The simply-laced large-odd-order tilting fusion quotient

By [F2] the only retained indecomposables are $T_{\zeta }(\nu )$ for $\nu \in A_{\mathrm{\ell }}$, and they are the stated simple Weyl modules. This set is finite: if $\theta ={\sum }_i{a}_i{\alpha }_i$ and $\nu ={\sum }_i{m}_i{\omega }_i$, then $m_i\ge 0$ and ${\sum }_i{a}_i{m}_i\le \mathrm{\ell }-h$, with every $a_i>0$. Each $m_i$ is therefore bounded. The generic quotient conclusion in [F2] now yields all the asserted fusion-category properties.

The Weyl characters are linearly independent over $\mathbb{Z}$: in a finite nonzero relation choose a maximal highest label; its own highest-weight coefficient is one and no lower or incomparable label contributes there, contradicting the relation. Fix $\nu \in A_{\mathrm{\ell }}$ and define an integer linear functional $b_{\nu }$ on their span by $b_{\nu }(\mathrm{ch}{\mathrm{\Delta }}_{\zeta }(\eta ))=\{\begin{array}{ll}\det (w),& \eta =w\cdot \nu \in P^{+},\\ 0,& \eta \mathrm{\notin }{W}_{\mathrm{\ell }}\cdot \nu \mathrm{.}\end{array}$ It is well defined by the independence just proved and uniqueness of $w$ in [F5]. Every character expansion used here is finite.

If $T_{\zeta }(\tau )$ belongs to an affine orbit different from $\nu$'s, [F3] gives $b_{\nu }(\mathrm{ch}{T}_{\zeta }(\tau ))=0$. If it is in that orbit, additivity and the dimension signs in [F3] give $d_{T_{\zeta }(\tau )}=d_{\nu }{b}_{\nu }(\mathrm{ch}{T}_{\zeta }(\tau ))\mathrm{.}$ Here $d_{\nu }\mathrm{\ne }0$. When $\tau \mathrm{\ne }\nu$, it cannot be another open-bottom label by [F5], so its dimension is zero by [F2]; the functional is therefore zero. When $\tau =\nu$, the tilting is ${\mathrm{\Delta }}_{\zeta }(\nu )$ and the functional is one. Thus $b_{\nu }$ of the character of any tilting counts precisely its $T_{\zeta }(\nu )$ summands, hence the multiplicity of the surviving simple in the quotient. Complex dimensions are used only in this equality, never as positive numbers.

For $\lambda ,\mu \in A_{\mathrm{\ell }}$ the tensor of the two simple tiltings equals ${\mathrm{\Delta }}_{\zeta }(\lambda )\otimes {\mathrm{\Delta }}_{\zeta }(\mu )$ and is tilting. Its Weyl-flag character coefficients are $c_{\lambda \mu }^{\eta }$: [F1] gives equality with the classical tensor character and the independence from step 1.2 determines the coefficients, as also expressed in [F4]. Applying $b_{\nu }$ and step 2.1 proves $N_{\lambda \mu }^{\nu }={\sum }_{\begin{array}{c}w\in W_{\mathrm{\ell }}\\ w\cdot \nu \in P^{+}\end{array}}\det (w)c_{\lambda \mu }^{w\cdot \nu }\mathrm{.}$ It is finite since only finitely many classical coefficients are nonzero, and [F5] gives at most one $w$ per label. It is a nonnegative integer because it counts actual summands.

Substitute [F4] into step 3.1. Every regular affine orbit point $a\cdot \nu$ has a unique finite Weyl conjugate in the strict shifted dominant chamber: the finite reflection action folds to the dominant chamber, and its walls have no regular points. Its shifted integral label there is dominant. Consequently the pairs $(w,v)$ with $w\cdot \nu$ dominant and $v\in W$ are in bijection with $a\in W_{\mathrm{\ell }}$ by $a=v^{-1}w$. Also $\det (v)\det (w)=\det (a)$. Regrouping gives $N_{\lambda \mu }^{\nu }={\sum }_{a\in W_{\mathrm{\ell }}}\det (a)\dim {\mathrm{\Delta }}_{\zeta }(\lambda {)}_{a\cdot \nu -\mu }\mathrm{.}$ This regrouping is finite directly: $a\cdot \nu$ must belong to $\mu$ plus the finite weight set of ${\mathrm{\Delta }}_{\zeta }(\lambda )$, and regularity gives at most one $a$ per such point. Thus it is valid independently of any rearrangement of infinite sums. Steps 1.1–4.1 establish all the conclusions using affine linkage and quantum-dimension signs, without an inversion of a monodromy matrix.