Proof A classification of three-dimensional non-Lie Leibniz algebras over complex numbers was obtained by Rikhsiboev and Rakhimov in Theorem 2 of [Reference Rikhsiboev and Rakhimov35], where it was claimed that any algebra from $\mathbf {L}_3$ belongs to the following list: $\mathcal {L}_1$ , $\mathcal {L}_2^{\lambda }$ , $\mathcal {L}_3$ , $\mathcal {L}_4^{\lambda }$ , $\mathcal {L}_5$ , $\mathcal {L}_6$ , $\mathcal {L}_7^{\lambda }$ , $\mathcal {L}_8$ , $\mathcal {L}_9$ with $\lambda \in \mathbb {F}$ . Unfortunately, this classification is not complete:

• • The algebra $\mathcal {L}_2^0$ is a Lie algebra, therefore, we should assume that $\lambda $ is non-zero in case $\mathcal {L}_2^{\lambda }$ .

• • On page 361 of the proof of Theorem 2 of [Reference Rikhsiboev and Rakhimov35] it was claimed that $\mathcal {L}_{10}$ is isomorphic to $\mathcal {L}_6$ , which is not the case. Actually, as it follows from Corollary 6.5 (see below), $\mathcal {L}_{10}$ is not isomorphic to any other algebra from the formulation of Theorem 4.1.

• • The algebra $\mathcal {L}_{11}$ was missed in Case 2.1 of the proof of Theorem 2 of [Reference Rikhsiboev and Rakhimov35]. To show that $\mathcal {L}_{11}$ is not isomorphic to any other algebra from the formulation of Theorem 4.1, we note that the only commutative algebras from the formulation of Theorem 4.1 are $\mathcal {L}_{5}$ and $\mathcal {L}_{11}$ , but $\dim \mathrm {Ann}^{\mathrm {R}}(\mathcal {L}_{5})=1$ and $\dim \mathrm {Ann}^{\mathrm {R}}(\mathcal {L}_{11})=2$ , where the right annihilator of an algebra $\mathcal {A}$ is

  $$ \begin{align*}\mathrm{Ann}^{\mathrm{R}}(\mathcal{A}) =\{a\in \mathcal{A} \,|\, ba=0 \text{ for all } b\in\mathcal{A}\}.\end{align*} $$

To complete the proof, we will prove that the classification from Theorem 4.1 is equivalent to the computer-based classification of three-dimensional non-Lie Leibniz algebras over complex numbers by Casas, Insua, Ladra, and Ladra [Reference Casas, Insua, Ladra and Ladra5]. For $\delta \in \mathbb {F}$ , we denote by

(4.1) $$ \begin{align} \mathcal{L}_{2a}^{\delta},\; \mathcal{L}_{2b},\; \mathcal{L}_{2c},\; \mathcal{L}_{2d},\; \mathcal{L}_{2e}^{\delta}\; (\delta\neq0),\; \mathcal{L}_{2f},\; \mathcal{L}_{2g},\; \mathcal{L}_{3a}^{\delta}\; (\delta\in\widetilde{\mathbb{F}}),\; \mathcal{L}_{3b},\; \mathcal{L}_{3c},\; \mathcal{L}_{3d} \end{align} $$

the three-dimensional Leibniz algebras from page 3752 of [Reference Casas, Insua, Ladra and Ladra5], where

$$ \begin{align*}\widetilde{\mathbb{F}}= \{\alpha\in\mathbb{C}\,|\, |\alpha|> 1\} \cup \{\alpha\in\mathbb{C} \,|\, |\alpha| = 1,\; 0\leqslant\mathrm{arg}(\alpha)\leqslant \pi\}.\end{align*} $$

Note that $\pm 1\in \widetilde {\mathbb {F}}$ , and for every $\alpha \in \mathbb {F}\backslash \{0,\pm 1\},$ either $\alpha $ or $\alpha ^{-1}$ lies in $\widetilde {\mathbb {F}}$ . In [Reference Casas, Insua, Ladra and Ladra5] it was proven that any algebra from $\mathbf {L}_3$ is isomorphic to one and only one algebra from the list (4.1). The following isomorphisms complete the proof, where

• • for each line, the algebra from the first column is isomorphic to the algebra from the second column;

• • the tableau of multiplication of the algebra from the first column considered with respect to the ordered basis from the third column is equal to the tableau of multiplication of the algebra from the second column.

$$ \begin{align*} \begin{array}{l|l|l} \hline \mathcal{L}_1 & \mathcal{L}_{2g} & \{e_1, e_1+e_2, -e_2-e_3\}\\ \hline \mathcal{L}_2^{\lambda\neq0} & \mathcal{L}_{2e}^{-\lambda} & \{e_1, e_2, -e_3\}\\ \hline \mathcal{L}_{3} & \mathcal{L}_{2f} & \{e_1, e_2, -e_3\}\\ \hline \mathcal{L}_{4}^{\lambda\neq0} & \mathcal{L}_{2a}^{\lambda} & \{e_1, e_3, e_2\}\\ \hline \mathcal{L}_{4}^0 & \mathcal{L}_{2a}^0 & \{e_1, e_3, e_2\}\\ \hline \mathcal{L}_{5} & \mathcal{L}_{2c} & \{e_1, e_2, e_3\}\\ \hline \mathcal{L}_{6} & \mathcal{L}_{3a}^{-1} & \{e_1+e_2, e_2-e_1, -e_3\}\\ \hline \mathcal{L}_{7}^{\lambda\neq -\frac{1}{4},0} & \mathcal{L}_{3a}^{\lambda'} & \{\xi_1 e_1+e_2, \xi_2 e_1+e_2, \nu e_3\}\\ \hline \mathcal{L}_{7}^{-\frac{1}{4}} & \mathcal{L}_{3b} & \{e_2, -\frac{e_1}{2}+e_2, 2e_3\}\\ \hline \mathcal{L}_{7}^0 & \mathcal{L}_{2d} & \{e_2, e_1-e_2, e_3\}\\ \hline \mathcal{L}_{8} & \mathcal{L}_{3c} & \{e_1, e_2, e_3\}\\ \hline \mathcal{L}_{9} & \mathcal{L}_{3d} & \{2e_1+e_2, 2e_1+2e_2, e_1+e_3\}\\ \hline \mathcal{L}_{10} & \mathcal{L}_{3a}^1 & \{e_1, e_2, e_3\}\\ \hline \mathcal{L}_{11} & \mathcal{L}_{2b} & \{e_1, e_2, e_3\}\\ \hline \end{array}\\ \end{align*} $$

Here,

• • $\xi _1,\xi _2\in \mathbb {F}$ are pairwise different roots of the equation $x^2+x-\lambda =0$ , which are uniquely defined by the condition that $\lambda ':=\frac {\xi _2}{\xi _1}$ belongs to $\widetilde {\mathbb {F}}\backslash \{\pm 1\}$ ;

• • $\nu =\frac {2\xi _1+1}{\xi _1-2\lambda }$ .

It is straightforward to verify all cases but the case of $\mathcal {L}=\mathcal {L}_{7}^{\lambda }$ with $\lambda \in \mathbb {F}\backslash \{-1/4,0\}$ . Since the discriminant of the equation $x^2+x-\lambda =0$ is non-zero, then its roots $\xi _1,\xi _2$ are pairwise different. Note that

(4.2) $$ \begin{align} \xi_1 \xi_2 = -\lambda \;\text{ and }\; \xi_1+\xi_2=-1. \end{align} $$

Since $\lambda \neq -1/4$ , we have

(4.3) $$ \begin{align} \xi_1\neq 2\lambda \;\text{ and }\; \xi_2\neq 2\lambda. \end{align} $$

Since $\xi _1$ and $\xi _2$ are non-zero, $\lambda '$ is well-defined and $\lambda '\neq 0$ . Permuting $\xi _1$ and $\xi _2$ , if necessary, we can assume that $\lambda '\in \widetilde {\mathbb {F}}$ . Obviously, if $\lambda '=-1$ , then $\xi _1=\xi _2$ ; a contradiction. Similarly, if $\lambda '=1$ , then $\lambda =-1/4$ ; a contradiction. Hence, $\lambda '\in \widetilde {\mathbb {F}}\backslash \{\pm 1\}$ .

Formulas (4.3) imply that $\nu $ is well-defined and $\nu \neq 0$ . Hence,

$$ \begin{align*}v_1=\xi_1 e_1+e_2,\;\; v_2=\xi_2 e_1+e_2,\;\; v_3=\nu e_3\end{align*} $$

is a basis for $\mathcal {L}$ .

Compute the tableau of multiplication for $\mathcal {L}$ with respect to the ordered basis $\{v_1,v_2,v_3\}$ :

• • Using formulas (4.2), it is not difficult to see that $\nu \lambda =\lambda ' \xi _1$ and $\nu (\xi _1+1)=\lambda '$ . Hence,

  $$ \begin{align*}v_1v_3=\xi_1\nu e_2 + \nu (\lambda e_1+e_2) = \nu \lambda e_1 + \nu (\xi_1+1) e_2 = \lambda' \xi_1 e_1 + \lambda' e_2 = \lambda' v_1.\end{align*} $$

• • Using formulas (4.2), it is not difficult to see that $\nu \lambda = \xi _2$ and $\nu (\xi _2+1)=1$ . Hence,

  $$ \begin{align*}v_2v_3=\xi_2\nu e_2 + \nu (\lambda e_1+e_2) = \nu \lambda e_1 + \nu (\xi_2+1) e_2 = \xi_2 e_1 + e_2 = v_2.\end{align*} $$

• • Others products $v_i v_j$ are zero.

Therefore, the tableau of multiplication for $\mathcal {L}$ with respect to the ordered basis $\{v_1,v_2,v_3\}$ is

Hence, $\mathcal {L}\simeq \mathcal {L}_{3a}^{\lambda '}$ . The following claim, which can be proven by straightforward computations, completes the proof:

$$ \begin{align*} & \kern-15pt \mathit{for\ every}\ \lambda'\in\widetilde{\mathbb{F}}\backslash\{\pm1\},\ \mathit{there\ exists\ a\ unique}\ \lambda\in\mathbb{F}\backslash\{0,-1/4\}\ \mathit{with}\ \lambda'=\frac{\xi_2}{\xi_1}, \\& \kern-15pt \mathit{where}\ \xi_1,\xi_2\in\mathbb{F}\ \mathit{are\ pairwise\ different\ roots\ of\ the\ equation}\ x^2+x-\lambda=0.\\[-37pt] \end{align*} $$

Proof Given $\mathcal {L}\in \mathbf {L}_3$ , consider an invertible linear map $g:\mathcal {L}\to \mathcal {L}$ , given by a matrix

with respect to the basis $\{e_1,e_2,e_3\}$ of $\mathcal {L}$ from Theorem 4.1, where ${\alpha _1,\ldots ,\alpha _9\in \mathbb {F}}$ . Then, g is an automorphism of $\mathcal {L}$ if and only if for each $1\leqslant i,j\leqslant 3,$ we have $g_{ij}:=g(e_ie_j)-g(e_i)g(e_j) = 0$ . It is straightforward to verify that matrices $g,g_i$ from the formulation of this theorem are automorphisms. So, we assume that g is an automorphism.

1. 1. Assume $\mathcal {L}=\mathcal {L}_1$ . Consequently considering $g_{13}=0$ , $g_{11}=0$ , $g_{23}=0,$ we obtain $\alpha _7 = 0$ , $\alpha _4 = 0$ , $\alpha _8 = 0$ , respectively. Since $\det (g)=\alpha _1 \alpha _5 \alpha _9$ , we have $\alpha _1\neq 0$ and consequently considering $g_{13}=0$ , $g_{22}=0$ , $g_{23}=0$ , $g_{33}=0,$ we obtain $\alpha _9 = 1$ , $\alpha _1 = \alpha _5^2$ , $\alpha _2 = \alpha _5 \alpha _6$ , $\alpha _3 = \alpha _6^2/2$ , respectively. The required statement is proven.

2. 2. Assume $\mathcal {L}=\mathcal {L}_2^{\lambda }$ for some non-zero $\lambda \in \mathbb {F}$ . Consequently considering $g_{13}=0$ , $g_{23}=0$ , $g_{32}=0,$ we obtain $\alpha _7 = 0$ , $\alpha _8 = 0$ , $\alpha _2 = 0$ , respectively. Since $\det (g)=\alpha _1 \alpha _5 \alpha _9\neq 0$ , we consequently consider $g_{31}=0$ , $g_{33}=0$ , $g_{13}=0$ to obtain $\alpha _4 = 0$ , $\alpha _3 = 0$ , $\alpha _9 = 1$ , respectively.

3. 3. Assume $\mathcal {L}=\mathcal {L}_3$ . Consequently considering $g_{11}=0$ , $g_{22}=0$ , $g_{23}=0$ , we obtain $\alpha _7 = 0$ , $\alpha _8 = 0$ , $\alpha _2 = 0$ , respectively. Since $\det (g)=\alpha _1 \alpha _5 \alpha _9\neq 0$ , we consequently consider $g_{13}=0$ , $g_{23}=0$ , $g_{33}=0$ to obtain $\alpha _4 = 0$ , $\alpha _9 = 1$ , $\alpha _1 = 1$ , respectively. The required statement is proven.

4. 4. Assume $\mathcal {L}=\mathcal {L}_4^{\lambda }$ for some $\lambda \in \mathbb {F}$ . Since $g_{22}=0$ , we have $\alpha _4 = 0$ and $\alpha _7 = 0$ . Considering equality $g_{22}=0$ once again, we can see that $\alpha _1 = \alpha _5^2 + \alpha _5 \alpha _8 + \lambda \alpha _8^2$ .

   1. (a) Let $\lambda \neq 0$ .

      1. (i) Assume $\alpha _6\neq 0$ . Then, equality $g_{32}=0$ implies that $\alpha _5 = -\alpha _8 (\alpha _6 + \lambda \alpha _9)/\alpha _6$ . Since $\det (g)\neq 0$ , we have $\alpha _6^2+\alpha _6 \alpha _9+\lambda \alpha _9^2\neq 0$ and $\alpha _8\neq 0$ . Thus, $g_{23}=0$ implies $\alpha _8 = -\alpha _6 / \lambda $ .

      2. (ii) Assume $\alpha _6= 0$ . Since $\det (g)\neq 0$ , we have $\alpha _5\neq 0$ and $\alpha _9\neq 0$ . Consequently considering $g_{32}=0$ and $g_{23}=0,$ we obtain that $\alpha _8 = 0$ and $\alpha _9 = \alpha _5$ , respectively.

   2. (b) Let $\lambda =0$ . Since $\det (g)\neq 0$ , we have $\alpha _5+\alpha _8\neq 0$ and $\alpha _5\neq 0$ . Consequently considering $g_{32}=0$ and $g_{23}=0,$ we obtain $\alpha _6 = 0$ and $\alpha _9 = \alpha _5 + \alpha _8$ , respectively. The required statement is proven.

5. 5. Assume $\mathcal {L}=\mathcal {L}_5$ . Equality $g_{22}=0$ implies $\alpha _4 = 0$ , $\alpha _7 = 0$ , and then equality $g_{33}=0$ implies $\alpha _1 = \alpha _6^2 + \alpha _9^2$ .

   1. (a) Let $\alpha _9\neq 0$ . Equality $g_{23}=0$ implies that $\alpha _8 = -\alpha _5 \alpha _6/\alpha _9$ . Since $\det (g)\neq 0$ , we have $\alpha _6^2+\alpha _9^2 \neq 0$ and $g_{22}=0$ implies that $\alpha _9 = \pm \alpha _5$ .

   2. (b) Let $\alpha _9= 0$ . Since $\det (g)=-\alpha _6^3 \alpha _8\neq 0$ , consequently considering $g_{23}=0$ , $g_{22}=0,$ we obtain $\alpha _5 = 0$ and $\alpha _8 = \pm \alpha _6$ , respectively. The required statement is proven.

6. 6. Assume $\mathcal {L}=\mathcal {L}_6$ . Consequently considering $g_{13}=0$ , $g_{23}=0,$ we obtain that ${\alpha _8 = 0}$ and $\alpha _7 = 0$ , respectively. Considering equality $g_{13}=0$ once again, we can see that $\alpha _5 = \alpha _1 \alpha _9$ and $\alpha _2 = \alpha _4 \alpha _9$ . Since $\det (g)=(\alpha _1^2 - \alpha _4^2) \alpha _9^2\neq 0$ , equality $g_{33}=0$ implies $\alpha _3 = \alpha _6 = 0$ .

   If $\alpha _9\neq \pm 1$ , then $g_{23}=0$ implies $\alpha _1=\alpha _4=0$ ; hence, $\det (g)=0$ ; a contradiction. Therefore, $\alpha _9 = \pm 1$ . The required statement is proven.

7. 7. Assume $\mathcal {L}=\mathcal {L}_7^{\lambda }$ for some $\lambda \in \mathbb {F}$ . Considering $g_{13}=0,$ we obtain $\alpha _8 = 0$ , ${\alpha _2 = \lambda \alpha _4 \alpha _9,}$ and $\alpha _5 = (\alpha _1 + \alpha _4) \alpha _9$ . Note that $\alpha _9\neq 0$ , since $\det (g)\neq 0$ . Hence, it follows from $g_{33}=0$ that $\alpha _6 = -\alpha _3$ .

   1. (a) Let $\lambda \neq 0$ . Consequently considering $g_{23}=0$ and $g_{33}=0$ , we obtain $\alpha _7 = 0$ and $\alpha _3 = 0$ , respectively.

      1. (i) Assume $\alpha _9 = -1$ . Since the characteristic of the field is zero, $g_{23}=0$ implies $\alpha _4 = 0$ and $\alpha _1 = 0$ . Thus, $\det (g)=0$ ; a contradiction.

      2. (ii) Assume $\alpha _9\neq \pm 1$ . Then, $g_{23}=0$ implies $\alpha _1 = -\alpha _4 \alpha _9/(1 + \alpha _9)$ . Considering $g_{23}=0$ once again, we obtain $\alpha _4(\alpha _9+\lambda (1+\alpha _9)^2)=0$ . Thus, $\det (g)=0$ ; a contradiction.

      Therefore, $\alpha _9 = 1$ .

   2. (b) Let $\lambda = 0$ . Note that $\det (g)=(\alpha _1+\alpha _4) \alpha _9 (-\alpha _3 \alpha _7+\alpha _1 \alpha _9)$ . Since $(\alpha _1+\alpha _4) \alpha _9 \neq 0$ , consequently considering $g_{11}=0$ and $g_{23}=0,$ we can see that $\alpha _7 = 0$ and $\alpha _9 = 1$ , respectively. The required statement is proven.

8. 8. Assume $\mathcal {L}=\mathcal {L}_8$ for some $\lambda \in \mathbb {F}$ . Equality $g_{13}=0$ implies that $\alpha _8 = 0$ , $\alpha _5 = \alpha _1 \alpha _9$ , $\alpha _2 = \alpha _7 \alpha _9$ . Hence, it follows from equality $g_{33}=0$ that $\alpha _7 = 0$ , $\alpha _4 = \alpha _3 \alpha _9$ , $\alpha _1 = \alpha _9^2$ . The required statement is proven.

9. 9. Assume $\mathcal {L}=\mathcal {L}_9$ . Equality $g_{33}=0$ implies that $\alpha _4 = \alpha _3 \alpha _9$ . Applying equality $g_{33}=0$ once again, we can see that $\alpha _7 = 0$ and $\alpha _1 = \alpha _9 (\alpha _3 + \alpha _9)$ . Then, equality $g_{13}=0$ implies $\alpha _8=0$ . Thus, $\alpha _9\neq 0$ , since $\det (g) = \alpha _9^2 (-\alpha _2 \alpha _3 + \alpha _5 (\alpha _3 + \alpha _9))$ . Consequently considering $g_{23}=0$ and $g_{13}=0,$ we obtain that $\alpha _2 = 0$ and $\alpha _5 = -\alpha _3 \alpha _9 + \alpha _3 \alpha _9^2 + \alpha _9^3$ , respectively.

   In case $\alpha _9 = -\alpha _3$ , we have $\det (g)=0$ ; a contradiction. Therefore, $\alpha _9\neq -\alpha _3$ and $g_{13}=0$ implies that $\alpha _9=1$ . The required statement is proven.

10. 10. Assume $\mathcal {L}=\mathcal {L}_{10}$ . Equality $g_{13}=0$ implies that $\alpha _7 = 0$ . Applying equality $g_{23}=0$ , we can see that $\alpha _8 = 0$ . Note that $\alpha _9\neq 0$ , since $\det (g)\neq 0$ . Hence, it follows from $g_{33}=0$ that $\alpha _3 = \alpha _6=0$ .

    1. (a) Let $\alpha _9\neq 1$ . Then, $g_{13}=0$ implies $\alpha _1 = \alpha _4 = 0$ ; hence, $\det (g)=0$ , a contradiction.

    2. (b) Let $\alpha _9=1$ . Then, the required statement holds.

11. 11. Assume $\mathcal {L}=\mathcal {L}_{11}$ . Equality $g_{11}=0$ implies that $\alpha _7 = 0$ . Applying equality $g_{22}=0$ , we can see that $\alpha _8 = 0$ . Hence, it follows from equality $g_{33}=0$ that $\alpha _4=0$ and $\alpha _1=\alpha _9^2$ . The required statement is proven.

Proof For each $\mathcal {L}\in \mathbf {L}_3$ , the group $\mathrm {Aut}(\mathcal {L})$ is the following affine variety:

$$ \begin{align*}\left \{ \Big(b_1,\ldots,b_d,f_1({\underline{b}} ),\ldots,f_r({\underline{b}} ) \Big)\,\Big|\, b_1,\ldots,b_d\in\mathbb{F} \text{ and } h_1({\underline{b}} )\neq0,\ldots, h_s({\underline{b}} )\neq0 \right\}\end{align*} $$

for some polynomials $f_1,\ldots ,f_r,h_1,\ldots ,h_s$ from $\mathbb {F}[x_1,\ldots ,x_d]$ with $r,s\geqslant 0$ , where ${\underline {b}} $ stands for $(b_1,\ldots ,b_d)$ . Then, the dimension of $\mathrm {Aut}(\mathcal {L})$ is d. Applying this reasoning together with Theorem 4.2, we complete the proof.

Proof Lemma 4.5 together with Remark 4.4 implies that for every $\lambda \in \mathbb {F}$ and $m>0,$ we have the following inclusions and equalities for algebras of invariants:

• • $I_m(\mathcal {L}_1)$ , $I_m(\mathcal {L}_2^{{\lambda \neq 0}})$ , $I_m(\mathcal {L}_6)$ , $I_m(\mathcal {L}_7)$ , $I_m(\mathcal {L}_{10})$ belong to $\mathbb {F}[z_1,\ldots ,z_m]$ ;

• • $I_m(\mathcal {L}_3)\subset \mathbb {F}[x_1,\ldots ,x_m,z_1,\ldots ,z_m]$ ;

• • $I_m(\mathcal {L}_4^{\lambda })=I_m(\mathcal {L}_5)=I_m(\mathcal {L}_8)={ I_m(\mathcal {L}_{11})}=\mathbb {F}$ .

Theorem 4.2 implies that $z_1,\ldots , z_m\in I_m(\mathcal {L})$ for every $\mathcal {L}$ from the following list: $\mathcal {L}_1$ , $\mathcal {L}_2^{{\lambda \neq 0}}$ , $\mathcal {L}_3$ , $\mathcal {L}_7$ , $\mathcal {L}_9$ , $\mathcal {L}_{10}$ . Thus, the statement of the theorem holds for algebras $\mathcal {L}_1$ , $\mathcal {L}_2^{{\lambda \neq 0}}$ , $\mathcal {L}_4^{\lambda }$ , $\mathcal {L}_5$ , $\mathcal {L}_7$ , $\mathcal {L}_8$ , $\mathcal {L}_{10}$ , $\mathcal {L}_{11}$ .

Consider the algebra $\mathcal {L}_3$ . The subspace $V=\mathbb {F}\text {-span}\{e_1,e_3\}$ is invariant with respect to $\mathrm { Aut}(\mathcal {L}_3)$ and the restriction of $\mathrm {Aut}(\mathcal {L}_3)$ to V is the set of all matrices

for all $\alpha \in \mathbb {F}$ . Then, applying Lemma 3.5, we prove the statement about $\mathcal {L}_3$ .

Consider the algebra $\mathcal {L}_6$ . Note that $z_r z_s \in I_m(\mathcal {L}_6)$ for all $1\leqslant r\leqslant s\leqslant m$ by Theorem 4.2. Then, a monomial in $z_1,\ldots ,z_m$ lies in $I_m(\mathcal {L}_6)$ if and only if it has even degree. Restricting the group $\mathrm {Aut}(\mathcal {L}_6)$ to $\mathbb {F}\text {-span}\{e_3\}$ and applying Lemma 3.4, we conclude the proof about $\mathcal {L}_6$ .

Consider the algebra $\mathcal {L}_9$ . Using formulas (5.1), we can see that $(x_r-y_r) z_s - z_r (x_s-y_s)$ lies in $I_m(\mathcal {L}_9)$ . For every $\alpha \in \mathbb {F}\backslash \{-1\},$ consider an automorphism

of $\mathcal {L}_9$ (see Theorem 4.2). For $e_1'=e_1-e_2$ , we have that

(5.2) $$ \begin{align} ge_1' = e_1' + \alpha e_3,\;\; g e_2 = \alpha e^{\prime}_1 + (\alpha+1) e_2,\;\; g e_3 = e_3. \end{align} $$

For $x_r'=x_r-y_r$ , we have that $\mathbb {F}[\mathcal {L}^m]=\mathbb {F}[x^{\prime }_r,y_r,z_r\,|\,1\leqslant r\leqslant m]$ , and until the end of this proof, we consider elements of $\mathbb {F}[\mathcal {L}^m]$ as linear combinations of monomials in $x^{\prime }_r$ , $y_r$ , $z_r$ .

For each $m\geqslant 0$ , denote by $\Omega _m$ the set of all monomials $w=w_1\dots w_m \in \mathbb {F}[\mathcal {L}^m]$ of multidegree $1^m$ for some $w_r\in \{x^{\prime }_r,y_r,z_r\}$ , where $1\leqslant r\leqslant m$ . Given $w\in \Omega _m$ , denote

$$ \begin{align*}\deg_{x'}(w) & = \deg_{x^{\prime}_1}(w) + \cdots + \deg_{x_m}(w), \\ \deg_{y}(w) & = \deg_{y_1}(w) + \cdots + \deg_{y_m}(w), \\ \deg_{z}(w) & = \deg_{z_1}(w) + \cdots + \deg_{z_m}(w).\end{align*} $$

As an example, for $m=3,$ we have $\deg _y(z_1 x^{\prime }_2y_3)=1$ and $\deg _y(x^{\prime }_1y_2y_3) =2$ . Introduce a complete order on $\Omega _m$ as follows: for $v=v_1\dots v_m$ and $w=w_1\dots w_m$ from $\Omega _m$ , we have $v< w$ if and only if

1. (a) $\deg _y(v)<\deg _y(w)$ , or

2. (b) $\deg _y(v)=\deg _y(w)$ and $\deg _x(v)<\deg _x(w)$ , or

3. (c) $\deg _y(v)=\deg _y(w)$ , $\deg _x(v)=\deg _x(w)$ , and there exists $1\leqslant r\leqslant m$ with

   • • $v_1=w_1, \ldots , v_{r-1}=w_{r-1}$ ,

   • • $(v_r,w_r)$ is one of the following pairs: $(x^{\prime }_r,y_r)$ , $(z_r,y_r)$ , $(z_r,x^{\prime }_r)$ .

As an example, for $m=3,$ we have

$$ \begin{align*}x_1' x_2' y_3 < z_1 y_2 y_3,\;\;\;\; x_1' x_2' y_3 < z_1 x^{\prime}_2 y_3,\;\; \text{ and }\;\; y_1 z_2 x^{\prime}_3 < y_1 x^{\prime}_2 y_3.\end{align*} $$

For a multilinear $f=\sum _j \beta _j w_j\in \mathbb {F}[\mathcal {L}^m]$ with non-zero $\beta _j\in \mathbb {F}$ and $w_j\in \Omega _m$ , we say that $w_{j_0}$ is the highest term for f, if there is no $j\neq j_0$ with $w_j>w_{j_0}$ . Note that the highest term for f always exists and is unique.

Consider a multilinear $f=\beta w + \sum _j \beta _j w_j\in I_m(\mathcal {L})$ with non-zero $\beta ,\beta _j\in \mathbb {F}$ and $w,w_j\in \Omega _m$ , where w is the highest term for f. Without loss of generality, we can assume that

$$ \begin{align*}w=x^{\prime}_1\dots x^{\prime}_r y_{r+1} \dots y_{s} z_{s+1} \dots z_m,\end{align*} $$

where $0\leqslant r\leqslant s\leqslant m$ . Formulas (5.1) and (5.2) imply that

$$ \begin{align*}g^{-1}f &= \beta (x_1' + \alpha z_1)\dots (x_r' + \alpha z_r) \big(\alpha x^{\prime}_{r+1} + (\alpha+1) y_{r+1}\big)\dots \big(\alpha x^{\prime}_s + (\alpha+1) y_s\big) z_{s+1} \dots z_m \\& \quad + \sum_j \beta_j g^{-1}w_j. \end{align*} $$

It is easy to see that w is also the highest term of $g^{-1}f$ with the coefficient $(\alpha +1)^{s-r}$ . Therefore, $(\alpha +1)^{s-r}=1$ for all $\alpha \in \mathbb {F}\backslash \{-1\}$ . Hence, $s=r$ , i.e., w does not contain letters $y_1,\ldots ,y_m$ . Hence, $f\in \mathbb {F}[x^{\prime }_r,z_r\,|\,1\leqslant r\leqslant m]$ .

Note that the subspace $V=\mathbb {F}\text {-span}\{e^{\prime }_1,e_3\}$ is invariant with respect to g and the restriction of g to V is

for all $\alpha \in \mathbb {F}$ . Since the coordinate ring of V is $\mathbb {F}[x^{\prime }_r,z_r\,|\,1\leqslant r\leqslant m]$ , applying Lemma 3.5, we conclude the proof of the statement about $\mathcal {L}_9$ .

Proof We take tableaux of multiplication for Leibniz algebras from Theorem 4.1. To calculate traces $\mathrm { tr}(\chi _r \chi _0)$ and $\mathrm {tr}(\chi _0 \chi _r),$ we apply equalities (6.1) and (6.2), respectively. To calculate the rest of the traces, we apply Proposition 3.1.

Proof We apply Proposition 3.1, Theorem 5.1, part (c) of Remark 2.1, and the fact that the algebra $I_m(\mathcal {L})$ has $\mathbb {N}^m$ -grading by multidegrees.

Proof For the algebras $\mathcal {L}_1$ , $\mathcal {L}_2^{{\lambda \neq 0}}$ , $\mathcal {L}_3$ , $\mathcal {L}_6$ , $\mathcal {L}_7^{\lambda }$ , the elements $(\cdots ((e_2 e_3) e_3) \cdots )e_3$ are non-zero. Similarly, $(\cdots ((e_1 e_3) e_3) \cdots )e_3\neq 0$ in $\mathcal {L}_9$ and in $\mathcal {L}_{10}$ . Since $(e_3 e_3)e_3=e_2\neq 0$ in $\mathcal {L}_8$ , we can see that $\mathrm { ncl}(\mathcal {L}_8)=4$ . The cases of $\mathcal {L}_4^{\lambda }$ , $\mathcal {L}_5$ , and $\mathcal {L}_{11}$ are trivial.

Proof The claim of the corollary follows from Proposition 6.3, Theorem 5.1, and Proposition 6.1.