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On identities of model algebras S. V. Pchelintsevab a Saint Petersburg State University b Financial University under the Government of the Russian Federation, Moscow
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2023, Volume 87, Issue 6, DOI: https://doi.org/10.4213/im9395 
Bibliographic databases:
    § 1. IntroductionLet $\Phi$ be an infinite field of characteristic different from $2$. In what follows, unless otherwise stated, only associative unital $\Phi$-algebras are considered. Let us introduce the following notation: $F$ a free $\Phi$-algebra over a countable set $X={\{x_1,x_2,\dots}\}$ of free generators; $[x_1,\dots,x_{n}]$ is a right-normed commutator of degree $n\geqslant 2$, that is, $[x_1,x_2]=x_1 x_2 -x_2 x_1 $ and, by induction, $[x_1, \dots,x_n]=[[x_1,\dots,x_{n-1}],x_n]$; $T^{(n)}$ is the $T$-ideal in $F$ generated by a right-normed commutator of degree $n$; $\mathrm{LN}(n)\colon [x_1,\dots,x_n]=0$ is the Lie identity of degree nilpotency $n$; $F^{(n)}$ is the relatively free algebra with identity $\mathrm{LN}(n)$; $F^{(n)}_{r}$ is a relatively free algebra with identity $\mathrm{LN}(n)$ on $r$ free generators; $A'$ is a commutator ideal (commutant) of the algebra $A$; $Z(A)$ is the centre of $A$; $Z^*(A)$ is the kernel of $A$ (the largest ideal of the algebra $A$ contained in the centre). If $T$ is an ideal of $F$, and $A$ is an associative algebra, then by $T(A)$ we denote the set of values of the ideal $T$ in the algebra $A$. An algebra is called a Lie nilpotent algebra of class $n-1$ if it satisfies the identity $\mathrm{LN}(n)$ but does not satisfy $\mathrm{ LN}(n-1)$. A polynomial $p\in F$ is called proper if it is contained in the subalgebra generated by the commutators
$$
\begin{equation*}
[x_1, x_2,\dots,x_m ],\quad \text{where }\ x_1,x_2,\dots,x_m \in X\ \text{ and }\ m\geqslant 2.
\end{equation*}
\notag
$$
Let us recall the concept of a model algebra, which first appeared in [1]. Let $E$ be the associative (unital) algebra over the field $\Phi$ given by the set of generators $\{e_m, \theta_{i,j}\mid m, i,j \in \mathbb{N},\, i\leqslant j\}$ and defining relations $e_i \circ e_j = \theta_{ij}$, $[\theta_{ij}, e_m] = 0$, where $x\circ y = xy + yx$ is the Jordan product of elements $x,y\in E$. Note that $e_i^2 = \frac{1}{2}\theta_{ii}$, and that the algebra $E$ is generated by the elements $\{e_m\mid m \in \mathbb{N}\}$. We let $\Theta$ denote the ideal in $E$ generated by the elements $\theta _{ij}$. A model algebra of multiplicity $m$ is the quotient algebra $E^{(m)} = E/\Theta ^m$. Note that $E^{(1)}$ is a Grassmann algebra. The original conjecture from [1] says the algebra $E^{(m)}$ generates a variety of associative algebras with the identity $\mathrm{LN}(2m+1)$ in the case of a field $\Phi$ of zero characteristic. Model algebras play an important role in the study of free Lie nilpotent algebras. In [2] some general properties of model algebras were identified; it was also proved that the algebras $F^{(2m+1)}$ for $m\geqslant 2$ contain non-zero kernel elements, which are identities of the model algebra $E^{(m)}$. In [3] and [4], it was proved that the converse assertion is also true for the algebra $F^{(5)}$. In addition, in these papers, a basis of identities for the algebra $E^{(2)}$ was found (first, over a field of characteristic $0$, and then in the general case, where the characteristic of an infinite field is different from $2$ and $3$). Some finitely generated subalgebras of model algebras were considered in [5]. In particular, it was proved that the nilpotency index of the commutator subalgebra $E_{2m+1}^{(m)}$ of the model algebra $E^{(m)}$ generated by the elements $\{e_1,\dots,e_{2m+1}\}$ is $2m$. For $n\leqslant 7$, a more general fact is proved: the nilpotency index of the commutator subgroup of the algebra $F_n^{(n)}$ is $n-1$. In [5], the conjecture on the validity of this result for all $n$ was also set forth. In the present paper, we study the model algebras and their identities, and, in particular, arbitrary 2-generated subalgebras and identities in 2-variable model algebras. The paper consists of 10 sections. In § 2, a sharp upper bound for the nilpotency index of the commutator ideal $A'$ 2-generated subalgebra $A$ of the model algebra $E^{(m)}$ is given; it is about half that in the case of 2-generated algebras with the identity $\mathrm{LN}(2m+1)$ (see [6] and [7]). In § 3, it is proved that each proper polynomial in 2 variables of degree $7$ is an identity of the algebra $E^{(3)}$. In [2], the kernel elements of the algebras $F^{(n)}$, $n \geqslant 4$, are indicated. For even $n$, the kernel element is the commutator of degree $n-1$, for odd $n$, this element is a weak Hall element of degree $n$. In § 4, for each $m\geqslant 3$, the kernel elements of the algebra $F^{(2m+1)}$ of degree $2m$ are given. Thus, the algebras $F^{(n)}$, $n \geqslant 4$, $n\neq 5$, contain kernel elements of degree $n-1$; this is the minimum possible degree of kernel elements, since a polynomial of degree $\leqslant n-2$ is not central. Note that the smallest degree of a kernel element of the algebra $F^{(5)}$ is $5$ (see [2]), and the algebra $F^{(3)}$ does not contain kernel elements. Finally, in this section, we indicate a non-zero element $p\in F^{(7)}$ which is an identity of the model algebra $E^{(3)}$, but is not a central element in $F^{(7)}$. Thus, the kernel of the algebra $F^{(7)}$ does not coincide with the ideal of identities of the algebra $E^{(3)}$. In this section, we also formulate a conjecture about the kernel of the algebra $F^{(2m+1)}$, $m\geqslant 3$. In § 5, the commutator identities of degree $6$ in $2$ variables that hold in the algebra $E^{(3)}$ are described. It is proved that all these identities are consequences of the identity $[a,b,b,b,[a,b]]=0$ (in § 8, this result is carried over to arbitrary proper $2$-identities of degree $6$). In § 6, a more simple algebra $C^{(m)}$ with the same ideals of identities as in the model algebra $E^{(m)}$ is specified. In § 7, it is proved that the model algebra $E^{(3)}$ has no identities of degree $5$ in $2$ variables. Let us recall that the model algebra $E^{(2)}$ has no identities of degree $4$. In § 8, it is proved that the conjecture about the kernel of the algebra $F^{(7)}$ is valid for elements in 2 variables. It is not yet known whether the kernel conjecture is valid in general even for the algebra $F^{(7)}$. Thus, a number of negative answers to the questions formulated in [2] is obtained. In § 9, it is proved that there is no non-trivial identity that holds in all model algebras. Moreover, it is shown that the model algebra of multiplicity $m$ does not satisfy an identity of degree $m$. Finally, in § 10, some open questions are formulated. § 2. $2$-generated subalgebras of model algebrasBefore proceeding with the main part of the paper, we note the following three identities, which hold in any associative algebra: Proof. Recall (see [2]) that the algebra $E$ has an additive basis of elements where $v(\theta _{kl})$ is the standard normalized associative-commutative monomial of variables of the form $\theta _{kl}$ equipped with the lexicographic order, and $1\leqslant i_1 <\dots<i_n$.
We put $\theta(w) = v$ and call it the $\theta$-coefficient of an element $w$. If $q$ is a linear combination of elements $w_k$ of the form (2.4) and $q\in \Theta^m$, then each $\theta$-coefficients $\theta(w_k)$ lies in $\Theta^m$, and hence so each $w_k\in \Theta^m$. Let us assume that $p=\sum_k \alpha_k w_k$, where $\alpha_k\in \Phi$, $w_k$ has the form (2.4). Assume that $p[x,y]\in \Theta^m$ for all $x,y\in E$. We choose a number $N$ that is greater than all indices $l$ such that $e_l$ is included in $p$. The element $p[e_N,e_{N+1}]$ is a linear combination of the elements $w_ke_Ne_{N+1}$ and $w_k\theta_{N,N+1}$, where the $\theta$-coefficients of the elements $w_ke_Ne_{N+1}$ and $w_k$ are the same. Hence $w_k\in \Theta^m$ for each $k$, that is, $p\in \Theta^m$. The lemma is proved. Since the algebra $F^{(2m+1)}$ has non-zero kernel (see [2]), the variety $\operatorname{var}(E^{(m)})$ generated by the algebra $E^{(m)}$ differs from that given by the identity $\mathrm{LN}(2m+1)$. In this section, we show that in the model algebra $E^{(m)}$ of multiplicities $m$ the nilpotency index of the commutant of any $2$-generated subalgebra is about half that in $2$-generated algebras with the Lie identity of nilpotency of class $2m$. Theorem 1. The commutant of a $2$-generated subalgebra of the model algebra $E^{(m)}$ of multiplicity $m$ is nilpotent of index $\leqslant m+1$. This estimate is sharp if $m$ is odd, or if the characteristic of the field is equal to $0$, or if it is $\geqslant m/2$ for even $m$. Proof. Let us first show that the above estimate is sharp, that is, $[a,b]^m \neq 0$ for suitable elements $a,b\in E^{(m)}$.
Assume that $m = 2k+1$. Let $w=[e_1,e_2]$. Then
$$
\begin{equation*}
w = 2e_1e_2 - \theta _{12},\qquad 4e_1e_2e_1e_2 - 4e_1e_2\theta _{12}= -4e_1e_2e_2e_1=-\theta _{11}\theta _{22}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
w^2 = (2e_1e_2 - \theta _{12})^2 = 4e_1e_2e_1e_2 - 4e_1e_2\theta _{12} + \theta _{12}^2 = \Delta,
\end{equation*}
\notag
$$
where $\Delta = \theta _{12}^2 - \theta _{11}\theta _{22}$. Finally,
$$
\begin{equation*}
w^m = w^{2k}w =\Delta ^k (2e_1e_2 - \theta _{12}) = 2 \Delta ^k e_1e_2 - \Delta ^k\theta _{12} \notin \Theta^{2k+1} = \Theta^m.
\end{equation*}
\notag
$$
Now let $m = 2k$. In what follows, when passing from $E$ to the quotient algebra, we keep the previous notation for the images of $e_i$ under the canonical homomorphism. Assume that the following relations hold in the quotient algebra:
$$
\begin{equation*}
\theta _{13}=\theta _{23}=\theta _{33}=\theta _{12}^2 =0.
\end{equation*}
\notag
$$
Let us put $w = [a,b]$, where $a= e_1, b = e_2 + e_2e_3$. Since
$$
\begin{equation*}
[e_1,e_2]= 2e_1e_2 - \theta _{12},\qquad [e_1, e_2e_3] = e_1e_2e_3 - e_2e_3 e_1 = (e_1\circ e_2)e_3 =\theta_{12}e_3,
\end{equation*}
\notag
$$
we get Let us now calculate $w^2$. First, we have
$$
\begin{equation*}
(2e_1e_2 - \theta _{12})^2 = 4e_1e_2e_1e_2 - 4e_1e_2\theta _{12} = - 4e_1^2e_2^2 = \Delta,
\end{equation*}
\notag
$$
where $\Delta = - \theta _{11}\theta _{22}$. Second, since $\theta _{13}=\theta _{23}=0$, it follows that $e_3$ anticommutes both with $e_1$ and with $e_2$, and so $e_1e_2e_3 = -e_1e_3e_2 = e_3e_1e_2$, that is, $[e_1e_2,e_3] = 0$. Hence
$$
\begin{equation*}
w^2 = \Delta + g,\quad \text{where}\quad g= 4e_1e_2e_3\theta_{12}.
\end{equation*}
\notag
$$
Since $\Delta$ is central and $g^2 = 0$, we have, modulo $\Theta^m$,
$$
\begin{equation*}
\begin{aligned} \, w^m &= w^{2k} = (w^2)^k = (\Delta + g)^k \equiv k \Delta^{k-1}g \\ &=4k e_1e_2e_3 \Delta^{k-1}\theta_{12} = 4k(-1)^{k-1}(\theta _{11}\theta_{22})^{k-1} \theta_{12}e_1e_2e_3. \end{aligned}
\end{equation*}
\notag
$$
Since the $\theta$-degree of the monomial $(\theta _{11}\theta _{22})^{k-1}\theta_{12}$ is $2k\,{-}\,1=m\,{-}\,1$, we have $w^m\,{\notin}\,\Theta^m$.
Let $A$ be a unital subalgebra of $E$ generated by elements $a$ and $b$. Let us prove that $(A')^{m+1} \subseteq \Theta^{m}$. By identity (2.1) and in view of the equality of the elements $a$ and $b$, it suffices to verify by induction on $m$ that $(A')^m[a,E] \subseteq \Theta^m$. The basis of induction for $m=1$ is true, since the identity $[x,t][t,y]=0$ holds in a Grassmann algebra. Moreover, $A'\equiv A[a,b]$ is true modulo $\Theta$. Note also that $[\Theta^m,[E,E]] \subseteq \Theta^{m+1}$. The induction step: let us show that $(A')^{m+1}[a,E] \subseteq \Theta^{m+1}$. Let $x,y,z\in E$. Then, modulo $\Theta^{m+1}$,
$$
\begin{equation*}
\begin{aligned} \, &(A')^{m+1}[a,x][y,z] \equiv [a,x](A')^{m+1}[y,z] = [a,x](A')^m [a,b][y,z] \\ &\qquad\equiv -[a,x](A')^m [a,y][b,z] \equiv -[a,y][a,x] (A')^m[b,z] \subseteq \Theta \cdot \Theta^m \equiv 0, \end{aligned}
\end{equation*}
\notag
$$
that is Hence, by Lemma 1, This proves the inclusion $(A')^m[a,E] \subseteq \Theta^m$, and, therefore, Theorem 1. Remark 1. The commutant of the algebra $F_{2}^{(2m+1)}$ is nilpotent of index $2m$ (see [6] and [7]). On the other hand, the commutants of each of the algebras $E_{2m+1}^{(m)}$ and $F_{2m+1}^{(2m+1)}$ for $m=1,2,3$ have the same nilpotency indices equal to $2m$ (see [5]), where $E_{2m+1}^{(m)}$ is a subalgebra of $E^{(m)}$ generated by $e_1,\dots,e_{2m+1}$. Remark 2. In § 8 of [5], the following conjecture was set forth § 3. Proper identities in 2 variables of degree $\geqslant 7$ of the algebra $E^{(3)}$In what follows, unless otherwise stated, the following notation is used: $\Theta$ is the ideal of the algebra $E$ generated by the elements $\theta _{ij}$; $A$ is the unital subalgebra in $E$ generated by the elements $a$ and $b$; $u_i$, $u'_i$ is the right-normed commutators of degree $i$ in $a$ and $b$; $w_i = [c_1,\dots,c_i]$ is the right-normed commutators in $c_j\in E$; $W_i$ is the subspace generated by the elements $w_i$; the notation $x\equiv_k y$, where $x,y\in E$, means congruence modulo the ideal $\Theta^k$, that is, $x-y\in \Theta^k$. We also note that, modulo $W_{i+j+1}E$, the following congruences hold:
$$
\begin{equation*}
\begin{gathered} \, [w_{i+1},w_j] \equiv 0, \\ 2[w_i,x][w_j,x] \equiv [w_i,x]\circ [w_j,x] = [w_i,x\circ [w_j,x]]-x\circ [w_i,[w_j,x]] \equiv 0. \end{gathered}
\end{equation*}
\notag
$$
In addition, In what follows, these remarks are used without special explanation. Proof. Note that $u_2=[a,b]$ and we can assume that $u_3=[a,b,b]$. By Lemma 1, it suffices to prove that, for all $x,y\in E$,
The proof is in several steps. $1^{\circ}$. We have $[x,y,z] \equiv_1 0$ and $[x,y][x,z] \equiv_1 0$ for all $x,y,z \in E$, since the Grassmann algebra $E^{(1)}$ satisfies the identities $[x,y,z]=[x,y][x,z] = 0$. $2^{\circ}$. We have $u_3u_2 \equiv_2 0$, since the weak Hall identity $[[a,b]^2,b] = 0$ holds in the algebra $E^{(2)}$ (see [2] and [4]). $3^{\circ}$. We have $[\Theta^k,E,E] + [\Theta^k,[E,E]] \subseteq \Theta^{k+1}$, since every element of $\Theta^k$ can be linearly expressed via $v(\theta _{ij})e$, where $v(\theta _{ij})$ is a monomial of degree $k$ on the variables $\theta _{ij}$, $e\in E$, and $v(\theta _{ij})\in Z(E)$ and $[E,E,E] \subseteq \Theta$. $4^{\circ}$. Let us prove that $[b,y]u_3 = [b,y][a,b,b] \equiv_2 [a,b][y,b,b]$. It suffices to linearize $a \to y$ the comparison $[a,b][a,b,b] \equiv_2 0$ from $2^{\circ}$. $5^{\circ}$. Let us prove that $[a,x][a,b]^2\equiv_2 0$. By $1^{\circ}$ and $3^{\circ}$,
$$
\begin{equation*}
[a,x][a,b]^2[y,z] \equiv_2 -[a,x][a,b][a,y][b,z] \equiv_2 -[a,y][a,x][a,b][b,z] \equiv_2 0.
\end{equation*}
\notag
$$
$6^{\circ}$. Let us prove that $u_3u_2u_2[x,y] \equiv_3 0$. We have
$$
\begin{equation*}
\begin{aligned} \, u_3u_2u_2[x,y] &= u_3u_2[a,b][x,y]\stackrel{\text{by }1^{\circ}, \text{ }2^{\circ}}{\equiv_3} - u_3u_2[a,x][b,y] \stackrel{\text{ by }2^{\circ}, \text{ }3^{\circ}}{\equiv_3} - [a,x][b,y]u_3u_2 \\ &\equiv_3 -[a,x][a,b][y,b,b]u_2 \stackrel{\text{ by }4^{\circ}}{\equiv_3} -[a,x][a,b]^2[y,b,b]\stackrel{\text{ by }5^{\circ}}{\equiv_3} 0. \end{aligned}
\end{equation*}
\notag
$$
Now the required result easily follows from $6^{\circ}$. Lemma 2 is proved. Proof. It suffices to show that $u_4u_3[x,y] \equiv_3 0$, where $u_3=[a,b,b]$. Assume first that $u_4=[u'_3,b]$. Arguing as Lemma 2, we have, modulo the ideal $\Theta^3$,
$$
\begin{equation*}
\begin{aligned} \, u_4u_3 [x,y] &\equiv u_3 u_4 [x,y] = u_3 [u'_3,b] [x,y] \equiv - u_3 [u'_3,x] [b,y] = - [u'_3,x] u_3 [b,y] \\ &= - [u'_3,x] [a,b,b] [b,y] \equiv [u'_3,x] [y,b,b] [b,a] \equiv [y,b,b] [u'_3,x] [b,a] \\ &= - [y,b,b] [u'_3,x]u_2 = [y,b,b] (-[u'_3 u_2,x] + u'_3 [u_2,x]) \equiv 0. \end{aligned}
\end{equation*}
\notag
$$
The case $u_4=[u'_3,a]$ is dealt with similarly:
$$
\begin{equation*}
\begin{aligned} \, u_4u_3 [x,y] &\equiv u_3 u_4 [x,y] = u_3 [u'_3,a] [x,y] \equiv - u_3 [u'_3,x] [a,y]=- [u'_3,x] u_3 [a,y] \\ &= - [u'_3,x] [a,b,b] [a,y] \equiv [u'_3,x] ([a,y,b] + [a,b,y])[a,b] \\ &\equiv ([a,y,b] + [a,b,y])[u'_3,x][a,b] = ([a,y,b] + [a,b,y])[u'_3,x]u_2 \\ &= ([a,y,b] + [a,b,y])([u'_3 u_2,x] - u'_3 [u_2,x]) \equiv 0. \end{aligned}
\end{equation*}
\notag
$$
Lemma 3 is proved. Proof. By identity (2.2), the mapping $D_x\colon r\mapsto [r,x]$, is a derivation of the algebra $A$, which is called the internal derivation. By the Leibniz formula,
$$
\begin{equation*}
(rs)D_xD_y = (rD_xD_y)s + r(sD_xD_y) + (rD_x)(sD_y) + (rD_y)(sD_x).
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
(u_3u_2)D_xD_y = (u_3D_xD_y)u_2 + u_3(u_2D_xD_y) + (u_3D_x)(u_2D_y) + (u_3D_y)(u_2D_x)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
(u_3D_xD_y)u_2 = - u_3(u_2D_xD_y) - (u_3D_x)(u_2D_y) - (u_3D_y)(u_2D_x)+ (u_3u_2)D_xD_y,
\end{equation*}
\notag
$$
that is, the element $u_5u_2$ is a linear combination of the elements $u_4u_3$, $u_3u_4$ and elements of the form $(u_3u_2)D_xD_y$, where $x, y \in \{a,b\}$. Now the required result follows from $2^{\circ}$ and $3^{\circ}$ in Lemma 2 and Lemma 3. This proves Lemma 4. Theorem 2. A proper polynomial $p(a,b)$ in two variables of degree $\geqslant 7$ is an identity of the algebra $E^{(3)}$. Proof. Let $A$ be the unital subalgebra of $E^{(3)}$ generated by $a$ and $b$. Let $T^{(n)}=T^{(n)}(A)$. From Lemmas 2–4 we have where $P\ast Q = PQ + QP$ for arbitrary subspaces $P$ and $Q$ in the algebra $E$. It remains to note that proving the theorem.
§ 4. On the kernel of the algebra $F^{(2m+1)}$A description of the kernel elements of the algebra $F^{(5)}$ is known, Namely, the kernel $Z^*(F^{(5)})$ coincides with the identity ideal of the model algebra $E^{(2)}$ of multiplicity $2$, and qua $T$-ideal is generated by a commutator of degree $5$ and a weak Hall element (see [3] and [4]). By Lemma 2, the element $[a,b,b] \cdot [a,b]^2$ is an identity of the algebra $E^{(3)}$, but it is not central in $F^{(7)}$ because its weight is $5$, and the elements of weight $\geqslant 6$ are the only central elements of $2$ variables (see [7]). The following conjecture seems plausible. It is known that $0 \neq Z^*(F^{(2m+1)}) \subsetneq Z(F^{(2m+1)})$. More precisely, for $m\geqslant 2$ the elements of degree $2m+1$ of the following form were introduced in [2]:
$$
\begin{equation*}
H_{m-2} = [h,x_1,y_1,\dots,x_{m-2},y_{m-2}],\qquad H'_{m-2} = [h',x_1,y_1,\dots,x_{m-2},y_{m-2}],
\end{equation*}
\notag
$$
where $h=[[a,b]^2,c]$ and $h'=[[a,b]^2,b]$ are Hall elements; it was also proved that a) $H_{m-2}\in Z(F^{(2m+1)})$; b) $0 \neq H'_{m-2} \in Z^*(F^{(2m+1)})$. Note that the minimum degree of the central element in the algebra $F^{(n)}$ is $n- 1$. It is easy to show that in the algebra $F^{(2m)}$ the commutator of degree $2m-1$ is kernel (see Theorem 2 in [2]). Now we indicate the kernel element of the algebra $F^{(2m+1)}$ of the minimum possible degree $2m$. Let $[a,b]_1=[a,b]$ and define by induction $[a,b]_{k+1}=[[a,b]_k,b]$. Theorem 3. Let $c_k=[a,b]_{k-1}$, $k\geqslant 2$. The polynomial is a kernel element of degree $2m$. In particular, $p_{2m}=0$ is an identity of degree $2m$ of the model algebra $E^{(m)}$ of multiplicity $m$. Proof. Let $x,y\in F^{(2m+1)}$. Since $p_{2m} \in Z(F^{(2m+1)})$, it suffices to show that $p_{2m}[x,y]=0$ in the algebra $F^{(2m+1)}$. We proceed in several steps.
$1^{\circ}$. First, we note that the element $[c_k,x]$ can be represented as a linear combination of right-normed commutators of degree $k+1$, that start with $x$ and end with either $a$ or $b$. Indeed, by the Jacobi identity (2.3), Each of the terms can be inducted on $k$.$2^{\circ}$. Hence, the commutator $[c_{2m-2},[x,y]]$ can be linearly expressed in terms of the commutators $[v_{2m-1},a]$ and $[v_{2m-1},b]$, where $v_{2m-1}$ is the commutator of degree $2m-1$. $3^{\circ}$. By Latyshev’s lemma (see [8]), where $v_k$ is a right-normed commutator of degree $k$.$4^{\circ}$. From $2^{\circ}$ and $3^{\circ}$ it follows that, in the algebra $F^{(2m+1)}$, $5^{\circ}$. By Volichenko’s lemma (see [9]),
$$
\begin{equation*}
T^{(k)} T^{(3)} \subseteq T^{(k+2)},\qquad [v_k,T^{(3)}] \subseteq T^{(k+3)},
\end{equation*}
\notag
$$
where $v_k$ is a right-normed commutator of degree $k$.
$6^{\circ}$. Using the product theorem for a 3-generated algebra (see [7]) and Lemma 2 from [2], we have in the algebra $F^{(2m+1)}$
$$
\begin{equation*}
\begin{gathered} \, [c_{2m-2} c_2, [x,y]] \in [T^{(2m-1)}, [x,y]] \subseteq T^{(2m+1)} =0, \\ [[ c_{2m-3},x] c_2, [b,y]] \in [T^{(2m-1)}, [b,y]] \subseteq T^{(2m+1)} =0. \end{gathered}
\end{equation*}
\notag
$$
$7^{\circ}$. We claim that $[[c_{2m-3},x] [b,y],c_2]=0$. Indeed, since $[b,y] c_2 = [b,y][a,b]\in T^{(3)}$, using $5^{\circ}$, $6^{\circ}$, and identity (2.1), we obtain in succession
$$
\begin{equation*}
[[c_{2m-3},x] [b,y],c_2]= [[c_{2m-3},x],[b,y] c_2] - [[c_{2m-3},x] c_2,[b,y]] = 0.
\end{equation*}
\notag
$$
$8^{\circ}$. Let us now transform the element
$$
\begin{equation*}
\begin{aligned} \, p_{2m} [x,y] &= [c_{2m-2},c_2] [x,y] = [c_{2m-2},c_2 [x,y]] - c_2 [c_{2m-2}, [x,y]] \\ &\!\!\stackrel{\text{by }4^{\circ}}{=} [c_{2m-2},c_2 [x,y]] \stackrel{(2.1)}{=} [c_{2m-2} c_2, [x,y]] + [[x,y] c_{2m-2},c_2] \\ &\!\!\stackrel{\text{by }6^{\circ}}{=} [c_{2m-2} [x,y],c_2] \stackrel{\text{by }3^{\circ}}{=} -[[c_{2m-3},x] [b,y],c_2]\stackrel{\text{by }7^{\circ}}{=} 0. \end{aligned}
\end{equation*}
\notag
$$
This proves Theorem 3. § 5. Commutator identities in 2 variables of degree $6$ of the algebra $E^{(3)}$Lemma 5. Let $w(a,b)$ be a homogeneous Lie polynomial in two variables $a$, $b$ of degree $6$. If $w(a,b)=0$ is an identity of the algebra $E^{(3)}$, then it is a consequence of the commutator identity: where $c_k=[c_{k-1},b]$, $k\geqslant 3$, $c_2=[a,b]$. Proof. By Theorem 3, identity (5.1)holds in the algebra $E^{(3)}$. The proof is in several steps.
$1^{\circ}$. Let $v=v(a,b)$ be a commutator (Lie monomial) and $\operatorname{deg}_a v=2$, $\operatorname{deg}_b v=4$. Then the element $v$ is proportional to the commutator $v_0=[[a,b]_4,a]$. Indeed, taking into account (2.3) and (5.1), we have
$$
\begin{equation}
[c_3,[b,a],b] = [c_3,[[b,a],b]] + [c_3,b,[b,a]] = -[c_3,c_3] - [c_4,c_2] =0,
\end{equation}
\tag{5.2}
$$
and hence, by equalities (2.3), (5.1) and (5.2),
$$
\begin{equation*}
v_0 = [c_4,b,a] = [c_4,a,b] = [c_3,a,b,b] + [c_3,[b,a],b] = [c_2,b,a,b,b] = [c_2,a,b,b,b].
\end{equation*}
\notag
$$
$2^{\circ}$. Let us prove that the identities hold in the algebra $E^{(3)}$. Indeed, by linearizing we have from (5.1)
$$
\begin{equation*}
0=[a,b,a,b,[a,b]] + [a,b,b,a,[a,b]] = 2[a,b,a,b,[a,b]].
\end{equation*}
\notag
$$
$3^{\circ}$. The space generated by commutators in $2$-variables (of degree $3$ in each of them) is $2$-dimensional with a basis of the elements Using (2.3) and (5.3) we obtain in succession
$$
\begin{equation*}
\begin{gathered} \, [c_3,a] = [a,b,b,a] = [a,b,a,b], \\ u:=[c_3,[a,b],a] = [c_3,[[a,b],a]] + [c_3,a,[a,b]] = - u_2, \\ v=[c_3,a,b,a]=[a,b,a,b,b,a] = [a,b,a,b,a,b]=u_1, \\ w:=[c_3,b,a,a] = [c_3,a,b,a] + [c_3,[b,a],a] = v - u = u_1 + u_2, \\ t:=[c_2,a,[a,b],b] = [c_2,a,[[a,b],b]]= -[b,a,a,[[a,b],b]] =u_2, \\ z:=[c_2,a,a,b,b] = [c_2,a,b,a,b] + [c_2,a,[a,b],b] = u_1 + t = u_1 + u_2. \end{gathered}
\end{equation*}
\notag
$$
The argument for the other elements is similar. Before proving the linear independence of the elements $u_1$ and $u_2$, some special calculations are required.
$4^{\circ}$. Assuming $\theta_{12} = 0$, check the equality: Indeed, since $[e_1,e_2] = 2e_1e_2$, we have
$$
\begin{equation*}
[e_1,e_2,e_2] = 2[e_1e_2,e_2] = 2[e_1,e_2]e_2 = 4 e_1e_2^2= 2 e_1 \theta_{22}.
\end{equation*}
\notag
$$
$5^{\circ}$. Let us now prove that the elements $u_1$ and $u_2$ are linearly independent. To this end, we calculate the elements $u_1(a,b)=[a,b,a,b,a,b]$ and $u_2(a,b)=[a,b,b,[b,a,a]]$ with $a= e_1$, $b=e_2$ and the condition $\theta_{12} = 0$. By (5.5),
$$
\begin{equation*}
\begin{gathered} \, u_1(e_1,e_2) = [e_1,e_2,e_1,e_2,e_1,e_2] = [e_1,e_2,e_2,e_1,e_1,e_2] = 2\theta_{11} [e_1,e_1,e_1,e_2] = 0, \\ u_2(e_1,e_2) = [e_1,e_2,e_2,[e_2,e_1,e_1]] = 2 \theta_{22}[e_1, 2\theta_{11}e_2] = 8\theta_{11}\theta_{22} e_1e_2 \neq 0. \end{gathered}
\end{equation*}
\notag
$$
We now claim that $u_1\neq 0$. On the contrary, if $u_1= 0$, then by linearizing $a\to b$, we obtain the equality for the $a$-linear component which is impossible by (5.5), because
$$
\begin{equation*}
[e_1,e_2]_5 = 2 \theta_{22} [e_1,e_2]_3 = 4 \theta_{22}^2 [e_1,e_2] = 8 \theta_{22}^2 e_1e_2\neq 0.
\end{equation*}
\notag
$$
Lemma 5 is proved. § 6. The Canonical algebra $C^{(m)}$ of multiplicity $m$Since it is rather difficult to perform calculations in the model algebra $E^{(m)}$, we introduce a simpler object — the canonical algebra $C^{(m)}$ of multiplicity $m$. This algebra is obtained by factorizing the model algebra $E^{(m)}$ with respect to the ideal generated by the elements $\theta_{ij}$, $i\neq j$. In what follows, we assume that $e_i^2 = \theta_i$. Thus, the canonical algebra $C^{(m)}$ is generated by the elements $e_i$, $\theta_j$, $i, j \in \mathbb{N}$, and satisfies the defining relations
$$
\begin{equation*}
e_i \circ e_j = 0, \quad i\neq j,\qquad e_i^2 = \theta_i,\qquad [\theta_i, e_j] = 0, \qquad \prod_{j=1}^m \theta_{i_j} =0.
\end{equation*}
\notag
$$
Moreover, the additive basis of this algebra consists of elements of the form where $v(\theta_j)$ is a standard normalized monomial on the indicated variables of degree $m$, and $1\leqslant i_1 <\dots<i_n$. Let us prove that the algebras $E^{(m)}$ and $C^{(m)}$ cannot be distinguished by any multilinear identity. Proof. Let $m=3$ (the argument in the general case follows the same lines). It suffices to show that if the multilinear polynomial $f(x_1,\dots,x_n)$ is not an identity in $E^{(3)}$, then it is also non-zero in the algebra $C^{(3)}$. A monomial of the form (6.1) is called normalized if its $\theta$-coefficient $v(\theta_j)$ is $1$. If $f(x_1,\dots,x_n)\neq 0$ in $E^{(3)}$, then there exist basis elements $w_i$ of the form (2.4) such that $f(w_1,\dots,w_n)\neq 0$. Without loss of generality we can assume that any the variables $e_k$ is not included in the two basic elements $w_i$, $i=1,\dots,n$. We have the representation
$$
\begin{equation*}
f(w_1,\dots,w_n) = \sum \beta_{ij,kl} \theta_{ij} \theta_{kl} \widehat{w}_{ijkl},
\end{equation*}
\notag
$$
where $\beta_{ij,kl}\in \Phi^{\times}=\Phi\setminus \{0\}$, $\widehat{w}_{ijkl}$ are normalized monomials of the form (2.4) not containing the characters $e_i$, $e_j$, $e_k$, $e_l$.
Of course, it may turn out that this representation does not involve some factor $\theta$. However, this case can easily be reduced to the one under consideration by multiplying both parts by $\theta_{ij}$ with suitable indices $i$ and $j$. Note that the normalized monomial $\widehat{w}_{ijkl}$ may arise only together with the following products: $\theta_{ij} \theta_{kl}$, $\theta_{ik} \theta_{jl}$, $\theta_{il} \theta_{jk}$. Let $\varphi$ denote the endomorphism of the algebra $E^{(3)}$ translating $e_j\mapsto e_i$, $e_l\mapsto e_k$ and fixing the other generators motionless (such an endomorphism exists because the images satisfy the same defining relations as for the original generators). We have
$$
\begin{equation*}
\theta_{ij}^ \varphi = (e_i\circ e_j)^\varphi = e_i\circ e_i = \theta_{ii},
\end{equation*}
\notag
$$
and hence
$$
\begin{equation*}
(f(w_1,\dots,w_n))^{\varphi} = \sum(\beta_{ij,kl} \theta_{ii} \theta_{kk} + \gamma \theta_{ik}^2) \widehat{w}_{ijkl}.
\end{equation*}
\notag
$$
So, in the canonical algebra $C^{(3)}$, we have
$$
\begin{equation*}
(f(w_1,\dots,w_n))^{\varphi} = 4\beta_{ij,kl} \theta_{i} \theta_{k}\widehat{w}_{ijkl} \neq 0.
\end{equation*}
\notag
$$
This proves Proposition 1. § 7. Proper polynomials in 2 variables of degree $5$ that are not identities in $E^{(3)}$ Proof. Indeed, we have $[e_1,e_2] = 2 e_1 e_2$ and so we get
$$
\begin{equation*}
\begin{gathered} \, [e_1,[e_1,e_2]] = 2[e_1,e_1e_2] = 2e_1[e_1,e_2]= 4 \theta_{1}e_2, \\ [e_2,[e_1,e_2]] = -[e_2,[e_2,e_1]] = - 4 \theta_2e_1, \qquad [[e_1, [e_1,e_2], [e_1,e_2]] = - 16 \theta_{1}\theta_2e_1 \neq 0. \end{gathered}
\end{equation*}
\notag
$$
This proves Lemma 6. Proof. Assume on the contrary that the identity $[[a,b]^2,a,b]= 0$ holds in $C^{(3)}$. We have $[[a,b],[a,b^2],a]=0$. Hence its complete linearization also holds, where $S_3$ and $A_3$ are symmetric and alternating groups of degree $3$, respectively.
Let $e_1$, $e_2$, $g=e_3$, $h=e_4$ be the generating elements of the algebra $C^{(3)}$. Let us calculate the value of the polynomial $f$ at the point:
$$
\begin{equation*}
a_1=e_1e_2,\quad a_2=e_1,\quad a_3=g,\qquad b_1=e_1e_2,\quad b_2=e_2,\quad b_3=h.
\end{equation*}
\notag
$$
First, the Jordan products of the elements $b_i$, $i=1,2,3$ are as follows:
$$
\begin{equation*}
\begin{gathered} \, b_1\circ b_2=(e_1e_2)\circ e_2 = e_1e_2e_2 + e_2e_1e_2=(e_1\circ e_2)e_2=0, \\ b_2\circ b_3=0,\qquad b_1\circ b_3=e_1e_2 h + h e_1e_2= 2he_1e_2. \end{gathered}
\end{equation*}
\notag
$$
The element $f$ has the form
$$
\begin{equation*}
\begin{aligned} \, f &= \sum_{\sigma\in S_3}\sum_{\tau\in A_3} [[a_{1\sigma},b_{1\tau}],[a_{2\sigma},b_{2\tau}\circ b_{3\tau}], a_{3\sigma}] \\ &=\sum_{\sigma\in S_3} [[a_{1\sigma},b_2],[a_{2\sigma},b_1\circ b_3], a_{3\sigma}] = 2\sum_{\sigma\in S_3} [[a_{1\sigma},e_2],[a_{\sigma},he_1e_2], a_{3\sigma}]. \end{aligned}
\end{equation*}
\notag
$$
Second, $[e_ie_j, e_k] =0$ if the indices $i$, $j$, $k$ are pairwise distinct. Hence
$$
\begin{equation*}
\begin{gathered} \, [a_1, he_1e_2] = [e_1e_2, he_1e_2] =0,\qquad [a_2, he_1e_2] = [e_1, he_1e_2] = -e_1[e_1,he_2]= 0, \\ [a_3, he_1e_2] = [g, he_1e_2] = 2gh e_1 e_2. \end{gathered}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\begin{aligned} \, f &= 2 \sum_{\sigma\in S_3} [[a_{1\sigma},e_2],[a_{2\sigma},he_1e_2], a_{3\sigma}] \\ &=2 \{[[a_1,e_2],[a_3,he_1e_2], a_2] + [[a_2,e_2],[a_3,he_1e_2], a_1]\} \\ &=4gh \{[[a_1,e_2], e_1e_2, a_2] + [[a_2, e_2], e_1e_2, a_1] \} \\ &=4gh \{[[e_1e_2,e_2], e_1e_2, e_1] + [[e_1, e_2], e_1e_2, e_1e_2] \} \\ &=4gh [[e_1e_2,e_2], e_1e_2, e_1]= 4gh\theta_2 [e_1, e_1e_2, e_1] \\ &=8gh \theta_1\theta_2 [e_2, e_1] = -16 \theta_1\theta_2 e_1e_2 gh \neq 0, \end{aligned}
\end{equation*}
\notag
$$
since by (6.1) the last element is a basic element. This proves Lemma 7. Proof. Let $f(a,b)=0$ be an identity of degree $5$ of the algebra $E^{(3)}$. Without loss of generality we can assume that $f(a,b)$ is a proper polynomial (see [10]). If $f$ is linear in $a$, then it is proportional to the commutator $[a,b,b,b,b]$, that is $f=\alpha [a,b,b,b,b]$. Since $[a,b,b,b,b]\neq 0$, we have $\alpha =0$.
Since the total degree of $f$ is $5$, it follows that $f$ can be considered quadratic in $a$. Hence using
$$
\begin{equation*}
[a,b,a,b]=[a,b,b,a],\qquad [a,b]\circ [a,b,b]=[[a,b]^2,b]
\end{equation*}
\notag
$$
for suitable scalars $\alpha, \beta, \gamma\in \Phi$, we have
$$
\begin{equation*}
f(a,b) = \alpha [a,b,b,a,b] + \beta [a,b,b,b,a] + \gamma [[a,b]^2,b] .
\end{equation*}
\notag
$$
Hence, after linearizing $a\to b$, we obtain $(\alpha + \beta) [a,b,b,b,b] = 0$, hence $\alpha + \beta = 0$ and the algebra $E^{(3)}$ satisfies the identity Here, by Lemma 5, we obtain $\gamma [[a,b]^2,b,b] = 0$. The linearization of $b\to a$ gives the identity $\gamma [[a,b]^2,a,b] = 0$. Taking into account Lemma 7, we get $\gamma = 0$ and $\alpha [a,b,b,[a,b]]=0$. Now, in view of Lemma 5, we have $\alpha =0$ and $f=0$. This proves Theorem 4. § 8. Proper identities in 2 variables of degree $6$ of the algebra $E^{(3)}$Theorem 5. Each proper identity of degree $6$ in two variables of the algebra $E^{(3)}$ is a consequence of the commutator identity (5.1). Proof. Let $f(a,b)$ be a homogeneous identity of degree $6$. By inspecting the proof of Theorem 4 we can assume that $\operatorname{deg}_a(f)\geqslant 2$. Let $f$ be quadratic in $a$. Then, by (5.1),
$$
\begin{equation*}
f = \alpha [a,b,b,a,b,b] + \beta [a,b,b,b,a,b] + \gamma [a,b,b,b][a,b] + \delta [a,b,b][a,b,b] = 0.
\end{equation*}
\notag
$$
By linearizing $a\to b$, we obtain Hence $\alpha + \beta = 0$, and so
$$
\begin{equation*}
\begin{aligned} \, 0 &= \alpha ([a,b,b,a,b,b] - [a,b,b,b,a,b]) + \gamma [a,b,b,b][a,b] + \delta [a,b,b][a,b,b] \\ &=\alpha [a,b,b,[a,b],b] + \gamma [a,b,b,b][a,b] + \delta [a,b,b][a,b,b] \\ &=\gamma [a,b,b,b][a,b] + \delta [a,b,b][a,b,b]. \end{aligned}
\end{equation*}
\notag
$$
An application of Lemma 5 shows that Since $[a,b,b]\circ [a,b,b] = [[a,b]^2,b,b] - [a,b,b,b]\circ [a,b]$, we have
Note that $[[a,b]^2,b,b] = [[a,b],[a,b^2],b] = 0$ if $b=e_2$. We claim that $[a,b,b,b]\circ [a,b]\neq 0$ for $b=e_2$. Indeed, if $a=e_1e_2 + g$, $b=e_2$ and $g=e_3$, then
$$
\begin{equation*}
\begin{aligned} \, [e_1e_2,b,b,b]\circ [g,b] + [g,b,b,b]\circ [e_1e_2,b] &=[e_1e_2,b,b,b]\circ [g,b] \\ &= 8\theta_2^2 e_1 \circ [g,b] = -32\theta_2^2 ge_1 e_2 \neq 0. \end{aligned}
\end{equation*}
\notag
$$
Hence, in view of (8.1), we have $\gamma = \delta = 0$. Thus, we have shown that if $\operatorname{deg}_a(f) = 2$, then $f=0$ is a consequence of identity (5.1).
Now let $\operatorname{deg}_a(f)= 3$. We can assume that $f=0$ has the form
$$
\begin{equation*}
g(a,b) + \gamma [a,b,a,b]\circ [a,b] + \delta [[a,b]^2,a,b] = 0,
\end{equation*}
\notag
$$
where $g(a,b)$ is a Lie polynomial. Linearizing $a\to b$, we obtain an identity of the form (8.1), which implies that $\gamma = \delta = 0$ and $g(a,b)=0$. However, by Lemma 5 the last identity is a consequence of (5.1). This proves Theorem 5. Proposition 2. Let $A$ be a subalgebra of $E^{(3)}$ generated by two elements $a,b$. Then an algebra $A$ satisfies the identity However, it does not satisfy any identity of degree $5$. Proof. Without loss of generality we can assume that $x=a$, $y=b$. Let $w=[a,b]$. Setting $T^{(3)}=T^{(3)}(A)$, we have We let $H'$ denote the $T$-ideal in $E$ generated by the weak Hall element $[w^2,a]$, let $\Theta_0$ denote the subalgebra in $E$ generated by the elements $\theta_{ij}$. Since the weak Hall element is an identity of the algebra $E^{(2)}$ (see [2]), we have
$$
\begin{equation}
H'\subseteq \Theta_0^2E,\qquad [H',[A,A]] \subseteq [\Theta_0^2E, [A,A]]\subseteq \Theta^3.
\end{equation}
\tag{8.3}
$$
Hence, for all $z,t,u,v\in A$, from (8.2) and (8.3) we have
$$
\begin{equation*}
\begin{aligned} \, [[x,y],[z,t],[u,v]] &\in [w,wA+T^{(3)},[u,v]] \subseteq [w,wA+T^{(3)},[u,v]] \\ &\subseteq [w,wA,[u,v]] \subseteq [w^2,A,[u,v]] \subseteq [H',[u,v]] = 0. \end{aligned}
\end{equation*}
\notag
$$
This proves Proposition 2. Let us show now that the kernel conjecture is valid in the algebra $F^{(7)}$ for elements of 2 variables. It is still unknown at present whether this conjecture is true in the general form even for the algebra $F^{(7)}$. Proposition 3. The central element $f(a,b)$ of the algebra $F^{(7)}$ is kernel if and only if $f(a,b)$ is an identity of the algebra $E^{(3)}$. Proof. The kernel element of the algebra $F^{(7)}$ is an identity of the algebra $E^{(3)}$. Conversely, let the element $f$ be central and $f=0$ be an identity of the algebra $E^{(3)}$. Using partial derivatives, we can assume that the element $f$ is proper. The element $f$ is central, and hence its weight is $\geqslant 6$ (see [7]). If its degree is $6$, then $f$ is a Lie polynomial. Hence by Lemma 5, the element $f$ is a consequence of the kernel element $p_6$ (see Theorem 3). If the degree of the proper polynomial $f$ is at least $7$, then the proof that $f$ is kernel is similar to that of Theorems 1–3. Proposition 3 is proved.
§ 9. On identities of the totality of all model algebrasProposition 4. There are no non-zero identities which hold in all model algebras. In particular, the algebra $E$ is not a $\mathrm{PI}$-algebra. Proof. The equality $\bigcap_m \Theta ^m = 0$ is true in the algebra $E$, and hence the algebra $E$ is approximated by the model algebras $E^{(m)}$. Hence, the algebra $E$ is a $\mathrm{PI}$-algebra if and only if all model algebras have some common identity.
The algebra $E$ has the following universal property. Let $V_n$ be a linear space over a field $\Phi$ with basis $b_1,\dots, b_n$ and let $f$ be a symmetric bilinear form on $V_n$. Then the mapping
$$
\begin{equation*}
e_m\mapsto b_m,\qquad \theta_{ij}\mapsto f(b_i,b_j),\quad \textit{where }\ m,i,j\leqslant n
\end{equation*}
\notag
$$
(the images of all other generators $e_m$, $\theta_{ij}$ are zero), can be extended to a homomorphism of the algebra $E$ onto the Clifford algebra $\mathrm{Cl}(V_n, f)$.
Since the algebra $\mathrm{Cl}(V_{2n}, f)$ for the non-degenerate form $f$ is central simple (see [11]), we have $\overline{\Phi} \otimes _{\Phi} \mathrm{Cl}(V_{2n}, f)$, where $\overline{\Phi}$ is the algebraic closure of the field $\Phi$, is a complete matrix algebra over the field $\overline{\Phi}$ of dimension $2^{2n}$, and, as it is well known, it contains no identities of degree $2^{2n}$ (see Ch. X in [12]). This proves that $E$ is not a $\mathrm{PI}$-algebra. Proposition 4 is proved. As an application of canonical algebras, we have the following result. Theorem 6. The model algebra $E^{(m)}$ of multiplicity $m$ does not satisfy any identity of degree $\leqslant m$. Proof. Assume that the algebra $E^{(m)}$ satisfies an identity of degree $m$. As usual, by $\operatorname{ad}(x)$ we denote the multiplication operator in the adjoint Lie algebra, that is $a\operatorname{ad}(x)= [a,x]$, $S_{k}$ is the permutation group, acting on the set $\{2,\dots,k\}$. Then the algebra $E^{(m)}$ satisfies the multilinear identity Assume first that the polynomial $f(x_1,\dots,x_m)$ is a Lie polynomial. It is known [13] that every Lie polynomial is a linear combination of the right-normed commutators, starting with the smallest variable $x_1$,
$$
\begin{equation*}
f(x_1,\dots,x_m) = \sum _{\sigma \in S_{m-1}} \alpha(\sigma) x_1 \operatorname{ad}(x_{2\sigma}) \cdots \operatorname{ad}(x_{m\sigma}),
\end{equation*}
\notag
$$
$\alpha(\sigma)\in \Phi$. Let us evaluate the polynomial $f(x_1,\dots,x_m)$ at the point. We have
$$
\begin{equation*}
x_1=e_1,\quad x_ 2=e_1e_2,\quad x_3 =e_2e_3,\quad\dots,\quad x_m = e_{m-1}e_m.
\end{equation*}
\notag
$$
Because the then only one term is non-zero, corresponding to the identical substitution $\varepsilon$. Therefore, we have a contradiction.
For the general case, we will employ the Specht basis free associative algebra (see [13]). Without loss of generality we can assume that the polynomial $f(x_1,\dots,x_m)$ is proper. Each proper polynomial can be represented as a linear combination of products of commutators $\pi = w_1w_2\cdots w_k$. Each commutator $w_i$ is right-normed and starts with the smallest variable included in this commutator; we denote this smallest variable by $x(w_i)$. In addition, in the product $\pi$, we have (the variables are arranged in order of increasing indices). The above products $\pi$ form an additive basis for the space of multilinear proper polynomials of a free associative algebra [13]. Let us show that the products $\pi$ are linearly independent on the canonical algebra $C^{(m)}$. Let the product
$$
\begin{equation*}
\pi = [x_1,x_2,\dots, x_{n_1}][y_1,y_2,\dots, y_{n_2}]\cdots [z_1,z_2,\dots, z_{n_k}]
\end{equation*}
\notag
$$
be such that $\sum_{i=1}^k n_i = m$ and $x_1<y_1<\dots<z_1$. In addition, the following restrictions hold: $x_1$ is the smallest variable, $y_1$ is the smallest variable among the variables included in all commutators except the first one, and so on. So, we will assume that $f(x_1,\dots,x_m)$ is a linear combination of products of the type $\pi$, where we also assume that the number of the factors $k$ is the greatest possible among all the products $\pi$, from $f$.
We assume that the set of generators of the algebra $C^{(m)}$ is the union of disjoint countable sets
$$
\begin{equation*}
\{e_1,e_2,\dots\},\quad \{g_1,g_2,\dots\},\quad \dots, \quad \{h_1,h_2,\dots\}.
\end{equation*}
\notag
$$
Consider the following change of variables
$$
\begin{equation*}
\begin{gathered} \, x_1=e_1,\quad x_ 2 =e_1e_2,\quad x_3 =e_2e_3,\quad \dots,\quad x_{n_1} = e_{n_1-1}e_{n_1}, \\ y_1=g_1,\quad y_2 =g_1g_2,\quad y_3 =g_2g_3,\quad\dots,\quad y_{n_2} = g_{n_2-1}g_{n_2}, \\ \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \\ z_1=h_1,\quad z_2 =h_1h_2,\quad z_3 =h_2h_3,\quad\dots,\quad z_{n_k} = h_{n_k-1}h_{n_k}. \end{gathered}
\end{equation*}
\notag
$$
Note that if $i$, $j$, $k$ are distinct, then $[e_i, e_je_k] = 0$ and $[e_i, e_ie_j] = 2 \theta_i e_j$. In addition, as noted earlier, the value of the commutator $[x_1,x_2,\dots, x_{n_1}]$ at this point is
$$
\begin{equation*}
2^{n_1-1}\biggl(\prod_{i=1}^{n_1-1}\theta_i\biggr)e_{n_1}.
\end{equation*}
\notag
$$
The values of the other commutators in the product $\pi$ have a similar form. Hence, the product $\pi$ at this point is obtained by multiplication of the basic element of the form
$$
\begin{equation*}
\biggl(\prod_{i=1}^{n_1-1}\theta_i\biggr)\biggl(\prod_{i=1}^{n_2-1}\lambda_i\biggr) \cdots \biggl(\prod_{i=1}^{n_k-1}\mu_i\biggr)e_{n_1}g_{n_2}\cdots h_{n_k}
\end{equation*}
\notag
$$
by some power of 2, where $g_i^2=\lambda_i$, $\dots$, $h_i^2=\mu_i$.
Note that if a permutation of variables takes place either within one commutator or from different commutators occurs in a $\pi$-type product, then the new product at this point will be equal to $0$. This proves Theorem 6. § 10. Open questionsIn conclusion, we formulate some open questions. 1. Is it true that the model algebra $E^{(m)}$ satisfies all the proper identities in $2$ variables of degree $\geqslant 2m+1$? 2. What is the minimum degree of an identity in 2 variables that holds in the model algebra $E^{(m)}$? 3. Does some identity of degree $2m-1$ hold in $E^{(m)}$ for $m\geqslant 3$? 4. Does the kernel conjecture hold for the algebra $F^{(7)}$?
Citation in format AMSBIB
\Bibitem{Pch23} |
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