Finite semisimple PBW and highest-weight construction

In a tensor word replace an adjacent pair $x_j{x}_i$ with $j>i$ by $x_i{x}_j+[x_j,x_i]$, expanding the bracket in the given finite basis. Order words first by length and then by the number of inverted pairs of indices. Each resulting word is smaller: swapping the adjacent inversion reduces its inversion number, and bracket terms have shorter length. Consequently recursive reduction terminates in a finite linear combination of ordered words. Here the recursion is finite because below any fixed length there are only finitely many basis words.

We verify uniqueness of the reduction by induction in the order from step 1.1. Two reductions at disjoint adjacent pairs commute after expansion. Two overlapping reducible pairs can only occur in $zyx$ with strictly descending basis indices $z>y>x$. Reducing its left pair and continuing the three-letter swaps yields $xyz+[y,x]z+y[z,x]+[z,y]x$; reducing its right pair yields $xyz+x[z,y]+[z,x]y+z[y,x]$. Their difference reduces, using the length-two relation and bilinearity, to $[[y,x],z]+[y,[z,x]]+[[z,y],x]=0$ by Jacobi. In any surrounding word these comparison terms have smaller length than the original, so the induction hypothesis applies to all their further reductions. Thus any two first reductions have the same final value, and induction proves a unique linear normal-form map on all words. The normal form of each contextual relation $u(x_j{x}_i-x_i{x}_j-[x_j,x_i])v$ is zero, since one of its two-letter orders is the permitted reduction; equality or reversed indices follow from antisymmetry. By linearity this holds for arbitrary linear $x,y$, so the entire defining ideal is killed. Conversely each reduction changes an element by that ideal. Ordered words are fixed by normal form, so their images are independent as well as spanning in the quotient. This proves PBW without importing a rewriting or symmetric-group presentation theorem. Evaluation of tensor words also proves the universal property: every representation of $\mathfrak{a}$ extends uniquely to a unital $U(\mathfrak{a})$-action.

F1 makes ${\mathfrak{n}}^{\pm }$ and $\mathfrak{b}$ subalgebras, since brackets add root weights. Order a basis of $\mathfrak{g}$ with negative roots first, Cartan second and positive roots last. Step 2.1 identifies multiplication $U({\mathfrak{n}}^{-})\otimes U(\mathfrak{b})\to U(\mathfrak{g})$ as a vector-space isomorphism and a right $U(\mathfrak{b})$-module isomorphism: both sides have exactly the corresponding ordered monomial basis, and right multiplication is respected by associativity. The one-dimensional action ${\mathbb{C}}_{\lambda }$ is well-defined because brackets in $\mathfrak{b}$ lie in ${\mathfrak{n}}^{+}$ and $\mathfrak{h}$ is abelian. Tensoring the displayed right-module decomposition gives the negative-root monomial basis of $M(\lambda )$. Evaluation $u\otimes 1\mapsto uv$ for any vector satisfying the highest relations proves the universal property and generation. Commuting $h$ through a negative monomial shows its weight is $\lambda -{\sum }_{\alpha >0}{a}_{\alpha }\alpha$. The positive roots are finite and have positive integer heights, so fixing that sum bounds every $a_{\alpha }$ by its total height and gives a finite-dimensional weight space. All of $Q_{+}$ occurs by monomials in the simple negative roots; height zero permits only the empty monomial.

In any weight module where $e_i,f_i$ are locally nilpotent, each weight vector $v$ generates a finite-dimensional module for that rank-one triple. Indeed PBW from step 2.1 applied to its three-dimensional algebra puts all acting words in the order $f_i^a{h}_i^b{e}_i^c$. There are only finitely many nonzero $e_i^{c}v$; each is an $h_i$-eigenvector, so varying $b$ only rescales it; for each of these finitely many vectors only finitely many powers $f_i^a$ are nonzero. Thus their span is finite-dimensional and invariant. If $\mu (h_i)=m\ge 0$ is an integer, F2 shows that $f_i^m$ is injective on weight-$m$ vectors in each such finite rank-one module: on every $V_n$ containing weight $m$, lowering $m$ steps reaches weight $-m$ with no zero coefficient. Therefore it is injective on the entire joint space $V_{\mu }$ and maps it into $V_{s_i\mu }$. For $m\le 0$, use $e_i^{-m}$ instead. Applying the same argument at $s_i\mu$ gives the reverse injection; for finite weight spaces their dimensions are equal. Since simple reflections generate $W$, multiplicities are $W$-invariant.

Every submodule of this weight module is a sum of its weight spaces. For an element with finitely many weight components choose $h\in \mathfrak{h}$ separating them and apply finite interpolation polynomials in $h$ to extract each component in the submodule. Such an $h$ exists by avoiding finitely many proper linear hyperplanes, for example using a polynomial curve in a basis and avoiding its finitely many scalar zeros. A proper submodule cannot contain the top vector, which generates $M(\lambda )$. The sum $J$ of all proper submodules is still missing the top component, hence is proper, and contains every proper submodule. Thus $L(\lambda )=M(\lambda )\mathrm{/}J$ is simple and is the unique simple quotient. Quotients retain the weight decomposition, finite weight spaces and one-dimensional top by these same projections. Two simple highest modules with highest weights $\lambda ,\mu$ can be isomorphic only if both $\lambda -\mu$ and $\mu -\lambda$ belong to $Q_{+}$, since each is a quotient of its Verma module. Independence of the simple roots then gives $\lambda =\mu$.

Suppose $\lambda$ is dominant integral and put $m_i=\lambda (h_i)\in {\mathbb{Z}}_{\ge 0}$. In $M(\lambda )$, the identity $e_i{f}_i^k{v}_{\lambda }=k(m_i-k+1)f_i^{k-1}{v}_{\lambda }$ follows by induction from the rank-one brackets. Hence $f_i^{m_i+1}{v}_{\lambda }$ is killed by $e_i$. It is killed by every $e_j$, $j\mathrm{\ne }i$, because $[e_j,f_i]=0$ by the root signs, and therefore by ${\mathfrak{n}}^{+}$ by simple generation. Its generated submodule has support below $\lambda -(m_i+1){\alpha }_i$ by the universal property in step 3.1. The finite sum $N$ of these submodules consequently misses the top weight and is proper. The nonzero quotient $Q=M(\lambda )\mathrm{/}N$ has $f_i^{m_i+1}{v}_{\lambda }=0$ for every $i$, and maps onto $L(\lambda )$ by step 4.1.

Both $e_i$ and $f_i$ act locally nilpotently on $Q$. For $e_i$, its powers shift a weight $\lambda -\sum n_j{\alpha }_j$ by $k{\alpha }_i$; if $k>n_i$ this leaves $\lambda -Q_{+}$, so gives zero. For $f_i$, F1 says ${\mathrm{ad}}_{f_i}$ is nilpotent on the finite-dimensional $\mathfrak{g}$. Its derivation on $U(\mathfrak{g})$ is locally nilpotent: on a fixed product of finitely many generators, the iterated product rule vanishes when one factor must have been differentiated beyond its finite bound, and the same holds for a finite sum. In $Q$ the finite binomial identity $f_i^{N}u{v}_{\lambda }={\sum }_{j=0}^N\left(\genfrac{}{}{0px}{}{N}{j}\right)({\mathrm{ad}}_{f_i}{)}^j(u)f_i^{N-j}{v}_{\lambda }$ then vanishes for sufficiently large $N$, by the bound for $u$ and the top-vector relation from step 5.1. Every vector of $Q$ is of this form or a finite sum, proving the assertion. It descends to $L(\lambda )$.

Apply step 3.2 to $L(\lambda )$ using step 6.1. Its weights lie in $\lambda -Q_{+}\subseteq P$. Every support weight has a dominant Weyl representative by F3, which stays in the support by multiplicity invariance and hence is at most $\lambda$. F3 makes these dominant points finite in number, and $W$ is finite. Thus the entire support is finite. Step 4.1 makes each weight space finite-dimensional, so $L(\lambda )$ is finite-dimensional. This establishes existence before using any finite-dimensional highest-weight classification.

Conversely let $V$ be a nonzero finite-dimensional simple $\mathfrak{g}$-module. F2 diagonalizes each $h_i$ with integral eigenvalues. They commute and span $\mathfrak{h}$, so successive eigenspace decomposition gives a finite set of joint weights in $P$. Choose a weight maximizing a real linear functional which is positive on every simple root, and choose a nonzero vector $v$ there. Every $e_i$ kills $v$, since otherwise it raises that functional, and therefore ${\mathfrak{n}}^{+}v=0$. For each rank-one triple, F2 shows a nonzero $h_i$-eigenvector in $\ker e_i$ has a nonnegative integral eigenvalue: in its direct sum of $V_m$ it is a sum of highest vectors with that same highest weight. Thus its joint weight $\lambda$ is dominant integral. Simplicity makes $v$ generate $V$, and steps 3.1–4.1 give $V\cong L(\lambda )$ with unique $\lambda$. For the zero algebra the root lists and basis are empty, $U=M(0)=L(0)=\mathbb{C}$, and every finite-dimensional simple module is one-dimensional with trivial action. All bases, reductions and projections were finite; the sums of all proper submodules are set-defined without selections, so no AC is used.