Lie Superalgebras and Super-Representations

Let $J$ consist of all finite sums ${\sum }_{k=1}^m{a}_k{s}_k{b}_k$, with $a_k,b_k\in A$ and $s_k\in S$, including the empty sum $0$. Concatenating sums and replacing $a_k$ by $-a_k$ show that $J$ is an additive subgroup; multiplying a sum on the left or right absorbs that factor into $a_k$ or $b_k$, using associativity from [F1], so [F2] makes $J$ a two-sided ideal. It contains every $s\in S$ as $1s1$. Every two-sided ideal containing $S$ contains each $a_k{s}_k{b}_k$ and every such finite sum. Thus $J$ is the smallest two-sided ideal containing $S$, and [F3] gives $I=J$.

For each summand $asb$ in a finite expression defining $J$, write $a=a_{\bar{0}}+a_{\bar{1}}$ and $b=b_{\bar{0}}+b_{\bar{1}}$. Then $asb={\sum }_{i,j\in \{0,1\}}{a}_{\bar{i}}s{b}_{\bar{j}}$; each term is homogeneous by [F1] and lies in $J$. Therefore every homogeneous component of every element of $J$ lies in $J$, proving $I=(I\cap A_{\bar{0}})\oplus (I\cap A_{\bar{1}})$. Every coset in $A\mathrm{/}I$ decomposes into its even and odd component classes; if a class lies in their intersection, its two homogeneous representatives differ by an element of the graded ideal $I$, so both representatives lie in $I$ and the class is zero. The quotient grading is direct, and the product respects it because the product in $A$ does.

Put $A=[x,[x,x]]$. The signed Jacobi identity in [F1], with $y=z=x$, gives $A=[[x,x],x]-A$, because $(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}=-1$.

The element $[x,x]$ is even, so graded skew symmetry in [F1] gives $[[x,x],x]=-[x,[x,x]]=-A$. Substituting into step 1.1 yields $A=-2A$, hence $3A=0$; since the scalar field is $\mathbb{C}$, $A=0$.

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.

The bracket table is homogeneous: even–even brackets lie in the even part, even–odd brackets in the odd part, and odd–odd brackets in the even part. Its odd–odd entries are symmetric and its even–odd reverse entries are their negatives, so it satisfies graded skew symmetry in [F2].

For homogeneous $x,y,z$, put $J(x,y,z)=(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }}[x,[y,z]]+(-1{)}^{\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}[y,[z,x]]+(-1{)}^{\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}[z,[x,y]]$. Graded skew symmetry gives $J(x,y,z)=(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }}([x,[y,z]]-[[x,y],z]-(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}[y,[x,z]])$ and $J(y,x,z)=-(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}J(x,y,z)$; cyclic rotation preserves $J$ by its formula. Hence the derivation Jacobi identity is equivalent to $J=0$, and it suffices to check representative basis triples up to permutation. For representatives $(h_i,h_j,x)$ with $x=e$ or $f$, the commuting diagonal actions of $h_1,h_2$ have weights $(1,-1)$ and $(-1,1)$, so $[h_i,[h_j,x]]=[h_j,[h_i,x]]$. All-even brackets vanish, so all-even triples also satisfy Jacobi.

Put $c=h_1+h_2$. By [F1], $[h_i,c]=[e,c]=[f,c]=0$. For a triple $(h_i,e,f)$, the Jacobi identity reduces to $[h_i,c]=[[h_i,e],f]+[e,[h_i,f]]$; the two terms on the right are opposite multiples of $c$, so their sum is zero. For $(h_i,e,e)$ or $(h_i,f,f)$, every term vanishes because $[e,e]=[f,f]=0$. This checks all basis triples with one even and two odd entries.

For three odd basis entries, each inner bracket is either zero or $c$, and every outer bracket of $c$ with $e$ or $f$ is zero by step 2.1. Hence every nested-bracket term vanishes. Together with the representative even and mixed checks from step 1.2 and step 2.1, permutation reduction and multilinearity give signed Jacobi for all homogeneous triples.

For any odd $x$, put $A=[x,[x,x]]$. Applying the signed Jacobi identity from [F2] to $(x,x,x)$ gives $A=[[x,x],x]-A$, while graded skew symmetry and the even parity of $[x,x]$ give $[[x,x],x]=-A$. Hence $3A=0$, and $A=0$ because $\mathbb{C}$ has characteristic zero. This proves the additional odd self-Jacobi clause, so the bracket satisfies the full complex Lie-superalgebra axioms.

The subspaces $\mathfrak{h},{\mathfrak{n}}^{+},{\mathfrak{n}}^{-}$ are graded subalgebras: $\mathfrak{h}$ is abelian, and $[e,e]=[f,f]=0$. Their direct sum is $E$. The table gives $[h_i,e]\in \mathbb{C}e$, so $\mathfrak{h}\oplus {\mathfrak{n}}^{+}$ is a subalgebra as well. These are precisely the conditions in [F3].

For adjacent homogeneous basis elements $x>y$, use [F2] to replace $xy$ by $(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}yx+[x,y]$; if $x=y$ is odd, use [F2] to replace $xx$ by $[x,x]\mathrm{/}2$. If $x=y$ is even, graded skew symmetry gives $[x,x]=0$ over $\mathbb{C}$, so no square rule is needed. Expand each bracket in the supplied basis.

The remaining overlaps have a repeated odd basis element. For $xxz$, where $x$ is odd and $x>z$, the two reductions differ after ordering shorter terms by $\frac{1}{2}[[x,x],z]-[x,[x,z]]=0$; the equality follows from [F1] with $(x,x,z)$. For $zxx$, where $x$ is odd and $z>x$, the difference is $\frac{1}{2}[[x,x],z]+(-1{)}^{\mathrm{\mid }z\mathrm{\mid }}[x,[z,x]]=0$: [F1] and graded skew give $[z,[x,x]]=2(-1{)}^{\mathrm{\mid }z\mathrm{\mid }}[x,[z,x]]$, while $[[x,x],z]=-[z,[x,x]]$. This also covers the pattern $xyy$ after renaming. For $xxx$, the two square reductions differ by $\frac{1}{2}[[x,x],x]=-\frac{1}{2}[x,[x,x]]=0$, using [F3]. Thus every one-step overlap resolves.

These replacements terminate: for a word $w=b_1\cdots b_n$, define $\mathrm{inv}(w)=\mathrm{\mid }\{(i,j):1\le i<j\le n,b_i>b_j\}\mathrm{\mid }$, and order words by $(n,\mathrm{inv}(w))$ lexicographically. Swapping an adjacent descending pair keeps length and reduces the inversion count by one; a bracket term has smaller length; and an odd-square replacement has length one. Each bracket has finite basis support, so each reduction is finite.

For a strict descending triple $x>y>z$, reducing the left pair first or the right pair first gives the same ordered cubic term. After using the length-two relation [F2] to order the bracket terms, the difference of the two results is $-[x,[y,z]]+(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}[y,[x,z]]+[[x,y],z]$, which is zero by [F1]. Disjoint reductions commute because they change disjoint adjacent pairs.

Termination and the verified overlap resolutions give a unique normal form by induction on the ordered pair (length, inversion count): compare the first reductions of any two reduction sequences; disjoint reductions commute, overlapping reductions have a common continuation, and all later terms have smaller measure. For $n=1$, $B$ is a basis of $\mathfrak{g}$. If words of length $n$ form a basis of ${\mathfrak{g}}^{\otimes n}$, apply [F4], using [F5], to that basis and $B$; under the recursive tensor-power identification, the resulting elementary tensors are exactly the words of length $n+1$. Thus words of length $n$ form a basis of ${\mathfrak{g}}^{\otimes n}$ for every $n\ge 1$. The $n=0$ summand is $\mathbb{C}$ with basis the unit, so the tensor words form a basis of $T(\mathfrak{g})$. Extend the normal-form operation linearly from those words to the vector space spanned by irreducible words.

In any left or right word context, a basis-pair defining relation has normal form zero: the local rewrite can be applied at that same adjacent pair, and confluence gives the same normal form. Relations for reversed pairs follow from graded skew symmetry; the equal-even relation is zero. Bilinearity gives the relation for arbitrary homogeneous pairs, so every generator of the two-sided ideal defining $U(\mathfrak{g})$, with arbitrary word contexts, lies in the kernel. Conversely, each reduction changes a word by an element of that ideal, so every word differs from its normal form by an element of the ideal. Thus the kernel is exactly the ideal. The normal-form map fixes each irreducible word, which is a tensor word in the basis from [F4]; these distinct words are linearly independent in its image and form a basis of the quotient.

A word is irreducible exactly when its basis letters are weakly increasing and no odd letter is repeated: every forbidden adjacent descending pair or odd square is reducible, while any nonordered word has an adjacent descent. Therefore the irreducible words are precisely the monomials in the Statement. The empty word is the unit and covers degree zero.

The tensor-algebra extension $\tilde{\varphi }:T(\mathfrak{g})\to A$ exists uniquely by [F3]; on a homogeneous pure tensor it is $\varphi (x_1)\cdots \varphi (x_n)$, whose parity is ${\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }$, so it preserves the supergrading.

For each homogeneous pair $x,y$, bracket preservation gives $\tilde{\varphi }(x\otimes y-(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}y\otimes x-[x,y])=[\varphi (x),\varphi (y){]}_s-\varphi ([x,y])=0$. Since $\tilde{\varphi }$ is an algebra homomorphism, it kills the two-sided ideal $I$ generated by these elements.

Therefore $\tilde{\varphi }$ factors through a unique homomorphism $\widehat{\varphi }:T(\mathfrak{g})\mathrm{/}I\to A$, which has $\widehat{\varphi }{\iota }_{\mathfrak{g}}=\varphi$. Any other such homomorphism would pull back to an extension of $\varphi$ to $T(\mathfrak{g})$; uniqueness in [F3] makes it equal to $\widehat{\varphi }$.

With pointwise addition and scalar multiplication, $\mathrm{End}(V)$ is a complex vector space: sums and scalar multiples of linear maps are linear, and its vector-space laws hold pointwise in $V$. Write $\mathrm{End}(V{)}_{\bar{0}}$ for the endomorphisms preserving $V_{\bar{0}},V_{\bar{1}}$ and $\mathrm{End}(V{)}_{\bar{1}}$ for those interchanging them. If $p_i$ is the projection onto $V_{\bar{i}}$, every $T\in \mathrm{End}(V)$ decomposes uniquely as $T_0+T_1$, where $T_0=p_{0}T{p}_0+p_{1}T{p}_1$ and $T_1=p_{1}T{p}_0+p_{0}T{p}_1$; hence these spaces give a direct-sum grading. Composition sends parities $i,j$ to $i+j$ because an operator of parity $j$ maps $V_{\bar{k}}$ into $V_{\overline{k+j}}$, and the identity is even. Composition is associative and complex-bilinear, so by [F4] this makes $\mathrm{End}(V)$ a unital associative superalgebra. Given a $\mathfrak{g}$-supermodule, each homogeneous $x$ acts as a parity-$\mathrm{\mid }x\mathrm{\mid }$ endomorphism $v\mapsto x\cdot v$. The module identity in [F1] says exactly that this map preserves the Lie bracket into the endomorphism supercommutator, so [F3] extends it uniquely to a unital graded algebra action of $U(\mathfrak{g})$.

Conversely, restrict a unital graded left $U(\mathfrak{g})$-module action along ${\iota }_{\mathfrak{g}}$. Applying the quotient relation [F2] to $v$ gives $[x,y]v=x(yv)-(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}y(xv)$; grading and bilinearity are inherited from the module action. Thus the restriction is a $\mathfrak{g}$-supermodule.

Starting from a supermodule, the extension in [F3] restricts to the original action by its defining property. Starting from a $U(\mathfrak{g})$-module, the extension of its restriction agrees with the original action on every generator ${\iota }_{\mathfrak{g}}(x)$ and on the unit; these generate the quotient algebra by [F2], so the actions agree on all of $U(\mathfrak{g})$.

A parity-preserving linear map intertwines the $\mathfrak{g}$-actions exactly when it intertwines the restrictions of the $U(\mathfrak{g})$-actions: one direction is restriction, and the other follows on products of generators and then on finite linear combinations. The correspondence is therefore mutually inverse on objects and morphisms.

The actions of $h_1,h_2$ preserve the parity of $v,w$, and the actions of $e,f$ reverse it. Thus the candidate action has the parity required in [F4].

The operators $h_1,h_2$ are diagonal in the basis $v,w$, so they commute, matching $[h_1,h_2]=0$. On $v$, both the bracket actions $[h_i,e]v$ and the operator commutators $(h_{i}e-e{h}_i)v$ vanish for $i=1,2$, since $[h_i,e]$ is a multiple of $e$, $ev=0$, and each $h_{i}v$ is a scalar multiple of $v$. On $w$, the commutators $h_{1}e-e{h}_1$ and $h_{2}e-e{h}_2$ give $(a-(a-1))(a+b)v=(a+b)v=[h_1,e]w$ and $(b-(b+1))(a+b)v=-(a+b)v=[h_2,e]w$. Likewise, $h_{1}f-f{h}_1$ and $h_{2}f-f{h}_2$ on $v$ give $((a-1)-a)w=-w=[h_1,f]v$ and $((b+1)-b)w=w=[h_2,f]v$, and both bracket actions and commutators vanish on $w$. Hence all even–odd bracket relations in [F1] hold.

Since $e^2=f^2=0$ on both basis vectors, the odd self-brackets act as zero. The anticommutator $ef+fe$ sends $v$ to $(a+b)v$ and $w$ to $(a+b)w$; this equals the action of $h_1+h_2$ on both vectors. Therefore the odd–odd relations in [F1] hold.