The Algebraic Classification of Nilpotent Tortkara Algebras1,2

Home , Lie algebra extension, Loop algebra, Malcev algebra

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The centrally extended Lie algebras play a dominant role in quantum field theory, particularly in conformal field theory, string theory and in M-theory. In the theory of Lie groups, Lie algebras and their representations, a Lie algebra extension is an enlargement of a given Lie algebra g by another Lie algebra h. Extensions arise in several ways. There is a trivial extension obtained by taking a direct sum of two Lie algebras. Other types are a split extension and a central extension. Extensions may arise naturally, for instance, when forming a Lie algebra from projective group representations. A central extension and an extension by a derivation of a polynomial loop algebra over finite-dimensional simple Lie algebra give a Lie algebra which is isomorphic to a non-twisted affine Kac-Moody algebra [4, Chapter 19]. Using the centrally extended loop algebra one may construct a current algebra in two spacetime dimensions. The Virasoro algebra is the universal central extension of the Witt algebra, the Heisenberg algebra is the central extension of a commutative Lie algebra [4, Chapter 18]. The algebraic study of central extensions of Lie and non-Lie algebras has a very long history [2,20,21,25,33,35]. Thus, Skjelbred and Sund used central extensions of Lie algebras for a classification of nilpotent Lie algebras [33]. After that, the method introduced by Skjelbred and Sund was used to describe all non-Lie central extensions of all 4-dimensional Malcev algebras [21], all non-associative central extensions of 3-dimensional Jordan algebras [20], all anticommutative central extensions of 3-dimensional anticommutative algebras [6], all central extensions of 2-dimensional algebras [7]. The method of central extensions was used to describe all 4-dimensional nilpotent associative algebras [12], all 4-dimensional nilpotent bicommutative algebras [26], all 4-dimensional nilpotent Novikov algebras [24], all 5-dimensional nilpotent Jordan algebras [19], all 5-dimensional nilpotent restricted Lie algebras [11], all 6-dimensional nilpotent Lie algebras [10,13], all 6-dimensional nilpotent Malcev algebras [22], all 6-dimensional nilpotent binary Lie algebras [3], all 6-dimensional nilpotent anticommutative CD-algebras [3] and some other.

1. PRELIMINARIES 1.1. Method of classification of nilpotent algebras. Throughout this paper, we use the notations and methods described in [7,20,21] and adapted for the Tortkara case with some modifications. From now on, we will give only some important definitions. Let (A, ) be a Tortkara algebra over C and V a vector space over the same base field. Then the C-linear space Z2 (A· , V) is defined as the set of all skew-symmetric bilinear maps θ : A A V, such that T × −→ θ(xy, zt)+ θ(xt, zy)= θ(J(x, y, z), t)+ θ(J(x, t, z),y). Its elements will be called cocycles. For a linear map f from A to V, if we define δf : A A V by 2 A V 2 A V A V × −→ δf (x, y) = f(xy), then δf ZT ( , ). Let B ( , ) = θ = δf : f Hom( , ) . One can easily 2 A V ∈ 2 A V { ∈ } check that B ( , ) is a linear subspace of ZT ( , ) whose elements are called coboundaries. We define 2 A V 2 A V 2 A V the second cohomology space HT ( , ) as the quotient space ZT ( , ) B ( , ). Let Aut (A) be the automorphism group of the Tortkara algebra A and let φ Aut (A). For θ 2 A V 2 A V A ∈ 2 A V ∈ ZT ( , ) define φθ (x, y)= θ (φ (x) ,φ (y)). Then φθ ZT ( , ). So, Aut ( ) acts on ZT ( , ). Itis 2 A V ∈ A A 2 A V easy to verify that B ( , ) is invariant under the action of Aut ( ), and thus Aut ( ) acts on HT ( , ). Let A be a Tortkara algebra of dimension m < n over C, and V be a C-vector space of dimension 2 n m. For any θ Z (A, V) define on the linear space Aθ := A V the bilinear product “[ , ]A ” − ∈ T ⊕ − − θ by [x + x′,y + y′]A = xy + θ (x, y) for all x, y A, x′,y′ V. Then Aθ is a Tortkara algebra called θ ∈ ∈ an (n m)-dimensional central extension of A by V. In fact, Aθ is a Tortkara algebra if and only if −2 θ ZT (A, V). ∈We also call the set Ann(θ)= x A : θ (x, A)=0 the annihilator of θ. We recall that the annihilator { ∈ } of an algebra A is defined as the ideal Ann (A) = x A : xA =0 and observe that Ann (Aθ) = Ann(θ) Ann (A) V. { ∈ } ∩ ⊕ 3

We have the next key result: Lemma 1. Let A be an n-dimensional Tortkara algebra such that dim(Ann(A)) = m = 0. Then there 6 exists, up to isomorphism, a unique (n m)-dimensional Tortkara algebra A′ and a bilinear map θ 2 A V A − V A A ∈ ZT ( , ) with Ann( ) Ann(θ)=0, where is a vector space of dimension m, such that = θ′ and A A A ∩ ∼ / Ann( ) ∼= ′.

Proof. Let A′ be a linear complement of Ann(A) in A. Define a linear map P : A A′ by P (x+v)= x −→ for x A′ and v Ann(A) and define a multiplication on A′ by [x, y]A′ = P (xy) for x, y A′. For x, y ∈A we have ∈ ∈ ∈ P (xy)= P ((x P (x)+ P (x))(y P (y)+ P (y))) = P (P (x)P (y)) = [P (x),P (y)]A′ − − A A A A A Since P is a homomorphism, then P ( )= ′ is a Tortkara algebra and / Ann( ) ∼= ′, which gives us the uniqueness of A′. Now, define the map θ : A′ A′ Ann(A) by θ(x, y)= xy [x, y]A′ . Then A A 2 A V A × −→ − θ′ is and therefore θ ZT ( , ) and Ann( ) Ann(θ)=0. ∈ ∩

However, in order to solve the isomorphism problem, we need to study the action of Aut (A) on 2 A V V 2 A V HT ( , ). Tothisend, let us fix e1,...,es a basis of , and θ ZT ( , ). Then θ can be uniquely written s ∈ 2 A C as θ (x, y)= θi (x, y) ei, where θi ZT ( , ). Moreover, Ann(θ) = Ann(θ1) Ann(θ2) Ann(θs). i=1 ∈ ∩ ···∩ 2 2 Further, θ BP(A, V) if and only if all θi B (A, C). ∈ ∈ Definition 2. Given a Tortkara algebra A, if A = I Cx is a direct sum of two ideals, then Cx is called an annihilator component of A. ⊕ Definition 3. A central extension of an algebra A without an annihilator component is called a non-split central extension.

It is not difﬁcult to prove (see [21, Lemma 13]) that, given a Tortkara algebra Aθ, if we write as above s 2 A V A A θ (x, y) = θi (x, y) ei ZT ( , ) and we have Ann(θ) Ann ( )=0, then θ has an annihilator i=1 ∈ ∩ 2 A C componentP if and only if [θ1] , [θ2] ,..., [θs] are linearly dependent in HT ( , ).

Let V be a ﬁnite-dimensional vector space over C. The Grassmannian Gk (V) is the set of all k- V 2 A C dimensional linear subspaces of . Let Gs (HT ( , )) be the Grassmannian of subspaces of dimen- 2 A C A 2 A C A sion s in HT ( , ). There is a natural action of Aut ( ) on Gs (HT ( , )). Let φ Aut ( ). 2 A C ∈ For W = [θ1] , [θ2] ,..., [θs] Gs (HT ( , )) deﬁne φW = [φθ1] , [φθ2] ,..., [φθs] . Then φW 2 A hC i ∈ 2 A C h Ai ∈ Gs (HT ( , )). We denote the orbit of W Gs (HT ( , )) under the action of Aut ( ) by Orb(W ). Since, given ∈ 2 W = [θ ] , [θ ] ,..., [θs] , W = [ϑ ] , [ϑ ] ,..., [ϑs] Gs H (A, C) , 1 h 1 2 i 2 h 1 2 i ∈ T s s we easily have that if W1 = W2, then Ann(θi) Ann (A)= Ann(ϑi) Ann (A), we can introduce i∩=1 ∩ i∩=1 ∩ the set

s T A 2 A C A s ( )= W = [θ1] , [θ2] ,..., [θs] Gs HT ( , ) : Ann(θi) Ann ( )=0 , h i ∈ i∩=1 ∩ which is stable undern the action of Aut (A). o Now, let V be an s-dimensional linear space and let us denote by E (A, V) the set of all non-split s- dimensional central extensions of A by V. We can write s

E (A, V)= Aθ : θ (x, y)= θi (x, y) ei and [θ ] , [θ ] ,..., [θs] Ts (A) . h 1 2 i ∈ ( i=1 ) X 4

Also, we have the next result, which can be proved as [21, Lemma 17]. s s Lemma 4. Let Aθ, Aϑ E (A, V) . Suppose that θ (x, y) = θi (x, y) ei and ϑ (x, y) = ϑi (x, y) ei. ∈ i=1 i=1 A A Then the Tortkara algebras θ and ϑ are isomorphic if and onlyP if P

Orb [θ ] , [θ ] ,..., [θs] = Orb [ϑ ] , [ϑ ] ,..., [ϑs] . h 1 2 i h 1 2 i Thus, there exists a one-to-one correspondence between the set of Aut (A)-orbits on Ts (A) and the set of isomorphism classes of E (A, V). Consequently, we have a procedure that allows us, given a Tortkara algebra A′ of dimension n, to construct all non-split central extensions of A′. This procedure is as follows: Procedure

(1) For a given (nilpotent) Tortkara algebra A′ of dimension n s, determine Ts(A′) and Aut(A′). − (2) Determine the set of Aut(A′)-orbits on Ts(A′). (3) For each orbit, construct the Tortkara algebra corresponding to one of its representatives. The above described method gives all (Malcev and non-Malcev) Tortkara algebras. But we are interested in developing this method in such a way that it only gives non-Malcev Tortkara algebras, because the classification of all Malcev-Tortkara algebras is a part of the classification of all Malcev algebras [22, 28]. Clearly, any central extension of a non-Malcev Tortkara algebra is non-Malcev. But a Malcev-Tortkara algebra may have extensions which are Malcev algebras. More precisely, let M be a Malcev algebra and 2 θ Z (M, C). Then Mθ is a Malcev algebra if and only if ∈ T θ (wy, xz)= θ ((wx)y, z)+ θ ((xy)z,w)+ θ ((yz)w, x)+ θ ((zw)x, y) for all x,y,z,w M. Define the subspace Z2 (M, C) of Z2 (M, C) by ∈ TM T θ Z2 (M, C): θ (wy, xz)= θ ((wx)y, z)+ θ ((xy)z,w)+ Z2 (M, C)= T . TM ∈ θ ((yz)w, x)+ θ ((zw)x, y) , for all x,y,z,w M ∈ Observe that B2 (M, C) Z2 (M, C). Let H2 (M, C)=Z2 (M, C) B2 (M, C). Then H2 (M, C) ⊆ TM TM TM TM is a subspace of H2 (M, C). Define T 2 Rs (M) = W Ts (M): W Gs H (M, C) , ∈ ∈ TM 2 Us (M) = W Ts (M): W / Gs H (M, C) . ∈ ∈ TM T M R M U M R M U M M Then s ( )= s ( ) s ( ). The sets s ( ) and s ( ) are stable under the action of Aut ( ). ∪· Thus, the Tortkara algebras corresponding to the representatives of Aut (M)-orbits on Rs (M) are Malcev algebras, while those corresponding to the representatives of Aut (M)-orbits on Us (M) are not Malcev algebras. Hence, we may construct all non-split non-Malcev Tortkara algebras A of dimension n with s-dimensional annihilator from a given Tortkara algebra A′ of dimension n s in the following way: − (1) If A′ is non-Malcev, then apply the Procedure. (2) Otherwise, do the following: (a) Determine Us (A′) and Aut(A′). (b) Determine the set of Aut(A′)-orbits on Us (A′). (c) For each orbit, construct the Tortkara algebra corresponding to one of its representatives.

1.2. Notations. Let us introduce the following notations. Let A be a Tortkara algebra with a basis e , e ,...,en. Then by ∆ij we will denote the bilinear form ∆ij : A A C with ∆ij (el, em)=0, 1 2 × −→ if i, j = l, m , and ∆ij (ei, ej) = ∆ij (ej, ei)=1. The set ∆ij :1 i < j n is a basis for the linear{ space} 6 { of skew-symmetric} bilinear− forms on A, so every θ { Z2 (A≤, V) can≤ be uniquely} written as ∈ T θ = cij∆ij, where cij C. We also denote by 1 i

1.3. Central extensions of nilpotent low dimensional Tortkara algebras. There are no nontrivial 1- and 2-dimensional nilpotent Tortkara algebras. Thanks to [6] we have the description of all non-split 3- and 4-dimensional nilpotent Tortkara algebras: T3 01 : e1e2 = e3; T4 02 : e1e2 = e3, e1e3 = e4. Also, it is easy to see that every anticommutative central extension of Ni is a Lie algebra.

2. CENTRALEXTENSIONSOFNILPOTENT 4-DIMENSIONAL TORTKARA ALGEBRAS 2.1. The algebraic classiﬁcation of 4-dimensional nilpotent Tortkara algebras. A 2 A 2 A multiplication table HTM( ) HT( ) T4 2 T4 01 e1e2 = e3 [∆13], [∆14], [∆23], [∆24], [∆34] HTM( 01) T4 e e = e , e e = e h[∆ ], [∆ ] i H2 (T4 ) [∆ ] 02 1 2 3 1 3 4 h 14 23 i TM 01 ⊕h 24 i T4 2 T4 2 T4 2.2. Central extensions of 02. Since HTM( 02) = HT( 02), we can ﬁnd some Tortkara (non-Malcev) algebras. Let us introduce the following notations 6 = [∆ ], = [∆ ], = [∆ ]. ∇1 14 ∇2 23 ∇3 24 T4 The automorphism group of 02 consists of the invertible matrices of the form x 00 0 z y 0 0 φ = . u v xy 0  h g xv x2y   Since   0 0 0 α1 0 α∗ α∗∗ α1∗ 0 0 α α α∗ 0 α∗ α∗ φt  2 3 φ =  − 2 3 , 0 α 0 0 α∗∗ α∗ 0 0 − 2 − − 2  α1 α3 0 0   α∗ α∗ 0 0  − −   − 1 − 3  where     2 α1∗ = (α1x + α3z)x y, α2∗ = (α2y + α3v)xy, 2 2 α3∗ = α3x y , 3 3 T4 we obtain that the action of Aut ( 02) on the subspace αi i is given by αi∗ i . i=1 ∇ i=1 ∇ D E D E T4 P P 2.2.1. 1-dimensional central extensions of 02. Consider a cocycle θ = α + α + α . 1∇1 2∇2 3∇3 To construct non-Malcev Tortkara algebras, we will be interested only in θ with α3 = 0. Then choosing 1 α1 α2 we have the representative It gives6 the following new x = 1,y = α3 , z = α3 , v = 3 [∆24] . √ − −√(α3) Tortkara algebra: T5 10 : e1e2 = e3, e1e3 = e4, e2e4 = e5. T4 2.2.2. 2-dimensional central extensions of 02. Consider the vector space generated by the following two cocycles θ1 = α1 1 + α2 2 + α3 3, θ = β ∇ + β ∇ . ∇ 2 1∇1 2∇2 We are interested in θ1 with α3 =0. Then 6 2 (1) if β = 0, β = 0, then choosing y = β1x , z = α1x , v = α2y , we have the representative 1 6 2 6 β2 − α3 − α3 1 + 2, 3 . (2)h∇ if β =0∇, β∇=0i , then choosing z = α1x , v = α2y , we have the representative , . 1 2 6 − α3 − α3 h∇2 ∇3i 6

(3) if β =0, β =0, then choosing z = α1x , v = α2y , we have the representative , . 1 6 2 − α3 − α3 h∇1 ∇3i It is easy to see that these three orbits are pairwise distinct. So, we have all non-split non-Malcev 2- T4 dimensional central extensions of 02 :

T6 01 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e2e3 = e5, e2e4 = e6; T6 02 : e1e2 = e3, e1e3 = e4, e2e3 = e5, e2e4 = e6; T6 03 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e2e4 = e6.

3. CENTRAL EXTENSIONS OF 5-DIMENSIONAL NILPOTENT TORTKARA ALGEBRAS 2 T4 3.1. The algebraic classification of 5-dimensional nilpotent Tortkara algebras. Since HT( 01) = 2 T4 T4 HTM( 01), each central extension of 01 is a Malcev algebra. On the other hand, by some easy calculations, using the algebraic classification of 5-dimensional nilpotent Malcev algebras [22], one can see that each 5-dimensional nilpotent Malcev algebra is a Tortkara algebra. Combining this with the result of previous section, we have the algebraic classification of all nontrivial 5-dimensional nilpotent Tortkara algebras:

A 2 A 2 A multiplication table HTM( ) HT( ) T5 [∆13], [∆14], [∆15], [∆23], [∆24], 2 T5 01 e1e2 = e3 HTM( 01) [∆25], [∆34], [∆35], [∆45] D E T5 e e = e , e e = e [∆ ], [∆ ], [∆ ], [∆ ], [∆ ] H2 (T5 ) [∆ ], [∆ ] 02 1 2 3 1 3 4 14 15 23 25 35 TM 02 ⊕h 24 45 i T5 D [∆14], [∆15], [∆23], [∆24], E 2 T5 03 e1e2 = e4, e1e3 = e5 HTM( 03) [∆25], [∆34], [∆35] D E T5 e1e2 = e3, e1e3 = e4, 2 T5 04 [∆14], [∆15] + [∆24], [∆25] HTM( 04) [∆15] e2e3 = e5 ⊕h i D E T5 [∆12], [∆13], [∆14], [∆15], [∆23], 2 T5 05 e1e2 = e5, e3e4 = e5 HTM( 05) [∆24], [∆25], [∆35], [∆45] D E T5 e1e2 = e3, e1e4 = e5, 2 T5 06 [∆13], [∆14], [∆24], [∆25], [∆34] HTM( 06) [∆15], [∆45] e2e3 = e5 ⊕h i T5 D E 2 T5 07 e1e2 = e3, e3e4 = e5 [∆13], [∆14], [∆23], [∆24] HTM( 07) T5 e1e2 = e3, e1e3 = e4, D E 2 T5 08 [∆15], [∆23] HTM( 08) [∆24], [∆25] e1e4 = e5 ⊕h i D E T5 e1e2 = e3, e1e3 = e4, 2 T5 09 [∆14], [∆15] + [∆24] HTM( 09) [∆15], [∆25] e1e4 = e5, e2e3 = e5 ⊕h i D E T5 e1e2 = e3, e1e3 = e4, 10 — [∆14], [∆23], [∆34] + [∆15] e2e4 = e5 h i

T5 3.2. 1-dimensional central extensions of 02. Let us use the following notations

= [∆ ], = [∆ ], = [∆ ], = [∆ ], = [∆ ], = [∆ ], = [∆ ]. ∇1 14 ∇2 15 ∇3 23 ∇4 25 ∇5 35 ∇6 24 ∇7 45 T5 The automorphism group of 02 consists of invertible matrices of the form

x 00 0 0 f y 0 0 0   φ = u v xy 0 0 . h r xv x2y l   t g 0 0 z     7

Notice also that det φ = x4y3z =0. Since 6 0 0 0 α1 α2 0 α∗ α∗∗ α1∗ α2∗ 0 0 α3 α6 α4 α∗ 0 α3∗ α6∗ α4∗ T    −  φ 0 α 0 0 α φ = α∗∗ α∗ 0 0 α∗ , − 3 5 − − 3 5  α1 α6 0 0 α7  α∗ α∗ 0 0 α∗ − −   − 1 − 6 7  α2 α4 α5 α7 0   α∗ α∗ α∗ α∗ 0  − − − −   − 2 − 4 − 5 − 7  where     3 2 2 α∗ = α x y + α fx y α tx y, 1 1 6 − 7 α2∗ = α1xl + α2xz + α4fz + α5uz + α6fl + α7(hz tl), 2 − α∗ = α xy α gxy + α vxy α gvx, 3 3 − 5 6 − 7 α∗ = α yz + α vz + α yl + α (rz gl), 4 4 5 6 7 − α5∗ = α5xyz + α7vxz, 2 2 2 α6∗ = α6x y α7gx y, 2 − α7∗ = α7x yz, 7 7 T5 we have that the action of Aut( 02) on the subspace αi i is given by αi∗ i . i=1 ∇ i=1 ∇ We are interested in the cocycles with (α6,α7) = (0D,P0). So, weE have the followingD P cases:E 6 (1) α =0, then choosing g = α6y , v = α5y ,l =0,r = α4y+α5v , h = α2x+α4f+α5u , t = α1x+α6f 7 6 α7 − α7 − α7 − α7 α7 we have the representative α3∗ 3 + α7∗ 7 . Here we have two orbits: αh5α6∇ ∇ i 1 α5α6 (a) If α∗ = 0, i.e. α = 0, then choosing y = z = 1, x = (α ) we have the 3 6 3 − α7 6 α7 3 − α7 representative 3 + 7 . h∇ α5∇α6 i (b) If α∗ =0, i.e. α =0, then we have the representative . 3 3 − α7 h∇7i (α4y+α5v)z α5g α3y α1x − (2) α7 =0,α6 =0, then choosing l = α6y , v = α6 , f = α6 , we have the representative 6 2 2 − − α2∗ 2 + α5xyz 5 + α6x y 6 . Here we have three orbits: h ∇ ∇ ∇ i 1 2 4 6 (a) If α =0, then choosing x = α5 , y = z =1, u = (α l+α )x+(α +α l)f , we have the represen- 5 6 α6 − α5 tative 5 + 6 . h∇ ∇ i α1α4 1 α1α4 (b) If α = 0,α∗ = 0, i.e. α = 0, then choosing y = 1, x = (α ), we have the 5 2 6 2 − α6 6 α6 2 − α6 representative 2 + 6 . h∇ ∇ i α1α4 (c) If α =0,α∗ =0, i.e. α =0, then we have the representative . 5 2 2 − α6 h∇6i Note that the representative 6 gives a Tortkara algebra with 2-dimensional annihilator, which was found in the previous section. A straightforward∇ computation shows that the rest of the representatives belong to different orbits. They give the following pairwise non-isomorphic Tortkara algebras: T6 04 : e1e2 = e3, e1e3 = e4, e1e5 = e6, e2e4 = e6; T6 05 : e1e2 = e3, e1e3 = e4, e2e3 = e6, e4e5 = e6; T6 06 : e1e2 = e3, e1e3 = e4, e2e4 = e6, e3e5 = e6; T6 07 : e1e2 = e3, e1e3 = e4, e4e5 = e6.

T5 3.3. 1-dimensional central extensions of 04. Let us use the following notations

= [∆ ], = [∆ ], = [∆ ] + [∆ ], = [∆ ]. ∇1 14 ∇2 25 ∇3 15 24 ∇4 15 T5 The automorphism group of 04 consists of invertible matrices of the form x y 0 0 0 v z 0 0 0   φ = u h xz yv 0 0 . −  l r xh yu x(xz yv) y(xz yv)  − − −  t g vh zu v(xz yv) z(xz yv)  − − −    8

Notice that det φ =(xz yv)5 =0. Since − 6 0 00 α1 α3 + α4 0 β1 β2 α1∗ α3∗ + α4∗ 0 00 α3 α2 β1 0 β3 α3∗ α2∗ T     φ 0 000 0 φ = −β β 00 0 , − 2 − 3  α1 α3 00 0   α∗ α∗ 00 0   − −   − 1 − 3   α3 α4 α2 00 0   α∗ α∗ α∗ 00 0  − − −  − 3 − 4 − 2  where     2 2 α1∗ = (xz yv)(α1x + α2v + (2α3 + α4)xv), − 2 2 α∗ = (xz yv)(α y + α z + (2α + α )yz), 2 − 1 2 3 4 α3∗ = (xz yv)(α1xy + α2vz +(α3 + α4)yv + α3xz), − 2 α∗ = α (xz yv) , 4 4 − 4 4 T5 we have that the action of Aut( 04) on the subspace αi i is given by αi∗ i . i=1 ∇ i=1 ∇ We are only interested in cocycles with α4 =0. NoteD thatP this conditionE is invariantD P underE automorphisms. We have the following cases: 6

(1) α1 =0. (a)6 (2α + α )2 4α α =0. Then for all ﬁxed non-zero v and z the equations 3 4 − 1 2 6 2 2 α1x + α2v + (2α3 + α4)xv =0, 2 2 α1y + α2z + (2α3 + α4)yz =0 have roots x = x and y = y . If x z = y v, then x z = y v. Thus, we may choose x,y,z,v 1 6 2 1 6 2 1 1 2 6 1 such that xz yv = 0 and α∗ = α∗ = 0. Note that α∗ = 0, so applying an automorphism − 6 1 2 4 6 to the cocycle α∗ + α∗ , we may also make α∗ = 1. This gives the representative of the 3∇3 4∇4 4 orbit of the form ξα = α + . Given a pair α, β, it is easy to see that there exists an h ∇3 ∇4i automorphism φ, such that φ(ξα)= ξβ, if and only if β = α 1. − − (b) (2α + α )2 4α α = 0. Then choosing y = α2 z and x = α2 v, we get 3 4 − 1 2 − α1 6 − α1 the representative α1∗ 1 + α3∗ 3 + α4∗ 4 . A routine calculationq shows thatq the condition (2α + α )2 4αh α ∇= 0 is∇ invariant∇ underi the action of an automorphism, so we have 3 4 − 1 2 α4∗ = 2α3∗. Since α4∗ =0, then there is an automorphism which sends α1∗ 1 + α3∗ 3 2α3∗ 4 to α −+ 2 for6 some α = 0. Then choosing x = 1 , y = 0, z∇= √α,∇ we− have∇ the ∇1 ∇3 − ∇4 6 √α representative + 2 . h∇1 ∇3 − ∇4i (2) α =0, α =0. Then choosing x = z =0 and y = v =1, we get α∗ = α =0, soweare in1. 1 2 6 1 − 2 6 (3) α1 =0, α2 =0. This case has already appeared in 1a.

It is easily checked that the orbit of 1 + 3 2 4 does not belong to the family of orbits of α 3 + 4 . Thus, we have the following algebras:h∇ ∇ − ∇ i h ∇ ∇ i T6 08 : e1e2 = e3, e1e3 = e4, e1e4 = e6, e1e5 = e6, e2e3 = e5, e2e4 = e6; T6 − 09(α): e1e2 = e3, e1e3 = e4, e1e5 =(α + 1)e6, e2e3 = e5, e2e4 = αe6, where T6 = T6 (α), and T6 (α) = T6 (β) if and only if β = α 1. 08 6∼ 09 09 ∼ 09 − − T5 3.4. 1-dimensional central extensions of 06. Let us use the following notations = [∆ ], = [∆ ], = [∆ ], = [∆ ], = [∆ ], = [∆ ], = [∆ ]. ∇1 13 ∇2 14 ∇3 24 ∇4 25 ∇5 34 ∇6 15 ∇7 45 T5 The automorphism group of 06 consists of invertible matrices of the form x p 0 00 0 y 0 00   φ = zt xy py 0 . − q r 0 y2 0    s h xr yz pq f xy2  − −    9

Notice that det φ = x3y6 =0. Since 6 0 0 α1 α2 α6 0 α∗ α1∗ α2∗ + α∗∗ α6∗ 0 0 0 α3 α4 α∗ 0 α∗∗ α3∗ α4∗ T    −  φ α 0 0 α 0 φ = α∗ α∗ 0 α∗ 0 , − 1 5 − 1 − 5  α2 α3 α5 0 α7  α∗ α∗∗ α∗ α∗ 0 α∗ − − −  − 2 − − 3 − 5 7  α6 α4 0 α7 0   α∗ α∗ 0 α∗ 0  − − −   − 6 − 4 − 7  where    

α1∗ = xy(α1x α5q)+(α6x + α7q)(rx yz pq), − 2 − − α2∗ = 2α1pxy + α2xy + α4y(pq rx + yz)+ α5y(pq + rx + yz) − − 2 +α6(p(pq rx + yz)+ fx)+ α7(r(pq rx + yz) sy + fq), 2 − 2 − − 2 α3∗ = y( α1p + α2py + α3y + α4f)+ α5y(pr + yt)+ α6fp + α7(fr y h), −2 − α4∗ = xy (α4y + α6p + α7r), 2 α5∗ = y (α5xy + α7(pq rx + yz)), 2 − α6∗ = xy (α6x + α7q), 4 α7∗ = α7y x,

7 7 T5 we have that the action of Aut( 06) on the subspace αi i is given by αi∗ i . We are only i=1 ∇ i=1 ∇ interested in the cocycles with (α6,α7) = (0, 0). ThereD areP the followingE cases: D P E 6 α6 α5xy+α7(pq rx) α4y+α6p − (1) α7 = 0. Then choosing q = α7 x, z = α7y , r = α7 , and h, s such that 6 α1α−7+α5α6 2 − − α2∗ = α3∗ =0, we have α1∗ = α7 x y. So, there are 2 subcases: 5 7 α7 7 α1α7+α5α6 (a) α1α7 + α5α6 = 0. Then choose x = 4 and y = 3 . This gives the 6 (α1α7+α5α6) α7 representative 1 + 7 . q q h∇ ∇ i 1 (b) α1α7 + α5α6 =0. Then choose x = α7 , y =1 to get the representative 7 . h∇ i α4 (2) α7 =0, α6 =0. Then choosing r and f, we may make α1∗ = α2∗ =0. Moreover, if p = α6 y, then 6 2 − 3 α1α4 α2α4 2 α4 α4∗ =0. For this value of p we have α3∗ = y (α3 2 )+ α5y (t r). So, we have two − α6 − α6 − α6 subcases: √α5 (a) α5 = 0. Then choosing the appropriate value of t, we may make α3∗ = 0. So, x = 4 3 and 6 √α6 4 y = √α6 give the representative + . √α5 h∇5 ∇6i (b) α5 =0. We have two subcases: 2 2 2 3 α6 (i) α3α6 α1α4 α2α4α6 = 0. Then choosing y = 2 2 and x = − − 6 α3α6 α1α4 α2α4α6 3 2 2 − − √α3α6 α1α4 α2α4α6 q − 6 − we get the representative + . √α6 h∇3 ∇6i (ii) α α2 α α2 α α α = 0. Then choosing x = 1, y = 1 , we get the representative 3 6 − 1 4 − 2 4 6 √α6 . h∇6i A straightforward veriﬁcation shows that the found orbits are distinct. Thus, we have new non-isomorphic Tortkara algebras: T6 10 : e1e2 = e3, e1e3 = e6, e1e4 = e5, e2e3 = e5, e4e5 = e6; T6 11 : e1e2 = e3, e1e4 = e5, e1e5 = e6, e2e3 = e5; T6 12 : e1e2 = e3, e1e4 = e5, e1e5 = e6, e2e3 = e5, e2e4 = e6; T6 13 : e1e2 = e3, e1e4 = e5, e1e5 = e6, e2e3 = e5, e3e4 = e6; T6 14 : e1e2 = e3, e1e4 = e5, e2e3 = e5, e4e5 = e6.

T5 3.5. 1-dimensional central extensions of 08. Let us use the following notations = [∆ ], = [∆ ], = [∆ ], = [∆ ]. ∇1 15 ∇2 23 ∇3 24 ∇4 25 10

T5 The automorphism group of 08 consists of invertible matrices of the form x 00 0 0 z y 0 0 0   φ = t p xy 0 0 . q r xp x2y 0    h s xr x2p x3y     Notice that det φ = x7y4 =0. Since 6

0 0 00 α1 0 β1 β2 β3 α1∗ 0 0 α2 α3 α4 β1 0 α2∗ α3∗ α4∗ T   −  φ 0 α 0 0 0 φ = β α∗ 0 0 0 , − 2 − 2 − 2  0 α3 0 0 0   β3 α∗ 0 0 0   −  − − 3   α1 α4 0 0 0   α∗ α∗ 0 0 0  − −  − 1 4  where     3 α1∗ = x y(α1x + α4z), α2∗ = xy(α2y + α3p + α4r), 2 α3∗ = x y(α3y + α4p), 3 2 α4∗ = α4x y , 4 4 T5 we have that the action of Aut( 08) on the subspace αi i is given by αi∗ i . We are interested i=1 ∇ i=1 ∇ in cocycles with (α3,α4) = (0, 0). So, we have the followingD P cases:E D P E 6 (1) α =0. Then choosing p = α3x ,r = α2y+α3p and z = α1x we have the representative . 4 6 − α4 − α4 − α4 h∇4i (2) α4 =0,α3 =0. We have two subcases: 2 (a) 6 . Then choosing α2y and α1x we have the representative α1 =0 p = α3 y = α3 1 + 3 . 6 − 2 h∇ ∇ i (b) α =0. Then choosing p = α y we have the representative . 1 − α3 h∇3i Note that the representative 3 gives a 6-dimensional Tortkara algebra with 2-dimensional annihilator. It is a 2-dimensional central extensionh∇ i of a 4-dimensional Tortkara algebra, and it was found in the previous section. The rest of the representatives give new non-isomorphic Tortkara algebras: T6 15 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e1e5 = e6, e2e4 = e6; T6 16 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e2e5 = e6.

T5 3.6. 1-dimensional central extensions of 09. Let us use the following notations = [∆ ], = [∆ ] + [∆ ], = [∆ ], = [∆ ]. ∇1 14 ∇2 24 15 ∇3 15 ∇4 25 T5 The automorphism group of 09 consists of invertible matrices of the form x 0 0 00 y x2 0 00   φ = z t x3 0 0 . p q xt x4 0    r s x2z + xq + yt x3y + x2t x5  −    Notice that det φ = x15 =0. Since 6 0 00 α1 α2 + α3 0 α∗ α∗∗ α∗∗∗ + α1∗ α2∗ + α3∗ 0 00 α2 α4 α∗ 0 α∗∗∗ α2∗ α4∗ T    −  φ 0 000 0 φ = α∗∗ α∗∗∗ 0 0 0 , − −  α1 α2 00 0   α∗∗∗ α∗ α∗ 0 0 0   − −  − − 1 2   α2 α3 α4 00 0   α∗ α∗ α∗ 0 0 0  − − −   − 2 − 3 − 4      11 where 3 2 2 α1∗ = x (α1x +2α2xy + α3(xy + t)+ α4(y + xz q)), 4 2 − α2∗ = x (α2x + α4(xy + t)), 4 2 α3∗ = x (α3x α4t), 7 − α4∗ = α4x ,

4 4 T5 we have that the action of Aut( 09) on the subspace αi i is given by αi∗ i . We are only i=1 ∇ i=1 ∇ interested in cocycles with (α3,α4) = (0, 0). So, we haveD P the followingE cases: D P E 6 2 2 2 2 α3x α2x +α4t α1x +2α2xy+α3(xy+t)+α4(y q) (1) α = 0. Then choosing t = ,y = and z = − we have 4 6 α4 − α4x − α4x the representative 4 . h∇ i 2 (2) α = 0,α = 0. Then choosing t = α1x +2α2xy+α3xy we have the family of representatives 4 3 6 − α3 α 2 + 3 . Observe that there exists an automorphism φ, such that φ α 2 + 3 = β 2 + 3 ifh and∇ only∇ ifi α = β. h ∇ ∇ i h ∇ ∇ i Thus, we have the following family of pairwise non-isomorphic Tortkara algebras: T6 17 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e2e3 = e5, e2e5 = e6; T6 18(α): e1e2 = e3, e1e3 = e4, e1e4 = e5, e1e5 =(α + 1)e6, e2e3 = e5, e2e4 = αe6.

T5 3.7. 1-dimensional central extensions of 10. Let us use the following notations = [∆ ], = [∆ ], = [∆ ] + [∆ ]. ∇1 14 ∇2 23 ∇3 15 34 T5 The automorphism group of 10 consists of invertible matrices of the form x 00 0 0 0 y 0 0 0   φ = z 0 xy 0 0 . p q 0 x2y 0    r h yp 0 y2x2  −    Notice that det φ = x6y5 =0. Since 6

0 0 0 α1 α3 0 α∗ α∗∗ α1∗ α3∗ 0 0 α2 0 0 α∗ 0 α2∗ 0 0 T    −  φ 0 α 0 α 0 φ = α∗∗ α∗ 0 α∗ 0 , − 2 3 − − 2 3  α1 0 α3 0 0   α∗ 0 α∗ 0 0  − −   − 1 − 3   α3 0 0 00   α∗ 0 0 00  −   − 3      where 2 α1∗ = x y(α1x + α3z), α2∗ = xy(α2y α3q), 3 2 − α3∗ = α3x y ,

3 3 T5 we have that the action of Aut( 10) on the subspace αi i is given by αi∗ i . i=1 ∇ i=1 ∇ Every element with α3 = 0 gives an algebra withD 2P-dimensionalE annihilator,D P whichE was found in the 1 α1x α2 previous section. If α = 0, then choosing x = 3 ,y = 1, z = , q = we have the representative 3 6 √α3 − α3 α3 3 . h∇Thus,i we obtain the following new Tortkara algebra:

T6 19 : e1e2 = e3, e1e3 = e4, e1e5 = e6, e2e4 = e5, e3e4 = e6. 12

4. MAIN RESULT The algebraic classiﬁcation of all 6-dimensional nilpotent Malcev algebras was obtained in [22]. As it was proved in [22], every 6-dimensional nilpotent Malcev algebra is metabelian (i.e. (xy)(zt)=0). In particular, every 6-dimensional nilpotent Lie algebra is metabelian, and hence it is a Tortkara algebra. By some easy veriﬁcation of all 6-dimensional nilpotent non-Lie Malcev algebras, one can see that all these algebras are also Tortkara. Now we have Lemma 5. Let T be a 6-dimensional nilpotent Malcev-Tortkara algebra over C with non-trivial product. Then T is isomorphic to one of the nilpotent Malcev algebras listed in [22]. Now we are ready to formulate the main result of our paper: Theorem 6. Let T bea 6-dimensional nilpotent non-Malcev Tortkara algebra over C. Then T is isomorphic to one of the following algebras: T6 00 : e1e2 = e3, e1e3 = e4, e2e4 = e5; • T6 01 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e2e3 = e5, e2e4 = e6; • T6 02 : e1e2 = e3, e1e3 = e4, e2e3 = e5, e2e4 = e6; • T6 03 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e2e4 = e6; • T6 04 : e1e2 = e3, e1e3 = e4, e1e5 = e6, e2e4 = e6; • T6 05 : e1e2 = e3, e1e3 = e4, e2e3 = e6, e4e5 = e6; • T6 06 : e1e2 = e3, e1e3 = e4, e2e4 = e6, e3e5 = e6; • T6 07 : e1e2 = e3, e1e3 = e4, e4e5 = e6; • T6 08 : e1e2 = e3, e1e3 = e4, e1e4 = e6, e1e5 = e6, e2e3 = e5, e2e4 = e6; • T6 − 09(α): e1e2 = e3, e1e3 = e4, e1e5 =(α + 1)e6, e2e3 = e5, e2e4 = αe6; • T6 10 : e1e2 = e3, e1e3 = e6, e1e4 = e5, e2e3 = e5, e4e5 = e6; • T6 11 : e1e2 = e3, e1e4 = e5, e1e5 = e6, e2e3 = e5; • T6 12 : e1e2 = e3, e1e4 = e5, e1e5 = e6, e2e3 = e5, e2e4 = e6; • T6 13 : e1e2 = e3, e1e4 = e5, e1e5 = e6, e2e3 = e5, e3e4 = e6; • T6 14 : e1e2 = e3, e1e4 = e5, e2e3 = e5, e4e5 = e6; • T6 15 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e1e5 = e6, e2e4 = e6; • T6 16 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e2e5 = e6; • T6 17 : e1e2 = e3, e1e3 = e4, e1e4 = e5, e2e3 = e5, e2e5 = e6; • T6 18(α): e1e2 = e3, e1e3 = e4, e1e4 = e5, e1e5 =(α + 1)e6, e2e3 = e5, e2e4 = αe6; • T6 : e e = e , e e = e , e e = e , e e = e , e e = e . • 19 1 2 3 1 3 4 1 5 6 2 4 5 3 4 6 All listed algebras are non-isomorphic, except T6 (α) = T6 ( α 1). 09 ∼ 09 − − Remark 7. It was proved that all 6-dimensional nilpotent Malcev algebras are metabelian (i.e. (xy)(zt)= 0) [22]. As follows from the main theorem, there is only one 6-dimensional nilpotent non-metabelian T6 Tortkara algebra: 19.

5. APPENDIX: THE CLASSIFICATION OF 3-DIMENSIONAL TORTKARA ALGEBRAS Recall that the complete algebraic, geometric and degeneration classiﬁcation of all anticommutative 3- dimensional algebras was given in [23]. Easy corollary gives the algebraic and geometric classiﬁcation of all 3-dimensional Tortkara algebras.

Theorem 8. Let Tort3 be the variety of all 3-dimensional Tortkara algebra over C and T be a nonzero algebra from Tort3, then (1) T is isomorphic to one of the following algebras: g : e e = e • 1 2 3 1 g2 : e1e3 = e1, e2e3 = e2 • α g3 : e1e3 = e1 + e2, e2e3 = αe2 • A : e e = e , e e = e • 2 1 2 1 2 3 2 13

A0 1 : e1e2 = e3, e1e3 = e1 + e3, e2e3 = αe2 (2)• The variety of 3-dimensional Tortkara algebras has one irreducible component deﬁned by the rigid A0 algebra 1.

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