SOLVABLE EXTENSIONS OF NEGATIVE RICCI CURVATURE OF FILIFORM LIE GROUPS 5 1 Y. NIKOLAYEVSKY 0 2 Abstract. We give necessary and sufficient conditions of the existence of a left- n invariant metric of strictly negative Ricci curvature on a solvable Lie group the nil- a J radicalofwhoseLiealgebragisafiliformLiealgebran. Itturnsoutthatsuchametric 9 alwaysexists,exceptforinthe twocases,whennis oneofthe algebrasofranktwo,Ln orQn,andgisaone-dimensionalextensionofn,inwhichcasestheconditionsaregiven ] in terms of certain linear inequalities for the eigenvalues of the extension derivation. G D . h t a m 1. Introduction [ The question of whether (and when) a given manifold admits a Riemannian metric 1 withaparticularsignofthecurvatureisoneofthefundamental inRiemanniangeometry. v Similarly, for homogeneous manifolds, the same question can be asked for left-invariant 8 2 metrics. In that case the curvature is entirely expressed in terms of the algebraic struc- 0 ture of the given homogeneous space and one expects the answer to be stated in both 2 0 topological and algebraic terms. . In this paper we continue the study of metric solvable Lie groups admitting a left- 1 0 invariant metric of negative Ricci curvature, which has been started in [NN], and we refer 5 the reader to the Introduction of that paper for a detailed overview of known results. 1 : At present, necessary and sufficient conditions for a homogeneous space to admit a left- v invariant metric with a particular sign of the sectional curvature are well understood, i X as well as the conditions for a homogeneous space to admit a left-invariant metric with r a positive or with zero Ricci curvature. By the result of Milnor [Mil] (for Lie groups) and Berestovskii [Ber] (in the general case), a homogeneous space admits a left-invariant metric with Ric > 0 if and only if it is compact and has a finite fundamental group (compare to the Myers Theorem). By the result of [AK], any Ricci-flat homogeneous space is flat. Note that flat homogeneous spaces were completely described in [Ale, BB]: every such space is isometric to a solvmanifold, the nilradical n of whose Lie algebra g is abelian, its orthogonal complement a = n is also abelian and the operators ad , Y a, ⊥ Y ∈ are skew-symmetric. 2010 Mathematics Subject Classification. 53C30,22E25. Keywordsandphrases. SolvableLiealgebra,filiformLiealgebra,nilradical,negativeRiccicurvature. The author was partially supported by ARC Discovery Grant DP130103485. 1 2 Y. NIKOLAYEVSKY Much less is known, however, about Riemannian homogeneous spaces of negative Ricci curvature. No unimodular solvable Lie group (in particular, no nilpotent group) admits a left-invariant metric with Ric < 0, by the following result. Theorem 1 ([DLM]). A unimodular Lie group which admits a left-invariant metric with Ric < 0 is noncompact and semisimple. Examples of left-invariant metrics with Ric < 0 were constructed on SL(n,R), n 3, ≥ in [LDM] and on some complex simple Lie groups in [DLM]. Thegeneral, non-unimodular, casehowever, seemstobewideopen, evenforLiegroups (leave alone homogeneous spaces). It is well known that the Ricci curvature of a left- invariant metric on a Lie group G can be entirely computed from the algebraic data: the structure of the Lie algebra g of G and the inner product , on g (see Section 2.1 for h· ·i details). From this point on we descend to the level of Lie algebras and, with a slight abuse of terminology, will speak of the Ricci curvature of the metric Lie algebra (g, , ). h· ·i In all the known cases, the necessary and sufficient condition for a given solvable Lie algebragtoadmitaninnerproductwithRic < 0hasthefollowingform: gisanextension of its nilradical n by some derivations, one of which satisfies certain linear inequalities imposed on the real parts of its eigenvalues. The precise nature of such inequalities depends on the structure of n and in the general case remains unknown. In [NN] the authors speculated that they may be related to the fact that (the real semisimple part of) the derivation belongs to a certain convex cone in the torus of derivations of n. For example, a solvable Lie algebra g with an abelian nilradical n admits an inner product of negative Ricci curvature if and only if there exists Y g such that all the ∈ eigenvalues of the restriction of ad to n have positive real part. This is a consequence Y of the following Theorem. Theorem 2 ([NN]). Suppose g is a solvable Lie algebra. Let n be the nilradical of g and z be the centre of n. Then (1) If g admits an inner product of negative Ricci curvature, then there exists Y g ∈ such that Trad > 0 and all the eigenvalues of the restriction of the operator ad to Y Y z have positive real part; (2) If there exists Y g such that all the eigenvalues of the restriction of ad to n have Y ∈ positive real part, then g admits an inner product of negative Ricci curvature. The necessary and sufficient conditions to admit an inner product with Ric < 0 for Lie algebras whose nilradical is a Heisenberg Lie algebra or is a standard filiform algebra have a similar “flavour” [NN]. In this paper we study solvable algebras whose nilradical belongs to another important class of the variety of nilpotent Lie algebras – the class of filiform algebras. Filiform Lie algebras are those nilpotent algebras which are “the least” nilpotent – they have the maximal possible number of nonzero terms in the lower central series for a given dimension, that is, n(n 2) = 0, where n = dimn [Ver]. From among these algebras, − 6 SOLVABLE EXTENSIONS OF NEGATIVE RICCI CURVATURE OF FILIFORM LIE GROUPS 3 there are two distinguished ones, L and Q , which have rank two; all the other filiform n n algebras have rank one or zero by [GK] – see Section 2.2 for details. The algebra L = Span(X ,...,X ) is defined by the relations [X ,X ] = X , i = n 1 n 1 i i+1 2,...,n 1, where from now on we adopt the convention that all the relations between − the basis elements of a Lie algebra which we do not list are zero (unless they follow from the given ones by the skew-symmetry). The codimension one abelian ideal i = Span(X ,...,X ) and the one-dimensional centre RX of L are both characteristic 2 n n n ideals of L (they are invariant under the action of any derivation on L ). If g is a n n solvable extension of L , define the one-forms ι and ι on g as follows: for Y g, n 2 n ∈ [Y,X ] = ι (Y)X and ι (Y) = Tr((ad ) ). n n n 2 Y i | The algebra Q = Span(X ,...,X ), n = 2m, is defined by the relations [X ,X ] = n 1 n 1 i X , i = 2,...,n 2, and [X ,X ] = ( 1)j+1X , j = 2,...,n 1. The members i+1 j n j+1 n ik = Q(nk−2) = Span−(Xk,...,Xn), k−= 3,...,n−, of the lower central se−ries are all charac- teristic ideals of Q , so for a solvable extension g is of Q , we can define the one-forms n n ι , k = 3,...,n, on g as follows: for Y g, ι (Y) = Tr((ad ) ). k ∈ k Y |ik Our main result is the following theorem. Theorem 3. Let g be a solvable non-nilpotent Lie algebra with the filiform nilradical n. Then g admits an inner product of negative Ricci curvature if and only if (a) either n = L and g is a one-dimensional extension of n by a vector Y such that n ι (Y),ι (Y) > 0 [NN, Theorem 4]; 2 n (b) or n = Q , n = 2m, m > 2, and g is a one-dimensional extension of n by a vector n Y such that ι (Y) > 0, k = m+1,...,n; k (c) or, with no restrictions, in all the other cases: either n is any other filiform algebra, or otherwise n is L or Q and g is an extension of n of dimension greater than one. n n Remark 1. Note that the set of inequalities in (b) is abundant – for all of them to hold it is necessary and sufficient that just two of them hold, namely, ι (Y) > 0 (the eigenvalue n of ad on the centre of Q ) and one other ι (Y) > 0. Due to the fact that this l does not Y n l have a nice description we postpone this question till the end of Section 3 (Theorem 4). Note also that Case (c) is not as “universal” as it may sound: the truth is that “the majority” of filiform algebras are characteristically nilpotent, hence admitting no non- nilpotent extensions at all – see Section 2.2. I would like to thank Yurii Nikonorov for his suggestion to study this topic and many useful ideas, Grant Cairns and Nguyen Thang Tung Le the work with whom stimulated my interest in the question, and Jorge Lauret for useful discussions, including the ones concerning the moment map (see the end of Section 3). 2. Preliminaries 2.1. Ricci tensor. Let G be a Lie group with a left-invariant metric Q obtained by the lefttranslationsfromaninnerproduct , ontheLiealgebragofG. LetB betheKilling h· ·i form of g, and let H g be the mean curvature vector defined by H,X = Trad . X ∈ h i 4 Y. NIKOLAYEVSKY The Ricci curvature ric of the metric Lie group (G,Q) at the identity is given by 1 1 1 (1) ric(X) = [H,X],X B(X,X) [X,E ] 2 + [E ,E ],X 2, i i j −h i− 2 − 2 ik k 4 i,jh i X X for X g, where E is an orthonormal basis for (g, , ) (see e.g. [Ale] or [Bes]). i ∈ { } h· ·i Equivalently, one can define the Ricci operator Ric of the metric Lie algebra (g, , ) h· ·i (the symmetric operator associated to ric) by 1 1 1 (2) Ric = adt ad + ad adt B (ad )s, −2 Ei Ei 4 Ei Ei−2 − H Xi Xi where (ad )s = 1(ad +adt ) is the symmetric part of ad . H 2 H H H In the case when g is solvable and is a one-dimensional extension of its nilradical n (this is the only case in this paper for which we will need an explicit formula for the Ricci tensor), one can simplify (2) further. Choose an orthonormal basis e for n and i { } a unit vector f n. Denote T = Trad . Note that g is unimodular if and only if T = 0. f ⊥ Otherwise changing the sign of f if necessary we get T = H > 0. k k Relative to the basis e ,...,e ,f , the matrix of the Ricci operator of the solvable 1 n { } metric Lie algebra (g, , ) has the form (see the proof of [NiN, Theorem 3]) h· ·i R R (3) Ric = 1 2 , (cid:18) Rt r (cid:19) 2 3 where 1 (4) R = Ricn+ [A,At] TAs, 1 2 − n 1 (5) (R ) = [f,e ],[e ,e ] , j = 1,...,n, 2 j i i j 2 h i Xi=1 (6) r = Tr((As)2), 3 − and Ricn is the matrix of the Ricci operator of the metric nilpotent Lie algebra (n, , ) n h· ·i relative to the basis e ,...,e . As H = 0 and B = 0 we get from (1) 1 n { } 1 1 (7) RicnX,Y = X,[e ,e ] Y,[e ,e ] [X,e ],e [Y,e ],e ] . i j i j i j i j h i 4 h ih i− 2 h ih i Xi,j Xi,j for X,Y n. ∈ 2.2. Filiform algebras and their extensions. Filiform algebras are “the least nilpo- tent” among nilpotent Lie algebras. They have been introduced by Michele Vergne in [Ver] and studied extensively since then. In this paper we are interested mostly not in filiform algebras assuch, but in the solvable extensions of them. These have been studied in depth and classified depending on the dimension of the (non-nilpotent) extension in [Sun] or [GK]. It turns out that “most” filiform algebras are characteristically nilpotent, so that any derivation of them is nilpotent. SOLVABLE EXTENSIONS OF NEGATIVE RICCI CURVATURE OF FILIFORM LIE GROUPS 5 Clearly, extending a nilpotent algebra by a nilpotent derivation (or more generally, extending by a subspace of derivations, some nonzero elements of which are nilpotent) increases the nilradical, and we do not allow this. Temporarily passing to the complexi- fication, by [Sun, Theorem 2], we get that a solvable algebra g having a filiform algebra n as its nilradical is an extension by commuting derivations. Furthermore, we define the rank of a Lie algebra g to be the dimension of a maximal abelian subalgebra in the algebra of derivations Der(g), all whose nonzero elements are semisimple (such a subal- gebra is called a maximal torus of derivations; all the maximal tori are conjugate by an automorphism and have the same dimension [Che]). By [GK, Th´eor`eme 2] we have: Lemma 1. Suppose n is a filiform algebra over C of positive rank. Then 1. Either rkn = 2, in which case (a) either n is isomorphic to the algebra L defined by the relations [X ,X ] = X , n 1 i i+1 2 i n 1, with a maximal torus of derivations Span(φ ,φ ), where φ (X ) = 1 2 1 i ≤ ≤ − iX , 1 i n, and φ (X ) = 0,φ (X ) = X , 2 i n; i 2 1 2 i i ≤ ≤ ≤ ≤ (b) or n is isomorphic to the algebra Q , n = 2m, defined by the relations n (8) [X ,X ] = X , 2 i n 2, [X ,X ] = ( 1)j+1X , 2 j n 1, 1 i i+1 j n j+1 n ≤ ≤ − − − ≤ ≤ − with a maximal torus of derivations Span(φ ,φ ), where 1 2 φ (X ) = iX , 1 i n 1, φ (X ) = (n+1)X 1 i i 1 n n (9) ≤ ≤ − φ (X ) = 0, φ (X ) = X , 2 i n 1, φ (X ) = 2X . 2 1 2 i i 2 n n ≤ ≤ − 2. Orrkn = 1, in whichcasen isisomorphicto analgebrafrom the familyAr(α ,...,α ), n 1 t 1 r n 4, defined by the relations of L and some other relations, or from the n ≤ ≤ − family Br(α ,...,α ), n = 2m, 1 r n 5, defined by the relations of Q and n 1 t ≤ ≤ − n some other relations. In the both cases, α ,...,α are parameters satisfying certain 1 t algebraic equations, and a maximal torus of derivations is Span(φ), where (10) φ(X ) = X ,φ(X ) = (i+r)X , 2 i n 1, φ(X ) = (n+2r)X . 1 1 i i n n ≤ ≤ − As usual, the relations between the basis elements for L and for Q which we do not n n list are zero (unless they follow from the given ones by the skew-symmetry). Note that in [GK] there is one other class of algebras of rank one, C , but as it is shown in [GV, n Remarque 1] all the algebras of class C are isomorphic to Q . n n We intentionally do not give all the details on the families Ar and Br — for our n n purposes we only need to know φ. Remark 2. Note that when considering the algebra Q , n = 2m, bothover R andover C, n we can assume that n 4, asQ is given by the relations [X ,X ] = X , [X ,X ] = X 4 1 2 3 2 3 4 ≥ − and thus is isomorphic to L with the basis X ,X ,X ,X . We also note that 4 2 1 3 4 {− } changing the basis X for Q to the basis Y = X + X , Y = X , i > 1, we get the i n 1 1 2 i i relations [Y ,Y ] = Y , 2 i n 1, [Y ,Y ] = ( 1)i+1Y , 2 i n 1 (which 1 i i+1 i n i+1 n ≤ ≤ − − − ≤ ≤ − differ from (8) just by [Y ,Y = Y ]) that might sometimes be more convenient to use. 1 n 1 n − 6 Y. NIKOLAYEVSKY Getting back to the real solvable Lie algebra g with the filiform nilradical n we need to exercise a certain caution, as two real nilpotent algebras can be isomorphic over C, without being isomorphic over R (this phenomenon can be observed even in the low dimension classification lists). The complexification nC of n must be one of the algebras from Lemma 1. Now if rknC = 1, then gC is the one-dimensional extension of nC by a complex nonzero multiple of the derivation φ given by (10). It follows that g is the one-dimensional extension of n by a complex nonzero multiple of the derivation φ. But as φ is a real linear operator and as all the eigenvalues of φ have different absolute value (so that no eigenvalues of cφ,c C 0 , can possibly be complex conjugate), we obtain ∈ \{ } that g is the extension of n by a derivation all of whose eigenvalues are positive (up to changing the sign of φ), that is, g admits an inner product of negative Ricci curvature by Theorem 2(2). We now proceed to the algebras of rank two. In the next lemma, L and Q are n n R R complex Lie algebras given by the relations of Lemma 1(1). We denote L and Q n n respectively the real Lie algebras defined by the same relations. Lemma 2. 1. Let n be a real, n-dimensional, filiform Lie algebra given by the relations [Y ,Y ] = Y , for 2 i n 1 and [Y ,Y ] = K Y for 2 i,j n 1, where 1 i i+1 i j ij n K = (K ) is a nonsin≤gula≤r, ske−w-symmetric matrix. Then n is≤isomo≤rphic−to QR. ij n 2. Let n be a real filiform Lie algebra whose complexification is isomorphic to L (resp. n Q ) over C. Then n is isomorphic over R to the real algebra LR (resp. QR). n n n Proof. 1. For K to possibly be nonsingular n must be even (note that the elements of K are labelled by i,j = 2,...,n 1). From the Jacobi identities it follows that − K +K = 0 for 2 i,j n 2 and K = 0 for 2 i n 2. From this and i,j+1 i+1,j i+1,n 1 ≤ ≤ − − ≤ ≤ − the skew-symmetry we obtain K = 0 when i+j > n+1 or when i+j is even. When ij k = i+j is odd and k n+1 we get K = K = K = = K and 2,k 2 3,k 3 4,k 4 k 2,2 ≤ − − − − ··· − − in particular, K = 0 as K is nonsingular. Introduce a new basis Y for n by putting 2,n−1 6 i′ Y = Y , Y = Y +a Y +a Y +..., Y = [Y ,Y ], 2 i n 1. Then Y = Y and 1′ 1 2′ 2 3 3 4 4 i′+1 1′ i′ ≤ ≤ − n′ n [Y ,Y ] = K Y for a skew-symmetric matrix K . The arguments similar to the ones i′ j′ i′j n′ ′ above show that K = 0 when i +j > n +1 or when i+ j is even and that K = i′j 2,k 2 − K = K = = K for odd k n + 1. Furthermore, K = K −and3,tkh−e3n K 4,k−4= 2a··K· − +k−P2,2(a ), K ≤= 2a K + P (a ,a ,a2′,n)−,1..., w2h,ne−re1 2′,n 3 4 2,n 1 3 3 2′,n 5 6 2,n 1 5 3 4 5 P ,P ,... are−certain polyn−omials. As K −= 0, we can−successively choose a ,a ,... 3 5 2,n 1 4 6 − 6 in such a way that K = 0 for all i < n 1. Then, relative to the basis Y , the algebra 2′i − i′ n is given by the relations [Y ,Y ] = Y , 2 i n 1, and [Y ,Y ] = ( 1)i+1aY 1′ i′ i′+1 ≤ ≤ − i′ n′+1 i − n′ for 2 i n 1 (and all the other brackets [Y ,Y ] with 2 i < j − n 1 vanish), ≤ ≤ − i′ j′ ≤ ≤ − where a = K = 0. Changing Y , i 2, to a 1Y we get the relations for QR (as − 2,n−1 6 i′ ≥ − i′ n given in Remark 2). 2. In the both cases, n can be viewed as a real subspace of nC which is closed under the Lie bracket. In the case nC = L choose two elements Y = a X , Y = b X n such that n 1 i i i 2 i i i ∈ a1 = 0 and a1b2 a2b1 = 0 (this is always possiblePas the complexPification of n must be 6 − 6 SOLVABLE EXTENSIONS OF NEGATIVE RICCI CURVATURE OF FILIFORM LIE GROUPS 7 the whole L ). Replacing Y by Y Re(b /a )Y if necessary we can assume that b /a is n 2 2 1 1 1 1 1 − purely imaginary. Nowthevectors Y = [Y ,Y ] = (a b a b )X +(...), Y = [Y ,Y ] = 3 1 2 1 2 2 1 3 4 1 3 − a (a b a b )X + (...),...,Y = [Y ,Y ] = an 3(a b a b )X (where (...) is a 1 1 2 − 2 1 4 n 1 n−1 1− 1 2 − 2 1 n linear combination of the X with higher subscripts) is a real basis for n. Moreover, by i construction, [Y ,Y ] = Y for 2 i n 1 and [Y ,Y ] = 0 for i,j 3. Furthermore, 1 i i+1 i j ≤ ≤ − ≥ if b = 0, then the vector [Y ,Y ] = b (a b a b )X +(...) = (b /a )Y + (...) does 1 2 3 1 1 2 2 1 4 1 1 4 6 − not belong to n, as b /a is a nonzero imaginary number. It follows that b = 0, hence 1 1 1 [Y ,Y ] = 0 for i 3, as required. 2 i In the case nC≥ = Q , we start as above by constructing the basis Y for n with n i [Y ,Y ] = Y , 2 i n 1, such that Y = c X + (...), where (...) is a linear 1 i i+1 i i i ≤ ≤ − combination of the X , j > i, and c = 0. In particular, Y = c X , c = 0, lies in j i n n n n 6 6 the centre of n. Then [Y ,Y ] = K Y for 2 i,j n 1. The matrix K = (K ) i j ij n ij ≤ ≤ − is real and skew-symmetric. It cannot be singular, as otherwise the real span of the vectors Y ,...,Y in n would be a subalgebra with the centre of dimension greater than 2 n one, hence the same would be true for their complex span which is the subalgebra Span(X ,...,X ) nC. But the centre of this subalgebra has dimension one. It follows 2 n that K is nonsingu⊂lar and so the claim follows from assertion 1. (cid:3) From Lemma 1(1) and Lemma 2(2) it follows that we only need to prove Theorem 3 R R R for solvable extensions of the filiform algebras L and Q . The proof for L is given in n n n [NN, Theorem 4], so it remains only to consider the case n = QR, n = 2m, where we can n assume that m > 2 by Remark 2 and where, from now on, we will drop the superscript R, so that Q denotes the real Lie algebra given by the relations (8). n Wewill need aslightly moredetailed (ascomparedto (9)) knowledge ofthederivations of Q given in the following Lemma. n Lemma 3. Relative to the basis X from (8), the matrix of any derivation of Q is given i n by a b 0 0 d   a+d   (11) M =  2a+d ,  ...     ∗ (n 3)a+d   −   (n 3)a+2d   −  for some a,d,b R and possibly some nonzero elements below the diagonal marked by ∈ the asterisk. Proof. As the members of the lower central series are invariant with respect to a deriva- tion we obtain that all the entries of M above the diagonal in the columns starting from the third are zeros. Taking M = a, M = d we can find all the diagonal entries from 11 22 8 Y. NIKOLAYEVSKY the relations (8). The fact that M = 0 (which we will not need) follows by applying 21 the derivation to [X ,X ] = 0. (cid:3) 1 n 1 − 3. Proof of Theorem 3 As we have seen in Section 2.2, we only need to consider the case when g is a one- dimensional extension of n = Q , n = 2m, m > 2, by a non-nilpotent derivation. n Otherwise, the case when g is a one-dimensional extension of L was treated in [NN, n Theorem 4] (this proves Case (a) of Theorem 3), and in all the other cases when g is a non-nilpotent solvable Lie algebra having a filiform nilradical n, it follows from Lemma 1 and the arguments in Section 2.2 that for some Y g by which we are extending n, ∈ the restriction of the derivation ad to n has all eigenvalues positive, and so an inner Y product with Ric < 0 exists by Theorem 2(2). From Lemma 3 we see that the eigenvalues of any derivation of Q are a,d,a+d,2a+ n d,...,(n 3)a+d,(n 3)+2d. Now if we extend n = Q by at least three derivations, n − − thenoneofthemwill benilpotent, whichwedonotallow. Ifweextend Q byexactlytwo n derivations no nonzero linear combination of which is nilpotent, then there will exist a linear combination ofthem with all the eigenvalues positive (say the onewith a = d = 1). Then an inner product with Ric < 0 exists by Theorem 2(2). This, combined with the argument in the previous paragraph, completes the proof of Case (c) of Theorem 3. We can therefore assume that g is an extension of n by exactly one derivation (of the form (11)), which is not nilpotent, that is, either a or d is nonzero. What is more, as a unimodular g does not admit an inner product with Ric < 0 by Theorem 1, we can assume (changing the sign if necessary) that T = TrM = 1(n 1)(n 2)a + nd > 0. 2 − − Computing the numbers ι from Case (b) of Theorem 3 for such a derivation we obtain k that the only claim which remains to be proved is equivalent to the following statement. Proposition 1. Let g be a one-dimensional extension of n = Q , n = 2m,m > 2, by a n derivation (11). There exists an inner product on g with Ric < 0 if and only if n 3 − 1 (12) ι = (n 3)+ i a+(n k+2)d = ((n 3)n (k 3)(k 2))a+(n k+2)d> 0, k − − 2 − − − − − (cid:16) X (cid:17) i=k 2 − for all k = m+1,...,n (up to changing the sign of M if necessary). Remark 3. Note that inequalities (12) in fact imply T > 0, as T = 2 ι + n2 3n 6ι , n 3 n 1 2(−n 3−) n with both coefficients being positive when n > 4. − − − Proof. Necessity. Let g be the one-dimensional extension of n = Q by a derivation n φ whose matrix relative to the basis X satisfying (8) is given by (11). Suppose that i , is an inner product on g for which Ric < 0. Let f be a unit vector orthogonal to h· ·i n. Note that relative to the (in general, nonorthonormal) basis X for n the matrix of i the restriction of ad to n is proportional to M (some of the entries below the diagonal f may change, but we don’t care). Up to scaling the inner product and changing the sign SOLVABLE EXTENSIONS OF NEGATIVE RICCI CURVATURE OF FILIFORM LIE GROUPS 9 of f we can assume that (ad ) = M with T = TrM = 1(n 1)(n 2)a+ nd > 0. If f n 2 − − a = d, then the inequalities (12) are satisfied and there is nothing to prove. Assuming a = d we can modify the basis X by changing X to X b(a d) 1X to eliminate b i 2 2 − 1 6 − − from M, still keeping the relations (8) unchanged. We keep the notation X for this new i basis. Finally, let e ,e ,...,e ,e be the orthonormal basis for n constructed by the n n 1 2 1 − Gram-Schmidt procedure from the basis X ,X ,...,X ,X . We have the following n n 1 2 1 − Lemma. Lemma 4. 1. For i 1, Span(X ,...,X ) = Span(e ,...,e ), in particular e is a nonzero mul- i n i n n ≥ tiple of X and spans the centre of n and Span(e ,...,e ) = Span(X ,...,X ). n 2 n 2 n 2. Relative to the orthonormal basis e for n, the matrix A of the restriction of ad to n is i f lower-triangular, with the same diagonal entries as M, that is, λ = a, λ = d+(i 2)a 1 i − for 2 i n 1, and λ = 2d+(n 3)a (from the top left to the bottom right corner). n ≤ ≤ − − 3. For 2 i n 2, [e ,e ] = c e +e , where c = 0 and e Span(e ,...,e ). ≤ ≤ − 1 i i i+1 ′i+1 i 6 ′i+1 ∈ i+2 n For 2 i,j n 1, [e ,e ] = Ke ,e e , where K is a skew-symmetric matrix, i j i j n ≤ ≤ − h i with [e ,e ] = 0 for 2 i,j n 1, i+j > n+1. i j ≤ ≤ − Proof. Assertion 1 follows directly from our construction. Assertion 2 follows from as- sertion 1 and from the fact that M is lower-triangular. The first two statements of assertion 1 also follow from assertion 1 and relations (8). The fact that [e ,e ] = 0 for i j 2 i,j n 1, i+j > n+1, can be proved similar to that in the proof of Lemma 2(1): ≤ ≤ − from the Jacobi identities it follows that K + K = 0 for 2 i,j n 2 and i,j+1 i+1,j K = 0 for 2 i n 2, which implies the claim. ≤ ≤ − (cid:3) i+1,n 1 − ≤ ≤ − By Lemma 4(1) the subspaces i defined before Theorem 3 coincide with the subspaces k Span(e ,...,e ). Denote π End(n) the orthogonal projection to i . The proof of the k n k k ∈ necessity is based on estimating Trπ R . From Lemma 4(2) we have k 1 n n m (13) Tr(π [A,At]) = ( Ate 2 Ae 2) = A2 0, k k jk −k jk ji ≥ Xj=k Xj=k Xi=1 as A is lower-triangular. Furthermore, again by Lemma 4(2), n (14) Tr(π As) = Tr(π A) = λ = ι . k k j k X j=k From (7) we obtain 1 Ricne ,e = [e ,e ],e 2 n n i j n h i 4 h i Xi,j (15) n 1 k 1 n 1 n 1 1 − 1 − 1 − − = [e ,e ],e 2 + [e ,e ],e 2 + [e ,e ],e 2, 1 j n i j n i j n 2 h i 4 h i 2 h i Xj=2 iX,j=2 Xi=k Xj=2 10 Y. NIKOLAYEVSKY where we used the fact that k m + 1 and that [e ,e ] = 0 for 2 i,j n 1 with i j ≥ ≤ ≤ − i+j > n+1 from Lemma 4(3). Again from (7), with l = k,...,n 1, we have − 1 1 Ricne ,e = [e ,e ] 2 + [e ,e ],e 2 l l i l i j l h i −2 k k 4 h i Xi Xi,j n 1 n 1 1 − 1 − = [e ,e ] 2 + [e ,e ] 2 + [e ,e ],e 2 1 l i l 1 j l −2 k k k k 2 h i (cid:16) Xi=2 (cid:17) Xj=2 n 1 1 1 − 1 = [e ,e],e 2 [e ,e ],e 2 + [e ,e ],e 2, 1 l j i l n 1 j l −2 h i − 2 h i 2 h i l<Xj n Xi=2 2Xj<l ≤ ≤ where in the last line we used Lemma 4(3). From this and (15) we now obtain n 1 k 1 − 1 − Trπ Ricn = Ricne ,e + Ricne ,e = [e ,e ],e 2 k n n l l i j n h i h i 4 h i Xl=k iX,j=2 n 1 n 1 n n 1 l 1 1 − 1 − 1 − − + [e ,e ],e 2 [e ,e ],e 2 + [e ,e ],e 2 1 j n 1 l j 1 j l 2 h i − 2 h i 2 h i Xj=2 Xl=k jX=l+1 Xl=k Xj=2 k 1 n 1 n n l 1 1 − 1 − 1 − = [e ,e ],e 2 [e ,e ],e 2 + [e ,e ],e 2 i j n 1 j l 1 j l 4 h i − 2 h i 2 h i iX,j=2 Xj=k l=Xj+1 Xl=k Xj=2 k 1 n k 1 1 − 1 − = [e ,e ],e 2 + [e ,e ],e 2 0, i j n 1 j k 4 h i 2 h i ≥ iX,j=2 Xl=k Xj=2 Using this together with (13, 14) we find from (4) 1 Trπ R = Trπ (Ricn+ [A,At] TAs) Tι . k 1 k 2 − 1 ≥ − k As Ric < 0, the left-hand side must be negative, which proves the necessity of (12). Sufficiency. Suppose g = RY n is a one-dimensional extension of n = Q by a n ⊕ derivation M (which must be of the form (11) relative to the basis X by Lemma 3) such i that inequalities (12) are satisfied. We want to construct an inner product of negative Ricci curvature on g. First, if a = d, then all the eigenvalues of M are positive (up to changing the sign of M) and the existence of a required inner product follows from Theorem 2(2). Assuming a = d, as in the proof of the necessity above, we can modify the basis X by changing X i 2 6 to X b(a d) 1X to eliminate b from M, still keeping the relations (8) unchanged. 2 − 1 − − We keep the notation X for this new basis and the notation M for the matrix of our i derivation, which is now lower-triangular, with the same diagonal entries, namely λ = a, 1 λ = d+(i 2)a for 2 i n 1, and λ = 2d+(n 3)a (from the top left to the bottom i n − ≤ ≤ − − right corner). Next, let N be a positive derivation of n which is diagonal relative to the