Finite Serre word forms descend and have classical PBW rank bounds

If $I=\mathrm{\varnothing }$, only degree zero occurs; both forms and Gram matrices are $(1)$ and $p_0=1$, so every assertion holds directly. Otherwise supply an order on the finite set $I$. On the free algebra $T=k⟨f_i⟩$ define $D_i$ by the stated recursion. For homogeneous $x,y$, direct induction on word length gives $D_i(xy)=D_i(x)y+t^{-({\alpha }_i,\mathrm{deg}x)}x{D}_i(y)$, where $({\alpha }_i,{\alpha }_j)=a_{ij}$. The recursion also gives $D_i(f_i^n)=t^{-(n-1)}[n{]}_t{f}_i^{n-1}$. These are finite sums, and $D_i$ lowers height by one.

Every $D_{\mathrm{\ell }}$ kills each Serre polynomial already in $T$. Put $a=a_{ij}$ and $m=1-a$. If $\mathrm{\ell }\mathrm{\notin }\{i,j\}$ this is immediate. For $\mathrm{\ell }=j$, its value is $t^{-am}{f}_i^m{\sum }_r(-1{)}^r{\left(\genfrac{}{}{0px}{}{m}{r}\right)}_t{t}^{ar}=0$, since $a=-(m-1)$ and [F1] applies. For $\mathrm{\ell }=i$, the coefficient of $f_i^{m-1-r}{f}_j{f}_i^r$, $0\le r<m$, is $(-1{)}^r{t}^{-(m-r-1)}{\left(\genfrac{}{}{0px}{}{m}{r}\right)}_t[m-r{]}_t+(-1{)}^{r+1}{t}^{-2(m-r-1)-a-r}{\left(\genfrac{}{}{0px}{}{m}{r+1}\right)}_t[r+1{]}_t=0$: the two exponents agree because $a=1-m$, and the remaining factors agree by [F1]. This calculation includes $a=0,m=1$.

Let $J$ be the two-sided Serre ideal in $T$. Applying the twisted product rule to a homogeneous contextual relation $u{s}_{ij}v$ shows $D_{\mathrm{\ell }}(u{s}_{ij}v)\in J$: differentiating the middle factor gives zero, and differentiating either outside factor leaves the Serre factor. Homogeneous contextual relations span $J$, so $D_{\mathrm{\ell }}J\subseteq J$. Thus each $D_i$ descends to $S$, with augmentation $ϵ:S\to k$ killing positive height.

For a free word $u=f_{i_1}\cdots f_{i_h}$ and $y\in S$ define $b(u,y)=ϵ(D_{i_h}\cdots D_{i_1}y)$, where $D_{i_1}$ acts first. Extend linearly in $u$. This is well-defined in its second argument by step 3.1. Repeated deletion of letters in a representative word $v$ gives exactly the matching-permutation formula in the Statement: the first deletion at position $\pi (1)$ crosses the preceding remaining letters, and these are precisely the inversion pairs involving the first matched letter; continue inductively. Unequal root degrees give zero.

Fix $\alpha ={\sum }_i{n}_i{\alpha }_i$. The words $W_{\alpha }$ are a finite ordered list of length $N=({\sum }_i{n}_i)!\mathrm{/}{\prod }_i{n}_i!$. Construct the finite $N$-row matrix $R_{\alpha }(t)$ whose columns are all contextual Serre polynomials $u{s}_{ij}v$ of degree $\alpha$, expanded in $W_{\alpha }$. This is a finite list: $i,j$ have finitely many possibilities and both context lengths are bounded by $\mathrm{ht}\alpha$. Include all possible context words of the required complementary degrees. The image of this matrix over $k$ is exactly $J[\alpha ]$, since every homogeneous ideal element is a finite sum of such contexts. Consequently ${\dim }_{k}S[\alpha ]=N-{\mathrm{rank}}_k{R}_{\alpha }(t)$.

The bijection $\pi \mapsto {\pi }^{-1}$ preserves the exponent after interchanging $u,v$, since $a$ is symmetric. Consequently $b(u,[v])=b(v,[u])$ for free words and hence for free polynomials. If $P\in J$, then $b(u,[P])=0$ for every $u$; symmetry implies $b(P,[u])=0$ for every $u$. Thus $b$ factors in its first argument too, defining the asserted symmetric bilinear form on $S$. Its definition gives the adjunction and vacuum value. Conversely these adjunctions and homogeneity evaluate any first word by successive deletion, proving uniqueness. Identification [F2] makes this a form on the actual generic half, rather than a form on an unrelated free word space.

All entries of $R_{\alpha }(t)$ lie in $A$. At $t=1$, $[n{]}_t$ evaluates to $n$ and each Gaussian coefficient evaluates to the ordinary binomial coefficient, by its polynomial identity or factorial quotient. Hence $R_{\alpha }(1)$ is exactly the contextual classical Serre matrix. Every nonzero minor of $R_{\alpha }(1)$ comes from a Laurent-polynomial minor nonzero at $1$, and that minor is therefore nonzero over $k$. Thus ${\mathrm{rank}}_k{R}_{\alpha }(t)\ge {\mathrm{rank}}_{\mathbb{Q}}{R}_{\alpha }(1)$ and ${\dim }_{k}S[\alpha ]\le N-{\mathrm{rank}}_{\mathbb{Q}}{R}_{\alpha }(1)$. This is an upper bound; no equality of generic and classical ranks is needed.

The finite permutation expression has Laurent-polynomial values. Every divided-power product is an ordinary word divided by a finite product of quantum factorials. Bilinearity and [F2] therefore give the asserted localization bound on $B_t(U_A^{-},U_A^{-})$. This uses the actual integral lattice's defining inclusion in the actual half, not a claim that a torsion-free module over $A$ is free.

The final rational classical quotient is $U({\mathfrak{n}}_{\mathbb{Q}}^{-})$: define ${\mathfrak{n}}_{\mathbb{Q}}^{-}$ as the rational Lie algebra on the finite generators $f_i$ modulo $(\mathrm{ad}{f}_i{)}^{1-a_{ij}}{f}_j=0$. A linear map from these generators into the commutator Lie algebra of an associative algebra factors through ${\mathfrak{n}}_{\mathbb{Q}}^{-}$ exactly when its images satisfy the ordinary associative Serre sums, since expanding the iterated commutator gives ${\sum }_r(-1{)}^r\left(\genfrac{}{}{0px}{}{m}{r}\right)f_i^{m-r}{f}_j{f}_i^r$. Therefore the enveloping algebra and the associative classical Serre quotient have the same universal property, giving mutually inverse generator maps. Enumerate finite homogeneous bracket words, degree by degree, retaining each first independent image to obtain an ordered homogeneous Lie basis. Each root degree has finitely many bracket words and hence finite dimension. PBW [F3] identifies the degree-$\alpha$ component dimension with the number $p_{\alpha }$ of ordered basis monomials of that degree. Extending the finite rational matrix $R_{\alpha }(1)$ to $\mathbb{C}$ preserves its rank: each nonzero rational minor stays nonzero, and each zero minor stays zero. Thus the rational and complex associative quotient dimensions agree. Over $\mathbb{C}$, [F4] identifies the Lie Serre half with the actual classical Kac–Moody negative half; the enveloping universal property just proved identifies its enveloping algebra with the complex associative Serre quotient. Consequently $p_{\alpha }$ is exactly the classical PBW rank of the original target, rather than the rank of a newly chosen different Lie algebra. At $\alpha =0$ there is one empty word, no contextual relation, $G_0=(1)$ and $p_0=1$.

By step 5.1, $G_{\alpha }(t)$ is the pullback of a bilinear form on the ${\dim }_{k}S[\alpha ]$-dimensional quotient, so its rank is at most that dimension. At $c>0$ the Laurent entries evaluate. If a specialized minor of $G_{\alpha }(c)$ is nonzero over $\mathbb{Q}(c)$, its Laurent-polynomial minor is nonzero in $k$. Thus ${\mathrm{rank}}_{\mathbb{Q}(c)}{G}_{\alpha }(c)\le {\mathrm{rank}}_k{G}_{\alpha }(t)$. Combine this with steps 5.2 and 6.2 to obtain the full chain. All bases and matrix ranks needed here are furnished by finite pivot scans or the explicit bracket-word enumeration. The recursion defines the form in all degrees at once; there is no family of arbitrary choices.