aa r X i v : . [ m a t h . G R ] A ug Lattices of oscillator groups
Mathias FischerMarch 1, 2018
This paper is concerned with discrete, uniform subgroups (lattices) ofoscillator groups, which are certain semidirect products of the Heisenberggroup and the additive group R of real numbers.The present paper rectifies the uncertainties in [Med85] of Medina andRevoy and gives a complete classification of the lattices of the 4-dimensionaloscillator group up to isomorphism.
1. Introduction
Oscillator groups are certain semidirect products of the Heisenberg group and theadditive group R . They are interesting because they are the only simply connectedsolvable Lie groups, besides the abelian ones, which have a biinvariant Lorentzianmetric. In addition, the quotient of an oscillator group by a lattice gives an example ofa compact homogeneous Lorentzian manifold.In [Med85], Medina and Revoy already classified the lattices of the oscillator group. Butunfortunately the given maps in [Med85, p. 92] aren’t necessarily automorphisms, dif-ferent from the assumption. Hence, theorem III and the following corollary is incorrect.Our goal is to classify the lattices of the 4-dimensional oscillator group.More precisely we consider groups Osc n ( ω, B ), i. e. R × R n × R with the groupmultiplication given by( z , ξ, t )( v , η, s ) = (cid:18) z + v + ω ( ξ, e tB η ) , ξ + e tB η, t + s (cid:19) . Here B is an invertible 2 n × n -matrix and ω a symplectic form on R n such that ω ( B ξ, η ) = − ω ( ξ, B η ) for all ξ, η ∈ R n and ω ( B · , · ) is definite. Every oscillator group is isomorphicto some Osc n ( ω, B ). In addition, these groups are isomorphic to the groups G k ( λ )considered in [Med85].We compute the automorphisms of Osc n ( ω, B ) in Theorem 1. The theorem also rectifiesthe assertion in [Med85, p. 92]. Afterwards we classify the lattices of Osc ( ω, B ) in threesteps (Theorem 2-4). In Theorem 2, which also holds for lattices in Osc n ( ω, B ), we getto know that we can always assume that a lattice of a 4-dimensional oscillator group is1. Oscillator groupsgenerated by { (1 , , , (0 , e i , | i = , } and some 4th element (0 , ξ ,
1) in an oscillatorgroup
Osc ( ω r , B ) with a certain standard symplectic form ω r . Furthermore, we see inTheorem 3 that there is also, besides ω r , a unique standard matrix, denoted by B = λ B x , y .Then, the last step is to describe all lattices of the special kind we get in Theorem 3 upto automorphisms in Theorem 4. This gives restrictions for ξ . Our main tools for thesesteps are special isomorphisms and automorphisms of oscillator groups preserving thesubgroup h (1 , , , (0 , e i , | i = , i , we characterize in Lemma 3.2 and Lemma 3.6.Finally, each lattice L in Osc ( ω, B ) gives a data ( r , λ, ( x , y ) , ξ ). Thereby r is a positiveinteger, λ a certain angle, ( x , y ) an element of the "half fundamental domain" for themodular group and ξ a vector from a finite set to be extracted from the list in AppendixA.This data describes the lattices of the oscillator group in the sense that two lattices L and L of Osc ( ω, B ) have the same data if and only if there is an automorphism in Osc ( ω, B )mapping L onto L .
2. Oscillator groups
For a symplectic form ω on R n , let H n ( ω ) denote the group R × R n with the multiplication given by( z , ξ )( v , η ) = (cid:16) z + v + ω ( ξ, η ) , ξ + η (cid:17) . Since each of these groups are isomorphic we call them Heisenberg groups.Let H be a Heisenberg group, h its Lie algebra and z the center of h . Suppose that p isa one-parameter subgroup of the automorphism group of H with trivial action on thecenter of the Heisenberg group and satisfying that the map A : h / z × h / z → z (cid:27) R , A ( h + z , h + z ) : = h ((d f )(1))( h ) , h i is definite, where f : R → Aut ( h ) is the di ff erential of the automorphism p ( t ) ∈ Aut ( H )at the point 0 ∈ h , i. e. f ( t ) : = d ( p ( t )). Then the semidirect product H ⋊ p R of H and R with respect to p is called oscillator group.Later, we will see that these groups are isomorphic to the groups G k ( λ ) introduced in[Med85]. Let p be a one-parameter group of the automorphisms of H n ( ω ) . Then p satisfiesthe conditions in Definition 2.1 if and only if there is a δ ∈ R n and a B ∈ gl (2 n , R ) satisfying ω ( B ξ, η ) = − ω ( ξ, B η ) for all ξ, η ∈ R n and ω ( B · , · ) is definite, such thatp ( t ) = exp t δ T B !! . If B satisfies ω ( B ξ, η ) = − ω ( ξ, B η ) for all ξ, η ∈ R n and ω ( B · , · ) is definite, then B has purelyimaginary eigenvalues and can be diagonalized over C .
2. Oscillator groups
Proof.
The equivalence follows directly from the definition and known results for auto-morphisms of the Heisenberg group (see for instance [Tol78]).Let ω ( B · , · ) be positive definite (otherwise we consider - ω ( B · , · )) and consider the com-plexification of B and the complex bilinear extension of ω . One can check that ω ( Bz , z ) ∈ R and ω ( Bz , z ) > z ∈ C n . Suppose z is an eigenvector of B with its correspondingeigenvalue λ ,
0, then we see R ∋ ω ( Bz , z ) = λω ( z , z ). On the other hand ω ( z , z ) ∈ i R .Hence, λ ∈ i R and finally the first part follows.Now, we prove the second part by induction over n . For n = n , we will prove it for n +
1. Let Z be an eigenvector of B over C with corresponding eigenvalue i λ , λ ,
0. Then ω ( Z , Z ) = λ ω ( BZ , Z ) , ω | span n Z , Z o is nondegenerate. Also Z is an eigenvector of B with corre-sponding eigenvalue − i λ , so B maps the subspace span n Z , Z o onto itself. Since B isantisymmetric with respect to ω , B maps span n Z , Z o ⊥ into itself. Making use of theinduction hypothesis on B (cid:12)(cid:12)(cid:12) span n Z , Z o ⊥ yields the assertion. (cid:3) The Lie algebra of the oscillator group H n ( ω ) ⋊ p R is the semidirect sum of h n ( ω ) and R with respect to the derivation s s (cid:16) δ T B (cid:17) . Hence, there is a basis { X , Y , . . . , X n , Y n } of R n satisfying ω ( X i , Y j ) = δ i , j , ( δ i , j denotes theKronecker symbol) and ω ( X i , X j ) = ω ( Y i , Y j ) = for i,j = n ( ω ) ⋊ p R is R Z ⊕ span { X , Y , . . . , X n , Y n } ⊕ R Twith the non-zero brackets of n Z , X , Y , . . . , X n , Y n , T o given by [ X i , Y i ] = Z , [ T , X i ] = BX i + δ i − Z , [ T , Y i ] = BY i + δ i Zfor i = , . . . , n . Here δ i denotes the i-th. component of δ . The Lie group
Osc n ( ω, B ) is the oscillator group H n ( ω ) ⋊ p R , where p ( t ) = (cid:16) tB (cid:17) , ω ( B · , · ) = − ω ( · , B · ) and ω ( B · , · ) definite. Each oscillator group H n ( ω ) ⋊ p R is isomorphic to some Osc n ( ω, B ) , since the map φ : T T + n X j = ( δ j X j − δ j − Y j ) , X i X i , Y i Y i , Z Zis an isomorphism from the Lie algebra of H n ( ω ) ⋊ p R to osc n ( ω, B ) , where p ( t ) = exp (cid:16) t (cid:16) δ T B (cid:17)(cid:17) , ω ( B · , · ) = − ω ( · , B · ) and ω ( B · , · ) definite. I. e., we can always assume that δ = . The group multiplication in
Osc n ( ω, B ) is given by( z , ξ, t )( v , η, s ) = (cid:18) z + v + ω ( ξ, e tB η ) , ξ + e tB η, t + s (cid:19) ,
3. Oscillator groupsand inversion by ( z , ξ, t ) − = ( − z , − e − tB ξ, − t ) , where ξ, η ∈ R n and z , v , s , t ∈ R . We also define for λ = ( λ , . . . , λ n ), λ i ∈ R the matrix N λ : = − λ λ . . . − λ n λ n and the symplectic form ω λ ( ξ, η ) : = ξ T N − λ η . We denote Osc n ( ω (1 ,..., , N λ ) by Osc ( λ , . . . , λ n ), where λ i λ i + > i = , . . . n − The oscillator groups Osc ( ω, B ) and Osc ( λ , . . . , λ n ) , where < λ ≤ · · · ≤ λ n denote the positive imaginary parts of the eigenvalues of B with multiplicity, are isomorphic. Note that Osc ( λ , . . . , λ n ) is isomorphic to G k ( λ ) in [Med85].Proof of the lemma. Suppose that ω ( B · , · ) is positive definite (similarly for ω ( B · , · ) negativedefinite).Let n Z , Z , . . . , Z n , Z n o be a basis of eigenvectors as in the proof of Lemma 2.2, such that BZ j = i λ j Z j with λ j > λ ≤ · · · ≤ λ n . We set µ j : = i2 ω ( Z j , Z j ) = i2 λ j ω ( BZ j , Z j ) > Z Z , √ µ j Re( Z j ) X j , √ µ j Im( Z j ) Y j , T
7→ − T is an isomorphism from osc n ( ω, B ) to osc ( λ , . . . , λ n ). (cid:3) (Medina / Revoy: [Med85, p. 91])Let < λ ≤ · · · ≤ λ n and < λ ′ ≤ · · · ≤ λ ′ n . Then Osc ( λ , . . . , λ n ) and Osc ( λ ′ , . . . , λ ′ n ) areisomorphic if and only if there is a k ∈ R \{ } , such that k λ ′ i = λ i for i = , . . . , n . Proof.
Let osc denote the Lie algebra of
Osc ( λ , . . . , λ n ). We choose b + [ osc , osc ] ∈ osc / [ osc , osc ] , b < [ osc , osc ], and assign to this element a linear operator ˆ B of [ osc , osc ] / Z , byˆ B ( x + Z ) = [ b , x ] + Z . The operator ˆ B and therefore also its eigenvalues are uniquely determined by the struc-ture of the Lie algebra up to a factor k ,
0. On the other hand, eigenvalues of ˆ B associatedwith b = (0 , , T are exactly the eigenvalues of N λ .Hence, if there is no k ∈ R \{ } , such that λ i = k λ ′ i for i = , . . . , n , , then Osc ( λ , . . . , λ n )and Osc ( λ ′ , . . . , λ ′ n ) are not isomorphic.For the other implication note that the map ϕ : X i X i , Y i Y i , Z Z , T kT
4. Oscillator groupsis an isomorphism of Lie algebras from osc ( λ , . . . , λ n ) to osc ( λ ′ , . . . , λ ′ n ), where { X , Y , . . . , X n , Y n , Z , T } is the standard basis of R n + . (cid:3) Thus, each oscillator group of dimension 2 n + = Osc (1).
Theorem 1.
A map φ : Osc n ( ω, B ) → Osc n ( ω, B ) is an automorphism if and only if there arenumbers m ∈ R , µ ∈ { + , − } and a ∈ R \{ } , a vector b ∈ R n and a matrix S ∈ GL (2 n , R ) with S ∗ ω = a ω and SB = µ BS, such that φ ( z , ξ, t ) = (cid:18) az + ω ( S ξ, e t µ B b + b ) + mt + ω (e t µ B b , b ) , S ξ + e t µ B b − b , µ t (cid:19) . (1) Proof.
One can check that a map satisfying condition (1) is an automorphism. So weonly verify the other implication. Let φ be an automorphism.First of all, we check how elements of { } × { } × R will be mapped. Therefore, suppose φ (0 , , t ) = ( z ( t ) , ξ ( t ) , µ ( t )). Then (cid:16) z ( s + t ) , ξ ( s + t ) , µ ( s + t ) (cid:17) = φ (0 , , s ) φ (0 , , t ) = (cid:16) z ( s ) + z ( t ) + ω ( ξ ( s ) , e µ ( s ) B ξ ( t )) , ξ ( s ) + e µ ( s ) B ξ ( t ) , µ ( s ) + µ ( t ) (cid:17) . We notice that µ is a linear mapping, i. e. µ ( t ) = µ t for some µ ∈ R \{ } .Di ff erentiating ξ ( s + t ) = ξ ( s ) + e µ sB ξ ( t ) with respect to s and setting s =
0, we get ξ ′ ( t ) = µ Bb + µ B ξ ( t ) , where µ Bb = ξ ′ (0), b ∈ R n . The solution of this ODE with ξ (0) = ξ ( t ) = e µ tB b − b . Finally, the comparison of the first components shows that z ( s + t ) + ω ( b , e ( s + t ) µ B b ) = z ( s ) + ω ( b , e s µ B b ) + z ( t ) + ω ( b , e t µ B b )and thus z ( t ) = mt − ω ( b , e t µ B b )for some m ∈ R .Since H n ( ω ) is the commutator subgroup of Osc n ( ω, B ), the automorphism φ maps H n ( ω )onto itself.Hence φ ( z , ξ, = ( az + δ T ξ, S ξ, S ∗ ω = a ω and δ ∈ R n (compare to [Tol78, p.294]). 5. LatticesThus φ ( z , e tB ξ, φ (0 , , t ) = φ (0 , , t ) φ ( z , ξ,
0) gives( δ T e tB ξ + az + mt + ω (e t µ B b , b ) + ω ( S e tB ξ, e t µ B b − b ) , S e tB ξ + e t µ B b − b , µ t ) = ( mt + ω (e t µ B b , b ) + δ T ξ + az + ω (e t µ B b − b , e µ tB S ξ ) , e t µ B b − b + e µ tB S ξ, µ t ) . From the second component it follows, that SB = µ BS . Since det( SB ) = µ n det( BS ), weget µ ∈ { + , − } .From the first component, we get δ T e tB ξ + ω ( S e tB ξ, e t µ B b − b ) = δ T ξ for all ξ ∈ R n , t ∈ R . Di ff erentiating and setting t = δ T B ξ − ω ( SB ξ, b ) = ξ .This completes the proof. (cid:3) So the automorphisms ϕ satisfying ϕ (cid:12)(cid:12)(cid:12) H n ( ω ) = id are of the form ( z , ξ, t ) ( z + mt , ξ, t )with m ∈ R . This contradicts the assertion in [Med85, p. 92].
3. Lattices
Now, we study the lattices of oscillator groups and begin with an example.
Let Γ ω denote the subgroup in Osc n ( ω, B ) generated by n (1 , , , (0 , e i , | i = , . . . , n o . For each z ∈ R and ξ ∈ R n the subgroup L : = D Γ ω ∪ { ( z , ξ , }i isa lattice in Osc n ( ω, B ) with L ∩ H n ( ω ) = Γ ω if and only if (cid:16) ω ( ξ , e B e i ) , e B e i , (cid:17) ∈ Γ ω for i = , . . . , n .To see this, one can check that( z , ξ, t )( v , η, z , ξ, t ) − = ( v + ω ( ξ, e tB η ) , e tB η,
0) (2)for each ( v , η, ∈ H n ( ω ) ∩ L and ( z , ξ, t ) ∈ L .If L = h Γ ω ∪ { (0 , ξ, }i defines a lattice in Osc n ( ω, B ) with L ∩ H n ( ω ) = Γ ω , then we just callit L ( ξ ). Furthermore, we set Γ r : = Γ ω r . Note that Γ r = ( z , ξ, (cid:12)(cid:12)(cid:12)(cid:12) ξ ∈ Z n , z ∈ n X i = r i ξ i − ξ i + Z . In particular for n =
1, we have Γ r = Z × { } for r even and Γ r = n ( z , ξ, | ξ ∈ Z , z ∈ ξ ξ + Z o for r odd. Moreover let Π denote the projection onthe last component and for a lattice L of Osc n ( ω, B ) we denote by Π ( L ) the set Π ( L ) : = n t ∈ R | ∃ z , ξ : ( z , ξ, t ) ∈ L o . Note that Π ( L ) is a non-trivial discrete subgroup of R foreach lattice L (compare to [Med85, p. 90]). Theorem 2.
Let L be a lattice in Osc n ( ω, B ) . Then there exists a uniquely determinedr ∈ N n satisfying r i divides r i + for i = , . . . , n − , a linear map ˜ B , an isomorphism
6. Lattices Φ : Osc n ( ω, B ) → Osc n ( ω r , ˜ B ) and a ξ ∈ R n , such that Φ ( L ) = L ( ξ ) . Proof.
At first, note that L ∩ H n ( ω ) is a lattice in H n ( ω ), see [Rag72, p. 50]. Furthermore,we know from theorem 1.10 in [Tol78, p. 303] that there is a uniquely determined r = ( r , . . . , r n ) where r i ∈ N \{ } , r i | r i + for i = , . . . , n − ϕ : H n ( ω ) → H n ( ω r ), ϕ ( z , ξ ) = ( δ T ξ + az , S ξ ), where S ∗ ω r = a ω , such that ϕ ( L ∩ Heis (cid:16) R n , ω ) (cid:17) = Γ r . Wechoose b , such that δ T ξ = ω r ( S ξ, b ) for all ξ ∈ R n . Then ϕ ( z , ξ, t ) = (cid:18) az + ω r ( S ξ, e tSBS − b + b ) + ω r (e tSBS − b , b ) , S ξ + e tSBS − b − b , t (cid:19) is an isomorphism from Osc n ( ω, B ) to Osc n ( ω r , SBS − ), mapping L ∩ H n ( ω ) onto Γ r . Let t denote the smallest positive element in Π ( ϕ ( L )). So there is a z and a ξ , such that( z , ξ , t ) ∈ ϕ ( L ). The map φ : ( z , ξ, t ) (cid:18) z − z t t , ξ, tt (cid:19) is an isomorphism from Osc n ( ω r , SBS − ) to Osc n ( ω r , t SBS − ) such that φ (cid:12)(cid:12)(cid:12) H n ( ω r ) = id and ( z , ξ , t ) maps to (0 , ξ , (cid:3) From now on we consider n = There is an isomorphism ϕ : Osc ( ω r , B ) → Osc ( ω r , ˜ B ) mapping Γ r onto Γ r andsatisfying Π ( ϕ (0 , , = ± if and only if B or − B is conjugate to ˜ B with respect to an integermatrix with determinant ± .Proof. At first, we will construct an isomorphism satisfying the conditions in the lemma.Let B and ˜ B be conjugate with respect to an integer matrix with determinant ±
1. Thisis su ffi cient to assume, since the map φ : ( z , ξ, t ) ( z , ξ, − t ) is an isomorphism from Osc ( ω r , B ) to Osc ( ω r , − B ), satisfying the conditions in the lemma. Let S = s s s s ! be the integer conjugation matrix with determinant ±
1, such that ˜ B = SBS − , a = det ( S )and b : = (0 , , s s and s s are even(0 , ) , s s is even and s s is odd( , , s s is even and s s is odd . We only get these three cases, since det( S ) is odd. Then ϕ : ( z , ξ, t ) (cid:16) az + ω r ( S ξ, e t ˜ B b + b ) + ω r (e t ˜ B b , b ) , S ξ + e t ˜ B b − b , t (cid:17)
7. Latticesis an isomorphism from
Osc ( ω r , B ) to Osc ( ω r , ˜ B ) , satisfying ϕ ( Γ r ) = Γ r and Π ( ϕ (0 , , = ±
1. Here we will verify this only for the case that s s is even and s s odd (the othercases run similar). It’s not hard to see that ϕ is an isomorphism and Π ( ϕ (0 , , = ϕ (0 , e , = (cid:16) s r , ( s , s ) T , (cid:17) ∈ Γ r , since s r ∈ r s s + Z and similarly ϕ (0 , e , = (cid:16) s r , ( s , s ) T , (cid:17) ∈ Γ r . In addition ( − r , a ( s , − s ) T ,
0) and (0 , a ( − s , s ) T ,
0) areelements of Γ r and will be mapped to (0 , e ,
0) respectively (0 , e , ϕ (1 , , = ( a , , = ( ± , , ϕ ( Γ r ) = Γ r .So one direction of the lemma is verified.Now, let ϕ be an isomorphism satisfying the conditions in the lemma. We know fromLemmas 2.7 and 2.9 that there is a k ∈ R \{ } and a T ∈ SL (2 , R ), such that TBT − = k ˜ B .So we can write ϕ = ϕ ◦ ϕ , where ϕ ( z , ξ, t ) = ( z , T ξ, kt )is an isomorphism from Osc ( ω r , B ) to Osc ( ω r , ˜ B ) and ϕ an automorphisms of Osc ( ω r , ˜ B )Thus ϕ ( z , ξ, t ) = ( az + ω r ( ˜ T ξ, e t µ ˜ B b + b ) + mt + ω r (e t µ ˜ B b , b ) , ˜ T ξ + e t µ ˜ B b − b , µ t ) , where det( ˜ T ) = a , ˜ T ˜ B ˜ T − = µ ˜ B and µ ∈ { + , − } . Hence ϕ ( z , ξ, t ) = ( az + ω r ( ˜ TT ξ, e kt µ ˜ B b + b ) + mt + ω r (e kt µ ˜ B b , b ) , ˜ TT ξ + e kt µ ˜ B b − b , µ kt ) . We get ˜
TTBT − ˜ T − = ± ˜ B , since TBT − = k ˜ B , ˜ T ˜ B ˜ T − = µ ˜ B and Π ( ϕ (0 , , = ± ϕ maps (1 , ,
0) to ( ± , ,
0) and (0 , e i ,
0) into Γ r for i = ,
2. Hence det( ˜ TT ) = ± TTe i ∈ Z for i = ,
2. Thus we obtain the assertion. (cid:3)
To classify the lattices of oscillator groups, we have to choose representatives for theconjugacy classes of matrices, which appeared in the previous lemma.
For y , x ∈ R we denote B x , y : = xy − x y − y y − xy . The set of all B x , y is equal to the set of all matrices which are conjugate to N , and to theset of all 2 × B + : = n B x , y | y > o and B − : = n B x , y | y < o are invariant under conjugation with elements of SL (2 , R ).Furthermore, conjugation with elements of GL (2 , R ) with determinant − B + to B − and reverse. We define F : = (cid:26) ( x , y ) ∈ R | ≤ x ≤ , y > , x + y ≥ (cid:27)
8. Latticesand F = F ∪ (cid:26) ( x , y ) ∈ R | − < x < , y > , x + y > (cid:27) . Note that the map ι : B x , y x + iy is a bijection from n B x , y | y > o to the upper half planeof C , satisfying ι ( AB x , y A − ) = A ( x + iy ) for all A = (cid:16) a bc d (cid:17) ∈ SL (2 , R ), where Az = az + bcz + d .With this in mind, we rewrite the theorem in [Koe98, p. 109]: For all B x ′ , y ′ , y ′ > there is a uniquely defined ( x , y ) ∈ F and an S ∈ SL (2 , Z ) ,such that SB x ′ , y ′ S − = B x , y . Theorem 3.
Let L be a lattice of Osc ( ω, B ) . Then there exist • a uniquely determined r ∈ N \{ } , • a uniquely determined λ = λ + k π with k ∈ N and λ ∈ n π, π, π, π o • and a uniquely determined ( x , y ) = ( , √ ) , λ ∈ { π, π } = (0 , , λ = π ∈ F , λ = π , and an isomorphism ϕ : Osc ( ω, B ) → Osc ( ω r , λ B x , y ) satisfying ϕ ( L ) ∩ H ( ω r ) = Γ r and Π ( ϕ ( L )) = Z .Conversely, for any such data ( r , λ, ( x , y )) there exists a lattice L in Osc ( ω r , λ B x , y ) satisfyingL ∩ H ( ω r ) = Γ r and Π ( L ) = Z .Proof. Because of Theorem 2 we can suppose that ω = ω r and the lattice L given in Osc ( ω r , B ) satisfies L ∩ H ( ω r ) = Γ r and Π ( L ) = Z . The procedure is to find a ˜ B ( = λ B x , y )for a given B , such that there is an isomorphism ϕ from Osc ( ω r , B ) to Osc ( ω r , ˜ B ),mapping Γ r onto Γ r and satisfying Π ( ϕ ( L )) = Z .Let λ ∈ R be the positive imaginary part of the eigenvalue of B . Then we see that B and N λ are conjugate. Hence e B and e N λ = (cid:16) cos λ − sin λ sin λ cos λ (cid:17) are conjugate. So tr(e B ) = λ andcos λ ∈ n − , − , , , o , since e B ∈ SL (2 , Z ). Thus λ = λ + k π ,
0, where λ ∈ n π , π , π , π o and k ∈ Z . There is an x ′ ∈ R and a y ′ ,
0, such that B = λ B x ′ , y ′ . Now we can usethat B x ′ , − y ′ = − B x ′ , y ′ to assume that y ′ >
0. Now we choose an S ∈ SL (2 , Z ), suchthat SB x ′ , y ′ S − = B x , y , where x and y satisfy the conditions in Remark 3.5. So we set˜ B : = | λ | B | x | , y . Because of Lemma 3.2 and the fact that (cid:16) − (cid:17) B x , y (cid:16) − (cid:17) = − B − x , y , there is anisomorphism Osc ( ω r , B ) → Osc ( ω r , ˜ B ) mapping Γ r onto Γ r and satisfying Π ( ϕ ( L )) = Z .Now we want to see that λ > x , y ) ∈ F are uniquely determined. Let λ, λ > x , y ) , ( x , y ) ∈ F . Suppose that there is an isomorphism ϕ : Osc ( ω r , λ B x , y ) → Osc ( ω r , λ B x , y ) mapping Γ r onto Γ r and satisfying Π ( ϕ (0 , , = ±
1. Then we will see9. Latticesthat λ B x , y = λ B x , y . At first Lemma 3.2 gives that ± λ B x , y and λ B x , y are conjugate.Since det( ± λ B x , y ) = det( λ B x , y ), we get λ = λ . Hence ± B x , y and B x , y are conjugate.For a better readability we write Z -conjugate, if two matrices are conjugate with respectto a matrix in SL (2 , Z ) and Z − -conjugate, if two matrices are conjugate with respect toan integer matrix, having determinant − − B x , y = B x , − y and B x , y are not Z -conjugate. In addition B x , y and B x , y are not Z − -conjugate. If B x , y and B x , y are Z -conjugate, then ( x , y ) = ( x , y ) becauseof Remark 3.5. Finally, if − B x , y and B x , y are Z − -conjugate, then B − x , y and B x , y are Z -conjugate. Using Remark 3.5 gives x = x ∈ { , } and y = y easily follows.At last, we want to see how λ and ( x , y ) fit together. Suppose B = λ B x , y , where ( x , y ) ∈ F and λ = λ + k π , satisfying k ∈ N and λ ∈ n π, π, π o . Let L be a lattice in Osc ( ω r , λ B x , y )such that L ∩ H ( ω r ) = Γ r and Π ( L ) = Z . Then, using equation (2) with t = B = cos λ + xy sin λ − x y sin λ − y sin λ y sin λ − xy sin λ + cos λ ∈ SL (2 , Z ) . (3)Thus y sin λ ∈ Z and y = | sin λ | . So x = if cos λ = ± , and x = λ =
0. We get B = λ B , , cos λ = λ B , √ , cos λ = ± . Conversely, it is obvious that e λ B x , y ∈ SL (2 , Z ) for the data ( r , λ, ( x , y )) described in thetheorem. Hence there is a lattice satisfying the claimed conditions (see Example 3.1,where ξ = (0 , ξ = (0 , ) or ξ = ( ,
0) depending on r and e B ). (cid:3) Let B = λ B x , y , where x ∈ R and y > . Then an automorphism ϕ of Osc ( ω r , B ) maps Γ r onto itself if and only if ϕ ( z , ξ, t ) = (cid:16) µ z + ω r ( S ξ, e t µ B b + b ) + mt + ω r (e t µ B b , b ) , S ξ + e t µ B b − b , µ t (cid:17) , (4) where m ∈ R , µ ∈ {± } , S is an integer matrixS = s s s s ! ∈ ( e tB , e tB − x − ! | t ∈ R ) with det( S ) = µ and b ∈ ( Z r , Z r ) : = r Z × r Z , if r is even, respectivelyb = ( b , b ) ∈ ( Z r , Z r ) , s s and s s are even ( Z r , r + Z r ) , s s is even and s s is odd ( r + Z r , Z r ) , s s is even and s s is odd , if r is odd. For short, we call such an automorphism Γ r -preserving.
10. Lattices
Proof.
It’s not hard to show that an automorphism as defined in the theorem maps Γ r onto Γ r . Let ϕ be an automorphism in Osc ( ω r , B ), given as in (1) , mapping Γ r ontoitself. Then a = ±
1. In addition det S = a , since det( S ) ω r = S ∗ ω r = a ω r . Since B = λ B x , y and SB x , y S − = µ B x , y = B x ,µ y , we get det( S ) = µ .Because ϕ (0 , e i , = ( ω r ( Se i , b ) , Se i , S has integer entries.For T = √ y x √ y √ y , it holds that TN T − = B x , y . Since SB = µ BS we see that STN T − = µ TN T − ST . Then T − ST and N (anti-)commute. Hence T − ST ∈ O (2 , R ). Thus T − ST = e t λ N (cid:16) µ (cid:17) and S = e tB T (cid:16) µ (cid:17) T − , for some t ∈ R . If µ =
1, then S = e tB , if µ = −
1, then S = e tB − x − ! . Hence S is an integer matrix in ( e tB , e tB − x − ! | t ∈ R ) . Finally we want to see how b looks like. Let r be even. Then Γ r = Z and hence ω r (cid:16) ( s , s ) , ( b , b ) (cid:17) , ω r (cid:16) ( s , s ) , ( b , b ) (cid:17) ∈ Z . Thus b ∈ ( r Z , r Z ).Now let r be odd, so Γ r = n ( z , ( ξ , ξ ) , | z ∈ ξ ξ + Z o .We see that there are the same three cases to consider as in the previous lemma. Here,we only check the case that s s is even and s s is odd, especially s is even (thecase that s is even runs similarly). Since det( S ) is odd, we get that s is odd. Then ω r (cid:16) ( s , s ) , ( b , b ) (cid:17) ∈ Z and ω r (cid:16) ( s , s ) , ( b , b )) ∈ Z + . Hence r b b ! = det( S ) − s s − s s ! k k + ! , for some k , k ∈ Z . Thus rb ∈ Z and rb ∈ Z + .Finally, for each case we obtain the assertion. (cid:3) Now we can completely classify the lattices of
Osc ( ω, B ) by using Theorem 3 and thefollowing one. Theorem 4.
Suppose B = λ B x , y , ( x , y ) ∈ F and λ = λ + k π , where k ∈ N ∪ { } and λ ∈ n π, π, π, π o . Let L = D Γ r ∪ { ( z , ξ, } E be a lattice in Osc ( ω r , B ) with L ∩ H ( ω r ) = Γ r .
11. Lattices
Then there is a uniquely defined ξ to extract from the list in Appendix A and an automorphism ϕ of Osc ( ω r , B ) , such that ϕ ( L ) = L ( ξ ) .Proof. We will say that ξ and η are equivalent if there is an automorphism mapping L ( ξ )onto L ( η ). Existence
We begin with showing the existence of a ξ in the list in Appendix A andan Γ r -preserving automorphism ϕ for each lattice L satisfying that L ∩ H ( ω r ) = Γ r , suchthat ϕ ( L ) = L ( ξ ).The proof will be divided into parts, dependent on the value of λ . We, always, use e B ,which we can compute with equation (3) in the proof of Theorem 3.First we notice, depending on r , which values are possible for ξ ∈ R , such that h Γ r ∪ { ( z , ξ, }i defines a lattice. Therefor we use example 3.1. Afterward, we giveautomorphisms given as in (4) in Lemma 3.6, which map D Γ r ∪ { ( z , ξ, } E onto L ( ξ ) forsome ξ from the list.It is clear that the automorphism ( z , ξ, t ) ( z + mt , ξ, t ) for some m maps ( z , ξ ,
1) to(0 , ξ , S how the last component will be mapped bythe automorphism (4), since µ = det( S ). Therefore we just give S and b and check, how(0 , ξ,
1) will be mapped. Each automorphism we give is Γ r -preserving, so we don’t makemention of that always. For further argumentation let ξ = ( ξ , ξ ) denote an arbitraryvector in R which satisfies the condition in Example 3.1. Now, we begin with showingthe existence of a vector of the list in Appendix A.Suppose λ = λ + k π , k ∈ N and λ ∈ { π , π } . Let r be even. Then ξ ∈ ( r Z , r Z ). In addition det(e B − E ) =
1. So the automorphism (4), where b : = (e B − E ) − ξ ∈ ( r Z , r Z ) and S : = E , maps (0 , ,
1) to (0 , ξ, ,
0) and ξ are equiv-alent. Now let r be odd. For λ = π one can check that ξ ∈ ( r Z + r , r Z ). Since, again,det(e B − E ) =
1, the automorphism (4) given by b : = (e B − E ) − ( ξ − r , ξ ) ∈ ( r Z , r Z )and S : = E shows that ( r ,
0) and ξ are equivalent. For λ = π an analogous argumen-tation holds, except having r in the other component.Suppose λ = λ + k π , k ∈ N and λ ∈ { π , π } . Then ξ ∈ ( r Z , r Z ). In additiondet(e B − E ) =
2. If r ξ − r ξ is even, then the automorphism defined by b : = (e B − E ) − ξ ∈ ( r Z , r Z ) and S : = E maps (0 , ,
1) to (0 , ξ, ,
0) and ξ are equivalent. If r ξ − r ξ is odd, then we set S : = E and b : = (e B − E ) − ( ξ , ξ − r ) ∈ ( r Z , r Z ). Thus (0 , r ) is equiva-lent to ξ . If additionally r is odd, we instead set b : = ( e B − E ) − (1 + ξ , ξ ) ∈ ( r Z , r Z ). Therelated automorphism shows that (0 ,
0) is equivalent to (1 , + ξ , which is equivalent to ξ .Suppose λ = λ + k π , k ∈ N and λ ∈ { π , π } . Let r be even. Then ξ ∈ ( r Z , r Z ). If r ξ + r ξ ≡ b : = (e B − E ) − ξ ∈ ( r Z , r Z ) and S : = E maps (0 , ,
1) to (0 , ξ, ,
0) and ξ are equivalent. If r ξ + r ξ ≡ b : = (e B − E ) − ( ξ − r , ξ ) ∈ ( r Z , r Z )and S : = E shows that ( r ,
0) and ξ are equivalent. If r ξ + r ξ ≡ b : = (e B − E ) − ( ξ − r , ξ − r ) ∈ ( r Z , r Z ) and S : = (cid:16) −
11 0 (cid:17) . Thus ( r ,
0) and ξ are equivalent.If additionally 3 is not a factor of r , we set b : = (e B − E ) − ( x + ξ , ξ ) ∈ ( r Z , r Z ), where rx + r ξ + r ξ ≡ S : = E . So we get that (0 ,
0) is equivalent to ( x , + ξ , which isequivalent to ξ .Now let r be odd. We obtain λ = π . Using equation (2), we get ξ ∈ ( r Z , r Z + r ).If r ξ + r ξ − / ≡ b : = (e B − E ) − ( ξ , ξ − r ) ∈ ( r Z , r Z ) and S : = E shows that (0 , r ) and ξ are equivalent. If r ξ + r ξ − / ≡ b : = (e B − E ) − ( ξ − r , ξ − r ) ∈ ( r Z , r Z ) and S : = E . The related automorphismshows that ( r , r ) and ξ are equivalent. If r ξ + r ξ − / ≡ b : = (e B − E ) − ( ξ , ξ + r ) ∈ ( r Z , r Z ), S : = − E and get that (0 , r ) and ξ are equivalent.If additionally 3 is not a factor of r , then we set b : = (e B − E ) − ( x + ξ , r + ξ ) ∈ ( r Z , r Z ),where rx + r ξ + r ξ − / ≡ S : = E . The related automorphism shows that(0 , r ) is equivalent to ( x , + ξ , which is equivalent to ξ .For λ = π there is an analogue argumentation, where r is in the other component.Suppose λ = π + π k , k ∈ N . Then ξ ∈ ( r Z , r Z ). Let r be odd. Sothere are x ′ , y ′ ∈ { , } , such that b : = (e B − E ) − ( ξ + x ′ , ξ + y ′ ) ∈ ( r Z , r Z ) andthe automorphism defined by S = E and the above given b shows that (0 ,
0) isequivalent to ξ + ( x ′ , y ′ ), which is equivalent to ξ . Now let r be even. We set η : = r ( − + ( − r ξ , − + ( − r ξ ). Then η ∈ { (0 , , ( r , , (0 , r ) , ( r , r ) } and the auto-morphism given by S = E and b : = (e B − E ) − ( ξ + η ) ∈ ( r Z , r Z ) maps (0 , η,
1) to(0 , ξ, x + y =
1, then the automorphism given by S = (cid:16) (cid:17) and b = (0 ,
0) maps (cid:16) , ( r , , (cid:17) to (cid:16) , (0 , r ) , − (cid:17) = (cid:16) , (0 , r ) , (cid:17) − . Thus ( r ,
0) and (0 , r ) are equivalent.If x = , then the automorphism given by S = (cid:16) − − (cid:17) and b = ( − r , − r ) maps (cid:16) , (0 , r ) , (cid:17) to (cid:16) , ( r , r ) , − (cid:17) = (cid:16) , ( r , r ) , (cid:17) − . Hence (0 , r ) and ( r , r ) are equivalent. If ( x , y ) = ( , √ ),then both automorphisms can be used. Thus every ξ ∈ ( r Z , r Z ) is equivalent to (0 , r , r ). We come to the last case: Suppose λ = π k , k ∈ N .Using equation (2) we get ( ξ , ξ ) ∈ ( r Z , r Z ).In this case we can neglect b , since S ξ + e B b − b = S ξ + b − b = S ξ . Instead, it’s moreimportant to consider all the finitely many integer matrices in ( e tB x , y , e tB x , y − x − ! | t ∈ R ) , for B x , y .At first we give some automorphisms which map Γ r onto itself for all x and y . After-wards, we restrict our observation to the di ff erent cases.For each (cid:16) , ( ξ , ξ ) , (cid:17) there are 0 ≤ k , l < r such that Γ r ∪ n(cid:16) , ( ξ , ξ ) , (cid:17)o and Γ r ∪
13. Lattices n(cid:16) , ( kr , lr ) , (cid:17)o generate the same lattice.The automorphism where S = − E maps (0 , ( kr , lr ) ,
1) to (0 , ( − , − , , ( r − kr , r − lr ) , ξ ∈ ( r Z , r Z ) an equivalent ξ ∈ M : = ((cid:16) kr , lr (cid:17) | ≤ k , l ≤ r ) ∪ ((cid:16) kr , lr (cid:17) | < k < r < l < r ) . Thus the first row is verified.If additionally x =
0, then the automorphism where S = (cid:16) − (cid:17) maps (cid:16) , ( kr , lr ) , (cid:17) to a (cid:16) (0 , (0 , − , , ( kr , r − lr ) , (cid:17) − . Thus the second row is verified.If x = instead, then the automorphism where S = (cid:16) − (cid:17) maps (cid:16) , ( kr , lr ) , (cid:17) , where k > l ,to (cid:16) , ( k − lr , r − lr ) , (cid:17) − . So we can narrow the set of all ξ down to ((cid:16) kr , lr (cid:17) | ≤ k ≤ l ≤ r ) ∪ ((cid:16) kr , lr (cid:17) | < k < r < l < r ) ∪ ((cid:16) kr , (cid:17) | ≤ k ≤ r ) . Furthermore S = − (cid:16) − (cid:17) gives an Γ r -preserving automorphism, which maps (cid:16) , ( kr , lr ) , (cid:17) ,where l ≥ k , to (cid:16) , ( l − kr , lr ) , (cid:17) − . Hence the third row follows.For further argumentation suppose x + y =
1. We know that there is for each ξ ∈ ( r Z , r Z ) an equivalent ξ ∈ M . We can still restrict this set. The automorphism where S = (cid:16) − − (cid:17) maps (cid:16) , ( kr , lr ) , (cid:17) to (cid:16) , ( − lr , − kr ) , − (cid:17) = (cid:16) , ( lr , kr ) , (cid:17) − . So there is for each ξ ∈ ( r Z , r Z ) an equivalent ξ ∈ ((cid:16) kr , lr (cid:17) | ≤ k ≤ l ≤ r ) ∪ ((cid:16) kr , lr (cid:17) | < k < r < l < r ) . Additionally the automorphism where S = (cid:16) (cid:17) maps (cid:16) , ( kr , lr ) , (cid:17) to (cid:16) , ( − , , (cid:17)(cid:16) , (0 , − , (cid:17)(cid:16) , ( r − lr , r − kr ) , (cid:17)! − . Hence for x + y = ξ ∈ ( r Z , r Z ) there is an equivalent ξ ∈ M : = ((cid:16) kr , lr (cid:17) | ≤ k ≤ l ≤ r ) ∪ ((cid:16) kr , lr (cid:17) | k + l ≤ r , < k < r < l < r ) . Now we want to see, which elements in M are equivalent, if ( x , y ) = (0 ,
1) or ( x , y ) = ( , √ ).So suppose ( x , y ) = (0 , S = (cid:16) − (cid:17) maps (cid:16) , ( kr , lr ) , (cid:17) to (cid:16)(cid:16) , ( kr , r − lr ) , (cid:17)(cid:16) , (0 , − , (cid:17)(cid:17) − . Hence the fourth row follows.14. LatticesAt last suppose ( x , y ) = ( , √ ). The automorphism where S : = (cid:16) − − (cid:17) maps (0 , ( kr , lr ) , (cid:16) , ( k − lr , − lr ) , − (cid:17) = (cid:16) , ( l − kr , lr ) , (cid:17) − . Hence we get for each ξ ∈ ( r Z , r Z ) an equivalent ξ ∈ ((cid:16) kr , lr (cid:17) | k + l ≤ r , < k < r < l < r , k ≤ l ) ∪ ((cid:16) kr , lr (cid:17) T | ≤ k ≤ l ≤ l ≤ r ) . Additionally the automorphism where S = (cid:16) − (cid:17) maps (cid:16) , ( kr , lr ) , (cid:17) to (cid:16) , (0 , − , (cid:17)(cid:16) , ( kr , r − l + kr ) , (cid:17)! − . Thus the last row follows.
Uniqueness
Now we want to see that ξ from the list is uniquely determined. There-for, we use a proof by contradiction.Assume that for an r ∈ N \{ } , ( x , y ) ∈ F and a λ = λ + k π , where λ = n π , π , π , π o and k ∈ N , there are ξ and ˜ ξ from the list, such that there is an automorphism ϕ of Osc ( ω r , λ B x , y ), mapping L ( ξ ) onto L ( ˜ ξ ). Then, moreover, this automorphism maps Γ r onto itself.So we can check each Γ r -preserving automorphism and will note that none of themmaps L ( ξ ) onto L ( ˜ ξ ), and we get our contradiction.Let us begin with λ = π + π k , k ∈ N and r even.Let ϕ be an automorphism, given as in (1) in Theorem 1, which maps L ( ξ ) onto L ( ˜ ξ ).Then the corresponding mapˆ ϕ : R → R , ξ S ξ + e B b − b = S ξ − b maps ξ = ( ξ , ξ ) to a vector η = ( η , η ), where r η / is even if and only if ˜ ξ / is even.But the corresponding map ˆ ϕ for an Γ r -preserving automorphism maps (0 ,
0) to a vectorin r Z . Thus there is a contradiction for ˜ ξ ∈ n ( r , , (0 , r ) , ( r , r ) o .In an analogue way, we find a contradiction if ξ = ( r , x and y .If x + y > x , , then each Γ r -preserving automorphism gives a correspondingmap ˆ ϕ , which maps ( r ,
0) to a vector in ( r Z + r , r Z ). Thus there is a contradiction forall ˜ ξ ∈ n (0 , r ) , ( r , r ) o . If x + y > x = , then each Γ r -preserving automorphismgives a corresponding map ˆ ϕ , which maps ( r ,
0) to a vector in ( r Z , r Z ). Thus there is acontradiction for ˜ ξ = ( r , r ).If x + y = x < n , o , then the corresponding map ˆ ϕ for every Γ r -preserving15. Latticesautomorphism maps ( r ,
0) to a vector with one entry in r Z and one in r Z + r . Thusthere is a contradiction for ˜ ξ = ( r , r ).At last suppose ξ = (0 , r ) for x + y > x , . Each Γ r -preserving automorphismgives a corresponding map ˆ ϕ , which maps (0 , r ) to a vector in ( r Z , r Z ). Thus there is acontradiction for ( r , r ).Altogether, we verified that the ξ from the list, which we refer to a lattice L ( ξ ) in Osc ( ω r , λ B x , y ), where λ = π + π k , k ∈ N , ( x , y ) ∈ F , is uniquely determined.Now, we consider the case that λ = π k , k ∈ N \{ } , and get the samecontradiction. But, first of all, we note that ξ ∈ R and ˜ ξ are equivalent if and only ifthere are t , t ∈ Z and an integer matrix S ∈ n e tB , e tB (cid:16) − x − (cid:17) | t ∈ R o with det S = µ , such that ˜ ξ = µ S ξ + t e + t e .So it su ffi ces to fix a t and t for each integer matrix S ∈ n e tB , e tB (cid:16) − x − (cid:17) | t ∈ R o and each ξ , such that µ S ξ + t e + t e ∈ [0 , and show that µ S ξ + t e + t e is equal to ξ or notan element in the set from the list.We consider the case that ( x , y ) = (0 , ± E , ± (cid:16) −
11 0 (cid:17) , ± (cid:16) (cid:17) and ± (cid:16) − (cid:17) arethe only integer matrices in n e tB , e tB (cid:16) − x − (cid:17) | t ∈ R o for this case.We will denote by M the set M : = n(cid:16) k ′ r , l ′ r (cid:17) | ≤ k ≤ l ≤ r o and set ξ = ( kr , lr ) ∈ M . • If S = E , then S ξ = ξ . • Let S = − E . For the cases that k = l = k = l = r , or k = l = r , weget S ξ = ξ , S ξ + e + e = ξ or S ξ + e = ξ respectively. If 0 < k , l < r , then S ξ + e + e = ( r − kr , r − lr ) ∈ [0 , , but r − kr > r . At last, if k = < l < r , then S ξ + e = (0 , r − lr ) ∈ [0 , , but r − lr > r . • Let S = (cid:16) −
11 0 (cid:17) . First of all, we see that S ξ + e = ξ , respectively S ξ = ξ for k = l = r or k = l =
0. If l < { , r } , then S ξ + e = ( r − lr , kr ) ∈ [0 , , but r − lr > . If l = r and k < r , then S ξ + e = ( , kr ) ∈ [0 , , but kr < . • Similar arguments apply to the case S = (cid:16) − (cid:17) . • Let S = − (cid:16) (cid:17) . It follows that − S ξ = ( lr , kr ) ∈ [0 , . Hence, we get ( lr , kr ) < M for k < l , and − S ξ = ξ , for k = l . • Let S = (cid:16) (cid:17) . We see again that − S ξ = ξ for k = l =
0. If k = l >
0, then − S ξ + e ∈ [0 , , but r − lr >
0. For k , l , − S ξ + e + e = ( r − lr , r − kr ) ∈ [0 , .If, additionally, k < r , then r − kr > r . If, however, k = l = r , then ( r − lr , r − kr ) = ξ . • Let S = (cid:16) − (cid:17) . If k < { , r } , then − S ξ + e = ( r − kr , lr ) ∈ [0 , , but r − kr > r . If k = l = r ,then − S ξ + e = ξ and if k =
0, then − S ξ = ξ .16. Lattices • The same reasoning applies to the case S = (cid:16) − (cid:17) .Finally, for ( x , y ) = (0 ,
1) it follows that ( kr , lr ) ∈ M and ( k ′ r , l ′ r ) ∈ M are equivalent, if andonly if k = k ′ and l = l ′ .The rest of the case λ = π k runs as before.For λ = λ + k π , where λ ∈ n π , π , π o and k ∈ N , we use some other way toprove the assertion. First of all we begin with a definition. A group, generated by four elements { α, β, γ, δ } is called O-lattice, if: • There is an r ∈ N \{ } , such that αβα − β − = γ r . • There is a k ∈ N \{ } , such that δ k and γ generate the center of the group and • δαδ − and δβδ − are both elements of h α, β, γ i .It is not hard to see that the computed lattices of Osc ( ω r , λ B x , y ), where λ , k π , k ∈ N satisfy Definition 3.7.Standard arguments yield the following lemma. Let G be an O-lattice as in Definition 3.7, H a group and ˜ ϕ : (cid:8) α, β, γ, δ (cid:9) → H amap. The map ϕ : G → H, defined by ϕ ( α x β y γ z δ t ) = ˜ ϕ ( α ) x ˜ ϕ ( β ) y ˜ ϕ ( γ ) z ˜ ϕ ( δ ) t for all x , y , z , t ∈ Z , is a homomorphism, if and only if: • ϕ ( δ ) ϕ ( α ) ϕ ( δ ) − = ϕ ( δαδ − ) , • ϕ ( δ ) ϕ ( β ) ϕ ( δ ) − = ϕ ( δβδ − ) , • ϕ ( γ ) is an element of the center of H and • ϕ ( α ) ϕ ( β ) ϕ ( α ) − ϕ ( β ) − = ϕ ( γ ) r . Let G be an O-lattice as in Definition 3.7. Let, furthermore, H be an O-lattice,generated by n ˜ α, ˜ β, ˜ γ, ˜ δ o , where ˜ γ and ˜ δ l generate the center of H and ˜ α ˜ β ˜ α − ˜ β − = ˜ γ s . Let ϕ : G → H be an isomorphism. Then ϕ ( γ ) = γ ± . To see this, note that ϕ ( γ ) ∈ h ˜ γ i , since γ is in the center of G and γ r in the commutator subgroup. Then using the bijectivity of ϕ yields the assertion. Furthermore, ϕ ( δ k ) = ˜ γ p ˜ δ ± l for some p ∈ Z . Indeed, since ϕ is anisomorphism, there are p , q , u , v ∈ Z , such that ϕ ( δ k ) = ˜ γ p ˜ δ ul and ϕ ( γ q δ vk ) = ˜ δ l . So we get δ k = ϕ − ( ϕ ( δ k )) = γ qu ± p δ uvk and the assertion holds. In addition, it is straight forward to seethat ϕ maps δ to ˜ α x ˜ β y ˜ γ z ˜ δ ± for some x , y , z ∈ Z and that k = l. For a fixed B = λ B , , where cos λ = and an even r, the lattices L ( ξ ) and L ( ˜ ξ ) of Osc ( ω r , B ) , where ξ , ˜ ξ from the list, are not isomorphic as abstract groups.
17. Lattices
Proof.
We first prove the assertion for λ = π + π k and k ∈ N . We set α : = (0 , e , β : = (0 , e , γ : = (1 , , δ : = (0 , ,
1) and δ : = (cid:16) , ( r , , (cid:17) . Using equation (2) ande B = (cid:16) −
11 0 (cid:17) we get δ αδ − = β, δ βδ − = α − , δ αδ − = βγ, δ βδ − = α − . Suppose that there is an isomorphism ϕ from h α, β, γ, δ i onto h α, β, γ, δ i . Then ϕ ( δ ) = α x β y γ z δ ± for some x , y , z ∈ Z . Furthermore ϕ ( α ) = α n β n γ n for some n , n , n ∈ Z ,since ϕ ( δ ) ϕ ( α ) ϕ ( δ ) − = ϕ ( α ) − and δ ηδ − ∈ h α, β, γ i for every η ∈ h α, β, γ i . On thesame way, we get that ϕ ( β ) = α n ′ β n ′ γ n ′ for some n ′ , n ′ , n ′ ∈ Z .Let ϕ ( δ ) = α x β y γ z δ (similar arguments apply to the case ϕ ( δ ) = α x β y γ z δ − ). Then α n ′ β n ′ γ n ′ = ϕ ( β ) = ϕ ( δ ) ϕ ( α ) ϕ ( δ ) − = α − n β n γ n + n + r O , for some O ∈ Z . Hence n ′ = − n , n ′ = n and n ′ = n + n + r O . Similar computationfor ϕ ( δ ) ϕ ( β ) ϕ ( δ ) − shows that − n = n ′ + n ′ + r O , for some O ∈ Z . Hence − n + n ′ + n and n + n − n ′ are even. Thus n + n is even. This contradicts the bijectivity of ϕ . Sothere is no isomorphism.We consider the second case now. For B = λ B , , λ = π + π k , k ∈ N and r even we set˜ α : = (0 , e , , ˜ β : = (0 , e , , ˜ γ : = (1 , , δ : = (0 , ,
1) and ˜ δ : = (cid:16) , (0 , r ) , (cid:17) .The maps defined by ˜ α α − , ˜ β β − , ˜ γ γ, ˜ δ i δ − i are isomorphisms from the O-lattice D ˜ α, ˜ β, ˜ γ, ˜ δ i E onto D α, β, γ, δ i E for i = ,
1. Hence, thelemma follows, by using the first case of the proof. (cid:3)
For a fixed B = λ B , √ , where cos λ = − and an r divisible by , the latticesL ( ξ ) and L ( ˜ ξ ) of Osc ( ω r , B ) , where ξ , ˜ ξ from the list, are not isomorphic as abstract groups.Proof. We subdivide the proof into two parts and argue as in the proof of the previouslemma.At first let B = λ B , √ , where cos λ = − , sin λ = √ and let r be divisible by 6. Weset α : = (0 , e , , β : = (0 , e , , γ : = (1 , , , δ : = (0 , ,
1) and δ : = (cid:16) , ( r , , (cid:17) . Usingequation (2) and e B = (cid:16) − − (cid:17) gives: δ αδ − = β, δ βδ − = α − β − γ − r , δ αδ − = βγ, δ βδ − = α − β − γ − r − . Suppose that there is an isomorphism ϕ from h α, β, γ, δ i onto h α, β, γ, δ i .Then ϕ ( δ ) ϕ ( α ) ϕ ( δ ) − = ϕ ( δ αδ − ) = ϕ ( β ) . Let ϕ ( δ ) = α x β y γ z δ (The case ϕ ( δ ) = α x β y γ z δ − runs similar).18. LatticesSince δ ηδ − ∈ h α, β, γ i for all η ∈ h α, β, γ i , one can check that ϕ ( α ) = α m β m γ m for some m , m , m ∈ Z .Thus ϕ ( β ) = ϕ ( δ ) ϕ ( α ) ϕ ( δ − ) = α − m β m − m γ m + m ( − r / − + m + r O , where O ∈ Z .Furthermore, we obtain γ ± r = ϕ ( α ) ϕ ( β ) ϕ ( α ) − ϕ ( β ) − = γ r ( m − m m + m ) . Hence, ϕ ( γ ) = γ and ( m , m ) ∈ n (1 , , ( − , , (0 , , (0 , − , (1 , , ( − , − o .In addition α m − m β − m γ − m − m ( − r − − m − r + r O = ϕ ( α ) − ϕ ( β ) − ϕ ( γ ) − r = ϕ ( δ ) ϕ ( β ) ϕ ( δ ) − = α x β y γ z α − m β m − m γ m + m ( − r − + m + r O δ − β − y α − x = α m − m β − m γ − m − r m + m + r O , where O , O ∈ Z . Thus 3 m = r m − m + r m + m − r + r O , for some O ∈ Z . Since m ∈ Z , the term − m + m must be divisible by 3.But, contrarily, for ( m , m ) ∈ n (1 , , (0 , , ( − , , (0 , − , (1 , , ( − , − o it follows that2 m − m is not divisible by 3. Finally, the first case is shown.For the second case let sin λ = − √ . We set ˜ α : = (0 , e , β : = (0 , e , γ : = (1 , , δ : = (0 , ,
1) and ˜ δ = (cid:16) , (0 , r ) , (cid:17) . The maps defined by˜ α β, ˜ β α, ˜ γ γ − , ˜ δ i δ i are isomorphisms from the O-lattice n ˜ α, ˜ β, ˜ γ, ˜ δ i o onto { α, β, γ, δ i } for i = ,
1. Using thefirst part brings the assertion. (cid:3)
For a fixed B = λ B , √ where cos λ = − and an odd r divisible by , the latticesL ( ξ ) and L ( ˜ ξ ) of Osc ( ω r , B ) , where ξ , ˜ ξ from the list, are not isomorphic as abstract groups.Proof. This follows by the same method as in Lemma 3.11. (cid:3)
Finally, Theorem 4 is verified. (cid:3)
19. Listof ξ A. List of ξ λ x , y ξ for an even r ξ for an odd r π + π k x = , y = √ { (0 , } { ( r , } π + π k x = , y = √ { (0 , } { (0 , r ) } π + π k x = , y = { (0 , , (0 , r ) } { (0 , } π + π k x = , y = { (0 , , (0 , r ) } { (0 , } π + π k x = , y = √ r ≡ { (0 , , ( r , } r ≡ { (0 , r ) , ( r , r ) } else: { (0 , } else: { (0 , r ) } π + π k x = , y = √ r ≡ { (0 , , ( r , } r ≡ { ( r , , ( r + r , } else: { (0 , } else: { ( r , } π + π k x + y > , { (0 , , ( r , , (0 , r ) , ( r , r ) } { (0 , } x , x + y > , { (0 , , ( r , , ( r , r ) } x = x + y = , { (0 , , ( r , , ( r , r ) } x , x = , y = √ { (0 , , ( r , r ) } π k x + y > { ( kr , lr ) | ≤ k , l ≤ r } ∪ { ( kr , lr ) | < k < r < l < r } x < { , } x + y > { ( kr , lr ) | ≤ k , l ≤ r } x = x + y > { ( kr , lr ) | ≤ , k ≤ l ≤ l ≤ r } ∪ { ( kr , | ≤ k ≤ r }∪ x = { ( kr , lr ) | < k < r < l < r , k ≤ l } x + y = { ( kr , lr ) | ≤ k ≤ l ≤ r } ∪ { ( kr , lr ) | < k < r < l < r , k + l ≤ r } x < { , } x = , y = { ( kr , lr ) | ≤ k ≤ l ≤ r } x = , y = √ { ( kr , lr ) | ≤ k ≤ l ≤ k + r } References [Koe98] Koecher, M; Krieg, A.
Elliptische Funktionen und Modulformen . Springer-VerlagBerlin Heidelberg New York, 1998.[Med85] Medina, A.; Revoy, P. “Les groups oscillateurs et leurs reseaux”. In:
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52 (1985), pp. 81–95.[Rag72] Raghunathan, M. S.
Discrete Subgroups of Lie Groups . Springer-Verlag NewYork Heidelberg Berlin, 1972.[Tol78] Tolimieri, R. “Heisenberg manifolds and theta functions”. In: