Semigenerated Carnot algebras and applications to sub-Riemannian perimeter

This is a self-archived version of an original article. This version may differ from the original in pagination and typographic details. Author(s): Title: Year: Version: Copyright: Rights: Rights url: Please cite the original version: CC BY 4.0 https://creativecommons.org/licenses/by/4.0/ Semigenerated Carnot algebras and applications to sub-Riemannian perimeter © The Author(s) 2021 Published version Le Donne, Enrico; Moisala, Terhi Le Donne, E., & Moisala, T. (2021). Semigenerated Carnot algebras and applications to sub- Riemannian perimeter. Mathematische Zeitschrift, 299(3-4), 2257-2285. https://doi.org/10.1007/s00209-021-02744-4 2021 Mathematische Zeitschrift https://doi.org/10.1007/s00209-021-02744-4 Mathematische Zeitschrift Semigenerated Carnot algebras and applications to sub-Riemannian perimeter Enrico Le Donne1,2 ·Terhi Moisala2 Received: 30 May 2020 / Accepted: 1 March 2021 © The Author(s) 2021 Abstract This paper contributes to the study of sets of finite intrinsic perimeter in Carnot groups. Our intent is to characterize in which groups the only sets with constant intrinsic normal are the vertical half-spaces. Our viewpoint is algebraic: such a phenomenon happens if and only if the semigroup generated by each horizontal half-space is a vertical half-space. We call semigenerated those Carnot groups with this property. For Carnot groups of nilpotency step 3 we provide a complete characterization of semigeneration in terms of whether such groups do not have any Engel-type quotients. Engel-type groups, which are introduced here, are the minimal (in terms of quotients) counterexamples. In addition, we give some sufficient criteria for semigeneration of Carnot groups of arbitrary step. For doing this, we define a new class of Carnot groups, which we call type (♦)and which generalizes the previous notion of type () defined by M. Marchi. As an application, we get that in type (♦)groups and in step 3 groups that do not have any Engel-type algebra as a quotient, one achieves a strong rectifiability result for sets of finite perimeter in the sense of Franchi, Serapioni, and Serra-Cassano. Keywords Carnot algebra ·Horizontal half-space ·Semigroup generated ·Lie wedge · Constant intrinsic normal ·Finite sub-Riemannian perimeter ·Engel-type algebras ·Tipe diamond ·Trimmed algebra Mathematics Subject Classification 22E15 ·53C17 ·22A15 ·22E25 ·28A75 ·49Q15 · 22A15 ·22E15 E.L.D. was partially supported by the Academy of Finland (grant 288501 ‘Geometry of subRiemannian groups’ and by grant 322898 ‘Sub-Riemannian Geometry via Metric-geometry and Lie-group Theory’) and by the European Research Council (ERC Starting Grant 713998 GeoMeG ‘Geometry of Metric Groups’). BEnrico Le Donne [email protected] Terhi Moisala [email protected] 1Dipartimento di Matematica, Università di Pisa, Largo B. Pontecorvo 5, 56127 Pisa, Italy 2Department of Mathematics and Statistics, University of Jyväskylä, P.O. Box 35 (MaD), 40014 Jyväskylä, Finland 123 E. Le Donne, T. Moisala Contents 1 Introduction ................................................ 2 Preliminaries ................................................ 2.1 Lemmata in arbitrary algebras .................................... 3 Sufficient criteria for semigeneration ................................... 3.1 Carnot groups of type (♦)....................................... 4 Some results and examples in low-step algebras ............................. 5 Engel-type algebras ............................................ 5.1 Definition and properties ....................................... 5.2 Proof of Theorem 1.2 ......................................... 5.2.1 Proof of Theorem 1.2 ...................................... References ................................................... 1 Introduction Carnot groups, which are by definition simply connected Lie groups with stratified Lie algebras, raised attention because of their natural occurrences in Geometric Measure Theory and Metric Geometry. In particular, subsets of Carnot groups whose intrinsic normal is constantly equal to a left-invariant vector field appear both in the development of a theory à la De Giorgi for sets of locally finite perimeter in sub-Riemannian spaces [1,2,8,9]andin the obstruction results for bi-Lipschitz embeddings into L1of non-abelian nilpotent groups [6]. The work [9] by Franchi, Serapioni and Serra-Cassano provides complete understanding of sets with constant intrinsic normal in the case of Carnot groups with nilpotency step 2 by proving that they are half-spaces when read in exponential coordinates. However, in higher step the study appears to be much more challenging due to the more complex underlying algebraic structure, and only in the case of type () groups and of filiform groups we have a satisfactory understanding of sets with constant intrinsic normal, see [4,15]. In a recent paper [5], Bellettini and the first-named author of this article related the property of having constant intrinsic normal to the containment of distinguished constant-normal sets, which are semigroups generated by the horizontal half-space defined by the normal, as we shall explain soon. We shall use the following terminology: a horizontal half-space of a stratified algebra gwith horizontal layer V1is the closure of either of the two parts into which a hyperplane divides V1.Avertical half-space is defined as the direct sum of a horizontal halfspace and the derived subalgebra [g,g].By[5, Corollary 2.31], in exponential coordinates a Carnot group has the property that all its constant-normal sets are equivalent to vertical halfspaces if and only if the closure of the semigroup generated by each horizontal half-space is a vertical half-space. In Carnot groups with this property, one has the intrinsic C1-rectifiability result for finite-perimeter sets à la De Giorgi. In arbitrary groups, the study of semigroups can still give some weaker rectifiability results, see [7]. In this paper, we continue the study of such semigroups from an algebraic viewpoint. In particular, we get to a complete characterization of those step-3 Carnot groups for which all constant-normal sets are vertical half-spaces. In addition, for Carnot groups of arbitrary nilpotency step, we give some sufficient criteria which generalize the previous work by Marchi [15]. Definition 1.1 Given a Carnot group Gwith exponential map exp :g→G, we say that a set W⊆gis semigenerating if the closure of the semigroup generated by exp(W)in Gcontains the commutator subgroup [G,G]. We say also that the Lie algebra gis semigenerated if every horizontal half-space Win gis semigenerating. 123 Semigenerated Carnot algebras and applications... We shall use the term Carnot algebra to denote the (stratified) Lie algebra of a Carnot group, which is completely determined by the Lie group, see [13]. By the work [9] of Franchi, Serapioni and Serra-Cassano, we know that step-2 Carnot algebras are semigenerated. Their work has been then extended by Marchi to a class of Carnot algebras, called of type (), which includes examples of arbitrarily large nilpotency step. However, the basic example given by the Engel Lie algebra is not semigenerated, see [4,9], and also Proposition 5.10. From this example, it is easy to generate more examples of non-semigenerated algebras, because of the observation that each quotient of a semigenerated Lie algebra is semigenerated, see Proposition 2.7. Thus, for example we have that no stratified Lie algebra of rank 2 and step ≥3 is semigenerated because each of them has the Engel Lie algebra as quotient as pointed out in Remark 2.8. Here, we mostly focus on step-3 Lie algebras, in which we discover a class of Lie algebras that are not semigenerated. Since they are a generalization of the Engel Lie algebra we call them Engel-type algebras. Our main result is that these algebras are the only obstruction to semigeneration. Theorem 1.2 Let gbe a stratified Lie algebra of step at most 3. Then gis not semigenerated if and only if it has one of the Engel-type algebras (as in Definition 1.3) as a quotient. Definition 1.3 For each n∈N,wecalln-th Engel-type algebra the 2(n+1)-dimensional Lie algebra (of step 3 and rank n+1) with basis {X,Yi,Ti,Z}n i=1, where the only non-trivial bracket relations are given by [Yi,X]=Tiand [Yi,Ti]=Zfor all i∈{1,...,n}. It is a challenge to understand how one can express in pure combinatorial terms the property of not having any Engel-type algebra as quotient. However, we have examples of step-3 Lie algebras that are not of type () but have no Engel-type quotients. Hence, our result is a strict generalization of [15]. It is possible that Theorem 1.2 holds also in case the nilpotency step is arbitrary; we have no counterexample. However, the situation in step greater than 3 is more technical. For this reason, we are only able to give a sufficient condition to ensure semigeneration in arbitrary step. Such criterion is not necessary (see Example 4.7); however, as for the type () condition, it is computable in terms of brackets of some particular basis. In the next result, we assume the existence of a basis with specific properties. We could restate the condition in other forms (see Lemma 3.1), which alas are just as technical. Definition 1.4 Let gbe a stratified Lie algebra. If, for each subalgebra hof gfor which h∩V1 has codimension 1 in V1, there exists a basis {X1,...,Xm}of V1such that ad2 XiXj∈hand ad2 adk XiXj(Xi)∈h,for all i,j=1,...,mand k≥2, then we say that gis of type (♦). Theorem 1.5 Every stratified Lie algebra that is of type (♦)(as in Definition 1.4) is semigenerated. To put the results in perspective, we remind the reader that by [9], we know that if a Carnot group has the property that every set with constant intrinsic normal is a vertical half-space, then every set of locally finite sub-Riemannian perimeter have a strong rectifiability property. Since semigroups generated by horizontal half-spaces are minimal constant-normal sets with respect to set inclusion according to [5], we obtain the following corollary. Corollary 1.6 If the Lie algebra of a Carnot group G is semigenerated (e.g., if it is of type (♦), see Theorem 1.5, or has step 3 and does not have any Engel-type algebra as a quotient, 123 E. Le Donne, T. Moisala see Theorem 1.2), then the reduced boundary of every set of locally finite perimeter in G is intrinsically C1-rectifiable. The structure of the article is the following. In Sect. 2we discuss some preliminaries. In addition to the notions of semigenerated and trimmed algebras, we introduce a useful set called the edge of a semigroup. In Sect. 3we analyze Lie algebras of type (♦)and prove Theorem 1.5, see Corollary 3.7. Section 4is devoted to both a list of examples and of results valid for Carnot algebras of step at most 4. In Sect. 5we study the Engel-type algebras. We show that they are the only non-semigenerated Carnot algebras with step 3 that are minimal with respect to quotient in a sense that will be made precise with a notion that we call trimmed (see Definition 2.10). We end with the proof of Theorem 1.2. 2 Preliminaries We start with a small list of notations. Then, in Definition 2.1, we define the edge esand the wedge wsof a semigroup s. The notion of edge will be in the core of the discussion, since understanding if a horizontal half-space is semigenerating reduces to calculating the edge of its generated semigroup. We provide several preliminary results regarding the size of such edges. In particular, we consider Lemma 2.3 extremely useful and we shall exploit it repeatedly. In Proposition 2.9 we provide equivalent conditions for the definition of trimmed algebra, a notion that is fundamental in our arguments in Sect. 5. In this paper, the Lie algebra gwill always be stratified with layers Vi=Vi(g).Wedenote by Z(g)the center of a Lie algebra g. Given an ideal iof gwe denote by π=πi:g→g/i the quotient map, and we shall interchangeably use the equivalent notations A/i=A+i= πi(A), for subsets Aof g. We denote for a subset Aof gby Ig(A)the ideal generated by A within g,byLie(A)the Lie algebra generated by A, and by Cl(A)or ¯ Athe closure of Ain g. We say that Wis a horizontal half-space of gif there exists a non-zero element λin the dual of V1such that W=λ−1([0,+∞)) ⊆V1.(2.1) If Wis a horizontal half-space defined by λ∈(V1)∗as in (2.1), then we define its (horizontal) boundary as ∂W:=λ−1({0}). Notice that Wis a closed subset of V1and ∂Wis its boundary within V1, which in our case will always contain 0 ∈g. Observe that ∂Wis a hyperplane in V1. Given a subset Wof g, which we shall usually assume to be a horizontal half-space, the semigroup SWgenerated by exp(W)is described as SW= ∞  k=1 (exp(W))k,(2.2) where (exp(W))k:={k i=1exp(wi)|w1,...,w k∈W}. Be aware that, even when Wis closed (within V1), the set SWmay not be closed within exp(g). A vector space hof a stratified Lie algebra gis said to be homogeneous if there exist subspaces hiof Vi(g)such that h=h1⊕...⊕hs.Equivalently,wehavethathis homogeneous 123 Semigenerated Carnot algebras and applications... if and only if δλh=hfor all λ>0, where δλis the Lie algebra automorphism such that δλ(v) =λv for v∈V1(g). We shall frequently use the fact that the center of a stratified Lie algebra is homogeneous: Z(g)=V1(g)∩Z(g)⊕...⊕Vs(g)∩Z(g).(2.3) 2.1 Lemmata in arbitrary algebras In this subsection, let gbe a Lie algebra of a simply connected Lie group G. We assume that gis stratified with nilpotency step equal to s.SinceGis consequently nilpotent and simply connected, the exponential map exp :g→Gis a bijection. We then have a correspondence between subsets s⊆gand subsets S=exp(s)⊆G. Definition 2.1 We associate with every subset s⊆gthe following two sets ws:= {X∈g:R+X⊆s}.(2.4) es:= ws∩(−ws)=ws∩w−s.(2.5) The set wsis known as the tangent wedge of sand esas the edge of the wedge ws,see[12, Page 2 and page 19]. For typographical reasons, we sometimes write e(s)instead of es.An equivalent definition for esis es={X∈g:RX⊆s}.(2.6) Regarding the next result, we claim very little originality. The arguments are mostly taken from [5,12]. Also, the notions of cone and convexity that we shall use are the usual ones with respect to the vector-space structure of the Lie algebra. Lemma 2.2 Let G be a Lie group whose exponential map exp :g→G is injective. Let s⊆gbe such that exp(s)is a semigroup. Then the sets esand ws, defined in (2.6)and (2.4), respectively, satisfy the following properties: (1) wsis the largest cone in s; (2) esis the largest subalgebra of gcontained in s; (3) for each X ∈s∩(−s), we have that s,ws, and esare invariant under eadX, i.e., eadXs=s,for all X such that ±X∈s,(2.7) eadXws=ws,for all X such that ±X∈s,(2.8) eadXes=es,for all X such that ±X∈s;(2.9) (4) if exp(s)is closed, then wsis closed and convex. Proof Point (1) is immediate from the definition. Regarding (2), to see that esis a Lie algebra, let hbe the Lie algebra generated by es.Let ˆ Sbe the semigroup generated by exp(es).Since esLie generates h,thenby[3, Theorem 8.1] the set ˆ Shas nonempty interior in exp(h). Since esis symmetric, then ˆ Sis closed under inversion, hence a group. Being a group with nonempty interior, ˆ Sis an open subgroup of exp(h).Since ˆ Sis an open subgroup of the connected group exp(h), then exp(h)equals ˆ S, which is a subset of exp(s). Therefore, his a subset of s, being exp injective. Since in addition his symmetric, we infer that h⊆es,which tells us that esis a subalgebra of gcontained in s. It is the largest since, if a Lie algebra his contained in s, then from Rh=hwe deduce that h⊆es. 123 E. Le Donne, T. Moisala To prove (2.7), take Xsuch that ±X∈s, so that for all Y∈swe have exp(eadXY)=exp(Adexp(X)Y) =exp((Cexp(X))∗Y) =Cexp(X)(exp(Y)) =exp(X)exp(Y)exp(−X)∈S, where we have used that ad is the differential of Ad, that Adgis the differential of Cg,that exp intertwines this differential with Cgand, finally, that Sis a semigroup. Hence, we have proved that eadXs=s. Consequently, since the map eadXis linear, it sends half-lines to half-lines and lines to lines. Thus, we have (2.8)and(2.9). We now prove (4). If exp(s)is closed, then also sis closed since exp is continuous and injective. Then the closure of wsis a cone in s. By maximality of ws, we deduce that wsis closed. Since wsis a cone, to check that wsis convex it is enough to show that X+Y∈s for all X,Y∈ws. Indeed, noticing that also R+X,R+Y⊆ws, this would imply that R+(X+Y)⊆sand so X+Y∈ws. To prove that X+Y∈sfor every X,Y∈ws, recall the formula, which holds in all Lie groups, exp(X+Y)=lim n→∞ exp 1 nXexp 1 nYn.(2.10) Set S=exp(s).SinceR+X,R+Y⊆s, then exp(1 nX), exp(1 nY)∈S,foralln∈N. Consequently, since Sis a semigroup, we have exp(1 nX)exp(1 nY)n∈S.BeingSclosed by assumption, we get from (2.10) that exp(X+Y)∈S. Since exp is injective, we infer that X+Y∈s. So the convexity of wsis proved.  We prove next a useful lemma, which states that if R+X⊆s,RY⊆s,and Rad2 YX⊆s then also R[X,Y]⊆s. Recall the notions of esand ws,definedin(2.6)and(2.4). Lemma 2.3 Let gbe a stratified Lie algebra. Let s⊆gbe a subset such that exp(s)is a closed semigroup. If X ∈wsand Y ∈esare such that ad2 YX∈es,then[X,Y]∈es. Proof One the one hand, since esis a Lie algebra by Lemma 2.2.(2) we have adk YX∈es, for all k≥2. On the other hand, from (2.8)wehavethateadtY X∈ws,forallt∈R. Hence, since wsis convex by Lemma 2.2.(4), for all t∈Rwe have X+t[Y,X]=eadtY X− k≥2 tk k!adk Y(X)∈ws. Hence 1 |t|(X+t[Y,X])∈ws,forallt∈R. Therefore, taking tto ±∞ and using that wsis closed and convex by Lemma 2.2.(4), we get [Y,X]∈es. In the rest of the paper, we focus on semigroups generated by horizontal half-spaces in stratified Lie algebras. For every horizontal half-space W,see(2.1), in a stratified Lie algebra g,wedenotebySWthe semigroup generated by exp(W)in exp(g),see(2.2), and by sW⊂g the set such that exp(sW)=SW,i.e., sW:= log(SW). (2.11) If s:=sW, we stress the following two immediate facts: for every X∈V1,either X∈wsor −X∈ws;(2.12) ∂W=es∩V1.(2.13) 123 Semigenerated Carnot algebras and applications... The semigeneration condition stated in the introduction can equivalently be defined as follows: A set Win gis semigenerating if [g,g]⊆Cl(sW), (2.14) and we say that gis semigenerated if every horizontal half-space Win gis semigenerating. Observe that, by (2.6)asetW⊆gis semigenerating if and only if [g,g]⊆esfor s=Cl(sW). We will exploit this fact several times. Remark 2.4 For a horizontal half-space W⊆gand for sequal to sWor Cl(sW),wehave that esis a homogeneous subalgebra of gcontained in s.Indeed,inLemma2.2.(2) we already proved everything except the homogeneity. In such a case, for all λ>0wehave that δλW=Wand, hence, δλs=s. Thus we infer that δλRX⊆sif and only if RX⊆s. Therefore esis homogeneous. Lemma 2.5 Let gbe a stratified Lie algebra and W ⊆ga horizontal half-space. Then the set s=Cl(sW)has the following two properties: X,Y∈V1with ad2 XY=ad2 YX=0implies [X,Y]∈es;(2.15) V2∩Z(g)⊆es.(2.16) Proof Regarding (2.15), we have, up to changing signs, that X,Y∈W. Moreover, since ∂Wis a codimension 1 subspace of V1, we have that, up to possibly swapping Xwith Y, there exists some a∈Rfor which Z:=Y−aX ∈∂W. Thus, we have Z∈∂W⊆esand X∈W⊆ws. Moreover, by the assumptions on Xand Y,wehavethat ad2 ZX=[Y−aX,[Y,X]]=ad2 YX+aad2 XY=0∈es. By Lemma 2.3, we obtain es[X,Z]=[X,Y]. Regarding (2.16), we choose {X1,...,Xm}to be a basis of V1such that X1∈W⊆ws and X2,...,Xm∈∂W⊆es.TakeZ∈V2∩Z(g)and express it, for some ai,bij ∈R,as Z= i≥2 ai[Xi,X1]+  i,j≥2 bij[Xi,Xj]=[Y,X1]+ ˜ Y,(2.17) where Y:=  i≥2 aiXiand ˜ Y:=  i,j≥2 bij[Xi,Xj]. Since esis a Lie algebra by Lemma 2.2.2, we have that the elements Y,˜ Y,[Y,˜ Y]belong to es.SinceZ∈Z(g), we get also 0=[Y,Z]=ad2 YX1+[Y,˜ Y], which implies that ad2 YX1∈es.SinceX1∈ws,Y∈es,andad 2 YX1∈esLemma 2.3 tells us that [Y,X1]∈es. Going back to (2.17), we finally infer that Z∈es, again because esis a Lie algebra by Lemma 2.2.2.  For the next lemma, recall that πi:g→g/iis the quotient map modulo an ideal i.We also recall the basic fact that in the Lie algebra gof a simply connected nilpotent Lie group G,asubseti⊆gis an ideal if and only if N:= exp(i)is a normal Lie subgroup of G;inthis case, the quotient g/iis canonically isomorphic to the Lie algebra of G/Nand we have the following commutative diagram: 123 E. Le Donne, T. Moisala gg/i GG/N. exp πi exp πN Moreover, if gis stratified, then g/icanonically admits a stratification if and only if iis homogeneous. We stress that we have the following fact for each subset W⊆gof a Lie algebra g: πi(sW)=sπi(W).(2.18) Indeed, setting N:= exp(i)anddenotingbyS(A)the semigroup generated by A,we need to show that πN(S(exp(W))) =S(πNexp(W)). In fact, on the one hand, since the homomorphic image of a semigroup is a semigroup, we have that πN(S(exp(W))) is a semigroup containing πN(exp(W)),soS(πNexp(W)) ⊆πN(S(exp(W))). On the other hand, the set πN(S(exp(W))) =S(exp(W))Nis contained in the semigroup generated by exp(W)N=πN(exp(W)), i.e., we have πN(S(exp(W))) ⊆S(πNexp(W)). Lemma 2.6 Let ibe a homogeneous ideal of a stratified Lie algebra gand let W ⊆g. (i) If W is semigenerating, then πi(W)is semigenerating. (ii) If i⊆Cl(sW)and πi(W)is semigenerating, then W is semigenerating. Proof Assume first that Wis semigenerating. Then from (2.18) we obtain that πi(W)is semigenerating by the following calculation: [πi(g), πi(g)]=πi([g,g]) (2.14) ⊆πi(Cl(sW)) ⊆Cl(πi(sW)) (2.18) =Cl(sπi(W)). Suppose then that πi(W)is semigenerating and that i⊆Cl(sW).Thenwealsohavethe containment N:= exp(i)⊆Cl(SW).SinceCl(SW)is a semigroup, we have Cl(SW)·N=Cl(SW). (2.19) Therefore from (2.18)weget Sπi(W)=exp(sπi(W))(2.18) =exp(πi(sW)) =πN(exp(sW)) =πN(SW). (2.20) Taking the closure and the preimage under πN, from the fact that πNis an open map (and hence π−1 Nand Cl commute) and from (2.19), we get that π−1 NCl(Sπi(W))(2.20) =π−1 NCl(πN(SW)) =Cl(SW·N)=Cl(SW)·N(2.19) =Cl(SW). (2.21) Consequently, taking the logarithm, π−1 iCl(sπi(W))=log(π−1 NCl(Sπi(W))) (2.21) =log Cl(SW)=Cl(sW). (2.22) Hence, since πi(W)is semigenerating, we infer [g,g]⊆[g,g]+i=π−1 i[g/i,g/i]⊆π−1 iCl(sπi(W))(2.22) =Cl(sW), proving that Wis semigenerating.  We keep reminding that a quotient algebra g/iof a Carnot algebra gis Carnot if and only if the ideal iis homogeneous. In such a case, we say that g/iis a Carnot quotient of g. 123 Semigenerated Carnot algebras and applications... Lemma 3.8 Let n ∈Nand let glbe a stratified Lie algebra for each l ∈{1,...,n}.Let g:= n l=1glwith projections πl:g→gland fix a basis {Xl 1,...,Xl ml}for each gl.Ifiis a homogeneous ideal of gsuch that ad2 Xl i Xl j∈πl(i)and ad2 adk Xl i Xl j (Xl i)∈πl(i), (3.6) for all i,j∈{1,...,ml},k≥2and l ∈{1,...,n},theng/iis of type (♦). Proof Let hbe a subalgebra of g/ifor which h∩V1(g/i)has codimension 1 in V1(g/i)and, denoting by πthe projection π:g→g/i,let hbe a subalgebra of gfor which π(  h)=h. Notice first that the set {π(Xl 1),...,π(Xl ml)}n l=1spans V1(g/i).Taking(3.1) into account, since [Xl i,Xk j]=0 whenever l= k, to prove semigeneration of g/iit is enough to check that ad2 π(Xl i)π(Xl j)=π(ad2 Xl i Xl j)∈hand ad2 adk π(Xl i)π(Xl j)π(Xl i)=π(ad2 adk Xl i Xl j (Xl i)) ∈h, (3.7) for all i,j∈{1,...,ml},k≥2andl∈{1,...,n}. To do this, let ·,· be a scalar product on V1(g)that makes the basis {Xl 1,...,Xl ml}n l=1 orthonormal. Let ν∈V1(g)\{0}be a vector that is orthogonal to  h∩V1(g), i.e., let ν∈  h⊥∩V1(g). Write νas ν= n  l=1 ml  i=1 al iXl i,al i∈R. Without loss of generality, assume that a1 1=1. Then, for every l=2,...,nand i= 1,...,mlwe have that Yl i=Xl i−al iX1 1∈ h, as now Yl i,ν=0. Since X1 1commutes with every glfor which l∈{2,...,n}, we immediately deduce that  h⊃Lie({Yl 1,...,Yl ml}n l=2)∩[g,g]=Lie({Xl 1,...,Xl ml}n l=2)∩[g,g]= n  l=2 [gl,gl]. This proves (3.7)foralli,j∈{1,...,ml},k≥2andl∈{2,...,n}.Itisthenlefttoshow that each term in (3.6) with l=1 is projected to h.LetZbe such a term. By (3.6), we have Z∈π1(i).Sinceiis homogeneous and Z∈[g1,g1], there exists some  Z∈n l=2[gl,gl] such that Z+ Z∈i. But since n l=2[gl,gl]⊆ h,wehavethatZ∈ h+iand therefore π(Z)∈h. 4 Some results and examples in low-step algebras In the following section we collect some lemmata that are valid in Carnot algebras of step at most 4 and which will be used later in Section 5. However, these lemmata can also be useful when proving semigeneration of specific examples in low step. In the end of this section we provide two examples in step 3 that show that, on the one hand, algebras of type (♦)form a strictly larger class than algebras of type () and, on the other hand, that yet being of type (♦)is not a necessary condition for semigeneration. The following result gives, for step ≤4, equivalent conditions for being a semigenerating horizontal half-space. 123 E. Le Donne, T. Moisala Lemma 4.1 Let gbe a stratified Lie algebra of step at most 4. For each horizontal half-space Wing, writing s=Cl(sW), the following are equivalent: (i) V2⊆es; (ii) ad2 YX∈esfor every X ∈ws∩V1and Y ∈es∩V1; (iii) ad2 YX∈es, for every X,Y∈V1; (iv) V3⊆es; (v) W is semigenerating. Proof Implications (v) ⇒ (iv) ⇒ (iii) ⇒ (ii) are immediate. Regarding (ii) ⇒ (i), recall that V2is spanned by elements of the form [Y,Y]and [Y,X],whereY,Y∈es∩V1 and X∈ws∩V1. Since, by Lemma 2.2.2, esis a Lie algebra, each term [Y,Y]∈esand, by Lemma 2.3, the terms [X,Y]belong to es. Let us finally prove (i) ⇒ (v). We claim that it is enough to show that adk XY∈esfor every X∈V1,Y∈esand k=1,2,3. Indeed, then by Lemma 2.12 we have [g,g]⊆e(s)⊆s and Wis semigenerating. Now ad1 XY=[X,Y]∈esby (i). Then, as X∈V1⊆ws∪(−ws) and ad2 [X,Y]X∈V5={0}⊆e(s), by Lemma 2.3 we have [[X,Y],X]=−ad2 XY∈es. Similarly, ad2 ad2 XYX=0 and therefore [ad2 XY,X]=ad3 XY∈es. So (v) follows.  Remark 4.2 Let us observe what happens to condition (3.4) in low step. Given k≥2, the vector ad2 adk XY(X)is in V2k+3. Hence, if gis of step s, it is enough to require the conditions (3.1)or(3.2)forallk≤(s−3)/2. In particular, if s≤6, then a horizontal half-space Wof gis semigenerated if there exists a basis {X1,...,Xm}of V1such that ad2 XiXj∈Lie(∂W) for all i,j=1,...,m. Here, we denote by Lie(∂W)the Lie subalgebra of ggenerated by the subset ∂W. Similarly to Lemma 2.3, the following lemma gives (in step at most 4) a method to deduce new directions that are contained in the edge of a semigroup generated by a horizontal half-space. Lemma 4.3 below will be used in Example 4.7 and again in the proof of Proposition 5.12. Lemma 4.3 Let gbe a stratified Lie algebra of step at most 4 and let W be a horizontal half-space in g.Lets:=Cl(sW).IfZ∈V2∩es,thenIg(Z)⊆es. Proof Observe that V1=RX⊕∂Wfor some X∈W⊆ws. The ideal i:=Ig(Z)is graded and, recalling that Z∈V2,wehavethatitslayersare V1(i)={0},V2(i)=RZ,V3(i)=span{[X,Z],[∂W,Z]}. V4(i)=span{[X,[X,Z]],[X,[∂W,Z]],[∂W,[X,Z]],[∂W,[∂W,Z]]}. We plan to show that Ig(Z)⊆es, where we recall that esis a Lie algebra by Lemma 2.2. On the one hand, by assumption, we have that Z∈es, so from ∂W⊆eswe get that [∂W,Z]∈es. On the other hand, with the aim of applying Lemma 2.3, we observe that, since Z∈V2,wehavead 2 ZX∈V5={0}∈es, and hence we also have [X,Z]∈es. Hence V3(i)⊆es. We also check that V4(i)⊆es.Sinceesis closed under bracket, we immediately have that [∂W,[X,Z]],[∂W,[∂W,Z]⊆es. Regarding [X,[X,Z]],[X,[∂W,Z]],we repeat the previous part of the argument of this proof with Z∈{[X,Z]} ∪ [∂W,Z].Indeed,wehave that ad2 ZX=0andZ∈es. Hence, by Lemma 2.3 we also have [X,Z]∈es. 123 Semigenerated Carnot algebras and applications... Next we prove that having a sufficiently large semigenerated subalgebra implies semigeneration. We shall exploit this fact later in the proof of Proposition 5.12. Lemma 4.4 Let gbe a stratified Lie algebra of step at most 4. If ghas a semigenerated proper subalgebra hsuch that V3(g)⊆h,thengis semigenerated. Proof Let Wbe a horizontal half-space and let us show that it is semigenerating. Set H:=V1(h).IfH⊂∂W,then V3(g)⊆h⊆Lie(∂W)⊆e(sW). Consequently, by Lemma 4.1 we deduce that Wis semigenerating. We then assume that H∂W. Observe that  W:=H∩Wis a horizontal half-space in H. Hence, since by assumption his semigenerated,  Wis semigenerating within h. In particular, denoting by ¯ sh  W the closure of the (log of the) semigroup generated by  Wwithin h,wehavethatV3(h)⊆¯ sh  W. Since hisassumedtocontainthethirdlayerofg, we get the inclusions V3(g)⊆V3(h)⊂¯ sh  W⊆¯ sW, where the last containment is a consequence of the inclusions h⊂gand  W⊂W.Since V3(g)is a vector subspace of g, then by definition of e(¯ sW)we have V3(g)⊂e(¯ sW). Hence Wis semigenerating again by Lemma 4.1. We remark that, actually, the above Lemma 4.4 has the following analogue in algebras of arbitrary step: if his a semigenerated subalgebra of gand there exists a basis {X1,...,Xm} of V1(g)such that the Diamond-terms (3.1)areinh,thengis semigenerated. The proof is the same, but in the final step one needs to use Theorem 3.6 instead of Lemma 4.1. Corollary 4.5 (of Proposition 2.11). Let gbe a stratified Lie algebra of step 3. If gis not semigenerated, then there exists a quotient algebra of gthat is trimmed and not semigenerated. Proof By Proposition 2.11, there exists a quotient algebra ˆ gof gfor which Z(ˆ g)∩V1(ˆ g)= Z(ˆ g)∩V2(ˆ g)={0}and dim(Z(ˆ g)∩V3(ˆ g)) ≤1. Since the center of a stratified Lie algebra is non-trivial and homogeneous (see (2.3)), we deduce that dim Z(ˆ g)=dim Z(ˆ g)∩V3(ˆ g)=1, proving that ˆ gis trimmed.  In the rest of this section we provide some examples. We first show a 7-dimensional Lie algebra of step 3 that is of type (♦)but that is not of type (), see Example 4.6. Then we provide a 6-dimensional Lie algebra that is semigenerated but not of type (♦), see Example 4.7. Example 4.6 Let h1,h2be two copies of the four-dimensional Engel algebra E1and consider their product Lie algebra h1×h2.DenotingbyZ1and Z2the generators of V3(h1)and V3(h2), respectively, and identifying h1and h2with the respective subalgebras of h1×h2, we have that V3(h1×h2)=span{Z1,Z2}.ThenR(Z1−Z2)is an ideal of h1×h2and the quotient algebra g:=(h1×h2)/R(Z1−Z2) is a 7-dimensional (trimmed) stratified Lie algebra of step 3 (which in the Gong’s classification [10, p. 57] is denoted by (137A)). We claim that gis of type (♦)but it is not of type (). Indeed, the fact that gis of type (♦)follows immediately from Lemma 3.8 with i=R(Z1−Z2),as now πl(i)=V3(hl)for both l=1andl=2. 123 E. Le Donne, T. Moisala We argue next that gis not of type ().Let{X1,...,X7}beabasisofgfor which {X1,...,X4}is a basis of V1(g)and the only nonzero brackets are [X1,X2]= X5,[X3,X4]=X6and [X1,[X1,X2]] = [X3,[X3,X4]] = X7, as presented in the diagram below. Then for a vector Y:= 4 i=1aiXi∈V1we have, for instance, that ad2 Y(X2)=a2 1X7. In particular, if Yis such that ad2 Y(V1)=0, then a1=0. Consequently, the set of vectors Y∈V1satisfying ad2 Y(V1)=0 is contained in a 3-dimensional subspace of V1(g).We conclude by Remark 3.5 that gis not of type (). The Lie brackets of this last example can be described by the following diagram, where each pair of lines forming the shape of a V have at the extremes two vectors; the symbol should be read as follows: the bracket of the vector at the left hand of the V with the vector at the right hand of the V gives the vector at the bottom (here, for instance, [X1,X2]=X5). For more uses of these type of diagrams see [14]. X1X2X3X4 X5X6 X7 Example 4.7 Let gbe the 6-dimensional step-3 Lie algebra (N6,2,6in [10,p.33]),wherethe only non-trivial brackets are given by [X1,X2]=X4,[X1,X3]=X5,[X1,X4]=[X3,X5]=X6. The Lie brackets can be described by the following diagram: X1X2X3 X4X5 X6 The above diagram should be read as the one in the previous example, except the fact that for the V-shape with an arrow going from right to left of the V, one should interpret that the bracket of the vector on the right with the vector on the left gives the bottom, i.e., [X1,X4]=X6. The Carnot algebra gof this example is semigenerated but it is not of type (♦). Indeed, to prove that it is semigenerated, let W⊆gbe a horizontal half-space and let us show that V3=RX6⊆es,wheres:=Cl(sW). This would show that Wis semigenerating by Lemma 4.1. Suppose first that X1/∈∂W. The rank of gis 3, so ∂Whas dimension 2. Then there exist some a,b∈Rsuch that Y2:=X2−aX1and Y3:=X3−bX1form a basis for ∂W.Since∂W⊆esand esis a Lie algebra by Lemma 2.2, we obtain [Y2,Y3]=[X2−aX1,X3−bX1]=bX4−aX5∈es. If a= 0orb= 0, we get V3⊆I([Y2,Y3])⊆es, 123 Semigenerated Carnot algebras and applications... where the last inclusion comes from Lemma 4.3. If instead a=b=0, then X2=Y2∈es. Since ad2 X2X1=0, by Lemma 2.3 we get that [X1,X2]=X4∈es.Again,sinceV3⊆ I(X4), by Lemma 4.3 we get that V3⊆es. The cases X2/∈∂Wand X3/∈∂Ware easier: if X2/∈∂Wwe find, like above, some a,b∈Rsuch that ∂W=span{X1−aX2,X3−bX2}. It then suffices to notice that X6∈Lie(X1−aX2,X3−bX2)⊆es, for all choices of a,b∈R. Similarly, the case X3/∈∂Wfollows from the fact that X6∈Lie(X1−aX3,X2−bX3)for all a,b∈R.We conclude that gis semigenerated. Finally, to justify that gis not of type (♦), observe that the span of X2and X3is an abelian stratified subalgebra of g.Ifgwere type of (♦),thenbyRemark3.4 it would be of type (). However, similarly to Example 4.6,wehaveforanarbitraryelementY= a1X1+a2X2+a3X3∈V1that ad2 Y(X2)=a2 1X6. Hence vectors Y∈V1for which ad2 Y(V1)=0 must satisfy a1=0, which proves the non-existence of type ()-basis by Remark 3.5. 5 Engel-type algebras In the rest of the paper we concentrate on a family of Carnot algebras that we call of Engel type. These algebras can be constructed through an iterative process from the classical 4-dimensional Engel algebra. Similarly to the Engel algebra, every Engel-type algebra is trimmed and non-semigenerated, as we shall show in Propositions 5.8 and 5.10. A more subtle result is that, at least in step 3, the Engel-type algebras are the only Carnot algebras with these properties. For this last part, see Proposition 5.12. The proof of Theorem 1.2 will then be straightforward. 5.1 Definition and properties Definition 5.1 (Engel-type algebra En). For each n∈N,wedenotebyEnand call it the n-th Engel-type algebra the 2(n+1)-dimensional Lie algebra (of step 3 and rank n+1) with basis {X,Yi,Ti,Z}n i=1, where the only non-trivial brackets are given by [Yi,X]=Tiand [Yi,Ti]=Z∀i∈{1,...,n}.(5.1) The first two Engel-type algebras are the following. The first is E1, and it is commonly known as Engel algebra, see [4], and we represent it by the diagram below. Y1X T1 Z The second Engel-type algebra E2is the six-dimensional algebra N6,3,1ain [10, p. 135] and its diagram is presented below. 123 E. Le Donne, T. Moisala Y1XY 2 T1T2 Z Each Engel-type algebra is a Lie algebra (see the simple verification in Remark 5.2)and admits a step-3 stratification: V1(En):=span{X,Y1,...,Yn}, V2(En):=span{T1,...,Tn}, V3(En):=span{Z}. We shall also give another equivalent definition for Enin Proposition 5.7. We first list some properties of such Lie algebras. Remark 5.2 Each Engel-type algebra given by the brackets (5.1) is indeed a Lie algebra. Namely, let us verify that the Jacobi identity is satisfied. Since the basis has a natural stratification, it is enough to check the identity for triples in the set {X,Y1,...,Yn}. Hence, we just consider the case X,Yi,Yjor Yi,Yj,Yk. In the first case, we have [X,[Yi,Yj]]+[Yi,[Yj,X]]+[Yj,[X,Yi]] = 0+[Yi,Tj,]+[Yj,−Ti] =δijZ−δijZ=0. In the second case, we have [Yi,[Yj,Yk]]+[Yj,[Yk,Yi]]+[Yk,[Yi,Yj]] = 0+0+0=0. Lemma 5.3 (Properties of Engel-type algebras) The n-th Engel-type algebra Enwith a basis satisfying (5.1)has the following properties. (i) If n ≥2,thenspan{Y1,...,Yn}is the unique abelian n-dimensional subspace of V1(En); (ii) the line RX is the unique horizontal line satisfying [RX,V2]={0}; (iii) for every nonzero Y ∈span{Y1,...,Yn},wehave ad2 Y(V1)=RZ. Proof (i) Obviously, the space span(Y1,...,Yn)is an abelian n-space. Vice versa, let Hbe an n-dimensional subspace of V1(En)such that H= span(Y1,...,Yn). Then there exists ν∈Hof the form ν:=X+ n  i=1 aiYi,with ai∈Rfor i=1,...,n, and a nonzero Y∈H∩span(Y1,...,Yn). Writing Y=n i=1biYifor some bi∈R, we obtain [Y,ν]= i biTi= 0, proving that His nonabelian. 123 Semigenerated Carnot algebras and applications... (ii) Let now ν:=aX + n  i=1 aiYi,a,ai∈R∀i=1,...,n, where ak= 0forsomek∈{1,...,n}.Then [ν, Tk]=akZ= 0, which shows that [ν, V2] = 0ifν/∈RX. (iii) Let again Y=n i=1biYifor some real numbers binot all identically zero. Since ad2 Y(RX)=Rad2 YX=R n  i=1 b2 iZ=RZ and also V3(En)=RZ,weget RZ=ad2 Y(RX)⊆ad2 Y(V1)⊆RZ.  We provide next the automorphism group of the Engel-type algebras, which will be used later in the proof of Lemma 5.5 where we characterize all stratified subalgebras of the Engeltype algebras. For the automorphism group of the Engel algebra, we refer to [4, Lemma 2.3]. Lemma 5.4 Consider the basis {X,Y1,...,Yn}of V1(En),n≥2, defined in (5.1).Fixa scalar product ·,· on V1(En)that makes {Y1,...,Yn}orthonormal. Then every linear transformation on V1(En)that in the basis {X,Y1,...,Yn}is given by the block matrix a0 0bA,a,b∈R\{0}and A ∈O(n), induces a Lie algebra automorphism of En. Moreover, every automorphism of Enis induced by such a transformation on V1(En). Proof By Lemma 5.3.(i) and (ii), every ∈Aut(En)must fix the subspaces span{Y1,...,Yn} and RX. Moreover, notice that any linear map on V1(En)fixing these subspaces satisfies, for some a= 0, the two equalities: [(Yi), (X)]=[(Yi), aX]= n  k=1 a(Yi), YkTkand [(Yi), Tk]=[ n  =1 (Yi), YY,Tk]=(Yi), YkZ. Therefore, using again that {Yj}jare orthonormal, we deduce that [(Yi), [(Yj), (X)]] = n  k=1 a(Yj), Yk(Yi), YkZ=a(Yi), (Yj)Z.(5.2) Recall that the basis vectors of Ensatisfy [Yi,[Yj,X]] = δijZ. Therefore, the map  induces a Lie algebra automorphism of Enif and only if there exists some b= 0 such that [(Yi), [(Yj), (X)]] = bδijZ,∀i,j∈{1,...,n} 123 E. Le Donne, T. Moisala According to (5.2), this is equivalent to saying that the map is, up to scaling, an orthogonal transformation on span{Y1,...,Yn}. For a good understanding of the rest of this section, we stress that we say that a subalgebra of a stratified Lie algebra is stratified if it is homogeneous and stratified with respect to the induced grading. Lemma 5.5 Let 1≤k≤n. If his a stratified rank-k subalgebra of the n-th Engel-type algebra En, then either his abelian or it is isomorphic to Ek−1. Proof If n=1, the claim is trivially true. Assume then that n≥2andlet{X,Y1,...,Yn} be a basis of V1(En)as in (5.1). We start by proving the case k=n. Assume first that his the subalgebra generated by {X+aYn,Y1,...,Yn−1}for some a∈R. Observe that then h is isomorphic to En−1for every value of a∈Rsince, for each i=1,...,n−1, we have that [Yi,X+aYn]=[Yi,X]=Tiand [X+aYn,Ti]=[X,Ti]=0. Recall that, by Lemma 5.3.i, the subspace span{Y1,...,Yn}is the unique abelian stratified subalgebra of En. Since every n-dimensional subspace of V1(En)which is not equal to span{Y1,...,Yn} can be realized from some subspace span{X+aYn,Y1,...,Yn−1},a∈R, by a rotation of V1(En)around the X-axis, we infer by Lemma 5.4 that any subalgebra generated by such subspace is isomorphic to En−1. Regarding the case k<n, fix a non-abelian stratified rank-ksubalgebra hkof En. Then we find a filtration hk⊂hk+1⊂ ··· ⊂ hn⊂Enof non-abelian stratified rank-l subalgebras hl,l=k+1,...,n. The claim follows now by the first part of this proof.  There are plenty of Lie algebras of arbitrarily large step whose first three layers coincide with the first Engel-type algebra, for example, the filiform algebras. The same phenomenon does not happen for the other Engel-type algebras. Since we need this latter fact in the proof of Proposition 5.7, we clarify such a phenomenon in the next remark. Remark 5.6 For each n≥2, the Lie algebra Encannot be ‘prolonged’ in the following sense: if gis a stratified Lie algebra for which g/g(4)is isomorphic to Enfor some n≥2, where g(4)=V4⊕···⊕Vs,theng=En. Proof Let {X,Yi}n i=1,{Ti}n i=1and {Z}be bases of V1(g),V2(g)and V3(g), respectively, satisfying the bracket relations (5.1) modulo g(4). Observe that, being gstratified, the subspace V4is spanned by elements of the form [ν, Z],whereν∈{X,Yi}n i=1. We need to show that V4={0}. Indeed, by the Jacobi identity, we have [X,Z]=[X,[Yi,[Yi,X]]]=−[Yi,[[Yi,X],X]] − [[Yi,X],[X,Yi]] = 0, where we deduced that [Yi,[[Yi,X],X]] = 0, since [X,V2]={0}by Lemma 5.3.ii. Moreover, for every j=1,...,nand i= j(which exists since n≥2) one has [Yj,Z]=[Yj,[Yi,[Yi,X]]]=−[Yi,[[Yi,X],Yj]]−[[Yi,X],[Yj,Yi]] = 0, where we used that [Yj,[Yi,X]] = [Yj,Ti]=0andthat[Yj,Yi]=0. This proves that V4={0}, in which case also g(4)={0}and hence g∼ =g/g(4)∼ =En. The Engel-type algebras have the following equivalent definition using induction. 123 Semigenerated Carnot algebras and applications... Proposition 5.7 (A characterization of Engel-type algebras) Let gbe a stratified Lie algebra of rank n +1≥4.Thengis isomorphic to Enif and only if ghas a unique abelian stratified subalgebra of rank n and every other stratified subalgebra of rank n is isomorphic to En−1. Moreover, this characterization holds for rank n +1=3if in addition dim V3(g)=1. Proof One direction is proven in Lemmata 5.3.i and 5.5. Regarding the other direction, let g bearankn+1 stratified Lie algebra, with n+1≥3, which has a unique abelian stratified subalgebra h0of rank nand every other rank-nstratified subalgebra is isomorphic to En−1. Our first aim is to show that the condition dim V3(g)=1 always holds as long as n+1≥4. We start by claiming the following property: If n+1≥4andl⊂V1(g)is an (n−1)-dimensional abelian subspace of V1(g), then l⊂h0. (5.3) Indeed, let ν∈V1(g)\l. On the one hand, if l⊕Rνis an abelian subalgebra of g,then l⊕Rν=h0by uniqueness of h0and so l⊂h0. On the other hand, if l⊕Rνgenerates a nonabelian subalgebra, then l⊕Rνis isomorphic to V1(En−1),wheren−1≥2. Since h0∩(l⊕Rν) is an abelian (n−1)-dimensional subspace of V1(En−1), we deduce that h0∩(l⊕Rν) =lby uniqueness of (n−1)-dimensional abelian subspaces of V1(En−1) because n−1≥2 (see Lemma 5.3.i). Then again l⊂h0and (5.3) is proven. Recall that V3(g)=span{[X1,[X2,X3]] | Xi∈V1,i=1,2,3} as gis stratified. We are going to show that vectors [X1,[X2,X3]] and [ X1,[ X2, X3]] are linearly dependent, for every choice of vectors Xi, Xi∈V1,i=1,2,3. So let Xi, Xi∈V1 for i=1,2,3 be such that [X1,[X2,X3]] and [ X1,[ X2, X3]] are nonzero and let h1⊇ {X1,X2,X3}and h2⊇{  X1, X2, X3}be rank-nsubalgebras of g.Sinceh1and h2have nonzero third layers, they are isomorphic to En−1. We may assume that V1(h1)= V1(h2), since otherwise [X1,[X2,X3]] and [ X1,[ X2, X3]] are linearly dependent. Hence, being it the intersection of two different hyperplanes, the space V1(h1)∩V1(h2)is a codimension 2 subspace of V1(g), i.e., it has dimension n−1. To prove that dim V3(g)=1, we have to show that V3(h1)=V3(h2), for all such h1and h2as above. The proof for the latter fact is divided into two cases depending on whether V1(h1)∩V1(h2)is closed under brackets (or equivalently, it is an abelian subalgebra) or not. Assume first that V1(h1)∩V1(h2)forms an abelian subalgebra. Then by claim (5.3)wehave that V1(h1)∩V1(h2)⊆h0. Let us fix a basis {Z1,...,Zn−1}for V1(h1)∩V1(h2)and let also Zn∈h0and X∈V1(g)be such that {Z1,...,Zn,X}is a basis of V1(g). Fix next Yi∈V1(hi)\(V1(h1)∩V1(h2)) for i=1,2 and write it in terms of this basis as Yi=aiX+ n  j=1 bi jZj for some ai,bi j∈R. Notice that, since hiis not abelian, we have ai= 0. Since now {Yi,Z1,...,Zn−1}is a basis of V1(hi)for i=1,2 and since h0=span{Z1,...,Zn}is abelian, we obtain V3(h1)=ad2 Z1(V1(h1)) =ad2 Z1(RY1)=ad2 Z1(RX)=ad2 Z1(RY2)=ad2 Z1(V1(h2)) =V3(h2), where the first and the last equality follow from Lemma 5.3 (iii). Since the third layers of h1 and h2are one dimensional, we deduce that [X1,[X2,X3]] and [ X1,[ X2, X3]] are linearly dependent. 123 E. Le Donne, T. Moisala Assume instead that V1(h1)∩V1(h2)is not a subalgebra. Then, by Lemma 5.5,theLie algebra generated by V1(h1)∩V1(h2)is isomorphic to En−2. In particular, it has step 3. Exploiting again the fact dim V3(h1)=dim V3(h2)=1, we get that V3(h1)=V3(Lie(V1(h1)∩V1(h2))) =V3(h2). Therefore [X1,[X2,X3]] and [ X1,[ X2, X3]] are linearly dependent as in the previous case. This concludes the proof for the fact that dim V3(g)=1whenn+1≥4. Hence, in what follows we may assume that V3(g)is one dimensional, and that n+1≥3. In the rest of this proof we are going to construct a basis of gthat satisfies the defining commutator relations (5.1) of the Engel-type algebra En.Leth∼ =En−1be some nonabelian rank-nsubalgebra of gand let {X,Yi,Ti,Z}n−1 i=1be a basis of hsatisfying relations (5.1). We aim to find a vector Yn∈V1which together with Tn:=[Yn,X]completes {X,Yi,Ti,Z}n−1 i=1 to the defining basis of En.SinceEncannot be prolonged (see Remark 5.6), this is enough to prove that gis isomorphic to En. Notice that h0∩h=span{Y1,...,Yn−1}since span{Y1,...,Yn−1}is the unique abelian subspace of V1(h)by Lemma 5.3 (i). Moreover, V3(g)=RZas V3(g)is one dimensional. Fix next ˆ Yn∈h0\hand write Yn:=aˆ Yn+ n−1  i=1 aiYi∈h0, where a,ai∈R,i=1,...,n−1, are values to be determined later. Now whenever a= 0 we have that span{Y1,...,Yn}=h0is abelian and {Y1,...,Yn,X}is a basis of V1(g).To conclude the proof of the proposition, we claim that it suffices to show that (i) there exist ai∈R,i=1,...,n−1, such that [Yn,Ti]=0foralli=1,...,n−1; (ii) there exists a∈Rsuch that [Yn,[Yn,X]] = Z; (iii) with the above choices of ai,a∈R, the vector Tn:=[Yn,X]is linearly independent of T1,...,Tn−1. Indeed, we stress again that, by Remark 5.6,theLiealgebraEncannot be prolonged and hence we indeed have that the set {X,Yi,Ti,Z}n−1 i=1is a basis of a step-3 Lie algebra isomorphic to En. To show (i), observe that for every i=1,...,n−1, [Yn,Ti]=a([ˆ Yn,Ti]+aiZ). Since V3(g)is one dimensional, the two vectors [ˆ Yn,Ti]and Zare linearly dependent. Hence for every i=1,...,n−1 there exists ai∈Rsuch that [Yn,Ti]=0, which proves (i). Regarding (ii), let then h:=Lie(Y2,...,Yn,X)∼ =En−1. Since span{Y2,...,Yn}is again the unique abelian (n−1)-dimensional subspace of V1(h), it holds [Yn,[Yn,RX]]=[Yn,[Yn,V1(h)]]. As [Yn,[Yn,V1(h)]] = 0 by Lemma 5.3 (iii) and since V3(g)=RZ, we may choose a∈R such that [Yn,[Yn,X]] = Z. Thus (ii) is proven. Regarding (iii), it is enough to notice that from (i) we have [Yn, n−1  i=1 biTi]= n−1  i=1 bi[Yn,Ti]=0, 123