Pseudo-Riemannian Symmetries on Heisenberg group $\mathbb{H}_{3}$

0.18 MB·English

by  Michel Goze

Pseudo-Riemannian Symmetries on Heisenberg
H
group
2 3
1
0
∗
2 Michel Goze , Paola Piu

n
January 4, 2012
a
J

2

] Abstract
G
D The notion of Γ-symmetric space is a natural generalization of the
classical notion of symmetric space based on Z2-grading of Lie algebras.
.
h In our case, we consider homogeneous spaces G/H such that the Lie al-
t gebra g of G admits a Γ-grading where Γ is a finite abelian group. In
a
m this work we study Riemannian metrics and Lorentzian metrics on the
Heisenberg group H3 adapted to thesymmetries of a Γ-symmetric struc-
[ tureonH3. Weprovethattheclassification ofZ22-symmetricRiemannian
1 and Lorentzian metrics on H3 corresponds to the classification of left in-
v variant Riemannian and Lorentzian metrics, up to isometries. This gives
7 examples of non-symmetricLorentzian homogeneous spaces.
4
4 Mathematics Subject Classification 2000: 22F30, 53C30, 53C35,17B70
0
Key words: Γ-symmetric spaces, Heisenberg group, Graded Lie algebras.
.
1
0
2 1 Γ-symmetric spaces
1
:
v Let Γ be a finite abelian group. A Γ-symmetric space is an homogeneous space
i G/H such that there exists an injective homomorphism
X
r
ρ:Γ Aut(G)
a
→
where Aut(G) is the groupof automorphisms of the Lie groupG, the subgroup
H satisfying GΓ H GΓ where GΓ = x G/ρ(γ)(x)=x, γ Γ and GΓ
e ⊂ ⊂ { ∈ ∀ ∈ } e
is the connected identity component of GΓ of G.
The notion of Γ-symmetric space is a generalization of the classical notion of
symmetric space by considering a general finite abelian group of symmetries Γ
instead of Z . The case Γ = Z , the cyclic group of order k, was considered
2 k
by A.J. Ledger, M. Obata [13], A. Gray, J. A. Wolf, [8] and O. Kowalski [11]
in terms of k-symmetric spaces. The general notion of Γ-symmetric spaces was
∗Thefirstauthor wassupportedby: Visitingprofessorprogram,RegioneAutonomadella
Sardegna-Italy. ThesecondauthorwassupportedbyGNSAGA(Italy)
1
introduced by R. Lutz [12] and was algebraically reconsidered by Y. Bahturin
and M. Goze [1]. In this last work the authors proved, in particular, that a Γ-
symmetricspaceM =G/H isreductiveandthe Liealgebrag ofGis Γ-graded,
that is,
g= g
γ
γ⊕∈Γ
with
[gγ,gγ′] gγγ′ γ,γ′ Γ.
⊂ ∀ ∈
Examples.
1. IfΓ=Z andgacomplexorrealLiealgebra,aΓ-gradingofgcorresponds
2
to the classical symmetric decomposition of g.
2. IfgisasimplecomplexLiealgebraandΓ=Z ,k 3,wehavethenotion
k
≥
of generalized symmetric spaces and the classification of Γ-gradings are
described by V. Kac in [10].
3. Let g = g be a Lie algebra Γ-graded. For any commutative as-
γ∈Γ γ
⊕
sociative algebra , the current algebra g (see [18]) also admits a
A A⊗
Γ-grading.
4. In [1], the Z Z grading on classical simple complex Lie algebras are
2 2
×
classified.
One proves also in [1] that the structure of Γ-symmetric space on G/H is,
when G is connected, completely determinate by the Γ-grading of g. Thus, if
G is connected, the classificationof the Γ-symmetric spaces is equivalent to the
classification of the Γ-graded Lie algebras. Many results of this last problem
concernmoreparticularlythe simple Lie algebras. For solvableornilpotent Lie
algebras,itisanopenproblem. Afirstapproachistostudyinducedgradingon
Borel or parabolic subalgebras of simple Lie algebras. In this work we describe
Γ-grading of the Heisenberg algebra h . Two reasons for this study
3
Heisenberg algebras are nilradical of some Borel subalgebras.
•
The Riemannian and Lorentziangeometries onthe 3-dimensionalHeisen-
•
berg group have been studied recently by many authors. Thus it is inter-
estingtostudytheRiemannianandLorentziansymmetrieswiththenatu-
ralsymmetriesassociatedwithaΓ-symmetricstructureontheHeisenberg
group. In this paper we prove that these geometries are entirely deter-
minate by Riemannian and Lorentzian structures adapted to (Z2 Z2)-
×
symmetric structures.
Recallthatthe Heisenbergalgebrah is the realLie algebrawhoseelements
3
are matrices
0 x z
0 0 y with x,y,z R
  ∈
0 0 0
 
2
The elements of h , X , X , X , corresponding to (x,y,z) = (1,0,0),(0,1,0)
3 1 2 3
and (0,0,1) form a basis of h and the Lie brackets are given in this basis by
3
[X ,X ]=X
1 2 3
([X1,X3]=[X2,X3]=0.
TheHeisenberggroupistherealLiegroupofdimension3consistingofmatrices
1 a c
0 1 b a,b,c R
  ∈
0 0 1
 
and its Lie algebra is h .
3
2 Finite abelian subgroups of Aut(h )
3
LetusdenotebyAut(h )thegroupofautomorphismsoftheHeisenbergalgebra
3
h . Every τ Aut(h ) admits, with regards to the basis X ,X ,X , the
3 ∈ 3 { 1 2 3}
following matricial representation:
α α 0
1 2
α α 0
3 4
 
α α ∆
5 6
 
with ∆=α α α α =0.
1 4 2 3
− 6
LetΓ be afinite abeliansubgroupofAut(h ). Itadmits acyclic decomposi-
3
tion. If Γ contains a cyclical component isomorphic to Z , then there exists an
k
automorphism τ satisfying τk = Id. The aim of this section is to determinate
the cyclic decomposition of any finite abelian subgroup Γ.
2.1 Subgroups of Aut(h ) isomorphic to Z
3 2
Let τ Aut(h ) satisfying τ2 =Id. If
∈ 3
α α 0
1 2
α α 0
3 4
 
α α ∆
5 6
 
isitsmatricialrepresentation,thentheinvolutioncanbewritteninmatrixform
α2+α α α α +α α 0 1 0 0
1 2 3 1 2 2 4
α α +α α α α +α2 0 = 0 1 0
 1 3 3 4 2 3 4   
α α +α α +∆α α α +α α +∆α ∆2 0 0 1
1 5 3 6 5 2 5 4 6 6
   
The resolution of the system can be done using formal calculation software.
Here we use Mathematica.
3
Proposition 2.1. Any involutive automorphism τ of Aut(h ) is equal to one
3
of the following automorphisms
1 0 0
1 0 0
−
α 1 0
Id, τ (α ,α )= 3 ,τ (α ,α )= α 1 0 ,
1 3 6 α3α6  2 3 5  3 − 
α 1 α 0 1
 2 6 −  5 −
   
α α 0
1 2
1 α2 1 0 0
τ3(α1,α2 6=0,α6)= −α2 1 −α1 0 ,τ4(α5,α6)=−0 −1 0.
(1+α )α α α 1
 1 6 α 1 5 6
 6 
 α2 −   
 
Corollary 2.2. Any subgroup of Aut(h ) isomorphic to Z is equal to one of
3 2
the following:
1. Γ (α ,α )= Id,τ (α ,α ) ,
1 3 6 1 3 6
{ }
2. Γ (α ,α )= Id,τ (α ,α ) ,
2 3 5 2 3 5
{ }
3. Γ (α ,α ,α )= Id,τ (α ,α ,α ), α =0 ,
3 1 2 6 3 1 2 6 2
{ 6 }
4. Γ (α ,α )= Id,τ (α ,α ) .
4 5 6 4 5 6
{ }
2.2 Subgroups of Aut(h ) isomorphic to Z
3 3
Let τ be an automorphism satisfying τ3 = Id. This identity is equivalent to
τ2 =τ−1. If we have
α α 0
1 2
τ = α α 0 , ∆=α α α α ,
3 4 1 4 2 3
  −
α α ∆
5 6
 
then
α α 0
4 2
τ−1 = 1 α3 −α1 0
∆α3α6− α4α5 α2α5 α1α6 
− − 1
 ∆ ∆ 
 
The condition τ2 = τ−1 implies ∆3 = 1 and the only real solution is ∆ = 1.
Thus τ2 =τ−1 is equivalent to
α α α α =1,
1 4 2 3
−
α2+α α =α ,
 1 2 3 4
ααα2423((+11α++2ααα113++=ααα441)),==00,, (1)
α (1+α +α )=0,
α56(1+α11+α44)=0.
4
If α +α = 1, then α =α =α =α =0 and α =α =1. In this case
1 4 2 3 5 6 1 4
6 −
τ =Id.
Let us assume α +α = 1. Then (1) is reduced to
1 4
−
α2+α +α α +1=0
1 1 2 3
If α α > 3, we have no solutions. Assume that α α 3. Then
2 3 −4 2 3 ≤−4
1 √ 3 4α α
2 3
α = − ± − − .
1
2
So we obtain
1 √ 3 4α α
2 3
− − − − α 0
2
2
τ = 1+√ 3 4α α ,
5 2 3
α − − − 0
3
 2 
 α α 1
 5 6 
 
and
1+√ 3 4α α
2 3
− − − α 0
2
2
τ′ = 1 √ 3 4α α .
5 α − − − − 2 3 0
3
 2 
 α α 1
 5 6 
 
Since
τ2(α ,α ,α ,α )=τ′( α , α ,α′,α′)
5 2 3 5 6 5 − 2 − 3 5 6
where
α √ 3 4α α α 2α α α +√ 3 4α α α 2α α
α′ = 5− − − 2 3 5− 3 6, α′ = 6 − − 2 3 6− 2 5
5 2 6 2
and
τ′2(α ,α ,α ,α )=τ ( α , α ,α′′,α′′)
5 2 3 5 6 5 − 2 − 3 5 6
where
α +√ 3 4α α α 2α α α √ 3 4α α α 2α α
α′′ = 5 − − 2 3 5− 3 6, α′′ = 6− − − 2 3 6− 2 5,
5 2 6 2
we deduce
Proposition 2.3. Any abelian subgroup of Aut(h ) isomorphic to Z is equal
3 3
to
Γ (α ,α ,α ,α )= Id,τ (α ,α ,α ,α ),τ′( α , α ,α′,α′), 4α α 3 .
5 2 3 5 6 { 5 2 3 5 6 5 − 2 − 3 5 6 2 3 ≤− }
5
2.3 Subgroups of Aut(h3) isomorphic to Zk, k > 3
If τ Aut(h ) satisfies τk = Id, k > 3, its minimal polynomial has 3 simple
∈ 3
roots and it is of degree 3. More precisely, it is written
m (x)=(x 1)(x µ )(x µ )
τ k k
− − −
where µ is a root of order k of 1. As τ has to generate a cyclic subgroup of
k
Aut(h ) isomorphic to Z , the root µ is a primitive root of 1. There exists m,
3 k k
2miπ
a prime number with k such that µ =exp( ). If
k
k
α α 0
1 2
τ = α α 0
3 4
 
α α ∆
5 6
 
2mπ
isthematricialrepresentationofτ,then∆=1andα +α =2cos . Thus
1 4
k
2mπ 2mπ
α =cos cos2 1 α α ,
1 2 3
k − k − −
 r
 α4 =cos2mπ + cos2 2mπ 1 α2α3,
k k − −
r
or 
2mπ 2mπ
α =cos + cos2 1 α α ,
1 2 3
k k − −
 r
 α4 =cos2mπ cos2 2mπ 1 α2α3.
k − k − −
r
If τ′ and τ′′ denote the automorphisms corresponding to these solutions, we
have, for a good choice of the parameters α , τ′ τ′′ = Id and τ′′ = (τ′)k−1.
i
◦
Thus these automorphisms generate the same subgroup of Aut(h ). Moreover,
3
with same considerations,we can choosem=1. Thus we havedeterminate the
automorphism τ (α ,α ,α ,α ) whose matrix is
6 2 3 5 6
2π 2π
cos + cos2 1 α α α 0
2 3 2
k k − −
 r 
2π 2π
α cos cos2 1 α α 0
 3 k − k − − 2 3 
 r 
 α α 1
 5 6 
 
Proposition 2.4. Any abelian subgroup of Aut(h ) isomorphic to Z is equal
3 k
to
2π
Γ (α ,α ,α ,α )= Id,τ (α ,α ,α ,α ), ,τk−1, α α 1+cos2 .
6,k 2 3 5 6 6 2 3 5 6 ··· 6 2 3 ≤− k
(cid:26) (cid:27)
6
2.4 General case
Now suppose that the cyclic decomposition of a finite abelian subgroup Γ of
Aut(h ) is isomorphic to Zk2 Zk3 Zkp with k 0.
3 2 × 3 ×···× p i ≥
Lemma 2.5. Let Γ be an abelian finite subgroup of Aut(h ) with a cyclic de-
3
composition isomorphic to
Zk2 Zk3 Zkp.
2 × 3 ×···× p
Then
If there is i 3 such that k =0, then k 1.
i 2
• ≥ 6 ≤
If k 2, then Γ is isomorphic to Zk2.
• 2 ≥ 2
Proof. Assume that there is i 3 such that k 1. If k 1, there exist
i 2
≥ ≥ ≥
two automorphisms τ and τ′ satisfying τ′2 = τ2 = Id and τ′ τ = τ τ′.
◦ ◦
Thus τ′ andτ canbe reduced simultaneouslyin the diagonalformand admita
common basis of eigenvectors. As for any σ Aut(h ) we have σ(X )= ∆X ,
∈ 3 3 3
X is an eigenvector for τ′ and τ associated to the eigenvalue 1 for τ′ and
3
1 for τ. As the two other eigenvalues of τ′ are complex conjugate numbers,
±
the corresponding eigenvectors are complex conjugate. This implies that the
eigenvalues of τ distinguished of ∆ = 1 are equal and from Proposition 2.1,
±
τ =τ (α ,α ). But
4 5 6
τ (α ,α ) τ (α′,α′)=τ (α′,α′) τ (α ,α ) α =α′, α =α′.
4 5 6 ◦ 4 5 6 4 5 6 ◦ 4 5 6 ⇔ 5 5 6 6
Thus, we have to determine, in a first step, the subgroups Γ of Aut(h )
3
isomorphic a (Z )k with k 2.
2
≥
Any involutive automorphism τ commuting with τ (α ,α ) and distinct
1 3 6
•
from it is equal to one of the following automorphisms
τ ( α ,α ) τ (α , α )
2 3 5 4 5 6
− −
Indeed, if we set [τ,τ′]=τ τ′ τ′ τ then
◦ − ◦
[τ (α ,α ),τ (α′,α′)] = 0 if and only if α′ =α andα′ =α
1 3 6 1 3 6 3 3 6 6
[τ (α ,α ),τ (α′,α′)] = 0 if and only if α′ = α
1 3 6 2 3 5 3 − 3
[τ (α ,α ),τ (α ,α ,α′)] = 0 whatever they are α ,α ,α′
1 3 6 3 1 2 3 6 1 2 3
[τ (α ,α ),τ (α ,α′)] = 0 if and only if α′ = α
1 3 6 4 5 6 6 − 6
These results follow directly from the matrix calculation. In addition we
have
α α
3 6
τ (α ,α ) τ ( α ,α )=τ α , α
1 3 6 2 3 5 4 5 6
◦ − − 2 − −
(cid:16) (cid:17)
7
and
α α
3 6
τ ( α ,α ),τ α , α =0.
2 3 5 4 5 6
− − 2 − −
h (cid:16) (cid:17)i
Thus
α α
3 6
Γ (α ,α ,α )= Id,τ (α ,α ),τ ( α ,α ),τ α , α
7 3 5 6 1 3 6 2 3 5 4 5 6
− − 2 − −
n (cid:16) (cid:17)o
is a subgroup of Aut(h ) isomorphic to Z2. Moreover it is the only sub-
3 2
group of Aut(h ) of type (Z )k, k 2, containing an automorphism of
3 2
≥
type τ (α ,α ).
1 3 6
Let us suppose that τ (α ,α ) Γ and that τ (α ,α ) / Γ. We have
2 3 5 1 3 6
• ∈ ∈
[τ (α ,α ),τ (α′,α′)] = 0 if and only if α′ =α andα′ =α
2 3 5 2 3 5 3 3 5 5
[τ (α ,α ),τ (α ,α ,α )] = 0 because by assumption α =0
2 3 5 3 1 2 6 2
6 6
α α
[τ (α ,α ),τ (α′,α )] = 0 if and only if α′ = α 3 6
2 3 5 4 5 6 5 − 5− 2
But
α α
3 6
τ (α ,α ) τ ( α ,α )=τ (α ,α ).
2 3 5 4 5 6 1 3 6
◦ − − 2
Thus everyabeliansubgroupΓ containing τ (α α ) are either isomorphic
2 3 5
to Z or is equal to Γ
2 7
Assume that τ (α ,α ,α ) Γ. We have
3 1 3 6
• ∈
[τ (α ,α ,α ),τ (α′,α′,α′)]=0 if and only if α′ = α andα′ = α .
3 1 2 6 3 1 2 6 1 − 1 2 − 2
Thus
[τ (α ,α ,α ),τ ( α , α ,α′)] = 0,
3 1 2 6 3 − 1 − 2 6
and
[τ (α ,α ,α ),τ (α ,α′)] = 0 if and only if α α +2α =(α 1)α′.
3 1 2 6 4 5 6 2 5 6 1− 6
Moreover
α′(1 α ) α (1+α )
τ (α ,α ,α ) τ ( α , α ,α′)=τ 6 − 1 − 6 1 , α α′
3 1 2 6 ◦ 3 − 1 − 2 6 4 α − 6− 6
(cid:18) 2 (cid:19)
because
α′(1 α ) α (1+α )
α 6 − 1 − 6 1 +2α +(1 α )( α α′)=0.
2 α 6 − 1 − 6− 6
(cid:18) 2 (cid:19)
The subgroup of Γ (α ,α ,α ,α′) of Aut(h ) equal to
8 1 2 6 6 3
α′(1 α ) α (1+α )
Id,τ (α ,α ,α ),τ ( α , α ,α′),τ 6 − 1 − 6 1 , α α′
3 1 2 6 3 − 1 − 2 6 4 α − 6− 6
(cid:26) (cid:18) 2 (cid:19)(cid:27)
is isomorphic to Z2.
2
8
We suppose that τ (α ,α ) Γ. If Γ is not isomorphic to Z , then Γ is
4 5 6 2
• ∈
one of the groups Γ ,Γ .
7 8
Theorem 2.1. Any finite abelian subgroup Γ of Aut(h ) isomorphic to (Z )k
3 2
is one of the following
1. k =1,Γ=Γ (α ,α ), Γ (α ,α ), Γ (α ,α ,α ), α =0, Γ (α ,α ),
1 3 6 2 3 5 3 1 2 6 2 4 5 6
6
2. k =2, Γ=Γ (α ,α ,α ), Γ (α ,α ,α ,α′).
7 3 5 6 8 1 2 6 6
Let us assume now that Γ is isomorphic to Zk3 with k 2. If τ Γ , its
3 3 ≥ ∈ 5
matricial representationis
1 √ 3 4α α
2 3
− − − − α 0
2
2
 1+√ 3 4α α .
2 3
α − − − 0
3
 2 
 α α 1
 5 6 
 
1 √ 3 4α α
To simplify, we put λ = − − − − 2 3. The eigenvalues of τ are 1,j,j2
2
and the corresponding eigenvectors X ,V,V with
3
λ j α α (λ j)
5 6
V =(1, − , + − )
− α −1 j α (1 j)
2 2
− −
if α =0. If τ′ is an automorphism of order 3 commuting with τ, then
2
6
τ′V =jV or j2V.
But the two first components of τ′(V) are
β λ′(λ j)
λ′ 2(λ j),β −
3
− α − − α
2 2
where β and λ′ are the corresponding coefficients of the matrix of τ′. This
i
implies
α λ′ β (λ j)=α j or α j2.
2 2 2 2
− −
Considering the real and complex parts of this equation, we obtain
α λ′ β λ=0,
2 2
−
β j =α j or α j2.
2 2 2
(cid:26)
Asα =0,weobtainα =β andλ=λ′. Letuscomparethesecondcomponent
2 2 2
6
of τ′(V).We obtain
β α λ′(λ j)= (λ j)j or (λ j)j2.
3 2
− − − − − −
Asλ=λ′,wehaveinthefirstcase2λj =j2andinthesecondcase2λj =j3 =1.
Inanycase,thisisimpossible. Thusα =0and,fromsection2.2,τ =Id. This
2
implies that k =1 or 0.
3
9
Theorem 2.2. Let Γ be a finite abelian subgroup of Aut(h ). Thus Γ is iso-
3
morphic to one of the following
1. Z Z ,
2 2
×
2. Zk2 Zk3 Zkp with k =0 or 1 for i=2, ,p.
2 × 3 ×···× p i ···
To prove the second part, we show identically to the case i =3 that k =1
i
as soon as k =0.
i
6
Example. The group
Γ (0,0) Γ (0,0,0,0) Γ (0,0,0,0)
4 5 6,k
× ×···×
satisfies the second property of the theorem.
Remark. We have determined the finite abelian subgroups of Aut(h ). There
3
are non-abelian finite subgroups with elements of order at most 3. Take for
example the subgroup generated by
1 0 0 1 α 0
−2
σ = 0 1 0 , σ = 3 1 0 α=0
1   2 −4α −2  6
0 0 1 0 0 1
−
   
The relations on the generators are
σ2 =Id,
1
σ3 =Id,
 2
σ1σ2σ1 =σ22.
Thus the group generated by σ and σ is isomorphic to the symmetric group
1 2
Σ of degree 3.
3
3 Γ-grading of h
3
3.1 Description of the Z and Z2-gradings
2 2
Let Γ be a finite abelian subgroup of Aut(h ). We consider a Γ-grading of h
3 3
h = h
3 3,γ
γ⊕∈Γ
such that h = 0 where e is the identity of Γ. In this case, the space Γ-
3,e
{ }
symmetric associated with this grading is isomorphic to the Heisenberg group
H andthenH canbe studiedinterms ofΓ-symmetric spaces. Inthis section,
3 3
we are particulary interested by the case Γ=Z or Γ=Z Z .
2 2 2
×
If Γ=Z then the grading of h is of the type
2 3
•
h =g g
3 0 1
⊕
10