On $\mathbb{Z}_{2}$-graded polynomial identities of $sl_{2}(F)$ over a finite field

TL;DR

This paper determines a basis for the $Z_2$-graded polynomial identities of the Lie algebra $sl_2(F)$ over a finite field with characteristic greater than 3, enhancing understanding of its algebraic structure.

Contribution

It provides the first explicit basis for the $Z_2$-graded identities of $sl_2(F)$ over finite fields, filling a gap in the algebraic theory.

Findings

Explicit basis for $Z_2$-graded identities of $sl_2(F)$

Characterization of identities over finite fields with char > 3

Advances understanding of graded polynomial identities in Lie algebras

Abstract

Let $F$ be a finite field of $char F > 3$ and $sl_{2}(F)$ be the Lie algebra of traceless $2\times 2$ matrices over $F$. In this paper, we find a basis for the $\mathbb{Z}_{2}$-graded identities of $sl_{2}(F)$.

Full text

Keywords: 16R10,17B01,15A72,17B70.

Mathematics Subject Classification 2010: Graded identities; Lie algebras; finite basis identities.

1. Introduction

The well-known Ado-Iwasawa’ theorem posits that any finite-dimensional Lie algebra over an arbitrary field has a faithful finite-dimensional representation. Briefly, any finite dimensional Lie algebra can be viewed as a subalgebra of a Lie algebra of square matrices under the commutator brackets. Thus, the study of Lie algebras of matrices is of considerable interest.

A task in PI-theory is to describe the identities of $sl_{2}(F)$, the Lie algebra of traceless $2\times 2$ matrices over a field $F$ of $charF\neq 2$. The first breakthrough in this area was made by Razmyslov [12], who described a basis for the identities of $sl_{2}(F)$ when $charF=0$. Vasilovsky [16] found a single identity for the identities of $sl_{2}(F)$ when $F$ is an infinite field of $charF\neq 2$, and Semenov [13] described a basis (with two identities) for the identities of $sl_{2}(F)$ when $F$ is a finite field of $charF>3$.

The Lie algebra $sl_{2}(F)$ can be naturally graded by $\mathbb{Z}_{2}$ as follows: $sl_{2}(F)=\newline (sl_{2}(F))_{0}\oplus(sl_{2}(F))_{1}$ where $(sl_{2}(F))_{0},(sl_{2}(F))_{1}$ contain diagonal and off-diagonal matrices respectively. A recent development in PI-theory is the description of the graded identities of $sl_{2}(F)$. Using invariant theory techniques, Koshlukov [8] described the $\mathbb{Z}_{2}$-graded identities for $sl_{2}(F)$ when $F$ is an infinite field of $charF\neq 2$. Several further papers on graded identities of $sl_{2}(F)$ have appeared in recent years (cf. e.g., [4] and [5]). In these studies, the ground field is of characteristic zero.

To date, no basis has been found for the $\mathbb{Z}_{2}$-graded identities of $sl_{2}(F)$ when $F$ is a finite field.

In this paper we give a basis for the graded identities $sl_{2}(F)$ when $F$ is a finite field of $charF>3$.

2. Preliminaries

Let $F$ be a fixed finite field of $charF>3$ and size $|F|=q$, let $\mathbb{N}_{0}=\{1,2,\ldots,\newline ,n,\ldots\}$, let $G=(\mathbb{Z}_{2},+)$, and let $L$ be a Lie algebra over $F$. In this study (unless otherwise stated), all vector spaces and Lie algebras are considered over $F$. The $+^{\cdot},\oplus,span_{F}\{a_{1},\ldots,a_{n}\},\langle a_{1},\ldots,a_{n}\rangle,(a_{1},\ldots,a_{n}\in L)$ signs denote the direct sum of Lie algebras, the direct sum of vector spaces, the vector space generated by $a_{1},\ldots,a_{n}$, and the ideal generated by $a_{1},\ldots,a_{n}$ respectively, while an associative product is represented by a dot: $``.^{\prime\prime}$. The commutator $([,])$ denotes the multiplication operation of a Lie algebra. We assume that all commutators are left-normed, i.e., $[x_{1},x_{2},\ldots,x_{n}]:=[[x_{1},x_{2},\ldots,x_{n-1}],x_{n}]\ \ n\geq 3$. We use the convention $[x_{1},x_{2}^{k}]=[x_{1},x_{2},\ldots,x_{2}]$, where $x_{2}$ appears $k$ times in the expanded commutator.

We denote by $gl_{2}(F)$ the Lie algebra of $2\times 2$ matrices over $F$. Let $sl_{2}(F)$ denote the Lie algebra of traceless $2\times 2$ matrices over $F$. Here, $e_{ij}\subset gl_{2}(F)$ denotes the unitary matrix unit whose elements are $1$ in the positions $(ij)$ and [math] otherwise.

The basic concepts of Lie algebra adopted in this study can be found in Chapters 1 and 2 of [6]. We denote the center of $L$ by $Z(L)$. If $x\in L$, we denote by $adx$ the linear map with the function rule: $y\mapsto[x,y]$. $L$ is said to be metabelian if it is solvable of class at most $2$. As is known, if $L$ ($L$ over a finite field of $charF>3$) is a three-dimensional simple Lie algebra, then $L\cong sl_{2}(F)$. $L$ is regarded as a Lie $A$-algebra if all of its nilpotent subalgebras are abelian.

A Lie algebra $L$ is said to be $G$-graded (a graded Lie algebra or graded by $G$) when there exist subspaces $\{L_{g}\}_{g\in G}\subset L$ such that $L=\bigoplus_{g\in G}L_{g}$, and $[L_{g},L_{h}]\subset L_{g+h}$ for any $g,h\in G$. $G$-graded associative algebras are defined in the same way. In that context, $\{L_{g}\}_{g\in G}$ is said to be a grading for $L$. An element $a$ is called homogeneous when $a\in\bigcup_{g\in G}L_{g}$. We say that $a\neq 0$ is a homogeneous element of $G$-degree $g$ when $a\in L_{g}$. A $G$-graded homomorphism of two $G$-graded Lie algebras $L_{1}$ and $L_{2}$ is a homomorphism $\phi:L_{1}\rightarrow L_{2}$ such that $\phi({L_{1}}_{g})\subset{L_{2}}_{g}$ for all $g\in G$. Two gradings on $L$ $\{L_{g}\}_{g\in G}$ and $\{L^{\prime}_{g}\}_{g\in G}$ on $L$ are called isomorphic when there exists a $G$-graded isomorphism $\phi:L\rightarrow L$ such that $\phi(L_{g})=L^{\prime}_{g}$ for all $g\in G$. An ideal $I\subset L$ is graded when $I=\bigoplus_{g\in G}(I\cap L_{g})$ (we define graded Lie subalgebras similarly). Likewise, if $I$ is a graded ideal of $L$, $C_{L}(I)=\{a\in L|[a,I]=\{0\}\}$ is also a graded ideal of $L$. Furthermore, $Z(L),L^{n}$ (the $n$-th term of descending central series), and $L^{(n)}$ (the $n$-th term of derived series) are graded ideals of $L$. We use the convention that $L^{(1)}=[L,L]$ and $L^{1}=L$.

Let $L$ be a finite-dimensional Lie algebra. Denote by $Nil(L)$ the greatest nilpotent ideal of $L$ and by $Rad(L)$ the greatest solvable ideal of $L$. Clearly, $Nil(L)$ is the unique maximal abelian ideal of $L$ when $L$ is a Lie $A$-algebra. Furthermore, every subalgebra and every factor algebra of $L$ is a Lie $A$-algebra when $L$ is also a Lie $A$-algebra (see Lemma 2.1 in [15] and Lemma 1 in [10]).

The next theorem is a structural result on solvable Lie $A$-algebras.

Theorem 2.1** (Towers, Theorem 3.5, [15]).**

Let $L$ be a (finite-dimensional) solvable Lie $A$-algebra (over an arbitrary field $F$) of derived length $n+1$ with nilradical $Nil(L)$. Moreover, let $K$ be an ideal of $L$ and $B$ a minimal ideal of $L$. Then we have the following:

**: **

$K=(K\cap A_{n})\oplus(K\cap A_{n-1})\oplus\ldots\oplus(K\cap A_{0})$;

**: **

$Nil(L)=A_{n}+^{\cdot}(A_{n-1}\cap Nil(L))+^{\cdot}\ldots+^{\cdot}(A_{0}\cap Nil(L))$;

**: **

$Z(L^{(i)})=Nil(L)\cap A_{i}$* for each $0\leq i\leq n$;*

**: **

$B\subseteqq Nil(L)\cap A_{i}$* for some $0\leq i\leq n$.*

$A_{n}=L^{(n)}$, $A_{n-1},\ldots,A_{0}$ are abelian subalgebras of $L$ defined in the proof of Corollary 3.2 in [15].

Remark 2.2**.**

Assuming Theorem 2.1, we can prove that, if $L=\bigoplus_{g\in G}L_{g}$ is a (finite-dimensional) solvable graded Lie $A$-algebra (over an arbitrary field $F$) of derived length $n+1$ with nilradical $Nil(L)$ , then $Nil(L)$ is a graded ideal. Moreover, if $L$ is finite-dimensional metabelian Lie $A$-algebra (over an arbitrary field), then $Nil(L)=[L,L]+^{\cdot}Z(L)$.

A finite-dimensional Lie algebra $L$ is called semisimple if $Rad(L)=\{0\}$. Recall that $L$ (finite-dimensional and non solvable) has a Levi decomposition when there exist a semisimple subalgebra $S\neq\{0\}$ (termed a Levi subalgebra) such that $L$ is a semidirect product of $S$ and $Rad(L)$. We now present a result.

Proposition 2.3** **(Premet and Semenov, Proposition 2, adapted,

[10]).

Let $L$ be a finite-dimensional Lie $A$-algebra over a finite field $F$ of $charF>3$. Then,

**: **

$[L,L]\cap Z(L)=\{0\}$.

**: **

$L$* has a Levi decomposition. Moreover, each Levi subalgebra $S$ is represented as a direct sum of $F$-simple ideals in $S$, each one of which splits over some finite extension of the ground field into a direct sum of the ideals isomorphic to $sl_{2}(F)$.*

A Lie algebra $L$ is said to be $G$-simple if $[L,L]\neq\{0\}$, and $L$ does not have any proper non-trivial graded ideals.

By mimicking the arguments of Zaicev et al. in [11] (Lemma 2.1; Section 3; Proposition 3.1, items i and ii), we have the following.

Proposition 2.4**.**

Let $L$ be a finite dimensional graded Lie algebra. The ideal $Rad(L)$ is a graded ideal. If $L$ is $G$-simple, then $L$ is a direct sum of simple Lie algebras. If $L$ is direct sum of simple Lie algebras, then $L$ is a direct sum of $G$-simple Lie algebras.

3. Graded identities and varieties of graded Lie algebras

Let $X=\{X_{g}=\{x_{1}^{g},\ldots,x_{n}^{g},\ldots\}|g\in G\}$ be a class of pairwise-disjoint enumerable sets, where $X_{g}$ denotes the variables of $G$-degree $g$. Let $F\langle X\rangle$ be the free associative unital algebra and let $L(X)$ be the Lie subalgebra of $F\langle X\rangle$ generated by $X$. $L(X)$ is known to be isomorphic to the free Lie algebra with a set of free generators $X$. The algebras $L(X)$ and $F\langle X\rangle$ have natural $G$-grading. A graded ideal $I\subset L(X)$ invariant under all graded endomorphisms is called a graded verbal ideal. Let $S\subset L(X)$ be a non empty set. The graded verbal ideal generated by $S$, $\langle S\rangle_{T}$, is defined as the intersection of all verbal ideals containing $S$. A polynomial $f\in L(X)$ is called a consequence of $g\in L(X)$ when $f\in\langle g\rangle_{T}$, and it is called a graded polynomial identity for a graded Lie algebra $L$ if $f$ vanishes on $L$ whenever the variables from $X_{g}$ are substituted by elements of $L_{g}$ for all $g\in G$. We denote by $Id_{G}(L)$ the set of all graded identities of $L$. The variety determined by $S\subset L(X)$ is denoted by $\mathcal{V}(S)=\{A\ \mbox{is a}\ \mbox{G-graded Lie algebra}|Id_{G}(A)\supset\langle S\rangle_{T}\}$. The variety generated by a graded Lie algebra $L$ is denoted by $var_{G}(L)=\{A\ \mbox{is a}\ \ \mbox{G-graded Lie algebra}|Id_{G}(L)\subset Id_{G}(A)\}$. We say that a class of graded Lie algebras $\{L_{i}\}_{i\in\Gamma}$, where $\Gamma$ is an index set, generates $\mathcal{V}(S)$ when $\langle S\rangle_{T}=\bigcap_{i\in\Gamma}Id_{G}(L_{i})$.

We denote by $Id(L)$ the set of all ordinary identities of a Lie algebra $L$, and by $var(L)$ the variety generated by $L$. The variety of metabelian Lie algebras over $F$ is denoted by $A^{2}$. A set $S\subset L(X)$ of ordinary polynomials (respectively graded polynomials) is called a basis for the ordinary identities (respectively graded identities) of a Lie algebra (respectively a graded Lie algebra) $A$ when $Id(A)=\langle S\rangle_{T}$ (respectively $Id_{G}(A)=\langle S\rangle_{T}$).

Example 1**.**

In 1990’s, Semenov (Proposition 1, [13]) proved that

$Sem_{1}(x_{1},x_{2})=(x_{1})f(ad(x_{2})),\ \ f(t)=t^{q^{2}+2}-t^{3}$,

$Sem_{2}(x_{1},x_{2})=[x_{1},x_{2}]-[x_{1},x_{2},(x_{1})^{q^{2}-1}]-[x_{1},(x_{2})^{q}]+[x_{1},x_{2},(x_{1})^{q^{2}-1},(x_{2})^{q-1}]+[x_{1},x_{2},((x_{1})^{q^{2}}-(x_{1})),[x_{1},x_{2}]^{q-2},(x_{2})^{q^{2}}-(x_{2})]-[x_{2},([(x_{1})^{q^{2}}-(x_{1}),x_{2}])^{q},((x_{2})^{q^{2}-2}-(x_{2})^{q-2})]$.

*are polynomial identities of $sl_{2}(F)$. *

A finite-dimensional ordinary (respectively graded) Lie algebra $L$ is critical if $var(L)$ (respectively $var_{G}(L)$) is not generated by all proper subquotients of $L$. It is monolithic if it contains a single ordinary (respectively graded) minimal ideal. This single ideal is termed a monolith. It is known that if $L$ is an ordinary (respectively graded) critical Lie algebra, then $L$ is monolithic. Notice that if $L=\oplus_{g\in G}L_{g}$ is a critical ordinary Lie algebra, then $L$ is critical as a $G$-graded Lie algebra.

Example 2**.**

If $L$ is a critical abelian (respectively graded) Lie algebra, then $dimL=1$. If $L$ is a two-dimensional (non abelian) metabelian Lie algebra, then $L$ is critical. Furthermore, $sl_{2}(F)$ is a critical Lie algebra.

Proposition 3.1**.**

Let $L=\bigoplus_{g\in G}L_{g}$ be a finite-dimensional (non abelian) metabelian graded Lie $A$-algebra over an arbitrary field $F$. If $L$ is monolithic, then $Nil(L)=[L,L]$.

Proof.

According to Theorem 2.1, $Nil(L)=[L,L]+^{\cdot}Z(L)$. By hypothesis, $L$ is monolithic. Thus, $Z(L)=\{0\}$ and $Nil(L)=[L,L]$. ∎

The next theorem describes the relationship between critical metabelian Lie $A$-algebras and monolithic Lie $A$-algebras.

Theorem 3.2** (Sheina, Theorem 1,[14]).**

A finite-dimensional monolithic Lie $A$-algebra $L$ over an arbitrary finite field is critical if, and only if, its derived algebra cannot be represented as a sum of two ideals strictly contained within it.

A locally finite Lie algebra is a Lie algebra for which every finitely generated subalgebra is finite. A variety of Lie algebras (respectively graded Lie algebras) is said to be locally finite when every finitely generated Lie algebra (respectively graded Lie algebra) has finite cardinality. It is known that a variety generated by a finite Lie algebra (respectively a graded finite Lie algebra) is locally finite. As in the ordinary case, if a variety of graded Lie algebras is locally finite, then it is generated by its critical algebras. For more details about varieties of Lie algebras, see Chapters 4 and 7 of [1].

The next result will prove useful for our purposes.

Theorem 3.3** **(Semenov, Proposition 2,

[13]).

Let $\mathcal{B}$ be a variety of ordinary Lie algebras over a finite field $F$. If there exists a polynomial $f(t)=a_{1}t+\ldots+a_{n}t^{n}\in F[t]$ such that $yf(adx):=a_{1}[y,x]+\ldots+a_{n}[y,x^{n}]\in Id(\mathcal{B})$, then $\mathcal{B}$ is a locally finite variety.

Let $L_{1}$ and $L_{2}$ be two graded Lie algebras (finite -dimensional), and $I_{1}\subset L_{1}$ and $I_{2}\subset L_{2}$ be graded ideals. We say that $I_{1}$ (in $L_{1}$) is similar to $I_{2}$ (in $L_{2}$) ($I_{1}\trianglelefteq A_{1}\sim I_{2}\trianglelefteq A_{2}$) if there exist isomorphisms $\alpha_{1}:I_{1}\rightarrow I_{2}$ and $\alpha_{2}:\frac{L_{1}}{C_{L_{1}}(I_{1})}\rightarrow\frac{L_{2}}{C_{L_{2}}(I_{2})}$ such that for all $a\in I_{1}$ and $b+C_{L_{1}}(I_{1})\in\frac{L_{1}}{C_{L_{1}}(I_{1})}$:

$\alpha_{1}([a,c])=[\alpha_{1}(a),d]$,

where $c+C_{L_{1}}(I_{1})=b+C_{L_{1}}(I_{1})\ \mbox{and}\ d+C_{L_{2}}(I_{2})=\alpha_{2}(b+C_{L_{1}}(I_{1}))$.

By proceeding as in [9] (cf. pages 162 to 166), we have the following.

Proposition 3.4**.**

If two critical graded Lie algebras $L_{1}$ and $L_{2}$ generate the same variety, then their monoliths are similar.

4. $\mathbb{Z}_{2}$-graded identities of $sl_{2}(F)$

From now on, we denote by $Y=\{y_{1},\ldots,y_{n},\ldots\}$ the even variables, by $Z=\{z_{1},\ldots,z_{n},\ldots\}$ the odd variables.

Lemma 4.1**.**

Let $sl_{2}(F)$ be the Lie algebra of traceless $2\times 2$ matrices over $F$ endowed with the natural grading. The following polynomials are graded identities of $sl_{2}(F)$

$[y_{1},y_{2}],[z_{1},y_{1}^{q}]=[z_{1},y_{1}]$.

Proof.

It is clear that $[y_{1},y_{2}]\in Id_{G}(sl_{2}(F))$, because the diagonal is commutative. Choose $a_{i}=\lambda_{11,i}e_{11}-\lambda_{11,i}e_{22}$ and $b_{j}=\lambda_{12,j}e_{12}+\lambda_{21,j}e_{21}$, so:

$[b_{j},a_{i}^{q}]=\lambda_{11,i}^{q}[b_{j},{h}^{q}]=\lambda_{11,i}^{q}((-2)^{q}\lambda_{12,j}e_{12}+2^{q}\lambda_{21,j}e_{21})=\lambda_{11,i}(-2\lambda_{12,j}e_{12}+2\lambda_{21,j}e_{21})=[b_{j},a_{i}]$.

Thus, $[z_{1},y_{1}^{q}]=[z_{1},y_{1}]\in Id_{G}(sl_{2}(F))$. The proof is complete. ∎

We now cite two papers. First we present a corollary of Bahturin et al. in [2].

Proposition 4.2** **(Bahturin et al., Corollary 1,

[2]).

Let $R=M_{n}(F)$, $charF=p>0$, $p\neq 2$. Let $G$ be an elementary abelian $p$-group. Suppose that $R=\bigoplus_{g\in G}R_{g}$ is a grading on $R^{(-)}$. Then $R=\bigoplus_{g\in G}R_{g}$ is a grading on $R$ if and only if $1\in R_{e}$.

Remark 4.3**.**

Here, $1$ denotes the identity matrix of $M_{n}(F)$, and $e$ denotes the identity element of $G$.

In this study, $F$ is a finite field of $charF=p>3$ and size $q$. So, there exists $b\in F-\{0\}$ that is not a perfect square.

Notice that if $sl_{2}(F)=(sl_{2}(F))_{0}\oplus(sl_{2}(F))_{1}$ is a $\mathbb{Z}_{2}$-grading on $sl_{2}(F)$, then $((sl_{2}(F))_{0}\oplus F(e_{11}+e_{22}))\oplus(sl_{2}(F))_{1}$ is a $\mathbb{Z}_{2}$-grading on $sl_{2}(F)$. By Proposition 4.2, $((sl_{2}(F))_{0}\oplus F(e_{11}+e_{22}))\oplus(sl_{2}(F))_{1}$ is a $\mathbb{Z}_{2}$-grading on $M_{2}(F)$. The next proposition describes the $\mathbb{Z}_{2}$-grading on $M_{2}(F)$.

Proposition 4.4** (Khazal et al., Theorem 1.1, adapted, [7]).**

Let $F$ be a field of $charF\neq 2$. Then any $\mathbb{Z}_{2}$-grading of $M_{2}(F)$ is isomorphic to one of the following.

**: **

$(M_{2}(F)_{0},M_{2}(F)_{1})=(M_{2}(F),0)$;

**: **

$(M_{2}(F)_{0},M_{2}(F)_{1})=(Fe_{11}\oplus Fe_{22},Fe_{12}\oplus Fe_{21})$;

**: **

$(M_{2}(F)_{0},M_{2}(F)_{1})=$**

$(F(e_{11}+e_{22})\oplus F(e_{12}+be_{21}),F(e_{11}-e_{22})\oplus F(e_{12}-be_{21}))$, where $b\in F-F^{2}$.

Remark 4.5**.**

It is well known that $(F-\{0\},.)$ is a cyclic group of order $q-1$. By elementary theory of groups, for every divisor $d$ of $q-1$, there exists a unique subgroup $H^{\prime}$ of $(F-\{0\},.)$ of order $d$. Let $H$ be the subgroup of order $\frac{q-1}{2}$. It is easy to see that there exists $b^{\prime}\in(F-F^{2})\cap(F-H)$. Finally, note that

$[(e_{11}-e_{22}),(e_{12}+b^{\prime}e_{21})]\neq[(e_{11}-e_{22}),(e_{12}+b^{\prime}e_{21}),\ldots,(e_{12}+b^{\prime}e_{21})]$,

where $(e_{12}+b^{\prime}e_{21})$ appears $q$ times in the expanded commutator.

Proposition 4.6**.**

Let $sl_{2}(F)=(sl_{2}(F))_{0}\oplus(sl_{2}(F))_{1}$ be a $\mathbb{Z}_{2}$-grading on $sl_{2}(F)$ having the following characteristics:

**: **

$dim\ (sl_{2}(F))_{0}=1$,

**: **

$[a,c^{q}]=[a,c]$* for all $a\in(sl_{2}(F))_{1}$ and $c\in F(e_{11}+e_{22})\oplus(sl_{2}(F))_{0}$.*

Then the $\mathbb{Z}_{2}$-gradings $((sl_{2}(F))_{0},(sl_{2}(F))_{1})$ and $(F(e_{11}-e_{22}),Fe_{12}\oplus Fe_{21})$ are isomorphic.

Proof.

First, note that $(((sl_{2}(F))_{0}\oplus F(e_{11}+e_{22}))\oplus(sl_{2}(F))_{1})$ is a $\mathbb{Z}_{2}$-grading on $gl_{2}(F)$. According to Proposition 4.2, $(((sl_{2}(F))_{0}\oplus F(e_{11}+e_{22}))\oplus(sl_{2}(F))_{1})$ is a $\mathbb{Z}_{2}$-grading on $M_{2}(F)$. It is clear that this grading on $M_{2}(F)$ is not isomorphic to the first grading presented in Proposition 4.4. Notice also that $(F(e_{11}+e_{22})\oplus(sl_{2}(F))_{0},(sl_{2}(F))_{1})$ cannot be isomorphic to the third grading presented in Proposition 4.4, because $[z_{1},y_{1}]=[z_{1},y_{1}^{q}]$ is not a polynomial identity for $M_{2}(F)$ endowed with third grading (Remark 4.5). According to Proposition 4.4, there exists a $G$-graded isomorphism $\phi:M_{2}(F)\rightarrow M_{2}(F)$ such that

$\phi((sl_{2}(F))_{0}\oplus F(e_{11}+e_{22}))=Fe_{11}\oplus Fe_{22}$ and $\phi((sl_{2}(F))_{1})=Fe_{12}\oplus Fe_{21}$.

Note that $\phi:gl_{2}(F)\rightarrow gl_{2}(F)$ is an isomorphism of Lie algebras and $\phi(sl_{2}(F))=sl_{2}(F)$. Thus, $((sl_{2}(F))_{0},(sl_{2}(F))_{1})$ and $(F(e_{11}-e_{22}),Fe_{12}\oplus Fe_{21})$ are isomorphic. The proof is complete. ∎

Henceforth we consider only $sl_{2}(F)$ and $Fe_{11}\oplus Fe_{12}$ endowed with the natural grading by $(\mathbb{Z}_{2},+)$. Recall that $Sem_{1}(x_{1},x_{2}),Sem_{2}(x_{1},x_{2})\in Id(sl_{2}(F))$. We denote by $S$ the set with following polynomials.

$Sem_{1}(y_{1}+z_{1},y_{2}+z_{2}),Sem_{2}(y_{1}+z_{1},y_{2}+z_{2}),[y_{1},y_{2}],\mbox{and}\ [z_{1},y_{1}^{q}]=[z_{1},y_{1}]$.

Corollary 4.7**.**

The variety $\mathcal{V}(S)$ is locally finite.

Proof.

Let $L=L_{0}\oplus L_{1}\in\mathcal{V}(S)$ be a finitely generated algebra. By definition of $S$, $Sem_{1}(y_{1}+z_{1},y_{2}+z_{2})\in Id_{G}(L)$. Hence, $Sem_{1}(x_{1},x_{2})\in Id(L)$. So, by Theorem 3.3, it follows that $L$ is a finite Lie algebra. ∎

Corollary 4.8**.**

Let $L\in\mathcal{V}(S)$ be a finite-dimensional Lie algebra. Then every nilpotent subalgebra of $L$ is abelian.

Proof.

From the definition of $S$, it follows that $Sem_{2}(x_{1},x_{2})\in Id(L)$. Let $M\neq\{0\}$ be a nilpotent (unnecessarily graded) subalgebra of $L$. If $M^{t}=\{0\}$ for a positive integer $t\leq q+1$, it is clear that $M$ is abelian. If the index of nilpotency is equal to $q+2$, then $\frac{M}{Z(M)}$ is abelian. Consequently, $M$ is abelian. Induction on the index of nilpotency will give the desired result. ∎

It is well known that a verbal ideal (and, respectively, a graded verbal ideal) over an infinite field is multi homogeneous. This fact can be weakened, as stated in the next lemma.

Lemma 4.9**.**

Let $I$ be a graded verbal ideal over a field of size $q$. If $f(x_{1},\ldots,x_{n})\in I$ and $0\leq deg_{x_{1}}f,\ldots,deg_{x_{n}}f<q$, then each multi homogeneous component of $f$ belongs to $I$ as well.

Lemma 4.10**.**

If $L=span_{F}\{e_{11},e_{12}\}\subset gl_{2}(F)$, then the $\mathbb{Z}_{2}$-graded identities of $L$ follow from

$[y_{1},y_{2}],[z_{1},z_{2}]$* and $[z_{1},y_{1}^{q}]=[z_{1},y_{1}]$.*

Proof.

It is clear that $L$ satisfies the identities $[y_{1},y_{2}],[z_{1},z_{2}]$ and $[z_{1},y_{1}^{q}]=[z_{1},y_{1}]$. We will prove that the reverse inclusion holds true. Let $f$ be a polynomial identity of $L$. We may write

$f=g+h$,

where $h\in\langle[y_{1},y_{2}],[z_{1},z_{2}],[z_{1},y_{1}^{q}]=[z_{1},y_{1}]\rangle$ and $g(x_{1},\ldots,x_{n})\in Id_{G}(L)$, with $0\leq deg_{x_{1}}g,\ldots,deg_{x_{n}}g<q$. In this way, we may suppose that $g$ is a multi homogeneous polynomial. If $g(y_{1})=\alpha_{1}.y_{1}$ or $g(z_{1})=\alpha_{2}z_{1}$ we can easily see that $\alpha_{1}=\alpha_{2}=0$. In the other case, we may assume that

$g(z_{1},y_{1},\ldots,y_{l})=\alpha_{3}.[z_{1},y_{1}^{a_{1}},\ldots,y_{l}^{a_{l}}],1\leq a_{1},\ldots,a_{l}<q$.

However, $g(e_{12},e_{11},\ldots,e_{11})$ is a non zero multiple scalar of $e_{12}$, and consequently, $\alpha_{3}=0$. Hence, $f=h$ and the proof is complete. ∎

Lemma 4.11**.**

Let $L=L_{0}\oplus L_{1}\in A^{2}\cap\mathcal{V}(S)$ be a critical Lie $A$-algebra. Then $\newline L\in var_{G}(span_{F}\{e_{11},e_{12}\})$.

Proof.

According to Lemma 4.10, it is sufficient to prove that $L$ satisfies the identity $[z_{1},z_{2}]$.

By assumption, $L$ is critical and therefore $L$ is monolithic. If $L$ is abelian, then $dimL=1$. In this case $L\cong span_{F}\{e_{11}\}$ or $L\cong span_{F}\{e_{12}\}$.

In the sequel, we suppose that $L$ is non abelian . From Proposition 3.1, we have $[L,L]=Nil(L)=[L_{1},L_{1}]\oplus[L_{0},L_{1}]$. From the identity $[z_{1},y_{1}]=[z_{1},y_{1}^{q}]$, $\{0\}=[L_{1},[L_{1},L_{1}]]=-[L_{1},L_{1},L_{1}]$. So, by the identity $Sem_{2}(y_{1}+z_{1},y_{2}+z_{2})$, we have $[z_{1},z_{2}]\in Id_{G}(span_{F}\{e_{11},e_{12}\})$, as required. The proof is complete. ∎

Corollary 4.12**.**

$A^{2}\cap var_{G}(sl_{2}(F)),A^{2}\cap\mathcal{V}(S)\mbox{and}\ \ var_{G}(span_{F}\{e_{11},e_{12}\})$* coincide.*

Proof.

First, notice that $A^{2}\cap var_{G}(sl_{2}(F))\subset A^{2}\cap\mathcal{V}(S)$ which is a locally finite variety. By Lemma 4.11, all critical algebras of $A^{2}\cap\mathcal{V}(S)$ belong to $var_{G}(span_{F}\{e_{11},e_{12}\})\newline \subset A^{2}\cap var_{G}(sl_{2}(F))$. Therefore, $A^{2}\cap\mathcal{V}(S)\subset var_{G}(span_{F}\{e_{11},e_{12}\})$. Thus, we have

$A^{2}\cap var_{G}(sl_{2}(F))=A^{2}\cap\mathcal{V}(S)=var_{G}(span_{F}\{e_{11},e_{12}\})$.

∎

Lemma 4.13**.**

Let $L$ be a critical solvable Lie $A$-algebra belonging to $\mathcal{V}(S)$. Then $L$ is metabelian.

Proof.

Let $L$ be a critical (non abelian) solvable Lie algebra that belongs to $\mathcal{V}(S)$ with monolith $W$. By Proposition 2.3, we have $[L,L]\cap Z(L)=\{0\}$. Consequently, $Z(L)=\{0\}$. Notice that $Z(C_{L}(Nil(L)))=Nil(L)$. If $(Nil(L))_{1}=L_{1}$, then $L$ is metabelian. Now, we suppose that $(Nil(L))_{1}\varsubsetneq L_{1}$. We assert that $(Nil(L))_{0}=\{0\}$. Suppose, on the contrary, that there exists $a\neq 0\in(Nil(L))_{0}$. Hence, there exists $b\in L_{1}-(Nil(L))_{1}$ such that $[b,a]\neq 0$, because $Z(L)=\{0\}$. However, $[b,a]=[b,a^{q}]=0$. This is a contradiction. Thus, $[L_{1},Nil(L)]=\{0\}$. Consequently $C_{L}(Nil(L))\supset L_{1}\cup[L_{1},L_{1}]$. By Proposition 2.3

$Z(C_{L}(Nil(L)))\cap[C_{L}(Nil(L)),C_{L}(Nil(L))]=\{0\}$.

Hence, $[C_{L}(Nil(L)),C_{L}(Nil(L))]=\{0\}$. So, $L^{(2)}=\{0\}$ and the proof is complete. ∎

Lemma 4.14**.**

Let $L$ be a critical non-solvable Lie $A$-algebra belonging to $\mathcal{V}(S)$. Then $L$ is $G$-simple.

Proof.

Let $W$ be the monolith of $L$. We claim that $L$ is semisimple. Suppose on the contrary that $Rad(L)\neq\{0\}$. Thus, $W\subset Rad(L)\cap[L,L]$. The non trivial subspace $W$ is an abelian ideal and it is contained in $L^{(n)}$, where $n$ is the least nonnegative integer such that $L^{(n)}=L^{(n+1)}$. According to Proposition 2.3, $[L,L]\cap Z(L)=\{0\}$. So $Z(L)=\{0\}$ and $[L,W]=W$. The identities $[y_{1},y_{2}]$ and $[z_{1},y_{1}]=[z_{1},y_{1}^{q}]$ mean that the subspace $W_{0}=\{0\}$. Notice that $[W,[L,L]]=\{0\}=[W,L^{(n)}]$ and $Z(L^{(n)})\supset W$. By Proposition 2.3, $Z(L^{(n)})\cap L^{(n)}=\{0\}$. This is a contradiction, so $L$ is semisimple. By Propositions 2.3 and 2.4, $L$ is a direct sum of $G$-simple Lie algebras. Given that $L$ is monolithic, we conclude that $L$ is $G$-simple. ∎

The next theorem was proved by Drensky in ([3] Lemma, page 991).

Lemma 4.15**.**

Let $V$ be a finite dimensional vector space over $F$ and let $A$ be an abelian Lie algebra of the linear transformations $\phi:V\rightarrow V$, where each has the equality

$\phi^{q}=\phi$.

Then, every $\phi\in A$ is diagonalizable.

Definition 4.16**.**

Let $L$ be a finite dimensional Lie algebra with a diagonalizable operator $T:L\rightarrow L$. We denote by $V(T)$ a basis of $L$ formed by the eigenvectors of $T$. Moreover, we denote $V(T)_{\lambda}=\{v\in V(T)|T(v)=\lambda.v\}$. We denote $EV(w)$ the eigenvalue associated with the eigenvector $w\in V(T)$.

Let $L\in\mathcal{V}(S)$ be a finite dimensional Lie algebra. It is not difficult to see that $ad(L_{0})=\{ada:L\rightarrow L|a\in L_{0}\}$ is an abelian subalgebra of linear transformations of $L$. Moreover, $(ada_{0})^{p}=ada_{0}$ for all $a_{0}\in L_{0}$. By Lemma 4.15, we have the following.

Corollary 4.17**.**

Let $L\in\mathcal{V}(S)$ be a finite dimensional Lie algebra. Let $a_{0}\in L_{0}$. Then there exists $V(ada_{0})\subset L_{0}\cup L_{1}$.

Proposition 4.18**.**

Let $L\in\mathcal{V}(S)$ be a finite dimensional $G$-simple algebra. Let $a_{0}\in L_{0}$. Then there exists $V(ada_{0})\subset L_{0}\cup L_{1}$. Moreover, $V(ada_{0})_{0}\cap L_{1}=\emptyset$ for any basis $V(ada_{0})\subset L_{0}\cup L_{1}$.

Proof.

According to Corollary 4.17, there exists $V(ada_{0})\subset L_{0}\cup L_{1}$.

Let $b_{1},b_{2}\in V(ada_{0})\cap L_{1}$. Notice that if $[b_{1},b_{2}]\neq 0$, then $EV(b_{1})=-EV(b_{2})$.

It is clear that $\langle a_{0}\rangle$ is a graded ideal, and that it is equal to $L$. Notice also that $L=span_{F}\{[a_{0},b_{1},\ldots,b_{n}]|b_{1},\ldots,b_{n}\in V(ada_{0}),n\geq 1\}$.

If there was a non zero element $b\in V(ada_{0})_{0}\cap L_{1}$, we could easily check that $[a_{0},b_{1},b]=0$ for any $b_{1}\in V(ad(a_{0}))$. More generally, by an inductive argument and routine calculations, we would have $[a_{0},b_{1},\ldots,b_{n},b]=0$ for any $n\geq 1$ and $b_{1},\ldots,b_{n}\in V(ada_{0})$. However, an element such as $b$ cannot be, because $Z(L)=\{0\}$. So, $V(ada_{0})_{0}\cap L_{1}=\emptyset$. ∎

Lemma 4.19**.**

Let $L\in\mathcal{V}(S)$ be a critical non solvable algebra, then $L\cong sl_{2}(F)$.

Proof.

First of all, notice that $dimL_{0}\geq 1$ and $dimL_{1}\geq 2$. According to Lemma 4.14 $L$ is $G$-simple. Let $a_{0}\in L_{0}$. By Proposition 4.18, there exists $V(ada_{0})=\{b_{1},\ldots,b_{n}\}\subset L_{0}\cup L_{1}$. Moreover, $V(ada_{0})_{0}\cap L_{1}=\emptyset$.

Let $-\lambda_{1}\leq\ldots\leq-\lambda_{m}<0<\lambda_{m}\leq\ldots\leq\lambda_{1}$ be the eigenvalues associated with the eigenvectors of $V(ada_{0})$. Notice that

$L_{0}=\sum\limits_{i=1}^{m}span_{F}\{[V(ada_{0})_{\lambda_{i}},V(ada_{0})_{-\lambda_{i}}]\}$.

Without loss of generality, suppose that $span_{F}\{[V(ada_{0})_{\lambda_{1}},V(ada_{0})_{-\lambda_{1}}]\}\neq\{0\}$. We assert that $span_{F}\{[V(ada_{0})_{\lambda_{1}},V(ada_{0})_{-\lambda_{1}}]\}\oplus span_{F}\{V(ada_{0})_{\lambda_{1}}\}$ is a subalgebra of $L$.

In fact, let $a\in V(ada_{0})_{\lambda_{1}}$ and $b\in span_{F}\{[V(ada_{0})_{\lambda_{1}},V(ada_{0})_{-\lambda_{1}}]\}$. Consider $[a,b]=\sum\limits_{i=1}^{n}\alpha_{i}b_{i}$. So,

$[a,b,a_{0}]=-\sum\limits_{i=1}^{n}\alpha_{i}.EV(b_{i})b_{i}$.

On the other hand,

$[a,b,a_{0}]=-\lambda_{1}[a,b]=-\lambda_{1}(\sum\limits_{i=1}^{n}\alpha_{i}b_{i})$.

Hence

$(-EV(b_{j}).\alpha_{j}+\lambda_{1}.\alpha_{j})b_{j}=0$.

Consequently, if $\alpha_{j}\neq 0$, then $\lambda_{1}=EV(b_{j})$.

Similarly, the subspace $span_{F}\{[V(ada_{0})_{\lambda_{1}},V(ada_{0})_{-\lambda_{1}}]\}\oplus span_{F}\{V(ada_{0})_{-\lambda_{1}}\}$ is a subalgebra. Notice that

$span_{F}\{[V(ada_{0})_{\lambda_{1}},V(ada_{0})_{-\lambda_{1}}]\}\oplus span_{F}\{V(ada_{0})_{\lambda_{1}}\}\oplus span_{F}\{V(ada_{0})_{-\lambda_{1}}\}$

is a graded ideal of $L$.

Therefore, $L_{0}=span_{F}\{V(ada_{0})_{0}\}=span_{F}\{[V(ada_{0})_{\lambda_{1}},V(ada_{0})_{-\lambda_{1}}]\}$ and the subspace $L_{1}$ is equal to $span_{F}\{V(ada_{0})_{\lambda_{1}}\}\oplus span_{F}\{V(ada_{0})_{-\lambda_{1}}\}$.

Notice that $span_{F}\{V(ada_{0})_{\lambda_{1}}\}$ is an irreducible $L_{0}$-module, because $L$ is $G$-simple. Moreover, it is not difficult to see that $L_{0}\oplus span_{F}\{V(ada_{0})_{\lambda_{1}}\}$ is a monolithic metabelian Lie algebra with monolith $span_{F}\{V(ada_{0})_{\lambda_{1}}\}$ when viewed as ordinary Lie algebra. Notice that

$[L_{0}\oplus span_{F}\{V(ada_{0})_{\lambda_{1}}\},L_{0}\oplus span_{F}\{V(ada_{0})_{\lambda_{1}}\}]=span_{F}\{V(ada_{0})_{\lambda_{1}}\}$

cannot be represented by the sum of two ideals strictly contained within it. By Theorem 3.2, $L_{0}\oplus span_{F}\{V(ada_{0})_{\lambda_{1}}\}$ is critical when viewed as an ordinary Lie algebra. Thus, it is critical when viewed as a graded algebra as well.

Following the arguments of Lemma 4.10, we can prove that

$Id_{G}(L_{0}\oplus span_{F}\{V(ada_{0})_{\lambda_{1}}\})=\langle[y_{1},y_{2}],[z_{1},y_{1}]=[z_{1},y_{1}^{q}],[z_{1},z_{2}]\rangle_{T}$.

Consequently, it follows from Proposition 3.4 that $span_{F}\{V(ada_{0})_{\lambda_{1}}\}$ is a one-dimensional vector space. Analogously, we have $dim(span_{F}\{V(ada_{0})_{-\lambda_{1}}\})=1$. Therefore, $L_{0}\oplus span_{F}\{V(ada_{0})_{\lambda_{1}}\}\oplus span_{F}\{V(ada_{0})_{-\lambda_{1}}\}$ is a three-dimensional $G$-simple Lie algebra. So, $L$ is simple and isomorphic to $sl_{2}(F)$ (as ordinary Lie algebras). Hence, by Proposition 4.6, $L\cong sl_{2}(F)$ (as graded Lie algebras), where $sl_{2}(F)$ is naturally graded by $\mathbb{Z}_{2}$. The proof is complete. ∎

5. Main theorem

We now prove the main theorem of this paper.

Theorem 5.1**.**

Let $F$ be a field of $char(F)>3$ and size $|F|=q$. The $\mathbb{Z}_{2}$-graded identities of $sl_{2}(F)$ follow from

$[y_{1},y_{2}],Sem_{1}(y_{1}+z_{1},y_{2}+z_{2}),Sem_{2}(y_{1}+z_{1},y_{2}+z_{2}),\mbox{and}\ [z_{1},y_{1}]=[z_{1},y_{1}^{q}]$.

Proof.

It is clear that $var_{G}(sl_{2}(F))\subset\mathcal{V}(S)$. To prove that the reverse inclusion holds, it is sufficient to prove that all critical algebras of $\mathcal{V}(S)$ are also critical algebras of $var_{G}(sl_{2}(F))$. According to Corollary 4.12, $A^{2}\cap\mathcal{V}(S)=A^{2}\cap var_{G}(sl_{2}(F))$. By Lemma 4.13, any critical solvable Lie algebra of $\mathcal{V}(S)$ is metabelian. By Lemma 4.19, any critical non solvable Lie algebra of $\mathcal{V}(S)$ is isomorphic to $sl_{2}(F)$. Therefore, $\mathcal{V}(S)\subset var_{G}(sl_{2}(F))$, and the theorem is proved. ∎

6. Acknowledgments

The author thanks CNPq for his Ph.D. scholarship. Moreover, the author thanks the reviewer for his/her comments.