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Corrigendum to “Novikov algebras carrying an invariant Lorentzian symmetric bilinear form” [J. Geom. Phys. 82 (2014) 132–144]
Journal of Geometry and Physics 99 (2016) 263–266
Contents lists available at ScienceDirect
Journal of Geometry and Physics journal homepage: www.elsevier.com/locate/jgp
Erratum
Corrigendum to ‘‘Novikov algebras carrying an invariant Lorentzian symmetric bilinear form’’ [J. Geom. Phys. 82 (2014) 132–144] M. Guediri Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh, 11451, Saudi Arabia
article
info
Article history: Received 24 August 2015 Accepted 30 August 2015 Available online 27 October 2015
abstract The purpose of this note is to correct the classification list of Novikov algebras admitting an invariant Lorentzian symmetric bilinear form in our paper mentioned above. © 2015 Elsevier B.V. All rights reserved.
MSC: 17A30 17B05 53C50 Keywords: Invariant Lorentzian forms Novikov algebras Extensions of left-symmetric algebras
Recall that a finite-dimensional algebra (A, ·) over a field F is called left-symmetric if it satisfies the identity (x, y, z ) = (y, x, z ), for all x, y, z ∈ A, where (x, y, z ) denotes the associator (x, y, z ) = (x · y) · z − x · (y · z ). We say that A is a Novikov algebra if it satisfies the identity (x · y) · z = (x · z ) · y, for all x, y, z ∈ A, or equivalently, if the right multiplications Rx commute. In [1], we have classified Novikov algebras admitting an invariant Lorentzian symmetric bilinear form. However, it turns out that the classification list is incomplete and that some algebras should be removed from the classification list. Consequently, Table 1 in [1] must be corrected and completed. More precisely, Theorem 3 in [1] should be stated as follows. Theorem 1. Let A be a real n-dimensional Novikov algebra provided with an invariant Lorentzian symmetric bilinear form. Then A is isomorphic to an orthogonal direct sum of the form A = A1 ⊕ A2 , where A1 is an algebra in Table 1 and A2 is a direct sum of the algebras Ak,0 and Fk+1,0 in the following precise sense: In the case A1 is nontrivial, A2 is an orthogonal direct sum of the algebras Ak,0 and Fk+1,0 . In that case, the restriction of the Lorentzian symmetric bilinear form to any nontrivial factor Ak,0 or Fk+1,0 is positive definite. In the case A1 is trivial, A2 is either an orthogonal or a pseudo-orthogonal direct sum of two degenerate factors of the algebras Ak,0 and Fk+1,0 .
DOI of original article: http://dx.doi.org/10.1016/j.geomphys.2014.04.007. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.geomphys.2015.08.024 0393-0440/© 2015 Elsevier B.V. All rights reserved.
264
M. Guediri / Journal of Geometry and Physics 99 (2016) 263–266 Table 1 LSA
Basis
Ak,0 A2,1
e1 , . . . , ek e1 , e2
Non-zero products
A2,2
e1 , e2
S2,1 A3,1 N3,1 N3,2
e1 , e2 e1 , e2 , e3 e1 , e2 , e3 e1 , e2 , e3
N3λ,3
e1 , e2 , e3
G3,1 Fk+1,0 k≥0
e1 , e2 , e3
e1 · e2 = e2 ,
e0 , . . . , ek
ei · e0 = ei ,
Lk+1
e0 , . . . , ek
Lλk+1
e0 , . . . , ek
e1 e1 e2 e1 e1 e1 e2 e1 e1
· e1 · e1 · e1 · e2 · e1 · e2 · e2 · e1 · e3
Lie algebra
= e2 = e1 , e1 · e2 = e2 = e2 , e2 · e2 = −e1 = e2 = e3 , e1 · e3 = e3 · e1 = e2 = e3 = e1 , e1 · e2 = e3 = e3 , = λe2 , e3 · e1 = e2 , with λ ̸= 1
e0 · e0 = e0 , ei · e0 = ei , e0 · e0 = e0 , ei · e0 = ei ,
e1 · e3 =
1 e 2 3
,
e3 · e3 = e2
0≤i≤k e0 · e1 1≤i e0 · e1 1≤i
= e2 , ≤k = λ e1 , ≤k
Rk R2 R2
G2 R3 H3 H3 H3 1
L32 R, if k = 0 Gk+1 , if k ≥ 1 Lk+1 Lλk+1
The structures of the Lie algebras appearing in Table 1 are explicitly described in the following table. Table 2 Lie algebra
Basis
R H3 Gk+1 , k ≥ 1
e1 , . . . , ek e1 , e2 , e3 e0 , . . . , ek
Lk+1 , k ≥ 2
e0 , . . . , ek
Lλk+1 , k ≥ 2
e0 , . . . , ek
k
Non-zero brackets [e1 , e2 ] = e3 [ei , e0 ] = ei , 1 ≤ i ≤ k [e1 , e0 ] = e1 − e2 , [ei , e0 ] = ei , 2 ≤ i ≤ k [e1 , e0 ] = (1 − λ) e1 , λ ̸= 0, [ei , e0 ] = ei , 2 ≤ i ≤ k
In what follows, we will explain briefly how to complete the classification list. In [1], Case 3 of Section 2 (i.e., Proof of Theorem 3), we claimed that since the operators Rx commute with each other, then all Rx must have the form
Rx =
0 λ (x)
0 0
⊕ 0n−2 ,
(1)
or all must have of the form
Rx =
0 0 λ (x)
0 0 0
0 λ (x) 0
⊕ 0n−3 ,
(2)
for some real function λ (x) . In other words, we claimed that all Rx should have the same minimal polynomial among the following three polynomials: t, (t − λ (x))2 t, or (t − λ (x))3 t. However, this is not true because if (t − λ (x))2 t is the minimal polynomial of Rx and (t − λ (y))3 t is the minimal polynomial of Ry , then Rx and Ry commute if and only if they have a common null eigenvector. This implies that the case when Rx has the form (1) and Ry has the form (2) can occur. For the sake of clarity, recall from [1] that we are given a Novikov real algebra A equipped with an invariant Lorentzian symmetric bilinear form ⟨, ⟩ such that A is decomposed into a direct sum of ideals A = i Ai , where each ideal Ai is either complete (i.e., right-nilpotent) or Ai /N (Ai ) is isomorphic to the field R or the algebra A2,2 : e1 · e1 = e1 , e1 · e2 = e2 · e1 = e2 , e2 · e2 = −e1 , where N (Ai ) is the so-called right radical of Ai . In fact, since Ai is Novikov, we have N (Ai ) = {a ∈ Ai : tr (Ra ) = 0} . In what follows, for a fixed i, let ⟨, ⟩|A denote the restriction of ⟨, ⟩ to Ai . In the missing case in [1], Ai is complete and i
⟨, ⟩|A is Lorentzian. In that case, if e1 , . . . , en is a basis of Ai , we know that the minimal polynomial of Rei is either of the form i
t, (t − λi )2 t, or (t − λi )3 t. Since the Rei commute with each other, then they have a common null eigenvector that we can assume to be e2 . In other words, if Rei is not diagonalizable then it takes either the form (1) or the form (2). As we can carefully check, there is only two cases which will lead to new types of complete algebras that can be added to the classification list. All other cases will lead to existing types of algebras. Case 1. Re1 and Re2 are both of the form (1) and Re3 is of the form (2). In this case, we have e1 · e1 = λ1 e2 , e1 · e2 = λ2 e2 , e1 · e3 = λ3 e3 , e3 · e3 = λ3 e2 . Let i ≥ 4. If Rei has the form (1), then by setting e′i = ei − λi e2 , we can assume that
M. Guediri / Journal of Geometry and Physics 99 (2016) 263–266
265
λi = 0. If Rei has the form (2), then e1 · ei = λi e3 , e3 · ei = λi e2 . By left-symmetry, we have (e1 · ei ) · ei − e1 · (ei · ei ) = (ei · e1 ) · ei − ei · (e1 · ei ) , which yields λi = 0. Taking i = 3 in (2), we get that λ3 (2λ3 − λ2 ) = 0. If λ2 = λ3 = 0, then this leads to the algebra defined λ
by e1 · e1 = λ1 e2 , an algebra that already exists in Table 1 of [1]. If λ3 = 0 and λ2 ̸= 0, then by setting e′1 = λ1 e1 − λ1 e2 , 2 2 we see that the structure simplifies to e1 · e2 = e2 , an algebra that also already exists in Table 1 of [1]. If λ2 ̸= 0, then λ2 = 2λ3 , and by setting e′1 = λ1 e1 − λ12 e2 and e′2 = λ3 e2 , we see that Ai is isomorphic to the algebra G3,1 ⊕ An−2,0 , where 4λ3
3
Ak,0 denote the real vector space Rk with zero multiplication and G3,1 is given by G3,1 : e1 · e2 = e2 ,
e1 · e3 =
1 2
e3 ,
e3 · e3 = e2 .
We notice that G3,1 is an algebra of type (A13) in Table 2 of [2]. It is also the algebra D3,1 (µ), with µ =
1 , 2
in Theorem 137
of [3]. Its Lie algebra is the non-unimodular Lie algebra given by [e1 , e2 ] = e2 , [e1 , e3 ] = . Case 2. Re2 and Re3 both have the same form (1) and Re1 has the form (2). It is easy to see that, in almost the same way as in the previous case, the present case leads to a direct sum of Ak,0 and one of the following algebras: 1 e 2 3
A3,1 : e1 · e1 = e3 ,
e1 · e3 = e3 · e1 = e2 ,
N3λ,3 : e1 · e1 = e3 ,
e1 · e3 = λe2 ,
or e3 · e1 = e2 ,
with λ ̸= 1.
The first algebra A3,1 is of type (A4) in Table 2 of [2]. The second one N3λ,3 is of type (A7) in Table 2 of [2]. Now, we will use the above new types of complete algebras to find the new types of non-complete algebras. For, assume that Ai is non-complete and let ⟨, ⟩|N A denote the restriction of ⟨, ⟩ to N (Ai ) . As mentioned above, we know that Ai /N (Ai ) is ( i)
isomorphic to the field R or the algebra A2,2 . Since any invariant nondegenerate symmetric bilinear form on A2,2 is necessarily Lorentzian (see [1, Lemma 18]), and since ⟨, ⟩|N A is Lorentzian it follows that Ai /N (Ai ) is isomorphic to the field R. In this ( i)
case, we have a short exact sequence of left-symmetric algebras 0 → N (Ai ) → Ai → R → 0. Setting Ai = N (Ai ) ⊕ R as a vector space, and letting 1 denote the unit element of the field R, we adopt the notation (x, a) to denote elements of Ai . In particular, setting e0 = (0, 1), we have e0 = e0 · e0 . According to the above short exact sequence, there exist two endomorphisms λ, ρ : N (Ai ) → N (Ai ) and a bilinear map ω : R2 → N (Ai ) such that the left-symmetric product of Ai can be written
(x, a) · (y, b) = (x · y + aλ (y) + bρ (x) + abω (1, 1) , ab) , for all x, y ∈ N (Ai ) and a, b ∈ R. Consequently, we have e0 = e0 · e0 = (ω (1, 1) , 0) + e0 , which yields ω = 0. Since ⟨, ⟩|N A is Lorentzian, we can write ⟨(x, a) , (y, b)⟩ = ⟨x, y⟩ + ab for all x, y ∈ N (Ai ) and a, b ∈ R, and in a similar ( i)
fashion as we did in [1], we deduce by the invariance of ⟨, ⟩ that λ = 0 and ρ is self-adjoint. It follows that the left-symmetric product of Ai reduces to
(x, a) · (y, b) = (x · y + bρ(x), ab) .
(3)
On the other hand, the invariance of ⟨, ⟩ also implies that ρ is self-adjoint, and by virtue of (iv) of Theorem 9 in [1] we conclude that ρ 2 = ρ . Since ρ 2 = ρ, then we can show that ρ is diagonalizable (see Claim 19 in [1]). In this case, by using the fact that ρ 2 = ρ , we may assume without of generality that, in an appropriate basis, we have ρ = Idm ⊕ 0n−m for some integer m ≤ n. Setting N (Ai ) = Vi ⊕ ker ρ , where Vi = ker (ρ − Id), we have dim Vi = m and ρ|V = Idm . i
According to the above discussion concerning the complete case, we have three cases to be considered in the noncomplete case. In the case when N (Ai ) is isomorphic to G3,1 , we proceed as follows. Based on what we have seen in Case 1 above, there exists a pseudo-orthonormal basis e1 , . . . , en of N (Ai ), with e1 and e2 null vectors satisfying ⟨e1 , e2 ⟩ = 1, such that e1 · e1 = µe2 , e1 · e2 = 2λe2 , e1 · e3 = λe3 , e3 · e3 = λe2 , with λ ̸= 0. With the notation above, set ρ ej = εj ej , 1 ≤ j ≤ n, where εj = 1 or 0 according to whether ej ∈ Vi or ej ̸∈ Vi , respectively. Claim 2. We have ε1 = ε2 = ε3 = 0. Proof. Onthe one side, the identity ⟨ρ (e1 ) , e2 ⟩ = ⟨e1 , ρ (e2 )⟩ yields ε1 = ε2 . On the other side, by setting e0 = (0, 1) and ej = ej , 0 , 1 ≤ j ≤ n, the identity (e3 · e0 ) · e3 = (e3 · e3 ) · e0 yields ε2 = ε3 and the identity (e3 · e0 ) · e3 − e3 · (e0 · e3 ) = (e0 · e3 ) · e3 − e0 · (e3 · e3 ) yields ε3 = 0, given that λ ̸= 0.
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M. Guediri / Journal of Geometry and Physics 99 (2016) 263–266
It follows that the structure of Ai is given by e0 · e0 = e0 ,
e1 · e1 = µe2 ,
e3 · e3 = λe2 ,
ej · e0 = εj ej , µ
e1 · e2 = 2λe2 ,
e1 · e3 = λ e3 ,
4 ≤ j ≤ n.
By setting e′1 = 21λ e1 − 2λ e2 and e2 = λe2 , we get
e0 · e0 = e0 ,
e′1 · e′2 = e′2 ,
e′1 · e3 =
1 2
e3 ,
e3 · e3 = e′2 ,
ej · e0 = εj ej ,
4 ≤ j ≤ n.
In other words, we have shown that Ai is isomorphic to G3,1 ⊕ Fk+1,0 ⊕ An−k−3,0 , for some k. It follows that, with the 1
notation of Table 2, the Lie algebra of Ai is isomorphic to the direct product L32 ⊕ Gk+1 ⊕ Rn−k−3 , for some integer k. In the cases when N (Ai ) is isomorphic to A3,1 or N3λ,3 we proceed just as with the case above to finally show that Ai is
isomorphic to either A3,1 ⊕ Fk+1,0 ⊕ An−k−3,0 or N3λ,3 ⊕ Fk+1,0 ⊕ An−k−3,0 , for some integer k and some real number λ ̸= 1. In
these cases, the Lie algebra of Ai is isomorphic to either the direct product Gk+1 ⊕ Rn−k or H3 ⊕ Gk+1 ⊕ Rn−k−3 , respectively. Here H3 denotes the Heisenberg group of dimension 3. We end this note by noting that in Subsubcases 2.2.2 and 2.2.3 of [1] we should have ε1 = ε2 = 0. Similarly, Claims 20 and 21 of [1] should read: ε1 = ε2 = ε3 = 0. References [1] M. Guediri, Novikov algebras carrying an invariant Lorentzian symmetric bilinear form, J. Geom. Phys. 82 (2014) 132–144. [2] C. Bai, D. Meng, The classification of Novikov algebras in low dimensions, J. Phys. A: Math. Gen. 34 (2001) 1581–1594. [3] K. Al-Balawi, Left-symmetric structures on solvable Lie algebras (Ph.D. thesis, King Saud University), 2014.
Contents lists available at ScienceDirect
Journal of Geometry and Physics journal homepage: www.elsevier.com/locate/jgp
Erratum
Corrigendum to ‘‘Novikov algebras carrying an invariant Lorentzian symmetric bilinear form’’ [J. Geom. Phys. 82 (2014) 132–144] M. Guediri Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh, 11451, Saudi Arabia
article
info
Article history: Received 24 August 2015 Accepted 30 August 2015 Available online 27 October 2015
abstract The purpose of this note is to correct the classification list of Novikov algebras admitting an invariant Lorentzian symmetric bilinear form in our paper mentioned above. © 2015 Elsevier B.V. All rights reserved.
MSC: 17A30 17B05 53C50 Keywords: Invariant Lorentzian forms Novikov algebras Extensions of left-symmetric algebras
Recall that a finite-dimensional algebra (A, ·) over a field F is called left-symmetric if it satisfies the identity (x, y, z ) = (y, x, z ), for all x, y, z ∈ A, where (x, y, z ) denotes the associator (x, y, z ) = (x · y) · z − x · (y · z ). We say that A is a Novikov algebra if it satisfies the identity (x · y) · z = (x · z ) · y, for all x, y, z ∈ A, or equivalently, if the right multiplications Rx commute. In [1], we have classified Novikov algebras admitting an invariant Lorentzian symmetric bilinear form. However, it turns out that the classification list is incomplete and that some algebras should be removed from the classification list. Consequently, Table 1 in [1] must be corrected and completed. More precisely, Theorem 3 in [1] should be stated as follows. Theorem 1. Let A be a real n-dimensional Novikov algebra provided with an invariant Lorentzian symmetric bilinear form. Then A is isomorphic to an orthogonal direct sum of the form A = A1 ⊕ A2 , where A1 is an algebra in Table 1 and A2 is a direct sum of the algebras Ak,0 and Fk+1,0 in the following precise sense: In the case A1 is nontrivial, A2 is an orthogonal direct sum of the algebras Ak,0 and Fk+1,0 . In that case, the restriction of the Lorentzian symmetric bilinear form to any nontrivial factor Ak,0 or Fk+1,0 is positive definite. In the case A1 is trivial, A2 is either an orthogonal or a pseudo-orthogonal direct sum of two degenerate factors of the algebras Ak,0 and Fk+1,0 .
DOI of original article: http://dx.doi.org/10.1016/j.geomphys.2014.04.007. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.geomphys.2015.08.024 0393-0440/© 2015 Elsevier B.V. All rights reserved.
264
M. Guediri / Journal of Geometry and Physics 99 (2016) 263–266 Table 1 LSA
Basis
Ak,0 A2,1
e1 , . . . , ek e1 , e2
Non-zero products
A2,2
e1 , e2
S2,1 A3,1 N3,1 N3,2
e1 , e2 e1 , e2 , e3 e1 , e2 , e3 e1 , e2 , e3
N3λ,3
e1 , e2 , e3
G3,1 Fk+1,0 k≥0
e1 , e2 , e3
e1 · e2 = e2 ,
e0 , . . . , ek
ei · e0 = ei ,
Lk+1
e0 , . . . , ek
Lλk+1
e0 , . . . , ek
e1 e1 e2 e1 e1 e1 e2 e1 e1
· e1 · e1 · e1 · e2 · e1 · e2 · e2 · e1 · e3
Lie algebra
= e2 = e1 , e1 · e2 = e2 = e2 , e2 · e2 = −e1 = e2 = e3 , e1 · e3 = e3 · e1 = e2 = e3 = e1 , e1 · e2 = e3 = e3 , = λe2 , e3 · e1 = e2 , with λ ̸= 1
e0 · e0 = e0 , ei · e0 = ei , e0 · e0 = e0 , ei · e0 = ei ,
e1 · e3 =
1 e 2 3
,
e3 · e3 = e2
0≤i≤k e0 · e1 1≤i e0 · e1 1≤i
= e2 , ≤k = λ e1 , ≤k
Rk R2 R2
G2 R3 H3 H3 H3 1
L32 R, if k = 0 Gk+1 , if k ≥ 1 Lk+1 Lλk+1
The structures of the Lie algebras appearing in Table 1 are explicitly described in the following table. Table 2 Lie algebra
Basis
R H3 Gk+1 , k ≥ 1
e1 , . . . , ek e1 , e2 , e3 e0 , . . . , ek
Lk+1 , k ≥ 2
e0 , . . . , ek
Lλk+1 , k ≥ 2
e0 , . . . , ek
k
Non-zero brackets [e1 , e2 ] = e3 [ei , e0 ] = ei , 1 ≤ i ≤ k [e1 , e0 ] = e1 − e2 , [ei , e0 ] = ei , 2 ≤ i ≤ k [e1 , e0 ] = (1 − λ) e1 , λ ̸= 0, [ei , e0 ] = ei , 2 ≤ i ≤ k
In what follows, we will explain briefly how to complete the classification list. In [1], Case 3 of Section 2 (i.e., Proof of Theorem 3), we claimed that since the operators Rx commute with each other, then all Rx must have the form
Rx =
0 λ (x)
0 0
⊕ 0n−2 ,
(1)
or all must have of the form
Rx =
0 0 λ (x)
0 0 0
0 λ (x) 0
⊕ 0n−3 ,
(2)
for some real function λ (x) . In other words, we claimed that all Rx should have the same minimal polynomial among the following three polynomials: t, (t − λ (x))2 t, or (t − λ (x))3 t. However, this is not true because if (t − λ (x))2 t is the minimal polynomial of Rx and (t − λ (y))3 t is the minimal polynomial of Ry , then Rx and Ry commute if and only if they have a common null eigenvector. This implies that the case when Rx has the form (1) and Ry has the form (2) can occur. For the sake of clarity, recall from [1] that we are given a Novikov real algebra A equipped with an invariant Lorentzian symmetric bilinear form ⟨, ⟩ such that A is decomposed into a direct sum of ideals A = i Ai , where each ideal Ai is either complete (i.e., right-nilpotent) or Ai /N (Ai ) is isomorphic to the field R or the algebra A2,2 : e1 · e1 = e1 , e1 · e2 = e2 · e1 = e2 , e2 · e2 = −e1 , where N (Ai ) is the so-called right radical of Ai . In fact, since Ai is Novikov, we have N (Ai ) = {a ∈ Ai : tr (Ra ) = 0} . In what follows, for a fixed i, let ⟨, ⟩|A denote the restriction of ⟨, ⟩ to Ai . In the missing case in [1], Ai is complete and i
⟨, ⟩|A is Lorentzian. In that case, if e1 , . . . , en is a basis of Ai , we know that the minimal polynomial of Rei is either of the form i
t, (t − λi )2 t, or (t − λi )3 t. Since the Rei commute with each other, then they have a common null eigenvector that we can assume to be e2 . In other words, if Rei is not diagonalizable then it takes either the form (1) or the form (2). As we can carefully check, there is only two cases which will lead to new types of complete algebras that can be added to the classification list. All other cases will lead to existing types of algebras. Case 1. Re1 and Re2 are both of the form (1) and Re3 is of the form (2). In this case, we have e1 · e1 = λ1 e2 , e1 · e2 = λ2 e2 , e1 · e3 = λ3 e3 , e3 · e3 = λ3 e2 . Let i ≥ 4. If Rei has the form (1), then by setting e′i = ei − λi e2 , we can assume that
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λi = 0. If Rei has the form (2), then e1 · ei = λi e3 , e3 · ei = λi e2 . By left-symmetry, we have (e1 · ei ) · ei − e1 · (ei · ei ) = (ei · e1 ) · ei − ei · (e1 · ei ) , which yields λi = 0. Taking i = 3 in (2), we get that λ3 (2λ3 − λ2 ) = 0. If λ2 = λ3 = 0, then this leads to the algebra defined λ
by e1 · e1 = λ1 e2 , an algebra that already exists in Table 1 of [1]. If λ3 = 0 and λ2 ̸= 0, then by setting e′1 = λ1 e1 − λ1 e2 , 2 2 we see that the structure simplifies to e1 · e2 = e2 , an algebra that also already exists in Table 1 of [1]. If λ2 ̸= 0, then λ2 = 2λ3 , and by setting e′1 = λ1 e1 − λ12 e2 and e′2 = λ3 e2 , we see that Ai is isomorphic to the algebra G3,1 ⊕ An−2,0 , where 4λ3
3
Ak,0 denote the real vector space Rk with zero multiplication and G3,1 is given by G3,1 : e1 · e2 = e2 ,
e1 · e3 =
1 2
e3 ,
e3 · e3 = e2 .
We notice that G3,1 is an algebra of type (A13) in Table 2 of [2]. It is also the algebra D3,1 (µ), with µ =
1 , 2
in Theorem 137
of [3]. Its Lie algebra is the non-unimodular Lie algebra given by [e1 , e2 ] = e2 , [e1 , e3 ] = . Case 2. Re2 and Re3 both have the same form (1) and Re1 has the form (2). It is easy to see that, in almost the same way as in the previous case, the present case leads to a direct sum of Ak,0 and one of the following algebras: 1 e 2 3
A3,1 : e1 · e1 = e3 ,
e1 · e3 = e3 · e1 = e2 ,
N3λ,3 : e1 · e1 = e3 ,
e1 · e3 = λe2 ,
or e3 · e1 = e2 ,
with λ ̸= 1.
The first algebra A3,1 is of type (A4) in Table 2 of [2]. The second one N3λ,3 is of type (A7) in Table 2 of [2]. Now, we will use the above new types of complete algebras to find the new types of non-complete algebras. For, assume that Ai is non-complete and let ⟨, ⟩|N A denote the restriction of ⟨, ⟩ to N (Ai ) . As mentioned above, we know that Ai /N (Ai ) is ( i)
isomorphic to the field R or the algebra A2,2 . Since any invariant nondegenerate symmetric bilinear form on A2,2 is necessarily Lorentzian (see [1, Lemma 18]), and since ⟨, ⟩|N A is Lorentzian it follows that Ai /N (Ai ) is isomorphic to the field R. In this ( i)
case, we have a short exact sequence of left-symmetric algebras 0 → N (Ai ) → Ai → R → 0. Setting Ai = N (Ai ) ⊕ R as a vector space, and letting 1 denote the unit element of the field R, we adopt the notation (x, a) to denote elements of Ai . In particular, setting e0 = (0, 1), we have e0 = e0 · e0 . According to the above short exact sequence, there exist two endomorphisms λ, ρ : N (Ai ) → N (Ai ) and a bilinear map ω : R2 → N (Ai ) such that the left-symmetric product of Ai can be written
(x, a) · (y, b) = (x · y + aλ (y) + bρ (x) + abω (1, 1) , ab) , for all x, y ∈ N (Ai ) and a, b ∈ R. Consequently, we have e0 = e0 · e0 = (ω (1, 1) , 0) + e0 , which yields ω = 0. Since ⟨, ⟩|N A is Lorentzian, we can write ⟨(x, a) , (y, b)⟩ = ⟨x, y⟩ + ab for all x, y ∈ N (Ai ) and a, b ∈ R, and in a similar ( i)
fashion as we did in [1], we deduce by the invariance of ⟨, ⟩ that λ = 0 and ρ is self-adjoint. It follows that the left-symmetric product of Ai reduces to
(x, a) · (y, b) = (x · y + bρ(x), ab) .
(3)
On the other hand, the invariance of ⟨, ⟩ also implies that ρ is self-adjoint, and by virtue of (iv) of Theorem 9 in [1] we conclude that ρ 2 = ρ . Since ρ 2 = ρ, then we can show that ρ is diagonalizable (see Claim 19 in [1]). In this case, by using the fact that ρ 2 = ρ , we may assume without of generality that, in an appropriate basis, we have ρ = Idm ⊕ 0n−m for some integer m ≤ n. Setting N (Ai ) = Vi ⊕ ker ρ , where Vi = ker (ρ − Id), we have dim Vi = m and ρ|V = Idm . i
According to the above discussion concerning the complete case, we have three cases to be considered in the noncomplete case. In the case when N (Ai ) is isomorphic to G3,1 , we proceed as follows. Based on what we have seen in Case 1 above, there exists a pseudo-orthonormal basis e1 , . . . , en of N (Ai ), with e1 and e2 null vectors satisfying ⟨e1 , e2 ⟩ = 1, such that e1 · e1 = µe2 , e1 · e2 = 2λe2 , e1 · e3 = λe3 , e3 · e3 = λe2 , with λ ̸= 0. With the notation above, set ρ ej = εj ej , 1 ≤ j ≤ n, where εj = 1 or 0 according to whether ej ∈ Vi or ej ̸∈ Vi , respectively. Claim 2. We have ε1 = ε2 = ε3 = 0. Proof. Onthe one side, the identity ⟨ρ (e1 ) , e2 ⟩ = ⟨e1 , ρ (e2 )⟩ yields ε1 = ε2 . On the other side, by setting e0 = (0, 1) and ej = ej , 0 , 1 ≤ j ≤ n, the identity (e3 · e0 ) · e3 = (e3 · e3 ) · e0 yields ε2 = ε3 and the identity (e3 · e0 ) · e3 − e3 · (e0 · e3 ) = (e0 · e3 ) · e3 − e0 · (e3 · e3 ) yields ε3 = 0, given that λ ̸= 0.
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It follows that the structure of Ai is given by e0 · e0 = e0 ,
e1 · e1 = µe2 ,
e3 · e3 = λe2 ,
ej · e0 = εj ej , µ
e1 · e2 = 2λe2 ,
e1 · e3 = λ e3 ,
4 ≤ j ≤ n.
By setting e′1 = 21λ e1 − 2λ e2 and e2 = λe2 , we get
e0 · e0 = e0 ,
e′1 · e′2 = e′2 ,
e′1 · e3 =
1 2
e3 ,
e3 · e3 = e′2 ,
ej · e0 = εj ej ,
4 ≤ j ≤ n.
In other words, we have shown that Ai is isomorphic to G3,1 ⊕ Fk+1,0 ⊕ An−k−3,0 , for some k. It follows that, with the 1
notation of Table 2, the Lie algebra of Ai is isomorphic to the direct product L32 ⊕ Gk+1 ⊕ Rn−k−3 , for some integer k. In the cases when N (Ai ) is isomorphic to A3,1 or N3λ,3 we proceed just as with the case above to finally show that Ai is
isomorphic to either A3,1 ⊕ Fk+1,0 ⊕ An−k−3,0 or N3λ,3 ⊕ Fk+1,0 ⊕ An−k−3,0 , for some integer k and some real number λ ̸= 1. In
these cases, the Lie algebra of Ai is isomorphic to either the direct product Gk+1 ⊕ Rn−k or H3 ⊕ Gk+1 ⊕ Rn−k−3 , respectively. Here H3 denotes the Heisenberg group of dimension 3. We end this note by noting that in Subsubcases 2.2.2 and 2.2.3 of [1] we should have ε1 = ε2 = 0. Similarly, Claims 20 and 21 of [1] should read: ε1 = ε2 = ε3 = 0. References [1] M. Guediri, Novikov algebras carrying an invariant Lorentzian symmetric bilinear form, J. Geom. Phys. 82 (2014) 132–144. [2] C. Bai, D. Meng, The classification of Novikov algebras in low dimensions, J. Phys. A: Math. Gen. 34 (2001) 1581–1594. [3] K. Al-Balawi, Left-symmetric structures on solvable Lie algebras (Ph.D. thesis, King Saud University), 2014.