Deformations and rigidity in varieties of Lie algebras
aa r X i v : . [ m a t h . R A ] J a n DEFORMATIONS AND RIGIDITY IN VARIETIESOF LIE ALGEBRAS
JOSEFINA BARRIONUEVO AND PAULO TIRAO
Abstract.
We present a novel construction of linear deformations forLie algebras and use it to prove the non-rigidity of several classes ofLie algebras in different varieties. We consider the family of Lie alge-bras with an abelian factor showing that, in general, they are not rigideven for the case of a 1-dimensional abelian factor. We also addressthe problem of k -rigidity for k -step nilpotent Lie algebras. Using ourconstruction and recent results in [BCC] we prove the that the k -stepfree nilpotent Lie algebras are k -rigid but not ( k + 1)-rigid and thatHeisenberg Lie algebras are 2-rigid but not 3-rigid. Introduction
Given a field K , the variety L of n -dimensional Lie algebras over K is theaffine algebraic variety of all antisymmetric bilinear maps µ : K n × K n → K n which satisfy the Jacobi identity, called Lie brackets. The orbits of thenatural action of GL( K n ) by change of basis are the isomorphism classes ofLie bracketsIn the case when K = R or K = C , besides the Zariski topology in L onemay also consider the Euclidean topology, which is finer. A real or complexLie bracket µ is called rigid if its orbit is Euclidean open; or equivalently µ is not rigid if and only if in any Euclidean neighborhood of it there is a nonisomorphic bracket λ .In the case K = C an orbit O ( µ ) of a bracket µ is Euclidean open ifand only if it is Zariski open. Also the Zariski closure and the Euclideanclosure of an orbit also coincide in this case. In general, for an arbitrary K ,a bracket is said to be rigid if its orbit is Zariski open. So that, for K = C both notions coincide.Determining all n -dimensional rigid Lie algebras is an enormous andhighly relevant problem which is out of reach today. There are a finite num-ber of them and the closure of the orbit of a rigid bracket is an irreduciblecomponent of L .In the understanding of the moduli space of n -dimensional Lie algebrasor Lie brackets (we shall use both indistinctly) it is worth investigating tworelevant phenomena that occur in L , which up to certain extend might beconsidered reciprocal, even though they are not.(1) Degenerations in L . A bracket ν is said to degenerate to another bracket µ , if µ ∈ O ( ν ) − O ( ν ), that is if µ lies in the boundary of the orbit of Date : January, 2021.2010
Mathematics Subject Classification.
Primary 17B30; Secondary 17B99.
Key words and phrases.
Nilpotent Lie algebras, deformations, rigidity. ν . As a first example one notices that any ν degenerates to the abelianbracket µ = 0.(2) Deformations in L . When K = R or K = C , by a deformation of abracket µ we understand a continuous curve µ t in L with µ = µ . [In K and in L we consider the Euclidean topology.] We call it non-trivialif for every 0 < | ǫ | , there is a t with | t | < | ǫ | such that µ µ t . For abracket µ in a subvariety L ′ one might consider deformations of µ in L ′ .Given any bracket ν = 0 in a given subvariety L ′ , the deformation in L ′ (1.1) µ t = tν of the abelian bracket µ = 0 is non-trivial, since for any t = 0, µ t ≃ ν and ν µ .Different problems concerning the variety of Lie algebras have been ad-dressed quite extensively since a long time. Understanding their irreduciblecomponents, determining their rigid points, and investigating orbit closures,degenerations and deformations have been some of the goals for many au-thors. The reader may look at the following very short list, that is far fromexhaustive in any sense, and the references therein: [BS, C, GA, GH1, GT1,GT2, S, TV, V]. The general picture becomes even more interesting if onelooks to different subvarieties or families of subvarieties of L and addressthe same problems in them.We consider with special interest the subvariety N of nilpotent Lie alge-bras and the descending chain of subvarieties N k , for k = n − , . . . ,
1, ofnilpotent Lie algebras with nilpotency index less than or equal to k . Noticethat N n − = N and that the complement of N n − inside N is the open sub-variety of filiform Lie algebras. Also we consider the subvariety S of solvableLie algebras and the corresponding chain S k of solvable Lie algebras withsolvability index less than or equal to k .A well known result states that if H ( g , g ) = 0, then g is rigid in L .The first example showing that the reciprocal does not hold was provided in[R]. In the same direction, recently in [BCC] it is shown that, the vanishingof a k -step second cohomology group for a k -step nilpotent Lie algebra n implies the rigidity of n in N k . In the opposite direction, in order to showthat a given Lie algebra in a given variety is not rigid, one can construct anon-trivial deformation of it.Semisimple Lie algebras are rigid by Whitehead’s Lemma. Also a semisim-ple Lie algebra g plus a 1-dimensional abelian factor a is rigid, since its sec-ond cohomology group vanishes. This fact follows from the Hochshild-Serrespectral sequence associated to the the ideal g of g ⊕ a . Also Borel subalge-bras of semisimple Lie algebras have vanishing second cohomology [LL] andhence are rigid. So that, there are solvable rigid Lie algebras in L .A natural and very interesting open question is: • Are there nilpotent rigid Lie algebras in L ?This question, known since 1970 as Vergne’s conjecture, has not been an-swered. We beleive that the answer is, in general, NO . The exceptionsshould only exist in small dimensions ( ≤ Heisenberg Lie algebra is rigid in N . This paper, in particular, adds supportto our believe.The following stronger versions of this question are also very challenging: • Are there k -step nilpotent rigid Lie algebras in N k ? • Are there k -step nilpotent rigid Lie algebras in N k +1 ?Regarding the first question, we show that the answer is yes , we provethat the free k -step nilpotent Lie algebras are rigid in N k and that theHeisenberg Lie algebras are rigid in N . Their rigidity follows by provingthat some cohomology introduced recently in [BCC] vanishes.Regarding the second question, up to our knowledge based on existingexamples, its answer is (in general) also no . In this paper we provide furtherclasses of examples by constructing non trivial deformations. In particularthe free k -step nilpotent and Heisenberg Lie algebras are prover to be non-rigid in N k +1 and N respectively.The construction of non trivial deformations is done by using sytematicalya novel constrction of linear deformations that we present in Section 3.Among the applications of it, we address the problem of deforming arbitraryLie algebras with an abelian factor. This is clearly possible in general, butturns out to be very interesting for very small abelian factors, in particularfor 1-dimensional factors.2. Some preliminaries
In this paper the field K will be either R or C and n will be a fixed naturalnumber. It will not be necessary to make any distinction in general.We will consider Lie algebras over K of dimension n . Through the wholepaper we will refer to a Lie algebra g or to its Lie bracket µ indistinctly,according to which notation fits the exposition better. We shall mainly use µ and use g when the underlying vector space is relevant, for instance torefer to a subalgebra. We may also write ( g , µ ).2.1. Multilinear maps.
Let V be a vector space and let C i ( V ) = Hom( V ⊗ i , V )be the space of i -multilinear maps from V to V . Given ϕ ∈ C i ( V ) and ψ ∈ C j ( V ) let ϕ ◦ ψ ∈ C i + j − ( V ) be the multilinear map defined by ϕ ◦ ψ ( x , . . . , x i + j − ) = ϕ ( ψ ( x , . . . , x j ) , x j +1 , . . . , x i + j − ) . Also we define inductively ϕ k = ϕ ◦ ϕ k − . Notice that if f is linear and ϕ is i -linear, then f ◦ ϕ is again i -linear.If µ is a bilinear map on V , then µ ◦ µ is the trilinear map defined by µ ◦ µ ( x, y, z ) = µ ( µ ( x, y ) , z ) . For a trilinear map ϕ we write (cid:9) ϕ ( x, y, z ) = ϕ ( x, y, z ) + ϕ ( y, z, x ) + ϕ ( z, x, y ) . Hence, for a bilinear map µ , we have that (cid:9) µ ◦ µ ( x, y, z ) = µ ( µ ( x, y ) , z ) + µ ( µ ( y, z ) , x ) + µ ( µ ( z, x ) , y ) . JOSEFINA BARRIONUEVO AND PAULO TIRAO
So that the Jacobi identity for µ is (cid:9) µ ◦ µ = 0 . Given three linear functions f, g, h ∈ C ( V ), f · g · h is the trilinear mapdefined by f · g · h ( x, y, z ) = f ( x ) g ( y ) h ( z ) . A direct computation yields the following result that we shall use in thenext section.
Lemma 2.1.
Given f, g, h ∈ C ( V ) , it holds that (cid:9) ( f · g − g · f ) · f = 0 , (cid:9) ( f · g − g · f ) · g = 0 and (cid:9) ( f · g − g · f ) · h = (cid:9) ( f · h − h · f ) · g . Varieties of Lie algebras.
Let V = K n . The variety L of Lie algebrasof dimension n over K , is the affine algebraic variety of all the antisymmetricmaps µ ∈ C ( V ) satisfying (cid:9) µ ◦ µ = 0.The subvariety N of nilpotent Lie algebras, is composite by those Liebrackets µ such that µ j = 0, for some j ≥
1. A Lie algebra µ is said to be k -step nilpotent, for k ≥
2, if µ k = 0 and µ k − = 0. We consider the abelianLie algebra µ = 0 as 1-step nilpotent. The subvariety N k of nilpotent Liealgebras at most k -step nilpotent is then composite by all Lie brackets µ such that µ k = 0. Notice that N k ⊂ N k +1 and that N n − = N . Remark . The subset of filiform
Lie algebras, ( n − N n − in N , so that it is also a subva-riety.The subvarieties of solvable and k -step solvable Lie algebras are definedanalogously by considering µ ( k ) instead of µ k , where µ (1) = µ and for k ≥ µ ( k ) = µ (cid:0) µ ( k − , µ ( k − (cid:1) . Notation . Given a Lie algebra ( g , µ ), we consider it as an element ofthe variety L with n = dim g . Analogously we say that µ ∈ L ′ , if it is anelement of the subvariety L ′ of L . Orbits and rigidity . The orbit O ( µ ) in L of a Lie algebra µ under the actionof GL n ( K ) given by change of basis is the isomorphism class of µ . Clearly,if µ is in any of the subvarieties described above, its orbit O ( µ ) is containedin it. A Lie algebra µ in any of these subvarieties is rigid in it, if its orbit isopen there. k -rigidity for nilpotent Lie algebras . Given a nilpotent Lie algebra µ , we saythat it is k -rigid , if it is rigid in N k .Given a k -step nilpotent Lie algebra g we may ask for the smallest j , j ≥ k , such that g is not rigid in N j , if such a j exists. According to theavailable evidence we have, in general the smallest j is just k + 1. That is,there is no known k -step nilpotent Lie algebra which is ( k + 1)-rigid. Thispaper adds more evidence to the negative answer of the question: • Are there k -step nilpotent Lie algebras which are ( k + 1)-rigid? Linear deformations.
Given a Lie algebra µ , we shall consider lineardeformations of µ , that is curves µ t = µ + tϕ , where t is a parameter in K .It is straightforward to verify that µ t is a Lie algebra for all t if and only if ϕ is a Lie algebra and a 2-cocycle for µ , that is µ ◦ ϕ + ϕ ◦ µ = 0. The Grunewald-O’Halloran construction . Given a Lie algebra g with bracket µ , an ideal h of codimension 1 and a derivation D of h , Grunewald andO’Halloran [GH2] considered the linear deformation of µµ t = µ + tϕ D , where ϕ D is the (Dixmier) 2-cocycle defined by ϕ D ( x, h ) = D ( h ) , ϕ D ( h, h ′ ) = 0 , for h, h ′ ∈ h and x a fixed element outside of h . Notice that h remains anideal of µ t , for all t .2.4. The nil-cohomology.
We recall briefly some results from [BCC] whichare relevant for our purposes.Even though K = R in [BCC], their main results hold also for K = C since they are based on the very general Nash-Moser Theorem of Hamilton[H, Theorem 3.1.1].Let ( g , µ ) be a nilpotent Lie algebra of dimension n over K . For k ≥ N k : Λ g ∗ ⊗ g → ( g ∗ ) ⊗ ( k +1) ⊗ g defined by N k ( µ ) = µ k . Its differential at µ , dN k | µ , is given by dN k | µ ( σ ) = k − X j =0 µ k − − j ◦ σ ◦ µ j . The second k -nilpotent cohomology group of g is then defined by H k − nil ( g , g ) = Ker( d µ ) ∩ Ker( dN k | µ )Im( d µ ) , where d µ and d µ are the usual Chevalley-Eilenberg differentials. Remark . If µ is 2-step nilpotent and σ ∈ Ker( dN | µ ), then µ ◦ σ + σ ◦ µ = 0,hence (cid:9) ( µ ◦ σ + σ ◦ µ ) = 0 and σ ∈ Ker( d µ ). Thus Ker( d µ ) ∩ Ker( dN | µ ) =Ker( dN | µ ). Theorem 2.5 (Brega, Cagliero and Chavez-Ochoa) . Let ( g , µ ) be a k -stepnilpotent Lie algebra over K . If H k − nil ( g , g ) = 0 , then ( g , µ ) is k -rigid.Notation . Given a nilpotent Lie algebra g , for simplicity we write H k ( g ) = H k − nil ( g , g ).3. A novel construction of linear deformations
In what follows, we construct linear deformations of a given Lie algebra g starting from a subalgebra h of g of codimension 2.Let a , a ∈ g be such that h a , a i is a complementary subspace to h , i.e. g = h a , a i ⊕ h . JOSEFINA BARRIONUEVO AND PAULO TIRAO
For a basis { h , . . . , h n − } of h , B = { a , a , h , . . . , h n − } is a basis of g . Let B ∗ = { a ∗ , a ∗ , h ∗ , . . . , h ∗ n − } be the dual basis of B . Hence,given x ∈ g , there is a unique x h ∈ h such that x = a ∗ ( x ) a + a ∗ ( x ) a + x h . We denote the linear map x x h by π h .By ad x we denote the adjoint of x ∈ g , ad x : g → g . For h ∈ h , we denoteby ad h h the adjoint of h corresponding to h , ad h : h → h .In addition, for y ∈ g , we shall consider the antisymmetric bilinear map ϕ = a ∗ ∧ a ∗ ⊗ y . Recall that, for all u, v ∈ g ,( a ∗ ∧ a ∗ ⊗ y )( u, v ) = ( a ∗ · a ∗ − a ∗ · a ∗ )( u, v ) y = ( a ∗ ( u ) a ∗ ( v ) − a ∗ ( u ) a ∗ ( v )) y. In particular ϕ ( h , g ) = 0. Finally notice that (cid:9) ϕ ◦ ϕ = 0, so that ϕ is a Liebracket. Theorem 3.1.
Let ( g , µ ) be a Lie algebra and h ֒ → g be a subalgebra ofcodimension 2 and let a , a ∈ g be such that h a , a i is a complementarysubspace to h . If a ∗ ◦ ad a + a ∗ ◦ ad a = 0 in h , then for any y ∈ Z g ( h ) µ t = µ + t ( a ∗ ∧ a ∗ ⊗ y ) is a linear deformation of µ .Proof. Let us denote µ = [ , ] and a ∗ ∧ a ∗ ⊗ y = ϕ . Since ϕ is already a Liebracket, it remains to show that ϕ is a 2-cocycle for µ , that is (cid:9) ([ , ] ◦ ϕ + ϕ ◦ [ , ]) = 0 . We show that moreover (cid:9) [ , ] ◦ ϕ = 0 and (cid:9) ϕ ◦ [ , ] = 0.On the one hand, let u, v, w ∈ g . ([ , ] ◦ ϕ ) ( u, v, w ) = [ ϕ ( u, v ) , w ] and[ ϕ ( u, v ) , w ] = [ ( a ∗ a ∗ − a ∗ a ∗ )( u, v ) y , a ∗ ( w ) a + a ∗ ( w ) a + w h ]= (( a ∗ a ∗ − a ∗ a ∗ ) a ∗ )( u, v, w ) [ y, a ]+(( a ∗ a ∗ − a ∗ a ∗ ) a ∗ )( u, v, w ) [ y, a ]+( a ∗ a ∗ − a ∗ a ∗ )( u, v ) [ y, w h ]= 0 . The first two terms are equal to 0 by Lemma 2.1 and the third one because y ∈ Z g ( h ).On the other hand, let u, v, w ∈ g , ( ϕ ◦ [ , ])( u, v, w ) = ϕ ([ u, v ] , w ). Writing u = a ∗ ( u ) a + a ∗ ( u ) a + u h v = a ∗ ( v ) a + a ∗ ( v ) a + v h , it follows that[ u, v ] = ( a ∗ ( u ) a ∗ ( v ) − a ∗ ( u ) a ∗ ( v )) [ a , a ] + [ u h , v h ] + a ∗ ( u )[ a , v h ]+ a ∗ ( u )[ a , v h ] + a ∗ ( v )[ u h , a ] + a ∗ ( v )[ u h , a ] , and since [ u h , v h ] ∈ h we have that ϕ ([ u, v ] , w ) = ( a ∗ ( u ) a ∗ ( v ) − a ∗ ( u ) a ∗ ( v )) ϕ ([ a , a ] , w ) + a ∗ ( u ) ϕ ([ a , v h ] , w )+ a ∗ ( u ) ϕ ([ a , v h ] , w ) − a ∗ ( v ) ϕ ([ a , u h ] , w ) − a ∗ ( v ) ϕ ([ a , u h ] , w ) . The first term is equal to S ( u, v, w ) = (( a ∗ · a ∗ − a ∗ · a ∗ ) · a ∗ )( u, v, w ) a ∗ ([ a , a ]) y − (( a ∗ · a ∗ − a ∗ · a ∗ ) · a ∗ )( u, v, w ) a ∗ ([ a , a ]) y ;the sum of the second and forth terms is equal to S ( u, v, w ) = ( a ∗ · ( a ∗ ◦ ad a ◦ π h ) − ( a ∗ ◦ ad a ◦ π h ) · a ∗ ) · a ∗ ( u, v, w ) y +(( a ∗ ◦ ad a ◦ π h ) · a ∗ − a ∗ · ( a ∗ ◦ ad a ◦ π h )) · a ∗ ( u, v, w ) y ;and the sum of the third and fifth terms is equal to S ( u, v, w ) = ( a ∗ · ( a ∗ ◦ ad a ◦ π h ) − ( a ∗ ◦ ad a ◦ π h ) · a ∗ ) · a ∗ ( u, v, w ) y +(( a ∗ ◦ ad a ◦ π h ) · a ∗ − a ∗ · ( a ∗ ◦ ad a ◦ π h )) · a ∗ ( u, v, w ) y. Now we have that, by Lemma 2.1, (cid:9) S ( u, v, w ) = 0 . Also by the same lemma, (cid:9) S ( u, v, w ) = (cid:9) ( a ∗ · ( a ∗ ◦ ad a ◦ π h ) − ( a ∗ ◦ ad a ◦ π h ) · a ∗ ) · a ∗ ( u, v, w ) y (cid:9) S ( u, v.w ) = (cid:9) (( a ∗ ◦ ad a ◦ π h ) · a ∗ − a ∗ · ( a ∗ ◦ ad a ◦ π h )) · a ∗ ( u, v, w ) y. Finally, by hypothesis a ∗ ◦ ad a ◦ π h = − a ∗ ◦ ad a ◦ π h . This implies, by usingLemma 2.1, that (cid:9) ( S + S ) = 0and therefore (cid:9) ϕ ◦ [ , ] = 0as we wanted to prove. (cid:3) Corollary 3.2.
Let g be a nilpotent Lie algebra with Lie bracket µ and h ⊆ g a subalgebra of codimension 2. Let a , a ∈ g be such that h a , a i is a complementary subspace to h , so that g = h a , a i ⊕ h . Then for any y ∈ Z g ( h ) µ t = µ + t ( a ∗ ∧ a ∗ ⊗ y ) is a linear deformation of µ .Proof. Both g and h are nilpotent Lie algebras, hence tr(ad x ) = 0, for all x ∈ g and tr(ad h h ) = 0, for all h ∈ h . Since tr (ad h ) = a ∗ ([ h, a ]) + a ∗ ([ h, a ]) + tr (ad h h ) , it follows that 0 = a ∗ ◦ ad a ( h ) a ∗ ◦ ad a ( h ) . (cid:3) Remark . If g is nilpotent, this construction is a particular case of theGrunewald-O’Halloran construction [GH2]. It follows from two easy argu-ments. First, all subalgebra of codimension 2 can be extended to an ideal ofcodimension 1. Then we can suppose that g = h a i ⊕ I , with h ⊆ I . Finally,the linear function a → y is a derivation of h .In general, our construction is different from that in [GH2], as the follow-ing two examples show. JOSEFINA BARRIONUEVO AND PAULO TIRAO
Example 3.4.
Let ( g = h a , a , y i , µ ) with Lie bracket µ ( a , y ) = 2 a µ ( a , y ) = 2 a µ ( a , a ) = 0If we take h = h y i , it satisfies the hypotheses of the Theorem 3.1. It resultthat µ t is isomorphic to sl ( C ) for all t = 0. Since sl ( C ) has no ideals, thedeformation µ t is not of Grunewald-O’Halloran type. Example 3.5.
Let ( h , ν ) be a non-perfect Lie algebra with non-trivial centerand let f : h → K be a non-zero linear map such that f ( ν ( h , h )) = 0. Definethe Lie algebra ( g , µ ) by taking g = h a , a i ⊕ h with µ defined by: µ | h × h = ν,µ ( a , h ) = f ( h ) a ,µ ( a , h ) = f ( h ) a ,µ ( a , a ) = 0 , for all h ∈ h . By taking y = 0 ∈ Z ( h ), the hypotheses of Theorem 3.1are satisfied. The corresponding linear deformation µ t is not of Grunewald-O’Halloran type.Indeed, if µ t = µ + t ( a ∗ ∧ a ∗ ⊗ y ) were of that type, it would exist an ideal I ⊳ g of codimension one, x ∈ g and D ∈ Der( I ) such that g = h x i ⊕ I and µ t ( i , i ) = µ ( i , i )(3.1) µ t ( x, i ) = µ ( x, i ) + tD ( i )(3.2)for all i, i , i ∈ I . Since I ⊳ g has codimension one, then µ ( g , g ) ⊆ I andthen a , a ∈ I . Now from (3.1) we get that µ t ( a , a ) = µ ( a , a ) . But by the definition of µ t , we have that µ t ( a , a ) = ( µ + t ( a ∗ ∧ a ∗ ⊗ y ))( a , a ) = µ ( a , a ) + ty, which is not possible for t = 0. Therefore µ t is not of Grunewald-O’Hallorantype.3.1. As an example of the previousconstruction, we deform 2-step nilpotent graph Lie algebras. Let G be thegraph with vertices V = { v , . . . , v m } and edges A = { a ij : ( i, j ) ∈ I } . Thegraph Lie algebra associated to G is the K -vector space generated by V ∪ A ,where the non-zero brackets of basis elements are[ v i , v j ] = a ij , if ( i, j ) ∈ I. Notice that if A = ∅ , then g = a m the m -dimensional abelian Lie algebra.But in general, if A = ∅ , it is a 2-step nilpotent Lie algebra. Example 3.6.
The Lie algebra associated to the graph with two verticesand one edge is the 3-dimensional Heisenberg Lie algebra h .In general, 2-step graph Lie algebras are not 3-rigid, as stated preciselyin the following theorem. Theorem 3.7.
Let g be a graph Lie algebra non isomorphic to h , a or a .Then g is non-rigid in N .Proof. Let [ , ] be the bracket of g . In all cases we construct a non-trivial3-step deformation of [ , ].If A = ∅ , g ≃ a n with n ≥ , ] = 0, then (1.1) provides non-trivial2-step deformations of [ , ].If | A | = 1, since g ≇ h , then m >
2. We may assume, by relabelingthe vertices if necessary, that A = { a } . The hypotheses of Corollary 3.2are fulfilled for a = v , a = a , h = h V − { v }i and y = v . Then[ , ] t = [ , ] + t ( v ∧ a ⊗ v ) is a linear deformation of [ , ], which is 3-stepnilpotent for all t = 0.If | A | >
1, we can assume that a , a ∈ A . The hypotheses of Corollary3.2 are fulfilled for a = v , a = a , h = h ( V − { v } ) ∪ ( A − { a } ) i and y = a . Then [ , ] t = [ , ] + t ( v ∧ a ⊗ a ) is a linear deformation of [ , ],which 3-step nilpotent for all t = 0. (cid:3) Free nilpotent Lie algebras
Let L ( k ) ( m ) be the k -step free nilpotent Lie algebra on m generators,where m ≥
2. In this section we explore the rigidity of L ( k ) ( m ) in thevarieties N k and N k +1 , showing that it is rigid in the first one, but not inthe second one. There is a single exception is L (2) (2), which is isomorphicto the 3-dimensional Heisenberg Lie algebra h . The Heisenberg Lie algebrais rigid not only in N = N (for n = 3), but also in N .Let us recall briefly the construction of L ( k ) ( m ). Given a set of generators X , one constructs one after the other, the free magma M ( X ), the freealgebra A ( X ), the free Lie algebra L ( X ) and finally the k -step free nilpotentLie algebra on X , L ( k ) ( X ). In the case X = { x , ..., x m } , we shall denote L ( k ) ( X ) by L ( k ) ( m ).4.1. The free magma on X . Is the set M ( X ) with an operation ‘ (cid:5) ’. M ( X ) = [ i ∈ N M i ( X ) , where M ( X ) = X and for i ≥ M i ( X ) = { p (cid:5) q | p ∈ M s ( X ) , q ∈ M t ( X ) , s + t = i } . The elements of M i ( X ) are said to be of length i .4.2. The free algebra on X . Is the algebra A ( X ) built on the linear spacegenerated by the set M ( X ) with the bilinear product induced by ‘ (cid:5) ’. Clearly,it is naturally graded A ( X ) = ∞ M i =1 A i ( X ) , where A i ( X ) = h M i ( X ) i . Notice that A i ( X ) (cid:5) A j ( X ) ⊆ A i + j ( X ). Theelements of A i ( X ) are said to be of degree i . The free Lie algebra on X . Is the Lie algebra ( L ( X ) , λ ), constructedas the quotient algebra L ( X ) = A ( X ) I , where I = (cid:10) { a (cid:5) b + b (cid:5) a | a, b ∈ A ( X ) } ∪ { (cid:9) (cid:5) ( a, b, c ) | a, b, c ∈ A ( X ) } (cid:11) . Thequotient projection is π : A ( X ) → L ( X ). Since the ideal I is homogeneous, L ( X ) is naturally graded L ( X ) = ∞ M i =1 L i ( X ) , where L i ( X ) = π ( A i ( X )).4.4. The k -step free nilpotent Lie algebra on X . Is the Lie algebra( L ( k ) ( X ) , [ , ]), constructed as the quotient algebra L ( k ) ( X ) = L ( X ) L k +1 ( X ) , where L k +1 ( X ) is the ( k + 1)-th term of the central descending series of L ( X ), L k +1 ( X ) = ∞ M i = k +1 L i ( X ) . The quotient projection is : L ( X ) → L ( k ) ( X ). Since the ideal L k +1 ( X ) ishomogeneous, L ( k ) ( X ) is naturally graded L ( k ) ( X ) = k M i =1 L i ( X ) , where, for i ≤ k , we identify L i ( X ) with its image in the quotient. Finally,we denote by π k = ◦ π the composition π k : A ( X ) −→ L ( X ) −→ L ( k ) ( X ) . Hall bases.
Let X = { x , · · · , x m } . Since L i ( X ) = π ( A i ( X )) and A i ( X ) = h M i ( X ) i , it follows that L i ( X ) = h π ( M i ( X )) i . Linear basis for each L i ( X ), and hence for the free Lie algebra L ( X ) and forthe k -step free nilpotent Lie algebras, can be chosen from π ( M i ( X )). Verywell known bases of this kind are the H all bases.(1) Starting from an ordered basis B of L ( X ) one constructs recursivelyordered basis B i of L i ( X ) as follows:(i) Let B be the set of generators X ordered by x < x < · · · < x m . (ii) Given ordered bases B , . . . , B k of L ( X ) , . . . , L k ( X ) respectively, u ∈ B i and v ∈ B j , with i = j , are ordered by u < v, if i < j. (iii) Now B k +1 is formed by all brackets λ ( u, v ), with u ∈ B i , v ∈ B j with i + j = k + 1, subject to the following restrictions: u > v and if u = λ ( w, z ) , v ≥ z. The elements of B k +1 are ordered lexicographically; that is λ ( u, v ) >λ ( w, z ) if u > w , or u = w and v > z .(iv) Given an order for the generators, the basis so constructed isuniquely determined.(2) Observe that each element v in the basis B i has a multidegree D v =( d , . . . , d n ), where d j is the number of occurrences of x j in v . Clearly d + · · · + d n = i and moreover the bracket is multigraded, that is D [ v,w ] = D v + D w .From the construction and the last observation three facts, that werecall for later use, follow:(i) The element c k = λ k − ( x , x , . . . , x ), of multidegree D = ( k − , , , . . . ,
0) is in B k , for every k ≥ v ∈ B i and w ∈ B j , λ ( v, w ) ∈ L i + j ( X ) is a linear com-bination of the those elements of B i + j with multidegree equal to D v + D w . The coefficient of c k , with k = i + j , in this linearcombination is almost always zero. The only exception is the case v = c k − and w = x , in which case λ ( v, w ) = c k .(iii) In general dim L i ( X ) ≥
2, with the only exception of L ( X ) if m = 2, in which case dim L ( X ) = 1.4.6. Main results.
Given X = { x , ..., x m } let us denote M ( X ), A ( X ), L ( X ) and L ( k ) ( X ) by M ( m ), A ( m ), L ( m ) and L ( k ) ( m ) respectively. Theorem 4.1.
The k -step free nilpotent Lie algebra on m generators L ( k ) ( m ) ,is non-rigid in N k +1 , if it is not isomorphic to L (2) (2) .Proof. We construct a ( k +1)-step linear deformation of L ( k ) ( m ) using Corol-lary 3.2.Let B be the Hall basis given in 4.5 associated to the set of generators { x , ..., x m } ordered by x < · · · < x m . Let a = x and a = c k = [[ . . . [[ x , x ] , x ] , . . . ] , x ] , and h = h B − { a , a }i . The subspace h is in fact a subalgebra, as it followsfrom 4.5(2)(ii). B k has at least 2 elements, as it follows from 4.5(2)(iii)and the fact that L ( k ) ( m ) L (2) (2). Then there exist y ∈ B k linearlyindependent with a ∈ B k . Since B k ⊆ Z ( L ( k ) ( m )), then y ∈ Z L ( k ) ( m ) ( h ).So that, by Corollary 3.2, we can consider the linear deformation of L ( k ) ( m )given by [ , ] t = [ , ] + t ( a ∧ a ⊗ y ) . If t = 0, [ a , a ] t = ty = 0, so that L k +1 t = h y i 6 = 0 and therefore L t is a( k + 1)-step nilpotent Lie algebra for all t = 0. (cid:3) The following fact is needed in our proof of the next theorem.
Remark . Given x i , . . . , x i k in X , consider the element λ k − ( x i , . . . , x i k ) = [[ . . . [[ x i , x i ] , x i ] . . . ] , x i k ] in L k ( m ). The set formed by these elements will be denoted by λ k − ( X k ).In general λ k − ( X k ) is not a linearly independent set, but it generates thehomogeneous component L k ( m ). This is clear for L ( m ) and L ( m ) andfollows inductively for L k ( m ), by the Jacobi identity. Theorem 4.3.
The k -step free nilpotent Lie algebra L ( k ) ( m ) is rigid in N k .Proof. According to Theorem 2.5, it suffices to prove that H k ( L ( k ) ( m )) =0. To this end given σ ∈ Z k ( L ( k ) ( m )), we construct a linear function f : L ( k ) ( m ) → L ( k ) ( m ) such that σ = d ( f ).We start by considering the linear function f : A ( m ) → L ( k ) ( m ) definedin the basis M ( m ) of A ( m ) recursively by f ( x ) = 0 , for all x ∈ X = M ( m ) f ( p (cid:5) q ) = [ f ( p ) , π k ( q )] + [ π k ( p ) , f ( q )] − σ ( π k ( p ) , π k ( q )) . A direct calculation shows that f satisfies, for all a, b, c ∈ A ( m ), f ( a (cid:5) b + b (cid:5) a ) = 0 ,f ( (cid:9) (cid:5) ( a, b, c )) = 0 . So that it induces a linear map f : L ( m ) = A ( m ) /I → L ( k ) ( m ). This mapsatisfies, for u, v ∈ L ( m ),(4.1) f ( λ ( u, v )) = [ f ( u ) , ¯ v ] + [¯ u, f ( v )] − σ (¯ u, ¯ v ) . Recall that ¯ : L ( m ) → L ( k ) ( m ) is the quotient projection. Moreover, wewill prove that f ( L ( m ) k +1 ) = 0 and therefore it induces a linear map f : L ( k ) ( m ) → L ( k ) ( m ) satisfying, for all u, v ∈ L ( k ) ( m ), f ([ u, v ]) = [ f ( u ) , v ] + [ u, f ( v )] − σ ( u, v ) . And thus σ = d ( f ).It remains to see that f ( L k +1 ( m )) = 0. To prove this, it suffices to showthat f ◦ λ i ( X i +1 ) = 0, for all i ≥ k , because λ i ( X i +1 ) generates L i +1 ( m )(see 4.2).That f ◦ λ i ( X i +1 ) = 0, for all i ≥ k is a consequence of the followingidentity of ( i + 1)-multilinear functions in X i +1 :(4.2) f ◦ λ i = − i − X j =0 µ i − − j ◦ σ ◦ µ j , where µ is the Lie bracket of L ( k ) ( m ).We prove this identity by induction. If i = 1 and ( x, y ) ∈ X , since x = x , y = y , and f ( x ) = 0 = f ( y ), by using (4.1) we have that f ◦ λ ( x, y ) = f ( λ ( x, y ))= [ f ( x ) , y ] + [ x, f ( y )] − σ ( x, y )= − X j =0 µ − j ◦ σ ◦ µ j ( x, y ) . If f ◦ λ i = − P i − j =0 µ i − − j ◦ σ ◦ µ j and x = ( x ′ , x ) ∈ X i +2 , for some x ′ ∈ X i +1 and x ∈ X , we have that f ◦ λ i +1 ( x ) = f ( λ ( λ i ( x ′ ) , x ))= [ f ( λ i ( x ′ )) , x ] + [ λ i ( x ′ ) , f ( x )] − σ ( λ i ( x ′ ) , x )= [ f ◦ λ i ( x ′ ) , x ] + σ ( µ i ( x ′ ) , x );the second identity follows from (4.1) and the third one follows because x = x , λ i ( x ′ ) = µ i ( x ′ ) = µ i ( x ′ ) and f ( x ) = 0. Now by the inductivehypothesis, we have that f ◦ λ i +1 ( x ) = (cid:2) − i − X j =0 µ i − − j ◦ σ ◦ µ j ( x ′ ) , x (cid:3) − σ ( µ i ( x ′ ) , x )= − i − X j =0 µ i − j ◦ σ ◦ µ j ( x ) − µ i − i ◦ σ ◦ µ i ( x )= − i X j =0 µ i − j ◦ σ ◦ µ j ( x ) . Finally, we show that f ◦ λ i = 0, for all i ≥ k . If i ≥ k , then i = k + c , forsome c ≥
0. So that f ◦ λ i = − i − X j =0 µ i − − j ◦ σ ◦ µ j = − k + c − X j =0 µ k + c − − j ◦ σ ◦ µ j = − k − X j =0 µ k + c − − j ◦ σ ◦ µ j − k + c − X j = k µ k + c − − j ◦ σ ◦ µ j = − µ c ◦ ( k − X j =0 µ k − − j ◦ σ ◦ µ j ) − k + c − X j = k µ k + c − − j ◦ σ ◦ µ j = 0 . The last identity holds because both terms are zero. The first one is zero,because σ ∈ H k − nil ( L ( k ) ( m )) and hence P k − j =0 µ k − − j ◦ σ ◦ µ j = 0. Thesecond one is equal to zero, because µ is k -step nilpotent Lie algebra andhence µ j = 0, for all j ≥ k . (cid:3) Heisenberg Lie algebras
The (2 m + 1)-dimensional Heisenberg Lie algebra is h m = V ⊕ Z = h x , y , . . . , x m , y m i ⊕ h z i , where the non-zero brackets of basis elements are[ x i , y i ] = z, for all 1 ≤ i ≤ m. h m is a 2-step nilpotent Lie algebra with center Z . Let { x ∗ , y ∗ , . . . , x ∗ m , y ∗ m , z ∗ } be the dual basis of the given one and for any x ∈ h m , denote by x V theunique element in V such that x = x V + z ∗ ( x ) z . Remark . Given x, y ∈ h m , ad x = ad y if and only if x V = y V or equiva-lently if there is v ∈ V such that x = v + z ∗ ( x ) z and y = v + z ∗ ( y ) z . Theorem 5.2.
The Heisenberg Lie algebra h m is non-rigid in N , for all m > .Proof. We construct a 3-step linear deformation of h m using Corollary 3.2.Let a = x , a = x , h = h B − { x , x } ) i and y = y . So that, we canconsider the linear deformation of h m [ , ] t = [ , ] + t ( a ∧ a ⊗ y ) , which is 3-step nilpotent, for all t = 0. (cid:3) Theorem 5.3.
The Heisenberg Lie algebra h m is rigid in N , for all m ∈ N .Proof. According to Theorem 2.5, it is enough to prove that H ( h m ) = 0.Given σ ∈ Z ( h m ), we prove that there exist a linear function f : h m → h m ,such that σ = d ( f ).Since [ , ] ◦ σ = − σ ◦ [ , ], then on the one hand(5.1) ad σ ( x i ,y i ) = − σ ( z, · ) = ad σ ( x j ,y j ) , for all i, j ∈ { , ..., m } , and hence there exist v ∈ V such that σ ( x i , y i ) = v + z ∗ ( σ ( x i , y i )) z, for all i = 1 , ..., m (see Remark 5.1). On the other hand(5.2) if [ x, y ] = 0 , then σ ( x, y ) ∈ Z. Now, d ( f ) = σ if and only if f satisfies the following system of linearequations: σ ( x i , y i ) =[ f ( x i ) , y i ] + [ x i , f ( y i )] − f ( z ) , for i = 1 , . . . , m ; σ ( x i , y j )=[ f ( x i ) , y j ] + [ x i , f ( y j )] , for 1 ≤ i, j ≤ m, i = j ; σ ( x i , x j )=[ f ( x i ) , x j ] + [ x i , f ( x j )] , for 1 ≤ i < j ≤ m ; σ ( y i , y j ) =[ f ( y i ) , y j ] + [ y i , f ( y j )] , for 1 ≤ i < j ≤ m ; σ ( z, x i ) =[ f ( z ) , x i ] , for i = 1 , . . . , m ; σ ( z, y i ) =[ f ( z ) , y i ] , for i = 1 , . . . , m. We start by defining f ( z ) = − v . Then the last two sets of equations aresatisfied because σ satisfies (5.1).To define f in V , we set that z ∗ ( f ( v )) = 0, for all v ∈ V , so that f ( x i ) = m X k =1 x ∗ k ( f ( x i )) x k + m X k =1 y ∗ k ( f ( x i )) y k ,f ( y i ) = m X k =1 x ∗ k ( f ( y i )) x k + m X k =1 y ∗ k ( f ( y i )) y k . It remains to determine a ik = x ∗ k ( f ( x i )), b ik = y ∗ k ( f ( x i )), c ik = x ∗ k ( f ( y i )) and d ik = y ∗ k ( f ( y i )), for k, i ∈ { , . . . , m } , which now must satisfy the system of linear equations σ ( x i , y i ) − v = ( a ii + d ii ) z, for i = 1 , . . . , m ; σ ( x i , y j ) = ( a ij + d ji ) z, for 1 ≤ i, j ≤ m, i = j ; σ ( x i , x j ) = ( − b ij + b ji ) z, for 1 ≤ i, j ≤ m ; σ ( y i , y j ) = ( c ij − c ji ) z, for 1 ≤ i, j ≤ m . Since this one is clearly consistent, we are done. (cid:3) Lie algebras with an abelian factor
In this section we explore the rigidity and non-rigidity of Lie algebraswith an abelian factor.Given a Lie algebra ( g , λ ) and an abelian Lie algebra ( a , µ ), the questionwe address in this section is whether is g ⊕ a rigid or not. Denote by λ ⊕ µ the bracket of g ⊕ a .The answer, which depends on the size of a and the framework variety, isin general no. However the situation for small abelian factors, in particularfor one dimensional factors is quite interesting.Given a Lie algebra ( g , λ ) and ( a l , µ ) the l -dimensional abelian Lie al-gebra, we observe that for any non-abelian Lie bracket ν , the non-trivialdeformation µ t = tν (recall (1.1)) gives rise to the non-trivial deformation( λ ⊕ µ ) t = λ ⊕ tν, of λ ⊕ µ . From this observation it follows that:(1) If l ≥
2, then for any g , g ⊕ a l is non-rigid in L .(2) If l ≥
2, then for any solvable s , s ⊕ a l is non-rigid in S . Moreover, if s is k -step solvable, s ⊕ a l is non-rigid in S k .(3) If l ≥
3, then for any nilpotent n , n ⊕ a l is non-rigid in N . Moreover, if n is k -step nilpotent, n ⊕ a l is non-rigid in N k .So that, the only cases remaining to consider are:(1’) g ⊕ a l ∈ L , where g is any and l = 1.(2’) s ⊕ a l ∈ S , where s is any solvable and l = 1. And its stronger versionfor s ⊕ a l ∈ S k .(3’) n ⊕ a l ∈ N , where n is any nilpotent and l = 1 or l = 2. And its strongerversion for n ⊕ a l ∈ N k .We will show that the answer for (2’) and for (3’) is the general one,namely they are not rigid, even in their strongest forms. The answer for (1’)turns out to be more intricate. Even though there is no a unified answerindependently of g , we shall see that being g perfect or not plays a role.Recall that g is called perfect if λ ( g , g ) = g . Example 6.1.
Given g a semisimple Lie algebra, let g = g ⊕ a . Then H ( g , g ) = 0 and hence g is rigid in L . The fact that the second adjointcohomology group of g vanishes, follows directly from the Hochshild-Serrespectral sequence associated to the ideal g of g and the fact that H ( g , g ) = H ( g , g ) = 0. We address now questions (1’), (2’) and (3’), where the abelian factor issmall, one at a time. For the sake of completeness, the statements are givenfor arbitrary abelian factors.
Notation . For convenience, we will denote by [ , ] the bracket of g ⊕ a l .6.1. Non-perfect Lie algebras.Theorem 6.3. If g is a non-perfect Lie algebra, then g ⊕ a l is non-rigid in L , for all l ∈ N .Proof. The only remaining case is l = 1. We construct a linear deformationof g ⊕ a in L , using Corollary 3.2.Given A = a any basis of a and B any basis of g , there exist b ∈ B suchthat b / ∈ [ g , g ]. Take a = a, a = b, h = h ( A ∪ B ) − { a, b }i and y = a . Sothat, from the Corollary 3.2, we can consider the linear deformation of g ⊕ a [ , ] t = [ , ] + t ( a ∧ a ⊗ y ) . This deformation is non-trivial, because the dimension of the commutatorcorresponding to t = 0 is larger than the dimension of the original one. (cid:3) Solvable Lie algebras.Theorem 6.4. If s ∈ S k , then s ⊕ a l is non-rigid in S k , for all l ∈ N .Proof. The only remaining case is l = 1. Since s is solvable, it is non-perfect.The deformation given in the proof of the Theorem 6.3 is a k -step solvablelinear deformation. (cid:3) Nilpotent Lie algebras.
In this section we complete the proof of thefact that a nilpotent Lie algebra plus an abelian factor is non-rigid, evenin the smaller subvariety N k , by considering the remaining cases where thedimension of the abelian factor is l = 2 or l = 1. It is worth saying thatthere is a single exception, namely h ⊕ a . Remark . There are only two, non isomorphic, nilpotent Lie algebras ofdimension 4 in N : a and h ⊕ a . Therefore, h ⊕ a is rigid in N . It isworth mentioning that h ⊕ a is not rigid in N = N .Nilpotent Lie algebras are non-perfect, so by Theorem 6.3, n ⊕ a l is non-rigid in L , for all l ∈ N . But to prove that n ⊕ a l is non-rigid in N k we mustwork harder, since the deformation constructed in the proof of that theoremis not nilpotent. Notation . We fix some notation for what follows. The bases for a and a will be A = { c } and A = { c , c } respectively. For n ∈ N k , we choose B k ⊆ ... ⊆ B such that B i is a basis of n i , the i -th term of the descendingcentral series of n , for all i = 1 , . . . , k . We denote, for i = 1 , . . . , k − B i − B i +1 = { x i , . . . , x in i } . Notice that B = B is a basis of n and B ∪ A isa basis of n ⊕ a l . Proposition 6.7. If n ∈ N k , then n ⊕ a is non-rigid in N k . Proof.
We construct a non-trivial linear deformation of n ⊕ a in N k usingCorollary 3.2.Take a = c , a = x , h = h ( A ∪ B ) − { c , x }i and y = c . So that wecan consider the linear deformation of n ⊕ a given by[ , ] t = [ , ] + t ( a ∧ a ⊗ y ) . It is easy to see that this deformation is k -step nilpotent. It is non-trivial,because the dimension of the commutator corresponding to t = 0 is largerthan the dimension of the original one. (cid:3) We come now to the most difficult case, that for l = 1. We look at the2-step nilpotent quotient ˜ n = nn ; recall that n = λ ( λ ( n , n ) , n ). And wesplit the proof into two propositions, according to whether this quotient isisomorphic to a free 2-step nilpotent Lie algebra or it is not.In general, given a nilpotent Lie algebra g with an adapted basis B as in6.6, by taking a , a ∈ B − B , h = h B − { a , a }i and y a central element,the hypotheses of our construction are trivially fulfilled. In the case weare dealing with, in which g = n ⊕ a , we may take y = c . Assumingthat ˜ n ≇ L (2) ( m ) we are able to prove that the resulting deformation isnon-trivial. In the case ˜ n ∼ = L (2) ( m ), we do something different. Proposition 6.8. If n ∈ N k and ˜ n ≇ L ( m ) , then n ⊕ a is non-rigid in N k .Proof. We construct a non-trivial linear deformation of n ⊕ a in N k , usingCorollary 3.2.Since n < ( n − ), the projection of the set { [ x j , x i ] : 1 ≤ i < j ≤ n } onto˜ n is a linearly dependent set. Hence, after relabeling if necessary, we mayassume that(6.1) [ x , x ] ∈ D(cid:16)(cid:8) [ x j , x i ] : 1 ≤ i < j ≤ n } − { [ x , x ] (cid:9)(cid:17) ∪ B E . Take a = x , a = x , h = h A ∪ ( B − { x , x }i and y = c . So that, fromCorollary 3.2, we can consider the linear deformation of n ⊕ a given by[ , ] t = [ , ] + t ( a ∧ a ⊗ y ) . It is easy to see that, this deformation is k -step nilpotent. Also it is non-trivial because the dimension of the commutator corresponding to t = 0 islarger than the dimension of the original one (6.1). (cid:3) For the last case, we shall assume without lost of generality, that n hasno abelian factor. In fact, if n has an abelian factor then n ⊕ a falls in thecase covered by Proposition 6.7. Proposition 6.9.
Let n ∈ N k be such that it has no abelian factor, n ≇ h and ˜ n ∼ = L (2) ( m ) . Then n ⊕ a is non-rigid in N k .Proof. We construct a non-trivial linear deformation of n ⊕ a in N k , usingCorollary 3.2.Consider the sets S and R , S = (cid:8) x ∈ n : [ x, n ] = 0 and dim([ x, n ]) ≤ (cid:9) ,R = (cid:8) r ∈ { , . . . , k } : S ∩ ( n r − n r +1 ) = ∅ (cid:9) . Let r = min R . Notice that if r = 1, then m = 2. If r ≥
2, there exist y = 0 such that y ∈ S ∩ n . Let us consider separately the cases r = 1and r ≥ Case 1: If r = 1, then ˜ n ∼ = L (2) (2). Since n ≇ L (2) (2) ≃ h , then k ≥ B = { x , x } ∪ { [ x , x ] } ∪ B , with x ∈ S . We take a = x , a = [ x , x ], h = h{ x } ∪ B i and y = c .So that, from the Corollary 3.2, we can consider the linear deformation of n ⊕ a [ , ] t = [ , ] + t ( a ∧ a ⊗ y ) . It is easy to see that this deformation is k -step nilpotent. Also it is non-trivial because dim(( n ⊕ a ) t ) = dim(( n ⊕ a ) ) + 1, for t = 0. Case 2: If r ≥
2, let 0 = y ∈ S ∩ ( n r − n r +1 ). Choose B such that[ y , b ] = 0 , for all b ∈ B − { x } , and take a = x , a = c , h = h B − { x }i and y = y . So that, fromCorollary 3.2, we can consider the linear deformation of n ⊕ a given by[ , ] t = [ , ] + t ( a ∧ a ⊗ y ) . This deformation is k -step nilpotent, because y ∈ n .In order to prove that this deformation is non-trivial, assume instead thatfor arbitrary small t , ( n ⊕ a ) t ∼ = n ⊕ a . Hence ( n ⊕ a ) t has an abelian factor h z i . Then [ z, b ] t = 0 , for all b ∈ B ;(6.2) z / ∈ ( n ⊕ a ) t = n . (6.3)Writing z = z n + αc , (6.2) implies that[ z n , b ] = 0 , for all b ∈ B − { x } ;(6.4) [ z n , x ] = [ z n , x ] t = tαy . (6.5)On the one hand, if α = 0, then z n ∈ Z ( n ) and z n / ∈ n (6.3) and therefore z n is an abelian factor of n . On the other hand, if α = 0, z n ∈ S (6.4) andthen by (6.5) there exist r ∈ R with r < r . (cid:3) Summarizing all we have proved, it follows that k -step nilpotent Lie al-gebras with an abelian factor are never k -rigid, except for h ⊕ a . Theorem 6.10. If n ∈ N k , then n ⊕ a l is rigid in N k if and only if n ∼ = h and l = 1 . The exceptional case.
The only exceptional case for which there is no a unified answer onwhether g ⊕ a l is rigid or not in L , is that for g a perfect Lie algebra and l = 1. Example 6.1 shows that the answer might be “rigid”. The followingexample shows that the answer might be “non rigid”. Example 6.11.
Let g be the 5-dimensional Lie with basis { a, b, c, d, e } andbracket defined by: λ ( a, b ) = 2 b, λ ( a, c ) = − c, λ ( b, c ) = a,λ ( a, d ) = d, λ ( a, e ) = − e, λ ( b, e ) = d, λ ( c, d ) = e. Take { f } as a basis for a and let g = g ⊕ a . Denote its bracket by [ , ].Consider the linear deformation of g given by[ , ] t = [ , ] + tϕ, where ϕ is the 2-cocycle ϕ := d ∗ ∧ e ∗ ⊗ f. Then [ , ] t is given by[ a, b ] t = 2 b, [ a, c ] t = − c, [ b, c ] t = a, [ a, d ] t = d, [ a, e ] t = − e, [ b, e ] t = d, [ c, d ] t = e, [ d, e ] t = tf. Clearly g t is perfect, for every t = 0, so that it has no abelian factor and thedeformation is non-trivial. So that g is non-rigid. Acknowledgements.
This paper is part of the PhD thesis of Josefina Bar-rionuevo, being carried out thanks to a Doctoral Fellowship from CONICET,Argentina. It was finished while Paulo Tirao was on a sabbatical year fromthe University of C´ordoba as Visiting Professor at the Technion - Israel In-stitute of Technology in Haifa, Israel and the Guangdong Technion - IsraelInstitute of Technology in Shantou, China.
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