Integral convex root PBW bases in finite simply-laced type

If $\mathrm{\ell }(w{s}_i)=\mathrm{\ell }(w)+1$, then $S_w(E_i)$ is positive. Prove this by induction on $\mathrm{\ell }(w)$. Choose a right descent $j$ of a nonidentity $w$; necessarily $j\mathrm{\ne }i$. Write $w=w^{\mathrm{\prime }}v$ with $w^{\mathrm{\prime }}$ the minimal right coset representative for $\{i,j\}$ and $v$ in that rank-two parabolic. Both $i,j$ are ascents of $w^{\mathrm{\prime }}$. Since $i$ is an ascent and $j$ a descent of $v$, its alternating reduced word has length at most $m_{ij}-1$. The rank-two calculations in [F1] therefore make $S_v(E_i)$ either a simple generator or one of the two neighboring positive polynomials $E_j{E}_i-q{E}_i{E}_j$ and $E_i{E}_j-q{E}_j{E}_i$. Induction makes both $S_{w^{\mathrm{\prime }}}(E_i),S_{w^{\mathrm{\prime }}}(E_j)$ positive, hence also $S_w(E_i)$. Its divided powers belong to $U_A$ by [F1], so [F2] puts them in $U_A^{+}$. Apply this to every prefix in the chosen reduced word.

Let $M_{\mathbf{i}}$ be the $A$-span of the ordered divided root monomials for the word $\mathbf{i}$. A braid move changes only its local root block; later factors agree by the braid relation. For commuting colors the block spans agree by commutation. For $A_2$ put $x=E_i$, $z=E_j$, $y=xz-q^{-1}zx$. Up to Laurent units its three root factors are $x,y,z$. The exact identity from [F1] is $z^{(c)}{x}^{(a)}={\sum }_{r=0}^{\min (a,c)}(-1{)}^r{q}^{(a-r)(c-r)+r}{x}^{(a-r)}{y}^{(r)}{z}^{(c-r)}\mathrm{.}$ It shows that the span of $x^{(a)}{y}^{(b)}{z}^{(c)}$ contains $1$ and is stable under every left divided power of $x$ and $z$: use $zy=q^{-1}yz$ and divided multiplication after the displayed formula. Thus it is the entire divided rank-two half. The reversed block has the same property, so both integral block spans agree. Applying their common preceding $S$ prefix preserves equality. By [F4] the space $M_{\mathbf{i}}$ is independent of the reduced word.

Every simple color $i$ starts some reduced word for $w_0$. In that word $E_i^{(r)}$ multiplies the first root factor by the Gaussian coefficient ${\left[\begin{array}{c}r+a_1\\ r\end{array}\right]}_q\in A$. Hence the common space from step 2.1 is stable under every divided simple generator. It contains $1$ and lies in $U_A^{+}$ by step 1.1, so it equals $U_A^{+}$. In any fixed weight $\nu$, the ordered root monomials number the Kostant partitions of $\nu$, exactly the generic dimension in [F3]. Since they span, they are linearly independent over the generic field and therefore over $A$. They form an integral basis, not just a torsion-free spanning set. The half exchange gives the negative result. Base change of these actual free bases retains the bases at every scalar parameter.

Repeat the block-span argument for ordinary powers over $A[(q-q^{-1}{)}^{-1}]$. In $A_2$ the relations $zx=qxz-qy$, $xy=qyx$ and $yz=qzy$ reorder ordinary words with coefficients in this ring and no divided factorial. Thus the ordinary block span is the entire ordinary rank-two algebra, and the common-span argument applies. A positive root vector belongs to the ordinary half: it lies in the ordinary full form by [F1], and projecting an ordinary mixed normal-order expansion onto negative degree zero and toral exponent zero maps each normal-ordered word into the ordinary positive half and fixes a pure positive element. Mixed reordering uses only the inverted $q-q^{-1}$. This justifies the positivity-intersection step for the ordinary form without assuming its integral PBW theorem. Its common root-monomial span is stable under all ordinary $E_i$, by choosing a word starting $i$, and hence equals its ordinary half. Exchanging halves completes the ordinary assertion.