The formal quantum shuffle Borel and its Cartan crossed product

For three word blocks, a shuffle of all letters is uniquely a shuffle of the first two blocks followed by a shuffle with the third, and also uniquely a shuffle of the last two followed by one with the first. In either description every inverted pair of letters occurs exactly once and contributes $q^{-⟨\mathrm{deg}{z}_i,\mathrm{deg}{z}_j⟩}$; scalar weights therefore agree, proving associativity. The color degree is additive, and a fixed color multidegree has only finitely many words, each an $R$-basis vector in the tensor-word space. Thus every graded component of $\mathrm{S}\mathrm{h}(V)$ is finite free. Also $R$ is a PID: a nonzero ideal has a least order $m$, an element of that order is ${\mathrm{\hslash }}^m$ times a unit by [F13], and every other element is divisible by ${\mathrm{\hslash }}^m$, so the ideal is $({\mathrm{\hslash }}^m)$. Each $⟨V⟩[\alpha ]$ is a submodule of the finite free word module and hence finite free by [F13]. Tensoring these finite bases with $A_C$ shows that each color-degree component of $\mathcal{S}$ and $\mathcal{V}$ is a finite direct sum of copies of the complete coefficient algebra $A_C$. Consequently each is $\mathrm{\hslash }$-adically complete, and continuous multiplication on a fixed degree split extends to its completion; the finitely many splits of a fixed total degree give the stated completed multiplication.

On a word, $h_i^V$ acts by the sum of the $a_{ij}$ weights of its letters and $D_j^V$ by the number of color-$j$ letters. Every shuffle term preserves these color counts, so each operator satisfies Leibniz for the shuffle product; their diagonal actions commute. Since each sends $v_i$ to a scalar multiple of itself, it preserves the subalgebra $⟨V⟩$.

Repeatedly commute each $X_s$ past a coefficient using $X_{s}x=x{X}_s+{\delta }_s(x)$; the commuting derivations and Leibniz rule give the finite multi-binomial formula in (b). Associativity follows from the two possible reductions of $X_s{X}_{t}x$, which agree because ${\delta }_s{\delta }_t={\delta }_t{\delta }_s$, and the two reductions of $X_{s}xy$, which agree because ${\delta }_s(xy)={\delta }_s(x)y+x{\delta }_s(y)$. Products of formal $\mathrm{\hslash }$-series have finite coefficient sums and each multi-index sum is finite, so multiplication is well-defined coefficientwise.

The principal minor gives $\mathrm{\mid }I\setminus J\mathrm{\mid }=n-r$; the minimal realization has $n$ independent simple coroots and dimension $2n-r$, so the $n$ coroot symbols and $n-r$ supplementary symbols give the required count. This uses no claim that $J$ alone spans the Cartan.

The exponent $Z={\sum }_{t\le k}{d}_{i_t}{H}_{i_t}$ is a polynomial in commuting Cartan variables. Apply [F5] in the commutative coefficient algebra with formal variable $\mathrm{\hslash }$: at order $m$ the coefficient is the finite polynomial $Z^m\mathrm{/}m!$. The coefficient of ${\mathrm{\hslash }}^m$ in $\exp (\mathrm{\hslash }Z)\exp (-\mathrm{\hslash }Z)$ is $Z^m{\sum }_{j=0}^m(-1{)}^{m-j}\left(\genfrac{}{}{0px}{}{m}{j}\right)\mathrm{/}m!$, equal to $0$ for $m>0$ and $1$ for $m=0$, so $G_k$ is invertible with inverse $\exp (-\mathrm{\hslash }Z)$. Since each Cartan variable is primitive, coefficientwise binomial expansion gives $\mathrm{\Delta }(G_k)=\exp (\mathrm{\hslash }(Z\otimes 1+1\otimes Z))=G_k\otimes G_k$.

Fix cuts $u=u_{+}{u}_{-}$ and $v=v_{+}{v}_{-}$, and write $G_{\alpha }=\exp (\mathrm{\hslash }{\sum }_i{\alpha }_i{d}_i{H}_i)$ for a prefix of color degree $\alpha$. If $w_{\beta }$ has color degree $\beta$, then $H_i{w}_{\beta }=w_{\beta }(H_i+{\sum }_j{\beta }_j{a}_{ij})$, so $G_{\alpha }{w}_{\beta }=q^{{\sum }_{i,j}{\alpha }_i{d}_i{\beta }_j{a}_{ij}}{w}_{\beta }{G}_{\alpha }=q^{⟨\alpha ,\beta ⟩}{w}_{\beta }{G}_{\alpha }$. A global shuffle with this cut consists of a shuffle of the prefixes and a shuffle of the suffixes; its cross-cut inversions contribute $q^{-⟨\mathrm{deg}{u}_{-},\mathrm{deg}{v}_{+}⟩}=q^{-⟨\mathrm{deg}{v}_{+},\mathrm{deg}{u}_{-}⟩}$ by [F1]. On the right, multiplying the second tensor factors gives $(G_{\mathrm{deg}{u}_{+}}{u}_{-})(G_{\mathrm{deg}{v}_{+}}{v}_{-})$; moving $u_{-}$ past $G_{\mathrm{deg}{v}_{+}}$ contributes the same cross-cut factor. Rewriting the resulting combined prefix exponential in word-times-Cartan normal form contributes $q^{⟨\mathrm{deg}{u}_{+}+\mathrm{deg}{v}_{+},\mathrm{deg}{u}_{-}+\mathrm{deg}{v}_{-}⟩}$. Rewriting the corresponding Cartan-left factor on the left-hand coproduct gives this identical normal-form factor. The prefix and suffix shuffle weights also agree, so every cut and pair of shuffles has the same coefficient on both sides. Thus $\mathrm{\Delta }$ preserves the shuffle product.

The operators $h_i^V,D_j^V$ are additive over a word cut and annihilate the Cartan symbols, so they fix each Cartan exponential. Hence $\mathrm{\Delta }({\delta }_s(w))=({\delta }_s\otimes \mathrm{i}\mathrm{d}+\mathrm{i}\mathrm{d}\otimes {\delta }_s)\mathrm{\Delta }(w)$, which is exactly the commutator of $\mathrm{\Delta }(X_s)=X_s\otimes 1+1\otimes X_s$ with $\mathrm{\Delta }(w)$. Thus the coproduct preserves every crossed relation and extends as an algebra map to $\mathcal{S}$.

Both iterated coproducts of a word sum over two cuts; the factors agree because each prefix exponential is group-like, so $\mathrm{\Delta }$ is coassociative. Define $\epsilon$ as projection to the empty-word, constant-Cartan coefficient. It is an algebra map: positive color degree maps to zero, and every crossed commutator ${\delta }_s(x)$ with a positive-degree word also maps to zero. The counit identities hold because only the empty-prefix term survives after applying $\epsilon$ to the first factor, only the all-letter prefix term survives after applying it to the second, and every Cartan exponential has counit $1$.

Let $w$ be a word and $P$ a Cartan monomial, and induct lexicographically on (word length, Cartan degree). On Cartan polynomials set $S(P(X))=P(-X)$; these are the antipode equations for the primitive commuting Cartan generators. For positive word length, in $\mathrm{\Delta }(wP)$ the unique term with the full word and full Cartan degree in the first tensor factor is $wP\otimes G_w$. Every other term has a shorter first word or smaller first Cartan degree. Since $G_w$ is invertible, the left convolution equation recursively determines $S(wP)$. The unique term with full word and full Cartan degree in the second tensor factor is $1\otimes wP$; all other terms have a shorter second word or smaller second Cartan degree, so the right convolution equation recursively determines a right inverse as well. The recursions terminate at every pair of finite word length and Cartan degree; each coefficient of every $G_w^{\pm 1}$ is a finite Cartan polynomial, so they extend coefficientwise to the formal $\mathrm{\hslash }$-series in the stated modules. The left and right convolution inverses coincide by associativity. For a letter $v_i$, $\mathrm{\Delta }(v_i)=1\otimes v_i+v_i\otimes G_i$, so the equations give $S(v_i)=-v_i{G}_i^{-1}=-[v_i]\exp (-\mathrm{\hslash }{d}_i{H}_i)$.

In ${\mathrm{Hom}}_R(\mathcal{S}{\widehat{\otimes }}_R\mathcal{S},\mathcal{S})$ with convolution from the completed tensor coalgebra, $S\circ m$ is a two-sided convolution inverse of $m$: since $\mathrm{\Delta }$ and $\epsilon$ are algebra maps, applying the two antipode equations to $ab$ gives both inverse equations. The map $T=m(S\otimes S)\circ \mathrm{f}\mathrm{l}\mathrm{i}\mathrm{p}$ is also a two-sided inverse, since $(T\ast m)(a\otimes b)=\sum S(b_{(1)})S(a_{(1)})a_{(2)}{b}_{(2)}=\epsilon (a)\epsilon (b)1$ and $(m\ast T)(a\otimes b)=\sum a_{(1)}{b}_{(1)}S(b_{(2)})S(a_{(2)})=\epsilon (a)\epsilon (b)1$. These equalities follow by first applying the left and right antipode equations to $a$ and $b$, respectively. Uniqueness of convolution inverses gives $S\circ m=T$, so $S$ is anti-multiplicative. Hence $S(X_s)=-X_s$ and $S([v_i])=-[v_i]\exp (-\mathrm{\hslash }{d}_i{H}_i)$ show $S(\mathcal{V})\subseteq \mathcal{V}$. The coproduct and counit also restrict, so $\mathcal{V}$ is a topological Hopf subalgebra.

The Cartan derivations preserve the two-sided ideal generated by the quantum Serre sums because each sum is homogeneous in the color grading. Under the stated vanishing condition, the free-algebra assignment $e_i\mapsto [v_i]$ kills every defining Serre generator; its Cartan commutators match the crossed-product derivations. By [F9] it therefore factors uniquely through $U_{\mathrm{\hslash }}{\mathfrak{b}}^{+}$ to the stated algebra map.