Free Access

Graded identities of the adjoint representation of ${𝔰 𝔩}_{2} (K)$

Cássia Sampaio

https://orcid.org/0009-0007-1666-7124

Department of Mathematics, State University of Campinas, 651 Sérgio Buarque de Holanda, 13083-859 Campinas, SP, Brazil

E-mail Address: cfsampaio@ime.unicamp.br

Corresponding author.

Search for more papers by this author

and

Plamen Koshlukov

https://orcid.org/0000-0002-1819-0100

Department of Mathematics, State University of Campinas, 651 Sérgio Buarque de Holanda, 13083-859 Campinas, SP, Brazil

E-mail Address: plamen@unicamp.br

Search for more papers by this author

https://doi.org/10.1142/S0218196725500055Cited by:0 (Source: Crossref)

Abstract

Let K be a field of characteristic zero and let ${𝔰 𝔩}_{2} (K)$ be the 3-dimensional simple Lie algebra over K. In this paper, we describe a finite basis for the $ℤ_{2}$ -graded identities of the adjoint representation of ${𝔰 𝔩}_{2} (K)$ , or equivalently, the $ℤ_{2}$ -graded identities for the pair $(M_{3} (K), {𝔰 𝔩}_{2} (K))$ . We work with the canonical grading on ${𝔰 𝔩}_{2} (K)$ and the only nontrivial $ℤ_{2}$ -grading of the associative algebra $M_{3} (K)$ induced by that on ${𝔰 𝔩}_{2} (K)$ .

Communicated: Ualbai Umirbaev

Keywords:

AMSC: 17B01, 17B70, 16R10, 16R50

1. Introduction

Determining a basis of polynomial identities for the matrix algebra of order n is a challenging problem that draws the attention of many researchers in the theory of algebras with polynomial identities. So far, there have been some results in this direction, but the problem, in general, remains widely open.

A finite basis for the identities of the matrix algebra of order 2, $M_{2} (K)$ , over a field of characteristic zero, was described by Razmyslov [17] and subsequently minimized by Drenski [5]. Later on Drensky in [6] described various numerical invariants of the relatively free algebras related to $M_{2} (K)$ , and to ${𝔰 𝔩}_{2} (K)$ . In positive characteristic, it is known that the Lie algebra $M_{2} (K)$ , $char (K) = 2$ , does not possess a finite basis of identities when K is infinite, a theorem proved by Vaughan-Lee in [22]. If $char (K) > 2$ and K is infinite, Koshlukov [13] exhibited a finite basis of identities for $M_{2} (K)$ . To achieve this, he crucially used the invariant theory of the orthogonal group, see [3] and weak identities [18]. Concerning the algebra $M_{n} (K)$ , $n \geq 3$ , little is known about its polynomial identities. For a finite field K, Genov and Siderov [7, 8] determined bases of identities for $M_{3} (K)$ and $M_{4} (K)$ . In the case of infinite fields, weak identities and graded identities are expected to be an essential tool, as it was the case for $M_{2} (K)$ , and also in the proof that the variety of associative algebras over fields of characteristic zero satisfies the Specht property, the celebrated theorem of Kemer.

Weak identities first appeared in Razmyslov’s work [17], where, in addition to exhibiting a finite basis of identities for $s l_{2} (K)$ and $M_{2} (K)$ , $char (K) = 0$ , he proved that the variety $var (M_{2} (K))$ satisfies the Specht property. The notions of weak identity and identity of a representation might be easily merged in the following sense. Let $K 〈 X 〉$ be the free associative algebra over K freely generated by a countable set $X = {x_{1}, x_{2}, \dots, x_{n}, \dots}$ . If L is a Lie algebra and $ρ : L \to 𝔤 𝔩 (V)$ is a Lie algebra representation of L on a vector space V, a polynomial identity of the representation $ρ$ is a polynomial $f (x_{1}, \dots, x_{n}) \in K <X>$ such that $f (ρ (e_{1}), \dots, ρ (e_{n})) = 0$ for every $e_{1}, \dots e_{n} \in L$ . Denoting by M the associative algebra generated by $ρ (L)$ , the problem of determining the identities of the representation $ρ$ of L is equivalent to finding the identities for the associative-Lie pair $(M, ρ (L))$ .

Identities of this type were also crucial in finding a basis of identities for the Lie algebra $s l_{2} (K)$ when the field K is infinite and of characteristic greater than 2 [21]. In the same context, Koshlukov [12] determined a basis of identities for the natural representation ${𝔰 𝔩}_{2} (F) \to M_{2} (F)$ . Here we recall that weak identities can be defined in a similar way for other classes of algebras that are not “very far” from associative ones, and they play an important role in studying polynomial identities for Lie, alternative and Jordan algebras.

We recall the notion of a graded identity. As we work with gradings by $ℤ_{2}$ , the group of order 2, we consider gradings by this group only. If $A = A_{0} ⨁ A_{1}$ is a $ℤ_{2}$ -graded algebra, then $f (y_{1}, \dots, y_{m}, z_{1}, \dots, z_{n})$ is a $ℤ_{2}$ -graded identity of A if $f (a_{1}, \dots, a_{m}, b_{1}, \dots, b_{n}) = 0$ for every $a_{i} \in A_{0}$ and $b_{i} \in A_{1}$ . Here we consider $X = Y \cup Z$ , Y and Z disjoint infinite sets, and the $ℤ_{2}$ -grading on $K 〈 X 〉$ is given by declaring the variables from Y of degree 0 (even ones), and those from Z of degree 1 (odd ones).

Graded identities have also proven to be an important tool in studying PI-algebras. They were essential, for example, in establishing that the variety of associative algebras satisfies the Specht property in characteristic 0 [10]. We emphasize that the relation between graded identities and ordinary ones established by Kemer [11, Theorem 1.1] is not completely valid in positive characteristic. We note that if two algebras satisfy the same graded identities they also satisfy the same ordinary identities. However, the relationship between graded and ordinary identities remains not that clear: for example, the knowledge of the graded identities of an algebra does not give much information about the ordinary ones satisfied by the same algebra.

Finite bases of identities for the $ℤ_{2}$ -graded algebras $M_{2} (K)$ , with $char K = 0$ , and for $M_{1, 1} (G)$ , where G is the Grassmann algebra over a field of characteristic 0 were found by Di Vincenzo [4]. Later on, these results were extended to infinite fields of characteristic different from 2, see [15]. The group gradings on the Lie algebra ${𝔰 𝔩}_{2} (K)$ , in characteristic 0, were studied by Repin in [19], where he also described the multilinear structure of the corresponding graded relatively free algebras. Once again, over an infinite field K, with $char K \neq 2$ , a basis for the $ℤ_{2}$ -graded identities of ${𝔰 𝔩}_{2} (K)$ was described in [14]. The Specht property of the graded identities for ${𝔰 𝔩}_{2} (K)$ , in characteristic 0, for every group grading, was obtained in [9]. The authors of [1] studied the growth of the homogeneous components of the relatively free algebras for graded Lie algebras, and computed the corresponding graded GK-dimensions. In order to obtain a finite basis of identities for the pair $(M_{2} (K), g l_{2} (K))$ , Koshlukov and Krasilnikov [16] found a basis for the $ℤ_{2}$ -graded identities for the pair $(M_{2} (D), g l_{2} (D))$ , where D is an infinite integral domain. Under the same assumptions for D, Códamo and Koshlukov [2] determined a basis of weak identities for the $ℤ_{2}$ -graded pair $(M_{2} (D), {𝔰 𝔩}_{2} (D))$ and proved that this pair possesses the Specht property. Trishin, in [20], found finite bases of identities for the irreducible representations of ${𝔰 𝔩}_{2} (K)$ under the assumption that K is an algebraically closed field of characteristic 0.

Here we combine ideas from [16] and [20] in order to describe a basis for the $ℤ_{2}$ -graded identities of the adjoint representation of ${𝔰 𝔩}_{2} (K)$ , the 3-dimensional simple Lie algebra over a field of characteristic zero. The image $ad (s l_{2} (K))$ of this representation generates $M_{3} (K)$ as an associative algebra. Hence, equivalently, we describe a basis of the graded identities for the pair $(M_{3} (K), {𝔰 𝔩}_{2} (K))$ . Clearly, if the field K is not algebraically closed there may exist nonisomorphic 3-dimensional simple Lie algebras over K. But a standard argument, using extensions of the scalars, shows that these will satisfy the same polynomial identities. Therefore, without loss of generality, one may assume K is algebraically closed.

2. Preliminaries

Throughout this paper, K stands for a field of characteristic zero, and all algebras and vector spaces are considered over K. As we stated above, we can (and will) assume K is algebraically closed. We use the additive notation for the cyclic group $ℤ_{2} = {0, 1}$ of order two. If A is an associative algebra we denote by $A^{(-)}$ the vector space of A together with the Lie bracket operation $[a, b] = a b - b a$ . It is well known that $A^{(-)}$ is a Lie algebra. We shall use the bracket notation for the product in a Lie algebra. The commutators without inner brackets are supposed to be left-normed, that is $[x_{1}, x_{2}, \dots, x_{n}] = [[x_{1}, x_{2}, \dots, x_{n - 1}], x_{n}]$ for every $n \geq 3$ .

An algebra A is said to be a $ℤ_{2}$ -graded algebra (also called simply graded algebra or, in the associative case, superalgebra) if $A = A_{0} ⨁ A_{1}$ , where $A_{0}$ and $A_{1}$ are vector subspaces, respecting $A_{i} A_{j} \subseteq A_{i + j}$ , $i, j \in ℤ_{2}$ . We call $(A, L)$ a $ℤ_{2}$ -graded pair if L is a Lie subalgebra of $A^{(-)}$ such that $L = L_{0} ⨁ L_{1}$ , $L_{i} = A_{i} \cap L$ , $i = 0$ , 1. Since we consider gradings by $ℤ_{2}$ only we will omit the group and we will often call these simply graded algebras or graded pairs.

Let $K 〈 X 〉$ be the free associative algebra freely generated by $X = Y \cup Z$ , where $Y = {y_{1}, y_{2}, \dots, y_{m}, \dots}$ and $Z = {z_{1}, z_{2}, \dots, z_{n}, \dots}$ are countable disjoint sets. We call even (respectively odd) variables the ones from Y (respectively Z). Setting the homogeneous degree of a monomial $m \in K 〈 X 〉$ as 0 if its degree with respect to the variables $z \in Z$ is even and as 1 otherwise, the algebra $K 〈 X 〉$ is $ℤ_{2}$ -graded with the natural grading $K 〈 X 〉 = K {〈 X 〉}_{0} ⨁ K {〈 X 〉}_{1}$ , where $K {〈 X 〉}_{i}$ is the vector subspace spanned by the monomials of homogeneous degree i, $i = 0$ , 1. Let $L 〈 X 〉$ be the Lie subalgebra of $K ł {〈 X 〉}^{(-)}$ generated by X. It is well known (Witt Theorem) that $L 〈 X 〉$ is the free Lie algebra freely generated by X. With the grading inherited from $K 〈 X 〉$ , $L 〈 X 〉$ is the free $ℤ_{2}$ -graded Lie algebra freely generated by X. Consequently, the pair $(K 〈 X 〉, L 〈 X 〉)$ is the free graded pair of countable rank in the class of all graded associative-Lie pairs. Hereafter, $K 〈 X 〉$ will always denote the free $ℤ_{2}$ -graded algebra.

We call an ideal I of $K 〈 X 〉$ a $T_{2}$ -ideal if I is invariant under all graded endomorphisms of $K 〈 X 〉$ , that is, if $ϕ (I) \subseteq I$ for every endomorphism $ϕ$ of $K 〈 X 〉$ such that $ϕ (y_{i}) \in K {〈 X 〉}_{0}$ and $ϕ (z_{i}) \in K {〈 X 〉}_{1}$ . An ideal I is called a weak $T_{2}$ -ideal if it is invariant under the graded endomorphisms of the pair $(K 〈 X 〉, L 〈 X 〉)$ . As in the ordinary case, we can define a (weak) $T_{2}$ -ideal generated by a set and a basis of (weak) graded identities.

A polynomial $f (y_{1}, \dots, y_{m}, z_{1}, \dots z_{n}) \in K 〈 X 〉$ is said to be a weak graded identity of the $ℤ_{2}$ -graded pair $(A, L)$ if $f (a_{1}^{(0)}, \dots, a_{m}^{(0)}$ , $a_{1}^{(1)}, \dots, a_{n}^{(1)}) = 0$ for all $a_{1}^{(0)}, \dots, a_{m}^{(0)} \in L_{0}$ and $a_{1}^{(1)}, \dots, a_{n}^{(1)} \in L_{1}$ . It can be easily proved that the set $T_{2} (A, L)$ of weak graded identities of a pair $(A, L)$ is a weak $T_{2}$ -ideal in $K 〈 X 〉$ . The converse also holds, its proof repeats word by word that of the case of ordinary (associative, ungraded) T-ideals.

Let $S \subset K 〈 X 〉$ be a nonempty set of polynomials, then the ideal of graded weak identities generated by S is the least ideal of graded weak identities I that contains S. Recall that in this case S is called a basis of I. A polynomial $f \in K 〈 X 〉$ is a consequence of the identities of S whenever $f \in I$ . Accordingly, f, $g \in K 〈 X 〉$ are equivalent as weak graded identities when they generate the same ideal of weak graded identities. (In other words, f is a consequence of g and g is a consequence of f.)

It is well known that over an infinite field every ideal of identities is generated by its multihomogeneous elements. When the base field is of characteristic 0, the ideals of identities are generated by the multilinear polynomials they contain. The proofs of these facts are standard, moreover, they transfer word by word to our case.

3. Graded Identities for the Pair $(M, S)$

Let ${𝔰 𝔩}_{2} (K)$ be the simple 3-dimensional Lie algebra over the field K and denote by S the image of the adjoint representation of ${𝔰 𝔩}_{2} (K)$ , $ad : g \in {𝔰 𝔩}_{2} \mapsto ad g \in 𝔤 𝔩 ({𝔰 𝔩}_{2} (K))$ . Then $𝔤 𝔩 ({𝔰 𝔩}_{2} (K)) ≅ M_{3} (K)$ , since the action of ${𝔰 𝔩}_{2}$ on the 3-dimensional vector space ${𝔰 𝔩}_{2} (K)$ is irreducible. Hence the image of ${𝔰 𝔩}_{2} (K)$ generates, as an associative algebra, the whole algebra of linear transformations in dimension 3. In this way, we can write simply $ad : {𝔰 𝔩}_{2} (K) \to M_{3} (K)$ .

Since ${𝔰 𝔩}_{2} (K)$ is a simple Lie algebra, the adjoint representation is faithful. Therefore, $S ≅ {𝔰 𝔩}_{2} (K)$ is also a simple Lie algebra. Taking the canonical basis of ${𝔰 𝔩}_{2}, {h, x, y}$ , with $[h, x] = 2 x, [h, y] = - 2 y$ and $[x, y] = h$ , we have

S = {(0 - n m - 2 m 2 a 0 2 n 0 - 2 a) | a, m, n \in K} . <math display="block" altimg="eq-00150.gif"><mrow><mi>S</mi><mo>=</mo><mfenced separators="" open="{" close="}"><mrow><mfenced separators="" open="" close="|"><mrow><mfenced separators="" open="(" close=")"><mrow><mtable displaystyle="true" equalrows="false" equalcolumns="false" class="array"><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mo>-</mo><mi>n</mi></mtd><mtd columnalign="center"><mi>m</mi></mtd></mtr><mtr><mtd columnalign="center"><mo>-</mo><mn>2</mn><mi>m</mi></mtd><mtd columnalign="center"><mn>2</mn><mi>a</mi></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mn>2</mn><mi>n</mi></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mo>-</mo><mn>2</mn><mi>a</mi></mtd></mtr></mtable></mrow></mfenced></mrow></mfenced><mi>a</mi><mo>,</mo><mi>m</mi><mo>,</mo><mi>n</mi><mo>\in</mo><mi>K</mi></mrow></mfenced><mspace width="-.17em" class="negativethinspace"></mspace><mo>.</mo></mrow></math>

Note that S generates $M_{3} (K)$ as an associative algebra. We have the following natural $ℤ_{2}$ -grading of the Lie algebra S.

S 0 = {(000 0 2 a 0 00 - 2 a)} and S 1 = {(0 - m n - 2 n 00 2 m 00)}, <math display="block" altimg="eq-00153.gif"><mrow><msub><mrow><mi>S</mi></mrow><mrow><mn>0</mn></mrow></msub><mo>=</mo><mfenced separators="" open="{" close="}"><mrow><mfenced separators="" open="(" close=")"><mrow><mtable displaystyle="true" equalrows="false" equalcolumns="false" class="array"><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>2</mn><mi>a</mi></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mo>-</mo><mn>2</mn><mi>a</mi></mtd></mtr></mtable></mrow></mfenced></mrow></mfenced><mspace width="1em" class="quad"></mspace><mstyle><mtext mathvariant="normal">and</mtext></mstyle><mspace width="1em" class="quad"></mspace><msub><mrow><mi>S</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><mfenced separators="" open="{" close="}"><mrow><mfenced separators="" open="(" close=")"><mrow><mtable displaystyle="true" equalrows="false" equalcolumns="false" class="array"><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mo>-</mo><mi>m</mi></mtd><mtd columnalign="center"><mi>n</mi></mtd></mtr><mtr><mtd columnalign="center"><mo>-</mo><mn>2</mn><mi>n</mi></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mn>2</mn><mi>m</mi></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr></mtable></mrow></mfenced></mrow></mfenced><mspace width="-.17em" class="negativethinspace"></mspace><mo>,</mo></mrow></math>

a, m,

$n \in K$ . It is well known that, up to a graded isomorphism, the above is the only nontrivial

$ℤ_{2}$ -grading on

${𝔰 𝔩}_{2} (K)$ , and on S.

Denote by M the matrix algebra $M_{3} (K)$ and consider the $ℤ_{2}$ -grading exhibited below which is, up to a graded isomorphism, the only nontrivial $ℤ_{2}$ -grading of $M_{3} (K)$ .

M 0 = {(K 00 0 K K 0 K K)} and M 1 = {(0 K K K 00 K 00)} . <math display="block" altimg="eq-00161.gif"><mrow><msub><mrow><mi>M</mi></mrow><mrow><mn>0</mn></mrow></msub><mo>=</mo><mfenced separators="" open="{" close="}"><mrow><mfenced separators="" open="(" close=")"><mrow><mtable displaystyle="true" equalrows="false" equalcolumns="false" class="array"><mtr><mtd columnalign="center"><mi>K</mi></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mi>K</mi></mtd><mtd columnalign="center"><mi>K</mi></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mi>K</mi></mtd><mtd columnalign="center"><mi>K</mi></mtd></mtr></mtable></mrow></mfenced></mrow></mfenced><mspace width="1em" class="quad"></mspace><mstyle><mtext mathvariant="normal">and</mtext></mstyle><mspace width="1em" class="quad"></mspace><msub><mrow><mi>M</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><mfenced separators="" open="{" close="}"><mrow><mfenced separators="" open="(" close=")"><mrow><mtable displaystyle="true" equalrows="false" equalcolumns="false" class="array"><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mi>K</mi></mtd><mtd columnalign="center"><mi>K</mi></mtd></mtr><mtr><mtd columnalign="center"><mi>K</mi></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mi>K</mi></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr></mtable></mrow></mfenced></mrow></mfenced><mspace width="-.17em" class="negativethinspace"></mspace><mo>.</mo></mrow></math>

We draw the readers’ attention that the latter grading is compatible with that on S: we have that

$S_{i} = M_{i} \cap S$ . This implies

$(M, S)$ is a graded pair.

Therefore, instead of considering the graded pair $(M_{3} (K), {𝔰 𝔩}_{2} (K))$ we can study the graded pair $(M, S)$ .

We denote by $s_{n} (x_{1}, \dots, x_{n}) = \sum_{σ \in S_{n}} {(- 1)}^{σ} x_{σ (1)}, \dots, x_{σ (n)}$ , the standard polynomial of degree n, where $S_{n}$ is the symmetric group of all permutations of $(1, 2, \dots, n)$ , and ${(- 1)}^{σ}$ stands for the sign of $σ \in S_{n}$ .

Lemma 3.1. The following polynomials are weak graded identities for the graded pair (M, S).

[y 1, y 2], <math display="block" altimg="eq-00171.gif"><mo stretchy="false">[</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo>,</mo></math> (3.1)

y 1 z y 2, <math display="block" altimg="eq-00172.gif"><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>z</mi><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo></math> (3.2)

z 1 y z 2 + z 2 y z 1, <math display="block" altimg="eq-00173.gif"><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>y</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mi>y</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo></math> (3.3)

s 3 (z 1, z 2, z 3), <math display="block" altimg="eq-00174.gif"><msub><mrow><mi>s</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">(</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">)</mo><mo>,</mo></math> (3.4)

y 1 z 1 z 2 y 2 - y 2 z 1 z 2 y 1, <math display="block" altimg="eq-00175.gif"><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo></math> (3.5)

z 1 y 1 y 2 z 2 - z 2 y 1 y 2 z 1, <math display="block" altimg="eq-00176.gif"><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo></math> (3.6)

s 3 (z 1, z 2, y) x - x s 3 (z 1, z 2, y), x = y i, z j, <math display="block" altimg="eq-00177.gif"><msub><mrow><mi>s</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">(</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mi>y</mi><mo stretchy="false">)</mo><mi>x</mi><mo>-</mo><mi>x</mi><msub><mrow><mi>s</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">(</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mi>y</mi><mo stretchy="false">)</mo><mo>,</mo><mspace width="1em" class="quad"></mspace><mi>x</mi><mo>=</mo><msub><mrow><mi>y</mi></mrow><mrow><mi>i</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>,</mo></math> (3.7)

y [z 1, z 2] z 3 + z 3 z 1 y z 2 . <math display="block" altimg="eq-00178.gif"><mi>y</mi><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>y</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>.</mo></math> (3.8)

Proof. The identities (3.1), (3.2), (3.3), (3.5) and (3.6) are verified by direct and easy computation. For example, two diagonal matrices commute, and hence (3.1) follows. Substituting distinct elements of $S_{0}$ and of $S_{1}$ in (3.2), (3.3), (3.5), (3.6) and (3.8), one gets that all of them vanish.

To see that (3.4) is an identity, note that $S_{1}$ is a 2-dimensional subspace and $s_{3} (z_{1}, z_{2}, z_{3})$ is a skew-symmetric polynomial in three odd variables. Therefore substituting in elements from the basis of $S_{1}$ we have to repeat some element, and the skew-symmetry yields that the polynomial vanishes. The identity (3.7) follows from the fact that $s_{3} (z_{1}, z_{2}, y)$ is a multiple of the Casimir element of the adjoint representation which belongs to the center of the universal enveloping algebra of ${𝔰 𝔩}_{2} (K)$ . □

As immediate consequences of (3.2) and (3.3), we have the following weak $ℤ_{2}$ -graded identities of $(M, S)$

[y, z 1, z 2] - y z 1 z 2 - z 2 z 1 y, <math display="block" altimg="eq-00188.gif"><mo stretchy="false">[</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo>-</mo><mi>y</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>y</mi><mo>,</mo></math> (3.9)

[z, y 1, y 2] - z y 1 y 2 - y 2 y 1 z, <math display="block" altimg="eq-00189.gif"><mo stretchy="false">[</mo><mi>z</mi><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo>-</mo><mi>z</mi><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>z</mi><mo>,</mo></math> (3.10)

[y 1, z, y 2] - [y 2, z, y 1], [z 1, y, z 2] - [z 2, y, z 1] . <math display="block" altimg="eq-00190.gif"><mo stretchy="false">[</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mi>z</mi><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo>-</mo><mo stretchy="false">[</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mi>z</mi><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo stretchy="false">]</mo><mo>,</mo><mspace width="1em" class="quad"></mspace><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo>-</mo><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo stretchy="false">]</mo><mo>.</mo></math> (3.11)

The next are consequences of the polynomial (3.4)

z 1 [z 2, z 3] - z 2 [z 1, z 3] + z 3 [z 1, z 2], <math display="block" altimg="eq-00191.gif"><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">]</mo><mo>-</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">]</mo><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo>,</mo></math> (3.12)

(1 - τ 13) z 1 z 2 z 3 - [z 1, z 3] \circ z 2, <math display="block" altimg="eq-00192.gif"><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mn>1</mn><mn>3</mn></mrow></msub><mo stretchy="false">)</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>-</mo><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">]</mo><mo>\circ</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo></math> (3.13)

where

$τ_{13}$ is the transposition of the variables

$z_{1}$ and

$z_{3}$ , and the symbol ∘ stands for the Jordan product, that is

$a \circ b = a b + b a$ .

Let I be the weak $T_{2}$ -ideal generated by the polynomials of the previous Lemma. We will prove that $I = T_{2} (M, S)$ .

Lemma 3.2. The following weak graded identities are valid in the pair (M,S).

z 1 y z 2 z 3 = z 3 [z 2, z 1] y, <math display="block" altimg="eq-00199.gif"><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>y</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>=</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo stretchy="false">]</mo><mi>y</mi><mo>,</mo></math> (3.14)

y z 1 z 2 z 3 = [y, z 2, z 3] z 1, <math display="block" altimg="eq-00200.gif"><mi>y</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>=</mo><mo stretchy="false">[</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">]</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo></math> (3.15)

z 1 z 2 z 3 y = z 3 [y, z 1, z 2] . <math display="block" altimg="eq-00201.gif"><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mi>y</mi><mo>=</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">[</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo>.</mo></math> (3.16)

Proof. To prove (3.14), we use (3.3), (3.7) and (3.8)

Now we prove (3.15).

Similarly, we prove (3.16). □

Lemma 3.3. The following identity holds in the pair (M,S)

for every

$k > 1$ and for every permutation

$σ \in S_{k}$ of the indices

$1, \dots, k$ .

In particular, we have

[y, z 1, z 2, \dots, z 2 k - 1, z 2 k] = [y, z 2 k - 1, z 2 k, z 2 k - 3, z 2 k - 2, \dots, z 1, z 2] . <math display="block" altimg="eq-00208.gif"><mo stretchy="false">[</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>-</mo><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>k</mi></mrow></msub><mo stretchy="false">]</mo><mo>=</mo><mo stretchy="false">[</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>-</mo><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>k</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>-</mo><mn>3</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>-</mo><mn>2</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo>.</mo></math> (3.18)

Proof. Note that it is sufficient to prove the assertion for $k = 2$ , since the polynomial $[y, z_{i_{1}}, z_{i_{2}}, \dots, z_{i_{2 l}}]$ lies in $L {〈 X 〉}_{0}$ . When $k = 2$ , Eq. (3.17) is equivalent to $[y, z_{1}, z_{2}, z_{3}, z_{4}] = [y, z_{3}, z_{4}, z_{1}, z_{2}]$ . Now we compute $[y, z_{1}, z_{2}, z_{3}, z_{4}]$ , it equals

To prove the general case, we apply iteratively the case considered above, using the fact that

$[y, z_{i_{1}}, z_{i_{2}}, \dots, z_{i_{2 l}}] \in L {〈 X 〉}_{0}$ . □

Lemma 3.4. The following polynomials belong to I, for every nonnegative integer ,k

y 1 z 1, \dots, z 2 k + 1 y 2, <math display="block" altimg="eq-00217.gif"><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo></math> (3.19)

z 1 y 1, \dots, y 2 k + 1 z 2 + z 2 y 1, \dots, y 2 k + 1 z 1, <math display="block" altimg="eq-00218.gif"><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo></math> (3.20)

z 1 y 1 y 2, \dots, y 2 k z 2 - z 2 y 1 y 2, \dots, y 2 k z 1, <math display="block" altimg="eq-00219.gif"><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>k</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>k</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo></math> (3.21)

y 1 z 1, \dots, z 2 k y 2 - y 2 z 1, \dots, z 2 k y 1, <math display="block" altimg="eq-00220.gif"><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>k</mi></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>k</mi></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo></math> (3.22)

Proof. First, we note that by (3.2), the identity (3.19) holds for $k = 0$ . Since $[y_{2}, z_{i}, z_{j}] \in S_{0}$ we have

Similarly, we prove (3.20) using (3.3), the fact that $[z, y_{i}, y_{j}] \in S_{1}$ and the weak graded identities (3.1), (3.2) and (3.10).

When $k = 1$ , (3.21) is equal to (3.6). Using (3.1), (3.10) and induction, we prove (3.21) for every k.

Finally, we use (3.15) and (3.18) to prove (3.22), observing that (3.22) is equal to (3.5), for $k = 1$ .

□

Lemma 3.5. The following weak graded identity holds in the graded pair (M, S)

y 1 y 2 z 1 z 2 - z 1 z 2 y 1 y 2 . <math display="block" altimg="eq-00228.gif"><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>.</mo></math> (3.23)

Proof. The proof consists in the following computation.

□

Inductively, for every integers m, $n \geq 0$ we have the following weak graded identity of the pair $(M, S)$ .

y 1 y 2, \dots, y 2 m z 1 z 2, \dots, z 2 n = z 1 z 2, \dots, z 2 n y 1 y 2, \dots, y 2 m . <math display="block" altimg="eq-00232.gif"><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>m</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>n</mi></mrow></msub><mo>=</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>n</mi></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>m</mi></mrow></msub><mo>.</mo></math> (3.24)

4. A Basis of Weak Graded Identities for $(M, S)$

In the previous section, we listed several weak $ℤ_{2}$ -graded identities of the pair $(M, S)$ . In this section, we shall describe the form of the polynomials in $K 〈 X 〉$ modulo I. We denote $F = K 〈 X 〉 / I$ .

Lemma 4.1. Let $f (y_{1}, y_{2}, \dots, y_{m}, z_{1}, z_{2}, \dots, z_{n})$ be a multihomogeneous polynomial in F. Then f is symmetric in its even variables $y_{1}, y_{2}, \dots, y_{m}$ .

Proof. It is sufficient to prove that, in an arbitrary monomial $m \in F$ , we can rearrange the indices of the even variables as we want, keeping the coefficient (and the sign) of the monomial. Thus we will show that transposing two of the even variables in a monomial does not change it, modulo I.

Let $Y_{i}$ and $Z_{j}$ denote words with only even and odd variables, respectively. If $m = Y_{1} Z_{1}$ or $m = Z_{1} Y_{1}$ , where $Z_{1}$ is allowed to be empty, by the identity (3.1), we can reorder the indices of the even variables in the way we want.

Suppose now that m has a factor of the form $y_{i} Z_{k} y_{j}$ . We can assume that $Z_{k}$ is of even length by (3.19). In this case, by (3.22), we can transpose $y_{i}$ and $y_{j}$ .

Now, let $m = Y_{1} Z_{1} Y_{2} Z_{2}, \dots, Y_{k} Z_{k}$ be a monomial, where $Y_{1}$ or $Z_{k}$ may be empty. Combining the two observations above, we can rearrange the indices of the y’s in the way we want. This finishes the proof. □

Since the characteristic of the ground field is zero, this lemma implies that if $f (y_{1}, y_{2}, \dots, y_{m}, z_{1}, z_{2}, \dots, z_{n}) \in K 〈 X 〉$ is a weak graded identity of the pair $(M, S)$ then f is equivalent to a polynomial in a single even variable, $g (y, z_{1}, z_{2}, \dots, z_{n})$ . In other words, we can symmetrize the symmetric variables. Pay attention that the degree of f in the new even variable y is equal to the sum of the degrees of f in $y_{1}, y_{2}, \dots, y_{m}$ .

Now we study the symmetry of the polynomials $f \in F$ with respect to the odd variables $z \in Z$ . Let $τ_{i j}$ denote the transposition of the two odd variables $z_{i}$ and $z_{j}$ and let ${deg}_{x} f$ denote the degree of f with respect to the variable $x \in X$ .

Proposition 4.2. Let $f (y, z, z_{1}, z_{2})$ be a multihomogeneous polynomial in F which is linear in $z_{1}$ and $z_{2}$ , ${deg}_{z} f > 0$ . Then $(1 - τ_{12}) f = φ {(y, y_{1}, z)}_{| y_{1} = [z_{1}, z_{2}]}$ , where $φ$ is a suitable polynomial that is linear in the “new” variable $y_{1}$ .

Proof. It is sufficient to prove the statement for a monomial $m = z_{1} g z_{2}$ , where g is a word depending only on y and z.

First, we consider the monomial $g = y^{a_{1}} z^{b_{1}} y^{a_{2}} z^{b_{2}}, \dots, y^{a_{m}} z^{b_{m}}$ , $a_{i}$ , $b_{i} \geq 0$ . By the identity (3.19), if $a_{i} > 0$ and $a_{i + 1} > 0$ , then $b_{i}$ can be considered even. From (3.20), we can assume that if $b_{i} > 0$ and $b_{i + 1} > 0$ then $a_{i}$ is even. Since (3.23) is valid on $(M, S)$ , after a rearrangement, we can suppose that g has one of the following forms:

y r, z s, y r z s, z s y r, z y 2 r z 2 s + 1, y r z 2 s y, <math display="block" altimg="eq-00282.gif"><msup><mrow><mi>y</mi></mrow><mrow><mi>r</mi></mrow></msup><mo>,</mo><mspace width="1em" class="quad"></mspace><msup><mrow><mi>z</mi></mrow><mrow><mi>s</mi></mrow></msup><mo>,</mo><mspace width="1em" class="quad"></mspace><msup><mrow><mi>y</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>z</mi></mrow><mrow><mi>s</mi></mrow></msup><mo>,</mo><mspace width="1em" class="quad"></mspace><msup><mrow><mi>z</mi></mrow><mrow><mi>s</mi></mrow></msup><msup><mrow><mi>y</mi></mrow><mrow><mi>r</mi></mrow></msup><mo>,</mo><mspace width="1em" class="quad"></mspace><mi>z</mi><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>r</mi></mrow></msup><msup><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>s</mi><mo>+</mo><mn>1</mn></mrow></msup><mo>,</mo><mspace width="1em" class="quad"></mspace><msup><mrow><mi>y</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>z</mi></mrow><mrow><mn>2</mn><mi>s</mi></mrow></msup><mi>y</mi><mo>,</mo></math> (4.1)

where r and s are nonnegative integers.

If $g = y^{2 k}$ , by (3.21), $(1 - τ_{12}) m = 0$ modulo I.

If $g = y^{2 k + 1}$ , we take into account the possible positions of z. By (3.2), in $m = g_{1} z_{1} y^{2 k + 1} z_{2} g_{2}$ we must have a variable z immediately to the right of $z_{2}$ or immediately to the left of $z_{1}$ . For $m = z z_{1} y^{2 k + 1} z_{2}$ , we have

z z 1 y 2 k + 1 z 2 = [z, z 1] y 2 k + 1 z 2 + z 1 z y 2 k + 1 z 2 (3.1), (3.20) = y 2 k + 1 [z, z 1] z 2 - z 1 z 2 y 2 k + 1 z . <math display="block" altimg="eq-00290.gif"><mrow><mi>z</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mo stretchy="false">[</mo><mi>z</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo stretchy="false">]</mo><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>z</mi><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mover><mrow><mo>=</mo></mrow><mrow><mo stretchy="false">(</mo><mn>3</mn><mo>.</mo><mn>1</mn><mo stretchy="false">)</mo><mo>,</mo><mo stretchy="false">(</mo><mn>3</mn><mo>.</mo><mn>2</mn><mn>0</mn><mo stretchy="false">)</mo></mrow></mover><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><mo stretchy="false">[</mo><mi>z</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo stretchy="false">]</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><mi>z</mi><mo>.</mo></mrow></math>

In the last summand of the last equality, permuting

$z_{1}$ and

$z_{2}$ we achieve the statement, while the first summand is in the form where

$g = z^{s}$ . The case where

$m = z_{1} y^{2 k + 1} z_{2} z$ is dealt with analogously.

For $g = z^{s}$ , we use induction on s. If $s = 1$ , then $z_{1} z z_{2} \overset{3.13}{=} z \circ [z_{1}, z_{2}]$ . For $s = 2$ ,

z 1 z 2 z 2 = [z 1, z] [z, z 2] + z z 1 z z 2 - z z 1 z 2 z + z 1 z z 2 z . <math display="block" altimg="eq-00299.gif"><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msup><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msup><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mi>z</mi><mo stretchy="false">]</mo><mo stretchy="false">[</mo><mi>z</mi><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo>+</mo><mi>z</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>z</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><mi>z</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mi>z</mi><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>z</mi><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mi>z</mi><mo>.</mo></mrow></math>

Since

$(1 - τ_{12}) [z_{1}, z] [z, z_{2}] \overset{(3.1)}{=} 0$ , the case

$s = 2$ is reduced to the cases

$s = 0$ (trivial) and

$s = 1$ (treated above).

For every $s > 1$ we have that

On the right-hand side, the first summand vanishes (a consequence of (3.19) or (3.22)) while in the remaining summands, the distance between

$z_{1}$ and

$z_{2}$ is less than s. By induction, we are done.

By applying (3.20) if r is odd, and (3.21) if r is even, we obtain that $z_{1} y^{r} z^{s} z_{2} = \pm z y^{r} z_{1} z^{s - 1} z_{2}$ . Thus the case $g = y^{r} z^{s}$ is reduced to the case $g = z^{s}$ . Similarly, we treat the case $g = z^{s} y^{r}$ .

Now, let $g = z y^{2 r} z^{2 s + 1}$ .

and we use induction and the previous cases.

If $g = y^{r} z^{2 s} y$ , then $m = z_{1} y^{r} z^{2 s} y z_{2} = - z_{1} y^{r} z^{2 s - 1} z_{2} y z$ and we are in a position to apply a case discussed above. This completes the proof. □

Remark 4.3. Since S is a simple Lie algebra, we have $[S, S] = S$ . As $char K = 0$ , the last proposition implies that if $f (y, z, z_{1}, z_{2})$ is a weak graded identity of $(M, S)$ that is linear in $z_{1}$ and $z_{2}$ and ${deg}_{z} f > 0$ , then the polynomial $φ$ of the statement is also a weak graded identity. Moreover, by Lemma 4.1, f is a consequence of a polynomial depending only on one even variable and one odd variable.

We denote $〈 ad a, ad b, ad c 〉 = s_{3} (ad a, ad b, ad c)$ . The identities in the next lemma can be proved by direct computation.

Lemma 4.4. The following identities hold in every associative algebra

[a, b, c, d] - [a, b, d, c] - [[a, b], [c, d]], <math display="block" altimg="eq-00325.gif"><mo stretchy="false">[</mo><mi>a</mi><mo>,</mo><mi>b</mi><mo>,</mo><mi>c</mi><mo>,</mo><mi>d</mi><mo stretchy="false">]</mo><mo>-</mo><mo stretchy="false">[</mo><mi>a</mi><mo>,</mo><mi>b</mi><mo>,</mo><mi>d</mi><mo>,</mo><mi>c</mi><mo stretchy="false">]</mo><mo>-</mo><mo stretchy="false">[</mo><mo stretchy="false">[</mo><mi>a</mi><mo>,</mo><mi>b</mi><mo stretchy="false">]</mo><mo>,</mo><mo stretchy="false">[</mo><mi>c</mi><mo>,</mo><mi>d</mi><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mo>,</mo></math> (4.2)

〈 ad a, ad b, ad c 〉 d - [[d, a], [b, c]] - [[d, b], [c, a]] - [[d, c], [a, b]] . <math display="block" altimg="eq-00326.gif"><mo stretchy="false">〈</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><mi>a</mi><mo>,</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><mi>b</mi><mo>,</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><mi>c</mi><mo stretchy="false">〉</mo><mi>d</mi><mo>-</mo><mo stretchy="false">[</mo><mo stretchy="false">[</mo><mi>d</mi><mo>,</mo><mi>a</mi><mo stretchy="false">]</mo><mo>,</mo><mo stretchy="false">[</mo><mi>b</mi><mo>,</mo><mi>c</mi><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mo>-</mo><mo stretchy="false">[</mo><mo stretchy="false">[</mo><mi>d</mi><mo>,</mo><mi>b</mi><mo stretchy="false">]</mo><mo>,</mo><mo stretchy="false">[</mo><mi>c</mi><mo>,</mo><mi>a</mi><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mo>-</mo><mo stretchy="false">[</mo><mo stretchy="false">[</mo><mi>d</mi><mo>,</mo><mi>c</mi><mo stretchy="false">]</mo><mo>,</mo><mo stretchy="false">[</mo><mi>a</mi><mo>,</mo><mi>b</mi><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mo>.</mo></math> (4.3)

In order to prove a kind of symmetry for the odd variables, we need the following results. Some of them are based on the results of [20].

Lemma 4.5. Modulo the identities (3.1), (3.2), the following identity on the $ℤ_{2}$ -graded pair (M,S) holds:

〈 ad z 2, ad z 3, ad y 〉 z 1 - 2 [[z 1, y], [z 2, z 3]] . <math display="block" altimg="eq-00328.gif"><mo stretchy="false">〈</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>,</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><mi>y</mi><mo stretchy="false">〉</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>-</mo><mn>2</mn><mo stretchy="false">[</mo><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mi>y</mi><mo stretchy="false">]</mo><mo>,</mo><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mo>.</mo></math> (4.4)

Proof. A computation shows that

□

Lemma 4.6. Let $f = v_{1} z_{i} v_{2} z_{j} v_{3}$ be a polynomial where $z_{i}$ and $z_{j}$ are odd variables and among the commutators $v_{1}$ , $v_{2}$ , $v_{3}$ , we can find one $v_{q} = [u_{1}, u_{2}]$ of length greater than 1. Then, for some multilinear polynomial $φ$ , the identity

(1 - τ i j) f = φ (y, ẑ i, ẑ j, \dots, ˆ v q, \dots, u 1, u 2) | y = [z i, z j] <math display="block" altimg="eq-00338.gif"><mrow><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo stretchy="false">)</mo><mi>f</mi><mo>=</mo><mi>φ</mi><msub><mrow><mo stretchy="false">(</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>ẑ</mi></mrow><mrow><mi>i</mi></mrow></msub><mo>,</mo><msub><mrow><mi>ẑ</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mover accent="true"><mrow><mi>v</mi></mrow><mo>̂</mo></mover></mrow><mrow><mi>q</mi></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">)</mo></mrow><mrow><mi>|</mi><mi>y</mi><mo>=</mo><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">]</mo></mrow></msub></mrow></math>

is a consequence of the identities (3.1), (3.2), (3.3) and (3.4).

Proof. Suppose $v_{2} = [u_{1}, u_{2}]$ . We will work with all the possible parities of $u_{1}$ and $u_{2}$ .

•

If both $u_{1}$ , $u_{2}$ are odd, we have

•

If both $u_{1}$ , $u_{2}$ are even, then $f \equiv 0$ by (3.1).

•

If $u_{1}$ and $u_{2}$ have distinct parity, we can write, without loss of generality $u_{1} = z \in L {〈 X 〉}_{1}$ and $u_{2} = y \in L {〈 X 〉}_{0}$ .

Now suppose $v_{1} = [u_{1}, u_{2}]$ . Since $(1 - τ_{i j}) f = (1 - τ_{i j}) v_{1} z_{i} z_{j} v_{3} = v_{1} [z_{i}, z_{j}] v_{3}$ , we can suppose that $v_{2}$ has length at least one.

•

If $v_{2}$ is odd, by (3.13), we have

(1 - τ i j) f = (1 - τ i j) [u 1, u 2] z i v 2 z j = [u 1, u 2] (v 2 \circ [z j, z i]) . <math display="block" altimg="eq-00357.gif"><mrow><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo stretchy="false">)</mo><mi>f</mi><mo>=</mo><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo stretchy="false">)</mo><mo stretchy="false">[</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>=</mo><mo stretchy="false">[</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">]</mo><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>\circ</mo><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mo stretchy="false">]</mo><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

•

If $v_{1}$ and $v_{2}$ are even, then $f = 0$ by (3.2).

•

If $v_{1}$ is odd and $v_{2}$ is even, we write $[u_{1}, u_{2}] = z$ and $v_{2} = y$ in order to simplify the notation.

and we can treat this case using the previous cases.

The case where $v_{3} = [u_{1}, u_{2}]$ is similar to the case where $v_{1} = [u_{1}, u_{2}]$ . □

Lemma 4.7. Let $f = v_{1} v_{2}, \dots, v_{k} z_{i} v_{k + 1,} \dots, v_{k + r} z_{j} v_{k + r + 1} \dots, v_{l}$ be a multilinear polynomial where $z_{i}$ and $z_{j}$ are odd variables and one of the $v_{1}, v_{2}, \dots, v_{l}$ is a nontrivial commutator $v_{q} = [u_{1}, u_{2}]$ . Then, for some multilinear polynomial $φ$ , the identity

(1 - τ i j) f = φ (y, ẑ i, ẑ j, \dots, ˆ v q, \dots, u 1, u 2) | y = [z i, z j] <math display="block" altimg="eq-00374.gif"><mrow><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo stretchy="false">)</mo><mi>f</mi><mo>=</mo><mi>φ</mi><msub><mrow><mo stretchy="false">(</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>ẑ</mi></mrow><mrow><mi>i</mi></mrow></msub><mo>,</mo><msub><mrow><mi>ẑ</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mover accent="true"><mrow><mi>v</mi></mrow><mo>̂</mo></mover></mrow><mrow><mi>q</mi></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>2</mn></mrow></msub><mo stretchy="false">)</mo></mrow><mrow><mi>|</mi><mi>y</mi><mo>=</mo><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">]</mo></mrow></msub></mrow></math>

is a consequence of the identities (3.1), (3.2), (3.3) and (3.4).

Proof. The distance between $z_{i}$ and $z_{j}$ in the element from the statements of the lemma equals r, the quantity of variables that appear between $z_{i}$ and $z_{j}$ . The idea of the proof consists of first reducing the distance between $z_{i}$ and $z_{j}$ , and then applying induction and the previous Lemma. We treat a nontrivial commutator $v_{q} = [u_{1}, u_{2}]$ as a “variable” when computing the distance. Indeed if S is a simple Lie algebra then $S = [S, S]$ and every element of S is a linear combination of commutators, thus we have to treat the commutators indeed as “variables”. For example, the distance between $z_{i}$ and $z_{j}$ in the polynomials $z_{i} [u_{1}, u_{2}] v_{2} v_{3} z_{j}$ , $z_{i} [u_{1}, u_{2}] [u_{2}, u_{3}] v_{1} z_{j}$ and $z_{i} [u_{1}, u_{2}, u_{3}] v_{2} v_{3} z_{j}$ is the same and equal to 3.

Case 1: $q \geq k + r + 1$ or $q \leq k$ .

By the equality $x_{1} x_{2} = [x_{1}, x_{2}] + x_{2} x_{1}$ and induction, we can suppose $q = k + r + 1$ or $q = k$ . First, we assume that $q = k + r + 1$ . As above we consider the parities of the commutators.

Note that, if $v_{k + r}$ and $v_{q}$ are both even, by the identity (3.2), $v_{k + r} z_{j} v_{q} = 0$ . If $v_{k + r}$ is even and $v_{q}$ is odd, by (3.9) and (3.11), we have

v k + r z j v q = [v k + r, z j, v q] - v q z j v k + r = [v k + r, v q, z j] - v q z j v k + r . <math display="block" altimg="eq-00399.gif"><mrow><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>=</mo><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>,</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo stretchy="false">]</mo><mo>-</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>=</mo><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">]</mo><mo>-</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>.</mo></mrow></math>

In this representation,

$v_{k + r} z_{j} v_{q}$ is written as summands in the form that will be treated in Case 2, but the distance between

$z_{i}$ and

$z_{j}$ was reduced. Similarly, if

$v_{k + r}$ is odd and

$v_{q}$ is even,

$v_{k + r} z_{j} v_{q} = [v_{k + r}, z_{j}, v_{q}] - v_{q} z_{j} v_{k + r} = [v_{q}, v_{k + r}, z_{j}] - v_{q} z_{j} v_{k + r}$ .

For the remaining case, we use

v k + r z j v q = v q z j v k + r - [v q, v k + r] z j - z j [v q, v k + r] + s 3 (v k + r, z j, v q) . <math display="block" altimg="eq-00406.gif"><mrow><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>=</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>-</mo><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>,</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo stretchy="false">]</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>-</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>,</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo stretchy="false">]</mo><mo>+</mo><msub><mrow><mi>s</mi></mrow><mrow><mn>3</mn></mrow></msub><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>,</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

On the right-hand side of the last equality, in the second and third summands, we reduced the distance between

$z_{i}$ and

$z_{j}$ . The last summand vanishes due to (3.4). The first summand will be treated in Case 2.

The case $q = k$ is handled similarly.

Case 2: $k < q < k + r + 1$ .

Once again, by the equality $x_{1} x_{2} = [x_{1}, x_{2}] + x_{2} x_{1}$ and induction, we have to deal only with the case $z_{i} g v_{q} v_{k + r} z_{j}$ , where $g = v_{k + 1}, \dots, v_{k + r - 2}$ .

If $v_{q}$ is odd and $v_{k + r}$ is even, by (3.3), $v_{q} v_{k + r} z_{j} = - z_{j} v_{k + r} v_{q}$ and we are back to Case 1, but the distance between $z_{i}$ and $z_{j}$ decreased.

Now we write

v q v k + r z j = v q z j v k + r + v k + r z j v q + z j [v q, v k + r] - z j v q v k + r + [v q, [v k + r, z j]] . <math display="block" altimg="eq-00419.gif"><mrow><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>=</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>,</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo stretchy="false">]</mo><mo>-</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>+</mo><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>,</mo><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mo>.</mo></mrow></math>

In this equality, we have reduced the distance between

$z_{i}$ and

$z_{j}$ in all summands, except the last one. We treat it separately, in accordance with the possible parities for

$v_{q}$ and

$v_{k + r}$ .

If $v_{q}$ is even and $v_{k + r}$ is odd, $[v_{q}, [v_{k + r}, z_{j}]] = 0$ , by identity (3.1).

Suppose both $v_{q}$ and $v_{k + r}$ are odd. Then, since we assumed that $v_{q}$ is a nontrivial commutator, we can write $v_{q} = [y, z]$ for some $y \in L {〈 X 〉}_{0}$ and some $z \in L {〈 X 〉}_{1}$ . Using the identity (4.4) and the the fact that the element $〈 ad x_{1}, ad x_{2}, ad x_{3} 〉$ , $x_{1}$ , $x_{2}$ , $x_{3} \in X$ , commutes with every element of the relatively free algebra F, we have

z i g [v q, [v k + r, z j]] = z i g [[y, z], [v k + r, z j]] = (1 / 2) z i 〈 ad v k + r, ad z j, ad y 〉 g z, <math display="block" altimg="eq-00437.gif"><mrow><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mi>g</mi><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>,</mo><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mo>=</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mi>g</mi><mo stretchy="false">[</mo><mo stretchy="false">[</mo><mi>y</mi><mo>,</mo><mi>z</mi><mo stretchy="false">]</mo><mo>,</mo><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mo>=</mo><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">/</mo><mn>2</mn><mo stretchy="false">)</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mo stretchy="false">〈</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>,</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><mi>y</mi><mo stretchy="false">〉</mo><mi>g</mi><mi>z</mi><mo>,</mo></mrow></math>

where the distance between

$z_{i}$ and

$z_{j}$ does not exceed 1.

If $v_{q}$ and $v_{k + r}$ are even, we can assume $v_{q} = [z, \tilde{z}]$ , where z, $\tilde{z} \in L {〈 X 〉}_{1}$ . Therefore,

z i g [v q, [v k + r, z j]] = z i g [[z, ˜ z], [v k + r, z j]] (4.4) = (1 / 2) z i 〈 ad v k + r, ad ˜ z, ad z j 〉 g z . <math display="block" altimg="eq-00444.gif"><mrow><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mi>g</mi><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>q</mi></mrow></msub><mo>,</mo><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mo>=</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mi>g</mi><mo stretchy="false">[</mo><mo stretchy="false">[</mo><mi>z</mi><mo>,</mo><mover accent="true"><mrow><mi>z</mi></mrow><mo>̃</mo></mover><mo stretchy="false">]</mo><mo>,</mo><mo stretchy="false">[</mo><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">]</mo><mo stretchy="false">]</mo><mover><mrow><mo>=</mo></mrow><mrow><mo stretchy="false">(</mo><mn>4</mn><mo>.</mo><mn>4</mn><mo stretchy="false">)</mo></mrow></mover><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">/</mo><mn>2</mn><mo stretchy="false">)</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mo stretchy="false">〈</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><msub><mrow><mi>v</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><mo>,</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><mover accent="true"><mrow><mi>z</mi></mrow><mo>̃</mo></mover><mo>,</mo><mstyle><mtext mathvariant="normal">ad</mtext></mstyle><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">〉</mo><mi>g</mi><mi>z</mi><mo>.</mo></mrow></math>

On the right-hand side of the last equality, the distance between

$z_{i}$ and

$z_{j}$ does not exceed 1.

Using induction and the previous lemma, it remains to prove the statement for the polynomial $[u, z_{i}, z_{j}, v]$ .

But $(1 - τ_{i j}) [u, z_{i}, z_{j}, v] = [u, z_{i}, v] |_{z_{i} = [z_{i}, z_{j}]}$ . □

Remark 4.8. We draw the readers’ attention to the interplay between Cases 1 and 2. We need Case 2 to work on Case 1 and vice versa, but the distance between $z_{i}$ and $z_{j}$ does not increase in Case 1 while it decreases in Case 2. Therefore eventually we use the base of the induction.

Lemma 4.9. Let f be a multilinear polynomial of degree $n \geq 4$ that is skew-symmetric in two different pairs of odd variables, and let us call them $z_{i}$ , $z_{j}$ and $z_{p}$ , $z_{q}$ . Then there exists a multilinear polynomial $φ$ of degree n-1, such that the $ℤ_{2}$ -graded identity $f = φ_{| y = [z_{i}, z_{j}]}$ belongs to the $ℤ_{2}$ -graded weak ideal I.

Proof. It suffices to prove the statement for the polynomial

f = (1 - τ i j) (1 - τ p q) x 1, \dots, x k z i x k + 1, \dots, x k + r z j x k + r + 1, \dots, x n - 2, <math display="block" altimg="eq-00460.gif"><mrow><mi>f</mi><mo>=</mo><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>p</mi><mi>q</mi></mrow></msub><mo stretchy="false">)</mo><msub><mrow><mi>x</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>x</mi></mrow><mrow><mi>k</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><msub><mrow><mi>x</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>x</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>x</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>x</mi></mrow><mrow><mi>n</mi><mo>-</mo><mn>2</mn></mrow></msub><mo>,</mo></mrow></math>

where

$z_{p}$ ,

$z_{q} \in {x_{1}, \dots, x_{n - 2}} \subseteq X = Y \cup Z$ , and

$τ_{u v}$ denotes the transposition of the corresponding odd variables.

If $k < p, q \leq k + r$ , using $a b = [a, b] + b a$ , we can suppose that

(1 - τ p q) x 1, \dots, x k z i x k + 1, \dots, x k + r z j x k + r + 1, \dots, x n - 2 <math display="block" altimg="eq-00466.gif"><mrow><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>p</mi><mi>q</mi></mrow></msub><mo stretchy="false">)</mo><msub><mrow><mi>x</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>x</mi></mrow><mrow><mi>k</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><msub><mrow><mi>x</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>x</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi></mrow></msub><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><msub><mrow><mi>x</mi></mrow><mrow><mi>k</mi><mo>+</mo><mi>r</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>x</mi></mrow><mrow><mi>n</mi><mo>-</mo><mn>2</mn></mrow></msub></mrow></math>

is a sum of polynomials containing nontrivial commutators. Then, by the previous lemma, there exists

$φ$ as in the statement. The cases where p,

$q \leq k$ or p,

$q > k + r$ are analogous.

If $p \leq k$ and $k < q \leq k + r$ , once again by the equality $a b = [a, b] + b a$ and Lemma 4.7, it is sufficient to analyze the case where $p = k$ and $q = k + 1$ . But then, by (3.4), $(1 - τ_{p q}) z_{p} z_{i} z_{q} = z_{i} [z_{q}, z_{p}] + [z_{q}, z_{p}] z_{i}$ , and once more we apply Lemma 4.7 to conclude this part of the result. The cases where $k < p \leq k + r$ and $q > k + r$ are analogous.

The remaining cases are similar to the previous ones, changing the role of the pairs $z_{i}$ , $z_{j}$ and $z_{p}$ , $z_{q}$ . □

Proposition 4.10. Let f be a multilinear monomial of degree $k + n \geq 4$ depending on k even variables and n odd variables with $k \geq 0$ , $n \geq 4$ . Among the odd variables, we single out two variables, $z_{i}$ and $z_{j}$ . Then, in the pair $(K 〈 X 〉, L 〈 X 〉)$ , the polynomial $(1 - τ_{i j}) f$ can be written, modulo I, as

(1−τij)(n−2)!∑σ∈Sn−2σf+φ(y,y1,…,yk,z1,…̂zi,…,̂zj,…,zn)|y=[zi,zj].<math display="block" altimg="eq-00489.gif"><mfrac><mrow><mo stretchy="false">(</mo><mn>1</mn><mo>−</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo stretchy="false">)</mo></mrow><mrow><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo><mo>!</mo></mrow></mfrac><munder><mrow><mo>∑</mo></mrow><mrow><mi>σ</mi><mo>∈</mo><msub><mrow><mi>S</mi></mrow><mrow><mi>n</mi><mo>−</mo><mn>2</mn></mrow></msub></mrow></munder><mi>σ</mi><mi>f</mi><mo>+</mo><mi>φ</mi><msub><mrow><mo stretchy="false">(</mo><mi>y</mi><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>…</mo><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mi>k</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>…</mo><mover accent="true"><mrow><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub></mrow><mo> ̂</mo></mover><mo>,</mo><mo>…</mo><mo>,</mo><mover accent="true"><mrow><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub></mrow><mo> ̂</mo></mover><mo>,</mo><mo>…</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>n</mi></mrow></msub><mo stretchy="false">)</mo></mrow><mrow><mi>|</mi><mi>y</mi><mo>=</mo><mo stretchy="false">[</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo stretchy="false">]</mo></mrow></msub><mo>.</mo></math>(4.5)

Here

$S_{n - 2}$ are the permutations of the set

${1, \dots, î, \dots, ĵ, \dots, n}$ ,

$φ$ is a multilinear polynomial of degree

$n + k - 1$ and

$σ f$ stands for

f (y 1, \dots, y k, z σ (1), \dots, z σ (i - 1), z i, z σ (i + 1), \dots, z σ (j - 1), z j, z σ (j + 1), \dots, z σ (n)) . <math display="block" altimg="eq-00495.gif"><mrow><mi>f</mi><mo stretchy="false">(</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>y</mi></mrow><mrow><mi>k</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>σ</mi><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>σ</mi><mo stretchy="false">(</mo><mi>i</mi><mo>-</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>i</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>σ</mi><mo stretchy="false">(</mo><mi>i</mi><mo>+</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>σ</mi><mo stretchy="false">(</mo><mi>j</mi><mo>-</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>j</mi></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>σ</mi><mo stretchy="false">(</mo><mi>j</mi><mo>+</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></msub><mo>,</mo><mo>\dots</mo><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mi>σ</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></msub><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

Proof. Since every permutation may be represented as a product of transpositions, for every $σ \in S_{n - 2}$ , the polynomial $(1 - τ_{i j}) (1 - σ) f$ satisfies the conditions of the previous lemma. Therefore, there exists a multilinear polynomial $φ_{σ}$ of degree $k + n - 1$ such that $(1 - τ_{i j}) (1 - σ) f = {φ_{σ}}_{| y = [z_{i}, z_{j}]}$ . Then,

\sum σ \in S n - 2 (1 - τ i j) (1 - σ) f = \sum σ \in S n - 2 φ σ, <math display="block" altimg="eq-00501.gif"><mrow><munder><mrow><mo>\sum</mo></mrow><mrow><mi>σ</mi><mo>\in</mo><msub><mrow><mi>S</mi></mrow><mrow><mi>n</mi><mo>-</mo><mn>2</mn></mrow></msub></mrow></munder><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><mi>σ</mi><mo stretchy="false">)</mo><mi>f</mi><mo>=</mo><munder><mrow><mo>\sum</mo></mrow><mrow><mi>σ</mi><mo>\in</mo><msub><mrow><mi>S</mi></mrow><mrow><mi>n</mi><mo>-</mo><mn>2</mn></mrow></msub></mrow></munder><msub><mrow><mi>φ</mi></mrow><mrow><mi>σ</mi></mrow></msub><mo>,</mo></mrow></math>

and this in turn implies

(n - 2)! (1 - τ i j) f - \sum σ \in S n - 2 (1 - τ i j) σ f = \sum σ \in S n - 2 φ σ . <math display="block" altimg="eq-00502.gif"><mrow><mo stretchy="false">(</mo><mi>n</mi><mo>-</mo><mn>2</mn><mo stretchy="false">)</mo><mo>!</mo><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo stretchy="false">)</mo><mi>f</mi><mo>-</mo><munder><mrow><mo>\sum</mo></mrow><mrow><mi>σ</mi><mo>\in</mo><msub><mrow><mi>S</mi></mrow><mrow><mi>n</mi><mo>-</mo><mn>2</mn></mrow></msub></mrow></munder><mo stretchy="false">(</mo><mn>1</mn><mo>-</mo><msub><mrow><mi>τ</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo stretchy="false">)</mo><mi>σ</mi><mi>f</mi><mo>=</mo><munder><mrow><mo>\sum</mo></mrow><mrow><mi>σ</mi><mo>\in</mo><msub><mrow><mi>S</mi></mrow><mrow><mi>n</mi><mo>-</mo><mn>2</mn></mrow></msub></mrow></munder><msub><mrow><mi>φ</mi></mrow><mrow><mi>σ</mi></mrow></msub><mo>.</mo></mrow></math>

Denoting

$φ = 1 / (n - 2)! \sum_{σ \in S_{n - 2}} φ_{σ}$ , the proof is completed. □

We recall that, since the base field is infinite, the multihomogeneous components of an identity are also identities. Then, as in Remark 4.3, we can conclude that the polynomial $φ$ of (4.5) is an identity of the $ℤ_{2}$ -graded pair $(M, S)$ . Inductively, $φ$ is a linear combination of polynomials which are symmetric in $l - 2$ of its $l \leq n - 1$ odd variables. Eventually, we conclude that $φ$ is a consequence of polynomials of the form $φ_{1} (y, z_{1}, z_{2})$ and $φ_{2} (y, z, z_{1}, z_{2})$ , and these are linear in $z_{1}$ and $z_{2}$ . The latter of these polynomials, in turn, is a consequence of a weak graded identity of the form $g (y, z)$ .

Likewise, the polynomial $(1 - τ_{i j}) \sum_{σ \in S_{n - 2}} σ f$ in (4.5) is a weak graded identity too, which is symmetric with respect to $n - 2$ odd variables and all its even variables. Therefore, it is equivalent to a weak graded identity of the form $g_{1} (y, z, z_{1}, z_{2})$ . Then, by Remark 4.3, $(1 - τ_{i j}) \sum_{σ \in S_{n - 2}} σ f$ is a consequence of a polynomial of the form $g (y, z)$ .

Therefore we can suppose that $(1 - τ_{i j}) f$ is a consequence of polynomials $g (y, z)$ and $\tilde{g} (y, z_{1}, z_{2})$ , where the latter polynomial is linear in $z_{1}$ and $z_{2}$ . Since every $σ \in S_{n}$ can be written as a product of transpositions, we have that $f - σ f$ is a consequence of polynomials of the form $g (y, z)$ and $\tilde{g} (y, z_{1}, z_{2})$ , for every $σ \in S_{n}$ . Summing $f - σ f$ over all $σ \in S_{n}$ we obtain the following corollary.

Corollary 4.11. Let $f (y_{1}, y_{2}, \dots, y_{m}, z_{1}, z_{2}, \dots, z_{n}) = 0$ be a multilinear $ℤ_{2}$ -graded identity of the pair (M,S), with $n \geq 3$ . Then f is a consequence of polynomials of the form $g (y, z)$ and $\tilde{g} (y, z_{1}, z_{2})$ , linear in $z_{1}$ and $z_{2}$ .

We will need the following analogues of the well known generic matrix algebras. Let $A : = K [a_{i}, b_{i}, c_{i} | i \geq 1]$ be the commutative polynomial algebra generated by the independent variables $a_{i}$ , $b_{i}$ , $c_{i}$ , $i \geq 1$ . Consider the associative algebra R and the Lie algebra L generated by the generic matrices

Observe that R and

$R^{(-)}$ are

$ℤ_{2}$ -graded associative and Lie algebras, respectively, with the

$ℤ_{2}$ -grading defined at the very beginning of the previous section. Also, L is a

$ℤ_{2}$ -graded Lie algebra, with

$L_{i} = R_{i}^{(-)} \cap L$ ,

$i = 0$ , 1.

Then a standard argument shows that the $ℤ_{2}$ -graded pair $(R, L)$ is the relatively free $ℤ_{2}$ -graded pair of countable rank in the variety of $ℤ_{2}$ -graded pairs generated by $T_{2} (M, S)$ (see for example [16, Lemma 3]). Therefore we shall work in the $ℤ_{2}$ -graded pair $(R, L)$ when necessary.

Recall that the weak $T_{2}$ -ideal I is contained in $T_{2} (M, S)$ and, as stated in the last corollary, the elements in $F = K <X> / I$ are consequences of polynomials in one even variable $y \in Y$ and at most two odd variables $z_{1}$ , $z_{2} \in Z$ . We draw the readers’ attention to the fact that these polynomials are multihomogeneous but they need not be linear, apart from the case of two odd variables where the polynomial is linear in both $z_{1}$ and $z_{2}$ .

Proposition 4.12. There are no nontrivial weak graded identities for the pair (M, S) modulo the $T_{2}$ -ideal I.

Proof. By Corollary 4.11, if we want to study the weak graded identities for $(M, S)$ modulo I it is sufficient to analyze the polynomials of the form $g (y, z)$ and $g (y, z_{1}, z_{2})$ , the latter is linear in $z_{1}$ and $z_{2}$ , since there are no weak graded identities depending only on y (that is of the form $y^{m}$ ) or only on z (that is of the form $z^{n}$ ), modulo I.

Thus, in order to prove that there is no nontrivial weak graded identity for the pair $(M, S)$ modulo I, it is enough to prove that the monomials in $K 〈 X 〉 / I$ depending on y and z or on y, $z_{1}$ and $z_{2}$ are linearly independent. To achieve this, in turn, we prove the linear independence for every set of multihomogeneous monomials of the same multidegree.

We work in the $ℤ_{2}$ -graded pair $(R, L)$ , setting $y = Y$ and $z_{i} = Z_{i}$ , where Y and $Z_{i}$ are the matrices introduced above.

Let us analyze first, according to their multihomogeneous degree, the polynomials modulo I that depend only on y and z which are, as we have seen above, $y^{r} z^{s}$ , $z^{s} y^{r}$ , $z y^{2 r} z^{s - 1}$ and $y^{r} z^{2 s} y$ .

•	If ${deg}_{y} f = r$ and ${deg}_{z} f = 2 k$ are both even, the possibilities for a monomial with such multidegree are $y^{r} z^{2 k}, z y^{r} z^{2 k - 1}$ and $y^{r - 1} z^{2 k} y$ . Then evaluating on the generic matrices we have By the form of the above matrices, it is clear that the second matrix cannot participate in a linear combination. Then a linear combination of the first and the third matrix equals 0 if and only if it is the trivial combination. Therefore no nontrivial linear combination of these elements is an identity.
•	If ${deg}_{y} f = r$ and ${deg}_{z} f = 2 k + 1$ are odd, the possible monomials are $y^{r} z^{s}$ and $z^{s} y^{r}$ . But $Y r Z s = 2 r + 2 k + 1 (000 - a r b k c k + 1 00 - a r b k + 1 c k 00), <math display="block" altimg="eq-00597.gif"><mrow><msup><mrow><mi>Y</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>Z</mi></mrow><mrow><mi>s</mi></mrow></msup><mo>=</mo><msup><mrow><mn>2</mn></mrow><mrow><mi>r</mi><mo>+</mo><mn>2</mn><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><mfenced separators="" open="(" close=")"><mrow><mtable displaystyle="true" equalrows="false" equalcolumns="false" class="array"><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi></mrow></msup></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr></mtable></mrow></mfenced><mspace width="-.17em" class="negativethinspace"></mspace><mo>,</mo></mrow></math>$ and $Z s Y r = 2 r + 2 k (0 - a r b k + 1 c k - a r b k c k + 1 000000) . <math display="block" altimg="eq-00598.gif"><mrow><msup><mrow><mi>Z</mi></mrow><mrow><mi>s</mi></mrow></msup><msup><mrow><mi>Y</mi></mrow><mrow><mi>r</mi></mrow></msup><mo>=</mo><msup><mrow><mn>2</mn></mrow><mrow><mi>r</mi><mo>+</mo><mn>2</mn><mi>k</mi></mrow></msup><mfenced separators="" open="(" close=")"><mrow><mtable displaystyle="true" equalrows="false" equalcolumns="false" class="array"><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi></mrow></msup></mtd><mtd columnalign="center"><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr></mtable></mrow></mfenced><mspace width="-.17em" class="negativethinspace"></mspace><mo>.</mo></mrow></math>$ Clearly, no nontrivial linear combination of these elements is a weak graded identity.
•	If ${deg}_{y} f = r$ is odd and ${deg}_{z} f = 2 k$ is even, we have the following possibilities: $Y r Z 2 k = 2 r + 2 k - 1 (000 0 a r b k c k - a r b k - 1 c k + 1 0 a r b k + 1 c k - 1 - a r b k c k), <math display="block" altimg="eq-00601.gif"><mrow><msup><mrow><mi>Y</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>Z</mi></mrow><mrow><mn>2</mn><mi>k</mi></mrow></msup><mo>=</mo><msup><mrow><mn>2</mn></mrow><mrow><mi>r</mi><mo>+</mo><mn>2</mn><mi>k</mi><mo>-</mo><mn>1</mn></mrow></msup><mfenced separators="" open="(" close=")"><mrow><mtable displaystyle="true" equalrows="false" equalcolumns="false" class="array"><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi></mrow></msup></mtd><mtd columnalign="center"><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi><mo>-</mo><mn>1</mn></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi><mo>-</mo><mn>1</mn></mrow></msup></mtd><mtd columnalign="center"><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi></mrow></msup></mtd></mtr></mtable></mrow></mfenced><mspace width="-.17em" class="negativethinspace"></mspace><mo>,</mo></mrow></math>$ and $Z 2 k Y r = 2 r + 2 k (000 0 a r b k c k a r b k - 1 c k + 1 0 - a r b k + 1 c k - 1 - a r b k c k) . <math display="block" altimg="eq-00602.gif"><mrow><msup><mrow><mi>Z</mi></mrow><mrow><mn>2</mn><mi>k</mi></mrow></msup><msup><mrow><mi>Y</mi></mrow><mrow><mi>r</mi></mrow></msup><mo>=</mo><msup><mrow><mn>2</mn></mrow><mrow><mi>r</mi><mo>+</mo><mn>2</mn><mi>k</mi></mrow></msup><mfenced separators="" open="(" close=")"><mrow><mtable displaystyle="true" equalrows="false" equalcolumns="false" class="array"><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi></mrow></msup></mtd><mtd columnalign="center"><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi><mo>-</mo><mn>1</mn></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup></mtd></mtr><mtr><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi><mo>-</mo><mn>1</mn></mrow></msup></mtd><mtd columnalign="center"><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>r</mi></mrow></msup><msup><mrow><mi>b</mi></mrow><mrow><mi>k</mi></mrow></msup><msup><mrow><mi>c</mi></mrow><mrow><mi>k</mi></mrow></msup></mtd></mtr></mtable></mrow></mfenced><mspace width="-.17em" class="negativethinspace"></mspace><mo>.</mo></mrow></math>$ No nontrivial linear combination of these elements is an identity.
•	If ${deg}_{y} f = r$ is even and ${deg}_{z} f = 2 k + 1$ is odd, the possibilities for monomials with such multidegrees are Clearly, there is no nontrivial linear combination in this case either.

Now we deal with the polynomials $f (y, z_{1}, z_{2})$ which are linear in $z_{1}$ and $z_{2}$ . First, we examine the possible monomials m that can appear in these polynomials. From (3.2), there is no factor of the type $y z_{i} y$ in m. This fact, along with identity (3.24) implies that m can only be in one of the following forms: $z_{i} z_{j} y^{2 r}$ , $y^{2 r - 1} z_{i} z_{j} y$ and $z_{i} y^{2 r} z_{j}$ if the total degree of f is even, or $y^{2 r + 1} z_{i} z_{j}$ , $z_{i} z_{j} y^{2 r + 1}$ and $z_{i} y^{2 r + 1} z_{j}$ otherwise, where $i \neq j \in {1, 2}$ .

Following the same argument as above, we have

•

If $deg f = 2 r + 2$ ,

Since

$i \neq j \in {1, 2}$ , we have the following linear combinations resulting zero:

$Y^{2 r} Z_{i} Z_{j} - Y^{2 r} Z_{j} Z_{i} - Y^{2 r - 1} Z_{i} Z_{j} Y + Y^{2 r - 1} Z_{j} Z_{i} Y = 0$ and

$Z_{i} Y^{2 r} Z_{j} - Z_{j} Y^{2 r} Z_{i} = 0$ , but these are consequences of (3.1) and (3.21), respectively. Consequently, they are trivial in R modulo I.

•

If $deg f = 2 r + 1$ ,

Here, we have

$Y^{2 r + 1} Z_{i} Z_{j} - Y^{2 r + 1} Z_{j} Z_{i} - Z_{i} Z_{j} Y^{2 r + 1} - Z_{j} Z_{i} Y^{2 r + 1} = 0$ and

$Z_{i} Y^{2 r + 1} Z_{j} + Z_{j} Y^{2 r + 1} Z_{i} = 0$ which are trivial in R modulo I by identities (3.1) and (3.20).

□

Therefore, we obtain the proof of the main result of this paper, summarized in the following theorem.

Theorem 4.13. Let K be a field of characteristic zero and let (M,S) be the $ℤ_{2}$ -graded pair relative to the adjoint representation of ${𝔰 𝔩}_{2} (K)$ equipped with the only nontrivial $ℤ_{2}$ -grading on $𝔰 𝔩_{2} (K)$ , and the induced grading on $M_{3} (K)$ . Then the following graded identities form a basis for the $ℤ_{2}$ -graded identities of the pair (M, S).

where

$x = y_{i}$ ,

$z_{i}$ .

One can naturally ask the following questions.

(1)

Let K be algebraically closed and of characteristic 0, and let $ρ : {𝔰 𝔩}_{2} (K) \to 𝔤 𝔩 (V)$ be an irreducible representation of dimension $m + 1$ . The natural $ℤ_{2}$ -grading on ${𝔰 𝔩}_{2} (K)$ induces a $ℤ_{2}$ -grading on V as follows. Take a maximal vector $v_{0} \in V$ and form the canonical basis $v_{0}, v_{1}, \dots, v_{m}$ of V. Then V is $ℤ_{2}$ -graded by assigning degree 0 to the $v_{2 i}$ and degree 1 to the $v_{2 i + 1}$ . This induces a $ℤ_{2}$ -grading on $M_{m + 1} (K)$ .

Describe a basis of the $ℤ_{2}$ -graded identities for $ρ$ .

(2)

If K is of positive characteristic p, the irreducible representations of ${𝔰 𝔩}_{2} (K)$ are finite-dimensional.

Describe their $ℤ_{2}$ -graded identities.

Acknowledgments

We are thankful to the anonymous referee whose numerous remarks helped us improve the exposition of the paper. C. Sampaio was financed in part by the Coordena de Aperfeioamento de Pessoal de Nvel Superior-Brasil (CAPES)-Finance Code 001.

P. Koshlukov was partially supported by CNPq Grant 307184/2023-4, and by FAPESP grant 2018/23690-6.

ORCID

Cássia Sampaio https://orcid.org/0009-0007-1666-7124

Plamen Koshlukov https://orcid.org/0000-0002-1819-0100

References

[1] L. Centrone and M. da Silva Souza , On the growth of graded polynomial identities of $s l_{n}$ , Linear Multilinear Algebra 65(4) (2017) 752–767. Crossref, Web of Science, Google Scholar
[2] R. Códamo and P. Koshlukov , Specht property for the graded identities of the pair $(M_{2} (D), s l_{2} (D))$ , Linear Algebra Appl. 698 (2024) 326–343. Crossref, Web of Science, Google Scholar
[3] C. De Concini and C. Procesi , A characteristic free approach to invariant theory, in Young Tableaux in Combinatorics, Invariant Theory, and Algebra (Elsevier, 1982), pp. 169–193. Crossref, Google Scholar
[4] O. M. Di Vincenzo , On the graded identities of $M_{1, 1} (E)$ , Israel J. Math. 80 (1992) 323–335. Crossref, Web of Science, Google Scholar
[5] V. S. Drenski , A minimal basis for identities of a second-order matrix algebra over a field of characteristic 0, Algebra Logika 20(3) (1981) 282–290. Crossref, Google Scholar
[6] V. Drensky , Codimensions of T-ideals and Hilbert series of relatively free algebras, J. Algebra 91 (1984) 1–17. Crossref, Web of Science, Google Scholar
[7] G. K. Genov , Basis for identities of a third-order matrix algebra over a finite field, Algebra Logika 20(4) (1981) 365–388. Crossref, Google Scholar
[8] G. K. Genov and P. N. Siderov , A basis of identities of the algebra of fourth-order matrices over a finite field II, Serdica 8 (1982) 313–323. Google Scholar
[9] A. Giambruno and M. da Silva Souza , Graded polynomial identities and Specht property of the Lie algebra $s l_{2}$ , J. Algebra 389 (2013) 6–22. Crossref, Web of Science, Google Scholar
[10] A. R. Kemer , Finite basis property of identities of associative algebras, Algebra Logic 26(5) (1987) 362–397. Crossref, Google Scholar
[11] A. R. Kemer , Ideals of Identities of Associative Algebras, Vol. 87 (American Mathematical Society, 1991). Crossref, Google Scholar
[12] P. Koshlukov , Weak polynomial identities for the matrix algebra of order two, J. Algebra 188(2) (1997) 610–625. Crossref, Web of Science, Google Scholar
[13] P. Koshlukov , Basis of the identities of the matrix algebra of order two over a field of characteristic $p \neq 2$ , J. Algebra 241(1) (2001) 410–434. Crossref, Web of Science, Google Scholar
[14] P. Koshlukov , Graded polynomial identities for the Lie algebra $s l_{2} (k)$ , Internat. J. Algebra Comput. 18(5) (2008) 825–836. Link, Web of Science, Google Scholar
[15] P. Koshlukov and S. S. de Azevedo , Graded identities for T-prime algebras over fields of positive characteristic, Israel J. Math. 128 (2002) 157–176. Crossref, Web of Science, Google Scholar
[16] P. Koshlukov and A. Krasilnikov , A basis for the graded identities of the pair $(M_{2} (K), g l_{2} (K))$ , Serdica Math. J. 38(1–3) (2012) 497–506. Google Scholar
[17] Yu. P. Razmyslov , The existence of a finite basis for the identities of the matrix algebra of order two over a field of characteristic zero, Algebra Logika 12(1) (1973) 83–113. Crossref, Google Scholar
[18] Yu. P. Razmyslov , Identities of Algebras and Their Representations, Vol. 138 (American Mathematical Society, 1994). Crossref, Google Scholar
[19] D. V. Repin , Graded identities of a simple three-dimensional Lie algebra, Vestn. Samar. Gos. Univ., Estestvennonauchn. Ser. 2004(2) (2004) 5–16. Google Scholar
[20] I. M. Trishin , Identities of irreducible representations of a three-dimensional simple Lie algebra, Algebra Logic 22(3) (1983) 229–250. Crossref, Google Scholar
[21] S. Yu. Vasilovskii , Basis of identities of a three-dimensional simple Lie algebra over an infinite field, Algebra Logic 28(5) (1989) 355–368. Crossref, Google Scholar
[22] M. R. Vaughan-Lee , Varieties of Lie algebras, Q. J. Math. 21(3) (1970) 297–308. Crossref, Google Scholar