Anosov automorphisms on nilmanifolds in dimensions 9 and 10

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by  Meera G. Mainkar

ANOSOV AUTOMORPHISMS ON NILMANIFOLDS IN
DIMENSIONS 9 AND 10.
9
0 MEERAMAINKAR, CYNTHIAE.WILL
0
2

n
a
Abstract. We study 9 and 10-dimensional Anosov Lie algebras, by using
J
the properties of very special algebraic numbers. We classifyk-step complex

3 AnosovLiealgebrasfork>2andinthetwostepcase,wegiveanexamplein
2 eachpossibletype,orwegiveanon-existence result.

]
S 1. Introduction
D
A diffeomorphisms f of a compact differentiable manifold M is called Anosov if
. the tangent bundle TM admits a continuous invariant splitting TM = E+ ⊕E−
h
t such that df expands E+ and contracts E− exponentially. In [19], S. Smale raised
a
the problem of classifying the compact nilmanifolds admitting Anosov diffeomor-
m
phisms, and up to now the only known examples of Anosov diffeomorphisms are
[
on nilmanifolds or manifolds finitely covered by nilmanifolds, which are known as

1 infranilmanifolds. It is even conjectured that any Anosov diffeomorphism is topo-
v logically conjugate to an Anosov automorphism of an infranilmanifold (see [18]).
9
This conjecture is known to be true in some cases (see [7], [14]). This highlights
3
the problem of classifying all compact nilmanifolds which admit an Anosov auto-
7
3 morphism.
. A rational Lie algebra nQ (i.e. with structure constants in Q) of dimension n is
1
saidto be Anosov if it admits a hyperbolic automorphismA (i.e. all its eigenvalues
0
9 have absolute value different from 1) which is unimodular, that is, [A]β ∈ GLn(Z)
0 for some basis β of nQ, where [A] denotes the matrix of A with respect to β.
β
v: We call a real or complex Lie algebra Anosov if it admits a rational form which is
i Anosov (see [10]).
X
Then, a simply connected n-dimensional nilpotent Lie group N, with its Lie
r algebra n, is a cover of a nilmanifold admitting an Anosov automorphism if and
a
only if there exists a Z-basis β (i.e. with integer structure constants) of n and a
hyperbolic automorphism A of n such that [A] ∈GL(n,Z).
β
Hence as far as the question of classifying nilmanifolds admitting Anosov auto-
morphismsisconcerned,oneisinterestedinrealAnosovLiealgebras. Theproblem
of deciding if a nilpotent Lie algebra n admits an Anosov automorphism is equiv-
alent to see if n admits a hyperbolic automorphism A such that [A] ∈ GL(n,Z),
β
where β is a Z-basis of n and n is the dimension of n.
Thefirstexample(due toBorel)ofanon-toralnilmanifoldadmittinganAnosov
automorphism was described by S. Smale ([19]). After that only relatively few
2000Mathematics Subject Classification. Primary: 37D20;Secondary: 22E25, 20F34.
Keywordsandphrases. Anosovdiffeomorphisms,nilmanifolds,nilpotentLiealgebras,hyperbolic
automorphisms.
SupportedbyCONICETandgrantsfromFONCyTandSeCyT..
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2 MEERAMAINKAR, CYNTHIAE.WILL
examples appeared in the literature. Recently, the families of nilmanifolds with
Anosov automorphisms have been constructed showing that a complete classifica-
tion does not seem to be possible (see [2], [3], [4], [5], [6], [10], [12], [15], [17]). In
[13], all real and rationalAnosov Lie algebrasof dimension ≤8 are classified up to
isomorphism and we note that curiously enough, there are quite few ones.
In this paper we will study Anosov complex Lie algebras. To classify Anosov
Lie algebras over R, one has to find all the others reals form of the given complex
one and study which ones are Anosov. More specifically, we study Anosov Lie
algebras of dimension 9 and 10 by using the properties of very special algebraic
numbers, following the ideas introduced in [10] and used also in [12],[13] and [17].
This algebraic numbers are actually the eigenvalues of the corresponding Anosov
automorphism,andtheyhaveveryspecialpropertiesthathavebeenstudiedin[16].
Usingallthisandnotingthattheexistenceofsuchanalgebraimpliestheexistence
of a lower dimensional one that can be found in the classification given in [13], we
classify k-step complex Anosov Lie algebras for k >2.
In the 2-step case, we consider every eventually possible type as in [13, Propo-
sition 2.5], and in each one we either give a non existence result, or we give an
exampleofAnosovLiealgebrashowingthatthiscaseisactuallypossible. Byusing
theresultsfrom[16],weprovethata9-dimensionalAnosovLiealgebrashouldbeof
type(6,3)or(3,3,3)andmoreoverweshowthatthereisonlyonecomplexAnosov
Lie algebra of type (3,3,3). We note that this is the first example of an Anosov
Lie algebra of type (3,3,∗) showing that the condition in [12, Proposition 2.3 (ii)]
n ≥3 is in fact attained. In the (6,3) case, we give a necessary condition related
3
tothePfaffianformandfromtherewegetalistofthepossibleAnosovLiealgebras
oftype(6,3),butforsomeofthemwearenotabletoconcludewhethertheyadmit
Anosov automorphism or not. In the last section we study 10-dimensional Anosov
Lie algebras. We prove that the possible types of 2-step Anosov Lie algebras are
(8,2),(6,4),(4,6)and(5,5). Therearealreadyknownexamplesoftype(8,2),(6,4)
and (4,6), and we give a new example of an Anosov Lie algebra of type (5,5). For
highersteps,weprovethatthereisonlyoneAnosovLiealgebraineachofthetypes
(6,2,2),(4,2,4),(4,2,2,2) and two of the type (4,4,2).
We would like to thanks J. Lauret for his constant help and advice.
2. Preliminaries
In this section, in order to introduce the framework in which we are going to
work, we will begin by recalling some facts about the eigenvalues of an Anosov
automorphism, proved in [12] that we are including for completeness.
Proposition 2.1. (see [12]) Let n be a real r-step nilpotent Lie algebra. We define
Ci(n) inductively by Ci(n) = [n,Ci−1(n)] for i ≥ 1, where C0(n) = n. If n is an
Anosov Lie algebra then there exist a decomposition n = n ⊕···⊕n satisfying
1 r
Ci(n)=n ⊕...⊕n , i=0,...,r, and a hyperbolic A∈Aut(n) such that
i+1 r
(i) An =n for all i=1,...,r.
i i
(ii) A is semisimple (in particular A is diagonalizable over C).
(iii) For each i, there exists a basis β of n such that [A ] ∈ SL (Z), where
i i i βi ni
ni =dimni and Ai =A|ni.
Ifwehaveadecompositionn=n ⊕···⊕n asinthe previousproposition,then
1 r
we say that n is of type (n ,n ,...,n ) where n = dim n for all i=1,...,r.
1 2 r i i
ANOSOV AUTOMORPHISMS ON NILMANIFOLDS 3
It is not hard to see that the eigenvalues of anAnosov automorphismare neces-
sarily algebraicunits, that is, an eigenvalue of an Anosov automorphismsatisfies a
monicpolynomialequationwithintegercoefficientsandunitconstantterm. Foran
algebraic number λ∈C, we denote by deg(λ) the degree of m (x), the irreducible
λ
monic polynomial over Q annihilated by λ and by the conjugated numbers to λ we
will denote the conjugated algebraic numbers of λ over Q, that is, the other roots
of m (x). We then have the following result (see [12]).
λ
Lemma 2.2. Let n be an Anosov real Lie algebra which is Anosov, and let A and
n = n1⊕n2⊕···⊕nr be as in Proposition 2.1. If Ai = A|ni then the eigenvalues
λ ,...,λ ofA ,arealgebraicunitssuchthat1<deg(λ )≤n forallj =1,··· ,n
1 ni i j i i
and we may assume that λ ...λ =1.
1 ni
From this, for each A as in the lemma, we can define the splitting of A by the
i i
tuple [k ;...;k ], k ∈ N, such that there exists a reordering of the eigenvalues
1 m i
of A , {λ ,...,λ } (counted with their multiplicities) such that for all l ≥ 1,
i 1 ni
l−1 l−1
degλ =k ifi= k +1,..., k +k ,whereforconsistencywedefinek =0.
i l t t l 0
t=0 t=0
X X
Morespecifically,the characteristicpolynomialofA canbewrittenasthe product
i
of m irreducible polynomials (over Z) f ,f ,...f such that degf = k for all i,
1 2 m i i
and the roots of f , l≥1, are
l
λPtl−=10 kt+1,...,λPtl−=10kt +kl.
Itiseasytoseethatk +···+k =n andsinceA ishyperbolicandunimodular,
1 m i i
we also have that k 6=1 for all j (see [12] Appendix). Also, since the order is not
j
relevant, we can order it from greatest to lowest.
We note that each eigenvalue of A−i is a product of certain eigenvalues of A
1
andeigenvaluesofAarealgebraicunits. Wearetheninterestedinhowtheproduct
ofsuchspecialalgebraicnumberscanbehave. Itisprovenin[16]thatthisbehavior
cannot be so wild. In fact, one has the following lemma that canbe deduced from
[16].
Lemma 2.3. Let A be an Anosov automorphism of a nilpotent Lie algebra n and
let α and β be two eigenvalues of A. Then the following hold:
1. If deg(α) and deg(β) are relatively prime then deg(αβ) cannot be a prime.
2. If αβ is also an eigenvalue of A and deg(α)≤deg(β) then
g.c.d.(deg(β),deg(αβ))6=1.
An abelian factor of n is an abelian ideal a for which there exists an ideal n˜ of n
suchthat n=n˜⊕a (i.e. [n˜,a]=0). Let m(n) denote the maximumdimension over
all abelian factors of n. By [12, Theorem 3.6 ] if n is a rational Lie algebra with
m(n)=r and n= n˜⊕Qr is any decomposition in ideals, that is, Qr is a maximal
abelian factor of n, then n is Anosov if and only if n˜ is Anosov and r ≥2.
Aswehavesaid,wearegoingtostartwithanAnosovLiealgebra,n,andbyusing
the properties we have mentioned of the eigenvalues of the Anosov automorphism
it admits, we are going to deduce properties of n. Therefore, if n is an Anosov Lie
algebra, with Anosov automorphism A, we are going to consider the complex Lie
algebra n = n⊗C which admits a decomposition n =(n ) ⊕...⊕(n ) , where
C C 1 C r C
4 MEERAMAINKAR, CYNTHIAE.WILL
(n ) = n ⊗C, as in Proposition 2.1, and a basis of eigenvectors of A for all i.
i C i i
Moreover, we also want to point out that if n has an abelian factor, then n must
C
also carry an abelian factor. In fact, z(n )∩[n ,n ]=(z(n)∩[n,n]) and hence if
C C C C
z(n )∩[n ,n ]6=z(n ) then (z(n)∩[n,n])6=z(n).
C C C C
Finally,wewillstatethefollowingobservationthatcanbeeasilyprovedbyusing
Proposition 2.1.
Lemma 2.4. Let n be an Anosov Lie algebra of type (n ,...,n ) and let n =
1 r
n ⊕...⊕n be the decomposition of n given in Proposition 2.1. Then n=n/n and
1 r r
n′ =[n,n] are both Anosov.
e
Infact,ifnisanAnosovLiealgebra,Aandn=n ⊕...⊕n areasinProposition
1 r
2.1 then, since A is an automorphism, it preserves n′ =[n,n]=n ⊕...⊕n . Also,
2 r
since An =n for all i=1,...,r, it is easy to see that it induces an automorphism
i i
of n = n/nr ≃ n1⊕...⊕nr−1. It is clear that, in both cases, these automorphisms
arehyperbolicandtoseethattheyalsoareunimodular,recallthatforeachi,there
exiestsabasisβi ofni suchthat[Ai]βi ∈SLni(Z),whereni =dimni andAi =A|ni.
Note that this argument is valid not only at real or complex level but also at
the rational case. Concerning the type of n or n′ we note that if n is a three step
nilpotent Anosov Lie algebra, then since [n ,n ]= 0 then n= n/n ≃ n ⊕n and
2 2 3 1 2
this decomposition gives the type of n. Onethe other hand, in this case, n′ = [n,n]
is an abelian Lie algebra. e
e
3. Dimension 9
InthissectionwewillstudycomplexAnosovLiealgebrasofdimension9.Infact
we will prove that any Anosov Lie algebra without an abelian factor has to be of
type (6,3)or (3,3,3). Moreover,there is only one complex AnosovLie algebra(up
to isomorphism) without an abelian factor of type (3,3,3). In the case of (6,3) we
give some new examples.
Inthe following,n, willbe acomplexAnosovLiealgebrawithno abelianfactor,
andA,A andn willbeasinProposition2.1. WewilldenotetheeigenvaluesofA
i i 1
byλ′s,eigenvaluesofA byµ′sandeigenvaluesofA byν′s,andthecorresponding
i 2 j 3 k
eigenvectors by X′s, Y′s and Z′s respectively.
i j k
ForsimplicityassumeforamomentthatnisatwostepnilpotentLiealgebra. We
mayassumethatforeachY ,thereexistX andX suchthatY =[X ,X ]. Onthe
i j k i j k
otherhand,iffor eachj,l,letak ∈C be the scalarssuchthat[X ,X ]= ak Y ,
j,l j l j,l k
since {Y } are linearly independent, for each k we obtain
k
P
(1) λ λ ak =µ ak .
j l j,l k j,l
Hence, if akj,l 6= 0, µk = λjλl, and therefore, if akj,l 6= 0 6= akj,′l, then µk = µk′. In
particular, if n =2, since µ 6= µ , for each j,l, there exists a unique k such that
2 1 2
[X ,X ] = a Y . If it is so, by (1), λ λ = µ . When n = 3, the same property
j l k k j l k 2
holds. Indeed, µ 6=µ foralli6=j sincedegµ >1 foralli.Ifn isa 3-stepAnosov
i j i
Lie algebra and if n = 2 or 3, then for each Z we have that Z = [X ,Y ] and
3 j j i k
therefore ν =λ µ for some i,k (see [13]).
j i k
ANOSOV AUTOMORPHISMS ON NILMANIFOLDS 5
UsingProposition[12,Prop. 2.3]wecanseethatthepossibletypesfora9dimen-
sional Anosov Lie algebra are (7,2), (6,3), (5,4), (4,5), (5,2,2), (4,3,2), (4,2,3)
and (3,3,3).
AsacorollaryofLemma2.4andthefactthatthereisnonon-toral7dimensional
AnosovLiealgebra(see [13]),there areno AnosovLiealgebrasoftype (5,2,2)and
(4,3,2).
We are now goingto show that a nilpotent Lie algebrawith no abelianfactor of
any of this types that actually admits an Anosov automorphismis of type (6,3)or
(3,3,3). For all the remaining cases but those, we will systematically use Lemma
2.3 to show that there must be an abelian factor. In these cases we will prove that
n have an abelian factor, and therefore n must also carry one. Note that in this
C
situation n = Cn⊕n˜ where n˜ is a lower dimensional Anosov Lie algebra with no
abelian factor, and therefore should be one of those in the classification given in
[13].
Case (7,2) Suppose that there exists an Anosov Lie algebra n of this type with
noabelianfactor,andlet{λ ,...,λ }and{µ ,µ }denotetheeigenvaluesofA and
1 7 1 2 1
A with eigenvectors X ,...X ,Y ,Y , respectively. Hence the splitting of A can
2 1 7 1 2 1
be[7], [5;2], [4;3]or[3;2;2]. Itisnothardtoseethatnoneofthiscasesarepossible.
Infact,ifthesplittingofA is[7]thatisdegλ =7forall1≤i≤7,thenµ =λ λ
1 i k i j
for some 1 ≤ i 6= j ≤ 7 with degµ = 2, and therefore (degλ λ ,degλ ) = 1, a
k i j i
contradiction to Lemma 2.3. In fact, we note that if max(degλ ,degλ ) is odd,
i j
then the corresponding eigenvectorsmust satisfy [X ,X ]=0 since λ λ can never
i j i j
be an eigenvalue of A . Therefore, the cases [5;2] and [3;2;2] are not possible
2
either, since [X ,X ]=0 for all i,j and the algebra will be abelian.
i j
Finally, since 2 is a prime number, Lemma 2.3 implies that if degλ = 4 and
i
degλ = 3, (or equivalently 3 and 3) then [X ,X ]=0 in the same way as before.
j i j
Therefore case [4;3] we will have an abelian factor, namely the ideal generated by
X ,X and X
5 6 7
Case (5,4)In this case we can arguein a similar way to the previous one, since
the possible splitting of A are [5] and [3;2]. In the first case, for each non zero
1
bracket among the eigenvectors of A , that is if [X ,X ]6=0 for some 1≤i,j ≤5,
1 i j
wethenhavethatλ λ =µ hasdegree4or2.Itiseasytoseethateitherwaythis
i j k
contradicts Lemma 2.3 since g.c.d(5,4)=g.c.d(5,2)=1. In the second one, by the
sameargument,weknowthat[X ,X ]=0fori,j ∈{1,2,3},andifµ =λ λ with
i j k i j
i ≤ 3 and j = 4 or 5, then g.c.d.(degλ ,degµ ) = 1, contradicting again Lemma
i k
2.3. Therefore, this case is not possible since [X ,X ]=0 for all i,j.
i j
Case (4,5) By using Lemma 2.3 and the previous techniques, we can see that
in this case also we would have an abelian factor. We also note that this case is
also considered in [6].
Case (4,2,3)As before, we start by noting that the possible splitting of A are
1
[4] or [2;2]. Let {λ ,λ ,λ ,λ ,µ ,µ ,ν ,ν ,ν } denote the eigenvalues of A with
1 2 3 4 1 2 1 2 3
correspondingeigenvectors{X ,X ,X ,X ,Y ,Y ,Z ,Z ,Z }. We note that since
1 2 3 4 1 2 1 2 3
n has no abelian factor, each ν =λ µ for some 1≤i,j ≤3 should have degree 3.
k i j
We then have a contradiction to Lemma 2.3, since if we denote by d the degree of
λ then either d=4, or d=2, d and 3 are coprime numbers in both cases.
i
6 MEERAMAINKAR, CYNTHIAE.WILL
Case (3,3,3) Let n be a nilpotent Lie algebra of type (3,3,3) that admits an
Anosov automorphism A. Let {λ ,λ ,λ ,µ ,µ ,µ ,ν ,ν ,ν } denote the eigenval-
1 2 3 1 2 3 1 2 3
ues of A with corresponding eigenvectors {X ,X ,X ,Y ,Y ,Y ,Z ,Z ,Z } such
1 2 3 1 2 3 1 2 3
that λ′s, ν′s and µ′s are eigenvalues of A , A and A respectively. It is easy to
i j k 1 2 3
see that we can assume that
[X ,X ]=Y , [X ,X ]=Y , [X ,X ]=Y ,
1 2 1 2 3 2 1 3 3
and since λ λ λ =±1 we also have that
1 2 3
[X ,Y ]=0, [X ,Y ]=0, [X ,Y ]=0.
3 1 1 2 2 3
Let n(a,b,c) denotes the Lie algebra given by
[X ,X ]=Y , [X ,X ]=Y , [X ,X ]=Y ,
1 2 1 2 3 2 1 3 3
(2)
[X ,Y ]=aZ , [X ,Y ]=bZ , [X ,Y ]=cZ ,
1 1 1 2 2 2 3 3 3
for a,b,c ∈ C. Moreover, since we are assuming that n has no abelian factor, we
have that abc 6= 0 and in this case, by changing the basis in the center it is easy
to see that n(a,b,c) is isomorphic to n(1,1,1). In the following we will show that
n(1,1,1) is the only Anosov Lie algebra with no abelian factor of type (3,3,3). To
do this, we will first show that n is isomorphic to n(1,1,1), and then we will give
anhyperbolic automorphismAanda Z-basisofn(1,1,1)preservedby Asuchthat
the matrix of A in that basis has integer coefficients.
To see that n is isomorphic to n(1,1,1), let’s suppose on the contrary that
(3) [X ,Y ]=Z , and [X ,Y]=Z
i j k i l r
for some i that we can assume to be 1. It is clear from (3) that j,l 6=2 so we may
assume that j = 1 and l = 3. Hence, if k = r, (3) implies that λ2λ = λ2λ and
1 2 1 3
thereforewe getto the contradictionλ =λ . Onthe other hand, if k 6=r, with no
2 3
loss of generality we can assume that
[X ,Y ]=aZ , [X ,Y ]=bZ .
1 1 1 1 3 2
Now,aswe havesaidbefore,onehas thatZ =[X ,Y ]for somei,j. Itisnothard
3 i j
to see that the possibilities for (i,j) are (2,1), (2,2), (3,2), and (3,3) but all of
themleadus to the same kindofcontradictionbycheckingthat the productofthe
eigenvalues in the center equals ν2, ν , ν , ν2 respectively. Indeed, if for example
1 1 2 2
we consider case (2,2), we obtain
[X ,Y ]=aZ , [X ,Y ]=bZ , [X ,Y ]=cZ ,
1 1 1 1 3 2 2 2 3
and therefore 1 = λ2λ .λ2λ .λ2λ = λ2λ = ν , contradicting the fact that A is
1 2 1 3 2 3 1 2 1 3
hyperbolic.
Concerningthe rationalformwhichisAnosov,letusbeginbynotingthatwedo
not have any restriction on the eigenvalues of A but the degree, so we can choose
now very specials ones. Consider the polynomial in Z[X] given by
(4) f(X)=X3−3X+1.
Then
(5) λ =ξ+ξ8, λ =ξ2+ξ7, λ =ξ4+ξ5,
1 2 3
ANOSOV AUTOMORPHISMS ON NILMANIFOLDS 7
are the roots of f(X), where ξ =e2iπ/9 a ninth root of unity. In this case we have
that the extension Q(λ ) is a cyclic extension of degree 3 over Q (see [1, p. 543]),
1
and moreover straightforwardcalculation shows that
(6) λ =λ2−2 λ =λ2−2, λ =λ2−2.
1 3 2 1 3 2
We consider now the automorphism A as above, corresponding to these λ ’s, that
i
is, defined by A(X )=λ X , A(Y )=µ Y and A(Z )=ν Z for i=1,2,3. Recall
i i i i i i i i i
that, from (2),
µ =λ λ , µ =λ λ , µ =λ λ ,
1 1 2 2 2 3 3 1 3
ν =λ µ , ν =λ µ , ν =λ µ .
1 1 1 2 2 2 3 3 3
Note that, due to our choice of λ , deg(µ )=3=deg(ν ) for all 1≤i≤3.
i i i
Concerning the new basis, for i=1,2,3 let us denote by
3
X = λi−1X ,
i j j
j=1
Y =λX1−i(λ −λ )Y +λ1−i(λ −λ )Y +λ1−i(λ −λ )Y ,
i 3 2 1 1 1 3 2 2 2 3 1 3
Z =(λ )i−1(λ −λ )Z +(λ )i−1(λ −λ )Z +(λ )i−1(λ −λ )Z .
i 1 2 1 1 2 3 2 2 3 3 1 3
Let β = {X , 1 ≤ i ≤ 3}, β = {Y , 1 ≤ i ≤ 3} and β = {Z , 1 ≤ i ≤ 3}.
1 i 2 i 3 i
It is easy to see that β′s are linearly independent sets. Moreover, we can see that
i
A(X )=X , A(Y )=Y for i=1,2,A(X )=3X −X and A(Y )=3Y −Y
i i+1 i i+1 3 2 1 3 3 1
by using that f(λ ) = 0 and therefore λ−1 = 3−λ2. Also, by using (6) one can
i i i
to see that A(Z ) is an integer linear combination of the Z′s. for all i. In fact, for
i i
example
A(Z )=λ2λ (λ −λ )Z +λ2λ (λ −λ )Z +λ2λ (λ −λ )Z
1 1 2 2 1 1 2 3 3 2 2 3 1 3 1 3
=λ2(λ2−2)(λ −λ )Z +λ2(λ2−2)(λ −λ )Z +λ2(λ2−2)(λ −λ )Z
1 1 2 1 1 2 2 3 2 2 3 3 3 1 3
=(λ2−λ )(λ −λ )Z +(λ2−λ )(λ −λ )Z +(λ2−λ )(λ −λ )Z
1 1 2 1 1 2 2 3 2 2 3 3 3 1 3
=Z −Z .
3 2
Therefore, β = {X ,Y ,Z : i,j,k = 1,2,3} is a basis of n(1,1,1) preserved by
i j k
A and moreover [A] ∈GL(9,Z).
β
To finish, it remains to show that β is also a Z-basis. We first note that since
λ λ λ =1, λ +λ λ =0 and λ−2 =3λ−1−λ , we have that
1 2 3 1 2 3 i i i
(7) [X ,X ]=Y , [X ,X ]=Y −3Y .
1 2 1 1 3 3 2
Moreover,by using (6) as above, we have
[X ,Y ]=Z , [X ,Y ]=Z −Z , [X ,Y ]=Z −2Z +Z .
1 1 1 1 2 2 1 1 3 3 2 1
Note thatsinceAinanautomorphismofnandbyourconstructionofthe basis,
these are the only brackets one need to check. Hence β is a Z-basis of n(1,1,1)
preserved by the hyperbolic automorphism A, such that [A] ∈ GL(9,Z), and
β
therefore n(1,1,1) is an Anosov Lie algebra, as we wanted to show.
Remark 3.1. We will use the above proof as a model proof to show that a given
Lie algebra is Anosov in some of the following cases where we will just exhibit the
automorphism and the Z-basis.
8 MEERAMAINKAR, CYNTHIAE.WILL
Case (6,3)We firstnote thatcomplex nilpotent Lie algebrasoftype (6,3)have
been studied in [8] sec 4. where they obtained a classificationup to SL(6)×SL(3)
action. Since we are going to use some of their results we will introduce now their
notation.
Let {X :1≤i≤9} be the canonical basis of R9 and for any t, s, r ∈C denote
i
by tU +sU +r U the 2-step nilpotent Lie algebra given by
1 2 3
[X ,X ]=tX [X ,X ]=tX [X ,X ]=tX
1 2 7 3 4 8 5 6 9
(8) [X ,X ]=sX [X ,X ]=sX [X ,X ]=sX
5 4 7 1 6 8 3 2 9
[X ,X ]=r X [X ,X ]=r X [X ,X ]=rX .
3 6 7 5 2 8 1 4 9
where U = (R9,[, ] ) is the 2-step nilpotent Lie algebra where [, ] is given by
i i i
the ith row of (8). It is easy to see that these are 2-step nilpotent Lie algebras
of type (6,3). The classification given in [8] splits the (6,3) nilpotent Lie algebras
in 7 families. The first 6, consists of Lie algebras with a semisimple part (closed
SL(6) × SL(3) orbit) given in terms of t U + s U + r U , for some values of
1 2 3
r,s,t plus a nilpotent part (non-closed SL(6)×SL(3) orbit) that are listed in a
table. The last family consists of 76 nilpotent elements. Note that since we are
interested in (6,3) Lie algebras up to isomorphism, we need orbits for the action
of GL(6)×GL(3), and therefore it is not hard to see that to this classification we
only have to add the action of scalar multiples of the identity. Then, for example
we have that sU +rU ≃ s′U +U for any r 6= 0 and the semisimple part of
2 3 2 3
Family 4 (and 5) are all isomorphic.
Coming back to the Anosov problem, let n be an Anosov Lie algebra of type
(6,3) with no abelian factor, and let A,A ,A , be as in Proposition 2.1. Denote
1 2
by λ ,...,λ the eigenvalues of A , by µ ,µ ,µ the eigenvalues of A and the
1 6 1 1 2 3 2
corresponding basis of eigenvectors by β = {X ,...,X ,Y ,Y ,Y }. Let f and g
1 6 1 2 3
be the characteristic polynomials of A and A respectively. By using Lemma 2.3
1 2
it can be seen that that g is irreducible overZ, and either f is irreducible or f has
two irreducible factors of degree 3 over Z. Note that this corresponds to the fact
that the spitting of A is [6] or [3;3]. We are going to study now each one of these
1
cases separately.
Case i: (f is irreducible over Z). We note that, in particular, this implies that
λ 6=λ for all i6=j. We begin by showing that we can assume that
i j
(9) [X ,X ]=Y , [X ,X ]=Y , [X ,X ]=Y .
1 2 1 3 4 2 5 6 3
In fact, if this is not the case we can reorder the basis so that
(10) [X ,X ]=Y , [X ,X ]=Y ,
1 2 1 1 3 2
and in this situation we have to consider three possibilities for Y ,
3
(11) (I)[X ,X ]=Y , (II)[X ,X ]=Y , (III)[X ,X ]=Y .
1 4 3 2 j 3 4 5 3
Note that each one of this situations represents a few others that are totally
equivalent to the considered one.
Case(III)isthesimplestonebecausefrom(10)itfollowsatoncethatλ2λ λ λ λ =
1 2 3 4 5
1 and therefore we obtain the contradiction λ =λ .
1 6
ANOSOV AUTOMORPHISMS ON NILMANIFOLDS 9
In case (I) we have that
(12) λ3λ λ λ =1 or, equivalenltly λ2 =λ λ .
1 2 3 4 1 5 6
Let us consider then the possibilities for [X ,X ] = Y , [X ,X ] = Y . With no
5 j k 6 l r
loss of generality we may assume that [X ,X ] = Y , and hence, λ λ = λ λ or
5 j 1 5 j 1 2
equivalently by (12), λ λ = λ λ . Therefore, by all this, it is easy to see that
1 j 6 2
j 6=1,2,5,6 and hence j =3 or 4. Since both cases are entirely analogous, we will
just consider j =4. That is,
[X ,X ]=Y , [X ,X ]=Y [X ,X ]=Y
1 2 1 1 3 2 1 4 3
(13)
[X ,X ]=Y .
5 4 1
From this, it is easy to see that r 6= 1, and if r = 3, then l = 2 since product of
all λ′s is 1 and then by rearrangingthe basis we would have (9) as desired. To see
i
thatr=2isnotpossibleeither,supposethat[X ,X ]=Y . Ifthisisthe case,itis
6 l 2
easy to see that l6=1,3,6,5,2,and so, we may assume that [X ,X ]=Y . Hence,
6 4 2
1=µ µ µ =λ λ3λ λ .
1 2 3 1 4 5 6
By using (12), (λ λ )3 = 1 which implies that µ3 = 1, contradicting the fact that
1 4 3
A is hyperbolic.
2
Finallyconcerningcase(II),notefirstthatwehavetodistinguishbetweenj =3
and j 6=3. If j 6=3, let us say j =4 (or equivalently j =5 or 6), we have
(14) [X ,X ]=Y , [X ,X ]=Y [X ,X ]=Y .
1 2 1 1 3 2 2 4 3
Therefore we obtain, equivalently,
(15) λ2λ2λ λ =1 or λ λ =λ λ .
1 2 3 4 5 6 1 2
It is clear from this that if [X ,X ]=aY or [X ,X ]=bY then l =6 and k =5
5 l 1 6 k 1
and we would have (9) as desired. In a very similar way we did in the previous
case, it is easy to see that this situation leads us to either case I or (9). Therefore,
we can assume (9), that is
[X ,X ]=Y , [X ,X ]=Y , [X ,X ]=Y .
1 2 1 3 4 2 5 6 3
If there are no more non trivial Lie bracketsbut these, then n≃h ⊕h ⊕h ≃U
3 3 3 1
which is known to be Anosov (see [13]). If there are more non trivial Lie brackets,
withoutanylossofgenerality,wecanassumethat[X ,X ]=cY .Moreover,using
3 2 3
that there is no abelian factor, that the λ ’s are all distinct and that λ ...λ =1,
i 1 6
one can see that n is isomorphic to n′(a,b,c), given by
C
[X ,X ]=Y , [X ,X ]=Y [X ,X ]=Y ,
1 2 1 3 4 2 5 6 3
(16)
[X ,X ]=aY , [X ,X ]=bY [X ,X ]=cY ,
5 4 1 1 6 2 3 2 3
for some a,b,c∈C. By the same reasonsgiven above, it is easy to see that we can
not add more nontrivial Lie brackets.
To study how many non isomorphic Lie algebras we obtain from (16), we start
by noting that if abc = 0, from [8] we get that we have only two non isomorphic
Lie algebras:
′ ′ ′ ′
n(0,0,c)≃n(0,0,1)≃U +n , and n(a,b,0)≃n(1,1,0)≃U +n ,
1 14 1 11
both in Family 6, where n denote the nilpotent part j given in Table 7.
j
10 MEERAMAINKAR, CYNTHIAE.WILL
On the other hand, if abc6=0, by changing the basis to
1 1 1
′
β ={X ,aX , X , X ,cX ,X ,aY , Y ,cY },
1 2 3 4 5 6 1 2 3
a c ac
we obtain
[X ,X ]=Y , [X ,X ]=Y [X ,X ]=Y ,
1 2 1 3 4 2 5 6 3
(17)
[X ,X ]=Y , [X ,X ]=abcY [X ,X ]=Y ,
5 4 1 1 6 2 3 2 3
showing that n′(a,b,c)≃n′(1,abc,1), and then (if abc6=0)
′ ′
n(a,b,c)≃n(s,s,s)≃U +sU ≃sU +U ,
1 2 2 3
where s3 =abc. Therefore, if s3 =s′3, sU +U ≃s′U +U . In the classification
2 3 2 3
given in [8] one has the following non isomorphic Lie algebras
• sU +U , corresponding to s3 6=0,±1 (Family 2)
2 3
• U +U , corresponding to s3 =1 (Family 4)
2 3
• −U +U , corresponding to s3 =−1 (Family 5)
2 3
• U ≃U , corresponding to s=0 (Family 6).
3 1
We arenot able to decide if the algebrasin Family 2 areAnosovor not, we note
that since it gives an uncountable family of nonisomorphic Lie algebras, not all of
them can be Anosov.
For s=0, as we have already said n≃h ⊕h ⊕h is an Anosov Lie algebra.
3 3 3
3
If s =1, (assumed to be 1) we will show that U +U (see (8) with t =0,r =
2 3
s=1), is an Anosov Lie algebra, by using the ideas of [17].
Let {λ ,λ ,λ } be the roots of the polynomial in (4) used in the (3,3,3)-case.
1 2 3
As we havenoted there, λ are degree 3 algebraicunits. Let µ be an algebraicunit
i
of degree 2, such that
|µ±1λ | =6 1, for j = 1,2,3, and let A be the automorphism of n such that
j
[A] = A1 , where
β A2
(cid:2) (cid:3) µλ3
µ−1λ3
A1 = µλ2 µ−1λ2 µλ1 , A2 =(cid:20)λ1λ2 λ1λ3 λ2λ3(cid:21).
 µ−1λ1
 ′ 
Due to our choice of µ and λ s, it is clear that A is hyperbolic, and moreover,it
i
is easy to see that A is also an (hyperbolic) automorphism for U = sU +U , for
2 3
any s∈C. Straightforwardcalculation shows that if
X =λl (µkX +µ−kX )+λl (µkX +µ−kX )+λl (µkX +µ−kX )
(k,l) 3 1 2 2 3 4 1 5 6
Y =(µ−1−µ)(λrY +λrY +λrY ),
r 3 1 2 2 1 3
then
′
β ={X ,Y |k =0,1, l =0,1,2 r =−1,0,1},
(k,l) r
is a Z-basis of n=U2+U3 preserved by A such that [A]β′ ∈GL(9,Z) (see [17]
Example 2.5 and Remark 2.6).
For the case when s3 =−1 let us change the basis of −U +U to
2 3
β ={X +X ,X +X ,X +X ,X −X ,X −X ,X −X ,2Y ,2Y ,−2Y }
1 2 3 4 5 6 1 2 3 4 5 6 3 1 2