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Two examples about zero torsion linear maps 0 1 on Lie algebras ∗ 0 2 n L. Magnin a J Institut Mathématique de Bourgogne † 5 1 [email protected] ] A January 15, 2010 R . h t a m Abstract [ The question of whether or not any zero torsion linear map on a 1 non abelian real Lie algebra g is necessarily an extension of some CR- v structureis considered andanswered in thenegative. Twoexamples are 3 5 provided, one in the negative and one in the positive. In both cases, the 6 computation up to equivalence of all zero torsion linear maps on g is 2 used for an explicit description of the equivalence classes of integrable . 1 complex structures on g×g. 0 0 1 : 1 Introduction. v i X Given a real Lie algebra g, the determination up to equivalence of zero torsion r a linear maps from g to g plays an important role in the computation of complex structures ondirect products involving g ([2]). In the present note, we consider the question of whether or not any such zero torsionlinear map for non abelian g is necessarily an extension of some CR-structure. We answer the question in the negative by computing (up to equivalence) all zero torsion linear maps from the real 3-dimensional Heisenberg Lie algebra n into itself. The result is then used to exhibit a complete set of representatives of equivalence classes of complex structures on n×n. We also compute all zero torsion linear maps ∗Math. Subj. Class. [2000] : 17B30. Key words : Complex structures, CR-structures, zero torsion, Heisenberg Lie algebra, sl(2,R). †UMR CNRS 5584,Université de Bourgogne,BP 47870, 21078 Dijon Cedex, France. 1 on sl(2,R). In that case they are extensions of CR-structures. We deduce a complete set of representatives of equivalence classes of complex structures on sl(2,R)×sl(2,R). 2 Preliminaries. Let G be a connected finite dimensional real Lie group, with Lie algebra g. A 0 linear map J : g → g is said to have zero torsion if it satisfies the condition [JX,JY]−[X,Y]−J[JX,Y]−J[X,JY] = 0 ∀X,Y ∈ g. (1) IfJ haszerotorsionandsatisfiesinadditionJ2 = −1,J isan(integrable) com- plexstructureong.ThatmeansthatG canbegiventhestructureofacomplex 0 manifold with the same underlying real structure and such that the canoni- cal complex structure on G is the left invariant almost complex structure Jˆ 0 associated to J (For more details, see [3]). To any (integrable) complex structure J is associated the complex subalgebra m = X˜ := X −iJX ;X ∈ g of n o the complexification gC of g. In that way, (integrable) complex structures can be identified with complex subalgebras m of gC such that gC = m ⊕ m¯, bar denoting conjugation. J is said to be abelian if m is. When computing the matrices of the zero torsion maps in some fixed basis (x ) of g, we will j 16j6n denote by ij|k (1 6 i,j,k 6 n) the torsion equation obtained by projecting on x the equation (1) with X = x ,Y = x . The automorphism group Aut g of k i j g acts on the set of all zero torsion linear maps and on the set of all complex structures on g by J 7→ Φ ◦ J ◦ Φ−1 ∀Φ ∈ Aut g. Two J,J′ on g are said ′ to be equivalent (notation: J ≡ J ) if they are on the same Aut g orbit. For complex structures and simply connected G , this amounts to the existence of 0 an f ∈ Aut G such that f : (G ,J) → (G ,J′) is biholomorphic. 0 0 0 R 3 Case of sl(2, ). Let G = SL(2,R) denote the Lie group of real 2×2 matrices with determinant 1 a b σ = , ad−bc = 1. (2) (cid:18)c d(cid:19) Its Lie algebra g = sl(2,R) consists of the zero trace real 2×2 matrices x y X = (cid:18)z −x(cid:19) = xH +yX+ +zX− 2 with basis H = (10 −01), X+ = (00 10), X− = (−01 00) and commutation relations [H,X+] = 2X+, [H,X−] = −2X−, [X+,X−] = H. (3) Beside the basis (H,X+,X−), we shall also make use of the basis (Y1,Y2,Y3) where Y1 = 21H, Y2 = 21(X+ − X−), Y3 = 21(X+ + X−), with commutation relations [Y ,Y ] = Y ,[Y ,Y ] = Y ,[Y ,Y ] = Y . (4) 1 2 3 1 3 2 2 3 1 The adjoint representation of G on g is given by Ad(σ)X = σXσ−1. The matrix Φ of Ad(σ) (σ as in (2)) in the basis (H,X+,X−) is 1+2bc −ac bd Φ =  −2ab a2 −b2. (5) 2cd −c2 d2   The adjoint group Ad(G) is the identity component of Aut g and one has Aut g = Ad(G)∪Ψ Ad(G) , Ψ = diag(1,−1,−1). (6) 0 0 The adjoint action of G on g preserves the form x2 +yz. The orbits are : (i) the trivial orbit {0}; (ii) the upper sheet z > 0 of the cone x2 +yz = 0 (orbit of X−); (iii) the lower sheet z < 0 of the cone x2 +yz = 0 (orbit of −X−); (iv) for all s > 0 the one-sheet hyperboloid x2 +yz = s2 (orbit of sH); (v) for all s > 0 the upper sheet z > 0 of the hyperboloid x2+yz = −s2 (orbit of s(−X+ +X−)); (vi) for all s > 0 the lower sheet z < 0 of the hyperboloid x2+yz = −s2 (orbit of s(X+ −X−)). The orbits of g under the whole Aut g are, beside {0}: (I) the cone x2 +yz = 0 (orbit of X−); (II) the one-sheet hyperboloid x2 +yz = s2 (orbit of sH) (s > 0); (III) the two-sheet hyperboloid x2 +yz = −s2 (orbit of s(X+ −X−)) (s > 0). Lemma 1. Let g = sl(2,R), and J : g → g any linear map. J has zero torsion if and only if it is equivalent to the endomorphism defined in the basis (Y1,Y2,Y3) (resp. (H,X+,X−)) by 0 0 −1 J∗(λ) = 0 λ 0  , λ ∈ R , (7) 1 0 0   J∗(λ) 6≡ J∗(µ) for λ 6= µ 3 (resp. 0 −1 −1 2 2 J(α) = 1 α −α , α ∈ R , (8) 1 −α α   J(α) 6≡ J(β) for α 6= β). Proof. Let J = (ξji)16i,j63 in the basis (H,X+,X−). The 9 torsion equations are in the basis (H,X+,X−): 12|1 2(ξ2 +ξ1)ξ1 +(ξ2 −ξ1)ξ3 −(ξ2 +2ξ1)ξ3 = 0, 2 1 2 2 1 1 1 3 2 12|2 2(ξ2ξ1 +1+(ξ2)2)−ξ3ξ2 −2ξ3ξ2 = 0, 1 2 2 1 1 2 3 12|3 (ξ3 +2ξ1)ξ3 −2(ξ2 +2ξ1)ξ3 +2ξ3ξ3 = 0, 1 2 1 2 1 2 3 2 13|1 (ξ2 −2ξ1)ξ1 +2ξ2ξ1 +ξ3ξ2 −(ξ2 +2ξ1)ξ3 = 0, 1 3 1 3 2 1 3 1 3 3 13|2 2(ξ2 −2ξ1)ξ2 +(ξ2 +2ξ1)ξ2 −2ξ3ξ2 = 0, 2 1 3 1 3 1 3 3 13|3 ξ3ξ2 −2ξ3ξ1 −2+2ξ3ξ2 −2(ξ3)2 = 0, 1 1 1 3 2 3 3 23|1 4ξ1ξ1 −1−ξ2ξ1 −ξ3ξ2 +(ξ2 −ξ1)ξ3 = 0, 3 2 2 1 2 3 2 1 3 23|2 4ξ2ξ1 −(ξ2 +ξ3)ξ2 = 0, 3 2 2 3 1 23|3 4ξ3ξ1 −(ξ2 +ξ3)ξ3 = 0. 2 3 2 3 1 J has at least one real eigenvalue λ. Let v ∈ g, v 6= 0, an eigenvector associated to λ. From the classification of the Aut g orbits of g, we then get 3 cases according to whether v is on the orbit (I),(II),(III) (in the cases (II), (III) one may choose v so that s = 1). Case 1. There exists ϕ ∈ Aut g such that v = ϕ(X−). Then, replacing J by ϕ−1Jϕ, we may suppose ξ1 = ξ2 = 0. That case is impossible from 13|2 and 3 3 13|3. Case 2. There exists ϕ ∈ Aut g such that v = ϕ(H). Then we may suppose ξ2 = ξ3 = 0. Then from 12|2, ξ2ξ3 6= 0, and 23|2, 23|3 yield ξ1 = ξ1 = 0. Then 1 1 3 2 2 3 12|3 and 13|2 successively give ξ3 = ξ2 + 2ξ1 and ξ1 = 0. Now 12|2 and 23|1 3 2 1 1 read resp. −ξ2ξ3 +(ξ2)2 +1 = 0, and ξ2ξ3 −(ξ2)2 +1 = 0. Hence that case is 3 2 2 3 2 2 impossible. Case 3. There exists ϕ ∈ Aut g such that v = ϕ(X+ − X−). Then we may suppose that v = X+−X−. Now instead of the basis (H,X+,X−), we consider the basis (Y ,Y ,Y ). The matrix of J in the basis (Y ,Y ,Y ) has the form 1 2 3 1 2 3 η1 0 η1 1 3 J∗ = η2 λ η2. 1 3 η3 0 η2  1 3 4 Then the 9 torsion equations ∗ij|k (the star is to underline that the new basis is in use) for J in that basis are: ∗12|1 (η3 +η1)λ−(η3 −η1)η1 = 0, 1 3 1 3 1 ∗12|2 (η1 +λ)η2 −η2η3 = 0, 1 3 1 1 ∗12|3 η1λ−1+(η3)2 −(η1 +λ)η3 = 0, 1 1 1 3 ∗13|1 η2η1 +η2η1 +η2η3 −η2η3 = 0, 3 3 1 1 3 1 1 3 ∗13|2 η1λ+1+(η2)2 +(η2)2 +η3η1 −(η1 −λ)η3 = 0, 1 1 3 1 3 1 3 ∗13|3 η2η1 −η2(η1 +η3)−η2η3 = 0, 3 1 1 3 1 3 3 ∗23|1 η1λ+1−(η1)2 +(η1 −λ)η3 = 0, 1 3 1 3 ∗23|2 η2η1 −(η3 +λ)η2 = 0, 3 3 3 1 ∗23|3 (η3 +η1)λ+(η3 −η1)η3 = 0. 1 3 1 3 3 From ∗12|1 and ∗23|3, η1(η3 −η1) = −η3(η3 −η1). (9) 1 1 3 3 1 3 1) Suppose first that η3 = η1. Then λη3 = 0. 1.1) Consider the subcase η3 = 0. 1 3 1 1 ∗13|1 and ∗13|3 read resp. (η3 −η1)η2 = 0,(η3 −η1)η2 = 0. Suppose η3 6= η1. 3 1 1 3 1 3 3 1 Then η2 = η2 = 0, and ∗13|2 gives η1λ+1 = (η1−λ)η3, which implies η3 = 0 1 3 1 1 3 3 by ∗23|1.As ∗12|3 then reads 1 = 0, this case η3 6= η1 is not possible. Now, the 3 1 case η3 = η1 is not possible either since then ∗23|1 would read (η1)2 +1 = 0. 3 1 1 We conclude that the subcase 1.1) is not possible. Hence we are in the subcase 1.2) η3 6= 0.Then λ = 0. From∗13|2, η3η1 6= 0. Then ∗23|1yields η3 = −1+(η13)2 1 3 1 3 η1 1 and ∗13|2 reads (η2)2+(η2)2+2 = 0. This subcase 1.2) is not possible either. 1 3 Hence case 1) is not possible, and we are necessarily in the case 2) η3 6= η1. 1 3 From (9), η3 = −η1. Then ∗13|2 reads (η1)2+(η2)2+(η2)2+1+η3η1 = 0 hence 3 1 1 1 3 1 3 η3 6= 0 and η1 = −(η11)2+(η12)2+(η32)2+1. From ∗12|2, η2 = η32(η11+λ). Then ∗23|2 1 3 η3 1 η3 1 1 reads η2(((η2)2+λ2+1)(η3)2+(η1+λ)2(η2)2) = 0, i.e. η2 = 0, which implies 3 3 1 1 3 3 η2 = 0. Now ∗12|1 reads λ(1 + (η1)2 − (η3)2) = −η1(1 + (η1)2 + (η3)2). The 1 1 1 1 1 1 subcase η1 6= 0 is not possible since then ∗12|3 would yield λ = −(η11)2+(η13)2−1 1 2η1 1 and ∗12|1 would read ((η1)2+(η3+1)2)((η1)2+(η3−1)2) = 0. Hence η1 = 0. 1 1 1 1 1 Then ∗12|3 reads (ξ3)2 = 1. The condition (ξ3)2 = 1 now implies the vanishing 1 1 of all the torsion equations. In that case 0 0 −ε J∗ = 0 λ 0  , ε = ±1. ε 0 0   5 Then in the basis (H,X+,X−) 0 −ε −ε 2 2 J = ε λ −λ 2 2 ε −λ λ  2 2  The cases ε = ±1 are equivalent under Ψ . 0 Remark 1. Recall that a rank r (r > 1) CR-structure on a real Lie algebra g can be defined ([4]) as (p,J ) where p is some 2r-dimensional vector subspace p of g and J : p → p is a linear map such that (a): J2 = −1, (b): [X,Y] − p p [J X,J Y] ∈ p ∀X,Y ∈ p, (c): (1) holds for J for all X,Y ∈ p. Then clearly p p p J∗(λ) is an extension of a CR-structure. R R 4 Case of sl(2, ) × sl(2, ). We consider the basis (Y(1),Y(1),Y(1),Y(2),Y(2),Y(2)) of sl(2,R) × sl(2,R), 1 2 3 1 2 3 with the upper index referring to the first or second factor. The automor- phisms of sl(2,R)×sl(2,R) fall into 2 kinds: the first kind is comprised of the diag(Φ ,Φ ), Φ ,Φ ∈ Autsl(2,R),andthesecond kind iscomprised ofthethe 1 2 1 2 Γ◦diag(Φ ,Φ ), withΓ theswitch between thetwo factorsofsl(2,R)×sl(2,R). 1 2 Lemma 2. Any integrable complex structure J on sl(2,R)×sl(2,R) is equivalent under some first kind automorphism to the endomorphism given in the basis (Y(1),Y(1),Y(1),Y(2),Y(2),Y(2)) by the matrix 1 2 3 1 2 3 0 0 −1 0 0 0 0 ξ2 0 0 ξ2 0  2 5 1 0 0 0 0 0 J˜∗(ξ22,ξ52) = 0 0 0 0 0 −1, ξ22,ξ52 ∈ R, ξ52 6= 0. (10)   0 −(ξ22)2+1 0 0 −ξ2 0   ξ2 2   5  0 0 0 1 0 0    J˜∗(ξ2,ξ2) is equivalent to J˜∗(ξ′2,ξ′2) under some first (resp. second) kind au- 2 5 2 5 tomorphism if and only if ξ′2 = ξ2, ξ′2 = ξ2 (resp. ξ′2 = −ξ2, ξ′2 = −(ξ22)2+1). 2 2 5 5 2 2 5 ξ2 5 J J Proof. Let J = (ξi) = 1 2 , (J ,J ,J ,J 3 × 3 blocks), an inte- j 16i,j66 (cid:18)J J (cid:19) 1 2 3 4 3 4 grable complex structure in the basis (Y(k)). From lemma 1, with some first ℓ 6 0 0 −1 0 0 −1 kindautomorphism, onemaysupposeJ = 0 ξ2 0 ,J = 0 ξ5 0 . 1 2 4 5 1 0 0 1 0 0     As Tr(J) = 0, ξ5 = −ξ2. Then one is led to (10) and the result follows (see 5 2 [1], CSsl22.red and its output). Remark 2. The complex subalgebra m associated to J˜∗(ξ2,ξ2) has basis 2 5 Y˜(1) = Y(1) − iY(1), Y˜(2) = Y(2) − iY(2), Y˜(2) = −iξ2Y(1) + (1 + iξ2)Y(2). 1 1 3 1 1 3 2 5 2 2 2 The complexification sl(2)×sl(2) of sl(2,R)×sl(2,R) has weight spaces de- composition with respect to the Cartan subalgeba h = CY(1) ⊕CY(2) : 2 2 h⊕C(Y(1) +iY(1))⊕C(Y(2) +iY(2))⊕CY˜(1) ⊕CY˜(2). 1 3 1 3 1 1 Then m = (h∩m)⊕CY˜(1)⊕CY˜(2) with h∩m = CY˜(2), which is a special case 1 1 2 of the general fact proved in [5] that any complex (integrable) structure on a reductive Lie group of class I is regular. 5 Case of n. Let n the real 3-dimensional Heisenberg Lie algebra with basis (x ,x ,x ) and 1 2 3 commutation relations [x ,x ] = x . 1 2 3 Lemma 3. Let J : n → n any linear map. J has zero torsion if and only if it is equivalent to one of the endomorphisms defined in the basis (x ,x ,x ) by: 1 2 3 0 −1 0 (i) S(ξ3) = 1 0 0 , ξ3 ∈ R (11) 3 3 0 0 ξ3  3 ξ1 0 0 1 (ii) D(ξ1) = 0 ξ1 0  , ξ1 ∈ R, ξ1 6= 0 (12) 1 1 1 1 (ξ1)2−1 0 0 1  2ξ1   1  0 −ab 0 (iii) T(a,b) = 1 b 0  , a,b ∈ R, b 6= 0 (13) 0 0 ab−1  b  Anytwo distinctendomorphismsin the precedinglistare nonequivalent. T(a,b) is equivalent to b −b 0 ′ T (a,b) = a 0 0  (14) 0 0 ab−1  b  7 Proof. Let J = (ξi) in the basis (x ,x ,x ). The 9 torsion equations are: j 16i,j63 1 2 3 12|1 ξ1(ξ2 +ξ1) = 0, 3 2 1 12|2 ξ2(ξ2 +ξ1) = 0, 3 2 1 12|3 ξ3(ξ2 +ξ1)−ξ2ξ1 +ξ2ξ1 +1 = 0, 3 2 1 2 1 1 2 13|1 ξ2ξ1 = 0, 3 3 13|2 (ξ2)2 = 0, 3 13|3 ξ2(ξ3 −ξ1)+ξ1ξ1 = 0, 3 3 1 2 3 23|1 (ξ1)2 = 0, 3 23|2 ξ2ξ1 = 0, 3 3 23|3 ξ1(ξ2 −ξ3)−ξ2ξ1 = 0. 3 2 3 3 2 Hence ξ1 = ξ2 = 0 , and we are left only with equation 12|3 which reads 3 3 ξ3 Tr(A) = det(A)−1 (15) 3 where A = ξ11 ξ21 . Suppose first Tr(A) = 0. Then A2 = −I, so that A is ξ2 ξ2 (cid:16) 1 2(cid:17) 0 −1 0 similar over C, hence over R, to (0 −1). Hence J ≡ 1 0 0 . Now, since ξ3 1 0 (cid:16)∗ ∗ ξ33(cid:17) 3 does not belong to the spectrum of (0 −1), taking the automorphism 10 01 00 1 0 α β 1 (cid:16) (cid:17) of n for suitable α,β ∈ R, one gets J ≡ S(ξ3). Suppose now Tr(A) 6= 0. Then 3 ξ1 0 0 1 ξ3 = det(A)−1. If A is a scalar matrix, i.e. A = ξ1I, then J =  0 ξ11 0  ≡ 3 Tr(A) 1 ∗ ∗ (ξ11)2−1  2ξ11  D(ξ1).IfAisnotascalarmatrix, thenAissimilar to 0 −ab forsomea,b ∈ R, 1 1 b and b 6= 0 from the trace. Then J ≡ T(a,b). Finally(cid:0), T′(a(cid:1),b) ≡ T(a,b) since the matrices 0 −ab and (b −b) are similar for they have the same spectrum 1 b a 0 and are no sc(cid:0)alar m(cid:1)atrices. Remark 3. S(ξ3) is an extension of a rank 1 CR-structure, however 3 D(ξ1),T(a,b) are not. 1 6 CR-structures on n. Lemma 4. (i) Any linear map J : n → n which has zero torsion and is an extension of a rank 1 CR-structure on n such that p is nonabelian is equivalent to a unique 0 −1 0 1 0 0 , ξ3 ∈ R. (16) 3 0 0 ξ3  3 8 (ii) Any linear map J : n → n which is an extension of a rank 1 CR-structure on n such that p is abelian is equivalent to a unique ξ1 0 0 1 0 0 1 , ξ1 ∈ R. (17) 1 0 −1 0   J has nonzero torsion. Proof. For any nonzero X ∈ p, (X,J X) is a basis of p. In case (i), [X,J X] 6= p p 0, since p is non abelian. Then [X,J X] = µx , µ 6= 0, and x 6∈ p since oth- p 3 3 erwise p would be abelian. One may extend J to n in the basis (X,J X,µx ) p p 3 as 0 −1 ξ1 3 J = 1 0 ξ2 (18) 3 0 0 ξ3  3 and J has zero torsion only if ξ1 = ξ2 = 0. In case (ii), necessarily x ∈ p 3 3 3 since p is abelian. Hence (x ,J x ) is a basis for p. Take any linear extension 3 p 3 J of J to n. There exists some eigenvector y 6= 0 of J associated to some p 1 eigenvalue ξ1 ∈ R. Then y 6∈ p, which implies [y ,Jx ] 6= 0, for otherwise y 1 1 1 3 1 would commute to the whole of n and then be some multiple of x ∈ p. Hence 3 [y ,Jx ] = λx , λ 6= 0, and dividing y by λ one may suppose λ = 1. In the 1 3 3 1 basis y ,y = Jx ,y = x one has 1 2 3 3 3 ξ1 0 0 1 J = 0 0 1 (19) 0 −1 0   and (ii) follows. 7 Complex structures on n × n. We will use for commutation relations [x ,x ] = x ,[x ,x ] = x . The au- 1 2 5 3 4 6 tomorphisms of n × n fall into 2 kinds. The first kind is comprised of the 9 matrices b1 b1 0 0 0 0 1 2  b2 b2 0 0 0 0  1 2 0 0 b3 b3 0 0 Φ =  3 4  ,  0 0 b4 b4 0 0   3 4   b5 b5 b5 b5 b1b2 −b1b2 0   1 2 3 4 1 2 2 1   b6 b6 b6 b6 0 b3b4 −b3b4   1 2 3 4 3 4 4 3  (b1b2 −b1b2)(b3b4 −b3b4) 6= 0. (20) 1 2 2 1 3 4 4 3 The second kind ones are Ψ = ΘΦ where Φ is first kind and 0 0 1 0 0 0  0 0 0 1 0 0  1 0 0 0 0 0 Θ =  . (21)  0 1 0 0 0 0     0 0 0 0 0 1     0 0 0 0 1 0    Lemma 5. Any integrable complex structure J on n × n is equivalent under some first kind automorphism to one of the following: 0 −1 0 0 0 0 1 0 0 0 0 0  0 0 0 −1 0 0 (i) S˜ (ξ5) =  , ε = ±1, ξ5 ∈ R. (22) ε 5 0 0 1 0 0 0  5   0 0 0 0 ξ5 −ε((ξ5)2 +1)  5 5  0 0 0 0 ε −ξ5   5  S˜ε′(ξ′55) is equivalent to S˜ε(ξ55) (ε,ε′ = ±1;ξ′55,ξ55 ∈ R) under some first (resp. second) kind automorphism if and only if ε′ = ε, ξ′5 = ξ5 (resp. ε′ = −ε, ξ′5 = 5 5 5 −ξ5 ). 5 ξ1 0 −((ξ1)2 +1) 0 0 0 1 1 0 ξ1 0 −((ξ1)2 +1) 0 0  1 1  1 0 −ξ1 0 0 0 1 (ii) D˜(ξ11) = 0 1 0 −ξ11 0 0 , 0 0 0 0 (ξ11)2−1 −((ξ11)2+1)2  2ξ1 2ξ1  0 0 0 0 11 1−(ξ111)2   2ξ11 2ξ11  ξ1 ∈ R\{0}. (23) 1 10