Serre presentation theorem

Let $\tilde{\mathfrak{g}}$ be the Lie algebra with generators $e_i,f_i,h_i$ and all relations of [L1] except the Serre relations. It is $\mathbb{Z}$-graded by $\mathrm{deg}{e}_i=1$, $\mathrm{deg}{f}_i=-1$, $\mathrm{deg}{h}_i=0$, and Jacobi and the mixed relations reduce every bracket to a linear combination of brackets only in the $e$'s, only in the $f$'s, or single $h$'s. More explicitly, $[e_i,{\tilde{\mathfrak{n}}}_{-}]\subseteq {\tilde{\mathfrak{n}}}_{-}+\tilde{\mathfrak{h}}$ follows by induction on bracket length; induction on positive bracket length then reduces $[{\tilde{\mathfrak{n}}}_{+},{\tilde{\mathfrak{n}}}_{-}]$ to the same three summands. Their nonzero degrees have opposite signs, so the sum is direct. Thus $\tilde{\mathfrak{g}}={\tilde{\mathfrak{n}}}_{+}\oplus \tilde{\mathfrak{h}}\oplus {\tilde{\mathfrak{n}}}_{-}$, where ${\tilde{\mathfrak{n}}}_{\pm }$ is the subalgebra generated by the $e_i$ respectively the $f_i$ and $\tilde{\mathfrak{h}}$ is spanned by the $h_i$. We claim that ${\tilde{\mathfrak{n}}}_{-}$ is free on the $f_i$, that ${\tilde{\mathfrak{n}}}_{+}$ is free on the $e_i$, and that the $h_i$ are linearly independent. For the first claim let $\mathfrak{a}={\mathfrak{h}}^{\mathrm{\prime }}⋉\mathbb{F}L(f_1^{\mathrm{\prime }},\dots ,f_r^{\mathrm{\prime }})$ be the semidirect product of the abelian Lie algebra with basis $h_1^{\mathrm{\prime }},\dots ,h_r^{\mathrm{\prime }}$ and the free Lie algebra on $f_1^{\mathrm{\prime }},\dots ,f_r^{\mathrm{\prime }}$, with $[h_i^{\mathrm{\prime }},h_j^{\mathrm{\prime }}]=0$ and $[h_i^{\mathrm{\prime }},f_j^{\mathrm{\prime }}]=-a_{ij}{f}_j^{\mathrm{\prime }}$; the universal properties in [L2] identify the enveloping algebra of the free Lie algebra with the free associative algebra (both represent arbitrary choices of the generator images in a unital associative algebra). PBW, with a basis of the free Lie ideal placed before the $h_i^{\mathrm{\prime }}$, then identifies multiplication $U(\mathbb{F}L(f_i^{\mathrm{\prime }}))\otimes U({\mathfrak{h}}^{\mathrm{\prime }})\to U(\mathfrak{a})$ as a vector-space isomorphism, so $U(\mathfrak{a})$ is identified with $\mathbb{C}⟨f_1^{\mathrm{\prime }},\dots ,f_r^{\mathrm{\prime }}⟩\otimes \mathbb{C}[h_1^{\mathrm{\prime }},\dots ,h_r^{\mathrm{\prime }}]$. Write ${\alpha }_j(h_i)=a_{ij}$ for the simple roots and put $\alpha ={\alpha }_{j_1}+\cdots +{\alpha }_{j_s}$ for the weight of a word $w=f_{j_1}^{\mathrm{\prime }}\cdots f_{j_s}^{\mathrm{\prime }}$, so that $\alpha (h_i)$ is the sum of the $a_{i{j}_k}$. Define endomorphisms of $U(\mathfrak{a})$ by $h_i(w\otimes P)=w\otimes (h_i^{\mathrm{\prime }}-\alpha (h_i))P,f_i(w\otimes P)=f_i^{\mathrm{\prime }}w\otimes P,$ $e_i(f_{j_1}^{\mathrm{\prime }}\cdots f_{j_s}^{\mathrm{\prime }}\otimes P)={\sum }_{k:j_k=i}{f}_{j_1}^{\mathrm{\prime }}\cdots \widehat{f_{j_k}^{\mathrm{\prime }}}\cdots f_{j_s}^{\mathrm{\prime }}\otimes (h_i^{\mathrm{\prime }}-({\alpha }_{j_{k+1}}+\cdots +{\alpha }_{j_s})(h_i))P,$ the hat marking an omitted letter. These satisfy the relations of $\tilde{\mathfrak{g}}$: the operators $h_i$ are multiplications by commuting polynomials, and passing from $w$ to $f_j^{\mathrm{\prime }}w$ lowers the weight by ${\alpha }_j$, so $[h_i,f_j]=-a_{ij}{f}_j$; each summand of $e_j$ has its left factor $(h_i^{\mathrm{\prime }}-\alpha (h_i))$ replaced by $(h_i^{\mathrm{\prime }}-(\alpha -{\alpha }_j)(h_i))$, which differs by $a_{ij}$, giving $[h_i,e_j]=a_{ij}{e}_j$; and $e_i{f}_j$ and $f_j{e}_i$ differ only by the summand in which the leading letter is removed, present exactly when $i=j$, where it equals $w\otimes (h_i^{\mathrm{\prime }}-\alpha (h_i))P$, giving $[e_i,f_j]={\delta }_{ij}{h}_i$. Hence $\tilde{\mathfrak{g}}$ acts on $U(\mathfrak{a})$. Let $\psi :\mathbb{F}L(f_i^{\mathrm{\prime }})\to {\tilde{\mathfrak{n}}}_{-}$ be the Lie homomorphism with $f_i^{\mathrm{\prime }}\mapsto f_i$, which is surjective, and let $\mathrm{\Theta }$ be the inclusion of $\mathbb{F}L(f_i^{\mathrm{\prime }})$ into $\mathbb{C}⟨f_i^{\mathrm{\prime }}⟩\subset U(\mathfrak{a})$, which exists by [L2]. The maps $z\mapsto$ (action of $\psi (z)$) and $z\mapsto$ (left multiplication by $\mathrm{\Theta }(z)$) are Lie homomorphisms that agree on the generators $f_i^{\mathrm{\prime }}$, hence on $\mathbb{F}L(f_i^{\mathrm{\prime }})$; evaluating both at $1$ gives $\psi (z)(1)=\mathrm{\Theta }(z)=z$. So $\psi (z)=0$ forces $z=0$, and $\psi$ is an isomorphism: ${\tilde{\mathfrak{n}}}_{-}$ is free on the $f_i$. Applying the same construction with the roles of $e_i$ and $f_i$ exchanged, that is, to the automorphism $e_i\mapsto f_i$, $f_i\mapsto e_i$, $h_i\mapsto -h_i$ of the presentation, which preserves the listed relations, shows that ${\tilde{\mathfrak{n}}}_{+}$ is free on the $e_i$. Finally $h_i(1)=h_i^{\mathrm{\prime }}$ for every $i$, and the $h_i^{\mathrm{\prime }}$ are linearly independent in $U(\mathfrak{a})$; a relation ${\sum }_i{c}_i{h}_i=0$ in $\tilde{\mathfrak{g}}$ therefore yields ${\sum }_i{c}_i{h}_i^{\mathrm{\prime }}=0$, hence $c_i=0$ for all $i$.

For later use, every root of a based root system is carried to a simple root by a product of simple reflections. For a positive nonsimple root $\gamma =\sum k_i{\alpha }_i$, positivity of $(\gamma ,\gamma )=\sum k_i(\gamma ,{\alpha }_i)$ supplies $i$ with $\gamma (h_i)>0$. Reflection subtracts the positive integer $\gamma (h_i)$ from the $i$-th coefficient. Another coefficient is positive, since reducedness excludes a nonsimple root on a simple-root line. By the one-sign property the reflected root stays positive and has smaller height. Induction reaches a simple root; negative roots are reduced to positive ones by a final sign-changing simple reflection.

Write $E_i=\mathrm{ad}{e}_i$, $F_i=\mathrm{ad}{f}_i$, $H_i=\mathrm{ad}{h}_i$ in the algebra before the Serre quotient. For $i\mathrm{\ne }j$ put $m=a_{ij}$ and $N=1-m$. The relations and induction give $F_i{E}_i^n{e}_j=-n(m+n-1)E_i^{n-1}{e}_j$, since $F_i{e}_j=0$. Hence $[f_i,S_{ij}^{+}]=0$ for $S_{ij}^{+}=E_i^N{e}_j$. For $k\mathrm{\notin }\{i,j\}$ the commutator with $f_k$ vanishes termwise. For $k=j$ it equals $-E_i^N{h}_j$, which is zero for $N\ge 2$, and for $N=1$ because then $a_{ij}=a_{ji}=0$. The involution exchanging $e,f$ and negating $h$ gives $[e_k,S_{ij}^{-}]=0$. Let $I_{\pm }$ be the ideals generated by these elements within the free positive and negative subalgebras. These are stable under $\mathfrak{h}$, since the generators and their iterated brackets are weight vectors. Jacobi induction on the number of positive generators bracketing $S_{ij}^{+}$ proves $[f_k,I_{+}]\subseteq I_{+}$: the base commutator is zero, and every new term $[f_k,e_l]=-{\delta }_{kl}{h}_k$ preserves the ideal. Similarly $[e_k,I_{-}]\subseteq I_{-}$. Thus $I_{+}+I_{-}$ is an ideal in the full algebra and is exactly the Serre ideal. It has no zero-degree part, so $\mathfrak{g}(A)={\mathfrak{n}}_{-}\oplus \mathfrak{h}\oplus {\mathfrak{n}}_{+}$ and the $h_i$ remain independent. The simple generators remain nonzero because $[e_i,f_i]=h_i$.

For fixed $i$ the three generators give a copy of ${\mathfrak{s}\mathfrak{l}}_2$, since their nonzero distinct weights and independent $h_i$ exclude linear relations. For $j\mathrm{\ne }i$, the span of $e_j,E_i{e}_j,\dots ,E_i^{-a_{ij}}{e}_j$ is a finite-dimensional module: $E_i$ kills its last vector by the Serre relation, $H_i$ acts by weights $a_{ij}+2n$, and the commutator formula in step 2.1 describes $F_i$. The analogous span generated by $f_j$ is also finite-dimensional. The span of $e_i,f_i,h_i,h_j$ is stable for every $j$. Every bracket of vectors in finite-dimensional modules is in a finite-dimensional module, because the bracket is an equivariant image of their tensor product. Since all elements are finite sums of iterated brackets, every element is in a finite-dimensional module for this triple. By [L3], $E_i,F_i$ act locally nilpotently.

The root-lattice grading assigns degrees ${\alpha }_i,-{\alpha }_i,0$ to $e_i,f_i,h_i$. Each homogeneous bracket has that $\mathfrak{h}$-weight by the defining relations. Nonsingularity of $A$ makes different lattice elements distinct functionals on $\mathfrak{h}$. The triangular decomposition implies that the only possible weights are $Q_{+}$ and $-Q_{+}$, where $Q_{+}={\sum }_i{\mathbb{Z}}_{\ge 0}{\alpha }_i$, and that ${\mathfrak{g}}_0=\mathfrak{h}$. Each nonzero weight space is finite-dimensional: it is spanned by the finitely many bracket words with its fixed multidegree. Furthermore ${\mathfrak{g}}_{m{\alpha }_i}=0$ for $\mathrm{\mid }m\mathrm{\mid }>1$, because a Lie algebra on a single generator has no bracket of length greater than one; ${\mathfrak{g}}_{{\alpha }_i}=\mathbb{C}{e}_i$ and ${\mathfrak{g}}_{-{\alpha }_i}=\mathbb{C}{f}_i$.

The finite sums $\exp (E_i)$ and $\exp (F_i)$ are automorphisms: for any derivation $D$, induction gives $D^n[x,y]={\sum }_{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right)[D^{k}x,D^{n-k}y]$, so exponentiation preserves brackets when $D$ is locally nilpotent, with inverse $\exp (-D)$. Define $w_i=\exp (E_i)\exp (-F_i)\exp (E_i)$. On the triple, direct substitution using $E_i{f}_i=h_i$, $E_i{h}_i=-2e_i$, $F_i{e}_i=-h_i$, $F_i{h}_i=2f_i$ gives $w_i(h_i)=-h_i$. It fixes every $h$ with ${\alpha }_i(h)=0$, so $w_i(h)=h-{\alpha }_i(h)h_i$ for all $h\in \mathfrak{h}$. Consequently $w_i$ maps ${\mathfrak{g}}_{\lambda }$ bijectively to ${\mathfrak{g}}_{s_i\lambda }$, where $s_i\lambda =\lambda -\lambda (h_i){\alpha }_i$. It also preserves every ideal, because such an ideal is preserved by $E_i,F_i$ and their exponentials.

Suppose ${\mathfrak{g}}_{\lambda }\mathrm{\ne }0$ with $\lambda =\sum k_i{\alpha }_i\in Q_{+}\setminus \{0\}$. If only one coefficient is positive, step 3.2 makes $\lambda$ a simple root. Otherwise $(\lambda ,\lambda )=\sum k_i(\lambda ,{\alpha }_i)>0$ supplies an $i$ with $c=\lambda (h_i)=2(\lambda ,{\alpha }_i)\mathrm{/}({\alpha }_i,{\alpha }_i)>0$. Here $c$ is an integer, not in general $k_i$. Step 4.1 gives a nonzero space at $s_i\lambda =\lambda -c{\alpha }_i$. At least one coefficient other than the $i$-th stays positive, so this weight is not in $-Q_{+}$. Step 3.2 therefore forces it into $Q_{+}$; its height is strictly smaller. Induction proves $s_i\lambda \in \mathrm{\Phi }(A)$ and hence $\lambda \in \mathrm{\Phi }(A)$. Negative weights follow by the same reflection argument or the presentation involution. Conversely every root occurs: step 1.2 carries it to a simple root, whose space is nonzero, and the automorphisms in step 4.1 carry that space back. The same isomorphisms show every root space has dimension one. Thus $\dim \mathfrak{g}(A)=r+\mathrm{\mid }\mathrm{\Phi }(A)\mathrm{\mid }$.

Suppose the diagram of $A$ is connected. If $J\mathrm{\ne }0$ is an ideal, invariance under $\mathrm{ad}\mathfrak{h}$ makes it a sum of weight spaces: projections onto the finitely many joint eigenspaces are polynomials in the commuting diagonal operators (choose an $h$ separating their finitely many weights and use interpolation). A nonzero root component, by steps 1.2 and 4.1, puts $e_i$ in $J$ for some $i$. A nonzero component $h\in J\cap \mathfrak{h}$ instead gives $e_i\in J$ from $[h,e_i]={\alpha }_i(h)e_i$ for some ${\alpha }_i(h)\mathrm{\ne }0$, by nonsingularity of $A$. Then $h_i=[e_i,f_i]\in J$ and $f_i=-[h_i,f_i]\mathrm{/}2\in J$. If $a_{ij}\mathrm{\ne }0$, $[h_i,e_j]=a_{ij}{e}_j$ puts the next generator in $J$. Connectedness propagates this to all generators, so the algebra is simple and nonabelian. For disconnected $A$, generators in distinct blocks commute: mixed $e,f$ brackets and $h$ brackets vanish by the initial relations, and the zero Cartan entries give $[e_i,e_j]=[f_i,f_j]=0$ by the Serre relations. Jacobi extends this to the block subalgebras. The block inclusions and projections supplied by the presentations are mutually inverse maps with their direct sum, proving semisimplicity. Finally $\mathfrak{h}$ is abelian, and its normalizer equals itself: a nonzero root component of a normalizing vector would give a nonzero root component in its bracket with some $h$, contrary to normalization. Hence it is a Cartan subalgebra by [L6], with precisely the root system and Cartan matrix already established.

Conversely let $\mathfrak{g}$ be finite-dimensional complex semisimple with the given base and triples. For $i\mathrm{\ne }j$, $[e_i,f_j]=0$ since ${\alpha }_i-{\alpha }_j$ has mixed signs and is not a root. The vector $f_j$ is killed by $E_i$ and has weight $-a_{ij}\ge 0$. In a finite-dimensional irreducible ${\mathfrak{s}\mathfrak{l}}_2$-module a vector killed by $e$ is a highest-weight vector: for a vector of weight $m$, the commutator formula $e{f}^{n}v=n(m-n+1)f^{n-1}v$ proves this and gives $f^{m+1}v=0$. Complete reducibility therefore gives $F_i^{1-a_{ij}}{f}_j=0$; exchange $e,f$ to get the other Serre relation. All other presentation relations follow from [L4]. The generated subalgebra contains $\mathfrak{h}$, since the simple coroots form a basis. It is preserved by the exponentials of $E_i,F_i$, which are nilpotent on this finite-dimensional adjoint module by [L3]. The calculation in step 4.1 therefore applies to the actual algebra too. Step 1.2 and one-dimensionality of its root spaces show that the generated subalgebra contains every root space. By the root-space decomposition it is all of $\mathfrak{g}$, giving a surjection $\mathfrak{g}(A)\to \mathfrak{g}$. Both dimensions equal $r+\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$ by step 5.1 and [L4], so the map is an isomorphism. For $r=0$ the presentation has no generators and is the zero algebra, with zero Cartan subalgebra and empty root system; the converse follows from [L4] as well.