aa r X i v : . [ m a t h . G R ] J a n Which Nilpotent Groups are Self-Similar? ∗ Olivier Mathieu † January 28, 2021
Abstract
Let Γ be a finitely generated torsion free nilpotent group, andlet A ω be the space of infinite words over a finite alphabet A . Weinvestigate two types of self-similar actions of Γ on A ω , namely thefaithfull actions with dense orbits and the free actions. A criterion forthe existence of a self-similar action of each type is established.Two corollaries about the nilmanifolds are deduced. The first in-volves the nilmanifolds endowed with an Anosov diffeomorphism, andthe second about the existence of an affine structure.Then we investigate the virtual actions of Γ, i.e. actions of asubgroup Γ ′ of finite index. A formula, with some number theoreticalcontent, is found for the minimal cardinal of an alphabet A endowedwith a virtual self-similar action on A ω of each type. Mathematics Subject Classification
Introduction
1. General introduction
Let A be a finite alphabet and let A ω be the topological space of infinitewords a a . . . over A , where the topology of A ω = lim − A n is the pro-finitetopology. ∗ Research supported by CNRS-UMR 5028 and Labex MILYON/ANR-10-LABX-0070. † Institut Camille Jordan, Universit´e de Lyon. Email: [email protected] action of a group Γ on A ω is called self-similar iff for any γ ∈ Γ and a ∈ A there exists γ a ∈ Γ and b ∈ A such that γ ( aw ) = bγ a ( w ) for any w ∈ A ω .The group Γ is called self-similar (respectively densely self-similar , re-spectively freely self-similar , respectively freely densely self-similar ) if Γ ad-mits a faithfull self-similar action (respectively a faithfull self-similar actionwith dense orbits, respectively a free self-similar action, respectively a freeself-similar action with dense orbits) on A ω for some finite alphabet A .Self-similar groups appeared in the early eighties, in the works of R.Grigorchuk [10] [11] and in the joint works of N. Gupta and S. Sidki [13] [14].See also the monography [24] for an extensive account before 2005 and [25][2] [9] [16] [12] for more recent works. A general question is which groups Γ are (merely, or densely ...) self-similar? This paper brings an answer for finitely generated torsion-free nilpotentgroups Γ, called FGTF nilpotent groups in the sequel. Then we will connectthe main result with topics involving differential geometry and arithmeticgroups.The systematic study of self-similar actions of nilpotent groups startedwith [4], and the previous question has been raised in some talks of S. Sidki.
2. The main results
A few definitions are now required. A grading of a Lie algebra m is a decom-position m = ⊕ n ∈ Z m n such that [ m n , m m ] ⊂ m n + m for all n, m ∈ Z . It iscalled special if m ∩ z = 0, where z is the center of m . It is called very special if m = 0.Let’s assume now that Γ is a FGTF nilpotent group. By Malcev Theory[18][26], Γ is a cocompact lattice in a unique connected, simply connected(or CSC in what follows) nilpotent Lie group N . Let n R be the Lie algebraof N and set n C = C ⊗ R n R .The main results, proved in Section 7, are the following Theorem 2.
The group Γ is densely self-similar iff the Lie algebra n C admitsa special grading. Theorem 3.
The following assertions are equivalent(i) The group Γ is freely self-similar,(ii) the group Γ is freely densely self-similar, and(iii) the Lie algebra n C admits a very special grading.
2s a consequence, let’s mention
Corollary 4.
Let M be a nilmanifold endowed with an Anosov diffeomor-phism. Then there a free self-similar action of π ( M ) with dense orbits on A ω , for some finite A . Corollary 8.
Let M be a nilmanifold. If π ( M ) is freely self-similar, then M is affine complete. Among FGTF nilpotent groups, some of them are self-similar but notdensely self-similar. Some of them are not even self-similar, since Theorem2 implies the next
Corollary 7.
Let M be one of the non-affine nilmanifolds appearing in [3].Then π ( M ) is not self-similar.3. A concrete version of Theorems 2 and 3 Let N be a CSC nilpotent Lie group, with Lie algebra n R . Let’s assumethat N contains some cocompact lattices Γ. By definition, the degree of aself-similar action of Γ on A ω is Card A . We ask the following question For a given cocompact lattice Γ ⊂ N , what is the minimal degreedegree of a faithfull (or a free) self-similar action with dense orbits? More notions are now defined. Recall that the commensurable class ξ ofa cocompact lattice Γ ⊂ N is the set of all cocompact lattices of N whichshare with Γ a subgroup of finite index. The complexity cp ξ (respectivelythe free complexity fcp ξ ) of the class ξ is the minimal degree of a self-similaraction of Γ with dense orbits (respectively a free self-similar action of Γ), forsome Γ ∈ ξ .For any algebraic number λ = 0, set d ( λ ) = Card O ( λ ) /π λ , where O ( λ )is the ring of integers of Q ( λ ) and π λ = { x ∈ O ( λ ) | xλ ∈ O ( λ ) } . For anyisomorphism h of a finite dimensional vector space over Q , setht h = Q λ ∈ Spec h/ Gal( Q ) d ( λ ) m λ ,where Spec h/ Gal( Q ) is the list of eigenvalues of h up to conjugacy by Gal( Q )and where m λ is the multiplicity of the eigenvalue λ .By Malcev’s Theory, the commensurable class ξ determines a canonical Q -form n ( ξ ) of the Lie algebra n R . Let S ( n ( ξ )) (respectively V ( n ( ξ ))) be theset of all f ∈ Aut n ( ξ ) such that Spec f | z C (respectively Spec f ) contains noalgebraic integer. 3 heorem 9. We have cp ξ = Min h ∈S ( n ( ξ )) ht h , and fcp ξ = Min h ∈V ( n ( ξ )) ht h . If, in the previous statement, S ( n ( ξ )) is empty, then the equality cp ξ = ∞ means that no Γ ∈ ξ admits a faithfull self-similar action with dense Γ-orbits.Theorem 9 answers the previous question only for the commensurableclasses ξ . For an individual Γ ∈ ξ , it provides some ugly estimates for theminimal degree of Γ-actions, and nothing better can be expected.The framework of nonabelian Galois cohomology shows the concretenessof Theorem 9. Up to conjugacy, the commensurable classes in N are classi-fied by the Q -forms of some classical objects with a prescribed R -form, seeCorollary 4 of ch. 9, and their complexity is an invariant of the arithmeticgroup Aut n ( ξ ).As an illustration of the previous vague sentence, we investigate a class N of CSC nilpotent Lie groups N , with Lie algebra n R . The commensurableclasses ξ ( q ) in N are classified, up to conjugacy, by the positive definitequadratic forms q on Q . Then, we havecp ξ ( q ) = F ( d ) e ( N ) where e ( N ) is an invariant of the special grading of C ⊗ n R , where − d is thediscriminant of q , and where F ( d ) is the norm of a specific ideal in Q ( √− d ),see Theorem 11 and Lemma 28.In particular, N contains some commensurable classes of arbitrarily highcomplexity. In a forthcoming paper [22], more complicated examples areinvestigated, but the formulas are less explicit.
4. About the proofs.
The proofs of the paper are based on different ideas.Theorem 1, which is a statement about rational points of algebraic tori, isthe key step in the proof of Theorems 2, 3 and 11. It is based on standardresults of number theory, including the Cebotarev’s Theorem. It is connectedwith the density of rational points for connected groups proved by A. Borel[6], see also [27].Also, the proof of Corollary 4 is based on a paper of A. Manning [20]about Anosov diffeomorphisms. The proof of Corollary 7 is based on verydifficult computations, which, fortunately, were entirely done in [3].4
Self-similar actions and self-similar data
Let Γ be a group. This section explains the correspondence between thefaithfull transitive self-similar Γ-actions and some virtual endomorphisms ofΓ, called self-similar data . Usually self-similar actions are actions on a rootedtree A ∗ , see [24]. Here the groups are acting on the boundary A ω of A ∗ . Thisequivalent viewpoint is better adapted to our setting. In addition of the definitions of the introduction, the following technicalnotion of transitivity will be used.A self-similar action of Γ on A ω induces an action of Γ on A . Indeed, for a, b ∈ A and γ ∈ Γ, we have γ ( a ) = b if γ ( aw ) = bγ a ( w ),for all w ∈ A ω . A self-similar action is called transitive if the induced actionon A is transitive. The group Γ is called transitive self-similar if it admits afaithfull transitive self-similar action.Similarly the self-similar action of Γ on A ω induces an action of Γ oneach level set A n . Such an action is often called level transitive if Γ actstransitively on each level A n . Obviously, the level transitive actions on A ∗ of[24] corresponds with the actions on A ω with dense orbits. f -core Let Γ be a group and Γ ′ be a subgroup. The core of Γ ′ is the biggest normalsubgroup K ⊳ G with K ⊂ Γ ′ . Equivalently the core is the kernel of the leftaction of Γ on Γ / Γ ′ .Now let f : Γ ′ ! Γ be a group morphism. By defintion the f -core is thebiggest normal subgroup K ⊳ G with K ⊂ Γ ′ and f ( K ) ⊂ K . Let Γ be a group. A virtual endomorphism of Γ is a pair (Γ ′ , f ), where Γ ′ is asubgroup of finite index and f : Γ ′ ! Γ is a group morphism. A self-similardatum is a virtual endomorphism (Γ ′ , f ) with a trivial f -core.Assume given a faithfull transitive self-similar action of Γ on A ω . Let a ∈ A , and let Γ ′ be the stabilizer of a . By definition, for each γ ∈ Γ ′ thereis a unique γ a ∈ Γ such that γ ( aw ) = aγ a ( w ),for any w ∈ A ω . Let f : Γ ′ ! Γ be the map γ γ a . Since the actionis faithfull, γ a is uniquely determined and f is a group morphism. Also it5ollows from Proposition 2.7.4 and 2.7.5 of [24] that the f -core of Γ ′ is thekernel of the action, therefore it is trivial. Hence (Γ ′ , f ) is a self-similardatum.Conversely, a virtual endomorphism (Γ ′ , f ) determines a transitive self-similar action of Γ on A ω , where A ≃ Γ / Γ ′ . Moreover the f -core is the kernelof the corresponding action, see ch 2 of [24] for details, especially subsection2.5.5 of [24]). In conclusion, we have Lemma 1.
Let Γ be a group. There is a correspondence between the self-similar data (Γ ′ , f ) and the faithfull transitive self-similar actions of Γ on A ω , where A ≃ Γ / Γ ′ . This correspondence is indeed a bijection up to conjugacy, see [24] for aprecise statement.
Let Γ be a group, and let (Γ ′ , f ) be a virtual endomorphism. Let Γ n be thesubgroups of Γ inductively defined by Γ = Γ, Γ = Γ ′ and for n ≥ n = { γ ∈ Γ n − | f ( γ ) ∈ Γ n − } Lemma 2.
The sequence n [Γ n : Γ n +1 ] is not increasing.Proof. For n >
0, the map f induces an injection of the set Γ n / Γ n +1 intoΓ n − / Γ n , thus we have [Γ n : Γ n +1 ] ≤ [Γ n − : Γ n ].The virtual endomorphism (Γ ′ , f ) is called good if [Γ n : Γ n +1 ] = [Γ / Γ ′ ] forall n .Let (Γ ′ , f ) be a self-similar datum, and let A ∗ be the corresponding treeon which Γ acts. If a is the distinguished point in A ≃ Γ / Γ ′ , then Γ n isthe stabilizer of a n . If the self-similar datum (Γ ′ , f ) is good, then [Γ : Γ n ] =Card A n and therefore Γ acts transitively on A n . Exactly as before, we have Lemma 3.
Let Γ be a group. There is a correspondence between the goodself-similar data (Γ ′ , f ) and the faithfull self-similar actions of Γ on A ω withdense orbits, where A ≃ Γ / Γ ′ .1.5 Fractal self-similar data Let Γ be a group. A self-similar datum (Γ ′ , f ) is called fractal (or recurrent)if f (Γ ′ ) = Γ. A self-similar action of Γ on some A ω is called fractal if it is6ransitive and the corresponding self-similar datum is fractal, see [24] section2.8. Obviously a fractal action has dense orbits.The group Γ is called fractal (respectively freely fractal ) if Γ admits afaithfull (respectively free) fractal action on some A ω . We are going to prove Theorem 1, about the rational points of algebraic tori.For the whole chapter, let H be an algebraic torus defined over Q andlet X ( H ) be the group of characters of H . For a number field K , let’sdenote by Gal( K ) := Gal( Q /K ) its absolute Galois group. The group X ( H )is a Gal( Q )-module which is isomorphic to Z r as an abelian group, where r = dim H . The splitting field of H is the smallest Galois extension L of Q such that Gal( L ) acts trivially on X ( H ), or, equivalently such that H is L -isomorphic to G rm , where G m denotes the multiplicative group. Moreover,we have χ ( h ) ∈ L ∗ for any χ ∈ X ( H ) and any h ∈ H ( Q ).Let O be the ring of integers of L . Recall that a fractional ideal is anonzero finitely generated O -submodule of K . A fractional ideal I is called integral if I ⊂ O . If the fractional ideal I is integral and I = O , then I ismerely an ideal of O .Let I be the set of all fractional ideals and I + be the subset of all integralideals. Given I and J in I , their product is the O -module generated by allproducts ab where a ∈ I and b ∈ J . Since O is a Dedekind ring, we have I ≃ ⊕ π ∈P Z [ π ] I + ≃ ⊕ π ∈P Z ≥ [ π ],where P is the set of prime ideals of O . Indeed the additive notation is usedfor for the group I and the monoid I + : view as an element of I the fractionalideal π m . . . π m n n will be denoted as m [ π ] + · · · + m n [ π n ].Since Gal( L/ Q ) acts by permutation on P , I is a Z Gal( L/ Q )-module.For S ⊂ P , set I S = ⊕ π ∈P\ S Z [ π ]. Lemma 4.
Let S ⊂ P be a finite subset and let r > be an integer.The Gal( L/ Q ) -module I contains a free Z Gal( L/ Q ) -module M ( r ) of rank r such that i) M ( r ) ∩ I + = { } , and(ii) M ( r ) ⊂ I S .Proof. Let S ′ be the set of all prime numbers which are divisible by some π ∈ S . Let Σ be the set of prime numbers p that are completely split in K ,i.e. such that O /p O ≃ F [ L : Q ] p . For p ∈ Σ, let π ∈ P be a prime ideal over p .When σ runs over Gal( L/ Q ) the ideals π σ are all distinct, and therefore [ π ]generates a free Z Gal( L/ Q )-submodule of rank one in I .By Cebotarev theorem, the set Σ is infinite. Choose r + 1 distinct primenumbers p , . . . p r in Σ \ S ′ , and let π , . . . , π r ∈ P such that O /π i = F p i .For 1 ≤ i ≤ r , set τ i = [ π i ] − [ π ] and let M ( r ) be the Z Gal( L/ Q )-modulegenerated by τ , . . . , τ r .Obviously, the Gal( L/ Q )-module M ( r ) is free of rank r and M ( r ) ⊂ I S .It remains to prove that M ( r ) ∩ I + = { } . Let A = P ≤ i ≤ r,σ ∈ Gal( L/ Q ) m i,σ τ σi be an element of M ( r ) ∩ I + . We have A = B − C , where B = P ≤ i ≤ r,σ ∈ Gal( L/ Q ) m i,σ [ π σi ], and C = P σ ∈ Gal( L/ Q ) ( P ≤ i ≤ r m σi )[ π σ ].Thus the condition A ∈ I + implies that m σi ≥
0, for any 1 ≤ i ≤ k and σ ∈ Gal( L/ Q ), and P ≤ i ≤ r m σi ≤
0, for any σ ∈ Gal( L/ Q ).Thus all the integers m σi vanish. Therefore M ( r ) intersects I + trivially.For π ∈ P , let v π : L ∗ ! Z be the corresponding valuation. Lemma 5.
Let S ⊂ P be a finite Gal( L/ Q ) -invariant subset and let r > be an integer.The Gal( L/ Q ) -module L ∗ contains a free Z Gal( L/ Q ) -module N ( r ) ofrank r such that(i) N ( r ) ∩ O = { } , and(ii) v π ( x ) = 0 for any x ∈ N ( r ) and any π ∈ S .Proof. Set L ∗ S = { x ∈ L ∗ | v π ( x ) = 0 , ∀ π ∈ S } and let θ : L ∗ S ! I S be themap x P π ∈P v π ( x ) [ π ].By Lemma 4, I S contains a free Z Gal( L/ Q )-module M ( r ) of rank r suchthat M ( r ) ∩ I + = { } . Let’s remark that Coker θ is a subgroup of the class8roup Cl( L ) of L . Since, by Dirichelet Theorem, Cl( L ) is finite, there is apositive integer d such that d.M ( r ) lies in the image of θ . Since M ( r ) isfree, there is a free Z Gal( K/ Q )-module N ( r ) ⊂ L ∗ S of rank r which is a liftof dM ( r ), i.e. such that θ induces an isomorphism N ( r ) ≃ d.M ( r ). Since θ ( O\
0) lies in I + , we have θ ( N ( r ) ∩O ) = { } . It follows that N ( r ) ∩O = { } .The second assertion follows from the fact that N ( r ) lies in L ∗ S .For π ∈ P , let O π and L π be the π -adic completions of O and L . Let x, y ∈ L and let n > x ≡ y modulo n O π means x π ≡ y π mod n O π , where x π and y π are the images of x and y in L π .The case n = 1 of the next statement will be used in further sections. Insuch a case, Assertion (ii) is tautological. Theorem 1.
Let H be an algebraic torus defined over Q , and let L be itssplitting field. Let n > be an integer and let S ⊂ P be the set of primedivisors of n .There exists h ∈ H ( Q ) such that(i) χ ( h ) is not an algebraic integer, for any non-trivial χ ∈ X ( H ) , and(ii) χ ( h ) ≡ n O π for any χ ∈ X ( H ) and any π ∈ S .Proof. Step 1. First an element h ′ ∈ H ( Q ) satisfying Assertion (i) and(iii) v π ( χ ( h ′ )) = 0, for any π ∈ S and any χ ∈ X ( H )is found.The abelian group X ( H ) is free of rank r where r = dim H . Therefore,the comultiplication ∆ : X ( H ) ! X ( H ) ⊗ Z Gal( L/ Q ) provides an embeddingof X ( H ) into a free Z Gal( L/ Q )-module of rank r . By lemma 5, there a free Z Gal( L/ Q )-module N ( r ) ⊂ L ∗ S of rank r with N ( r ) ∩ O = { } . Let µ : X ( H ) ⊗ Z Gal( L/ Q ) ! N ( r ),be an isomorphism of Z Gal( L/ Q )-modules, and set h ′ = µ ◦ ∆.Since H ( Q ) = Hom Gal( L/ Q ) ( X ( H ) , L ∗ ), the embedding h ′ is indeed anelement of H ( Q ). Viewed as a map from X ( H ) to L ∗ , h ′ is the morphism χ ∈ X ( H ) χ ( h ′ ).Since Im h ′ ∩ O = 1 and h ′ is injective, χ ( h ′ ) is not an algebraic integer if χ is a non-trivial character. Since Im h ′ ⊂ L ∗ S , we have v π ( χ ( h ′ )) = 0 for any χ ∈ X ( H ). Therefore h ′ satisfies Assertions (i) and (iii). Step 2.
Let χ , . . . , χ r be a basis of X ( H ). Since v π ( χ i ( h ′ )) = 0 for any π ∈ S , the element χ i ( h ′ ) mod n O π is an inversible element in the finite ring O π /n O π . Therefore there are positive integers m i,π such that9 i ( h ′ ) m i,π ≡ n O π ,for all 1 ≤ i ≤ r and all π ∈ S . Set m = lcm( m i,π ) and set h = h ′ m .Obviously h satisfies Assertion (i). Moreover we have χ i ( h ) ≡ n O π ,for all π ∈ S and all 1 ≤ i ≤ r , and therefore h satisfies Assertion (ii) as well. Let n be a finite dimensional Lie algebra defined over Q and let z be itscenter. The relations between the gradings of C ⊗ n and the automorphismsof n are investigated now. The following important definitions will be used in the whole paper . Let S ( n ) (respectively V ( n )) be the set of all f ∈ Aut n such that Spec f | z (re-spectively Spec f ) contains no algebraic integers. Moreover let F ( n ) be theset of all f ∈ S ( n ) such that all eigenvalues of f − are algebraic integers. Alsoset F + ( n ) = F ( n ) ∩ V ( n ). Here, by eigenvalues of a Q -linear endomorphism F , we always mean the eigenvalues of F in Q .For any field K of characteristic zero, set n K = K ⊗ n and z K = K ⊗ z . Let G = Aut n be the algebraic group of automorphisms of n . By definition, G isdefined over Q , and we have G ( K ) = Aut n K for any field K of characteristiczero. The notation n underlines that n can be viewed as the functor in Liealgebras K n K . Let H ⊂ G be a maximal torus defined over Q , whoseexistence is proved in [7], see also [6], Theorem 18.2.By definition, a K -grading of n is is a decomposition of n K n K = ⊕ n ∈ Z n Kn such that [ n Kn , n Km ] ⊂ n Kn + m for all n, m ∈ Z . A grading is called special (respectively very special ) if z K ∩ n K = 0 (respectively if n K = 0). A gradingis called non-negative (respectively positive ) if n Kn = 0 for n < n Kn = 0 for n ≤ K of characteristic zero, a K -grading of n can be identifiedwith an algebraic group morphism ρ : G m ! G defined over K , where G m denotes the multiplicative group.Consider the following two hypotheses( H K ) The Lie algebra n admits a special K -grading,( H K ) The Lie algebra n admits a very special K -grading.10 emma 6. Let K be the splitting field of H . Up to conjugacy, any gradingof n C is defined over K . In particular(i) The hypotheses H C and H Q are equivalent.(ii) The hypotheses H C and H Q are equivalent.Proof. Let n C = ⊕ n ∈ Z n C n be a grading of n C and let ρ : G m ! G be the corresponding algebraic groupmorphism. Since any maximal torus of G is G ( C )-conjugate to H , it can beassumed that ρ ( G m ) ⊂ H .Let X ( H ) be the character group of H . The group morphism ρ is deter-mined by the dual morphism L : X ( H ) ! Z = X ( G m ). However, Gal( K )acts trivially on X ( H ). Thus ρ is automaticaly defined over K . Lemma 7.
Let Λ be a finitely generated abelian group and let S ⊂ Λ bea finite subset containing no element of finite order. Then there exists amorphism L : Λ ! Z such that L ( λ ) = 0 for any λ ∈ S .Proof. Let F be the subgroup of finite order elements in Λ. Using Λ /F instead of Λ, it can be assumed that Λ = Z d for some d an 0 / ∈ S . Let’schoose a positive integer N such that S ⊂ ] − N, N [ d and let L : Λ ! Z bethe function defined by L ( a , . . . , a d ) = P ≤ i ≤ d a i N i − .For any λ = ( a , . . . , a d ) ∈ S , there is a smallest index i with a i = 0. We have L ( λ ) = a i N i − modulo N i . Since | a i | < N , it follows that L ( λ ) = 0 mod N i and therefore L ( λ ) = 0. Lemma 8.
Let f ∈ G ( Q ) . There is a f -invariant Z -grading of n Q such thatall eigenvalues of f on n Q are roots of unity.In particular, if Spec f contains no root of unity, then n Q admits a veryspecial grading.Proof. Let Λ ⊂ Q ∗ be the subgroup generated by the Spec f . For any λ ∈ Λdenote by E ( λ ) ⊂ n Q the corresponding generalized eigenspace of f . Let R be the set of all roots of unity in Spec f and set S = Spec f \ R .By Lemma 7, there is a morphism L : Λ ! Z such that L ( λ ) = 0 for any λ ∈ S . Let G be the decomposition 11 Q = ⊕ k ∈ Z n Q k of n Q defined by n Q k = ⊕ L ( λ )= k E ( λ ) . Since [ E ( λ ) , E ( µ ) ] ⊂ E ( λµ ) and L ( λµ ) = L ( λ ) + L ( µ ) for any λ, µ ∈ Λ, it follows that G is a grading of the Lie algebra n Q . Moreover we have n Q = ⊕ λ ∈ R E ( λ ) ,from which the lemma follows. Lemma 9.
With the previous notations(i) the Lie algebra n C admits a special grading iff S ( n ) = ∅ .(ii) the Lie algebra n C admits a very special grading iff V ( n ) = ∅ .Proof. In order to prove Assertion (i), let’s consider the following assertion( A ) H ( H ( Q ) , z Q ) = 0.The proof is based on the following ”cycle” of implications n C has a special grading ⇒ ( A ) ⇒ S ( n ) = ∅ ⇒ n C has a special grading. Step 1: the existence of a special grading of n C implies ( A ). By hypothesisand Lemma 6, n Q admits a special grading. Let ρ : G m ! G be the corre-sponding group morphism. Since all maximal tori of G are conjugate to H ,we can assume that ρ ( G m ) ⊂ H . Therefore we have H ( H ( Q ) , z Q ) ⊂ H ( ρ ( Q ∗ ) , z Q ) = 0.Thus Assertion A is proved. Step 2: proof that ( A ) implies S ( n ) = ∅ . By Theorem 1, there exists f ∈ H ( Q ) such that χ ( f ) is not an algebraic integer for any non-trivial character χ ∈ X ( H ). If we assume ( A ), then Spec f | z contains no algebraic integersand therefore S ( n ) = ∅ . Step 3: proof that S ( n ) = ∅ implies the existence of a special grading. For any f ∈ S ( n ), Since Spec f | z contains no roots of unity. It follows from Lemma 8that the Lie algebra n Q (and therefore n C ) admits a special grading. Therefore S ( n ) = ∅ implies the existence of a special grading.The proof of Assertion (ii) is almost identical. Instead of ( A ), the ”cycle”of implications uses the following assertion( A ) H ( H ( Q ) , n Q ) = 0. Lemma 10.
The following are equivalent:(i) the Lie algebra n Q admits a non-negative special grading,(ii) the Lie algebra n C admits a non-negative special grading, and(iii) The set F ( n ) is not empty. roof. Proof that ( ii ) ⇒ ( iii ) . Let n C = ⊕ k ≥ n C k be a non-negative specialgrading of n C and let ρ : G m ! G be the corresponding group morphism.Up to conjugacy, we can assume that ρ ( G m ) ⊂ H . It follows that the gradingis defined over the splitting field K of H .Let g ∈ H ( K ) be the isomorphism defined by g x = 2 k x if x ∈ n C k . Set n = [ K : Q ] and let g , g . . . g n be the Gal( L/ Q )-conjugates of g . Sinceall g i belongs to H ( K ), the automorphisms g i commute. Hence the product g := g . . . g n is well defined and g belongs to H ( Q ). By hypotheses, alleigenvalues of g i are power of 2, and all eigenvalues of g i | z C are distinct from1. Therefore all eigenvalues of g are integers, and all eigenvalues of g | z C are = ±
1. It follows that g − belongs to F ( n ). Therefore F ( n ) = ∅ Proof that ( iii ) ⇒ ( i ) . Let f ∈ F ( n ) and set g = f − . Set K = Q (Spec g )and let L : K ∗ ! Z be the map defined by L ( x ) = P p v p ( N K/ Q ( x ))where the sum runs over all prime numbers p and where N K/ Q : K ∗ ! Q ∗ denotes the norm map.For any integer k , set n Q k = L L ( x )= k E ( x ) where E ( x ) ⊂ n Q denotes the generalized eigenspace associated to x ∈ Spec g .We have [ E ( x ) , E ( y ) ] ⊂ E ( xy ) and L ( xy ) = L ( x ) + L ( y ), for any x, y ∈ K .Therefore the decomposition n K = ⊕ k ∈ Z n Q k is a grading G of the Lie algebra n Q . Since the function L is Gal( Q )-invariant,the grading G is indeed defined over Q . It remains to prove that G is non-negative and special.Since any x ∈ Spec g is an algebraic integer, we have L ( x ) ≥ x ∈ Spec g | z is an algebraic unit, we have N K/ Q ( x ) = ± L ( x ) >
0. Thus the grading is special, what proves that( iii ) = ⇒ ( i ). Lemma 11.
The following are equivalent:(i) the Lie algebra n Q admits a positive grading,(ii) the Lie algebra n C admits a positive grading, and(iii) The set F + ( n ) is not empty. Since the proof is almost identical to the previous proof, it will be skipped.The equivalence ( i ) ⇔ ( ii ) also appears in [8].13 Height and relative complexity
For the whole chapter, V denotes a finite dimensional vector space over Q . In this section, we define the notion of the height of the isomorphisms h ∈ GL ( V ) and the notion of a minimal lattice . Let h ∈ GL ( V ). Recall that a lattice of V is a finitely generated subgroupΛ which contains a basis of V . Let D ( h ) be the set of all couple of lattices(Λ , E ) of V such that E ⊂ Λ and h ( E ) ⊂ Λ. By definition, the height of h ,is the integer ht( h ) := Min (Λ ,E ) ∈D ( h ) [Λ : E ].Let D min ( h ) be the set of all couples (Λ , E ) ∈ D ( h ) such that [Λ : E ] = ht( h ).Similarly, for a lattice Λ of V , the h -complexity of Λ is the integercp h (Λ) := Min (Λ ,E ) ∈D ( h ) [Λ : E ].It is clear that cp h (Λ) = [Λ : E ], where E = Λ ∩ h − (Λ). The lattice Λ iscalled minimal relative to h if cp h (Λ) = ht( h ).For the proofs, a technical notion of relative height is needed. Let End h ( V )be the commutant of h and let let A ⊂ C ( h ) ⊂ End h ( V ) be a subring. Bydefinition, an A -lattice Λ means a lattice Λ which is an A -module. Let D A ( h )be the set of all couple of A -lattices (Λ , E ) in D ( h ). The A -height of h is theinteger ht A ( h ) := Min (Λ ,E ) ∈D A ( h ) [Λ : E ].Obviously, we have ht A ( h ) ≥ ht( h ) = ht Z ( h ). Let D Amin ( h )) be the set of allcouples (Λ , E ) ∈ D A ( h )) such that [Λ : E ] = ht A ( h ). Let V be a finite dimensional vector space over Q and let h ∈ GL ( V ). Let A be a subring of End h ( V ) and let A [ h ] be the subring of End h ( V ) generatedby A and h . Lemma 12.
Let V ⊂ V ⊂ · · · ⊂ V n = V be a fitration of V , whereeach vector space V i is a A [ h ] -submodule. For i = 1 to n , set h i = h V i /V i − .Then we have ht A ( h ) ≥ Q ≤ i ≤ n ht A ( h i ) .Moreover if V ≃ ⊕ V i /V i − as a A [ h ] -module, we have ht A ( h ) = Q ≤ i ≤ n ht A ( h i ) . roof. Clearly it is enough to prove the lemma for n = 2. Let (Λ , E ) ∈D Amin ( h ). Set Λ = Λ ∩ V , E = E ∩ V , Λ = Λ / Λ and E = E/E . Wehave [Λ : E ] = [Λ : E ][Λ : E ].Since (Λ , E ) ∈ D A ( h ) and (Λ , E ) ∈ D A ( h ), we haveht A ( h ) ≥ ht A ( h ) ht A ( h ),what proves the first assertion.Next, we assume that V ≃ V ⊕ V as a A [ h ]-module. Let (Λ , E ) ∈D Amin ( h ), (Λ , E ) ∈ D Amin ( h ) and set Λ = Λ ⊕ Λ and E = E ⊕ E . Wehave [Λ : E ] = [Λ : E ][Λ : E ] = ht A ( h ) ht A ( h ).Therefore ht A ( h ) ≤ ht A ( h ) ht A ( h ). Hence ht A ( h ) = ht A ( h ) ht A ( h ).Let h ∈ GL ( V ) as before. Its Chevalley decomposition h = h s h u isuniquely defined by the following three conditions: h s and h u commutes, h s is semi-simple and h u is unipotent. Lemma 13.
We have ht( h ) = ht( h s ) .Proof. By Lemma 12, it can be assumed that the Q [ h ]-module V is indecom-posable. Therefore there is a vector space V , a semi-simple endomorphism h ∈ End( V ) and an isomorphism V ≃ V ⊗ Q [ t ] / ( t n ),relative to which h s acts as h ⊗ h u acts as 1 ⊗ t . Let (Λ , E ) ∈ D min ( h )and set Λ = Λ ⊗ Z [ t ] / ( t n ) and E = E ⊗ Z [ t ] / ( t n ). By Lemma 12, we haveht( h ) ≥ ht( h s ) = ht( h ) n . Since (Λ , E ) ∈ D ( h ) and[Λ : E ] = [Λ : E ] n = ht( h ) n ,it follows that ht( h ) = ht( h s ) O ( h ) -lattices For any algebraic number λ , let O ( λ ) be the ring of integers of the numberfield Q ( λ ). Set π λ = { x ∈ O ( λ ) | xλ ∈ O ( λ ) } . Then π λ is an integral idealand its norm is the integer d ( λ ) := N Q ( λ ) / Q ( π λ ) = Card O ( λ ) /π λ .Let h ∈ GL ( V ) be semi-simple. Let P ( t ) be its minimal polynomial, let P = P . . . P k be its factorization into irreducible factors. For 1 ≤ i ≤ k , set15 i = Q [ t ] / ( P i ( t )) and let O i be the ring of integers of the number field K i .Set O ( h ) = ⊕ ≤ i ≤ k O i .For each λ ∈ Spec h , let m λ be its multiplicity. Note that the functions λ m λ and λ d ( λ ) are Gal( Q )-invariant, so they can be viewed asfunctions defined over Spec h/ Gal( Q ). Lemma 14.
Let Λ be an O ( h ) -lattice of V . Then cp h (Λ) = Q d ( λ ) m λ ,where the product runs over Spec h/ Gal( Q ) .Proof. With the previous notations, let e i be the unit of O i and set Λ i = e i Λ.Since Λ = ⊕ ≤ i ≤ k Λ i , it is enough to prove the lemma for k = 1, i.e. whenthe minimal polynomial of h is irreducible.Let λ be one eigenvalue of h . With these new hypotheses, we have Q [ h ] / ( P ( t )) ≃ Q ( λ ), O ( h ) ≃ O ( λ ) and V is a vector space of dimension m λ over Q ( λ ), relative to which h is identified with the multiplication by λ .We have r λ Λ = Λ ∩ h − Λ.Since Λ /r λ I ≃ ( O ( λ ) /r λ ) m λ , it follows that cp h (Λ) = d ( λ ) m λ . Let h ∈ GL ( V ) be semi-simple. Lemma 15.
We have ht( h ) = Q d ( λ ) m λ ,where the product runs over Spec h/ Gal( Q ) .Proof. Using Lemmas 13 and Lemma 12, we can be assumed V is a simple Q [ h ]-module, and let n be its dimension. The eigenvalues λ , . . . , λ n of h areconjugate by Gal( Q ). Under these simplifying hypotheses, the formula to beproved is ht( h ) = d ( λ ). Step 1: scalar extension.
Set K = Q ( λ , . . . , λ n ), let U = K ⊗ V , let ˜ h = 1 ⊗ h be the extension of h to U and let { v , . . . , v n } be a K basis of U such that˜ h.v i = λ i v i . We have U = ⊕ ≤ i ≤ n U i , where U i = K v i .Let O be the ring of integers of K . For each 1 ≤ i ≤ n , set ˜ h i = h | U i .Since each U i is a O [˜ h ]-module, Lemma 12 shows thatht O (˜ h ) = Q ≤ i ≤ n ht O (˜ h i ).16ext, the integers ht O (˜ h i ) are computed. Let Λ i ⊂ U i be any O -lattice.Since O contains O ( λ i ) = O (˜ h i ), it follows from Lemma 14 thatcp ˜ h i (Λ i ) = d ( λ i ) r where r = rk O (Λ i ) = [ K : Q ( λ i )]. Hence we have ht O (˜ h i ) = d ( λ i ) [ K : Q ( λ i )] . Itfollows that ht O (˜ h ) = Q ≤ i ≤ n d ( λ i ) [ K : Q ( λ i )] = d ( λ ) [ K : Q ] Step 2: end of the proof.
Now let (Λ , E ) ∈ D min ( h ). Set ˜Λ = O ⊗
Λ and˜ E = O ⊗ E . Since ˜ E is an O -module, we have [ ˜Λ : ˜ E ] ≥ ht O (˜ h ). It followsthat ht( h ) [ K : Q ] = [Λ : E ] [ K : Q ] = [ ˜Λ : ˜ E ] ≥ ht O (˜ h ) = d ( λ ) [ K : Q ] .Thus we have d ( λ ) ≤ ht( h ). By Lemma 14, we have ht O ( h ) ( h ) = d ( λ ). Itfollows that d ( λ ) ≤ ht( h ) ≤ ht O ( h ) ( h ) = d ( λ ),what proves the formula. Remark:
In number theory, the Weil height of an algebraic number λ is H ( λ ) = θd ( λ ) /n , where θ involves the norms at infinite places. Thereforeht( h ) is essentially the Weil’s height of h , up to the factor at infinite places. An obvious consequence of Lemmas 14 and 15 is
Lemma 16.
Let h ∈ GL ( V ) be semi-simple and let Λ be an O ( h ) -lattice of V . Then Λ is minimal relative to h . In this chapter, we recall Malcev’s Theorem. Then we collect some relatedresults, which are due to Malcev or viewed as folklore results. Then it is easyto characterize the self-similar data for FGTF nilpotent groups.
Let n be a finite dimensional be a nilpotent Lie algebra over Q . The Liealgebra n is endowed with two group structures, the addition and the theCampbell-Hausdorff product. To avoid confusion, the Campbell-Hausdorffproduct is called the multiplication and it is denoted accordingly.17 multiplicative subgroup Γ of n means a subgroup relative to the Campbell-Hausdorff product. In general, a multiplicative subgroup Γ is not an additivesubgroup of n . However, notice that Z .x ⊂ Γ for any x ∈ Γ, because x n = nx for any n ∈ Z .A finitely generated multiplicative subgroup Γ is called a multiplicativelattice if Γ mod [ n , n ] generates the Q -vector space n / [ n , n ], or, equivalently,if Γ generates the Lie algebra n . Let N be the CSC nilpotent Lie group withLie algebra n R = R ⊗ n . A discrete subroup Γ of N is called a cocompactlattice if N/ Γ is compact.It should be noted that three distinct notions of lattices will be used inthe sequel: the additive lattices, the multiplicative lattices and the cocom-pact lattices. When it is used alone, a lattice is always an additive lattice.This very commoun terminology could be confusing: the reader should read”multiplicative lattice” or ”cocompact lattice” as single words.
Any multiplicative lattice Γ of a finite dimensional nilpotent Lie algebra over Q is a FGTF nilpotent group. Conversely, Malcev proved in [18] Malcev’s Theorem.
Let Γ be a FGTF nilpotent group.1. There exists a unique nilpotent Lie algebra n over Q wich contains Γ as a multiplicative lattice.2. There exists a unique CSC nilpotent Lie group N which contains Γ asa cocompact lattice.3. The Lie algebra of N is R ⊗ n . The Lie algebra n of the previous theorem will be called the Malcev Liealgebra of Γ.
From now on, let n will be a finite dimensional nilpotent Lie algebra. Thecoset index, which is defined now, generalizes the notions of indices for ad-ditive lattices and for multiplicative lattices.A subset X of n is called a coset union if X is a finite union of Λ-cosetfor some additive lattice Λ.Recall that the nilpotency index of n is the smallest integer n such that C n +1 n = 0, where ( C n n ) n ≥ is its descending central series. The followinglemma is easily proved by induction on the nilpotency index of n . Lemma 17.
Any multiplicative lattice Γ of n is a coset union. X ⊃ Y be two coset unions in n . Obviously, there is a lattice Λ suchthat X and Y are both a finite union of Λ-coset. The coset index of Y in X is the number [ X : Y ] coset = Card X/ ΛCard Y/ Λ The numerator and denominator of the previous expression depends on thechoice of Λ, but [ X : Y ] coset is well defined. In general, the coset index is not an integer. Obviously if Λ ⊃ Λ ′ are additive lattices in n , we have[Λ : Λ ′ ] coset = [Λ : Λ ′ ].Similarly, for multiplicative lattices there is Lemma 18.
Let Γ ⊃ Γ ′ be multiplicative lattices in n , we have [Γ : Γ ′ ] coset = [Γ : Γ ′ ] . The proof, done by induction on the nipotency index of n , is skipped. Lemma 19.
Let Γ , Γ ′ ⊂ n be multiplicative lattices in n and let f : Γ ′ ! Γ be a group morphism. Then f extends uniquely to a Lie algebra morphism ˜ f : n ! n .Moreover ˜ f is an isomorphism if f is injective. When f is an isomorphism, the result is due to Malcev, see [18], Theorem5. In general, the lemma is a folklore result and it is implicitely used inHomotopy Theory, see e.g. [1]. Since we did not found a precise reference, aproof, essentially based on Hall’s collecting formula (see Theorem 12.3.1 in[15]), is now provided. Proof.
Let x ∈ n . Since Γ contains an additive lattice by Lemma 17, we have m Z x ⊂ Γ for some m >
0. Thus there is a unique map ˜ f : n ! n extending f such that ˜ f ( nx ) = n ˜ f ( x ) for any x ∈ n and n ∈ Z . It remains to provethat ˜ f ( x + y ) = ˜ f ( x ) + ˜ f ( y ), and ˜ f ([ x, y ]) = [ ˜ f ( x ) , ˜ f ( y )],for any x, y ∈ n .Let n be the nilpotency index of n . Set L (2 , n ) = L (2) /C n +1 L (2), where L (2) denotes the free Lie algebra over Q freely generated by X and Y . LetΓ(2 , n ) ⊂ L (2 , n ) be the multiplicative subgroup generated by X and Y .As before, m ( X + Y ) and m [ X, Y ] belongs to Γ(2 , n ) for some m > w , w in the free group over two generators, such that w ( X, Y ) = m ( X + Y ) and w ( X, Y ) = m [ X, Y ].19ince L (2 , n ) is a free in the category of nilpotent Lie algebras of nilpo-tency index ≤ n , we have w ( x, y ) = m ( x + y ) and w ( x, y ) = m [ x, y ]for any x, y ∈ n . From this it follows easily that ˜ f is a Lie algebra morphism. Let z be the center of n . Recall that S ( n ) (respectively V ( n )) is the set ofall f ∈ Aut n such that Spec f | z (respectively Spec f ) contains no algebraicintegers. Let Γ ⊃ Γ ′ be multiplicative lattices of n , let f : Γ ′ ! Γ be amorphism and let ˜ f : n ! n be its extension. Lemma 20.
Let’s assume that f is injective. Then(i) (Γ ′ , f ) is a self-similar datum iff ˜ f belongs to S ( n ) ,(ii) (Γ ′ , f ) is a free self-similar datum iff ˜ f belongs to V ( n ) (iii) if (Γ ′ , f ) is a fractal datum, then f belongs to F ( n ) .Proof. Let V be a finite dimensional vector space over Q and let f ∈ GL ( V ).We will repeatedly use the fact that Spec f contains an algebraic integer iff V contains a finitely generated subgroup E = 0 such that f ( E ) ⊂ E . Proof of Assertion (i).
Since Γ ′ contains a set of generators of n , the subgroup Z (Γ ′ ) := Γ ′ ∩ z is the center of Γ ′ . Let K be the f -core of the virtualendomorphism (Γ ′ , f ).Let’s assume that (Γ ′ , f ) is not a self-similar datum. Since K = 1, theadditive group K ∩ Z (Γ ′ ) is non-trivial, finitely generated and ˜ f -invariant.Therefore ˜ f / ∈ S ( n ).Conversely let’s assume that ˜ f / ∈ S ( n ). Then there is a nonzero finitelygenerated subgroup E ⊂ z such that ˜ f ( E ) ⊂ E . By Lemma 17, Z (Γ ′ ) is anadditive lattice of z . Therefore we have mE ⊂ Z (Γ ′ ) for some m >
0. Since K contains mE , it follows that (Γ ′ , f ) is not a self-similar datum. Proof of Assertion (ii).
Let A ⊂ Γ be a set of representatives of Γ / Γ ′ . Let’sconsider the action of Γ on A ω associated with the virtual endomorphism(Γ ′ , f ).Let’s assume that ˜ f / ∈ V ( n ). Then there is a nonzero finitely generatedabelian subgroup F ⊂ n such that ˜ f ( F ) ⊂ F . As before, it can be assumed F lies in Γ ′ . Let e ∈ A be the representative of the trivial coset and let e ω = ee . . . be the infinite word over the single letter e . Since f ( F ) ⊂ F , itfollows that γ ( e ω ) = e ω for any γ ∈ F . Hence Γ does not act freely on A ω .20onversely, let assume that Γ does not act freely on A ω . Let’s defineinductively the subsets H ( n ) ⊂ Γ by H (1) = ∪ a ∈ A a Γ ′ a − and H ( n + 1) = { γ ∈ Γ | ∃ a ∈ A : aγa − ∈ Γ ′ ∧ f ( aγa − ) ∈ H ( n ) } ,for n ≥
1. Indeed H ( n ) is the set of all γ ∈ Γ which have at least one fixedpoint on A n . It follows easily that H := ∩ n ≥ H ( n ) is the set of all γ ∈ Γwhich have at least one fixed point on A ω . There is an integer k such that H ⊂ C k n but H 6⊂ C k +1 n .Let H be the image of H in C k n /C k +1 n and let F be the additive subgroup of C k n /C k +1 n generated by H . Since Γ lies in a lattice, F is finitely generated.Moreover we have axa − ≡ x mod C k +1 n , for any x ∈ C k n and a ∈ A .It follows that ˜ f k ( H ) ⊂ H , where ˜ f k is the linear map induced by ˜ f on C k n /C k +1 n . Hence ˜ f k ( F ) ⊂ F and Spec ˜ f k contains an algebraic integer.Therefore ˜ f / ∈ V ( n ). Proof of Assertion (iii).
Let (Γ ′ , f ) be a fractal datum. Let Λ be the additivelattice generated by Γ. Since ˜ f − (Λ) ⊂ Λ,all x ∈ Spec ˜ f − are algebraic integers. Therefore ˜ f belongs to F ( n ). This chapter is the mutiplicative analogue of ch. 4. The main result is therefined criterion of minimality. Together with Theorem 1, it is the mainingredient of the proof of Theorem 2 and 3.Throughout the whole chapter, n is finite dimensional nilpotent Lie alge-bra over Q , and z is its center. Let f ∈ Aut n and let Γ be a multiplicative lattice of n . The complexity of Γrelative to f is the integer cp f (Γ) = [Γ : Γ ′ ],where Γ ′ = Γ ∩ f − (Γ). The multiplicative lattice Γ is called minimal rel-ative to f if cp f (Γ) = ht( f ). Thanks to Lemma 18 the notation cp f (Γ) isunambiguous. Lemma 21.
Let Γ be multiplicative lattices of n . Then we have cp f (Γ) ≥ ht( f ) . roof. The proof goes by induction on the nilpotency index of n .Let Z be the center of Γ. Set Γ ′ = Γ ∩ f − (Γ), Z ′ = Z ∩ f − ( Z ),Γ = Γ /Z , Γ ′ = Γ ′ /Z ′ . Also set n = n / z and let f : n ! n and f : z ! z bethe isomorphisms induced by f .By induction hypothesis, we have cp f (Γ) ≥ ht( f ) and therefore[Γ : Γ ′ ] ≥ ht( f ).By definition, we have [ Z : Z ′ ] = cp f Z ≥ ht( f ). Moreover by Lemma 15we have ht( f ) = ht( f )ht( f ). It follows thatcp f Γ = [Γ : Γ ′ ] = [ Z : Z ′ ] [Γ : Γ ′ ] ≥ ht( f )ht( f ) = ht( f ),and the statement is proved. Let Γ be a multiplicative lattice of n and let h ∈ Aut n . For simplicity, let’sassume that h is semi-simple. Lemma 22.
The following assertions are equivalent(i) Γ is minimal relative to h , and(ii) the virtual morphism (Γ ′ , h ) is good, where Γ ′ = Γ ∩ h − (Γ) .In particular, there is a multiplicative lattice ˜Γ ⊂ Γ which is minimalrelative to h .Proof. By Lemma 17, Γ is a coset union. Any additive lattice contains a O ( h )-module of finite index. Therefore there is an O ( h )-lattice Λ such thatΓ is an union of Λ-cosets.Let Γ , Γ , . . . be the multiplicative lattices inductively defined by Γ = Γ,Γ = Γ ′ and Γ n +1 = Γ n ∩ h − (Γ n ) for n ≥
1. Similarly let Λ , Λ , . . . be theadditive lattices defined by Λ = Λ, and Λ n +1 = Λ n ∩ h − (Λ n ) for n ≥ n : Γ n +1 ] is not increasing. By Lemma21, we have [Γ n : Γ n +1 ] ≥ ht( f ). Moreover, it follows from Lemma 16 that[Λ n : Λ n +1 ] = ht( h ) for all n .Let’s assume now that Γ is minimal relative to h . We have [Γ n : Γ n +1 ] =ht( f ) for all n , and therefore the virtual morphism (Γ ′ , h ) is good.Conversely, let’s assume that the virtual morphism (Γ ′ , h ) is good. Byhypotheses we have [Γ : Γ n ] = [Γ : Γ ] n and [Λ : Λ n ] = ht( h ) n for all n ≥ : Λ n ] coset = [Γ : Λ ] coset ht( h ) n .Since Γ n ⊃ Λ n , we have [Γ : Γ n ] ≤ [Γ : Λ n ] coset .and therefore 22Γ : Γ ] n ≤ [Γ : Λ ] coset ht( h ) n , for all n ≥ : Γ ] ≤ ht( f ). It follows from Lemma 21 that [Γ : Γ ] = ht( f ),thus Γ is minimal relative to h .In order to prove the last assertion, notice that the sequence [Γ n : Γ n +1 ] isstationary for n ≥ N , for some N >
0. Therefore (Γ N +1 , h ) is a good virtualmorphism of Γ N . Thus the subgroup ˜Γ = Γ N is minimal relative to h . A refined version of Lemma 16 is now provided. Let Γ be a multiplicativelattice in n and let h ∈ Aut n be semi-simple. Let L be the field generatedby Spec h , let O be its ring of integers and let P be the set of prime idealsof O .Let Λ be an O ( h )-lattice and let n > ⊃ Γ and Γ is an union of n Λ-cosets.
Lemma 23.
Let S be the set of divisors of n in P . Assume that λ ≡ n O π ,for any λ ∈ Spec h and any π ∈ S . Then Γ is minimal relative to h .Proof. Step 1. Since Spec h lies in O π for all π ∈ S , there exists a positiveinteger d , which is prime to n , such that dλ ∈ O for all λ ∈ O . Moreover wecan assume that d ≡ n .Let λ ∈ Spec h . We have dλ ≡ n O π for all π ∈ S . Therefore wehave dλ ∈ n O ,for all λ ∈ Spec h . Set H = dh . Since Spec dH and Spec ( H − /n lie in O , it follows that H ∈ O ( h ) and H ∈ n O ( h ). Step 2.
Set Λ ′ = Λ ∩ h − Λ. Since all eigenvalues of h are units in O π whenever π divides n , the height of h is prime to n . By Lemma 16, we have[Λ : Λ ′ ] = ht( h ). Therefore we getΛ = Λ ′ + n Λ.It follows that Γ = ` ≤ i ≤ k g i + n Λfor some g , ..., g k ∈ Λ ′ , where k = [Γ : n Λ] and where ` is the symbol of thedisjoint union. Since H ( g i ) ≡ g i mod n Λ, we get that h ( g i ) ∈ g i + n Λ ⊂ Γ.Therefore we have 23 ′ ⊃ ` ≤ i ≤ k g i + n Λ ′ ,Therefore we have [Γ ′ : n Λ ′ ] ≥ k = [Γ : n Λ]. It follows that[Γ : Γ ′ ] ≤ [ n Λ : n Λ ′ ] = ht( h ).By Lemma 21, we have [Γ : Γ ′ ] = ht( h ). Thus Γ is minimal relative to h . Let n be a finite dimensional nilpotent Lie algebra over Q and let z be itscenter and let Γ be a multiplicative lattice of n . Theorem 2.
The following assertions are equivalent(i) The group Γ is transitive self-similar,(ii) the group Γ is densely self-similar, and(iii) the Lie algebra n C admits a special grading.Proof. Let’s consider the following assertion( A ) S ( n ) = ∅ .The implication ( ii ) ⇒ ( i ) is tautological. Together with the Lemmas 6(i)and 9(i), the following implications are already proved( ii ) ⇒ ( i ) ⇒ ( A ) ⇔ ( iii ).Therefore, it is enough to prove that ( A ) ⇒ ( ii ). Step 1. Definition of some h ∈ G ( Q ) . Let’s assume that S ( n ) = ∅ , and let f ∈ S ( n ). Since the semi-simple part of f is also in G ( Q ), it can be assumedthat f is semi-simple. Let K ⊂ G be the Zariski closure of the subgroupgenerated by f and set H = K .Let Λ be the O ( f )-module generated by Γ. By Lemma 17, Γ is a cosetunion. Therefore Λ is a lattice and Γ is an union of n Λ-coset for some positiveinteger n .Let X ( H ) be the group of characters of H , let K be the splitting field of H , let O be the ring of integers of K , let P be the set of prime ideals of O and let S be set set of all π ∈ P dividing n .By Theorem 1, there exists h ∈ H ( Q ) such that, for any non-trivial χ ∈ X we have(i) χ ( h ) is not an algebraic integer, and(ii) χ ( h ) ≡ n O π for any π ∈ S .24 tep 2. Let Γ ′ = Γ ∩ h − (Γ). We claim that the virtual morphism (Γ ′ , h ) isa good self-similar datum.Since K ⊂ G is the Zariski closure of the subgroup generated by f , wehave Q [ h ] ⊂ Q [ f ] and therefore Λ is a O ( h )-lattice. It follows from Lemma23 that the virtual endomorphism (Γ ′ , h ) is good.Moreover, let Ω be the set of weights of H over z Q . There is an integer l such that f l ∈ K = H . The spectrum of f l on z Q are the numbers χ ( f l )when χ runs over Ω . Thus it follows that Ω does not contain the trivialcharacter, hence h belongs to S ( n ).Therefore by Lemma 20, the virtual endomorphism (Γ ′ , h ) is a good self-similar datum. Thus by Lemma 3, Γ is a densely self-similar group. Theorem 3.
The following assertions are equivalent(i) The group Γ is freely self-similar,(ii) the group Γ is freely densely self-similar, and(iii) the Lie algebra n C admits a very special grading.Proof. Let’s assume Assertion (i). Let’s consider a free self-similar action ofΓ on some A ω and let A ′ be any Γ-orbit in A . Then the action of Γ on A ′ ω is free transitive self-similar, thus Γ is freely transitive self-similar.The rest of the proof is identical to the previous proof, except that1) the assertion ( A ) is replaced by ( A ′ ): V ( n ) = ∅ ,2) the Lemmas 6(ii) and 9(ii) are used instead of Lemmas 6(i) and 9(i)in order to prove that ( ii ) ⇒ ( i ) ⇒ ( A ′ ) ⇔ ( iii ),3) the proof that A ′ ⇒ ( ii ) uses the weights of H and the eigenvalues of f on n instead of z . Let N be a CSC nilpotent Lie group N and let Γ be a cocompact lattice.The manifold M = N/ Γ is called a nilmanifold .A diffeomorphism f : M ! M is called an Anosov diffeomorphism if(i) there is a continuous splitting of the tangent bundle
T M as T M = E u ⊕ E s which is invariant by df , and(ii) there is a Riemannian metric relative to which df | E s and df − | E u arecontracting.For any x ∈ M , f induces a group automorphism f ∗ of Γ ≃ π ( M ).By Lemma 19, f ∗ extends to an isomorphism ˜ f ∗ : n R ! n R , where n R is25he Lie algebra of N . Strictly speaking, ˜ f ∗ is only defined up to an innerautomorphism. Since f ∗ is well defined modulo the unipotent radical ofAut n R , the set Spec f ∗ is unambiguously defined. Manning’s Theorem.
The set
Spec f ∗ contains no root of unity. See [18]. Later on, A. Manning proved a much stronger result. NamelySpec f ∗ contains no eigenvalues of absolute value 1, and f is topologicallyconjugated to an Anosov automorphism, see [19]. Corollary 4.
Let M be a nilmanifold endowed with an Anosov diffeomor-phism. Then π ( M ) is freely densely self-similar.Proof. By definition, we have M = N/ Γ, where N is a CSC nilpotent Liegroup and Γ ≃ π ( M ) is a cocompact lattice. Set n R = Lie N and n C = C ⊗ n R . By Manning’s Theorem and Lemma 8, n C has a very special grading.Therefore Γ is freely densely self-similar by Theorem 3. For completeness purpose, we will now investigate the non-negative gradingsof n C . Unlike Theorems 2 and 3, the proof of Propositions 5 and 6 are quiteobvious.Let n Q be a finite dimensional nilpotent Lie algebra and let Γ be a mul-tiplicative lattice in n . Set n C = C ⊗ n Q . Proposition 5.
The following assertions are equivalent(i) The group Γ is fractal(ii) n C admits a non-negative special grading.(iii) n Q admits a non-negative special grading.Proof. It follows from Lemma 10 that Assertions (ii) and (iii) are equivalent.
Proof that (i) ⇒ (ii). By assumption, there is a fractal datum (Γ ′ , f ). Let g : Γ ! Γ ′ be the inverse of f and let ˜ g ∈ Aut n be its unique extension.Let Λ ⊂ n be the additive subgroup generated by Γ. By Lemmas 17, Λis an additive lattice. Since we have ˜ g (Λ) ⊂ Λ, it follows that all eigenvaluesof ˜ g are algebraic integers.Moreover (Γ ′ , g − ) is a self-similar datum, thus Spec ˜ g − | z contains noroot of unity. Therefore, by Lemma 10, Assertion (ii) holds.26 roof that (iii) ⇒ (i). Let’s assume Assertion (iii) and let n Q = ⊕ k ≥ n Q k be a non-negative special grading of n Q .By Lemma 17, Γ lies in a lattice Λ. Since it is possible to enlarge Λ, wecan assume that Λ = ⊕ k ≥ Λ k ,where Λ k = Λ ∩ n Q k . Since Γ is a coset union, there is an integer d ≥ d Λ-cosets.Let g be the automorphism of n Q defined by g ( x ) = ( d + 1) k x if x ∈ n Q k .We claim that g (Γ) ⊂ Γ. Let x ∈ Γ and let x = P k ≥ x k be its decompositioninto homogenous components. We have g ( x ) = x + P k ≥ (( d + 1) k − x k .By hypothesis each homogenous component x k belongs to Λ. Since ( d +1) k − d , we have g ( x ) ∈ x + d Λ ⊂ Γ and the claim is proved.Set Γ ′ = g (Γ) and let f : Γ ′ ! Γ be the inverse of g . It is clear that(Γ ′ , f ) is a fractal datum for Γ, what proves Assertion (i). Proposition 6.
The following assertions are equivalent(i) The group Γ is freely fractal(ii) n C admits a positive grading.(iii) n Q admits a positive grading. Since the proof is strictly identical, it will be skipped.
This section provides an example of a FGTF nilpotent group which is noteven self-similar, see subsection 8.6. The end of the section is about theMilnor-Scheuneman conjecture.
Let Γ be a FGTF nilpotent group and let Z (Γ) be its center. Lemma 24.
Let’s asssume that Γ is self-similar and Z (Γ) ≃ Z . Then Γ istransitive self-similar. roof. Assume that Γ admits a faithful self-similar action on some A ω , where A is a finite alphabet. Let a , . . . , a k be a set of representatives of A/ Γ, where k is the number of Γ-orbits on A . For each 1 ≤ i ≤ k , let Γ i be the stabilizerof a i . For any h ∈ Γ i , there is h i ∈ Γ such that h ( a i w ) = a i h i ( w ),for all w ∈ A ω . Since the action is faithfull h i is uniquely determined andthe map f i : Γ i ! Γ , h h i is a group morphism.Let n Q be the Malcev Lie algebra of Γ, and let z be its center, and let z = 0 be a generator of ∩ i Z (Γ i ). By Lemma 19, the group morphism f i extends to a Lie algebra morphism ˜ f i : n Q ! n Q . Since z = Q ⊗ Z (Γ) is onedimensional, it follows that either ˜ f i is an isomorphism or ˜ f i ( z ) = 0. In anycase, we have ˜ f i ( z ) = x i z , for some x i ∈ Q . However Z z is not invariant byall ˜ f i , otherwise it would be in the kernel of the action. It follows that atleast one x i is not an integer.For such an index i , the f i -core of Γ i is trivial, and the virtual morphism(Γ i , f i ) is a self-similar datum for Γ. Thus Γ is transitive self-similar. Let N be a CSC nilpotent Lie group with Lie algebra n R and let Γ be acocompact lattice. Lemma 25. If Γ is transitive self-similar, then there exists a faithfull n R -module of dimension n R .Proof. By hypothesis, Γ is transitive self-similar. By Theorem 2, z C admitsa special grading n C = ⊕ n ∈ Z n C n .Let δ : n C ! n C be the derivation defined by δ ( x ) = nx if x ∈ n n . Since δ | z C is injective, it follows that there is some ∂ ∈ Der n R such that ∂ | z R isinjective.Set m R = R ∂ ⋉ n R . Relative to the adjoint action, m R is a faithfull z R -module. Therefore m R is a faithfull n R -module with the prescribed dimension. Let n be a nilpotent Lie algebra over Q . Let C n n be the decreasing centralseries, which is inductively defined by C n = n and C n +1 n = [ n , C n n ]. The28ilpotent Lie algebra n is called filiform if dim C n /C n = 2 and dim C k n /C k +1 n ≤ k >
1. Set n = dim n . It follows from the definition thatdim C k n /C k +1 n = 1 for any 0 < k ≤ n − C k n = 0 for any k ≥ n . Lemma 26.
Let n be a filiform nilpotent Lie algebra over Q , with dim n ≥ .Then its center z has dimension one.Proof. Let z ∈ n be nonzero. Let k be the integer such that z ∈ C k n \ C k +1 n .Since C k n = C k +1 n ⊕ Q zC k +1 n = [ n , C k n ] = [ n , C k +1 n ] + [ n , z ] = C k +2 n .It follows that C k +1 n = 0. Therefore z lies in C k n , which is a one dimensionalideal. Benoist’s Theorem.
There is a nilpotent Lie algebra n R B of dimension over R , with the following properties(i) The Lie algebra n R B has no faithfull representations of dimension ,(ii) the Lie algebra n R B is defined over Q , and(iii) the Lie algebra n R B is filiform. The three assertions appear in different places of [3]. Indeed Assertion(i), which is explicitely stated in Theorem 2 of [3], hold for a one-parameterfamily of eleven dimensional Lie algebras, which are denoted a − , ,t in section2.1 of [3]. These Lie algebras are filiform by Lemma 4.2.2 of [3]. Moreover,when t is rational, a − , ,t is defined over Q . Therefore the Benoist Theoremholds for the Lie algebras n B = a − , ,t where t is any rational number. Let N B the CSC nilpotent Lie group with Lie algebra n R B . Since n R B is definedover Q , N B contains some cocompact lattice. Corollary 7.
Let Γ be any cocompact lattice in N B . Then Γ is not self-similar.Proof. Let’s assume otherwise. By Benoist Theorem and Lemma 26, thecenter of n R B is one dimensional. Thus the center of Γ has rank one, and byLemma 24, Γ is transitive self-similar. By Lemma 25, n R B admits a faithfullrepresentation of dimension 12, which contradicts Benoist Theorem.Therefore Γ is not self-similar. 29 .6 On the Scheuneman-Milnor conjecture A smooth manifold M is called affine if it admits a torsion-free and flatconnection. Scheuneman [28] and Milnor [23] asked the following question is any nilmanifold M affine? The story of the Scheuneman-Milnor conjecture is quite interesting. Formany years, there are been a succession of proofs followed by refutations,but there was no doubts that the conjecture should be ultimalely proved...until a counterexample has been found by Benoist [3]. Indeed it is an easycorollary of his previously mentionned Theorem.The following question is a refinement of the previous conjecture if π ( M ) is densely self-similar, is the nilmanifold M affine? A positive result in that direction is
Corollary 8.
Let M be a nilmanifold. If π ( M ) is freely self-similar, then M is affine complete.Proof. Set M = N/ Γ, where N is a CSC nilpotent Lie group and Γ is acocompact lattice. Let n R be the Lie algebra of N . By Theorem 3, C ⊗ n R admits a very special grading, what implies that a generic derivation isinjective. Therefore there is a derivation δ of n R which is injective. Set m R = R δ ⋉ n R . Then N is equivariantly diffeomorphic to the affine space δ + n R ⊂ m R . Therefore M is affine complete. For the whole chapter, N will be a CSC nilpotent Lie groups, with Lie algebra n R . Let’s assumethat that N contains some cocompact lattices.Under the condition of Theorem 2 or 3, any cocompact lattice Γ in N admits a transitive or free self-similar action on some A ω . In this section, wetry to determine the minimal degree of these actions. The complexity of a cocompact lattice Γ ⊂ N , denoted by cp Γ, is the smallestdegree of a faithfull transitive self-similar action of Γ on some A ω , with theconvention that cp Γ = ∞ if Γ is not transitive self-similar. Similarly, the freecomplexity of Γ, denoted by fcp Γ, is the smallest degree of a free self-similaraction of Γ. Two cocompact lattices are called commensurable if they share a30ommoun subgroup of finite index. The complexity and the free complexityof a commensurable class ξ are the integerscp ξ = Min Γ ∈ ξ cp Γ, andfcp ξ = Min Γ ∈ ξ fcp Γ.Then, the complexity of the nilpotent group N iscp N = Max ξ cp ξ ,where ξ runs over all commensurable classes in N . In what follows, we willprovide a formula for the complexity of commensurable classes. The question under which condition cp N < ∞ ? is not solved, but it is a deep question. In chapter 10, a class of CSC nilpotentLie groups of infinite complexity is investigated. Let ξ be a commensurable class of cocompact lattices in N , and let Γ ∈ ξ .The Malcev Lie algebra Γ is a Q -form of the Lie algebra n R . Since it dependsonly on ξ , it will be denoted by n ( ξ ). Theorem 9.
We have cp ξ = Min h ∈S ( n ( ξ )) ht( h ) , and fcp ξ = Min h ∈V ( n ( ξ )) ht( h ) .Proof. Let h ∈ S ( n ( ξ )) be an isomorphism of minimal height. In order toshow that cp ξ = ht( h ), we can assume that h is semi-simple, by Lemma 13.Further, let Γ be any cocompact lattice in ξ . By Lemma 20, we havecp Γ = Min f ∈S ( n ( ξ )) cp f Γ. By lemma 21, we have cp f Γ ≥ ht( f ), therefore wehave cp Γ ≥ ht( h ). In particular cp ξ ≥ ht( h ).By Lemma 22, Γ contains a finite index subgroup ˜Γ which is minimalrelative to h . Since cp h ˜Γ = ht( h ), it follows that cp ξ ≤ ht( h ).Therefore cp ξ = ht( h ) and the first assertion is proved.For the second assertion, let’s notice that an free action of minimal degreeis automatically transitive, see the proof of Theorem 3. Then the rest of theproof is strictly identical to the previous proof. Obviously Malcev’s Theorem implies the following
Malcev’s Corollary.
The map ξ n ( ξ ) establishes a bijection between thecommensurable classes of lattices and the Q -forms of the Lie algebra n R . Q -forms of classical objects.Set G = Aut n C , let U be its unipotent radical and set G = G / U . Fromnow on, fix once for all a commensurable class ξ of cocompact lattices. Then n ( ξ ) is a Q -form of n C , what provides a Q -form of the algebraic groups G and G . It induces an action of Gal( Q ) over G ( Q ).Set Q re = Q ∩ R and let π : H (Gal( Q ) , G ( Q )) ! H (Gal( Q re ) , G ( Q ))be the natural map. Recall that these two non-abelian cohomologies arepointed sets, where the distinguished point ∗ comes from the given Q -formand the induced Q re -form. Denote by Ker π the kernel of π , i.e. the fiber π − ( ∗ ) of the distinguished point.Let L ( N ) be the set of all commensurable classes classes of lattices of N ,up to conjugacy. Corollary 10.
There is a natural identification L ( N ) ≃ Ker π .Proof. For any field K ⊂ C , set n K = K ⊗ n ( ξ ). For any two fields K ⊂ L ⊂ C , let F ( L/K ) be the set of K -forms of n L , up to conjugacy. Then F ( L/K )is a pointed set, whose distinguished point is the K -form n K .By the Lefschetz principle, the Q -forms of n C (up to conjugacy) are inbijection with the Q -forms of n Q . Similarly by the Tarski-Seidenberg principlethe real forms (up to conjugacy) of n C are in bijection with the Q re -forms of n Q . So we have F ( C / Q ) ≃ F ( Q / Q ) and F ( C / R ) ≃ F ( Q / Q re )Since a Lie algebra is a vector space endowed with a tensor (its Liebracket), it follows from [29], III-2, Proposition 1 that F ( Q / Q ) = H (Gal( Q ) , G ( Q )), and F ( Q / Q re ) = H (Gal( Q re ) , G ( Q )).Moreover since U is unipotent, we have H (Gal( Q ) , G ( Q )) ≃ H (Gal( Q ) , G ( Q )), and H (Gal( Q re ) , G ( Q )) ≃ H (Gal( Q re ) , G ( Q )).There is a commutative diagram of pointed sets32 ( C / Q ) θ −! F ( C / R ) F ( Q / Q ) θ ′ −! F ( Q / Q re ) H (Gal( Q ) , G ( Q )) π −! H (Gal( Q re ) , G ( Q )) H (Gal( Q ) , G ( Q )) π −! H (Gal( Q re ) , G ( Q ))where θ is the map R ⊗ Q − , θ ′ is the map Q re ⊗ Q − , π and π are restrictionsmaps. It is tautological that L ( N ) = Ker θ . Since all vertical maps arebijective, it follows that L ( N ) is isomorphic to Ker π .
10 Some Nilpotent Lie groups of infinite com-plexity.
This chapter is devoted to the analysis to a class of CSC nilpotent Lie groups N , for which the classification of commensurable classes and the computationof their complexity are very explicitely connected with the arithmetic ofcomplex quadratic fields.For K = R or C , let O (2 , K ) be the group of linear automorphisms of R preserving the quadratic form x + y . Let L be the class of nilpotent Liealgebras n R over R satisfying the following properties(i) n R has a Q -form(ii) n R / [ n R , n R ] ≃ R has dimension two(iii) the Lie algebra n C := C ⊗ n R has a special grading(iv) for K = R or C , the image of Aut n K in GL ( n K / [ n K , n K ]) is O (2 , K ).Let be the class of CSC nilpotent Lie groups N whose Lie algebra n R is in L .It should be noted that the class N is not empty. There is one Lie group N ∈ N of dimension 112, see [22]. Indeed [22] contains a general methodto find nilpotent Lie algebras with a prescribed group of automorphisms,modulo its unipotent radical. For the group O (2 , R ), N is the Lie groupof minimal dimension obtained with this method. However it is difficult toprovide more details, without going to very long explanations.From now on, N will be any Lie group in class N , ξ will be one com-mensurable class of lattices in N and n := n ( ξ ) will be the correspondingcorresponding Q form of n R . As before, set n K = K ⊗ n for any field K ⊂ C .33et G = Aut n the algebraic automorphism group of n , let U be its unipo-tent radical and set G = G / U . By hypothesis, G is the algebraic group O (2). Z -grading of n C Since G ( C ) = O (2 , C ), a maximal torus H of G ( C ) has dimension 1. There-fore n C has a Z -grading n C = ⊕ k ∈ Z n C k ,satisfying the following properties(i) the grading is essentially unique, namely any other grading is a mul-tiple of the given grading,(ii) dim n C k = dim n C − k for any k . In particular n C does not admit a (non-trivial) non-negative grading, and(iii) the grading is not defined over R .Indeed since G ( C ) = O (2 , C ), the normalizer K ( C ) of H ( C ) has two con-nected components, and any σ ∈ K ( C ) \ K ( C ) exchanges n C k and n C − k , whatshows Assertion (ii). Since G ( R ) = O (2 , R ), no torus of G ( R ) is split, whatimplies Assertion (iii).Moreover, the grading is not very special, so fcp( ξ ) = ∞ for any com-mensurable class ξ . For the forthcoming computation of cp( ξ ), the followingquantity will be involved e ( N ) = P k> k dim n C k .For example, for the Lie group N of [22], we have e ( N ) = 126. N Lemma 27.
Let N ∈ N . Up to conjugacy, there is a bijection between(i) the commensurable class of cocompact lattices in N , and(ii) the positive definite quadratic form on Q .Proof. Let q be a given definite quadratic form on Q . It determines a Q -form of the algebraic group O (2), and H (Gal( Q ) , O (2 , Q )) classifies thequadratic forms on Q , while the kernel of H (Gal( Q ) , O (2 , Q )) ! H (Gal( Q re ) , O (2 , Q ))classifies the positive definite quadratic forms on Q . Thus the lemma followsfrom Corollary 10.The classification of positive definite quadratic forms q on Q is wellknown. Up to conjugacy, q can be written as34 ( x, y ) = ax + ady ,where a, d are positive and d is a square-free integer. Then q is determinedby the following two invariants(i) its discriminant − d , viewed as an element of Q ∗ / Q ∗ ,(ii) the value a , viewed as an element in Q ∗ /N K/ Q ( K ∗ ), where K = Q ( √− d ). Equivalently, this means that q ( Q \
0) = aN K/ Q ( K ∗ ).For any positive definite quadratic forms q on Q , let ξ ( q ) be the cor-responding commensurable class (or more precisely, the conjugacy class ofthe commensurable class). By Theorem 9, cp ξ ( q ) only depends on O ( q ),therefore it only depends on the discriminant − d . F ( d )Let d be a positive square-free integer. Set K = Q ( √− d ), let O be its ring ofintegers, let R be te set of roots of unity in K and set K = { z ∈ K | zz = 1 } .For z ∈ K ∗ , recall that the integer d ( z ) is defined by d ( z ) = N K/ Q ( π z ) =Card O /π z , where π z is the ideal π z = { a ∈ O| az ∈ O} . Set F ( d ) = Min z ∈ K \ R d ( z ).We will now show two formulas for F ( d ). Indeed F ( d ) is the norm of somespecific ideal in K = Q ( √− d ), and it is also the minimal solution of somediohantine equation.Let J be the set of all ideals π of O such that π and π are coprime and π is principal. Lemma 28.
We have F ( d ) = Min π ∈J N k/ Q ( π ) .In particular, we have F (1) = 5 and F (3) = 7 .Proof. The map z π z induces a bijection ( K \ R ) /R ≃ J , from which thefirst assertion follows. Moreover, if Cl( K ) = { } , then F ( d ) is the smallestsplit prime number. Therefore F (1) = 5 and F (3) = 7.Let’s consider the following diophantine equation( E ) 4 n = a + db , with n > a > b = 0.A solution ( n, a, b ) of ( E ) is called primitive if gcd( n, a ) = 1. Let Sol ( E )(respectively Sol prim ( E )) be the set of solutions (respectively of primitivesolutions) of ( E ).Let π ∈ J . Since π is principal, there are integers a ( π ) > b ( π )such that a ( π ) + b ( π ) √− d is a generator of 4 π . Moreover, let’s assume that35 = 1 or 3. Then R = {± } and the integers a ( π ) and b ( π ) are uniquelydetermined. Thus there is a map θ : J ! Sol ( E ) defined by θ ( π ) = ( N K/ Q ( π ) , a ( π ) , b ( π )). Lemma 29.
Under the hypothesis that d = 1 or , the map θ induces abijection from J to Sol prim ( E ) . In particular F ( d ) = Min ( n,a,b ) ∈ Sol ( E ) n .Proof. Step 1: proof that θ ( J ) ⊂ Sol prim ( E ) . An algebraic integer z ∈ O iscalled primitive if there are no integer d > z/d is an algebraicinteger. Equivalently, there are no integer d > d | z + z and d | zz .Let π ∈ J and set z = 1 / a ( π ) + b ( π ) √− d ). Since z + z = a ( π ) and z.z = N K/ Q ( π ), z is an algebraic integer which is a generator of π . Since π and π are coprime, z is primitive. Since z.z = N K/ Q ( π ) , it follows that N K/ Q ( π ) and a ( π ) are coprime. Hence θ ( π ) ∈ Sol prim ( E ) and the claim isproved. Step 2: proof that θ ( J ) = Sol prim ( E ) . Let ( n, a, b ) ∈ Sol prim ( E ) and z =1 / a ( π ) + b ( π ) √− d ). Since z = z , z + z = a and zz = n , the number z isan algebraic integer. Set τ = z O and let τ = π m . . . π m k k be the factorization of τ into a product of prime ideals of O , where, as usualwe assume that π i = π j for i = j and all m i are positive.For 1 ≤ i ≤ k , let p i be the characteristic of the field O /π i . Since n and a are coprime, τ and τ are coprime. It follows that π i does not divide τ . Inparticular π i = π i and N K/ Q ( π i ) = p i . Since π i and π i are the only two idealsover p i , we have m i = v p i ( N K/ Q ( τ )) = v p i ( n ). Since each m i is even, wehave τ = π for some ideal π ∈ J . Therefore θ ( π ) = ( n, a, b ), and the claimis proved. Step 3.
It follows easily that θ is a bijection from J to Sol prim ( E ). Inparticular F ( d ) = Min ( n,a,b ) ∈ Sol prim ( E ) n , from which the lemma follows. Theorem 11.
Let q be a positive definite quadratic form on Q of discrimi-nant − d . Then we have cp ξ ( q ) = F ( d ) e ( n C ) roof. Step 1. Let G ⊂ End Q ( K ) be the group generated by the multiplica-tion by elements in K and by the complex conjugation. We have G ≃ O (2)and SO (2) ≃ K . As a O ( q )-module, there is an isomorphism V ≃ Q ( √− d ),where V = n ( ξ ( q )) / [ n ( ξ ( q )) , n ( ξ ( q ))]. Step 2.
Let S ( n ( ξ ( q )) be the image of S ( n ( ξ ( q )) in O ( q ). We claim that S ( n ( ξ ( q )) = K \ R .Indeed O ( q ) can be identified with a Levi factor of G ( Q ) and let ρ : O ( q ) ! G ( Q ) a corresponding lift. Any element in R ∪ O ( q ) \ SO ( q ) has finite order,hence we have S ( n ( ξ ( q )) ⊂ K \ R .Let z ∈ K \ R . It is clear that z is not an algebraic integer. Since thegrading is special, we have z C = ⊕ k =0 z C k .Since the eigenvalues of ρ ( z ) on z k is z k , it follows that z belongs to S ( n ( ξ ( q )),what proves the point. Step 3.
Let z ∈ K \ R . We have z = z − . Therefore by Lemma 15 we haveht ρ ( z ) = Q k ≥ d ( z k ) dim n C k = d ( z ) e ( N ) .Therefore Theorem 4 implies Theorem 5.Since F ( d ) ≥ √ d , it follows that Corollary 12.
The group N has infinite complexity. Since F (7) = F (15) = 2 and F ( d ) ≥ Corollary 13.
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