Separation of the negative half by iterated skew derivations

The normalized pairing on the half. Apply the substitution $z\mapsto q$ of [F5] to the pairing of [F4], getting a nondegenerate $\mathbb{Q}(q)$-valued pairing $⟨\cdot ,\cdot ⟩:H^{+}\times H^{-}\to \mathbb{Q}(q)$, still zero on unequal opposite degrees, with $⟨e_i,f_j⟩={\delta }_{ij}\mathrm{/}(q_i-q_i^{-1})$, $⟨1,1⟩=1$, $⟨1,y⟩=\epsilon (y)$, dual opposite graded components, and the two adjunctions of [F4]. Define the degree character $c:Q_{+}\to \mathbb{Q}(q{)}^{\times }$ by $c_{\nu }={\prod }_i(q_i-q_i^{-1}{)}^{n_i}$ for $\nu ={\sum }_i{n}_i{\alpha }_i$, and the rescaled pairing $⟨x,y{⟩}_c:=c_{\nu }⟨x,y⟩$ for $x\in H^{+}[\nu ]$, $y\in H^{-}[-\nu ]$, extended by zero on unequal opposite degrees, where $H^{\pm }[\pm \nu ]$ are the graded components of [F4]. Then $c$ is multiplicative because $\nu \mapsto n_i$ is additive and the two coproduct factors of a homogeneous element have degrees adding to its own; each nonzero term of the adjunctions therefore acquires the scalar $c_{{\nu }_{(1)}}{c}_{{\nu }_{(2)}}=c_{\nu }$, so $⟨x,y{y}^{\mathrm{\prime }}{⟩}_c=\sum ⟨x_{(1)},y{⟩}_c⟨x_{(2)},y^{\mathrm{\prime }}{⟩}_c$ and $⟨x{x}^{\mathrm{\prime }},y{⟩}_c=\sum ⟨x,y_{(1)}{⟩}_c⟨x^{\mathrm{\prime }},y_{(2)}{⟩}_c$. Moreover $⟨e_i,f_j{⟩}_c=(q_i-q_i^{-1}){\delta }_{ij}\mathrm{/}(q_i-q_i^{-1})={\delta }_{ij}$ and $⟨1,1{⟩}_c=1$, and $⟨\cdot ,\cdot {⟩}_c$ is nondegenerate on every pair of opposite graded components because each $c_{\nu }$ is a nonzero scalar. AC is inherited here from [F4] and used nowhere else in this item; the remaining steps are choice-free.

Generators pair only with height one from the left. If $u=ab\in H^{-}$ is a product of two elements of positive degree, then $⟨e_i,u⟩=0$ for every $i$. Indeed $e_i$ is primitive in $H^{+}$ [F4], so the first adjunction gives $⟨e_i,ab⟩=⟨e_i,a⟩⟨1,b⟩+⟨1,a⟩⟨e_i,b⟩$, and $⟨1,a⟩=\epsilon (a)$ and $⟨1,b⟩=\epsilon (b)$ both vanish because $a$ and $b$ have positive degree [F4]. Every homogeneous element of $H^{-}[-\mu ]$ with $\mathrm{ht}(\mu )\ge 2$ is a linear combination of such products: on the free algebra its defining word splits after the first letter, and the product of the two positive-degree classes is the class of that word, so the class lies in the span of products of two positive-degree elements. Hence $⟨e_i,u⟩=0$ whenever $u\in H^{-}[-\mu ]$ and $\mathrm{ht}(\mu )\ge 2$; by the same argument, scaled by $c_{{\alpha }_i}\mathrm{\ne }0$, also $⟨e_i,u{⟩}_c=0$.

The $e_i^{\mathrm{\prime }}$-adjunction. Define $B(P,Q):=⟨{\phi }_{+}(P),{\phi }_{-}(Q){⟩}_c$ on $U_q^{-}\times U_q^{-}$, where ${\phi }_{\pm }$ are the isomorphisms of [F5] and the pairing is extended by zero when the degrees of $P$ and $Q$ differ. Then $B$ is bilinear and homogeneous, and $B(1,1)=⟨1,1{⟩}_c=1$. We prove $B(F_{k}P,Q)=B(P,e_k^{\mathrm{\prime }}Q)$ for all $k$, $P$ and all $Q$ the class of a word $w=F_{j_1}\cdots F_{j_m}$; since the classes of words span $U_q^{-}$ [F1] and both sides are $\mathbb{Q}(q)$-linear in $Q$ (using [F2] for $e_k^{\mathrm{\prime }}$), this gives the adjunction for all $Q$. Let $y:={\phi }_{-}(w)=f_{j_1}\cdots f_{j_m}$. Because ${\phi }_{-}$ is an algebra homomorphism [F5] and $\mathrm{\Delta }$ is an algebra map into the braided tensor square with primitive generators [F4], $\mathrm{\Delta }(y)={\sum }_S\gamma (S){\phi }_{-}(w_S)\otimes {\phi }_{-}(w_{S^c})$ over all subsets $S\subseteq \{1,\dots ,m\}$, where $w_S$ is the subword with letters in $S$ and $\gamma (S)={\prod }_{s<t,s\mathrm{\notin }S,t\in S}{q}^{-⟨{\alpha }_{j_s},{\alpha }_{j_t}⟩}$ records precisely the braided cross-scalars of the product expansion. The second rescaled adjunction and multiplicativity of ${\phi }_{+}$ then give $B(F_{k}P,w)={\sum }_S\gamma (S)⟨e_k,{\phi }_{-}(w_S){⟩}_c⟨{\phi }_{+}(P),{\phi }_{-}(w_{S^c}){⟩}_c\mathrm{.}$ By step 1.2 the factor $⟨e_k,{\phi }_{-}(w_S){⟩}_c$ vanishes when $\mathrm{\mid }S\mathrm{\mid }\ge 2$; it vanishes when $\mathrm{\mid }S\mathrm{\mid }=0$ because ${\phi }_{-}(w_{\mathrm{\varnothing }})=1$ and $⟨e_k,1⟩=0$ [F4]; and when $S=\{t\}$ it equals ${\delta }_{j_{t}k}⟨e_k,f_k{⟩}_c={\delta }_{j_{t}k}$ by [F4] and step 1.1, while for $j_t\mathrm{\ne }k$ it vanishes by the degree support of the pairing. Hence $B(F_{k}P,w)={\sum }_{t:j_t=k}{q}^{-{\sum }_{s<t}⟨{\alpha }_{j_s},{\alpha }_k⟩}B(P,w\setminus t),$ and the right-hand side is $B(P,e_k^{\mathrm{\prime }}(w))$ by the deletion formula of [F2], because $q_i^{-a_{i{j}_s}}=q^{-⟨{\alpha }_{j_s},{\alpha }_i⟩}$. This proves the adjunction, which is axiom (R) of The Lusztig-Kashiwara bilinear form on the negative half: the recursion; by the uniqueness proved there, $B$ is the normalized Lusztig–Kashiwara form determined by (R), (H) and (N).

The matching-permutation formula and symmetry. For words $u=F_{i_1}\cdots F_{i_k}$ and $v=F_{j_1}\cdots F_{j_{k^{\mathrm{\prime }}}}$ one has $B(u,v)=0$ unless $k=k^{\mathrm{\prime }}$ and the multisets of letters agree, in which case $B(F_{i_1}\cdots F_{i_k},F_{j_1}\cdots F_{j_k})={\sum }_{\sigma }{q}^{-{\sum }_{s<t,\sigma (s)>\sigma (t)}⟨{\alpha }_{i_s},{\alpha }_{i_t}⟩},$ where $\sigma$ runs over the bijections $\{1,\dots ,k\}\to \{1,\dots ,k\}$ with $j_{\sigma (t)}=i_t$ for all $t$ and the inner sum is over the pairs $1\le s<t\le k$ with $\sigma (s)>\sigma (t)$. The vanishing statement is the homogeneity (H) of step 1.3, and the formula is proved by induction on $k$: for $k=0$ both sides equal $\epsilon$ of the argument, giving $1$ on the empty word and $0$ otherwise; for $k\ge 1$, step 1.3 and the deletion formula expand $B(F_{i_1}{u}^{\mathrm{\prime }},v)$ as ${\sum }_{t:j_t=i_1}{q}^{-{\sum }_{s<t}⟨{\alpha }_{j_s},{\alpha }_{i_1}⟩}B(u^{\mathrm{\prime }},v\setminus t)$, and for each $t$ the induction hypothesis and the bijection $\sigma \leftrightarrow (\sigma (1)=t,\sigma {\mathrm{\mid }}_{\{2,\dots ,k\}})$ exhibit the terms with $\sigma (1)=t$, the cross-pairs $(1,s)$ with $\sigma (s)<t$ contributing ${\prod }_{r<t}{q}^{-⟨{\alpha }_{j_r},{\alpha }_{i_1}⟩}$ because the letters $j_r$ with $r<t$ are exactly the letters $i_s$ with $\sigma (s)<t$. Finally, $\sigma \mapsto {\sigma }^{-1}$ bijects the matching permutations of $(u,v)$ onto those of $(v,u)$, and inversion pairs correspond with the same two letters in the reversed roles while $⟨{\alpha }_{i_s},{\alpha }_{i_t}⟩=⟨{\alpha }_{i_t},{\alpha }_{i_s}⟩$, so each term matches: $B$ is symmetric on words, hence on $U_q^{-}$ by bilinearity.

Part (i): one-step separation. Let $x\in U_q^{-}[-\nu ]$ satisfy $e_i^{\mathrm{\prime }}(x)=0$ for all $i$, and let $w=F_{i_1}\cdots F_{i_k}$ be any word of degree $\nu$ (then $k=\mathrm{ht}(\nu )\ge 1$). Iterating step 1.3 along the letters of $w$ and using (H) with $B(1,z)=\epsilon (z)$ gives $B(w,x)=\epsilon (e_{i_k}^{\mathrm{\prime }}\cdots e_{i_1}^{\mathrm{\prime }}(x))=\epsilon (0)=0$. Words span the finite-dimensional piece $U_q^{-}[-\nu ]$ [F1], so $B(y,x)=0$ for every $y\in U_q^{-}[-\nu ]$. The restriction of $B$ to $U_q^{-}[-\nu ]$ has zero left kernel by the nondegeneracy of step 1.1, so $x=0$.

Part (ii): iterated separation. Let $0\mathrm{\ne }x\in U_q^{-}[-\nu ]$. By the nondegeneracy of step 1.1 there is $y$ with $B(y,x)\mathrm{\ne }0$; decomposing $y$ into graded components and using the homogeneity of $B$, some word $w$ of degree $\nu$ satisfies $B(w,x)\mathrm{\ne }0$. Write $w=F_{i_1}\cdots F_{i_k}$; iterating step 1.3 as in step 2.1 gives $B(w,x)=\epsilon (e_{i_k}^{\mathrm{\prime }}\cdots e_{i_1}^{\mathrm{\prime }}(x))$, hence $e_{i_k}^{\mathrm{\prime }}\cdots e_{i_1}^{\mathrm{\prime }}(x)\mathrm{\ne }0$, and $\nu ={\alpha }_{i_1}+\cdots +{\alpha }_{i_k}$ with the multiset unique by [F1]. This is the displayed claim with the sequence $i_1,\dots ,i_k$, the first applied derivation $e_{i_1}^{\mathrm{\prime }}$ being the rightmost factor; and if a vector were killed by every composite $e_{i_1}^{\mathrm{\prime }}\cdots e_{i_k}^{\mathrm{\prime }}$ over all such sequences, then applying this to a single reversed order would force $B(w,x)=0$ for every word $w$ and hence $x=0$ by nondegeneracy, so the common kernel is zero. This proves (ii).

The $e_i^{\mathrm{\prime }\mathrm{\prime }}$ version. Let $0\mathrm{\ne }x\in U_q^{-}[-\nu ]$. By [F3], $e_i^{\mathrm{\prime }\mathrm{\prime }}=\psi e_i^{\mathrm{\prime }}\psi$, $\psi$ is an involutive algebra automorphism of $U_q^{-}$ preserving the grading, and $\psi (x)\mathrm{\ne }0$. Apply step 3.1 to $\psi (x)$: there is a sequence $i_1,\dots ,i_k$ with $\nu ={\alpha }_{i_1}+\cdots +{\alpha }_{i_k}$ and $e_{i_k}^{\mathrm{\prime }}\cdots e_{i_1}^{\mathrm{\prime }}(\psi (x))\mathrm{\ne }0$. Since $\psi (e_i^{\mathrm{\prime }}(y))=e_i^{\mathrm{\prime }\mathrm{\prime }}(\psi (y))$ for all $i,y$ and ${\psi }^2=\mathrm{i}\mathrm{d}$, induction on the number of factors gives $\psi (e_{i_k}^{\mathrm{\prime }}\cdots e_{i_1}^{\mathrm{\prime }}(\psi (x)))=e_{i_k}^{\mathrm{\prime }\mathrm{\prime }}\cdots e_{i_1}^{\mathrm{\prime }\mathrm{\prime }}(x)$, which is therefore nonzero because $\psi$ is injective; so (ii) holds with $e_i^{\mathrm{\prime }\mathrm{\prime }}$ in place of $e_i^{\mathrm{\prime }}$. The same computation with step 2.1 proves the one-step statement for $e_i^{\mathrm{\prime }\mathrm{\prime }}$. No canonical basis, integral form, freeness or character statement is used: the argument is the full field-level separation claim on the actual Serre quotient.