On Gradings Modulo 2 of Simple Lie Algebras in Characteristic 2

The ground field in the text is of characteristic 2. The classification of modulo 2 gradings of simple Lie algebras is vital for the classification of simple finite-dimensional Lie superalgebras: with each grading, a simple Lie superalgebra is associated, see arXiv:1407.1695. No classification of gradings was known for any type of simple Lie algebras, bar restricted Jacobson-Witt algebras (i.e., the first derived of the Lie algebras of vector fields with truncated polynomials as coefficients) on not less than 3 indeterminates. Here we completely describe gradings modulo 2 for several series of Lie algebras and their simple relatives: of special linear series, its projectivizations, and projectivizations of the derived Lie algebras of two inequivalent orthogonal series (except for ${\mathfrak{o}}_\Pi(8)$). The classification of gradings is new, but all of the corresponding superizations are known. For the simple derived Zassenhaus algebras of height $n>1$, there is an $(n-2)$-parametric family of modulo 2 gradings; all but one of the corresponding simple Lie superalgebras are new. Our classification also proves non-triviality of a deformation of a simple $3|2$-dimensional Lie superalgebra (new result).

1. Introduction 1.1. Basic definitions. Hereafter K is an algebraically closed field of characteristic p = 2, unless otherwise specified; all algebras are finite-dimensional; for a review of simple vectorial Lie (super)algebras over K and basic background, see [BGLLS2].
1.1.1. Lie superlagebras in characteristic 2. A Lie superalgebra is a superspace g = g0 ⊕g1 such that the even part g0 is a Lie algebra, the odd part g1 is a g0-module, and on g1, a squaring (roughly speaking, the halved bracket) is defined as a map x → x 2 such that (ax) 2 = a 2 x 2 for any x ∈ g1 and a ∈ K, and (x + y) 2 − x 2 − y 2 is a bilinear form on g1 with values in g0.
Then the bracket of odd elements is defined to be [x, y] := (x + y) 2 − x 2 − y 2 .
The Jacobi identity involving odd elements takes the following form: (1) [x 2 , y] = [x, [x, y]] for any x ∈ g1, y ∈ g0, [x 2 , x] = 0 for any x ∈ g1. Since any derivation D of a given algebra is determined by the values of D on the generators, we see that der(O(m; N )) has more than m functional parameters if N i = 1 for at least one i. Define distinguished partial derivatives ∂ i := ∂ u i by setting The Lie algebra of all derivations der(O(m; N)) turns out to be not so interesting (it is not transitive, and hence can not be simple) as the general vectorial Lie algebra of distinguished derivations: 1 vect(m; N ) = Span K (f ∂ k | f ∈ O(m; N ), k = 1, . . . , m) .
1.1.3. p-structure. For every x ∈ g, the operator (ad x ) p is a derivation of g. If this derivation is an inner one for every x ∈ g, then the Lie algebra g is said to be restricted or having a p-structure. More specifically, a p-structure on g is a map [p] : g → g, x → x [p] such that (2) [x [p] , y] = (ad x ) p (y) for any x, y ∈ g, (ax) [p] = a p x [p] for any a ∈ K, x ∈ g; (x + y) [ where s i (x, y) is the coefficient of λ i−1 in (ad λx+y ) p−1 (x).
1.1.4. Remarks. 1) If the Lie algebra g is without center, then the last two conditions of (2) follow from the first one. There might be several p-structures on one Lie algebra; all of them are equal modulo center. Hence, on any simple Lie algebra, there is at most one p-structure.
2) For every x ∈ g, the operator (ad x ) p is a derivation of g. If this derivation is an inner one for every x ∈ g, then the following condition is sufficient for a Lie algebra g to possess a p-structure: for a basis {g i } i∈I of g, there exist elements g [p] i such that [g [p] i , y] = (ad g i ) p (y) for any y ∈ g. 1 For p > 0, the Lie algebra vect(m; N ) is called Zassenhaus algebra if N = ½; it is usually denoted W (m; N ) for any m and N ; if N = ½, it is called Witt algebra. Zassenhaus algebras are simple if m > 1; there is no special name for the simple derived vect (1) (1; n), see a discussion in [GZ]. The Lie algebra of divergence-free, or "special", vector fields is denoted svect(m; N ), usually abbreviated to S(m; N ).
1.2. Problem. We consider the following problem, whose solution is well-known for p = 0; for a clear exposition of the solution, see the book [H] around p.500: (3) For any finitely generated commutative group G, classify G-gradings of simple finite-dimensional Lie algebras over K.
Lie algebras for which a solution of Problem (3) is known. Although we are interested in a particular case of Problem (3), let us briefly review the known general results.
For p = 2, [Ko] is a very clear review of the cases where the solution of Problem (3) has been found; for further details, see [BK, KPS]; for examples of applications of certain G-gradings, see [Kos].
For p = 2, the result of [BK] is as follows: all G-gradings of the Witt algebra W (m; ½) (it is vect(m; ½) in our notation) for m ≥ 3 are given by G-gradings of the corresponding algebra of divided powers O(m; ½) due to an isomorphism of their automorphism group schemes [Sk01]. The classification of such gradings is given. Any G-grading of O(m; ½) is equivalent, up to an algebra automorphism, to one that can be described as follows. For a given s such that 0 ≤ s ≤ m, and a 1 , . . . , a m ∈ G, set where a 1 , . . . , a m ∈ G are the respective degrees of the indeterminates generating O(m; ½), i.e., For p > 0 and the restricted Lie algebras considered in [BK] (for p = 2: Witt algebras; for p > 2: Witt algebras, simple relatives of Hamiltonian algebras and algebras of divergence-free vector fields), the only possible Z/2-gradings are generated by those described in eq. (4). For non-restricted Lie algebras, we know several examples of Z/2-gradings of Kaplansky algebras, see [BLLS].
1.3. Problem (3) for G = Z/2 and p = 2: an application of the solution. In [BLLS1], there are offered two methods for constructing a simple finite-dimensional Lie superalgebra from each simple finite-dimensional Lie algebra over a field of characteristic 2; it is proved that every simple finite-dimensional Lie superalgebra can be obtained by one of these two methods (queerification and "method 2"). The "method 2" depends on the Z/2-gradings of the simple Lie algebra which is being superized. Let us recall the method: Let g = g0 ⊕ g1 be a simple Lie algebra with a Z/2-grading gr. Let the 1-step restricted closure of g associated with the grading gr be (5) g <1> := the minimal Lie subalgebra of the restricted closure g containing g and all the elements x [2] , where x ∈ g1.
Clearly, there is a single way to extend the grading gr from g to g <1> ; we assume this extension performed. On the space of g <1> , define the structure of a Lie superalgebra denoted (in what follows, we often omit indicating the grading gr since it enters the definition of g <1> ) (6) S(g <1> , gr) by setting x 2 := x [2] for any x ∈ g1, and retaining the bracket of any even elements with any other element. The Lie superalgebra S(g <1> ) is simple, see [BLLS1]. Our strategic goal is classification of simple Lie (super)algebras, so we formulate our description of Z/2-gradings gr of g in terms of superizations S(g <1> ) whenever we can.
1.4. How we seek Z/2-gradings. Let g be a Lie algebra, and g = g0 ⊕ g1 its Z/2-grading. For an arbitrary x ∈ g, we denote by x0 its even part and by x1 its odd part. Finally, let U ∈ End(g), where g is considered as a vector space, be the projection on the odd part, i.e., Ux = x1 and (I − U)x = x0, where I is the identity operator. A linear operator U is such a projection on the odd part of g in some Z/2-grading if and only if it satisfies the following conditions for all x, y ∈ g: Having fixed a basis in g, we express these conditions in terms of structure constants c ij k as (summation over repeated indices is assumed) For p = 2, eq. (8b) becomes linear and easy (for example, for Mathematica-based computer package SuperLie, see [Gr]) to solve. There remains, however, a problem to be solved: we need equivalence classes of solutions U mod Aut(g), not individual operators. (9b) 1.4.1. A comment: gradings and derivations. For a general discussion of relation between gradings and derivations, and interesting examples for p = 2, see [BLLS]. For p = 3, an example illustrating the said discussion is the Z/4-grading of the Skryabin algebra by, see [GL].
The Lie algebra der vect(m; N) of all derivations of vect(m; N ) coincides with the p-envelop of vect(m; N). Hence, the problem of classification of equivalence classes of Z/p-gradings reduces to the classification of equivalence classes of toral elements in der vect(m; N ). The equivalence classes of maximal tori with respect to the group of automorphisms are described in [Kuz, T] and the answer depends on N 1 − 2 parameters for m = 1.
1.5. Our results. 1) We classify the Z/2-gradings of the Lie algebras of series sl, their simple subquotients, and simple derived or subquotients of both orthogonal series o I and o Π , (under an assumption of the structure of Lie algebras of their derivations in general case; this assumption is proved for ranks ≤ 8 in [BGLL2]) except for psl(4), see subsec. 1.6.1. In all cases considered, these gradings yield the known superizations of the corresponding Lie algebras.
Note that one of the gradings of o (1) (3) has no analogs among the Z/2-gradings of the simple 3-dimensional Lie algebras for p = 2. This unusual grading has analogs among Z/2-gradings of the Zassenhaus algebras vect (1) (1; n).
1.5.1. Examples where Z/2-gradings of "the same algebra" for p = 2 and p = 2 differ or where previously unknown Z/2-gradings have been found. These are the most interesting of our results. 1) For any p, there is only one simple Lie algebra of dimension 3: for p = 2, it is o(3) ≃ sl(2); for p = 2, it is o (1) (3), the derived of o(3).
For p = 2, there is only one nontrivial Z/2-grading (as always, up to an automorphism), and hence if the superization procedure had been defined for p = 2, the superization of sl(2) would have been unique, sl(1|1).
For p = 2, there are 2 inequivalent nontrivial Z/2-gradings of o (1) (3), as we will see below. The corresponding non-isomorphic, see [Leb], Lie superalgebras are oo    2) For p = 2, the Z/2-gradings of the Zassenhaus algebra vect(1; n) were only known for n = 1, i.e., for the restricted algebras, see [BK] (and, for n > 1, obvious Z/2-gradings with some of either x or x + 1 declared odd). We have found new (n − 2)-parametric family of Z/2gradings in the non-restricted case. These gradings yeild superizations which have no known analogs in the case p = 2.
1.6. Open questions. For G = Z/2, the gradings (4) with s = 0 of vect(m; ½) for m > 2 yield vect(k; ½|m − k), where the vectors ½ have m and k coordinates, respectively, i.e., known superizations. If s = 0 in (4) and a i =1 for some of the indices i ≤ s, then the corresponding indeterminates 1 + x i are odd; at the moment we are unable to identify the resulting Lie superalgebra.
The list of simple Lie algebras for which Problem (3) is open: • For p > 3, the simple vectorial Lie algebras for the shearing vector N = ½, and the deforms (results of deformations) thereof (for their classification and description, see [S, Kos, SkH]); • For p = 3, the simple vectorial Lie algebras; in particular, exceptional ones, mainly discovered by Skryabin and lucidly described in [GL], and their deforms to be described (for a review, see [BGL1]); the deforms of o(5) and Brown algebras described in [BLW]; • For p = 2, there are many examples of simple Lie algebras of types not existing for p = 2, see [SkT1,Ei,BGLL2,BLLS,GZ,BGLLS1,BGLLS2], and [BGL,BGL1].
1.6.2. A technical hypothesis. In §2, §3, §4, and §5, we use the conjectural description of all derivations of Lie algebras sl(n), psl(2n), as well as o Π (2n) and o I (n) and their simple relatives, see [BGLL2]. This description is proved if rk g ≤ 8, and sometimes for any n.
Then the general solution of the linear equation (7b) is U = ad A , where A ∈ sl(2n + 1). The nonlinear equation (7a) reduces then to ad A = (ad A ) 2 . Now recall that on sl(2n + 1) there is the 2-structure given by A [2] = A 2 , see [BGL,BLLS1]. Therefore and since sl(2n + 1) has no center, we have A = A 2 , i.e., A is a projection. Since its trace is equal to the dimension of the subspace onto which the projection projects (just look at the normal form of the matrix of the projection), it follows that (10) a given projection A belongs to sl if and only if it is a projection onto a subspace of even dimension.
If A ∈ sl(2n + 1) is a projection, then the Lie superalgebra obtained from it by "method 2" of [BLLS1] is isomorphic to So we see there are n + 1 equivalence classes of Z/2-gradings of sl(2n + 1) and n + 1 of its nonisomorphic superizations (including the trivial purely even one). We can enumerate them either as or, more simply, as sl(k|2n + 1 − k), where k = 0, . . . , n. So the answer is the same as for p = 2.
2.1.2. Proof: the sl(2n) series, n > 2. The Lie algebra der sl(2n) can be identified, see [BGLL2], with pgl(2n) in the sense that for any D ∈ der sl(2n), there is A D ∈ gl(2n) such that D coincides with the restriction of ad A D on sl(2n). These elements A D are defined modulo center; in particular, one can take A D 2 to be (A D ) 2 . Therefore D satisfies the condition (7a) if and only if (A D ) 2 = A D + cI 2n for some c ∈ K. Let d ∈ K be a root of the equation So we see that an operator U on sl(2n) satisfies the conditions (7a) if and only if it can be represented as a restriction of ad A on sl(2n) for some projection A ∈ gl(2n), and then the Lie superalgebra obtained from U by "method 2" is isomorphic to sl(dim Ker A| dim Im A) ∼ = sl(dim Im A| dim Ker A). So there are n+1 nonisomorphic superizations of sl(2n) (including the trivial purely even one): sl(k|2n−k), where k = 0, . . . , n. Again, the answer is the same as for p = 2.
2.1.4. Remark. A Z/2-grading of gl(2n+1) for n > 2 can produce a Lie superalgebra which is not isomorphic to any superalgebra of the form gl(k|2n+1−k). More specifically, such a superalgebra would be isomorphic to the direct sum of a Lie superalgebra of the form sl(k|2n + 1 − k), where 0 ≤ k ≤ n, and 0|1-dimensional center (i.e., 1-dimensional odd center). But since it does not result in any new simple Lie superalgebra, we do not consider gradings of gl here. (1) (1; 2), so this case is considered in Subsec. 6.3. In [BGL], the deforms of the superization of o (1) (3) associated with one of the three Z/2-gradings of o (1) (3) are described; but until now it was unclear if these deforms are true ones, i.e., the deformed algebra is not isomorphic to the initial algebra, as is the case for semitrivial deformations corresponding to certain integrable cocycles from nontrivial cohomology class, for examples, see [BLLS, Ri]. The results of Subsec. 6.3 prove that one of the deforms found in [BGL] is a true one.

Theorem. For any
Proof. The algebra of derivations of o (1) (2n + 1) is o(2n + 1)/c, the quotient of o(2n + 1) modulo center, or the subalgebra of traceless elements of o(2n + 1), see [BGLL2]. In what follows, we assume that A is a projection. It is easy to check that (11) is automatically satisfied if x, y ∈ Im A or if x, y ∈ Ker A. If x ∈ Im A and y ∈ Ker A, then (11) is equivalent to B(x, y) = 0. So A ∈ o B (V ) if and only if Im A and Ker A are orthogonal with respect to B. For A to be traceless, dim Im A must be even, see condition (10). Let A be a projection belonging to der o (1) Since B is nondegenerate, Im A ⊕ Ker A = V and Im A ⊥ Ker A with respect to B, it follows that these restrictions are nondegenerate.
Since dim Ker A = 2n + 1 − 2k is odd, B Ker A is equivalent to I 2n+1−2k . As dim Im A = 2k is even, B Im A may be equivalent to either I 2k or Π 2k . In the latter case, the resulting Lie superalgebra is isomorphic to oo (1) IΠ (2n + 1 − 2k|2k); in the former case, it is isomorphic to oo (1) II (2k|2n + 1 − 2k). Note that the collection "oo (1) II (2k|2n + 1 − 2k), where 1 ≤ k ≤ n" can be described more simply as "oo (1) So we get 2n + 1 nonisomorphic Lie superalgebras from o (1) (2n + 1), including the purely even case. To see that all those cases are really realizable with some projection A, the following approach can be used. Recall that in an odd-dimensional space, all nondegenerate symmetric forms are equivalent, so for any such form B and any symmetric invertible matrix, there is a basis in which the matrix of B is equal to the given one. In particular, for a given k, there are bases in which the Gram matrix of B is either I 2n+1 (call it the first basis) or diag(I 2n+1−2k , Π 2k ) (call it the second basis for this value of k). Take the operator A whose matrix in the first basis is diag(0 2n+1−2k , I 2k ); clearly, A 2 = A and dim Im A = 2k. Relative the form B, the space Im A, which is spanned by the last 2k vectors of the basis, is, clearly, orthogonal to Ker A, which is spanned by the first 2n + 1 − 2k vectors of the basis. So A ∈ der o B , and hence determines a grading of o (1) B . It is easy to see that the Lie superalgebra obtained by "method 2" of [BLLS1] is oo (1) Analogously, an operator A with the same matrix diag(0 2n+1−2k , I 2k ) in the second basis (it does not matter what is the matrix of A in the first basis) determines a grading that corresponds to the Lie superalgebra oo (1) It is possible to show that there are 2n + 1 equivalence classes of Z/2-gradings. It could have happened that there were more gradings than nonisomorphic superizations if two inequivalent gradings yielded isomorphic Lie superalgebras. This, however, does not happen; we skip the details.

The o
(1) Proof. If a bilinear form B on vector space V of dimension 2n is equivalent to I 2n , then the algebra der o (1) B (2n) for some projection A ∈ o B (2n). By arguments we used in the previous section, a projection A belongs to o B (2n) if and only if Im A and Ker A are orthogonal with respect to B. We denote restrictions of B to Im A and Ker A as B Im A and B Ker A , respectively. These restrictions have to be nondegenerate.
If dim Im A = 2k + 1, where 0 ≤ k ≤ n − 1, then these restrictions are equivalent to I 2k+1 and I 2n−2k−1 , respectively, and the resulting Lie superalgebra is isomorphic to can be equivalent to either I 2k or Π 2k , and B Ker A can be equivalent to either I 2n−2k or Π 2n−2k . However, it is impossible for B Im A to be equivalent to Π 2k while B Ker A is equivalent to Π 2n−2k at the same time, because the direct sum of these two forms is equivalent to Π 2n . All the other combinations are possible, and, depending on them, the resulting Lie superalgebra can be isomorphic to either o (1) IΠ (2n − 2k|2k). If dim Im A = 0 or 2n, then the resulting Lie superalgebra is purely even. So, as described above, o Proof. If n is odd, then the center of o Π (2n) is trivial, so in this case o Π (2n). Keeping this in mind, we will denote the simple relative of o Π (2n) as o (2) Π (2n)/c for both even and odd values of n. Similarly, we will denote the simple relatives of oo ΠΠ (2k|2n − 2k) and pe (2) (n) as oo (2) ΠΠ (2k|2n − 2k)/c and pe (2) (n)/c for both even and odd values of n. Consider the Lie algebra As is not difficult to see, for B ∼ Π 2n we havẽ Then, according to [BGLL2] (13) der (o c A =0: In this case, the definition (12) is equivalent to the statement that Im A and Ker A are orthogonal with respect to B. It means that the restrictions of B on Im A and Ker A have to be nondegenerate. Since B is anti-symmetric (i.e., B(x, x) = 0 for any x ∈ V ), these restrictions are anti-symmetric as well, which means that dimensions of Im A and Ker A are even. The resulting Lie superalgebra in this case is isomorphic to In this case, the condition (12) is equivalent to the statement that both Im A and Ker A are isotropic with respect to B. Since Im A ⊕ Ker A = V , this means that dim Im A = dim Ker A = n and there is an invertible linear map f : Ker A → (Im A) * such that B(x, y) = (f (y))(x) for all x ∈ Im A, y ∈ Ker A.
The resulting Lie superalgebra in this case is isomorphic to pe (2) (n)/c. c A =0,1: This is impossible, because in this case, (12) would mean that Im A and Ker A are both isotropic and orthogonal to each other with respect to B, i.e., B(x, y) = 0 for all x, y ∈ V .
Proof. Let us prove that qv n−1 is not isomorphic to any Lie superalgebra of the form q(g), where g is a Lie algebra. A superalgebra of the form q(g) possesses the following property: if x ∈ q(g)0 acts nilpotently on q(g)0, then it acts nilpotently on q(g)1 as well. Indeed, if (ad x ) k | q(g)0 = 0 for some positive integer k, then (ad x ) k Πy = Π (ad x ) k y = 0 for any y ∈ g.
Proof. Let u ∈ O(1; n) be a linear combination of only even powers of the indeterminate x, and let a be its constant term 2 . In other words, let u, a ∈ O(1; n) be such that (15) x · (∂u) = 0; ∂a = 0; u 2 = a 2 .
Consider derivation D u ∈ der O(1; n) of the form Properties (15) imply (after rather lengthy calculations we omit) that (D u ) 2 = D u , so D u describes a Z/2-grading of O(1; n). Consider the linear map ad Du on the Lie algebra der O(1; n) given by D → [D u , D]. Due to the Jacobi identity, we see that ad Du ∈ der(der O(1; n)), where the outer der is for derivations of the Lie algebra der O(1; n). Since the Lie algebra der O(1; n) has a 2-structure given by D [2] = D 2 , we have (ad Du ) 2 = ad Du . Thus, ad Du describes a Z/2-grading of der O(1; n).
The linear map ad Du sends elements of vect(1; n) to vect (1) (1; n) because D u is a linear combination of a vector field and elements of the form ∂ 2 i , and This means that the restriction (ad Du )| vect(1;n) describes a Z/2-grading of vect(1; n) and the restriction (ad Du )| vect (1) (1;n) describes a Z/2-grading of vect (1) (1; n). In what follows, we will denote the latter restriction by D u as well. Different polynomials u correspond to different gradings, as [D u , x∂] = u∂. We need, however, not individual gradings, but their equivalence classes. We were unable to solve this problem completely so far.
In notation of [T], any torus of the restricted closure of vect(1; n) lying in the maximal subalgebra of elements of non-negative degree (assuming deg x = 1) is called an inner one, the other tori are called outer ones.
Then D u = (u + x + xu∂ 2 u)∂ = (u + x + u∂u)∂. Desuperization of any superization of vect (1) (1; n) can be considered as a Lie subalgebra in v n+1 := (vect (1) (1; n)) <1> , see (14). The operator ad Du preserves v n+1 , so the grading it defines on vect (1) (1; n) can be extended to v n+1 with the same definition: D → [D u , D]. This extended operator describes a Z/2-grading of v n+1 . The elements of v n+1 are said to be u-even or u-odd if they are, respectively, even or odd in this grading.
Consider the linear maps T u , A u : v n+1 −→ v n+1 defined as follows (their action on other elements being extended by linearity): For any X ∈ v n+1 , one can check that T u X is u-even (resp. u-odd) if and only if X is 0-even (resp. 0-odd). We omit the calculations that show it, but the idea is as follows: ∂ is 0-odd, while (1 + ∂u + u∂ 2 u)∂ is u-odd. Besides, if we extend the concept of u-evenness/u-oddness to O(1; n), then f + u(1 + ∂u)∂f is u-even (resp. u-odd) for any f ∈ O(1; n) if and only if f is 0-even (resp. 0-odd).
Also, the following is true for any X, Y ∈ vect(1; n): Note also that the operator T u is invertible, so its image is the whole v n+1 . The invertibility follows from the fact that the matrix of T u is upper triangular with1's on the main diagonal in the basis ∂ 2 , ∂, x∂, . . . , x (2 n −1) ∂.
This means that the superization given by any function u such that u(0) = 0 can be considered as a deform of the superization given by u = 0 with the deformation parametrized by the polynomial u or, which is the same, the coefficients of u as follows: defined by linearity and anti-symmetry in other cases; (X 2 ) u = A u X 2 for 0-odd X.
This deformation is filtered relative the decreasing filtration in which L −2 = S(v n+1 ), see (6), while L k−1 for k ≥ −1 consists of vector fields of the form f ∂, where f ∈ O(1; n) does not contain term of degree < k. Such superizations are listed in heading 2) of Theorem.
Finally, observe that these formulas do not capture the trivial Z/2-grading given by the derivation U = 0; the even part g0 of this grading is the whole Lie algebra vect (1) (1; n). Since any torus in der(vect (1) (1; n)) has a form D u , see [Kuz, T] and footnote 3 on p. 12, we completely described all Z/2-gradings of vect (1) (1; n) and the corresponding superizations.
6.2.1. Remark. Observe an unpredicted fact: according to (20), if X, Y ∈ vect(1; n), then [X, Y ] u can be expressed in terms of [X, Y ] and u. In particular, concerning the 0-even part of vect(1; n), this implies the following fact: 6.2.2. Corollary. If u(0) = 0, then the even part of the superization corresponding to D u is a solvable Lie algebra.
For n = 2, any outer torus is conjugate to an inner one, see (22). This is not so for n > 2.
6.2.3. Lemma. Let n > 2, consider the superization qv n−1 = g0 ⊕ g1 corresponding to the outer torus D 1 . Its even part g0 ≃ v n is spanned by the elements (16), and g1 is spanned by the elements (17). The odd part g1 is a reducible g0-module with no lowest weight vector and with the two highest weight vectors Πo 2 n−1 −3 and Πo 2 n−1 −2 with respect to the standard Z-grading of g0, namely (g0) k = Ke k for k = −2, . . . , 2 n − 1.
6.2.3a. Remark. Put v = αΠo 2 n−1 −3 + βΠo 2 n−1 −2 , where α, β ∈ K. Construct a new basis in g1 by the rule . For n = 4, the coordinates of o ′ k in the old basis o k are defined by the following matrix  For α = 0, the g0-action on g1 in the new basis is defined by the following table In what follows we consider only generating functions of the form . . , c n−1 ) is a set of free parameters.
For a basis of its odd part we take O 1 = e 1,2 + e 2,1 , O 2 = e 1,3 + e 3,1 . The commutation relations are given by the following formulas: The squaring in S(g <1> ) is given by the following formulas: The (g <1> )0-module structure is as follows: The multiplication by elements E 1 and E 3 is given by the following formulas:

The isomorphism between oo
(1) II (1|2) and g <1> is given by the following formulas: 6.3.3. Remarks. 1) Observe that the even parts of both superizations are solvable Lie algebras, but the corresponding Lie superalgebras are simple. For more details and examples of this phenomenon indigenous to p = 2, see Shchepochkina's comment (the last section) in [BGL].
2) Observe that due to Theorem 6.2 the Lie superalgebra oo  6.4. Superizations corresponding to inner tori. Consider a superization of vect (1) (1; n) given by a generating function u such that u(0) = 0. For such a function u, the derivation D u spans an inner torus. According to [T], the derivation D u is conjugate to the derivation D f in vect (1) (1; n), where f = 1≤i≤n−1 c i x (2 i ) and the c i x (2 i ) are the corresponding terms of u, so the superizations corresponding to D u and D f are isomorphic. For this reason, in the rest of this Subsection we consider only generating functions of the form 1≤i≤n−1 c i x (2 i ) . This gives us an (n − 1)-parametric family of Z/2-gradings.
6.4.1. Conjecture. Let u 1 = 1≤k≤n−1 c k x (2 k ) and u 2 = 1≤k≤n−1 b k x (2 k ) be generating functions, then the superizations corresponding to D u 1 and D u 2 are isomorphic if and only if there exists ε ∈ K × such that c k = ε 2 k −1 b k for all k = 1, . . . , n − 1.
The following is a sketch of a proof of this conjecture, including a complete computer-aided proof for 2 ≤ n ≤ 6.
First of all, if there is an ε = 0 such that c k = ε 2 k −1 b k for all k = 1, . . . , n − 1, then D u 1 and D u 2 are conjugate by the automorphism of vect (1) (1; n) given by x (m) ∂ → ε m−1 x (m) ∂ (this automorphism is generated by the automorphism of O(1; n) given by x (m) → ε m x (m) ).
By Theorem 6.2, the superization given by u 1 is isomorphic to a deform of k(1; n − 1|1). Consider the even part of of such a deform. According to (20), it contains a commutative subalgebra of codimension 1, which is the 0-even part of vect(1; n). The element ∂ 2 acts on this commutative subalgebra, and its action is given by (26) v∂ → (1 + u∂ 2 u)∂ 2 v∂ for any 0-odd v ∈ O(1; n).
6.4.2. Conjecture (Proved by computer for n = 2, 3, 4, 5, 6). The characteristic polynomial of the linear operator (26) on the 0-even part of vect(1; n) for n ≥ 2 is defined by the formula For two deforms given by u 1 and u 2 to be isomorphic to each other, their even parts have to be isomorphic, which means that the two actions of ∂ 2 on the 0-even part of vect(1; n) have to be conjugate (similar) up to a non-zero scalar factor, i.e., if A and A ′ are such operators, then A ′ = αMAM −1 for some non-zero α ∈ K and an invertible linear map M. If two operators are conjugate up to a scalar multiple α, the roots of their characteristic polynomials differ by the same multiple, so if the dimension of the space they act on is d (in our case, d = 2 n−1 ) and one polynomial is equal to 0≤i≤d z i λ i , where z i ∈ K and z d = 1, then the other one would have the form 0≤i≤k z i α k−i λ i . So, if Conjecture 6.4.2 is correct, then for the superizations given by u 1 and u 2 to be isomorphic, there must exist non-zero α ∈ K such that or, equivalently, c k = ε 2 k −1 b k , where ε = √ α, for all k = 1, . . . , n − 1.
So the set of equivalence classes of superizations of vect (1) (1; n) corresponding to generating functions u such that u(0) = 0 consist of the following two types: A) the superization corresponding to u = 0, which is isomorphic to k(1; n − 1|1), B) an (n − 2)-parametric family of its pairwise non-isomorphic deforms. Note, though, that it is not a result of (n − 2)-parametric deformation of k(1; n − 1|1); to obtain all these deforms, an (n − 1)-parametric deformation is needed, but the deforms obtained from some sets of parameters are isomorphic: parameters (c 1 , . . . , c n−1 ) and (b 1 , . . . , b n−1 ) produce isomorphic deformations if and only if there exists ε ∈ K × such that c k = ε 2 k −1 b k for all k ∈ 1, n − 1.