )is constant. Since ϕ ( t ) commutes with X , this map θ is a one-parametermorphism with zero derivative. Hence, the kernel of ψ is an opensubgroup and the matrices θ ( t ), t ∈ Q p , have finite order. Since theyare unipotent, they equal e , which should be proved. (cid:3) Weakly regular and regular p -adic Lie groups. -ADIC LIE GROUPS 5 Definition 2.3 (Ratner, [13]) . A p -adic Lie group G is said to beweakly regular if any two one-parameter morphisms Q p → G with thesame derivative are equal. Note that any closed subgroup of a weakly regular p -adic Lie groupis also weakly regular. Example . Every closed subgroup ofGL( d, Q p ) is weakly regular. Proof.
This follows from Lemma 2.2. (cid:3)
Definition 2.5 (Ratner, [13]) . A p -adic Lie group G is said to be Ad -regular if the kernel of the adjoint map Ker(Ad g ) is equal to the center Z ( G ) of G . It is said to be regular if it is Ad -regular and if the finitesubgroups of G have uniformly bounded cardinality. Note that any open subgroup of a regular p -adic Lie group is alsoregular.This definition is motivated by the following example. Example . a ) The finite subgroups of a compact p -adic Lie group K have uniformly bounded cardinality. b ) The finite subgroups of a p -adic linear group have uniformlybounded cardinality. c ) The Zariski connected linear algebraic p -adic Lie groups G areregular. Proof of Example 2.6. (see [13]) a ) Since K contains a torsion free open normal subgroup Ω, for everyfinite subgroup F of K , one has the bound | F | ≤ | K/ Ω | . b ) We want to bound the cardinality | F | of a finite subgroup of agroup G ⊂ GL( d, Q p ). This follows from a ) since F is included in aconjugate of the compact group K = GL( d, Z p ). c ) It remains to check that G is Ad-regular. Let g be an elementin the kernel of the adjoint map Ad g . This means that the centralizer Z g of g in G is an open subgroup of G . This group Z g is also Zariskiclosed. Hence it is Zariski open. Since G is Zariski connected, onededuces Z g = G and g belongs to the center of G . (cid:3) We want to relate the two notions “weakly regular” and “regular”.We first recall the following implication in [13, Cor. 1.3].
Lemma 2.7.
Any regular p -adic Lie group is weakly regular.Proof. Let ϕ : Q p → G and ϕ : Q p → G be one-parameter mor-phisms of G with the same derivative. We want to prove that ϕ = ϕ . YVES BENOIST AND JEAN-FRANC¸ OIS QUINT
According to Lemma 2.2, the one-parameter morphisms Ad g ϕ andAd g ϕ are equal. Since G is Ad-regular, this implies that, for all t in Q p , the element θ ( t ) := ϕ ( t ) − ϕ ( t ) is in the center of G . Hence θ isa one-parameter morphism of the center of G with zero derivative. Itsimage is either trivial or an infinite p -torsion group. This second caseis excluded by the uniform bound on the finite subgroups of G . Thisproves the equality ϕ = ϕ . (cid:3) The aim of this text is to prove Theorem 5.12 which is a kind ofconverse to Lemma 2.7.3.
Algebraic unipotent p -adic Lie group In this chapter, we study the algebraic unipotent sub-groups of a weakly regular p -adic Lie group. The resultsin this chapter are mainly due to Ratner.3.1. Definition and closedness.
We first focus on a special class of p -adic Lie groups.We say that an element g of a p -adic Lie group G admits a logarithmif one has g p n −−−→ n →∞ e : indeed, for such a g , the map n g n extendsas a continuous morphism Z p → G and one can define the logarithmlog( g ) ∈ g as being the derivative at 0 of this morphism. Definition 3.1. A p -adic Lie group N is called algebraic unipotent ifits Lie algebra is nilpotent, if every element g in N admits a logarithm log( g ) , and if the logarithm map log : N → n is a bijection. This implies that every non trivial element g of N belongs to a uniqueone-parameter subgroup of N . By definition these groups N are weaklyregular. The inverse of the map log is denoted by exp. These mapsexp and log are Aut( N )-equivariant.The following lemma is in [13, Prop. 2.1]. Lemma 3.2.
Let N be an algebraic unipotent p -adic Lie group. Thenthe map n × n → n ; ( X, Y ) log(exp X exp Y ) is polynomial and is given by the Baker-Campbell-Hausdorff formula.In other words, a p -adic Lie group N is algebraic unipotent if and onlyif it is isomorphic to the group of Q p -points of a unipotent algebraicgroup defined over Q p . These groups have been studied in [13, Sect.2] (where they are called quasiconnected ).In particular, we have -ADIC LIE GROUPS 7
Corollary 3.3.
Let N be an algebraic unipotent p -adic Lie group. Thenthe exponential and logarithm maps of N are continuous.Proof of Lemma 3.2. In this proof, we will say that a p -adic Lie groupis strongly algebraic unipotent if it is isomorphic to the group of Q p -points of a unipotent algebraic group defined over Q p . Such a group isalways algebraic unipotent.The aim of this proof is to check the converse. Let N be an algebraicunipotent p -adic Lie group. We want to prove that N is strongly alge-braic unipotent. Its Lie algebra n contains a flag 0 ⊂ n ⊂ · · · ⊂ n r = n of ideals with dim n i = i . We will prove, by induction on i ≥
1, that theset N i := exp( n i ) is a closed subgroup of N which is strongly algebraicunipotent.By the induction assumption, the set N i − is a strongly algebraicunipotent closed subgroup. Since n i − is an ideal, this subgroup N i − is normal. Let X i be an element of n i r n i − and N ′ i the semidirectproduct N ′ i := Q p ⋉ N i − where the action of t ∈ Q p by conjugationon N i − is given by t exp( X ) t − = exp( e t ad X i X ), for all X in n i − .By construction this group N ′ i is strongly algebraic unipotent and themap ψ : N ′ i → N ; ( t, n ) exp( tX i ) n is a group morphism. Since N ′ i =exp( n ′ i ), the set N i = exp( n i ) is equal to the image ψ ( N ′ i ). Hence, byLemma 3.4 below, the set N i is a closed subgroup which is isomorphicto N ′ i and hence N i is strongly algebraic unipotent. (cid:3) The following lemma tells us that an algebraic unipotent Lie sub-group of a p -adic Lie group is always closed. Lemma 3.4.
Let G be a totally discontinuous locally compact topologi-cal group, N be an algebraic unipotent p -adic Lie group and ϕ : N → G be an injective morphism. Then ϕ is a proper map. In particular, ϕ ( N ) is a closed subgroup of G and ϕ is is an isomorphism of topologicalgroups from N onto ϕ ( N ) .Proof. If ϕ was not proper, there would exist a sequence Y n in the Liealgebra n of N such thatlim n →∞ ϕ (exp( Y n )) = e and lim n →∞ k Y n k = ∞ . We write Y n = p − k n X n with integers k n going to ∞ and k X n k = 1.Since the group G admits a basis of compact open subgroups, one alsohas lim n →∞ ϕ (exp( X n )) = e . Let X be a cluster point of the sequence X n .One has simultaneously, ϕ (exp( X )) = e and k X k = 1. This contradictsthe injectivity of ϕ . (cid:3) YVES BENOIST AND JEAN-FRANC¸ OIS QUINT
Lifting one-parameter morphisms.
We now explain how to lift one-parameter morphisms.
Lemma 3.5.
Let G be a p -adic Lie group, N ⊂ G a normal algebraicunipotent closed subgroup, G ′ := G/N , and π : G → G ′ the projection.Then, for any one-parameter morphism ϕ ′ of G ′ , there exists a one-parameter morphism ϕ of G which lifts ϕ ′ , i.e. such that ϕ ′ = π ◦ ϕ .If ϕ ′ has zero derivative, one can choose ϕ to have zero derivative.Proof. According to Lemma 3.4 the image ϕ ′ ( Q p ) is a closed subgroupof G ′ . Hence we can assume that ϕ ′ ( Q p ) = G ′ . First Case : N is central in G . For k ≥
1, we introduce the subgroup Q ′ k of G ′ spanned by the element g ′ k := ϕ ′ ( p − k ). Since Q ′ k is cyclic and N is central, the group Q k := π − ( Q ′ k ) is abelian. Since the increasingunion of these groups Q k is dense in G , the group G is also abelian.Since N is infinitely p -divisible, one can construct, by induction on k ≥
0, a sequence ( g k ) k ≥ in G such that π ( g k ) = g ′ k and p g k +1 = g k . We claim that g p k −−−→ k →∞ e . Indeed, since π ( g ) p k −−−→ k →∞ e and sinceevery element h of a p -adic Lie group that is close enough to the identityelement satisfies h p k −−−→ k →∞ e , one can find ℓ ≥ n in N such that( g p ℓ n − ) p k −−−→ k →∞ e . As N is algebraic unipotent, we have n p k −−−→ k →∞ e ,and the claim follows, since N is central.Now, the formulae ϕ ( p − k ) = g k , for all k ≥
0, define a unique one-parameter morphism ϕ of G which lifts ϕ ′ .Note that when ϕ ′ has zero derivative, one can assume, after a repara-metrization of ϕ ′ , that ϕ ′ ( Z p ) = 0 and choose the sequence g k so that g = e . Then the morphism ϕ has also zero derivative. General Case : The composition of ϕ ′ with the action by conjugationon the abelianized group N/ [ N, N ] ≃ Q dp is a one-parameter morphism ψ : Q p → GL( d, Q p ). According to Lemma 2.2, there exists a nilpotentmatrix X such that ψ ( t ) = exp( tX ) for all t ∈ Q p . The image ofthis matrix X corresponds to an algebraic unipotent subgroup N with[ N, N ] ⊂ N ( N which is normal in G and such that N/N is a centralsubgroup of the group G/N . According to the first case, the morphism ϕ ′ can be lifted as a morphism ϕ ′ of G/N . By an induction argumenton the dimension of N , this morphism ϕ ′ can be lifted as a morphismof G . (cid:3) -ADIC LIE GROUPS 9 Let G be a p -adic Lie group. We recall the notation(3.1) g G := { X ∈ g derivative of a one-parameter morphism of G } . Note that this set g G is invariant under the adjoint action of G .The following lemma tells us various stability properties by extensionwhen the normal subgroup is algebraic unipotent. Lemma 3.6.
Let G be a p -adic Lie group, N a normal algebraic unipo-tent subgroup of G , and G ′ := G/N . a ) One has the equivalence G is algebraic unipotent ⇐⇒ G ′ is algebraic unipotent. b ) Let X in g and X ′ its image in g ′ = g / n . One has the equivalence X ∈ g G ⇐⇒ X ′ ∈ g ′ G ′ . c ) One has the equivalence G is weakly regular ⇐⇒ G ′ is weakly regular. Later on in Corollary 5.7 we will be able to improve this Lemma.
Proof.
We denote by π : G → G/N the natural projection. a ) The implication ⇒ is well-known. Conversely, we assume that N and G/N are algebraic unipotent and we want to prove that G isalgebraic unipotent. Arguing by induction on dim G/N , we can assumedim
G/N = 1, i.e. that there exists an isomorphism ϕ ′ : Q p → G/N .According to Lemma 3.5, one can find a one-parameter morphism ϕ of G that lifts ϕ ′ . By Lemma 3.4, the image Q := ϕ ( Q p ) is closed and G is the semidirect product G = Q ⋉ N . By Lemma 2.2, the one-parameter morphism t Ad n ϕ ( t ) is unipotent, and hence the group G is algebraic unipotent. b ) The implication ⇒ is easy. Conversely, we assume that X ′ is thederivative of a one-parameter morphism ϕ ′ of G ′ . When X ′ = 0, theelement X belongs to n and, since N is algebraic unipotent, X is thederivative of a one-parameter morphism ϕ of N . We assume now that X ′ = 0 so that the group Q ′ := ϕ ′ ( Q p ) is algebraic unipotent andisomorphic to Q p . According to point a ), the group H := π − ( Q ′ ) isalgebraic unipotent. Since the element X belongs to the Lie algebra h of H , it is the derivative of a one-parameter morphism ϕ ′ of H . c ) ⇒ We assume that G is weakly regular. Let ϕ ′ and ϕ ′ be one-parameter morphisms of G ′ with the same derivative X ′ ∈ g ′ . We wantto prove that ϕ ′ = ϕ ′ . If this derivative X ′ is zero, by Lemma 3.5,we can lift both ϕ ′ and ϕ ′ as one-parameter morphisms of G withzero derivative. Since G is weakly regular, both ϕ and ϕ are trivialand ϕ ′ = ϕ ′ . We assume now that the derivative X ′ is non-zero.As above, for i = 1, 2, the groups Q ′ i := ϕ ′ i ( Q p ) and H i := π − ( Q ′ i )are algebraic unipotent. Since G is weakly regular, and its algebraic unipotent subgroups H and H have the same Lie algebra, one getssuccessively H = H , Q ′ = Q ′ , and ϕ ′ = ϕ ′ . ⇐ We assume that
G/N is weakly regular. Let ϕ and ϕ be one-parameter morphisms of G with the same derivative X ∈ g . We wantto prove that ϕ = ϕ . Since G/N is weakly regular the one-parametermorphisms π ◦ ϕ and π ◦ ϕ are equal and their image Q ′ is a unipotentalgebraic subgroup of G ′ . According to point a ), the group H := π − ( Q ′ ) is algebraic unipotent. Since ϕ and ϕ take their values in H ,one has ϕ = ϕ . (cid:3) Unipotent subgroups tangent to a nilpotent Lie algebra.
Proposition 3.7 below describes the nilpotent Lie sub-groups of a weakly regular p -adic Lie group G which arespanned by one-parameter morphisms.The following proposition is due to Ratner in [13, Thm. 2.1]. Proposition 3.7.
Let G be a weakly regular p -adic Lie group and n ⊂ g be a nilpotent Lie subalgebra. Then the set n G := n ∩ g G is an ideal of n and there exists an algebraic unipotent subgroup N G of G with Liealgebra n G . This group N G is unique. It is a closed subgroup of G . By construc-tion, it is the largest algebraic unipotent subgroup whose Lie algebrais included in n . Proof.
We argue by induction on dim n . We can assume n G = 0. First case : n is abelian. Let X , . . . , X r be a maximal family oflinearly indepent elements of n G and, for i ≤ r , let ϕ i be the one-parameter morphism with derivative X i . Since G is weakly regular,the group spanned by the images ϕ i ( Q p ) is commutative and the map ψ : Q dp → G ; ( t , . . . , t r ) ϕ ( t ) . . . ϕ r ( t r )is an injective morphism. Its image is a unipotent algebraic subgroup N G of G whose Lie algebra is n G . Second case : n is not abelian. Let z be the center of n and z theideal of n such that z / z is the center of n / z . If n G is included in thecentralizer n ′ of z , we can apply the induction hypothesis to n ′ . Weassume now that n G is not included in n ′ , i.e. there exists X ∈ n G and Y ∈ z such that [ X, Y ] = 0 . This element Z := [ X, Y ] belongs to the center z . -ADIC LIE GROUPS 11 We first check that Z belongs also to g G . Indeed, let m be the2-dimensional Lie subalgebra of n with basis X, Z . This Lie algebrais normalized by Y . For ε ∈ Q p small enough, there exists a groupmorphism ψ : ε Z p → G whose derivative at 0 is Y , and one hasAd( ψ ( ε )) Y = e ε ad Y X = X − εZ . Since X belongs to g G , the element X − εZ also belongs to g G . By the first case applied to the abelian Liesubalgebra m , the element Z belongs to g G .This means that there exists a one-parameter subgroup U of G whoseLie algebra is u = Q p Z . Let C be the centralizer of U in G . Accordingto Lemma 3.6, the quotient group C/U is also weakly regular. We applyour induction hypothesis to this group C ′ := C/U and the nilpotentLie algebra n / u . There exists a largest algebraic unipotent subgroup N ′ C ′ in C ′ whose Lie algebra is included in n ′ . Hence, using againLemma 3.6, there exists a largest algebraic unipotent subgroup N C of C whose Lie algebra is included in n . Since G is weakly regular, anyone-parameter subgroup of G tangent to n is included in C and N C isalso the largest algebraic unipotent subgroup of G whose Lie algebrais included in n . (cid:3) Largest normal algebraic unipotent subgroup.
We prove in this section that a weakly regular p -adicLie group contains a largest normal algebraic unipotentsubgroup.Let G be a p -adic Lie group. We denote by G u the closure of thesubgroup G u of G generated by all the one-parameter subgroups of G .This group G u is normal in G . We denote by g u the Lie algebra of G u .It is an ideal of g .We recall that the radical r of g is the largest solvable ideal of g and that the nilradical n of g is the largest nilpotent ideal of g . Thenilradical is the set of X in r such that ad X is nilpotent and one has[ g , r ] ⊂ n (see [5]).When G is weakly regular, we denote by R u the largest algebraicunipotent subgroup of G whose Lie algebra is included in n . It existsby Proposition 3.7.The following proposition is mainly in [13, Lem. 2.2]. Proposition 3.8.
Let G be a weakly regular p -adic Lie group. a ) The group R u is the largest normal algebraic unipotent subgroup of G . b ) Its Lie algebra r u is equal to r u = n ∩ g G = r ∩ g G . c ) One has the inclusion [ g u , r ] ⊂ r u . d ) Let G ′ = G/R u . Let X be in g and X ′ be its image in g ′ = g / r u . One has the equivalence X ∈ g G ⇐⇒ X ′ ∈ g ′ G ′ .Proof. a) We have to prove that any normal algebraic unipotent sub-group U of G is included in R u . Indeed the Lie algebra u of U is anilpotent ideal of g , hence it is included in n and U is included in R u . b ) We already know the equality r u = n ∩ g G from Proposition 3.7. Itremains to check the inclusion r ∩ g G ⊂ n . Indeed, let X be an elementin r ∩ g G . Since X is the derivative of a one-parameter morphism, byLemma 2.2, the endomorphism ad X is nilpotent. Since X is also inthe radical r , X has to be in the nilradical n . c ) We want to prove that the adjoint action of G u on the quotientLie algebra r / r u is trivial. That is, we want to prove that, for all X ∈ g G and Y ∈ r , one has [ X, Y ] ∈ r u . By Lemma 2.2, the endomorphism ad X is nilpotent. Since Y is in r ,the bracket [ X, Y ] belongs to n and the vector space m := Q p X ⊕ n isa nilpotent Lie algebra normalized by Y . Hence, by Proposition 3.7,the set m ∩ g G is a nilpotent Lie algebra. This set is normalized by Y since it is invariant by e ε ad Y for ε ∈ Q p small enough. In particular theelement [ X, Y ] belongs to g G and hence to r u . d ) This is a special case of Proposition 3.7. (cid:3) Derivatives of one-parameter morphisms
In this chapter, we describe the set g G of derivatives ofone-parameter morphisms of a weakly regular p -adic Liegroup G .4.1. Construction of one-parameter subgroups.
We explain first a construction of one-parameter mor-phisms of G borrowed from [1] and [2]. Lemma 4.1.
Let G be a p -adic Lie group and g ∈ G . Then the vectorspace g + g := { v ∈ g | lim n →∞ Ad g − n v = 0 } is included in g G . Note that g + g is a nilpotent Lie subalgebra of g .The proof relies on the existence of compact open subgroups of G for which the exponential map satisfies a nice equivariant property.We need some classical definition (see [8]). A p -adic Lie group Ω issaid to be a standard group if there exists a Q p -Lie algebra l and acompact open sub- Z p -algebra O of l such that the Baker-Campbell-Hausdorff series converges on O and Ω is isomorphic to the p -adicLie group O equipped with the group law defined by this formula. -ADIC LIE GROUPS 13 In this case, l identifies canonically with the Lie algebra of Ω, everyelement of Ω admits a logarithm and the logarithm map induces anisomorphism Ω → O . If G is any p -adic Lie group, it admits a standardopen subgroup (see [8, Theorem 8.29]). If Ω is such a subgroup andif O is the associated compact open sub- Z p -algebra of g , we denoteby exp Ω : O → Ω the inverse diffeomorphism of the logarithm mapΩ → O .Note that if Ω and Ω ′ are standard open subgroups of G , the mapsexp Ω and exp Ω ′ coincide in some neighborhood of 0 in g . Lemma 4.2.
Let G be a p -adic Lie group, Ω ⊂ G a standard opensubgroup and exp Ω : O → Ω the corresponding exponential map. Forevery compact subset K ⊂ G , there exists an open subset O K ⊂ g whichis contained in O and in all the translates Ad g − ( O ) , g ∈ K , and suchthat one has the equivariance property exp Ω (Ad g ( v )) = g exp Ω ( v ) g − for any v ∈ O K , g ∈ K. Proof.
We may assume that K contains e . The intersection Ω K := ∩ g ∈ K g − Ω g is an open neighborhood of e in G . We just choose O K tobe the open set O K := log(Ω K ). (cid:3) Proof of Lemma 4.1.
This is [1, Lem. 5.4]. For the sake of complete-ness, we recall the proof. Fix g ∈ G . Let Ω be a standard open sub-group of G with exponential map exp Ω : O → Ω. By Lemma 4.2, thereexists an open additive subgroup U ⊂ O ∩ g + g such that Ad g − U ⊂ O and that exp Ω ( u ) = g exp Ω (Ad g − u ) g − for any u in U .
After eventually replacing U by T k ≥ Ad g k U , we can assume Ad g − U ⊂ U . Now, for k ≥
0, let U k := Ad g k U and define a continuous map ψ k : U k → G by setting ψ k ( u ) = g k exp Ω (Ad g − k u ) g − k for any u in U k . We claim that, for any k , one has ψ k = ψ k − on U k − = Ad g − U k .Indeed, let u be in U k . As u k := Ad g − k u belongs to U , we have ψ k +1 ( u ) = g k ( g exp Ω (Ad g − u k ) g − ) g − k = g k exp Ω ( u k ) g − k = ψ k ( u ) . Therefore, as g + g = S k ≥ U k , one gets a map ψ : g + g → G ′ whoserestriction to any U k , k ≥
0, is ψ k . For every v in g + g , the map t ψ ( tv )is a one-parameter morphism of G whose derivative is equal to v . (cid:3) The group G nc and its Lie algebra g nc . We introduce in this section a normal subgroup G nc of G which contains G u .Let G be a p -adic Lie group and g nc be the smallest ideal of the Liealgebra g such that the Lie algebra s c := g / g nc is semisimple and suchthat the adjoint group Ad s c ( G ) is bounded in the group Aut( s c ) ofautomorphisms of s c . Let G nc be the kernel of the adjoint action in s c ,i.e. G nc := { g ∈ G | for all X in g , Ad g ( X ) − X ∈ g nc } By construction G nc is a closed normal subgroup of G with Lie algebra g nc . Lemma 4.3.
Let G be a p -adic Lie group. Any one-parameter mor-phism of G takes its values in G nc In other words, the group G u is included in G nc . Proof.
Let ϕ be a one-parameter morphism of G . Then Ad s c ◦ ϕ is aone-parameter morphism of Aut( s c ) whose image is relatively compact.By Lemma 2.2, this one-parameter morphism is trivial and ϕ takes itsvalues in G nc . (cid:3) Derivatives and Levi subalgebras.
We can now describe precisely which elements of g aretangent to one-parameter subgroups of G .We recall that an element X of a semisimple Lie algebra s is said tobe nilpotent if the endomorphism ad s X is nilpotent. In this case, forany finite dimensional representation ρ of s , the endomorphism ρ ( X )is also nilpotent.We recall that a Levi subalgebra s of a Lie algebra g is a maximalsemisimple Lie subalgebra, and that one has the Levi decomposition g = s ⊕ r .The following proposition is proven in [13, Th. 2.2] under the addi-tional assumption that G is Ad-regular. Proposition 4.4.
Let G be a weakly regular p -adic Lie group, r be theradical of g , s a Levi subalgebra of g and s u := s ∩ g nc . One has theequality : g G = { X ∈ s u ⊕ r u | ad X is nilpotent } . It will follow from Lemma 4.7 that the Lie algebra s u ⊕ r u does notdepend on the choice of s .The key ingredient in the proof of Proposition 4.4 will be Lemma4.1. We will begin by three preliminary lemmas. The first two lemmasare classical. -ADIC LIE GROUPS 15 Lemma 4.5.
Let V = Q dp and G be a subgroup of GL( V ) such that V is an unbounded and irreducible reprentation of G . Then G containsan element g with at least one eigenvalue of modulus not one.Proof. Let A be the associative subalgebra of End( V ) spanned by G .Since V is irreducible, the associative algebra A is semisimple and thebilinear form ( a, b ) tr ( ab ) is non-degenerate on A (see [9, Ch. 17]).If all the eigenvalues of all the elements of G have modulus 1, thisbilinear form is bounded on G × G . Since A admits a basis includedin G , for any a in A , the linear forms b tr ( ab ) on A are bounded onthe subset G . Hence G is a bounded subset of A . (cid:3) Lemma 4.6.
Let s be a simple Lie algebra over Q p and S := Aut( s ) .All open unbounded subgroups J of S have finite index in S .Proof. Since J is unbounded, by Lemma 4.5, it contains an element g with at least one eigenvalue of modulus not one. Since J is open, theunipotent Lie subgroups U + = { g ∈ S | lim n →∞ g − n gg n = e } and U − = { g ∈ S | lim n →∞ g n gg − n = e } are included in J . By [4, 6.2. v and 6.13], J has finite index in S . (cid:3) The third lemma contains the key ingredient.
Lemma 4.7.
Let G be a weakly regular p -adic Lie group, r the radicalof g , s a Levi subalgebra of g and s u := s ∩ g nc . a ) One has the inclusion [ s u , r ] ⊂ r u . b ) Every nilpotent element X in s u belongs to g G . Note that Lemma 4.7. a does not follow from Proposition 3.8. c , sincewith the definitions of g u and s u that we have given, we do not knowyet that s u = s ∩ g u . Proof. a) By Proposition 3.8, we can assume r u = 0. We want toprove that [ s u , r ] = 0. Let s i , i = 1 , . . . , ℓ be the simple ideals of s u .Replacing G by a finite index subgroup, we can also assume that theideals s i ⊕ r are G -invariant. Similarly, let r j , j = 1 , . . . , m be thesimple subquotients of a Jordan-H¨older sequence of the G -module r .On the one hand, by assumption, for all i ≤ ℓ , the group Ad s i ⊕ r / r ( G )is unbounded. Hence, by Lemma 4.5, there exists an element g i in G and X i in g + g i ∩ ( s i ⊕ r ) whose image in g / r is non zero. By Lemma 4.1,there exists a one-parameter morphism ϕ i of G whose derivative is X i .On the other hand, since r u = 0, by the same Lemma 4.1, for ev-ery g in G , all the eigenvalues of Ad r ( g ) have modulus 1. By Lemma4.5, for all j ≤ m , the image Ad r j ( G ) of G in any simple subquotient r j is bounded. In particular, the one-parameter morphisms Ad r j ◦ ϕ i are bounded. Hence, by Lemma 2.2, one has ad r j ( X i ) = 0. Since r j is a simple g -module, the Lie algebra ad r j ( g ) is reductive and con-tains ad r j ( s i ) as an ideal. Since s i is a simple Lie algebra, this impliesad r j ( s i ) = 0. Since the action of s u on r is semisimple, this implies theequality [ s u , r ] = 0. b ) As in a ), we can assume that G preserves the ideals s i ⊕ r andthat r u = 0. According to this point a ), the Lie algebras s u and r commute and hence s u is the unique Levi subalgebra of g nc (see [5, § g in G , one has Ad g ( s u ) = s u . Let X be a nilpotent element of s u . We want to prove that X is thederivative of a one parameter morphism of G . By Jacobson-Morozovtheorem, there exists an automorphism ψ of s u such that ψ ( X ) = p − X .Since, for every simple ideal s i of s , the subgroup Ad s i ⊕ r / r ( G nc ) ⊂ Aut( s i ) is unbounded and open, this subgroup has finite index. Hence,remembering also (4.1), there exists k ≥ g in G nc such thatAd g ( X ) = p − k X . Then, by Lemma 4.1, X is the derivative of a one-parameter morphism of G . (cid:3) Proof of Proposition 4.4.
We just have to gather what we have provedso far. By Proposition 3.8, we can assume r u = 0. Let X ∈ g . Wewrite X = X s + X r with X s ∈ s and X r ∈ r .Proof of the inclusion ⊂ . Assume that X is the derivative of a one-parameter morphism ϕ of G . By Lemma 4.3, X belongs to g nc andhence X s belongs to s u . By Lemma 2.2, the endomorphism ad g X isnilpotent, and hence X s is a nilpotent element of the semisimple Liealgebra s . According to Lemma 4.7, X s and X r commute and X s isthe derivative of a one-parameter morphism ϕ s of G . Then X r is alsothe derivative of a one-parameter morphism ϕ r of G , the one given by t ϕ s ( t ) − ϕ ( t ). Hence X r belongs to r u .Proof of the inclusion ⊃ . Assume that X s belongs to s u , X r belongsto r u and ad X is nilpotent. By Lemma 4.7, X s and X r commuteand X s is the derivative of a one-parameter morphism ϕ s of G . Byassumption X r is the derivative of a one-parameter morphism ϕ r of G .Hence X is also the derivative of a one-parameter morphism ϕ of G ,the one given by t ϕ s ( t ) ϕ r ( t ). Hence X belongs to g G . (cid:3) -ADIC LIE GROUPS 17 Groups spanned by unipotent subgroups
In this chapter, we prove the two main results Propo-sition 5.5 and Theorem 5.12 that we announced in theintroduction.5.1.
Semisimple regular p -adic Lie groups. We recall first a nice result due to Prasad-Raghunathanwhich is an output from the theory of congruence sub-groupsLet s be a semisimple Lie algebra over Q p , Aut( s ) the group ofautomorphisms of s and S + := Aut( s ) u ⊂ Aut( s ) the subgroup spannedby the one-parameter subgroups of Aut( s ). We will say that s is totallyisotropic if s is spanned by nilpotent elements. In this case S + is anopen finite index subgroup of Aut( s ), see [4, 6.14]. Since s = [ s , s ],the group S + is perfect i.e. S + = [ S + , S + ]. In particular this groupadmits a universal central topological extension e S + , i.e. a group whichis universal among the central topological extension of S + . In [12]Prasad-Raghunathan were able to describe this group e S + (special caseswere obtained before by Moore, Matsumoto and Deodhar). We willonly need here the fact that this group is a finite extension of S + . Proposition 5.1. (Prasad–Raghunathan)
Let s be a totally isotropicsemisimple p -adic Lie algebra. Then the group S + admits a universaltopological central extension −→ Z −→ e S + π −→ S + −→ and its center Z is a finite group. The word universal means that for all topological central extension1 −→ Z E −→ E π −→ S + −→ E is a locally compact group and Z E is a closed central subgroup,there exists a unique continuous morphism ψ : e S + → E such that π = π ◦ ψ . Proof.
See [12, Theorem 10.4]. (cid:3)
Remark . This result does not hold for real Lie groups: indeed, thecenter Z of the universal cover of SL(2 , R ) is isomorphic to Z . Corollary 5.3.
Let s be a totally isotropic semisimple p -adic Lie al-gebra. For every topological central extension −→ Z E −→ E π −→ S + −→ with E = [ E, E ] , the group Z E is finite. Proof of Corollary 5.3.
Let ψ : e S + → E be the morphism given by theuniversal property. Since, by Proposition 5.1, the group Z is finite,the projection π = π ◦ ψ is a proper map, hence ψ is also a proper mapand the image ψ ( e S + ) is a closed subgroup of E . Since E = ψ ( e S + ) Z E ,one has the inclusion [ E, E ] ⊂ ψ ( e S + ), and the assumption E = [ E, E ]implies that the morphism ψ is onto. Hence the group Z E = ψ ( Z ) isfinite. (cid:3) Remark . For real Lie groups, the center Z E might even be nondiscrete. Such an example is given by the quotient E of the product R × ^ SL(2 , R ) by a discrete subgroup of R × Z whose projection on R is dense.5.2. The Levi decomposition of G u . We prove in this section that in a weakly regular p -adicLie group G , the subgroup G u is closed and admits aLevi decomposition.Let G be a weakly regular p -adic Lie group and r be the solvableradical of g . We recall that g nc is the smallest ideal of g containing r such that the group Ad g / g nc ( G ) is bounded. Let s be a Levi subalgebraof g and s u := s ∩ g nc .We recall that G u is the subgroup of G spanned by all the one-parameter subgroups of G , that R u is the subgroup of G spanned byall the one-parameter subgroups of G tangent to r , and we define S u as the subgroup of G spanned by all the one-parameter subgroups of G tangent to s . Note that we don’t know yet, but we will see it in thenext proposition, that S u is indeed a closed subgroup with Lie algebraequal to s u . Proposition 5.5.
Let G be a weakly regular p -adic Lie group. a ) The group R u is closed. It is the largest normal algebraic unipotentsubgroup of G . b ) The group S u is closed. Its Lie algebra is s u , and the morphism Ad s u : S u → (Aut s u ) u is onto and has finite kernel. c ) The group G u is closed. One has G u = S u R u and S u ∩ R u = { e } .Remark . In a real Lie group, the group tangent to a Levi subalgebrais not necessary closed, as for example, if G = ( S × T ) /Z where S isthe universal cover of SL(2 , R ), T = R / Z and Z is the cyclic subgroupspanned by ( z , α ) with z a generator of the center of S and α anirrational element of T . Proof of Proposition 5.5. a ) This follows from Proposition 3.8. -ADIC LIE GROUPS 19 b ) Let E := S u be the closure of S u and S + := (Aut s u ) u . Note thatthis group E normalizes s u . We want to apply Corollary 5.3 to themorphism E π −→ S + where π is the adjoint action π := Ad s u .We first check that the assumptions of Corollary 5.3 are satisfied.Since E is weakly regular the kernel Z E of this morphism π commuteswith all the one-parameter subgroups tangent to s u . Hence Z E is equalto the center of E . Since S + is spanned by one-parameter subgroups,and since by Proposition 4.4 any nilpotent element X of s u is tangentto a one parameter subgroup ϕ of E , this morphism π is surjective.Now, by Jacobson Morozov Theorem, for any nilpotent element X of s u , there exists an element H in s u such that [ H, X ] = X . Let ϕ : Q p → G be the one-parameter subgroup tangent to X , which existsby Proposition 4.4. Since G is weakly regular, one has, for t in Q p and g ε := exp( εH ) with ε small, g ε ϕ ( t ) g − ε ϕ ( t ) − = ϕ ( e ε t ) ϕ ( − t ) = ϕ (( e ε − t ) . This proves that ϕ ( Q p ) is included in the derived subgroup [ E, E ]. Inparticular one has E = [ E, E ].According to Corollary 5.3, the kernel Z E is finite. In particular,one has dim E = dim s u . Since s u is totally isotropic, one can find abasis of s u all of whose elements are nilpotent. By Lemma 4.7, all theseelements are in g G . Hence, by the implicit function theorem, the group S u is open in E . Therefore, S u is also closed and S u = E . c ) Since the adjoint action of R u on g nc / r is trivial, the intersection S u ∩ R u is included in the kernel Z E of the adjoint map Ad s u . Since,by b ), this kernel is finite, and since the algebraic unipotent group R u does not contain finite subgroups, one gets S u ∩ R u = { e } .It remains to check that S u R u is closed and that G u = S u R u . Thanksto Propositions 3.8 and 4.4, we can assume that R u = { e } . In this case,we know from point b ) that S u is closed and from Proposition 4.4 that G u = S u . (cid:3) Here are a few corollaries. The first corollary is an improvement ofLemma 3.6.
Corollary 5.7.
Let G be a p -adic Lie group, H a normal weakly regularclosed subgroup of G such that H = H u , and G ′ := G/H . a ) One has the equality G ′ u = G u /H . b ) Let X in g and X ′ its image in g ′ = g / h . One has the equivalence X ∈ g G ⇐⇒ X ′ ∈ g ′ G ′ . c ) One has the equivalence G is weakly regular ⇐⇒ G ′ is weakly regular.Remark . The assumption that H = H u is important. For instancethe group G = Q p and its normal subgroup H = Z p are weakly regularwhile the quotient G/H is not weakly regular.
Proof.
We prove these three statements simultaneously. Since H = H u ,according to Proposition 5.5, the group H admits a Levi decomposition H = SR where R is a normal algebraic unipotent Lie subgroup andwhere S is a Lie subgroup with finite center Z whose Lie algebra issemisimple, totally isotropic, and such that the adjoint map Ad s : S → Aut( s ) u is surjective. Note that R is also a normal subgroup of G . Using Lemma 3.6, we can assume that R = { e } .Let C be the centralizer of H = S in G . Since H is normal in G , since H = H u and H is weakly regular, C is also the kernel of the adjointaction of G on h = s u . Therefore, by Proposition 5.5, the image of thegroup morphism H × C → G ; ( h, c ) hc has finite index in G . Its kernel is isomorphic to H ∩ C = Z and henceis finite. When this morphism H × C → G is an isomorphism, ourthree statements are clear. The general case reduces to this one thanksto Lemma 5.9 below. (cid:3) Lemma 5.9.
Let G be a locally compact topological group, Z be a finitecentral subgroup of G and ϕ : Q p → G/Z be a continuous morphism.Then ϕ may be lifted as a continuous morphism e ϕ : Q p → G .Proof. Let H be the inverse image of ϕ ( Z p ) in G . Then H is totallydiscontinuous. In particular it contains an open compact subgroup U such that U ∩ Z = { e } , so that U maps injectively in G/Z . Let ℓ bean integer such that ϕ ( p ℓ Z p ) ⊂ U Z/Z . After rescaling, we can assumethat ℓ = 0. We let g be the unique element of U such that ϕ (1) = g Z .Since U map injectively in G/Z , we have g p k −−−→ k →∞ e .Let X be the group of elements of p -torsion in Z and Y be the groupof elements whose torsion is prime to p . For any k ≥ g k in G such that ϕ ( p ℓ − k ) = g k Z and let x k and y k be the elements of X and Y such that g p k k = g x k y k . We let z k be the unique element of Y k such that z p k k = y k . Replacing g k by g k z − k , we can assume that z k = e .Since x k only takes finitely many values, we can find a x in X and anincreasing sequence ( k n ) such that, for any n , x k n = x . Now, since x isa central p -torsion element, one has ( g x ) p k −−−→ k →∞ e . Since, for any n , -ADIC LIE GROUPS 21 g p n k n = g x , there exists a unique morphism e ϕ : Q p → G such that, forany n , e ϕ ( p − k n ) = g k n and e ϕ clearly lifts ϕ . (cid:3) The second corollary is an improvement of Proposition 4.4.
Corollary 5.10.
Let G be a weakly regular p -adic Lie group. One hasthe equality : g G = { X ∈ g u | ad X is nilpotent } , where g u is the Liealgebra of G u .Proof. This follows from Proposition 4.4 since, by Proposition 5.5, onehas the equality g u = s u ⊕ r u . (cid:3) The last corollary tells us that a weakly regular p -adic Lie group G with G = G u is “almost” an algebraic Lie group. Corollary 5.11.
Let G be a weakly regular p -adic Lie group such that G = G u . Then there exists a Lie group morphism ψ : G → H withfinite kernel and cokernel where H is the group of Q p -points of a linearalgebraic group defined over Q p .Proof. According to Proposition 4.4, G = G u is a semidirect product G u = S u ⋉ R u . We choose H to be the semi direct product H := S ′ ⋉ R u where S ′ is the Zariski closure of the group Ad( S u ) in Aut( g u ). Notethat, since G is weakly regular, any automorphism of g u induces anautomorphism of R u . We define the morphism ψ : G u → H by ψ ( g ) =(Ad( s ) , r ) for g = sr with s ∈ S u , r ∈ R u . Proposition 4.4 tells us alsothat this morphism ψ has finite kernel and cokernel. (cid:3) Regular semiconnected component.
We are now ready to prove the following theorem whichwas the main motivation of our paper.
Theorem 5.12.
Let G be a weakly regular p -adic Lie group. Then,there exists an open regular subgroup G Ω of G which contains all theone-parameter subgroups of G .Remark . Let Ω be a standard open subgroup of G . We define theΩ -semiconnected component of G as its open subgroup G Ω := Ω G u (see[13]). In this language, Theorem 5.12 states that, the Ω-semiconnectedcomponent of a weakly regular p -adic Lie group is regular, if the stan-dard subgroup Ω is small enough. Proof of Theorem 5.12.
We will need some notations. Let s be a Levisubalgebra of g , s u := s ∩ g nc , s ′ the centralizer of s u in s , r the radicalof g , and r ′ the centralizer of s u in r .We can choose a standard subgroup Ω ′ S of G with Lie algebra s ′ ,and a standard subgroup Ω ′ R of G with Lie algebra r ′ such that Ω ′ S normalizes Ω ′ R and the semidirect product Ω ′ := Ω ′ S Ω ′ R is a standardsubgroup of G with Lie algebra s ′ ⊕ r ′ . Since G is weakly regular, byshrinking Ω ′ , we can assume that it commutes with S u and normalizes R u .We claim that, if Ω ′ is small enough, the group G Ω := Ω ′ G u is an open regular subgroup of G . First step : Openness. One has the equalities s = s ′ ⊕ s u and,according to Proposition 3.8, r = r ′ + r u . Hence G Ω is open in G . Second step : Ad-regularity. Let J ⊂ G Ω be the kernel of Ad g .We want to prove that J is the center of G Ω . Since J acts trivially onthe quotient g / r ≃ s ′ ⊕ s u , by Proposition 5.5, one has the inclusion J ⊂ J ′ := Z E Ω ′ R R u where Z E is a finite subgroup of S u . One has J u = J ∩ R u and this group is algebraic unipotent. Hence, by Lemma3.6, the quotient R u /J u is also algebraic unipotent. By Lemma 3.4,the group R u /J u is closed in the group J ′ /J . This means that R u J is closed in J ′ , hence that J/J u is closed in the compact group J ′ /R u .In particular, J/J u is compact. Therefore, there exists a compact set K ⊂ J such that J = KJ u . We can now prove that if Ω ′ is small enough the group J commuteswith G Ω . This is a consequence of the following three facts.( i ) Since J is the kernel of Ad g and G is weakly regular, J commuteswith G u .( ii ) Since J u is a subgroup of R u whose Lie algebra j u is included inthe center of g , if we choose Ω ′ small enough, one has Ad j u (Ω ′ ) = { e } ,and the group J u commutes with Ω ′ .( iii ) Since K is compact and Ad g ( K ) = { e } , by Lemma 4.2, if wechoose Ω ′ small enough, the group K commutes with Ω ′ . Third step : Size of finite subgroups. We want a uniform upperbound on the cardinality of the finite subgroups of G Ω . This followsfrom the inclusions R u ⊂ G u ⊂ G Ω of normal subgroups and from thefollowing three facts.( i ) Since the group R u is algebraic unipotent, it does not containfinite groups.( ii ) Since, by Proposition 5.5, the group G u /R u is a finite extensionof a linear group, by Example 2.6. b , its finite subgroups have boundedcardinality. -ADIC LIE GROUPS 23 ( iii ) Since the group G Ω /G u is a compact p -adic Lie group, by Ex-ample 2.6. a , its finite subgroups have bounded cardinality. (cid:3) Non weakly regular p -adic Lie groups. Not every p -adic Lie group is weakly regular. Here is asurprising example. Example . There exists a p -adic Lie group G with G = G u whichdoes not contain any open weakly regular subgroup H with H = H u .We will give the construction of such a group G with Lie algebra g = sl (2 , Q p ), but we will leave the verifications to the reader.We recall that the group G := SL(2 , Q p ) is an amalgamated product G = K ⋆ I K where K := SL(2 , Z p ) and I := { k ∈ K | k ≡ p } is an Iwahori subgroup of K . We define G as the amalgamated product G = K ⋆ I K where I ⊂ I is an open subgroup such that I = I .The morphism G → G is a non central extension. Using theconstruction in Lemma 4.1 one can check that G is spanned by one-parameter subgroups. However, one can check that the universal cen-tral extension e G → G can not be lifted as a morphism e G → G . References [1] Y. Benoist, J.-F. Quint, Stationary measures and invariant subsets of homoge-neous spaces (II),
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