Simply-laced braid symmetries preserve the full divided-power form

The Serre relation gives $XY=qYX$ and $YZ=qZY$. Mixed reordering gives $[F_i,Y]=E_j{K}_i^{-1}$ and $[Y,F_j]=K_j{E}_i$. For the first, expand $-H_i{E}_j+q^{-1}{E}_j{H}_i$ and move the torus to the right; its coefficient is $[-(q^{-1}{K}_i-q{K}_i^{-1})+q^{-1}(K_i-K_i^{-1})]\mathrm{/}(q-q^{-1})=K_i^{-1}$. The second similarly has coefficient $H_j(q{K}_j)-q^{-1}{H}_j(K_j)=K_j$. Put $B=-qY$ and $C=-K_i{F}_i$. These identities give $CB-qBC=E_j$, $C{E}_j=q^{-1}{E}_{j}C$ and $E_{j}B=q^{-1}B{E}_j$. Thus $C^{2}B-(q+q^{-1})CBC+B{C}^2=q^{-1}[-CBC+qB{C}^2+E_{j}C]=0$, and the other Serre relation follows by using $E_{j}B=q^{-1}B{E}_j$. These are exactly the positive relations involving color $i$ after the proposed assignment. Applying $\psi$ gives the negative relations with their stated orders and coefficients.

Here are all other positive Serre checks. On a homogeneous element use $d(u)=E_{i}u-q^{⟨\mathrm{w}\mathrm{t}(u),h_i⟩}u{E}_i$, with $d(uv)=d(u)v+q^{⟨\mathrm{w}\mathrm{t}(u),h_i⟩}ud(v)$. For a neighbor, $d(E_j)=Y_{ij}$ and $d^2(E_j)=0$; for a nonneighbor $d(E_j)=0$. If commuting colors $j,k$ both neighbor $i$, apply $d^2$ to their commutator, obtaining $(q+q^{-1})[Y_{ij},Y_{ik}]=0$; if just one neighbors $i$, apply $d$ once. If $j,k$ are neighbors, the ADE diagram has no triangle, so at most one neighbors $i$. Apply $d^2$ to the Serre relation with that color repeated and $d$ to the other Serre relation. For example the three terms with repeated $j$ become respectively $(q+q^{-1})Y_{ij}^2{E}_k$, $(q+q^{-1})Y_{ij}{E}_k{Y}_{ij}$ and $(q+q^{-1})E_k{Y}_{ij}^2$. Their Serre combination vanishes. These cases exhaust the pairs away from $i$; applying $\psi$ gives all negative pairs.

For the mixed checks put $Z_{ik}=F_i{F}_k-q^{-1}{F}_k{F}_i$. For distinct neighbors $j,k$, expansion using their commuting opposite colors gives $[Y_{ij},Z_{ik}]=H_i{E}_j{F}_k-q^{-1}{F}_k{H}_i{E}_j-q^{-1}{E}_j{H}_i{F}_k+q^{-2}{F}_k{E}_j{H}_i$. Move the torus left. Since $E_j{F}_k$ has $i$-weight zero, the coefficient is $(1+q^{-2})H_i-q^{-1}(H_i(q^{-1}{K}_i)+H_i(q{K}_i))=0$. For $j=k$, step 1.1 instead gives $[Y_{ij},Z_{ij}]=-q^{-1}(H_j{K}_i^{-1}+K_j{H}_i)=-q^{-1}(K_i{K}_j-(K_i{K}_j{)}^{-1})\mathrm{/}(q-q^{-1})$, exactly the required mixed relation after multiplying $Y$ by $-q$. Pairs involving $i$ commute because $F_i{Z}_{ij}=q{Z}_{ij}{F}_i$ and $K_i{Z}_{ij}=q^{-1}{Z}_{ij}{K}_i$; the exchanged calculation handles the other half. The repeated $i$ mixed relation becomes $[-K_i{F}_i,-E_i{K}_i^{-1}]=-H_i$. Pairs with fewer neighboring colors commute immediately. Toral relations follow from the reflected weights. Thus the assignment preserves the full presentation.

We record an exact $A_2$ normal-order calculation. Write $x=E_i$, $z=E_j$, $y=xz-q^{-1}zx$ and $x^{(a)}=x^a\mathrm{/}[a]!$, including $y^{(r)}=y^r\mathrm{/}[r]!$ in the generic algebra. Then $z^{(c)}{x}^{(a)}={\sum }_{r=0}^{\min (a,c)}(-1{)}^r{q}^{(a-r)(c-r)+r}{x}^{(a-r)}{y}^{(r)}{z}^{(c-r)}\mathrm{.}$ Indeed $z{x}^{(p)}=q^p{x}^{(p)}z-q{x}^{(p-1)}y$ and $z{y}^{(r)}=q^{-r}{y}^{(r)}z$. Writing $C_r(a,c)=(-1{)}^r{q}^{(a-r)(c-r)+r}$, multiplication by $z$ gives $[c+1]C_r(a,c+1)=q^{a-2r}[c-r+1]C_r(a,c)-q[r]C_{r-1}(a,c)$, where nonexistent terms are zero. Substitution reduces this to $[c+1]=q^{-r}[c-r+1]+q^{c-r+1}[r]$. This proves the formula by induction including both endpoints. In particular $x^{(a)}{z}^{(a+b)}{x}^{(b)}=z^{(b)}{x}^{(a+b)}{z}^{(a)}$: both expand as the sum over $0\le r\le b$ on $x^{(a+b-r)}{y}^{(r)}{z}^{(a+b-r)}$ with coefficient $(-1{)}^r{q}^{(a+b-r)(b-r)+r}{\left[\begin{array}{c}a+b-r\\ a\end{array}\right]}_q$.

Its inverse assignment is $E_i\mapsto -F_i{K}_i^{-1}$, $F_i\mapsto -K_i{E}_i$, $K_h\mapsto K_{s_{i}h}$, and for a neighbor $E_j\mapsto E_i{E}_j-q{E}_j{E}_i$, $F_j\mapsto F_j{F}_i-q^{-1}{F}_i{F}_j$. The same checks with orders reversed prove it is a homomorphism. On a neighboring positive generator one composition is $CB-qBC=E_j$ from step 1.1. For the other put $C^{\mathrm{\prime }}=-F_i{K}_i^{-1}$, $B^{\mathrm{\prime }}=E_i{E}_j-q{E}_j{E}_i$; mixed reordering gives $[F_i,B^{\mathrm{\prime }}]=E_j{K}_i$, hence $C^{\mathrm{\prime }}{B}^{\mathrm{\prime }}=q^{-1}{B}^{\mathrm{\prime }}{C}^{\mathrm{\prime }}-q^{-1}{E}_j$ and $B^{\mathrm{\prime }}{C}^{\mathrm{\prime }}-q{C}^{\mathrm{\prime }}{B}^{\mathrm{\prime }}=E_j$. The negative checks apply $\psi$, and the $i$ and toral compositions cancel directly. Both compositions are the identity on all generators.

On a finite generic rank-one string of highest label $m$ define $Q_i(F_i^{(r)}u)=(-1{)}^r{q}^{r(m-r+1)}{F}_i^{(m-r)}u$. It is invertible and conjugates its one-color action by $S_i$. For a neighbor, $E_j$ commutes $F_i$, and $E_i^2{E}_{j}u=0$. By [F2], $E_{j}u$ splits into a highest vector $a$ of label $m-1$ and $F_{i}b$, with $b$ highest of label $m+1$. On the $a$ component, with $s=m-r$, the coefficient of $Y{F}_i^{(s)}u$ is $[m-s]-q^{-1}[m-s+1]=-q^{-m+s-1}$. Multiplication by $-q$ gives $q^{-r}$, exactly the ratio between the two $Q_i$ string coefficients. On the $F_{i}b$ component it is $[m-s+1]([s+1]-q^{-1}[s])=[r+1]q^{m-r}$; multiplication by $-q$ and the original string coefficient gives $[r+1](-1{)}^{r+1}{q}^{(r+1)(m-r+1)}$, again the coefficient of $Q_i{E}_j$. Therefore $Q_i{E}_j=S_i(E_j)Q_i$.

From step 2.3 and quantum Pascal induction one gets $y^{(n)}={\sum }_{r+s=n}(-1{)}^r{q}^{-r}{x}^{(s)}{z}^{(n)}{x}^{(r)}$. Multiplying by $(-q{)}^n$ and exchanging $r,s$ gives the stated divided formula for $S_i(E_j^{(n)})$; applying $\psi$ gives the negative formula. On color $i$ ordinary multiplication gives $S_i(E_i^{(n)})=(-1{)}^n{q}^{n(n-1)}{K}_i^n{F}_i^{(n)}$ and $S_i(F_i^{(n)})=(-1{)}^n{q}^{-n(n-1)}{E}_i^{(n)}{K}_i^{-n}$. Inverses have the same integral formulas with order reversed. Thus they preserve the subalgebra $B$ generated by divided $E,F$ powers and $K_h$.

For the negative generator start at a lowest vector $l$ of label $-m$. Then $Q_i(E_i^{(r)}l)=d_r(m)E_i^{(m-r)}l$, where $d_r(m)=(-1{)}^{m-r}{q}^{(m-r)(r+1)}$. Now $F_{j}l$ splits into a lowest $a$ of label $-(m-1)$ and $E_{i}b$, with $b$ lowest of label $-(m+1)$. The coefficient of $S_i(F_j)E_i^{(s)}l$ on the first is $-q^{-m+s-1}=-q^{-r-1}$, and $d_r(m)(-q^{-r-1})=d_r(m-1)$. On the second it is $[r+1]q^{m-r}$, and $d_r(m)[r+1]q^{m-r}=[r+1]d_{r+1}(m+1)$. Thus $Q_i{F}_j=S_i(F_j)Q_i$. Nonneighbors commute with the full rank-one action and hence with its explicit $Q_i$. We have proved $Q_{i}u=S_i(u)Q_i$ on every generator.

In $A_2$ the identity $CB-qBC=E_j$ also gives $S_i{S}_j(E_i)=E_j$ and its exchanged version. Applying $\psi$ gives the negative identities. Thus the two three-term products of the $S$ maps agree on the rank-two algebra; both send $E_i$ to $-K_j{F}_j$ and $E_j$ to $-K_i{F}_i$. On its highest module of label $(a,b)$ the corresponding $Q$ products send the highest vector to $F_i^{(b)}{F}_j^{(a+b)}{F}_i^{(a)}u$ and $F_j^{(a)}{F}_i^{(a+b)}{F}_j^{(b)}u$, equal by step 2.3. Each intermediate vector is highest for the next color: for example $E_i{F}_j^{(a+b)}{F}_i^{(a)}u=F_j^{(a+b)}{F}_i^{(a-1)}u=0$, since the latter intermediate highest label for color $j$ is $a+b-1$. Their conjugations agree, so the $Q$ products agree on the entire cyclic module. By [F2] this holds on every finite module after rank-two restriction. For nonneighbors, restrictions are direct sums of tensor products of two independent strings, on which the two $Q$ operators commute.

Finite generic integrable modules separate the algebra. For a nonzero element in triangular form choose an integral toral label $\xi$ at which one nonzero Laurent coefficient stays nonzero; a Laurent polynomial cannot vanish on every grid $(q^{n_1},\dots ,q^{n_r})$. On a lowest-highest tensor of labels $-a,b$, $b-a=\xi$, an ordered positive-negative half product has the corresponding pure tensor as leading term, with a Laurent unit coefficient and corrections of smaller total half degree, by the triangular coproduct. Choose $a,b$ large enough that the exact highest/lowest half kernels in [F3], generated by divided singular powers beyond their limits, have degrees greater than every half degree in this finite expression. The leading term survives, giving a finite module on which the element acts nontrivially. Therefore the operator braid relations from step 6.1, together with the conjugations in steps 4.1 and 5.1 and invertibility of $Q_i$, imply the algebra braid relations on every generator, including colors outside the rank-two pair.

This $B$ is the full form. The toral product gives the adjacent-shift identity $T_{i,c,t}=q^{-t}{T}_{i,c-1,t}+q^{c-t}{K}_i{T}_{i,c-1,t-1}$ for $t\ge 1$. Indeed, factor the common $t-1$ numerator factors with shifts $c-1,\dots ,c-t+1$ and compare the remaining two terms; their difference is $q^{c-t}{K}_i(q^t-q^{-t})$, which cancels the last denominator. Induct on $t$, starting with $T_{i,c,0}=1$ for every $c$. In the equal-exponent reordering $E_i^{(t)}{F}_i^{(t)}={\sum }_{r=0}^t{F}_i^{(t-r)}{T}_{i,2r-2t,r}{E}_i^{(t-r)}$, all terms except $T_{i,0,t}$ belong to $B$ by the induction hypothesis, so $T_{i,0,t}\in B$. The shift identity gives every positive shift successively. Solving it as $T_{i,c-1,t}=q^t{T}_{i,c,t}-q^c{K}_i{T}_{i,c-1,t-1}$ gives every negative shift successively; all coefficients lie in $A$. Hence every toral binomial belongs to $B$. Preservation of $B$ really is preservation of $U_A$, so specialization induces inverse automorphisms of the actual algebra. Their generator formulas also preserve the ordinary form and its inverse map over the stated localization. This proves every assertion, with exactly the inherited Choice in [F4].