Metric nilpotent Lie algebras of dimension 5

Metric nilpotent Lie algebras of dimension 5 Ágota Figula and Péter T. Nagy University of Debrecen, University of Óbuda 16-17.06.2017, GTG, University of Trento

Preliminaries Denition Let g be a Lie algebra and G be the corresponding connected and simply connected Lie group. A metric Lie algebra (g,.,. ) is a Lie algebra g together with a Euclidean inner product.,. on g. This inner product.,. on g induces a left invariant Riemannian metric on the Lie group G in a natural way. Let (n,.,. ) be a nilpotent metric Lie algebra. The corresponding nilpotent Lie group N endowed with the left-invariant metric arising from.,. is a Riemannian nilmanifold. Denote OA(n) the group of orthogonal automorphisms of (n,.,. ), which preserve the inner product. E. Wilson in Isometry groups on homogeneous nilmanifolds, Geom. Dedicata 12 (1982), 337-346, has been proved that the group I(N) of isometries of (N,.,. ) (distance preserving bijections) is the semi-direct product OA(n) N of the group OA(n) and the group N itself, which is normal in the group of isometries.

E. Wilson described a classication procedure for the isometry equivalence classes of Riemannian nilmanifolds. This is applied by J. Lauret in Homogeneous Nilmanifolds of Dimensions 3 and 4, Geometriae Dedicata 68, (1997), 145-155, for the determination of the 3- and 4-dimensional Riemannian nilmanifolds up to isometry and their isometry groups. Sz. Homolya and O. Kowalski have classied in Simply connected two-step homogeneous nilmanifolds of dimension 5, Note Math. 26 (2006), 69-77, all 5-dimensional 2-step nilpotent Riemannian nilmanifolds and their isometry groups. Together with P.T. Nagy we want to determine explicitly all 5-dimensional non 2-step Riemannian nilmanifolds and the groups of their isometries.

A subalgebra h of a metric Lie algebra g is totally geodesic if for all Y, Z h, X h one has [X, Y ], Z + [X, Z], Y = 0. This denition is chosen so that the corresponding Lie subgroup H of h is a totally geodesic submanifold relative to the left invariant Riemannian metric dened by the inner product on the simply connected Lie group G of g. This notion is motivated by the fact that the left cosets of H dene a totally geodesic foliation on G. P.T. Nagy and Sz. Homolya in Geodesic vectors and subalgebras in two-step nilpotent metric Lie algebras, Adv. Geometry 15 (2015) have been proved that for 2-step nilpotent metric Lie algebras the linear structure of at totally geodesic subalgebras does not depend on the choice of the inner product only on the isomorphism class of the nilpotent Lie algebra n.

Moreover, in G. Cairns, A. Hini Gali, Yu. Nikolayevsky, Totally geodesic subalgebras of liform nilpotent Lie algebras, J. Lie Theory, 23 (2013) the authors determine the maximal dimension of totally geodesic subalgebras of liform nilpotent metric Lie algebras and show that this bound is attained. Together with P.T. Nagy we wish to determine the standard liform metric Lie algebras.

Canonical basis of non two-step nilpotent Lie algebras of dimension 5 We consider the non two-step nilpotent Lie algebras of dimension 5 which are not direct products of Lie algebras of lower dimension. According to W. A. de Graaf: Classication of 6-dimensional nilpotent Lie algebras over elds of characteristic not 2, J. Algebra 309 (2007), 640 653, the list of these Lie algebras is given up to isomorphism by the non-vanishing commutators with respect to a distinguished basis {E 1, E 2,... }: l 5,5 : [E 1, E 2 ] = E 4, [E 1, E 4 ] = E 5, [E 2, E 3 ] = E 5 ; l 5,6 : [E 1, E 2 ] = E 3, [E 1, E 3 ] = E 4, [E 1, E 4 ] = E 5, [E 2, E 3 ] = E 5 ; l 5,7 : [E 1, E 2 ] = E 3, [E 1, E 3 ] = E 4, [E 1, E 4 ] = E 5 ; l 5,9 : [E 1, E 2 ] = E 3, [E 1, E 3 ] = E 4, [E 2, E 3 ] = E 5. This basis we call the canonical basis of the corresponding Lie algebra.

Heuristic procedure for the classication of metric Lie algebras We use the following heuristic procedure for the classication of metric Lie algebras up to isometric isomorphisms: 1 Select a basis E = {E 1, E 2,..., E n } of the Lie algebra g, such that the commutation relations have a simple form. 2 For an inner product.,. on g let F = {F 1, F 2,..., F n } be the orthonormal basis of the form F i = n k=i a ike k with a ik R and a ii > 0 obtained from E = {E 1, E 2,..., E n } by the Gram-Schmidt process. These bases parametrize the inner products on g. 3 Compute the Lie bracket expressions with respect to the basis F and nd their general shapes depending on real parameters. Find for each Lie bracket operation given by these real parameters all possible inner products. The class of these inner products determines the class of isometric isomorphic metric Lie algebras.

1 Choose for any class of isometric isomorphic metric Lie algebras a representing inner product expressed by its orthonormal basis F. These inner products together with the Lie bracket operation give a classication of metric Lie algebras. We illustrate this classication method on the Lie algebra l 5,5 : [E 1, E 2 ] = E 4, [E 1, E 4 ] = E 5, [E 2, E 3 ] = E 5. We nd the following series i 5 < i 4 < < i 1 < l 5,5 = i 0 of ideals of l 5,5 with dim(i k /i k+1 ) = 1: the center Z(l 5,5 ) of l 5,5 is R E 5, the commutator subalgebra l 5,5 is R E 4 + R E 5. The preimage π 1 (Z(l 5,5 /Z(l 5,5 ))) of the center Z(l 5,5 /Z(l 5,5 )) of the factor algebra l 5,5 /Z(l 5,5 ) in l 5,5 is R E 3 + R E 4 + R E 5 and the centralizer C l5,5 (l 5,5 ) of l 5,5 is R E 2 + R E 3 + R E 4 + R E 5. The Gram-Schmidt process applied to the ordered canonical basis (E 5, E 4, E 3, E 2, E 1 ) yields an orthonormal basis {F 1, F 2, F 3, F 4, F 5 } of l 5,5, where the vector F i is a positive multiple of E i modulo the subspace E j ; j > i generated by {E j ; j > i} and orthogonal to E j ; j > i.

We have F 5 Z(l 5,5 ), F 4 l, F 5,5 3 π 1 (Z(l 5,5 /Z(l 5,5 ))) and F 2 C l5,5 (l ). Hence l 5,5 5,5 has the orthogonal direct sum decomposition l 5,5 = RF 1 RF 5 into one-dimensional subspaces RF 1,..., RF 5 which is determined uniquely by the algebraic and metric structure of (l 5,5,.,. ). Since F i = 5 k=i a ike k with a ii > 0 we have (1) [F 1, F 2 ] = af 4 +bf 5, [F 1, F 3 ] = cf 5, [F 1, F 4 ] = df 5, [F 2, F 3 ] = ff 5, a, d, f > 0, a, b, c, d, f R, where a = a 11a 22 a 44, b = a 44 (a 11a 24 + a 12a 23 a 13a 22) a 11a 22a 45 a 44 a 55, c = a 11a 34 + a 12 a 33 a 55, d = a 11a 44 a 55, f = a 22a 33 a 55.

Denition Let {G 1, G 2, G 3, G 4, G 5 } be an orthonormal basis in the Euclidean vector space E 5 and a, b, c, d, f real numbers with a, d, f 0. Let n 5,5 (a, b, c, d, f ) denote the metric Lie algebra dened on E 5 by the non-vanishing commutators [G 1, G 2 ] = ag 4 +bg 5, [G 1, G 3 ] = cg 5, [G 1, G 4 ] = dg 5, [G 2, G 3 ] = fg 5. The map E 1 E 1, E 2 ade 2 + be 4, E 3 f ad E 3 + ce 4, E 4 de 4, E 5 E 5 is an isomorphism l 5,5 n 5,5 (a, b, c, d, f ). Changing the orthonormal basis: F 1 = F 1, F 2 = F 2, F 3 = F 3, F 4 = F 4, F 5 = F 5 we obtain [ F 1, F 2 ] = a F 4 b F 5, [ F 1, F 3 ] = c F 5, [ F 1, F 4 ] = d F 5, [ F 2, F 3 ] = f F 5.

Similarly, with the change of the basis: F 1 = F 1, F 2 = F 2, F 3 = F 3, F 4 = F 4, F 5 = F 5 one has [ F 1, F 2 ] = a F 4 +b F 5, [ F 1, F 3 ] = c F 5, [ F 1, F 4 ] = d F 5, [ F 2, F 3 ] = f F 5. Hence there is an orthonormal basis such that in the commutators (1) the coecients b and c are non-negative. Theorem Let.,. be an inner product on the 5-dimensional three-step nilpotent Lie algebra l 5,5. There is a unique metric Lie algebra n 5,5 (a, b, c, d, f ) with a, d, f > 0, b, c 0, which is isometrically isomorphic to the metric Lie algebra (l 5,5,.,. ).

It turns out that up to one exceptional class all higher-step nilpotent metric Lie algebras of dimension 5 have such orthogonal direct sum decomposition: Denition Let (g,.,. ) be a metric Lie algebra of dimension n. An orthogonal direct sum decomposition g = V 1 V n on one-dimensional subspaces V 1,..., V n will be called a framing of (g,.,. ), if it is determined uniquely by the algebraic and metric structure of (g,.,. ). The metric Lie algebra (g,.,. ) is called framed, if it has a framing.

Let (g,.,. ) and (g,.,. ) be framed metric Lie algebras of dimension n with framings g = R e 1 R e n and g = R e R 1 e n, where (e 1,..., e n ), respectively (e,..., 1 e n) are orthonormal bases. If Φ : g g is an isometric isomorphism, it maps R e i R ei, i = 1,..., n, i.e. Φ(e i ) = ε i ei with ε i = ±1. Hence Φ[e i, e j ] = Σ n k=1 ck i,jε k e k = [Φ(e i), Φ(e j )] = ε i ε j Σ n k=1 c k i,je k, consequently c k i,j = c k i,j. Lemma Assume that the commutators [.,.] in g and [.,.] in g are given in the form [e i, e j ] = Σ n k=1 ck i,j e k, [e i, e j ] = Σ n k=1 c k i,je k, i, j, k = 1,..., n. Then one has c k i,j = ±c k i,j for all i, j, k = 1,..., n. Particularly, if c k i,j 0 and c k i,j 0 then c k i,j = c k i,j.

Theorem The connected component of the isometry group I(N) of a simply connected Riemannian nilmanifold (N,.,. ) corresponding to the framed metric Lie algebra (n,.,. ) is isomorphic to the Lie group N. Proof. Lemma 1 yields that any orthogonal automorphism Φ : n n is given by Φ(e i ) = ε i e i with ε i = ±1. Hence the group OA(n) of orthogonal automorphisms of (n,.,. ) is a subgroup of Z 2 Z 2, where the number of factors dim n. Since I(N) = N OA(n) the assertion follows. Hence the Lie algebra n 5,5 (a, b, c, d, f ), a > 0, b 0, c 0, d > 0, f > 0 is isometrically isomorphic to n 5,5 (a, b, c, d, f ), a > 0, b 0, c 0, d > 0, f > 0 precisely if a = a, b = b, c = c, d = d, f = f.

n 5,5 (a, b, c, d, f ): [G 1, G 2 ] = ag 4 + bg 5, [G 1, G 3 ] = cg 5, [G 1, G 4 ] = dg 5, [G 2, G 3 ] = fg 5. If T : n 5,5 (a, b, c, d, f ) n 5,5 (a, b, c, d, f ) is an orthogonal automorphism of n 5,5 (a, b, c, d, f ) then TG i = ε i G i and [TG i, TG j ] = [ε i G i, ε j G j ] = T [G i, G j ] for i, j = 1,..., 5, where ε i, ε j = ±1. Let b = c = 0. From the Lie brackets [ε 1 G 1, ε 2 G 2 ] = aε 4 G 4, [ε 1 G 1, ε 4 G 4 ] = dε 5 G 5, [ε 2 G 2, ε 3 G 3 ] = f ε 5 G 5 we obtain ε 4 = ε 1 ε 2, ε 5 = ε 1 ε 4 = ε 2 ε 3. Hence ε 2 = ε 5 = ε 2 ε 3, and ε 3 = 1, ε 1 ε 4 = ε 2 = ε 5. It follows that the group of orthogonal automorphisms of n (a,0,0,d,f ) is isomorphic to the group Z 2 Z 2. If b = 0, c > 0, then we have in addition [ε 1 G 1, ε 3 G 3 ] = cε 5 G 5, which yields that ε 1 ε 3 = ε 5. Hence we get ε 1 = ε 2 = ε 5, ε 3 = 1 = ε 4. The group of orthogonal automorphisms of n (a,0,c,d,f ) is isomorphic to the group Z 2.

Theorem Let.,. be an inner product on the 5-dimensional three-step nilpotent Lie algebra l 5,5. (1) There is a unique metric Lie algebra n 5,5 (a, b, c, d, f ) with a, d, f > 0, b, c 0, which is isometrically isomorphic to the metric Lie algebra (l 5,5,.,. ). (2) The group of orthogonal automorphisms of n 5,5 (a, b, c, d, f ) is isomorphic to the matrix group: 1 for b = c = 0: (2) ε 1 0 0 0 0 0 ε 1 ε 4 0 0 0 0 0 1 0 0 0 0 0 ε 4 0 0 0 0 0 ε 1 ε 4, ε 1, ε 4 = ±1 = Z 2 Z 2,

1 for b = 0, c > 0: (3) 2 for b > 0, c = 0: (4) ε 1 0 0 0 0 0 ε 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 ε 1 1 0 0 0 0 0 ε 2 0 0 0 0 0 1 0 0 0 0 0 ε 2 0 0 0 0 0 ε 2 3 if b > 0, c > 0, then it is trivial with respect to the basis {G 1, G 2, G 3, G 4, G 5 }., ε 1 = ±1 = Z 2,, ε 2 = ±1 = Z 2,

The Lie algebra l 5,9 having two-dimensional center Denition Let {G 1, G 2, G 3, G 4, G 5 } be an orthonormal basis in the Euclidean vector space E 5. Let n 5,9 (l, m, n, p, q), l, m, n, p, q R with l, p, q 0, denote the metric Lie algebra dened on E 5 by the non-vanishing commutators (5) [G 1, G 2 ] = lg 3 + mg 4 + ng 5, [G 1, G 3 ] = pg 4, [G 2, G 3 ] = qg 5. If n = 0, p = q we denote ñ 5,9 (l, m, p) = n 5,9 (l, m, 0, p, p). The map G 1 G 1 + n q G 3, G 2 G 2 m p G 3, G 3 lg 3, G 4 plg 4, G 5 qlg 5 is an isomorphism n 5,9 (l, m, n, p, q) to l 5,9. Theorem The metric Lie algebra (l 5,9,.,. ) is isometrically isomorphic to a unique n 5,9 (l, m, n, p, q) with l, m, n, p, q R such that l > 0, q > p > 0 and m, n 0, or to a unique ñ 5,9 (l, m, p) with l, m, p R such that l, p > 0 and m 0.

The Lie algebra l 5,9 is dened by: [E 1, E 2 ] = E 3, [E 1, E 3 ] = E 4, [E 2, E 3 ] = E 5. The center Z(l 5,9 ) of l 5,9 is R E 4 + R E 5 and the commutator subalgebra l is R E 5,9 3 + R E 4 + R E 5. Let be given an inner product.,. on l 5,9 and apply the Gram-Schmidt process to the ordered canonical basis (E 5, E 4, E 3, E 2, E 1 ) of l 5,9. We obtain an orthonormal basis Ei = 5 k=i a ike k, i = 5,..., 1 of l 5,9 with a ii > 0 such that [E 1, E 2 ] = le 3 +me 4 +ne 5, [E 1, E 3 ] = pe 4 +re 5, [E 2, E 3 ] = qe 5, where l = a 11a 22 a 33 > 0, p = a 11a 33 a 44 > 0, q = a 22a 33 a 55 > 0.

To distinguish a 1-dimensional subspace of the center Z(l 5,9 ) = E, E 4 5 and a 1-dimensional subspace of the (l 5,9 ) = E, E 1 2 we consider the orthogonal unit vectors F (t), F (t + π ) in 2 (l 5,9 ) : F (t) = cos te 1 + sin te 2, F (t + π 2 ) = sin te 1 + cos te 2. We have [F (t), F (t + π )] = [E, E ] and 2 1 2 Φ(t) = [F (t), E ] = a 11a 33 3 a 44 cos te + 4 [ ] a 33 a12 a 44 a 11 a 45 cos t + a 22 sin t E 5, a 55 a 44 and Φ(t + π ) = [F (t + π ), E ] = a 11a 33 2 2 3 a 44 sin te + 4 [ ] a 33 a12 a 44 + a 11 a 45 sin t + a 22 cos t E 5. a 55 a 44

Φ(t 0 ) and Φ(t 0 + π ) are orthogonal if 2 1 (a 222a 244 a 211a ) 255 (a 12 a 44 a 11 a 45 ) 2 sin 2t 0 + 2 Lemma a 22 a 44 (a 12 a 44 a 11 a 45 ) cos 2t 0 = 0. The vectors Φ(t) and Φ(t + π ) are orthogonal for any t R if and 2 only if (6) a 12 a 44 = a 11 a 45 and a 11 a 55 = a 22 a 44. Otherwise, there exists a unique 0 t 0 < π such that Φ(t) and 2 Φ(t + π ) are orthogonal if and only if t = t 2 0 + k π, k Z. 2

In the second case we have either Φ(t 0 ) < Φ(t 0 + π 2 ) and hence we dene F 1 = F (t 0 ), F 2 = F (t 0 + π 2 ), F 3 = E 3, F 4 = Φ(t 0) Φ(t 0 ), (7) F 5 = Φ(t 0 + π 2 ) Φ(t 0 + π 2 ), or if Φ(t 0 ) = Φ(t 0 + π) > Φ(t 0 + π ), then we dene 2 F 1 = F (t 0 + π 2 ), F 2 = F (t 0 + π), F 3 = E 3, F 4 = Φ(t 0 + π 2 ) Φ(t 0 + π 2 ), (8) F 5 = Φ(t 0 + π) Φ(t 0 + π).

From this construction we obtain that (9) [F 1, F 2 ] = lf 3 + mf 4 + nf 5, [F 1, F 3 ] = pf 4, [F 2, F 3 ] = qf 5 with l > 0, q > p > 0. Using the basis change F 1 = F 1, F 2 = F 2, F 3 = F 3, F 4 = F 4, F 5 = F 5 one has [ F 1, F 2 ] = l F 3 m F 4 + n F 5, [ F 1, F 3 ] = p F 4, [ F 2, F 3 ] = q F 5. With the basis change F 1 = F 1, F 2 = F 2, F 3 = F 3, F 4 = F 4, F 5 = F 5 one has [ F 1, F 2 ] = l F 3 + m F 4 n F 5, [ F 1, F 3 ] = p F 4, [ F 2, F 3 ] = q F 5. Hence we can choose an orthonormal basis such that in the commutators (9) the coecients satisfy m, n 0. The one-dimensional subspaces R F 1, R F 2, R F 3, R F 4, R F 5 form a framing of (l 5,9,.,. ) since the subspace RF 5 Z(l 5,9 ) is generated by the vector in {Φ(t 0 ), Φ(t 0 + π )} having greater norm. 2 The subspace RF 4 Z(l 5,9 ) is orthogonal to RF 5. The subspace R F 3 is contained in the commutator subalgebra and orthogonal to the center. The orthogonal one-dimensional subspaces R F 1, R F 2 are orthogonal to the commutator subalgebra and the subspace R[F 2, F 3 ] is contained in RF 5.

If Φ(t) and Φ(t + π ) are orthogonal for any t R, then 2 Φ(t) = Φ(t + π ) = a 11a 33 2 a 44 = const. For E, E 4 5 we obtain E = a ( 44 4 a 11 a 33 cos tφ(t) sin tφ(t + π )), 2 E = a ( 44 5 a 11 a 33 sin tφ(t) + cos tφ(t + π )). The Lie bracket 2 [F (t), F (t + π )] = [E, E ] = le + me + ne 2 1 2 3 4 5, l > 0, can be written into the form l E + 3 + a 44 a 11 a 33 { (m cos t + n sin t)φ(t) + ( m sin t + n cos t)φ(t + π 2 ) }. If m = n = 0 we put F i = Ei. For (m, n) (0, 0) there is unique t 0 ( π, π ] such that the solutions of the equation 2 2 m sin t + n cos t = 0 are t 0 + k π, k Z. Then we dene 2 F 1 = F (t 0 ), F 2 = F (t 0 + π ), 2 F 3 = E 3, F 4 = a 44 a 11 a 33 Φ(t 0 ), F 5 = a 44 a 11 a 33 Φ(t 0 + π 2 ).

We obtain that the non-vanishing brackets [F i, F j ], i, j = 1,..., 5, have the shape (10) [F 1, F 2 ] = lf 3 + mf 4, [F 1, F 3 ] = pf 4, [F 2, F 3 ] = pf 5, with some m R, l, p > 0. Using the isometric isomorphism F 1 = F 1, F2 = F 2, F3 = F 3, F4 = F 4, F5 = F 5 we obtain [ F 1, F 2 ] = l F 3 m F 4, [ F 1, F 3 ] = p F 4, [ F 2, F 3 ] = p F 5. Hence we can assume m 0. If m 0, then the subspace R F 5 Z(ñ 5,9 (l, m, p) is orthogonal to the Lie bracket of any two vectors contained in R F 1 R F 2. The subspace RF 4 Z(ñ 5,9 (l, m, p)) is orthogonal to RF 5. The subspace R F 3 is contained in the commutator subalgebra and orthogonal to the center. The orthogonal one-dimensional subspaces R F 1, R F 2 are orthogonal to the commutator subalgebra and the subspace R[F 2, F 3 ] is contained in RF 5. Hence the subspaces R F 1, R F 2, R F 3, R F 4, R F 5 form a framing of the metric Lie algebra (ñ 5,9 (l, m, p),.,. ).

If m = 0 then we have (11) [F 1, F 2 ] = lf 3, [F 1, F 3 ] = pf 4, [F 2, F 3 ] = pf 5. From this we can see that Z(ñ 5,9 (l, 0, p)) is orthogonal to the Lie bracket of any two vectors contained in R F 1 R F 2. Hence the metric Lie algebra (ñ 5,9 (l, 0, p),.,. ) is not framed. In this case for any isometric isomorphism Φ : ñ 5,9 (l, 0, p) ñ 5,9 (l, m, p ) one has Φ(F 1 ) = cos tf 1 ± sin tf 2, Φ(F 2 ) = sin tf 1 + cos tf 2, Φ(F 3 ) = εf 3, Φ(F 4 ) = cos tf 4 ± sin tf 5, Φ(F 5 ) = sin tf 4 + cos tf 5 from which easily follows m = 0, l = l, p = p.

Theorem Let.,. be an inner product on the 5-dimensional three-step nilpotent Lie algebra l 5,9. (1) The metric Lie algebra (l 5,9,.,. ) is isometrically isomorphic to a unique n 5,9 (l, m, n, p, q) with l, m, n, p, q R such that l > 0, q > p > 0 and m, n 0, or to a unique ñ 5,9 (l, m, p) with l, m, p R such that l, p > 0 and m 0. (2) The groups of orthogonal automorphisms are the following matrix groups with respect to the basis {G 1, G 2, G 3, G 4, G 5 }: (A) for n 5,9 (l, m, n, p, q) (i) if m = n = 0: (12) ε 1 0 0 0 0 0 ε 2 0 0 0 0 0 ε 1ε 2 0 0 0 0 0 ε 2 0 0 0 0 0 ε 1, ε1, ε2 = ±1 = Z 2 Z 2,

(ii) if m = 0, n > 0: (13) (iii) if m > 0, n = 0: (14) ε 1 0 0 0 0 0 1 0 0 0 0 0 ε 1 0 0 0 0 0 1 0 0 0 0 0 ε 1 1 0 0 0 0 0 ε 2 0 0 0 0 0 ε 2 0 0 0 0 0 ε 2 0 0 0 0 0 1 (iv) if m, n > 0, then it is trivial;, ε 1 = ±1 = Z 2,, ε 2 = ±1 = Z 2,

For the Lie algebra ñ 5,9 (l, m, p): (i) if m = 0: (15) cos t ε 2 sin t 0 0 0 sin t ε 2 cos t 0 0 0 0 0 ε 2 0 0 0 0 0 ε 2 cos t sin t, ε 2 = ±1, t [0, 2π), 0 0 0 ε 2 sin t cos t (ii) if m > 0, then it is the group (16) 1 0 0 0 0 0 ε 2 0 0 0 0 0 ε 2 0 0 0 0 0 ε 2 0 0 0 0 0 1, ε 2 = ±1 = Z 2.

Theorem Every liform metric Lie algebra is a framed metric Lie algebra. Proof. Every liform nilpotent Lie algebra g has a basis {E 1,, E n }: (17) [E 1, E i ] = E i+1, for all i 2, (dening relations of the standard liform Lie algebras) (18) [E i, E j ] g i+j = Span(E i+j,, E n ), for all i, j : i+j n+1 and there exists α R with (19) [E i, E n i+1 ] = ( 1) i αe n, for all 2 i n 1. If n is odd, then α = 0. One set E i = 0 for i > n (cf. Theorem 4.1 in G. Cairns, A. Hini Gali, Yu. Nikolayevsky, Totally geodesic subalgebras of nilpotent Lie algebras, J. Lie Theory, 23 (2013), 1023 1049).

Proof. The lower central series C 0 (g) = g,..., C i (g) = [g, C i 1 (g)] = R E i+2 + + R E n,..., C n 2 (g) of g forms a series of invariant ideals with dim(c i (g)/c i+1 (g)) = 1. The Gram-Schmidt process applied to the ordered basis (E n, E n 1,..., E 2, E 1 ) yields an orthonormal basis F = {F 1, F 2,..., F n } such that the commutator subalgebra g = C 1 (g) has a framing g = R F 3 R F n. The orthogonal complementer of the commutator subalgebra has dimension 2. If g is a standard liform Lie algebra, then the centralizer C g (g ) of the commutator algebra g is the ideal R E 2 + + R E n. Hence the decomposition g = R F 1 R F n is a framing of the metric standard liform algebra (g,.,. ).

Proof. From the commutator relations (17-19) we have [E 2, C 1 (g)] C 3 (g) = E 5,, E n. The factor Lie algebra g/c 3 (g) = Ē 1, Ē 2, Ē 3, Ē 4 is determined by the non-vanishing Lie brackets [Ē 1, Ē 2 ] = Ē 3, [Ē 1, Ē 3 ] = Ē 4. Hence it is isomorphic to the 4-dimensional standard liform Lie algebra s 4. The centre Z(g/C 3 (g)) of the factor Lie algebra g/c 3 (g) is R Ē4, the commutator subalgebra (g/c 3 (g)) of g/c 3 (g) is R Ē 3 + R Ē 4 and the centralizer C g/c 3 (g)(g/c 3 (g)) of the commutator subalgebra (g/c 3 (g)) of the factor algebra g/c 3 (g) is R Ē 2 + R Ē 3 + R Ē 4. Therefore the preimage of the centralizer C g/c 3 (g)(g/c 3 (g)) in g is R E 2 + R E 3 + R E 4 + R E 5 + + R E n. Applying the Gram-Schmidt process to the ordered basis (E n,, E 1 ) we obtain an orthonormal basis {F 1,, F n } such that the decomposition R F 1 R F n is a framing of the metric liform algebra (g,.,. ).

s n : [E 1, E 2 ] = E 3,..., [E 1, E i ] = E i+1,..., [E 1, E n 1 ] = E n. [G 1, G 2 ] = c 2,2 G 3 + c 2,3 G 4 + + c 2,n 1 G n [G 1, G i ] = c i,i G i+1 + + c i,n 1 G n [G 1, G n 1 ] = c n 1,n 1 G n. Denition Let {G 1,..., G n } be an orthonormal basis in the n-dimensional Euclidean vector space E n and let C = {c j,k R; 2 k j n 1} be a lower triangular n 2 n 2 matrix with positive diagonal elements. We denote by n C the Lie algebra and by [.,.] C its Lie bracket dened on E n by the non-vanishing commutators (20) [G 1, G i ] C = [G i, G 1 ] C = n 1 t=i c t,i G t+1, i = 2,..., n 1. The Lie algebra n C equipped with the inner product.,. C induced by the Euclidean inner product of E n is a metric Lie algebra (n C,.,. C ).

Using an isometric isomorphism we achive that the sign of the coecients c k,i such that k i is odd simultaneously changed. If the set P = {(k, i) : c k,i 0 and k i is odd} is not empty, then there is a suitable orthonormal basis satisfying c k0,i 0 > 0 for the minimal element (k 0, i 0 ) of P with respect to the anti-lexicographic ordering of pairs. Theorem Let.,. be an inner product on the n-dimensional standard liform nilpotent Lie algebra s n. There is a unique metric Lie algebra (n C,.,. C ) satisfying 1 (n C,.,. C ) is isometrically isomorphic to (s n,.,. ), 2 if the set P = {(k, i) : c k,i 0 and k i is odd} is not empty then c k0,i 0 > 0 for the minimal element (k 0, i 0 ) of P with respect to the anti-lexicographic ordering of pairs.

Theorem The group of orthogonal automorphisms of n C is the matrix group 1 if {(i, k) : c k,i 0 and k i is odd} = : (21) ε 1 0 0... 0 0 ε 2ε 1 2 0... 0 0 0 ε 3ε 1 2... 0. 0 0 0.. 0 0 0 0 0 ε nε 1 2, ε 1, ε 2 = ±1 = Z 2 Z 2, 2 if {(i, k) : c k,i 0 and k i is odd} : 1 0... 0 0 ε 2... 0 (22). 0 0.. 0, ε 2 = ±1 = Z 2, 0 0 0 ε 2 with respect to the basis {G 1,..., G n }.

l 5,6 : [E 1, E 2 ] = E 3, [E 1, E 3 ] = E 4, [E 1, E 4 ] = E 5, [E 2, E 3 ] = E 5 Denition Let {G 1, G 2, G 3, G 4, G 5 } be an orthonormal basis in the Euclidean vector space E 5 and a, b, c, d, f, g, h real numbers with a, d, g, h 0. The metric Lie algebra dened on E 5 by the non-vanishing commutators [G 1, G 2 ] = ag 3 + bg 4 + cg 5, [G 1, G 3 ] = dg 4 + fg 5, [G 1, G 4 ] = gg 5, [G 2, G 3 ] = hg 5 will be denoted by n(a, b, c, d, f, g, h). The map E 1 E 1, E 2 w(adg E 2 + bg E 3 + (c bf d )E 4), E 3 wdg E 3, E 4 w(g E 4 f d E 5), E 5 we 5, is an isomorphism l 5,6 n(a, b, c, d, f, g, h), where w = h ad 2 g 2.

Theorem Let.,. be an inner product on the 5-dimensional liform but not standard liform nilpotent Lie algebra l 5,6. (1) The metric Lie algebra (l 5,6,.,. ) is isometrically isomorphic to a unique metric Lie algebra n(a, b, c, d, f, g, h) with a, b, c, d, f, g, h R such that either a, b, d, g, h > 0, or a, d, g, h > 0, b = 0, f 0. (2) The groups of orthogonal automorphisms of n(a, b, c, d, f, g, h) are the following matrix groups with respect to the basis {G 1, G 2, G 3, G 4, G 5 }: (i) for b = f = 0: (23) ε 1 0 0 0 0 0 1 0 0 0 0 0 ε 1 0 0 0 0 0 1 0 0 0 0 0 ε 1, ε 1 = ±1 = Z 2, (ii) for b 2 + f 2 0, it is trivial.