We show that every unimodular Lie algebra, of dimension at most 4, equipped
with an inner product, possesses an orthonormal basis comprised of geodesic
elements. On the other hand, we give an example of a solvable unimodular Lie
algebra of dimension 5 that has no orthonormal geodesic basis, for any inner
product

Cairns, Grant

Le, Nguyen Thanh Tung

Nielsen, Anthony

Nikolayevsky, Yuri

English

arXiv

We show that every unimodular Lie algebra, of dimension at most 4, equipped with an inner product, possesses an orthonormal basis comprised of geodesic elements. On the other hand, we give an example of a solvable unimodular Lie algebra of dimension 5 that has no orthonormal geodesic basis, for any inner product

ESE - Salento University Publishing

Note di Matematica ISSN 1123-2536, e-ISSN 1590-0932Note Mat. 33 (2013) no. 2, 11–18. doi:10.1285/i15900932v33n2p11On the existence of orthonormal geodesicbases for Lie algebrasGrant CairnsiDepartment of Mathematics and Statistics, La Trobe University, Melbourne, Australia 3086g.cairns@latrobe.edu.auNguyen Thanh Tung LeiiDepartment of Mathematics and Statistics, La Trobe University, Melbourne, Australia 3086thanhtung1303@gmail.comAnthony NielsenDepartment of Mathematics and Statistics, La Trobe University, Melbourne, Australia 3086anthony.nielsen@latrobe.edu.auYuri NikolayevskyiiiDepartment of Mathematics and Statistics, La Trobe University, Melbourne, Australia 3086y.nikolayevsky@latrobe.edu.auReceived: 8.1.2013; accepted: 6.2.2013.Abstract. We show that every unimodular Lie algebra, of dimension at most 4, equippedwith an inner product, possesses an orthonormal basis comprised of geodesic elements. On theother hand, we give an example of a solvable unimodular Lie algebra of dimension 5 that hasno orthonormal geodesic basis, for any inner product.Keywords: geodesic vector, unimodular Lie algebraMSC 2010 classification: Primary 53C22, 17B301 IntroductionLet g be a Lie algebra equipped with an inner product 〈·, ·〉. Consider thecorresponding simply connected Lie group G equipped with the left invariantRiemannian metric determined by 〈·, ·〉. A nonzero element Y ∈ g is said tobe a geodesic vector if the corresponding left invariant vector ﬁeld on G is ageodesic vector ﬁeld. In terms of the Levi-Civita connection ∇, this means that∇Y Y = 0. This has a simple equivalent expression in terms of the Lie bracket[9, 10, 12, 2], which we state as a deﬁnition.iPart of this work was supported by ARC Discovery grant DP130103485iiPart of this work was conducted while the second author was an AMSI summer vacationscholar.iiiPart of this work was supported by ARC Discovery grant DP130103485http://siba-ese.unisalento.it/ c© 2013 Universita` del Salento12 Cairns, Le, Nielsen, NikolayevskyDefinition 1. Let g be a Lie algebra equipped with an inner product 〈·, ·〉.Then a nonzero element Y ∈ g is a geodesic vector if [X,Y ] is perpendicular toY for all X ∈ g.Note that some authors use the term homogeneous geodesic, to distinguishthem from general geodesics on the underlying Lie group. Some authors insistfurther that Y have unit length. For the more general case of totally geodesicsubalgebras of Lie algebras, see [12, 2, 3].Remark 1. A useful equivalent reformulation of the deﬁnition of a geodesicvector is as follows: Y ∈ g is geodesic if and only if Im(ad(Y )) is contained inthe orthogonal complement of Y .Every Lie algebra possesses at least one geodesic vector [8, 10, 2]. In [10] it isshown that semisimple Lie algebras possess an orthonormal basis comprised ofgeodesics vectors, for every inner product. Results for certain solvable algebrasare given in [5]. In the present paper we examine the existence of geodesic basesin algebras of low dimension, see [4] for further results on geodesic base.2 Preliminary observationsFirst notice that if a Lie algebra g, equipped with an orthonormal innerproduct, has a basis {X1, . . . , Xn} of geodesic vectors, then for all i, j, the ele-ment [Xi, Xj ] is orthogonal to both Xi and Xj . Consequently, the adjoint mapsad(Xi) have zero trace, and hence g is unimodular. So the natural problem isto determine which unimodular Lie algebras possess an orthonormal basis com-prised of geodesic vectors, for some inner product. We begin with two elementaryobservations.Proposition 1. Let g be a nilpotent Lie algebra equipped with an inner prod-uct 〈·, ·〉. Then there is an orthonormal basis of (g, 〈·, ·〉) comprised of geodesicvectors.Proof. The proof is by induction on the dimension of g. The proposition istrivial when dim(g) = 1. Suppose that dim(g) = n+ 1. Since g is nilpotent, itscentre is nonzero. Let Z be a central element of length one and set z := Span(Z).Let π : g→ h := g/z be the natural quotient map and give h the inner productfor which the restriction of π to the orthogonal complement z⊥ of z is an isometry.By the inductive hypothesis, h possesses an orthonormal basis {X1, . . . , Xn}comprised of geodesic vectors. So, by deﬁnition, for each pair i, j, the element[Xi, Xj ] is perpendicular to both Xi and Xj . For each i, let X¯i denote the uniqueelement of z⊥ with π(X¯i) = Xi. So for each pair i, j, the element [X¯i, X¯j ]is perpendicular to both X¯i and X¯j . Thus, as Z is central, the elements X¯iare geodesic vectors. Furthermore, as Z is central, Z is also geodesic. HenceOrthonormal Geodesics Bases 13{X¯1, . . . , X¯n, Z} is an orthonormal basis of geodesic vectors. QEDProposition 2. Let g be a unimodular Lie algebra having a codimension oneabelian ideal h. Then, for every inner product 〈·, ·〉 on g, there is an orthonormalbasis of (g, 〈·, ·〉) comprised of geodesic vectors.Proof. We make use of the following linear algebra result.Lemma 1 ([15, Theorem 10]). Suppose that A is a real square matrix withzero trace. Then there is an orthogonal matrix Q such that QAQ−1 has zerodiagonal.Let X ∈ g be a unit vector orthogonal to h. Note that X is geodesic since hcontains the derived algebra g′ of g. Choose an orthonormal basis for h and letA denote the matrix representation of the restriction adh(X) to h of the adjointmap ad(X). Since g is unimodular, the matrix A has zero trace. By Lemma 1,there is an orthonormal basis {Y1, Y2, . . . , Yn} for h relative to which adh(X)has zero diagonal. Consequently for each i, the element [X,Yi] is perpendicu-lar to Yi. Thus since h is abelian, the elements Yi are geodesic vectors. Hence{X,Y1, Y2, . . . , Yn} is a geodesic basis for g. QED3 Main ResultsIt is well known that in dimension less than or equal to three, there areonly 5 nonabelian unimodular real Lie algebras; see [17]. Milnor’s classiﬁcation[11] proceeds by considering, for an orthonormal basis {X1, X2, X3}, the linearmap L(Xi × Xj) := [Xi, Xj ], where × denotes the vector cross product. Thematrix of L relative to {X1, X2, X3} is symmetric and so its eigenvectors forman orthonormal basis {Y1, Y2, Y3} with, by construction,[Y2, Y3] = λ1Y1,[Y3, Y1] = λ2Y2,[Y1, Y2] = λ3Y3,for real coeﬃcients λi. So the basis elements Y1, Y2, Y3 are geodesic vectors.In dimension 4, there are inﬁnitely many isomorphism classes of unimodularreal Lie algebras; see [13, 6, 16]. Instead of using the classiﬁcation, we will arguedirectly. The following result resolves a question raised in [14].Theorem 1. Let g be a unimodular Lie algebra of dimension 4 equipped withan inner product 〈·, ·〉. Then there is an orthonormal basis of (g, 〈·, ·〉) comprisedof geodesic vectors.Proof. Let g be as in the statement of the theorem. First observe that ifg is not solvable, then from Levi’s Theorem and the classiﬁcation of semisimple14 Cairns, Le, Nielsen, Nikolayevskyalgebras [7], there is a Lie algebra isomorphism g ∼= h ⊕ R where h is a simpleLie algebra of dimension three; so h is isomorphic to either so(3,R) or sl(2,R)though we won’t need this fact. Let W ∈ g be a unit vector orthogonal to h.Since h = g′, the element W is geodesic. Note that W can be written uniquelyin the form W = X + Z, where Z is in the centre of g and X ∈ h. As wediscussed above, h has an orthonormal basis {Y1, Y2, Y3} of elements that aregeodesic vectors of h. Then for each i = 1, 2, 3,[W,Yi] = [X,Yi] ∈ Im(ad(Yi)|h),which is perpendicular to Yi by Remark 1. Hence Yi is geodesic in g and thus{Y1, Y2, Y3,W} is an orthonormal basis of geodesic vectors. So we may assumethat g is solvable.Consider the derived algebra g′ := [g, g]. As g is solvable, g′ is nilpotent [1,Chap. 1.5.3]. If dim(g′) = 0, the algebra g is abelian and the theorem holdstrivially. We consider three cases according to the remaining possibilities for thedimension of g′.If dim(g′) = 1, let g′ = Span(W ). For each X ∈ g we have [X,Y ] ∈ g′ for allY ∈ g and hence [X,W ] = tr(ad(X))W = 0. Thus g′ is contained in the centreof g. Consequently g is nilpotent, and the theorem holds by Proposition 1.If dim(g′) = 3, then either g′ is abelian or g′ is isomorphic to the HeisenbergLie algebra h1. If g′ is abelian, the required result follows from Proposition2. So suppose that g′ is isomorphic to the Heisenberg Lie algebra. Choose anorthonormal basis {X,Y, Z} of g′ with Z in the centre of g′ and [X,Y ] = λZfor some λ 6= 0. Let W ∈ g be a unit vector orthogonal to g′. Note that W isgeodesic. Let A denote the matrix representation of the restriction adg′(W ) tog′ of the adjoint map ad(W ). Since the centre of g′ is a characteristic ideal [1,Chap. 1.1.3], it is left invariant by ad(W ). Hence A has the formA =a b 0c d 0e f g ,for reals a, b, c, d, e, f, g. By the Jacobi identity,[ad(W )X,Y ] + [X, ad(W )Y ] = ad(W )λZ = gλZ.Thus[aX + cY + eZ, Y ] + [X, bX + dY + fZ] = (a+ d)λZ = gλZ,and so a + d = g. Since ad(W ) has zero trace, a + d + g = 0, and hencea + d = g = 0. So, by Lemma 1, we can make an orthonormal change of basisOrthonormal Geodesics Bases 15for the space Span(X,Y ) so that relative to the new basis, a = d = 0. Then{W,X, Y, Z} is an orthonormal geodesic basis for g.Finally, if dim(g′) = 2, then g′ is abelian. Let {X,Y } be an orthonormalbasis for the orthogonal complement g′⊥ to g′. Note that [X,Y ] ∈ g′ and soas g′ is abelian, the adjoint map ad([X,Y ]) acts trivially on g′. Thus, sincead(X)◦ad(Y )−ad(Y )◦ad(X) = ad([X,Y ]) by the Jacobi identity, we have thatthe restrictions adg′(X) and adg′(Y ) commute. Since adg′(X) and adg′(Y ) havezero trace, relative to any choice of orthonormal basis for g′, they have matrixrepresentations in sl(2,R). However, it is well known and easy to see that themaximum abelian subalgebras of sl(2,R) have dimension one. Consequently, byorthonormal change of basis for g′⊥ if necessary, we may assume that adg′(Y ) ≡0. Then Span(Y, g′) is a codimension one abelian ideal and the result followsfrom Proposition 2. QEDNote that in Propositions 1, 2 and Theorem 1, the required orthonormalgeodesic basis is obtained for every inner product. The following example givesa solvable unimodular Lie algebra that has no orthonormal geodesic basis forany inner product.Example 1. Consider the 5-dimensional algebra g with basis B ={X1, . . . , X5} and relations[X1, X2] = 3X2 [X2, X3] = X4[X1, X3] = −4X3 [X2, X4] = X5[X1, X4] = −X4[X1, X5] = 2X5.The ideal n generated by X2, . . . , X5 is the (unique) 4-dimensional ﬁliformnilpotent Lie algebra. The adjoint map ad(X1) is obviously a Lie derivationof n, so the Jacobi identities of g hold. Clearly, g is solvable and unimodu-lar, but not nilpotent. We will show that g has no orthonormal basis com-prised of geodesic vectors, for any inner product. Equip g with an inner productand suppose it has a basis (not necessarily orthonormal) of geodesic elementsY1, . . . , Y5. Consider an arbitrary geodesic vector Y = a1X1+· · ·+a5X5 ∈ g. LetVi := Span(Xi, Xi+1, . . . , X5). Relative to the basis B, the matrix representationof ad(Y ) isA =0 0 0 0 0−3a2 3a1 0 0 04a3 0 −4a1 0 0a4 −a3 a2 −a1 0−2a5 −a4 0 a2 2a1 . (3.1)A priori, there are ﬁve cases:16 Cairns, Le, Nielsen, Nikolayevsky(1) Y ∈ V5; that is, a1 = a2 = 0, a3 = 0, a4 = 0, a5 6= 0,(2) Y ∈ V4\V5; that is, a1 = a2 = 0, a3 = 0, a4 6= 0,(3) Y ∈ V3\V4; that is, a1 = a2 = 0, a3 6= 0,(4) Y ∈ V2\V3; that is, a1 = 0, a2 6= 0,(5) Y ∈ V1\V2; that is, a1 6= 0.In fact, case (1) is impossible as otherwise Y is a nonzero multiple of X5 butIm(ad(Y )) = V5, which must be orthogonal to Y by Remark 1. Similarly, case(2) is impossible as otherwise Y ∈ V4 but Im(ad(Y )) = V4, which has to beorthogonal to Y by Remark 1.In case (3), Im(ad(Y )) = Span(X3 +a44a3X4 − 2a54a3X5, X4 + a4a3X5).In case (4), Im(ad(Y )) = Span(X2 − 4a33a2X3, X4, X5) and in particular, Y isorthogonal to V4.In case (5), Im(ad(Y )) = V2, which is the orthogonal complement of Yby Remark 1. It follows that, up to a constant multiple, there is at most onegeodesic vector Y with a1 6= 0. Thus, there is at most one basis element, Y1 say,with Y1 6∈ V2.So we have established thatY2, Y3, Y4, Y5 ∈ V2\V4.But if Yi ∈ V2\V3, then as we saw in case (4), Yi is orthogonal to V4. So fordimension reasons, there are at most two of the Yi in V2\V3.Now suppose the Yi are orthogonal. If two of the Yi are in V2\V3, theywould both be orthogonal to V4. This would force the remaining two Yi to bein V4, which we have seen is impossible. So only one of the Yi, say Y2, can bein V2\V3. So Y3, Y4, Y5 are in V3\V4. Note that V3 is left invariant by ad(X1).Relative to the orthonormal basis {Y3, Y4, Y5} for V3, the map f := ad(X1)|V3has zero diagonal, because the elements Y3, Y4, Y5 are geodesic vectors. So fhas zero trace. But relative to the basis X3, X4, X5, it is clear that f has trace−4− 1 + 2 = −3. This is a contradiction.Remark 2. Although the algebra of the above example does not possesan orthonormal basis of geodesic vectors, it does posses an inner product forwhich there is a (nonorthonormal) basis of geodesic vectors. That is, we claimthat there exists an inner product such that the span of all the geodesic vectorsis the whole algebra. Using the above notation, regardless of the choice of aninner product, there is exactly one geodesic vector Y1 (up to scaling) not lyingin g′ = V2, namely any nonzero vector from (g′)⊥. We therefore want to showOrthonormal Geodesics Bases 17that the span of all the geodesic vectors from g′ covers g′. We will now specifythe inner product. First choose X2, X3 to be orthonormal and orthogonal to V4.Then from case (4), a vector Y = a2X2 + a3X3 + a4X4 + a5X5 (with a2 6= 0) isgeodesic if it is orthogonal to V4 (so a4 = a5 = 0) and√3a2 = ±2a3. This givestwo linearly independent geodesic vectors Y2, Y3 ∈ g′ ∩ V ⊥4 , neither of whichlies in V3. It remains to show that there exist at least two geodesic vectorsin V3 whose projections to V4 are linearly independent. Deﬁne the remainingcomponents of the inner product by requiring that ‖X4‖ = ‖X5‖ = 1 and〈X4, X5〉 = ε ∈ (0, 1). Then by case (3), a vector Y = a3X3 + a4X4 + a5X5(with a3 6= 0) is geodesic if and only if the following two conditions hold.4a23 + a24 − 2a25 − εa4a5 = 0, (3.2)a3(a4 + εa5) + a4(εa4 + a5) = 0. (3.3)Note that a4 + εa5 6= 0, as otherwise from (3.3) either a4 = 0 or εa4 + a5 = 0and in both cases we obtain a4 = a5 = 0 and then (3.2) would give a3 = 0,which is a contradiction. Solving (3.3) for a3 and substituting to (3.2) we get4a24(εa4 + a5)2 + (a24 − 2a25 − εa4a5)(a4 + εa5)2 = 0. (3.4)Note that as a3 6= 0 and a4 + εa5 6= 0, (3.3) gives a4 6= 0. Dividing (3.4) by a44and taking t = a5a−14 we obtain−2ε2t4 − (4ε+ ε3)t3 + (2− ε2)t2 + 9εt+ (1 + 4ε2) = 0.The polynomial on the left-hand side has at least one positive root t+ andat least one negative root t−. It follows that V3 contains at least two geodesicvectors Y± = (a3)±X3+X4+t±X5 (where (a3)± are determined by (3.3)). Theirprojections to V4 are linearly independent, hence Span(Y1, Y2, Y3, Y+, Y−) = g.Note that for an arbitrary inner product on this algebra, the geodesic vectorsmay not span the entire algebra. For example, choosing an inner product withall the Xi’s orthogonal we obtain that the span of all the geodesic vectors is theproper subspace X⊥4 of g.We conclude this paper with two questions that have arisen from this study:Question 1. Is it true that every unimodular Lie algebra possesses an innerproduct for which the geodesic vectors span the algebra?Question 2. Apart from nilpotent Lie algebras, are there natural families ofunimodular Lie algebras that possess an inner product for which there is anorthonormal basis of geodesic vectors?Acknowledgements. The authors are very grateful to Ana Hinic´ Galic´who detected a number of typos in the ﬁrst version of this paper.18 Cairns, Le, Nielsen, NikolayevskyReferences[1] Nicolas Bourbaki: Lie groups and Lie algebras. Chapters 1–3, Elements of Mathematics(Berlin), Springer-Verlag, Berlin, 1998, Translated from the French, Reprint of the 1989English translation.[2] Grant Cairns, Ana Hinic´ Galic´, and Yuri Nikolayevsky: Totally geodesic subal-gebras of nilpotent Lie algebras I, J. Lie Theory, 23 (2013), 1023–1049.[3] : Totally geodesic subalgebras of nilpotent Lie algebras II, J. Lie Theory, 23 (2013),1051–1074.[4] Grant Cairns, Ana Hinic´ Galic´, Yuri Nikolayevsky, and Ioannis Tsartsaflis:Geodesic bases for Lie algebras, preprint.[5] G. Calvaruso, O. Kowalski, and R. A. Marinosci: Homogeneous geodesics in solvableLie groups, Acta Math. Hungar., 101 (2003), no. 4, 313–322.[6] Serena Cicalo`, Willem A. de Graaf, and Csaba Schneider: Six-dimensional nilpo-tent Lie algebras, Linear Algebra Appl., 436 (2012), no. 1, 163–189.[7] Nathan Jacobson: Lie algebras, Dover Publications Inc., New York, 1979, Republicationof the 1962 original.[8] V. V. Ka˘ızer: Conjugate points of left-invariant metrics on Lie groups, Soviet Math. (Iz.VUZ), 34 (1990), no. 11, 32–44.[9] Oldrˇich Kowalski, Stana Nikcˇevic´, and Zdeneˇk Vla´sˇek: Homogeneous geodesicsin homogeneous Riemannian manifolds—examples, in Geometry and topology of subman-ifolds, X (Beijing/Berlin, 1999), World Sci. Publ., River Edge, NJ, 2000, pp. 104–112.[10] Oldrˇich Kowalski and Ja´nos Szenthe: On the existence of homogeneous geodesicsin homogeneous Riemannian manifolds, Geom. Dedicata, 81 (2000), no. 1-3, 209–214,Erratum, Geom. Dedicata, 84 (2001), no. 1-3, 331–332.[11] John Milnor: Curvatures of left invariant metrics on Lie groups, Advances in Math.,21 (1976), no. 3, 293–329.[12] Klas Modin, Matthew Perlmutter, Stephen Marsland, and Robert McLach-lan:On Euler-Arnold equations and totally geodesic subgroups, J. Geom. Phys., 61 (2011),no. 8, 1446–1461.[13] G. M. Mubarakzjanov: On solvable Lie algebras, Izv. Vyssˇ. Ucˇehn. Zaved. Matematika,1963 (1963), no. 1 (32), 114–123.[14] Anthony Nielsen: Fundamental differential geometry and geodesic flows on Lie groups,MSc Thesis, La Trobe University, 2001.[15] W. V. Parker: Sets of complex numbers associated with a matrix, Duke Math. J., 15(1948), 711–715.[16] H. Strade: Lie algebras of small dimension, in Lie algebras, vertex operator algebras andtheir applications, Contemp. Math., vol. 442, Amer. Math. Soc., Providence, RI, 2007,pp. 233–265.[17] Hiroyuki Tasaki and Masaaki Umehara: An invariant on 3-dimensional Lie algebras,Proc. Amer. Math. Soc., 115 (1992), no. 2, 293–294.

On the existence of orthonormal geodesic bases for Lie algebras

Università del Salento: ESE - Salento University Publishing

Note di Matematica ISSN 1123-2536, e-ISSN 1590-0932Note Mat. 33 (2013) no. 2, 11–18. doi:10.1285/i15900932v33n2p11On the existence of orthonormal geodesicbases for Lie algebrasGrant CairnsiDepartment of Mathematics and Statistics, La Trobe University, Melbourne, Australia 3086g.cairns@latrobe.edu.auNguyen Thanh Tung LeiiDepartment of Mathematics and Statistics, La Trobe University, Melbourne, Australia 3086thanhtung1303@gmail.comAnthony NielsenDepartment of Mathematics and Statistics, La Trobe University, Melbourne, Australia 3086anthony.nielsen@latrobe.edu.auYuri NikolayevskyiiiDepartment of Mathematics and Statistics, La Trobe University, Melbourne, Australia 3086y.nikolayevsky@latrobe.edu.auReceived: 8.1.2013; accepted: 6.2.2013.Abstract. We show that every unimodular Lie algebra, of dimension at most 4, equippedwith an inner product, possesses an orthonormal basis comprised of geodesic elements. On theother hand, we give an example of a solvable unimodular Lie algebra of dimension 5 that hasno orthonormal geodesic basis, for any inner product.Keywords: geodesic vector, unimodular Lie algebraMSC 2010 classification: Primary 53C22, 17B301 IntroductionLet g be a Lie algebra equipped with an inner product 〈·, ·〉. Consider thecorresponding simply connected Lie group G equipped with the left invariantRiemannian metric determined by 〈·, ·〉. A nonzero element Y ∈ g is said tobe a geodesic vector if the corresponding left invariant vector field on G is ageodesic vector field. In terms of the Levi-Civita connection ∇, this means that∇Y Y = 0. This has a simple equivalent expression in terms of the Lie bracket[9, 10, 12, 2], which we state as a definition.iPart of this work was supported by ARC Discovery grant DP130103485iiPart of this work was conducted while the second author was an AMSI summer vacationscholar.iiiPart of this work was supported by ARC Discovery grant DP130103485http://siba-ese.unisalento.it/ c© 2013 Universita` del Salento12 Cairns, Le, Nielsen, NikolayevskyDefinition 1. Let g be a Lie algebra equipped with an inner product 〈·, ·〉.Then a nonzero element Y ∈ g is a geodesic vector if [X,Y ] is perpendicular toY for all X ∈ g.Note that some authors use the term homogeneous geodesic, to distinguishthem from general geodesics on the underlying Lie group. Some authors insistfurther that Y have unit length. For the more general case of totally geodesicsubalgebras of Lie algebras, see [12, 2, 3].Remark 1. A useful equivalent reformulation of the definition of a geodesicvector is as follows: Y ∈ g is geodesic if and only if Im(ad(Y )) is contained inthe orthogonal complement of Y .Every Lie algebra possesses at least one geodesic vector [8, 10, 2]. In [10] it isshown that semisimple Lie algebras possess an orthonormal basis comprised ofgeodesics vectors, for every inner product. Results for certain solvable algebrasare given in [5]. In the present paper we examine the existence of geodesic basesin algebras of low dimension, see [4] for further results on geodesic base.2 Preliminary observationsFirst notice that if a Lie algebra g, equipped with an orthonormal innerproduct, has a basis {X1, . . . , Xn} of geodesic vectors, then for all i, j, the ele-ment [Xi, Xj ] is orthogonal to both Xi and Xj . Consequently, the adjoint mapsad(Xi) have zero trace, and hence g is unimodular. So the natural problem isto determine which unimodular Lie algebras possess an orthonormal basis com-prised of geodesic vectors, for some inner product. We begin with two elementaryobservations.Proposition 1. Let g be a nilpotent Lie algebra equipped with an inner prod-uct 〈·, ·〉. Then there is an orthonormal basis of (g, 〈·, ·〉) comprised of geodesicvectors.Proof. The proof is by induction on the dimension of g. The proposition istrivial when dim(g) = 1. Suppose that dim(g) = n+ 1. Since g is nilpotent, itscentre is nonzero. Let Z be a central element of length one and set z := Span(Z).Let π : g→ h := g/z be the natural quotient map and give h the inner productfor which the restriction of π to the orthogonal complement z⊥ of z is an isometry.By the inductive hypothesis, h possesses an orthonormal basis {X1, . . . , Xn}comprised of geodesic vectors. So, by definition, for each pair i, j, the element[Xi, Xj ] is perpendicular to both Xi and Xj . For each i, let X¯i denote the uniqueelement of z⊥ with π(X¯i) = Xi. So for each pair i, j, the element [X¯i, X¯j ]is perpendicular to both X¯i and X¯j . Thus, as Z is central, the elements X¯iare geodesic vectors. Furthermore, as Z is central, Z is also geodesic. HenceOrthonormal Geodesics Bases 13{X¯1, . . . , X¯n, Z} is an orthonormal basis of geodesic vectors. QEDProposition 2. Let g be a unimodular Lie algebra having a codimension oneabelian ideal h. Then, for every inner product 〈·, ·〉 on g, there is an orthonormalbasis of (g, 〈·, ·〉) comprised of geodesic vectors.Proof. We make use of the following linear algebra result.Lemma 1 ([15, Theorem 10]). Suppose that A is a real square matrix withzero trace. Then there is an orthogonal matrix Q such that QAQ−1 has zerodiagonal.Let X ∈ g be a unit vector orthogonal to h. Note that X is geodesic since hcontains the derived algebra g′ of g. Choose an orthonormal basis for h and letA denote the matrix representation of the restriction adh(X) to h of the adjointmap ad(X). Since g is unimodular, the matrix A has zero trace. By Lemma 1,there is an orthonormal basis {Y1, Y2, . . . , Yn} for h relative to which adh(X)has zero diagonal. Consequently for each i, the element [X,Yi] is perpendicu-lar to Yi. Thus since h is abelian, the elements Yi are geodesic vectors. Hence{X,Y1, Y2, . . . , Yn} is a geodesic basis for g. QED3 Main ResultsIt is well known that in dimension less than or equal to three, there areonly 5 nonabelian unimodular real Lie algebras; see [17]. Milnor’s classification[11] proceeds by considering, for an orthonormal basis {X1, X2, X3}, the linearmap L(Xi × Xj) := [Xi, Xj ], where × denotes the vector cross product. Thematrix of L relative to {X1, X2, X3} is symmetric and so its eigenvectors forman orthonormal basis {Y1, Y2, Y3} with, by construction,[Y2, Y3] = λ1Y1,[Y3, Y1] = λ2Y2,[Y1, Y2] = λ3Y3,for real coefficients λi. So the basis elements Y1, Y2, Y3 are geodesic vectors.In dimension 4, there are infinitely many isomorphism classes of unimodularreal Lie algebras; see [13, 6, 16]. Instead of using the classification, we will arguedirectly. The following result resolves a question raised in [14].Theorem 1. Let g be a unimodular Lie algebra of dimension 4 equipped withan inner product 〈·, ·〉. Then there is an orthonormal basis of (g, 〈·, ·〉) comprisedof geodesic vectors.Proof. Let g be as in the statement of the theorem. First observe that ifg is not solvable, then from Levi’s Theorem and the classification of semisimple14 Cairns, Le, Nielsen, Nikolayevskyalgebras [7], there is a Lie algebra isomorphism g ∼= h ⊕ R where h is a simpleLie algebra of dimension three; so h is isomorphic to either so(3,R) or sl(2,R)though we won’t need this fact. Let W ∈ g be a unit vector orthogonal to h.Since h = g′, the element W is geodesic. Note that W can be written uniquelyin the form W = X + Z, where Z is in the centre of g and X ∈ h. As wediscussed above, h has an orthonormal basis {Y1, Y2, Y3} of elements that aregeodesic vectors of h. Then for each i = 1, 2, 3,[W,Yi] = [X,Yi] ∈ Im(ad(Yi)|h),which is perpendicular to Yi by Remark 1. Hence Yi is geodesic in g and thus{Y1, Y2, Y3,W} is an orthonormal basis of geodesic vectors. So we may assumethat g is solvable.Consider the derived algebra g′ := [g, g]. As g is solvable, g′ is nilpotent [1,Chap. 1.5.3]. If dim(g′) = 0, the algebra g is abelian and the theorem holdstrivially. We consider three cases according to the remaining possibilities for thedimension of g′.If dim(g′) = 1, let g′ = Span(W ). For each X ∈ g we have [X,Y ] ∈ g′ for allY ∈ g and hence [X,W ] = tr(ad(X))W = 0. Thus g′ is contained in the centreof g. Consequently g is nilpotent, and the theorem holds by Proposition 1.If dim(g′) = 3, then either g′ is abelian or g′ is isomorphic to the HeisenbergLie algebra h1. If g′ is abelian, the required result follows from Proposition2. So suppose that g′ is isomorphic to the Heisenberg Lie algebra. Choose anorthonormal basis {X,Y, Z} of g′ with Z in the centre of g′ and [X,Y ] = λZfor some λ 6= 0. Let W ∈ g be a unit vector orthogonal to g′. Note that W isgeodesic. Let A denote the matrix representation of the restriction adg′(W ) tog′ of the adjoint map ad(W ). Since the centre of g′ is a characteristic ideal [1,Chap. 1.1.3], it is left invariant by ad(W ). Hence A has the formA =a b 0c d 0e f g ,for reals a, b, c, d, e, f, g. By the Jacobi identity,[ad(W )X,Y ] + [X, ad(W )Y ] = ad(W )λZ = gλZ.Thus[aX + cY + eZ, Y ] + [X, bX + dY + fZ] = (a+ d)λZ = gλZ,and so a + d = g. Since ad(W ) has zero trace, a + d + g = 0, and hencea + d = g = 0. So, by Lemma 1, we can make an orthonormal change of basisOrthonormal Geodesics Bases 15for the space Span(X,Y ) so that relative to the new basis, a = d = 0. Then{W,X, Y, Z} is an orthonormal geodesic basis for g.Finally, if dim(g′) = 2, then g′ is abelian. Let {X,Y } be an orthonormalbasis for the orthogonal complement g′⊥ to g′. Note that [X,Y ] ∈ g′ and soas g′ is abelian, the adjoint map ad([X,Y ]) acts trivially on g′. Thus, sincead(X)◦ad(Y )−ad(Y )◦ad(X) = ad([X,Y ]) by the Jacobi identity, we have thatthe restrictions adg′(X) and adg′(Y ) commute. Since adg′(X) and adg′(Y ) havezero trace, relative to any choice of orthonormal basis for g′, they have matrixrepresentations in sl(2,R). However, it is well known and easy to see that themaximum abelian subalgebras of sl(2,R) have dimension one. Consequently, byorthonormal change of basis for g′⊥ if necessary, we may assume that adg′(Y ) ≡0. Then Span(Y, g′) is a codimension one abelian ideal and the result followsfrom Proposition 2. QEDNote that in Propositions 1, 2 and Theorem 1, the required orthonormalgeodesic basis is obtained for every inner product. The following example givesa solvable unimodular Lie algebra that has no orthonormal geodesic basis forany inner product.Example 1. Consider the 5-dimensional algebra g with basis B ={X1, . . . , X5} and relations[X1, X2] = 3X2 [X2, X3] = X4[X1, X3] = −4X3 [X2, X4] = X5[X1, X4] = −X4[X1, X5] = 2X5.The ideal n generated by X2, . . . , X5 is the (unique) 4-dimensional filiformnilpotent Lie algebra. The adjoint map ad(X1) is obviously a Lie derivationof n, so the Jacobi identities of g hold. Clearly, g is solvable and unimodu-lar, but not nilpotent. We will show that g has no orthonormal basis com-prised of geodesic vectors, for any inner product. Equip g with an inner productand suppose it has a basis (not necessarily orthonormal) of geodesic elementsY1, . . . , Y5. Consider an arbitrary geodesic vector Y = a1X1+· · ·+a5X5 ∈ g. LetVi := Span(Xi, Xi+1, . . . , X5). Relative to the basis B, the matrix representationof ad(Y ) isA =0 0 0 0 0−3a2 3a1 0 0 04a3 0 −4a1 0 0a4 −a3 a2 −a1 0−2a5 −a4 0 a2 2a1 . (3.1)A priori, there are five cases:16 Cairns, Le, Nielsen, Nikolayevsky(1) Y ∈ V5; that is, a1 = a2 = 0, a3 = 0, a4 = 0, a5 6= 0,(2) Y ∈ V4\V5; that is, a1 = a2 = 0, a3 = 0, a4 6= 0,(3) Y ∈ V3\V4; that is, a1 = a2 = 0, a3 6= 0,(4) Y ∈ V2\V3; that is, a1 = 0, a2 6= 0,(5) Y ∈ V1\V2; that is, a1 6= 0.In fact, case (1) is impossible as otherwise Y is a nonzero multiple of X5 butIm(ad(Y )) = V5, which must be orthogonal to Y by Remark 1. Similarly, case(2) is impossible as otherwise Y ∈ V4 but Im(ad(Y )) = V4, which has to beorthogonal to Y by Remark 1.In case (3), Im(ad(Y )) = Span(X3 +a44a3X4 − 2a54a3X5, X4 + a4a3X5).In case (4), Im(ad(Y )) = Span(X2 − 4a33a2X3, X4, X5) and in particular, Y isorthogonal to V4.In case (5), Im(ad(Y )) = V2, which is the orthogonal complement of Yby Remark 1. It follows that, up to a constant multiple, there is at most onegeodesic vector Y with a1 6= 0. Thus, there is at most one basis element, Y1 say,with Y1 6∈ V2.So we have established thatY2, Y3, Y4, Y5 ∈ V2\V4.But if Yi ∈ V2\V3, then as we saw in case (4), Yi is orthogonal to V4. So fordimension reasons, there are at most two of the Yi in V2\V3.Now suppose the Yi are orthogonal. If two of the Yi are in V2\V3, theywould both be orthogonal to V4. This would force the remaining two Yi to bein V4, which we have seen is impossible. So only one of the Yi, say Y2, can bein V2\V3. So Y3, Y4, Y5 are in V3\V4. Note that V3 is left invariant by ad(X1).Relative to the orthonormal basis {Y3, Y4, Y5} for V3, the map f := ad(X1)|V3has zero diagonal, because the elements Y3, Y4, Y5 are geodesic vectors. So fhas zero trace. But relative to the basis X3, X4, X5, it is clear that f has trace−4− 1 + 2 = −3. This is a contradiction.Remark 2. Although the algebra of the above example does not possesan orthonormal basis of geodesic vectors, it does posses an inner product forwhich there is a (nonorthonormal) basis of geodesic vectors. That is, we claimthat there exists an inner product such that the span of all the geodesic vectorsis the whole algebra. Using the above notation, regardless of the choice of aninner product, there is exactly one geodesic vector Y1 (up to scaling) not lyingin g′ = V2, namely any nonzero vector from (g′)⊥. We therefore want to showOrthonormal Geodesics Bases 17that the span of all the geodesic vectors from g′ covers g′. We will now specifythe inner product. First choose X2, X3 to be orthonormal and orthogonal to V4.Then from case (4), a vector Y = a2X2 + a3X3 + a4X4 + a5X5 (with a2 6= 0) isgeodesic if it is orthogonal to V4 (so a4 = a5 = 0) and√3a2 = ±2a3. This givestwo linearly independent geodesic vectors Y2, Y3 ∈ g′ ∩ V ⊥4 , neither of whichlies in V3. It remains to show that there exist at least two geodesic vectorsin V3 whose projections to V4 are linearly independent. Define the remainingcomponents of the inner product by requiring that ‖X4‖ = ‖X5‖ = 1 and〈X4, X5〉 = ε ∈ (0, 1). Then by case (3), a vector Y = a3X3 + a4X4 + a5X5(with a3 6= 0) is geodesic if and only if the following two conditions hold.4a23 + a24 − 2a25 − εa4a5 = 0, (3.2)a3(a4 + εa5) + a4(εa4 + a5) = 0. (3.3)Note that a4 + εa5 6= 0, as otherwise from (3.3) either a4 = 0 or εa4 + a5 = 0and in both cases we obtain a4 = a5 = 0 and then (3.2) would give a3 = 0,which is a contradiction. Solving (3.3) for a3 and substituting to (3.2) we get4a24(εa4 + a5)2 + (a24 − 2a25 − εa4a5)(a4 + εa5)2 = 0. (3.4)Note that as a3 6= 0 and a4 + εa5 6= 0, (3.3) gives a4 6= 0. Dividing (3.4) by a44and taking t = a5a−14 we obtain−2ε2t4 − (4ε+ ε3)t3 + (2− ε2)t2 + 9εt+ (1 + 4ε2) = 0.The polynomial on the left-hand side has at least one positive root t+ andat least one negative root t−. It follows that V3 contains at least two geodesicvectors Y± = (a3)±X3+X4+t±X5 (where (a3)± are determined by (3.3)). Theirprojections to V4 are linearly independent, hence Span(Y1, Y2, Y3, Y+, Y−) = g.Note that for an arbitrary inner product on this algebra, the geodesic vectorsmay not span the entire algebra. For example, choosing an inner product withall the Xi’s orthogonal we obtain that the span of all the geodesic vectors is theproper subspace X⊥4 of g.We conclude this paper with two questions that have arisen from this study:Question 1. Is it true that every unimodular Lie algebra possesses an innerproduct for which the geodesic vectors span the algebra?Question 2. Apart from nilpotent Lie algebras, are there natural families ofunimodular Lie algebras that possess an inner product for which there is anorthonormal basis of geodesic vectors?Acknowledgements. The authors are very grateful to Ana Hinic´ Galic´who detected a number of typos in the first version of this paper.18 Cairns, Le, Nielsen, NikolayevskyReferences[1] Nicolas Bourbaki: Lie groups and Lie algebras. Chapters 1–3, Elements of Mathematics(Berlin), Springer-Verlag, Berlin, 1998, Translated from the French, Reprint of the 1989English translation.[2] Grant Cairns, Ana Hinic´ Galic´, and Yuri Nikolayevsky: Totally geodesic subal-gebras of nilpotent Lie algebras I, J. Lie Theory, 23 (2013), 1023–1049.[3] : Totally geodesic subalgebras of nilpotent Lie algebras II, J. 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On the existence of orthonormal geodesic bases for Lie algebras

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