The complexified N=1 super-Poincare bracket data satisfy super-Jacobi

By [F7], $\mathbb{C}$ is commutative. For fixed $A,B$, the pairings $(u,\varphi )\mapsto Au\otimes \varphi$ and $(u,\varphi )\mapsto -u\otimes \varphi B$ are complex-bilinear and balanced, so [F4] induces additive endomorphisms of $V$. By [F5], $c(u\otimes \varphi )=(cu)\otimes \varphi$; since $A$ and right multiplication by $B$ are complex-linear, each induced endomorphism is complex-linear on elementary tensors and hence on all of $V$ by [F6]. Thus the displayed tensor action is well-defined and complex-linear, and [F5] gives the complex-vector-space structure on $V$.

The action $A\cdot u=Au$ is a representation by matrix multiplication. For the dual action, $[{\rho }_B,{\rho }_{B^{\mathrm{\prime }}}]\varphi =-\varphi [B,B^{\mathrm{\prime }}]$, so it is also a representation. On each elementary tensor, the left and right actions commute, and their commutator sends $u\otimes \varphi$ to $[A,A^{\mathrm{\prime }}]u\otimes \varphi -u\otimes \varphi [B,B^{\mathrm{\prime }}]$. By [F6], elementary tensors span $V$, so this proves the displayed $\mathfrak{l}$-action on all of $V$ is a representation. To organize the Jacobi checks, set $J(x,y,z)=(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }}[x,[y,z]]+(-1{)}^{\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}[y,[z,x]]+(-1{)}^{\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}[z,[x,y]]$. From graded skew symmetry, $J(x,y,z)=(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }}([x,[y,z]]-[[x,y],z]-(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}[y,[x,z]])$, so $J=0$ is equivalent to the derivation-form Jacobi identity; the same skew relation gives $J(y,x,z)=-(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}J(x,y,z)$, while cyclic rotation preserves $J$. Thus one representative ordering for each parity pattern suffices.

The even bracket is the semidirect bracket for the representation of $\mathfrak{l}$ on $V$, so its Jacobi identity has $\mathfrak{l}$-component equal to Jacobi in [F1] and $V$-component equal to the representation identity verified in step 2.1. Thus ${\mathfrak{g}}_{\bar{0}}$ is a Lie algebra.

The action of ${\mathfrak{g}}_{\bar{0}}$ on $S_{+}\oplus S_{-}$ is a representation: its restriction to $\mathfrak{l}$ is the direct sum of the two actions in step 2.1, and the ideal $V$ acts by zero, consistently with $[\mathfrak{l},V]\subseteq V$. Therefore Jacobi holds for triples with two even and one odd entry.

The odd bracket is symmetric and is $\mathfrak{l}$-equivariant because $(A,B)\cdot (u\otimes \varphi )=Au\otimes \varphi +u\otimes (-\varphi B)$; it is $V$-equivariant since both $[V,S_{+}\oplus S_{-}]$ and $[V,V]$ vanish. This is exactly Jacobi for one even and two odd entries.

For three odd entries, each inner bracket lies in $V$, and $[V,S_{+}\oplus S_{-}]=0$; every nested-bracket term is zero. Hence Jacobi holds for this triple type.

The even bracket is skew because reversing $[(L,v),(L^{\mathrm{\prime }},v^{\mathrm{\prime }})]$ negates both $[L,L^{\mathrm{\prime }}]$ and $L\cdot v^{\mathrm{\prime }}-L^{\mathrm{\prime }}\cdot v$. Mixed brackets are defined by the even action and graded skew symmetry; the odd bracket is symmetric by its formula. Each bracket lands in the parity-sum component. The earlier even, mixed, and all-odd checks cover every homogeneous Jacobi triple type, so the signed Jacobi identity holds. For any odd $x$, put $A=[x,[x,x]]$. Applying that identity to $(x,x,x)$ gives $A=[[x,x],x]-A$, while graded skew symmetry and the even parity of $[x,x]$ give $[[x,x],x]=-A$. Thus $3A=0$, and $A=0$ because $\mathbb{C}$ has characteristic zero. The separate odd self-Jacobi clause in [F3] therefore holds, so the data satisfy all the Lie-superalgebra axioms.