Proof. Since $(\mathcal{L},[\cdot ,\cdot ],\alpha )$ is a left Hom–Leibniz algebra, then for all $x,y,z \in \mathcal{L}$ ,

\begin{align*} [[y,z],\alpha (x)]-[[y,x],\alpha (z)]-[\alpha (y),[z,x]]&=[[y,z],\alpha (x)]-[[y,x],\alpha (z)]-[[y,z],\alpha (x)]-[\alpha (z),[y,x]]\\ &= [[y,x],\alpha (z)]+[\alpha (z),[y,x]]). \end{align*}

Proof. Simple computation.

Proof. Let us assume that $\mathcal{L}^{-}$ is a regular Hom–Lie algebra, $(\mathcal{L}^{+})^{3}=\{0\},\,\,(\mathcal{L}^{+})^{2}\subseteq \mathcal{Z}(\mathcal{L}^{-})$ and $\{\mathcal{L},\mathcal{L}\}\subseteq Ann(\mathcal{L}^{+})$ . Then for all $x,y,z\in \mathcal{L}$ , we have $[\alpha (x),[y,z]]=\{\alpha (x),\{y,z\}\},\,[[x,y],\alpha (z)]=\{\{x,y\},\alpha (z)\}$ and $[\alpha (y),[x,z]]=\{\alpha (y),\{x,z\}\}$ . Therefore, $[\alpha (x),[y,z]]-[[x,y],\alpha (z)]-[\alpha (y),[x,z]]=\{\alpha (x),\{y,z\}\}-\{\{x,y\},\alpha (z)\}-\{\alpha (y),\{x,z\}\}=0$ because $\{\cdot ,\cdot \}$ is a Hom–Lie structure on $\mathcal{L}$ . Now, $[\alpha (x),[y,z]]+[[y,z],\alpha (x)]=\{\alpha (x),\{y,z\}\}+\{\{y,z\},\alpha (x)\}=0$ . We conclude that $(\mathcal{L},[\cdot ,\cdot ],\alpha )$ is a regular symmetric Hom–Leibniz algebra.

Proof. Assume that $\Omega \in Z^2_{\textrm {SLeib}}(\mathcal{L},\mathcal{V}))$ . Let $x,y,z\in \mathcal{L}$ , we have $\Omega (\alpha (x),[y,z])=-\Omega ([y,z],\alpha (x))$ if and only if $\Omega _{s}(\alpha (x),[y,z])=0$ . Consequently, $\Omega _{a}$ satisfies

\begin{equation*}\Omega _{a}(\alpha (x),[y,z])=\Omega _{a}([x,y],\alpha (z))+\Omega _{a}(\alpha (y),[x,z]),\,\forall x,y,z\in \mathcal{L}.\end{equation*}

Let $x,y,z\in \mathcal{L}$ . Since $\Omega (\alpha (x),[y,z]+[z,y])=0$ and $\Omega _{s}(\alpha (x),[y,z])=0$ , then $\Omega _{a}(\alpha (x),[y,z] +[z,y])=0$ . Consequently, the fact that $\Omega _{a}(\alpha (x),[y,z])=\Omega _{a}([x,y],\alpha (z))+\Omega _{a}(\alpha (y),[x,z])$ implies that $\Omega _{a}(\alpha (x),[y,z])+\Omega _{a}(\alpha (y),[z,x])+\Omega _{a}(\alpha (z),[x,y])=\Omega _{a}(\alpha (x),\{y,z\})+ \Omega _{a}(\alpha (y),\{z,x\})+\Omega _{a}(\alpha (z),\{x,y\})=0$ .

Conversely, if we suppose that $\Omega _{a}\in Z^2_{\textrm {Lie}}(\mathcal{L}^{-},\mathcal{V})$ and $\Omega _{s}(\alpha (x),[y,z])=0,\,\Omega _{a}(\alpha (x),y\bullet z)=0,\,\forall x,y,z\in \mathcal{L}$ , a simple computation proves that $\Omega \in Z^2_{\textrm {SLeib}}(\mathcal{L},\mathcal{V})$ .

Proof. $\mathcal{L}$ and $\mathcal{L}'$ are equivalent if and only if there exists $\Phi \,:\,\mathcal{L}\rightarrow \mathcal{L}'$ an isomorphism of symmetric Hom–Leibniz algebras such that $\Phi (x+u)=\Psi (x)+\mathfrak{q}(x)+\varphi (u),\,\forall x\in \mathcal{G},\,u\in \mathcal{V}$ , where $\Psi \,:\,\mathcal{G}\rightarrow \mathcal{G}, \,\mathfrak{q}\,:\,\mathcal{G}\rightarrow \mathcal{V}$ and $\varphi \,:\,\mathcal{V}\rightarrow \mathcal{V}$ are linear maps. Consequently, $\mathcal{L}$ and $\mathcal{L}'$ are equivalent if and only if for all $\bar {x}=x+u,\bar {y}=y+v\in \mathcal{L}$ , we have $\Phi \big (\bar {x}\star _{\Omega }\bar {y}\big )=\Phi (\bar {x})\star _{\Omega '}\Phi (\bar {y})$ and $\gamma \circ \Phi (\bar {x})=\Phi \circ \gamma (\bar {x})$ . So the fact that $\Phi \big (\bar {x}\star _{\Omega }\bar {y}\big )=\Phi (\bar {x})\star _{\Omega '}\Phi (\bar {y})$ entails that $\Psi (\{x,y\})+\mathfrak{q}(\{x,y\})+\varphi \big (\Omega (x,y)\big )=\{\Psi (x),\Psi (y)\}+\Omega '\big (\Psi (x),\Psi (y)\big )$ and the fact that $\gamma \circ \Phi (\bar {x})=\Phi \circ \gamma (\bar {x})$ leads to $\alpha \circ \Psi (x)+\xi \Big (\Psi (x)+\mathfrak{q}(x)+\varphi (u)\Big )=\Psi \circ \alpha (x)+\mathfrak{q}\big (\alpha (x)\big )+\varphi \big (\xi (\bar {x})\big )$ . Consequently, $\mathcal{L}$ and $\mathcal{L}'$ are equivalent if and only if

1. 1. $\Psi (\{x,y\})=\{\Psi (x),\Psi (y)\}$ and $\alpha \circ \Psi (x)=\Psi \circ \alpha (x)$ ;

2. 2. $\Omega '_{a}\big (\Psi (x),\Psi (y)\big )-\varphi \big (\Omega _{a}(x,y)\big )=\mathfrak{q}(\{x,y\})$ ;

3. 3. $\Omega '_{s}\big (\Psi (x),\Psi (y)\big )=\varphi \big (\Omega _{s}(x,y)\big )$ ;

4. 4. $\xi \Big (\Psi (x)+\mathfrak{q}(x)+\varphi (u)\Big )=\mathfrak{q}\big (\alpha (x)\big )+\varphi \big (\xi (\bar {x})\big )$ .

Proof. Suppose that $(\mathcal{L},[\cdot ,\cdot ],\alpha )$ is a left Hom–Leibniz algebra. Then for all $z\in \mathcal{L},\,i=[x,y]+[y,x]\in \mathcal{I}_{\mathcal{L}},\,\,[i,\alpha (z)]=0$ . So for all $u\in \mathcal{L},\,\,0= \mathfrak{B}([[x,y]+[y,x],\alpha (z)],\alpha (u))=\mathfrak{B}([x,y]+[y,x],[\alpha (z),\alpha (u)])= \mathfrak{B}(\alpha ([x,y]+[y,x]),[z,u])=\mathfrak{B}(\alpha (x),[\alpha (y),[z,u]]+[[z,u],\alpha (y)]),\,\forall x,y,z,u\in \mathcal{L}$ . The fact that $\mathfrak{B}$ is non-degenerate implies that $[\alpha (y),[z,u]]+[[z,u],\alpha (y)]=0,\,\forall y,z,u\in \mathcal{L}$ . Consequently, invoking Proposition 2.1, we infer that $(\mathcal{L},[\cdot ,\cdot ],\alpha )$ is a symmetric Hom–Leibniz algebra. By the same way, we prove the result for right Hom–Leibniz algebra.

Proof. ${(a)}\,$ Let $x\in \mathcal{U}^{\bot },\,y\in \mathcal{L},\,z\in \mathcal{U}$ . It is clear that $[\mathcal{U}^{\bot },\mathcal{L}]\subseteq \mathcal{U}^{\bot }$ and $[\mathcal{L},\mathcal{U}^{\bot }] \subseteq \mathcal{U}^{\bot }$ . Moreover, since $\alpha (\mathcal{U})\subseteq \mathcal{U}$ , then $\mathfrak{B}(\alpha (x),z)=\mathfrak{B}(x,\alpha (z))=0$ . It follows that $\mathcal{U}^{\bot }$ is an ideal of $\mathcal{L}$ . Further, $\mathfrak{B}([x,z],y)=\mathfrak{B}(x,[z,y])=0$ , since $\mathfrak{B}$ is non-degenerate, then $[x,z]=0$ . Thus, $[\mathcal{U}^{\bot },\mathcal{U}]={0}$ . The rest of the proof is clear.

Proof. We have $[\mathcal{L},\mathcal{Z}(\mathcal{L})]=[\mathcal{Z}(\mathcal{L}),\mathcal{L}]=\{0\}\subseteq \mathcal{Z}(\mathcal{L})$ . Further, let $x\in \mathcal{Z}(\mathcal{L})$ and $y,\, z\in \mathcal{L}$ , then the invariance and the symmetry of $\mathfrak{B}$ lead to $\mathfrak{B}([\alpha (x),y],z)=\mathfrak{B}(\alpha (x),[y,z])=\mathfrak{B}(x,\alpha ([y,z]))=\mathfrak{B}(x,[\alpha (y),\alpha (z)])=\mathfrak{B}([x,\alpha (y)],\alpha (z))=0$ . So, $[\alpha (x),y]=0$ . Similarly, we prove that $[y,\alpha (x)]=0$ .

Proof. Let $x,y\in \mathcal{L},\,i\in \mathcal{I}_{\mathcal{L}},\,u\in \mathcal{R}_{\mathcal{L}}$ and $z\in \mathcal{Z}(\mathcal{L})$ .

1. 1. It is clear that $\mathcal{I}_{\mathcal{L}}\subseteq [\mathcal{L},\mathcal{L}]$ . Moreover, since $[\alpha (y),i]=0$ , then $\mathfrak{B}([x,y],\alpha (i))=\mathfrak{B}(\alpha ([x,y]),i)=\mathfrak{B}(\alpha (x),[\alpha (y),i]) = 0$ . Consequently, $[\mathcal{L},\mathcal{L}]\subseteq \mathcal{I}_{\mathcal{L}}^{\bot }$ , because $\alpha (\mathcal{I}_{\mathcal{L}})\subseteq \mathcal{I}_{\mathcal{L}}$ .

2. 2. The assertion follows from Proposition 3.1 and Theorem 4.1.

3. 3. The fact that $\mathfrak{B}(x,[y,z])=\mathfrak{B}([x,y],z)=0$ , for all $x,y,z\in \mathcal{L}$ , proves the result.

4. 4. It is easy to see that $\mathcal{Z}(\mathcal{L})\subseteq \mathcal{R}_{\mathcal{L}}$ . So $\mathcal{I}_{\mathcal{L}}\subseteq \mathcal{R}_{\mathcal{L}}$ and $\mathcal{R}_{\mathcal{L}}^{\bot }\subseteq \mathcal{Z}(\mathcal{L})$ . Moreover, if $i=[x,y]+[y,x]\in \mathcal{I}_{\mathcal{L}}$ and $u\in \mathcal{R}_{\mathcal{L}}$ , then $\mathfrak{B}([x,y]+[y,x],u) = \mathfrak{B}(x,[y,u]+[u,y])=0$ . Therefore, $\mathcal{I}_{\mathcal{L}}\subseteq \mathcal{R}_{\mathcal{L}}^{\bot }$ .

Proof. The proof is the same as for Proposition 2.13 [Reference Benayadi and Hidri16].

Proof. Since $\beta (b)\cdot (a\cdot c)=\beta (a)\cdot (b\cdot c)=(a\cdot b)\cdot \beta (c),\,\forall a,b,c\in \mathscr{A}$ , then for all $x,y,z\in \mathcal{L}$ ,

\begin{align*} -[&\gamma (x\otimes a),[y\otimes b,z\otimes c]]+[[x\otimes a,y\otimes b],\gamma (z\otimes c)]+ [\gamma (y\otimes b),[x\otimes a,z\otimes c]]\\ &=-[\alpha (x),[y,z]]\otimes \beta (a)\cdot (b\cdot c)+[[x,y],\alpha (z)]\otimes (a\cdot b)\cdot \beta (c)+[\alpha (y),[x,z]]\otimes \beta (b)\cdot (a\cdot c)=0. \end{align*}

So $(\mathcal{L}\otimes \mathscr{A},[\cdot ,\cdot ],\gamma )$ is a left Hom–Leibniz algebra. Moreover, since $(b\cdot c)\cdot \beta (a)=(b\cdot a)\cdot \beta (c)=(a\cdot b)\cdot \beta (c)=\beta (a)\cdot (b\cdot c)$ , then

\begin{align*} [\gamma (x\otimes a),[y\otimes b,z\otimes c]]&+[[y\otimes b,z\otimes c],\gamma (x\otimes a)]\\ &=[\alpha (x),[y,z]]\otimes \beta (a)\cdot (b\cdot c)+[[y,z],\alpha (x)]\otimes (b\cdot c)\cdot \beta (a)=0. \end{align*}

So $(\mathcal{L}\otimes \mathscr{A},[\cdot ,\cdot ],\gamma )$ is a symmetric Hom–Leibniz algebra. Now, suppose that $(\mathcal{L},[\cdot ,\cdot ]_{\mathcal{L}},\alpha _{\mathcal{L}})$ is a Hom–Lie algebra. If $(\mathscr{A},\cdot ,\beta )$ is a commutative Hom–algebra, then $[x\otimes a,y\otimes b]+[y\otimes b,x\otimes a] =([x,y]+[y,x])\otimes a\cdot b= 0$ . So $(\mathcal{L}\otimes \mathscr{A},[\cdot ,\cdot ],\gamma )$ is a Hom–Lie algebra. Conversely, if $(\mathcal{L}\otimes \mathscr{A},[\cdot ,\cdot ],\gamma )$ is a Hom–Lie algebra, then $0=[x\otimes a,y\otimes b]+[y\otimes b,x\otimes a]=[x,y]\otimes (a\cdot b-b\cdot a)$ . Therefore, $(\mathscr{A},\cdot ,\beta )$ is a commutative Hom–algebra.

Proof. It is straightforward to show that $\mathscr{A}_{n}^{\alpha }$ is a commutative Hom–associative algebra. In addition, for any $k\geqslant 2,\,(\mathscr{A}_{n}^{\alpha })^{(k)}$ is spanned by the set $\{a_{2k-1},\ldots ,a_{n}\}$ . Therefore, if $n=2p$ , then $(\mathscr{A}_{n}^{\alpha })^{(p)}$ is spanned by the set $\{a_{n-1},a_{n}\}$ . So $(\mathscr{A}_{n}^{\alpha })^{(p+1)}=\{0\}$ . If $n=2p+1$ , then $(\mathscr{A}_{n}^{\alpha })^{(p+1)}$ is spanned by the set $\{a_{n}\}$ . Consequently, $(\mathscr{A}_{n}^{\alpha })^{(p+2)}=\{0\}$ .

Proof. Since $(\mathscr{A}_{n}^{\alpha })^{(2)}$ is spanned by the set $\{e_{3},\ldots ,e_{n}\}$ , so $\Omega ((\mathscr{A}_{n}^{\alpha })^{(2)},\mathscr{A}_{n}^{\alpha }) = \Omega (\mathscr{A}_{n}^{\alpha },(\mathscr{A}_{n}^{\alpha })^{(2)}) = \{0\}$ . Then $\Omega (a\cdot b,\alpha (c)) = \Omega (\alpha (a),b\cdot c),\,\forall a,b,c\in \mathscr{A}_{n}^{\alpha }$ . Further, $\Omega (a_{1},a_{2}) =\eta _{1}\neq 0=\Omega (a_{2},a_{1})$ . So $\Omega$ is not symmetric. The Proposition 4.7 yields that $(\overline {\mathscr{A}_{n}^{\alpha }},\bullet ,\tau )$ is a non-commutative Hom–associative Hom–LR algebra. Clearly, $(\overline {\mathscr{A}_{n}^{\alpha }})^{(2)}\subseteq (\mathscr{A}_{n}^{\alpha })^{(2)}+\Omega (\mathscr{A}_{n}^{\alpha },\mathscr{A}_{n}^{\alpha })e$ . Moreover, since $\Omega (\mathscr{A}_{n}^{\alpha },(\mathscr{A}_{n}^{\alpha })^{(2)})=\Omega ((\mathscr{A}_{n}^{\alpha })^{(2)},\mathscr{A}_{n}^{\alpha }) = \{0\}$ , then $(\overline {\mathscr{A}_{n}^{\alpha }})^{(3)}\subseteq (\mathscr{A}_{n}^{\alpha })^{(3)}$ . Therefore, for all $k\geqslant 3,\,(\overline {\mathscr{A}_{n}^{\alpha }})^{(k)}\subseteq (\mathscr{A}_{n}^{\alpha })^{(k)}$ , for $n\geqslant 3$ . So, according to Proposition 4.8, $\mathscr{A}_{n}^{\alpha }$ is nilpotent with index of nilpotency grater or equal to 3. We conclude that $\overline {\mathscr{A}_{n}^{\alpha }}$ is nilpotent and has the same index of nilpotency as $\mathscr{A}_{n}^{\alpha }$ .

Proof.

1. (1) Suppose that $\mathcal{L}$ is solvable of length $k$ . Since $(T^{*}_{\Omega }(\mathcal{L}))^{n}/\mathcal{L}^{*}\cong \mathcal{L}^{n}$ and $\mathcal{L}^{k}=\{0\}$ , we have $(T^{*}_{\Omega }(\mathcal{L}))^{k}\subseteq \mathcal{L}^{*}$ , which implies that $(T^{*}_{\Omega }(\mathcal{L}))^{k+1}=\{0\}$ . Therefore, $T^{*}_{\Omega }(\mathcal{L})$ is solvable of length $k$ or $k+1$ .

2. (2) Suppose that $\mathcal{L}$ is nilpotent of index of nilpotency $k$ . Since $(T^{*}_{\Omega }(\mathcal{L}))^{(n)}/\mathcal{L}^{*}\cong \mathcal{L}^{(n)}$ and $\mathcal{L}^{(k)}=\{0\}$ , we have $(T^{*}_{\Omega }(\mathcal{L}))^{(k)}\subseteq \mathcal{L}^{*}$ . Let $\xi \in (T^{*}_{\Omega }(\mathcal{L}))^{(k)}\subseteq \mathcal{L}^{*},\,\,y\in \mathcal{L}$ and $x_{1}+f_{1},\ldots ,x_{k-1}+f_{k-1}\in T^{*}_{\Omega }(\mathcal{L})$ , we have

   \begin{eqnarray*} [\ldots ,[\xi ,x_{1}+f_{1}]_{T^{*}_{\Omega }(\mathcal{L})},\ldots ,x_{k-1}+f_{k-1}]_{T^{*}_{\Omega }(\mathcal{L})}(y)&=&\xi \circ L_{x_{1}}\circ \ldots \circ L_{x_{k-1}}(y)\\ &=& \xi ([x_{1},\ldots ,[x_{k-1},y]_{\mathcal{L}}\ldots ]_{\mathcal{L}})\,\,\,\,\in \xi (\mathcal{L}^{(k)})\\ &=& \{0\}. \end{eqnarray*}
   This proves that $(T^{*}_{\Omega }(\mathcal{L}))^{\langle 2k-1\rangle }=\{0\}$ . Therefore, $T^{*}_{\Omega }(\mathcal{L})$ is nilpotent of index of nilpotency at least $k$ and at most $2k-1$ . Now consider the case of the trivial $\mathrm{T}^*$ -extension $T^{*}_{0}(\mathcal{L})$ of $\mathcal{L}$ . For $x_{i}+f_{i}\in T^{*}_{0}(\mathcal{L})$ , where $1\leqslant i\leqslant k$ , we have
   \begin{align*} [x_{1}+&f_{1},[x_{2}+f_{2},\ldots ,[x_{k-1}+f_{k-1},x_{k}+f_{k}]_{T^{*}_{0}(\mathcal{L})} \ldots ]_{T^{*}_{0}(\mathcal{L})}]_{T^{*}_{0}(\mathcal{L})}\\ &=[x_{1},[x_{2},\ldots , [x_{k-1},x_{k}]_{\mathcal{L}}\ldots ]_{\mathcal{L}}]_{\mathcal{L}} +\sum _{j=1}^{k} f_{j}\circ L_{([x_{j+1},[x_{j+2},\ldots , [x_{k-1},x_{k}]_{\mathcal{L}}\ldots ]_{\mathcal{L}}]_{\mathcal{L}})}\circ R_{x_{j-1}}\circ \ldots \circ R_{x_{1}}\\ &=0. \end{align*}
   Therefore, $(T^{*}_{0}(\mathcal{L}))^{(k)}=\{0\}$ .

Proof. $(i)\,$ Let $\Phi \,:\,T^{*}_{\Omega _{1}}(\mathcal{L})\rightarrow T^{*}_{\Omega _{2}}(\mathcal{L})$ be an isomorphism of Hom–Leibniz algebras such that $\Phi |_{\mathcal{L}^{*}}=id_{\mathcal{L}^{*}}$ and $\Phi (x)=x+\mathscr{P}(x),\,\forall x\in \mathcal{L}$ , where $\mathscr{P}\,:\,\mathcal{L}\rightarrow \mathcal{L}^{*}$ is a linear map. Then for all $x+f,y+g\in T^{*}_{\Omega _{1}}(\mathcal{L})$ , we have

\begin{align*} \Phi ([x+f,y+g]_{\Omega _{1}})&=\Phi ([x,y]_{\mathcal{L}}+\Omega _{1}(x,y)+f\circ L_{y}+g\circ R_{x})\\ &=[x,y]_{\mathcal{L}}+\mathscr{P}([x,y]_{\mathcal{L}})+\Omega _{1}(x,y)+f\circ L_{y}+g\circ R_{x}. \end{align*}

In the other hand,

\begin{align*} [\Phi (x+f),\Phi (y+g)]_{\Omega _{2}}&=[x+\mathscr{P}(x)+f,y+\mathscr{P}(y)+g]_{\Omega _{2}}\\ &=[x,y]_{\mathcal{L}}+\Omega _{2}(x,y)+\mathscr{P}(x)\circ L_{y}+f\circ L_{y}+\mathscr{P}(y)\circ R_{x}+g\circ R_{x}. \end{align*}

Since $\Phi$ is an isomorphism of Hom–Leibniz algebras, then (5.2) holds.

Moreover by $\gamma \circ \Phi =\Phi \circ \gamma$ , we may obtain $\mathscr{P}(x)\circ \alpha =\mathscr{P}(\alpha (x)),\,\forall x\in \mathcal{L}$ .

Conversely, suppose that there exists a linear map $\mathscr{P}\,:\,\mathcal{L}\rightarrow \mathcal{L}^{*}$ satisfying (5.2) and $\mathscr{P}(x)\circ \alpha =\mathscr{P}(\alpha (x)),\forall x\in \mathcal{L}$ , then we can define $\Phi :T^{*}_{\Omega _{1}}(\mathcal{L})\rightarrow T^{*}_{\Omega _{2}}(\mathcal{L})$ by $\Phi (x+f)=x+\mathscr{P}(x)+f$ . It is easy to prove that $\Phi$ is an isomorphism of Hom–Leibniz algebras. Thus, $T^{*}_{\Omega _{1}}(\mathcal{L})$ is equivalent to $T^{*}_{\Omega _{2}}(\mathcal{L})$ .

Now, consider the symmetric bilinear form $\mathfrak{B}_{\mathcal{L}}\,:\,\mathcal{L}\times \mathcal{L}\rightarrow \mathbb{K};\,(x,y)\mapsto \mathscr{P}_{s}(x)(y)$ induced by $\mathscr{P}_{s}$ . For all $x,y,z\in \mathcal{L}$ , note that

\begin{align*} \Omega _{2}(x,y)(z)-\Omega _{1}(x,y)(z)&=\mathscr{P}([x,y]_{\mathcal{L}})(z)-\mathscr{P}(x)\circ L_{y}(z)-\mathscr{P}(y)\circ R_{x}(z)\\ &=\mathscr{P}([x,y]_{\mathcal{L}})(z)-\mathscr{P}(x)([y,z]_{\mathcal{L}})-\mathscr{P}(y)([z,x]_{\mathcal{L}}) \end{align*}

and

\begin{align*} \Omega _{2}(y,z)(x)-\Omega _{1}(y,z)(x)&=\mathscr{P}([y,z]_{\mathcal{L}})(x)-\mathscr{P}(y)\circ L_{z}(x)-\mathscr{P}(z)\circ R_{y}(x)\\ &=\mathscr{P}([y,z]_{\mathcal{L}})(x)-\mathscr{P}(y)([z,x]_{\mathcal{L}})-\mathscr{P}(z)([x,y]_{\mathcal{L}}). \end{align*}

Since both $\Omega _{1}$ and $\Omega _{2}$ satisfy (5.1), then the right-hand sides of the above equations are equal. Hence,

\begin{equation*}\mathscr{P}([x,y]_{\mathcal{L}})(z)-\mathscr{P}(x)([y,z]_{\mathcal{L}})= \mathscr{P}([y,z]_{\mathcal{L}})(x)-\mathscr{P}(z)([x,y]_{\mathcal{L}}).\end{equation*}

Since ch $\mathbb{K}\neq 2$ , then $\mathfrak{B}_{\mathcal{L}}(x,[y,z])=\mathfrak{B}_{\mathcal{L}}([x,y],z)$ , which proves the invariance of the symmetric bilinear form $\mathfrak{B}_{\mathcal{L}}$ induced by $\mathscr{P}_{s}$ . Moreover,

\begin{align*} \mathfrak{B}_{\mathcal{L}}\big (\alpha (x),y\big )&=\frac {1}{2}\Big (\mathscr{P}(\alpha (x))(y)+\mathscr{P}(y)(\alpha (x))\Big )\\ &=\frac {1}{2}\Big (\mathscr{P}(x)(\alpha (y))+\mathscr{P}(\alpha (y))(x)\Big )=\mathfrak{B}_{\mathcal{L}}\big (x,\alpha (y)\big ). \end{align*}

Consequently, $\alpha$ is $\mathfrak{B}_{\mathcal{L}}$ -symmetric.

$(ii)$ Let the isomorphism $\Phi$ be defined as in (i). Then for all $x+f,y+g\in T^{*}_{\Omega _{1}}(\mathcal{L})$ , we have

\begin{align*} \mathfrak{B}(\Phi (x+f),\Phi (y+g))&=\mathfrak{B}(x+\mathscr{P}(x)+f,y+\mathscr{P}(y)+g)\\ &=\mathscr{P}(x)(y)+f(y)+\mathscr{P}(y)(x)+g(x)\\ &=2 \mathscr{P}_{s}(x)(y)+\mathfrak{B}(x+f,y+g). \end{align*}

Thus $\Phi$ is an isometry if and only if $\mathscr{P}_{s}=0$ .

Proof.

1. (i) $\Omega ([x,y],\alpha (z))+\Omega (\alpha (y),[x,z])-\Omega (\alpha (x),[y,z])=\mathfrak{B}(\alpha \circ \delta ([x,y])+[\delta \circ \alpha (y),x]-[\delta \circ \alpha (x),y],z)$ .

2. (ii) $\Omega (\alpha (x),[y,z])+\Omega ([y,z],\alpha (x))=\mathfrak{B}(y,[z,(\delta +\delta ^{*})\circ \alpha (x)]),\,\forall x,y,z\in \mathcal{L}$ .

Proof. By a simple computation, we can easily check that $(\widetilde {\mathcal{L}},\ast ,\beta )$ is a left Hom–Leibniz algebra verifying

\begin{equation*}\beta (\mathscr{X})\ast (\mathscr{Y}\ast \mathscr{Z})+ (\mathscr{Y}\ast \mathscr{Z})\ast \beta (\mathscr{X})=0\,\,\mbox{ for all}\,\mathscr{X},\mathscr{Y},\mathscr{Z}\in \widetilde {\mathcal{L}},\end{equation*}

if and only if the conditions $(p_{1})$ and $(p_{2})$ of (6.2) are satisfied. Moreover, $\beta (\mathscr{X}\ast \mathscr{Y})=\beta (\mathscr{X})\ast \beta (\mathscr{Y})$ if and only if the conditions $(p_{3})$ and $(p_{4})$ of (6.2) are verified.

Proof. The Proposition 6.1 entails that the map $\Phi \,:\,\mathcal{L}\times \mathcal{L}\rightarrow \mathcal{V}^{*};(x,y)\longmapsto \Phi (x,y)=\mathfrak{B}(\delta (x),y)d^{*}$ is a Leibniz 2-cocycle of $\mathcal{L}$ on the trivial $\mathcal{L}-$ module $\mathcal{V}^{*}$ . Then $\mathcal{L}_{1}=\mathcal{L}\oplus \mathcal{V}^{*}$ endowed with the product defined by

\begin{equation*}(x+f)\star _{\Omega }(y+g)=[x,y]_{\mathcal{L}}+\Phi (x,y)\,\,\textrm {for all} \, x,y\in \mathcal{L},\,f,g\in \mathcal{V}^{*},\end{equation*}

and the twisted map $\gamma \,:\,\mathcal{L}_{1}\rightarrow \mathcal{L}_{1}$ given by $\gamma (x)=\alpha (x)+\mathfrak{B}(a_{1},x)d^{*}$ and $\gamma (d^{*})=\lambda d^{*}$ is a regular symmetric Hom–Leibniz algebra, central extension of $\mathcal{L}$ by $\mathcal{V^{*}}$ by means of $\Phi$ .

Now, we define two endomorphisms $\delta _{1}$ and $\delta _{2}$ of $\mathcal{L}_{1}$ as follows:

\begin{equation*}\delta _{1}(x+\mu d^{*})=\delta (x)+\mathfrak{B}(x,a_{0})d^{*},\,\,\,\,\,\,\delta _{2}(x+\mu d^{*})=\delta ^{*}(x)+\mathfrak{B}(x,a_{0})d^{*},\,\forall x\in \mathcal{L},\,\mu \in \mathbb{K}.\end{equation*}

Let us consider $b_{0}=a_{0}+\lambda _{0}d^{*}$ and $b_{1}=a_{1}+\lambda _{1}d^{*}$ two elements of $\mathcal{L}_{1}$ , where $\lambda _{0},\lambda _{1}$ are two fixed scalars in $\mathbb{K}$ . So $b_{0}\in \mathcal{Z}(\mathcal{L}_{1})$ because $a_{0}\in \mathcal{Z}(\mathcal{L})$ and $\delta (a_{0})=\delta ^{*}(a_{0})=0$ .

Since the conditions (6.3) are satisfied, then a simple computation proves that $(\delta _{1},\delta _{2},b_{0},b_{1})$ is an admissible quadruple of $\mathcal{L}_{1}$ . Consequently, we can consider the regular symmetric Hom–Leibniz algebra $\overline {\mathcal{L}}=\mathcal{L}_{1}\oplus \mathcal{V}=\mathcal{V}^{*}\oplus \mathcal{L}\oplus \mathcal{V}$ , generalized semi-direct product of $\mathcal{L}_{1}$ by $\mathcal{V}$ by means of $(\delta _{1},\delta _{2},b_{0},b_{1})$ , provided with the product defined by (6.4) and the linear map given by (6.5). Furthermore, it is straightforward to demonstrate that the symmetric bilinear form $\overline {\mathfrak{B}}:\overline {\mathcal{L}}\times \overline {\mathcal{L}}\rightarrow \mathbb{K}$ given by $\overline {\mathfrak{B}}|_{\mathcal{L}\times \mathcal{L}}=\mathfrak{B},\,\overline {\mathfrak{B}}(d,d^{*})=1$ is an invariant scalar product on $\overline {\mathcal{L}}$ .

Proof. The proof is a straightforward but lengthy computation so we omit it.

Proof. Since $(\mathcal{L},[\cdot ,\cdot ],\alpha ,\mathfrak{B})$ is not a Hom–Lie algebra. Then $\{0\}\neq \mathcal{I}_{\mathcal{L}}\subseteq \mathcal{Z}(\mathcal{L})$ follows that there exist $e\in \mathcal{I}_{\mathcal{L}}\setminus \{0\}$ and $\lambda \in \mathbb{K}$ such that $\alpha (e)=\lambda e$ . Denote by $\mathcal{J}=\mathbb{K}e$ and $\mathcal{J}^{\bot }$ the orthogonal of $\mathcal{J}$ with respect to $\mathfrak{B}$ , then $\mathcal{J}\subseteq \mathcal{J}^{\bot }$ . Since $\mathfrak{B}$ is non-degenerate, then there exists $d\in \mathcal{L}\setminus \{0\}$ such that $\mathfrak{B}(e,d)=1$ and $\mathfrak{B}(d,d)=0$ . Let $\mathcal{V}=\mathbb{K}d$ and $\mathcal{H}=(\mathcal{J}\oplus \mathcal{V})^{\bot }$ , then $\mathcal{L}=\mathcal{J}\oplus \mathcal{H}\oplus \mathcal{V}$ and $\mathcal{J}^{\bot }=\mathcal{J}\oplus \mathcal{H}$ is an ideal of $\mathcal{L}$ with $\mathfrak{B}(e,e)=\mathfrak{B}(d,d)=0$ , $\mathfrak{B}(e,d)=1$ , and $\mathfrak{B}(\mathcal{H},e)=\mathfrak{B}(\mathcal{H},d)=\{0\}$ .

There exist $\lambda _{0},\gamma _{0}\in \mathbb{K},\,a_{0}\in \mathcal{H}$ such that $[d,d]=\lambda _{0} e+a_{0}+\gamma _{0} d$ and there exist $\lambda _{1},\gamma _{1}\in \mathbb{K},\,a_{1}\in \mathcal{H}$ such that $\alpha (d)=\lambda _{1}e+a_{1}+\gamma _{1} d.$ The fact that $\mathcal{J}^{\bot }$ is an ideal of $\mathcal{L}$ implies that

1. (1) There exist a bilinear map $[\cdot ,\cdot ]_{\mathcal{H}}\,:\,\mathcal{H}\times \mathcal{H}\rightarrow \mathcal{H}$ and a bilinear form $\phi \,:\,\mathcal{H}\times \mathcal{H}\rightarrow \mathbb{K}$ such that

   \begin{equation*}[x,y]=\phi (x,y)e+[x,y]_{\mathcal{H}},\,\forall x,y\in \mathcal{H}.\end{equation*}
   Moreover, for any $x\in \mathcal{H},\,[x,d]=\varphi (x)e+D(x)$ and $[d,x]=\psi (x)e+\delta (x)$ , where $D, \, \delta \in End(\mathcal{H})$ and $\varphi ,\, \psi \in \mathcal{H}^{*}$ .

2. (2) There exist a linear map $\alpha _{\mathcal{H}}\,:\,\mathcal{H}\rightarrow \mathcal{H}$ and a linear form $f\,:\,\mathcal{H}\rightarrow \mathbb{K}$ such that $\alpha (x)=f(x)e+\alpha _{\mathcal{H}}(x).$

It is easy to show that $(\mathcal{H},[\cdot ,\cdot ]_{\mathcal{H}},\alpha _{\mathcal{H}})$ is a regular symmetric Hom–Leibniz algebra and that the bilinear form $\mathfrak{B}_{\mathcal{H}}=\mathfrak{B}|_{\mathcal{H}\times \mathcal{H}}$ is an invariant scalar product on $(\mathcal{H},[\cdot ,\cdot ]_{\mathcal{H}},\alpha _{\mathcal{H}})$ . Therefore, the bilinear map $\phi \in Z^2_{\textrm {SLeib}}(\mathcal{H},\mathbb{K})$ . Let $x,y\in \mathcal{H}$ , then $\phi (x,y)=\mathfrak{B}([x,y],d)=\mathfrak{B}(x,[y,d])=\mathfrak{B}(y,[d,x])$ . It follows that $\mathfrak{B}(D(y),x)=\mathfrak{B}(y,\delta (x))$ . So $D=\delta ^{*}$ . Then $[x,y]=[x,y]_{\mathcal{H}}+\mathfrak{B}(\delta (x),y)e$ . Moreover, $\varphi (x)=\mathfrak{B}([x,d],d)=\mathfrak{B}(d,[d,x])=\psi (x)=\mathfrak{B}(x,a_{0})$ and $\gamma _{0}=\mathfrak{B}([d,d],e)=\mathfrak{B}(d,[d,e])=0$ . Then $[d,d]=\lambda _{0} e+a_{0}$ . Consequently, $a_{0}\in \mathcal{Z}(\mathcal{L})$ . So $\mathfrak{B}(a_{0},a_{0})=0$ and $\delta (a_{0})=D(a_{0})=0$ . In addition, for all $x\in \mathcal{H}$ we have, $[a_{0},x]_{\mathcal{H}}=[a_{0},x]-\mathfrak{B}(\delta (a_{0}),x)e=0,\,[x,a_{0}]_{\mathcal{H}}=[x,a_{0}]-\mathfrak{B}(x,D(a_{0}))e=0$ . Then $a_{0}\in \mathcal{Z}(\mathcal{H})$ .

Let $x,y\in \mathcal{H}$ , then the fact that $\mathfrak{B}(\alpha (x),y)=\mathfrak{B}(x,\alpha (y))$ implies that $\mathfrak{B}(\alpha _{\mathcal{H}}(x),y)=\mathfrak{B}(x,\alpha _{\mathcal{H}}(y))$ . Moreover, $\mathfrak{B}(\alpha (d),x)=\mathfrak{B}(d,\alpha (x))$ which is equivalent to $\mathfrak{B}(a_{1},x)=\mathfrak{B}(d,\alpha _{\mathcal{H}}(x)+f(x)e)=f(x).$ Then $f(x)=\mathfrak{B}(a_{1},x)$ . It follows that $\alpha (x)=\alpha _{\mathcal{H}}(x)+\mathfrak{B}(a_{1},x)e$ . Also $\gamma _{1}=\mathfrak{B}(\alpha (d),e)=\mathfrak{B}(d,\alpha (e))=\lambda$ . Then $\alpha (d)=\lambda _{1}e+a_{1}+\lambda d$ . Since $(\mathcal{L},[\cdot ,\cdot ],\alpha )$ is a left Hom–Leibniz algebra satisfying $[\alpha (\mathscr{X}),[\mathscr{Y},\mathscr{Z}]]+[[\mathscr{Y},\mathscr{Z}],\alpha (\mathscr{X})]=0,\,\forall \mathscr{X},\mathscr{Y},\mathscr{Z}\in \mathcal{L}$ , then a simple computation proves that

\begin{eqnarray*} &\,& D=\delta ^{*},\,\delta ^{2}+D\circ \delta =0,\,\delta \circ D=D\circ \delta ,\,(\delta +D)([\mathcal{H},\mathcal{H}]_{\mathcal{H}})=\{0\},\\ &\,& \delta (\mathcal{H})\subseteq \mathcal{R}_{\mathcal{H}},\,\delta ^{*}(\mathcal{H})\subseteq \mathcal{R}_{\mathcal{H}},\,\textrm {where}\,\mathcal{R}_{\mathcal{H}}=\{x\in \mathcal{H}/\,[x,y]_{\mathcal{H}}+[y,x]_{\mathcal{H}}=0,\,\forall y\in \mathcal{H}\}. \end{eqnarray*}

Moreover, for all $x,y\in \mathcal{H}$ , we have

\begin{align*} 0&=[[d,x],\alpha (y)]+[\alpha (x),[d,y]]-[\alpha (d),[x,y]]\\ &=[\delta (x),\alpha _{\mathcal{H}}(y)]+ [\alpha _{\mathcal{H}}(x),\delta (y)]-[a_{1},[x,y]_{\mathcal{H}}]-\lambda [d,[x,y]_{\mathcal{H}}]\\ &=[\delta (x),\alpha _{\mathcal{H}}(y)]_{\mathcal{H}}+\mathfrak{B}(\delta ^{2}(x),\alpha _{\mathcal{H}}(y))e+ [\alpha _{\mathcal{H}}(x),\delta (y)]_{\mathcal{H}}+\mathfrak{B}(\delta \circ \alpha _{\mathcal{H}}(x),\delta (y))e\\ &-[a_{1},[x,y]_{\mathcal{H}}]_{\mathcal{H}}-\mathfrak{B}(\delta (a_{1}),[x,y]_{\mathcal{H}})e-\lambda \delta ([x,y]_{\mathcal{H}}). \end{align*}

Therefore,

(7.1) \begin{equation} [a_{1},[x,y]_{\mathcal{H}}]_{\mathcal{H}}+\lambda \delta ([x,y]_{\mathcal{H}})=[\delta (x),\alpha _{\mathcal{H}}(y)]_{\mathcal{H}}+ [\alpha _{\mathcal{H}}(x),\delta (y)]_{\mathcal{H}} \end{equation}

and

(7.2) \begin{equation} [\delta (a_{1}),x]_{\mathcal{H}}=\alpha _{\mathcal{H}}\circ \delta ^{2}(x)+D\circ \delta \circ \alpha _{\mathcal{H}}(x), \end{equation}

thus

(7.3) \begin{equation} [\delta (a_{1}),\cdot ]_{\mathcal{H}}=\alpha _{\mathcal{H}}\circ \delta ^{2}+D\circ \delta \circ \alpha _{\mathcal{H}}. \end{equation}

Similarly, we prove that the identity $[\alpha (x),[y,d]]=[[x,y],\alpha (d)]+[\alpha (y),[x,d]]$ implies that

(7.4) \begin{equation} [[x,y]_{\mathcal{H}},a_{1}]_{\mathcal{H}}+\lambda D([x,y]_{\mathcal{H}})=[\alpha _{\mathcal{H}}(x),D(y)]_{\mathcal{H}}-[\alpha _{\mathcal{H}}(y),D(x)]_{\mathcal{H}}, \end{equation}

and

(7.5) \begin{equation} [D(a_{1}),x]_{\mathcal{H}}=\delta ^{2}\circ \alpha _{\mathcal{H}}(x)-\alpha _{\mathcal{H}}\circ D^{2}(x). \end{equation}

In addition, the fact that $[\alpha (y),[x,d]]=[[y,x],\alpha (d)]+[\alpha (x),[y,d]]$ implies that

(7.6) \begin{equation} [[y,x]_{\mathcal{H}},a_{1}]_{\mathcal{H}}+\lambda D([y,x]_{\mathcal{H}})=[\alpha _{\mathcal{H}}(y),D(x)]_{\mathcal{H}}-[\alpha _{\mathcal{H}}(x),D(y)]_{\mathcal{H}}, \end{equation}

and

(7.7) \begin{equation} [x,D(a_{1})]_{\mathcal{H}}=\alpha _{\mathcal{H}}\circ D^{2}(x)-\delta ^{2}\circ \alpha _{\mathcal{H}}(x), \end{equation}

Therefore, the identities (7.2), (7.5), and (7.7) lead to

(7.8) \begin{equation} [\delta (a_{1}),x]_{\mathcal{H}}=-[D(a_{1}),x]_{\mathcal{H}}=[x,D(a_{1})]_{\mathcal{H}}. \end{equation}

Besides, the identities (7.1) and (7.4) lead to $[a_{1},[x,y]_{\mathcal{H}}]_{\mathcal{H}}=-[[x,y]_{\mathcal{H}},a_{1}]_{\mathcal{H}}$ which is obviously satisfied by virtue of (2.1) and the fact that $\alpha _{\mathcal{H}}$ is invertible on $\mathcal{H}$ . Now, let $x,y\in \mathcal{H}$ , we have $\alpha ([x,y])=\alpha ([x,y]_{\mathcal{H}}+\mathfrak{B}(\delta (x),y)e) =\alpha _{\mathcal{H}}([x,y]_{\mathcal{H}})+\mathfrak{B}(a_{1},[x,y]_{\mathcal{H}})e+\mathfrak{B}(\lambda \delta (x),y)e$ . In the other hand, $[\alpha (x),\alpha (y)]=[\alpha _{\mathcal{H}}(x),\alpha _{\mathcal{H}}(y)]_{\mathcal{H}}+\mathfrak{B}(\delta \circ \alpha _{\mathcal{H}}(x),\alpha _{\mathcal{H}}(y))e$ . Then $\alpha _{\mathcal{H}}([x,y]_{\mathcal{H}})=[\alpha _{\mathcal{H}}(x),\alpha _{\mathcal{H}}(y)]_{\mathcal{H}}$ and $\mathfrak{B}(a_{1},[x,y]_{\mathcal{H}})+\mathfrak{B}(\lambda \delta (x),y)=\mathfrak{B}(\delta \circ \alpha _{\mathcal{H}}(x),\alpha _{\mathcal{H}}(y))$ . Since $\mathfrak{B}$ is non-degenerate, then the previous identity leads to $[a_{1},x]_{\mathcal{H}}=\alpha _{\mathcal{H}}\circ \delta \circ \alpha _{\mathcal{H}}(x)-\lambda \delta (x)$ , which may be written

(7.9) \begin{equation} [a_{1},\cdot ]_{\mathcal{H}}=\alpha _{\mathcal{H}}\circ \delta \circ \alpha _{\mathcal{H}}-\lambda \delta . \end{equation}

Similarly, the fact that $\alpha ([d,x])=[\alpha (d),\alpha (x)]$ implies that

(7.10) \begin{equation} \alpha _{\mathcal{H}}\circ \delta (x)=[a_{1},\alpha _{\mathcal{H}}(x)]_{\mathcal{H}}+\lambda \delta \circ \alpha _{\mathcal{H}}(x) \end{equation}

and $\mathfrak{B}(a_{1},\delta (x))+\mathfrak{B}(\lambda x,a_{0})=\mathfrak{B}(\delta (a_{1}),\alpha _{\mathcal{H}}(x))+\mathfrak{B}(\lambda \alpha _{\mathcal{H}}(x),a_{0})$ . Then the previous identity entails that

(7.11) \begin{equation} D(a_{1})+\lambda a_{0}=\alpha _{\mathcal{H}}\circ \delta (a_{1})+\lambda \alpha _{\mathcal{H}}(a_{0}). \end{equation}

Furthermore, $\alpha ([d,d])=[\alpha (d),\alpha (d)]$ follows that

(7.12) \begin{equation} \alpha _{\mathcal{H}}(a_{0})=[a_{1},a_{1}]_{\mathcal{H}}+\lambda ^{2} a_{0}+\lambda (\delta +D)(a_{1}), \end{equation}

also

(7.13) \begin{equation} \lambda \lambda _{0}+\mathfrak{B}(a_{1},a_{0})=\mathfrak{B}(\delta (a_{1}),a_{1})+\lambda ^{2} \lambda _{0}+2 \mathfrak{B}(\lambda a_{1},a_{0}). \end{equation}

Therefore, using the identities (7.3), (7.8), (7.11), and (7.13), we can consider the quadratic regular Hom–Leibniz algebra $\overline {\mathcal{L}}=\mathcal{V}^{*}\oplus \mathcal{H}\oplus \mathcal{V}$ double extension of $(\mathcal{H},[\cdot ,\cdot ]_{\mathcal{H}},\alpha _{\mathcal{H}},\mathfrak{B}_{\mathcal{H}})$ by $\mathcal{V}$ by means of the special admissible $(\delta ,a_{0},a_{1})$ .

It is clear that the map $\nabla :\mathcal{L}\rightarrow \overline {\mathcal{L}},\,\lambda e+x+\lambda 'd\longmapsto \lambda d^{*}+x+\lambda 'd$ is an isomorphism of Hom–Leibniz algebras. Then $\mathcal{L}$ is isomorphic to the double extension of $\mathcal{H}$ by $\mathcal{V}$ .

Proof. We proceed by induction on the dimension $n\in \mathbb{N}$ of $\mathcal{L}$ . If $n=2$ , then $\mathcal{L}$ is the “Leibniz” double extension of $\{0\}$ by the one-dimensional Lie algebra.

Assume that the result is true for every $k\lt n$ . Then, invoking the preceding theorem, $\mathcal{L}$ is a double extension of a quadratic regular Hom–Leibniz algebra $\mathcal{H}$ by the one-dimensional Lie algebra. Since $dim(\mathcal{H})=n-2\lt n$ , then applying the induction hypothesis to $\mathcal{H}$ , we conclude that $\mathcal{L}$ is obtained as in Corollary 7.1.

Proof. According to Corollary 7.1, Proposition 6.7, and Theorem 6.13 in [Reference Benayadi and Makhlouf19], $\mathcal{L}$ satisfies the Corollary 7.2.