Quantum Groups at Roots of Unity and Tilting Quotients

By [F1], transport $C$ to a fundamental alcove by an affine isometry, and fix an interior point $p$. For $z=gy$ outside $C$, reflect $z$ in any defining wall $H$ of $C$ whose interior halfspace it violates. No orbit point of $y$ lies on a wall. This step strictly decreases $\mathrm{\parallel }z-p\mathrm{\parallel }$, since $p$ is strictly on the opposite side of $H$. The orbit $W_{\mathrm{\ell }}y$ is a finite union of translates of the discrete lattice $\mathrm{\ell }Q$, so every bounded ball contains finitely many of its points. The strictly decreasing procedure therefore terminates in $C$, at $y$: the regular orbit representative in $C$ is unique by simple transitivity and disjoint alcove interiors in [F1].

Let $d_H$ be signed distance to $H$, positive on the interior side of $C$. Orthogonal decomposition normal to $H$ gives $\mathrm{\parallel }z-x{\mathrm{\parallel }}^2-\mathrm{\parallel }{r}_{H}z-x{\mathrm{\parallel }}^2=-4d_H(z)d_H(x)$. At each folding step $d_H(z)<0$ and $d_H(x)\ge 0$, so distance to $x$ never increases, strictly decreasing unless $x\in H$. Exactly one defining wall contains $x$, namely the wall of $F$. Thus equality of the initial and terminal distances permits only $s_F$ steps. After one such step the point is on the interior side of that wall, so a second consecutive $s_F$ is impossible. Therefore $\mathrm{\parallel }x-gy\mathrm{\parallel }\ge \mathrm{\parallel }x-y\mathrm{\parallel }$, with equality exactly when $gy=y$ or $gy=s_{F}y$. Conversely both these points attain equality because $s_{F}x=x$.

Apply step 2.1 with $g=u^{-1}v$. In the first equality case take $\tau =u$; in the second take $\tau =u{s}_F$, using $s_{F}x=x$. This proves the asserted common-isometry equality description. The stabilizer of $x$ is $\{1,s_F\}$: only two alcoves have $x$ in their closures, and simple transitivity in [F1] identifies the two possible images of either incident alcove. Consequently the minimizing regular points paired with $gx$ are precisely $gy$ and $g{s}_{F}y$.

Suppose $F$ is lower. Its defining reflection subtracts the positive multiple $((y,\alpha )-k\mathrm{\ell })\alpha$, so $s_{F}y<y$. Track both $y_0^{+}=y$ and $y_0^{-}=s_{F}y$, together with $x_0=x$, through the prescribed downward reflections. At each stage the regular points lie in the two alcoves adjacent to the unique facet containing the current $x_j$. Any reflecting hyperplane missing $x_j$ has both adjacent interiors on the same side as $x_j$: an alcove interior cannot meet a wall, and if its interior lay on the other side from a point in its closure that wall would contain the closure point. A strict downward step uses a hyperplane missing $x_j$. Hence both transported regular points satisfy its same strict downward inequality and are reflected by subtracting positive multiples of its positive finite root. Induction yields $gy\le y$ and $g{s}_{F}y\le s_{F}y<y$, where $g=r_N\cdots r_1$.

For a regular orbit point $y^{\mathrm{\prime }}$ with $\mathrm{\parallel }{x}^{\mathrm{\prime }}-y^{\mathrm{\prime }}\mathrm{\parallel }\le \mathrm{\parallel }x-y\mathrm{\parallel }$, step 3.1 forces equality and $y^{\mathrm{\prime }}=gy$ or $g{s}_{F}y$. Both are at most $y$ by step 4.1. If the points are integral weights and $\mathrm{\ell }$ is an integer, every subtracted positive-root coefficient is an integer by the weight pairings, including the initial facet reflection; this gives the asserted integral root order. The argument uses only finite reflection sequences and finite choices from fixed lists of defining walls, hence no Axiom of Choice.

For a homogeneous positive element $x$ of $i$-weight $w$, define $\delta (x)=E_{i}x-v^{w}x{E}_i$. It has weight $w+2$, and multiplication directly gives $\delta (xy)=\delta (x)y+v^{{\mathrm{wt}}_i(x)}x\delta (y)$. Induction using this rule and Gaussian Pascal gives ${\delta }^m(x)={\sum }_{r=0}^m(-1{)}^r{v}^{rw+r(r-1)}G(m,r;v^2)E_i^{m-r}x{E}_i^r$: applying $\delta$ to a summand raises the total weight from $w+2m$ to $w+2m+2$, and if $c_{m,r}$ is the coefficient of $E_i^{m-r}x{E}_i^r$, its two contributions give $c_{m+1,r}=c_{m,r}-v^{w+2m}{c}_{m,r-1}$. Substitution of the displayed coefficient reduces this to $G(m+1,r;v^2)=G(m,r;v^2)+v^{2(m-r+1)}G(m,r-1;v^2)$, exactly Pascal. Equivalently, substituting the Gaussian conversion in [F2], its coefficient is $(-1{)}^r{v}^{r(w+m-1)}{\left[\begin{array}{c}m\\ r\end{array}\right]}_v$.

The ordinary Serre relation is exactly ${\delta }^{a+1}(E_j)=0$, up to its nonzero scalar and reversal of the summation index: here $w=-a$, so the exponent $r(w+m-1)$ is zero when $m=a+1$. Applying the skew Leibniz rule repeatedly to $E_j^n$, every summand of ${\delta }^m(E_j^n)$ is a scalar times ${\delta }^{r_1}(E_j)\cdots {\delta }^{r_n}(E_j)$ with $\sum r_k=m$. If $m>an$, some $r_k>a$, so that factor vanishes by the Serre relation and further applications of $\delta$. For $n=0$, $\delta (1)=0$ gives the same conclusion. Thus ${\delta }^m(E_j^n)=0$ under the asserted inequality.

Divide this zero identity generically by $[m{]}_i![n{]}_j!$ and replace the index $r$ in step 1.1 by $m-r$. The common nonzero scalar is $(-1{)}^m{v}^{m(-an+m-1)}$; the remaining sum is exactly the one in the statement. Each of its factors is a divided power in the actual integral subalgebra, so this generic equality is already an equality there. Tensoring an equality with any specialization retains it; no specialized vanishing factorial is inverted.

The proof for $F$ takes place in the negative half with its positive-height grading: assign degree ${\alpha }_k$ to $F_k$ for this argument and define ${\delta }_{-}(x)=F_{i}x-v^{{\mathrm{wt}}_i(x)}x{F}_i$. Its same-sign Serre relations and the same multiplication computation give every preceding step verbatim, with $E$ replaced by $F$. This completes both integral identities by finite algebraic computations.

In types $A_n,D_n,E_6,E_7$ there is a nonzero minuscule fundamental weight $\omega$ with $(\omega ,\theta )=1$. Here is the finite root-data check: the highest-root marks are respectively $(1,\dots ,1)$, $(1,2,\dots ,2,1,1)$, $(1,2,2,3,2,1)$ and $(2,2,3,4,3,2,1)$ in the standard diagram numbering. The $A,D$ vectors follow directly from $e_1-e_{n+1}$ and $e_1+e_2$ in their coordinate models. For $E_6,E_7$ use the first six or seven simple vectors of the $E_8$ realization specified in the Statement: summing with the stated marks gives respectively $(1,1,1,1,1,-1,-1,1)\mathrm{/}2$ and $-e_7+e_8$, actual roots in that realization, and pairing with the respective simple vectors is nonnegative. They are therefore the highest roots, since a simply-laced irreducible root system has a single dominant root in its root orbit. Every positive root is at most this highest root in the simple-root order, so a fundamental weight at a mark-one node has all positive-root pairings at most one, proving the minuscule assertion and $(\omega ,\theta )=1$.

For a finite facet $i$ form the full-rank finite root subsystem with base $\{-\theta \}\cup \{{\alpha }_j:j\mathrm{\ne }i\}$. Independence follows from $a_i>0$ in the highest-root relation; pairwise off-diagonal inner products are nonpositive, and these roots generate a finite reflection subgroup of $W$. Choose a vector $z$ pairing positively with all these independent roots, and minimize $(w\omega ,z)$ over their finite generated reflection subgroup. Call a minimizer ${\eta }_i$. If it paired positively with one of those roots, reflecting in that root would strictly lower its pairing with $z$, a contradiction. Thus $({\eta }_i,{\alpha }_j)\le 0$ for $j\mathrm{\ne }i$ and $({\eta }_i,\theta )\ge 0$. The remaining pairing $({\eta }_i,{\alpha }_i)$ is positive: if it were nonpositive, all simple pairings and their positive-mark sum would force every pairing zero, hence ${\eta }_i=0$, impossible. By [F2] it is exactly one. Set $r=m_i+1$ and $\lambda =\mu -r{\eta }_i$. Its $i$-th Dynkin coordinate is $-1$, all other finite coordinates are nonnegative, and its highest-root pairing has not increased. Therefore $\lambda +\rho$ lies on exactly the $i$-th finite facet, and $\lambda$ is in its relative interior. The dominant displacement is $r\omega$, whose highest-root pairing is $r\le k+1$.

For the upper facet in the same types put $r=k+1-(\mu ,\theta )>0$ and $\lambda =\mu +r\omega$. Every finite Dynkin coordinate stays nonnegative and the highest-root pairing becomes $k+1$, so $\lambda$ is in exactly that facet's relative interior. The dominant displacement is the dominant representative of $-r\omega$, whose highest-root pairing is again $r$ (the Weyl involution $-w_0$ preserves the highest root). Since $r\le k+1$, it lies in the required closed alcove.

It remains to handle $E_8$, where $h=30$, $k$ is positive odd, and the marks are $(2,3,4,6,5,4,3,2)$. Use its root model consisting of $\pm e_i\pm e_j$ and half sums with an even number of minus signs. Fix ${\alpha }_1=(e_1-e_2-e_3-e_4-e_5-e_6-e_7+e_8)\mathrm{/}2$, ${\alpha }_2=e_1+e_2$, ${\alpha }_3=e_2-e_1$, and ${\alpha }_j=e_{j-1}-e_{j-2}$ for $4\le j\le 8$. Their Cartan pairings give the $E_8$ diagram and the marked sum is $\theta =e_7+e_8$, whose Dynkin coordinates are $(0,0,0,0,0,0,0,1)$. For finite facets $i=1,\dots ,8$ take respectively the roots ${\eta }_1=-e_7+e_8,{\eta }_2=(e_1+e_2+e_3+e_4+e_5-e_6-e_7+e_8)\mathrm{/}2,{\eta }_3=(-e_1+e_2+e_3+e_4-e_5-e_6-e_7+e_8)\mathrm{/}2,{\eta }_4=-e_1-e_2,{\eta }_5=e_4+e_8,{\eta }_6=e_5+e_8,{\eta }_7=e_6+e_8,{\eta }_8=e_7+e_8\mathrm{.}$ Their Dynkin coordinate vectors, obtained by the displayed dot products, are respectively $(1,0,0,0,0,0,0,-1)$, $(0,1,0,0,0,0,-1,0)$, $(0,0,1,0,0,-1,0,0)$, $(0,-2,0,1,0,0,0,0)$, $(0,0,0,0,1,-1,0,0)$, $(0,0,0,0,0,1,-1,0)$, $(0,0,0,0,0,0,1,-1)$ and $(0,0,0,0,0,0,0,1)$. Thus each has pairing one at node $i$, nonpositive pairings elsewhere, and nonnegative highest-root pairing (respectively $0,0,0,0,1,1,1,2$). Set $r=m_i+1$ and $\lambda =\mu -r{\eta }_i$. The same coordinate inequalities as step 2.1 place $\lambda$ in the required finite facet's relative interior. Every root is Weyl conjugate to $\theta$, so the dominant displacement is $r\theta$. Because $a_i\ge 2$ and $k$ is odd, $m_i\le (k-1)\mathrm{/}2$, hence $(r\theta ,\theta )=2r\le k+1$.

For the $E_8$ upper facet let $r=k+1-(\mu ,\theta )$. If $r$ is even, take $\lambda =\mu +(r\mathrm{/}2)\theta$. It is dominant, is in that facet's relative interior, and its dominant displacement has highest-root pairing $r\le k+1$. If $r$ is odd, $(\mu ,\theta )$ is odd since $k+1$ is even, so $m_j>0$ for some odd-mark node $j\in \{2,5,7\}$. Choose respectively ${\beta }_2=-e_1+e_8$, ${\beta }_5=e_3+e_8$, ${\beta }_7=e_5+e_8$. Their Dynkin coordinate vectors are $(0,-1,1,0,0,0,0,0)$, $(0,0,0,1,-1,0,0,0)$ and $(0,0,0,0,0,1,-1,0)$, and each has highest-root pairing one. With $u=(r-1)\mathrm{/}2$ put $\lambda =\mu +u\theta +{\beta }_j$. Its only potentially negative coordinate is repaired by $m_j\ge 1$, and its highest-root pairing increases by $2u+1=r$, placing it exactly in the upper facet's relative interior. For every root $\alpha$, $\mathrm{\mid }(u\theta +{\beta }_j,\alpha )\mathrm{\mid }\le 2u+2=r+1$, since all root pairings have absolute value at most two. The highest-root pairing of the dominant displacement is the maximum of its pairings with roots, hence at most $r+1$. Oddness of $r$ gives $r\le k$, so this is at most $k+1$. This proves the upper-facet assertion too.

For an arbitrary affine alcove write $\mu =w\cdot {\mu }_0$ with ${\mu }_0$ in the bottom alcove and transport the constructed facet point by the same dot action. The difference of the two transported points is the finite linear Weyl part of $w$ applied to their original difference, so its ordinary dominant representative is unchanged. If the transported facet is inside the strict shifted dominant chamber, each shifted simple pairing at its relative-interior integral point is a positive integer, and subtracting $\rho$ leaves nonnegative Dynkin coordinates. Thus the transported point is dominant exactly as required. No tensor decomposition conclusion has been used or asserted.

If $x\in M_A$ is bar fixed and belongs to $b+vL$ for $b\in B$, then $x=b$; if it belongs to $vL$, then $x=0$. Indeed write $x={\sum }_{c\in B}{a}_{c}c$ with $a_c\in A$. Being in $L$ means each coefficient is regular at zero, so it lies in $\mathbb{Z}[v]$; being bar fixed means $a_c(v)=a_c(v^{-1})$, so each such polynomial is constant. Its residue therefore determines the entire coefficient. This proves both assertions without a positivity or norm assumption.

Choose a maximal label $\lambda$ in the finite weight support of $M$. Every vector of its basis subset $B_{\lambda }$ is highest. Generic rank-one integrability makes $\lambda$ dominant, and [F2] gives for each $b\in B_{\lambda }$ an injective homomorphism ${\varphi }_b:V(\lambda )\to M$ carrying its highest vector to $b$. Its integral lattice maps into $M_A$ because it is generated by integral divided powers; its bar commutes with the target bar since both commute with the $U$ action and fix the highest vector. The map also commutes with all crystal operators by the unique string decomposition, and therefore maps the source crystal lattice into $L$. Thus every source global vector maps to an integral bar-fixed vector congruent either to a member of $B$ or to zero modulo $vL$. Step 1.1 and field injectivity show that every image is a member of $B$, and never zero.

The basis images for distinct $b,c\in B_{\lambda }$ are disjoint. Otherwise a common image basis vector is congruent to a nonzero crystal lowering path from each of $b,c$. Apply the reversed raising path from $c$ to that common vector; it recovers $c$ in the residue crystal. Because ${\varphi }_b$ commutes with crystal operators, this produces a source crystal vector in $V(\lambda )$ at weight $\lambda$ whose image has residue $c$. The source highest-weight space has basis the highest vector alone, hence its image residue is $b$, forcing $b=c$. Consequently the direct sum of the ${\varphi }_b$ identifies the direct sum of actual integral highest lattices with the basis-subset lattice spanned by their disjoint images. This lattice is saturated, $U_A$-stable and has a free complementary basis quotient. No equality of arbitrary tensor-product Hom ranks before and after specialization has been used.

Remove that basis-subset submodule and apply [F1] to its based quotient. Its dimension is strictly smaller, and induction on the finite dimension constructs a finite filtration with sections equal to direct sums of integral highest lattices; refine each such section by its finitely many copies. The initial zero module has the empty filtration. In every short exact sequence of this filtration the quotient is free as an $A$-module, so choosing its displayed basis gives an $A$-linear splitting and tensoring with any specialized field preserves injectivity as well as surjectivity. The specialized sections are exactly the root Weyl lattices by definition. This proves the complete assertion, with Choice confined to [F4].

Expanding a $q$-commuting binomial by induction gives the two displayed divided coproducts: the induction recurrence is the quantum Pascal recurrence in [F3], and the toral orders shown in the statement are retained. Directly applying the anti-map in [F1], $(E_i{K}_i{)}^r=q_i^{r(r-1)}{E}_i^r{K}_i^r$ and $(K_i^{-1}{F}_i{)}^r=q_i^{-r(r-1)}{K}_i^{-r}{F}_i^r$ give the displayed antipodes, before specialization. Thus these maps preserve every divided generator and every toral group element.

Here is the toral calculation, including arbitrary shifts. For nonnegative integers $m,n$ the coefficient identity obtained by splitting the product ${\prod }_{j=0}^{m+n-1}(1+q^{m+n-1-2j}z)$ into blocks of lengths $m,n$ is ${\left[\begin{array}{c}m+n\\ t\end{array}\right]}_q={\sum }_{a+b=t}{q}^{-bm+an}{\left[\begin{array}{c}m\\ a\end{array}\right]}_q{\left[\begin{array}{c}n\\ b\end{array}\right]}_q$. Regard this as an identity of Laurent polynomials in independent $X,Y$ over $\mathbb{Q}(q)$, replacing $q^m,q^n$ by $X,Y$. It holds for every sufficiently large pair $m,n$, and a Laurent polynomial vanishing on all pairs $(q^m,q^n)$ is zero: fixing one variable gives infinitely many distinct roots, then applying this again to the coefficients proves the assertion. Substituting $X=q^{c}K\otimes 1$ and $Y=1\otimes K$ yields exactly the displayed $\mathrm{\Delta }{T}_{i,c,t}$. Its right side belongs to $U_A{\otimes }_A{U}_A$ with no denominator.

Substituting $K\mapsto K^{-1}$ into the product defining $T_{i,c,t}$, negating each numerator and reversing the $t$ factors gives $S(T_{i,c,t})=(-1{)}^t{T}_{i,t-c-1,t}$. Substituting $K=1$ gives its stated counit; negative upper indices have the polynomial expression stated above, by reversing and negating the numerator factors. Hence the counit has values in $A$ and the antipode preserves all toral binomials.

By steps 1.1–1.3 all generic structure maps restrict to the integral form. Since its tensor powers inject into the generic tensor powers by [F2], the generic coassociativity, counit and antipode identities hold as integral identities. Moreover $S^2(E_i^{(r)})=q_i^{-2r}{E}_i^{(r)}$, $S^2(F_i^{(r)})=q_i^{2r}{F}_i^{(r)}$ and $S^2$ fixes the torus and all its binomials. These formulas and their inverses preserve the integral generators, so $S^2$ and hence $S$ are bijective on $U_A$. Base change preserves every identity and the inverse antipode. This proves all assertions without division by any vanishing root factorial.

For an indeterminate $Q$ put $P_N(t)={\prod }_{j=0}^{N-1}(1+Q^{j}t)$ for $N\ge 0$ and $P_N(t)={\prod }_{j=N}^{-1}(1+Q^{j}t{)}^{-1}$ for $N<0$, with inverses expanded as formal power series. Define $G(N,r;Q)$ by $P_N(t)={\sum }_{r\ge 0}{Q}^{r(r-1)\mathrm{/}2}G(N,r;Q)t^r$. Multiplying the last factor gives the Gaussian Pascal recurrence, so for $N\ge 0$ this is the Gaussian of [F1]. For $N<0$, the same recurrence downward from $N=0$ agrees with the generalized Gaussian product; each fixed coefficient uses finitely many terms of finitely many geometric series and is a Laurent polynomial in $Q$. Thus evaluation at a root of unity is legitimate, including negative $N$.

Set $Q=z^2$, which has exact order $\mathrm{\ell }$. Since $\mathrm{\ell }$ is odd, ${\prod }_{j=0}^{\mathrm{\ell }-1}(1+Q^{j}t)=1+t^{\mathrm{\ell }}$. Periodicity of the factors gives $P_N(t)=(1+t^{\mathrm{\ell }}{)}^{N_1}{P}_{N_0}(t)$, also for $N_1<0$ by formal inversion. The coefficient of $t^r$ is $\left(\genfrac{}{}{0px}{}{N_1}{r_1}\right)Q^{r_0(r_0-1)\mathrm{/}2}G(N_0,r_0;Q)$: the second factor has degree less than $\mathrm{\ell }$, so its only eligible degree is $r_0$. The difference $r(r-1)\mathrm{/}2-r_0(r_0-1)\mathrm{/}2$ is divisible by $\mathrm{\ell }$, because $r-r_0$ is divisible by the odd integer $\mathrm{\ell }$. Consequently $G(N,r;Q)=\left(\genfrac{}{}{0px}{}{N_1}{r_1}\right)G(N_0,r_0;Q)$.

In the symmetric conversion of [F1], the difference $r(N-r)-r_0(N_0-r_0)$ is divisible by $\mathrm{\ell }$. Therefore the factors $z^{-r(N-r)}$ and $z^{-r_0(N_0-r_0)}$ agree, proving the displayed quantum Lucas identity. For $0<r<\mathrm{\ell }$, $[r{]}_z\mathrm{\ne }0$, since otherwise $z^{2r}=1$ would contradict the exact order of $z^2$. Hence $[r{]}_z!\mathrm{\ne }0$ for $r<\mathrm{\ell }$.

Lucas gives ${\left[\begin{array}{c}a\mathrm{\ell }+r\\ r\end{array}\right]}_z=1$ and ${\left[\begin{array}{c}(a+1)\mathrm{\ell }\\ \mathrm{\ell }\end{array}\right]}_z=a+1$. Apply [F2] first repeatedly to $X^{(\mathrm{\ell })}$, obtaining $(X^{(\mathrm{\ell })}{)}^a=a!X^{(a\mathrm{\ell })}$, and then to $X^{(a\mathrm{\ell })}{X}^{(r)}$, obtaining $X^{(a\mathrm{\ell }+r)}=X^{(a\mathrm{\ell })}{X}^{(r)}$. Finally $X^{(r)}=X^r\mathrm{/}[r{]}_z!$ for $r<\mathrm{\ell }$. These prove the generation and quotient assertions, including $r=0$ and $a=0$, using only finite computations.