Outside-alcove rank-one tiltings have zero pivotal dimension

We first prove the needed zero/nonzero Gaussian rule. Put $z=v^2$; then ${\left[\begin{array}{c}N\\ J\end{array}\right]}_v=v^{-J(N-J)}{\left[\genfrac{}{}{0px}{}{N}{J}\right]}_z$. The generating identity ${\prod }_{r=0}^{N-1}(1+z^{r}t)={\sum }_J{z}^{J(J-1)\mathrm{/}2}{\left[\genfrac{}{}{0px}{}{N}{J}\right]}_z{t}^J$ follows by multiplying one factor and applying Pascal. At primitive $\mathrm{\ell }$th $z$, a full block has product $1-(-t{)}^{\mathrm{\ell }}$. For $N=A\mathrm{\ell }+B$ and $J=C\mathrm{\ell }+D$, $0\le B,D<\mathrm{\ell }$, the product is $(1-(-t{)}^{\mathrm{\ell }}{)}^A{\prod }_{r=0}^{B-1}(1+z^{r}t)$. Its degree-$J$ coefficient is zero when $D>B$, and otherwise is a nonzero root-of-unity factor times $\left(\genfrac{}{}{0px}{}{A}{C}\right){\left[\genfrac{}{}{0px}{}{B}{D}\right]}_z$, provided $0\le C\le A$. The residual Gaussian is nonzero because its numerator and denominator involve only nonzero factors of indices below $\mathrm{\ell }$. Thus ${\left[\begin{array}{c}N\\ J\end{array}\right]}_q$ has exactly that zero/nonzero rule; no phase convention is needed below.

The Weyl module $\mathrm{\Delta }(n)$ is simple for $0\le n<\mathrm{\ell }$ and for $n=a\mathrm{\ell }-1$, $a\ge 1$. Every nonzero submodule contains a nonzero weight vector because full toral characters separate weights. Its weight spaces are one-dimensional, so it contains some $v_j$. The coefficient in $E^{(j)}{v}_j={\left[\begin{array}{c}n\\ j\end{array}\right]}_q{v}_0$ is nonzero: for $n<\mathrm{\ell }$ use the product formula, and for $n=a\mathrm{\ell }-1$ use step 1.1 with residual digit $\mathrm{\ell }-1$. Thus it contains $v_0$ and every $F^{(k)}{v}_0=v_k$. The dual is simple with the same highest weight; the bounded-highest-vector map in [F2] is a nonzero isomorphism $\mathrm{\Delta }(n)\to \mathrm{\nabla }(n)$. These modules are therefore tilting and equal to $T(n)$ by [F2].

Fix $a\ge 1$, $0\le b\le \mathrm{\ell }-2$, and put $m=a\mathrm{\ell }+b$, $m^{\mathrm{\prime }}=a\mathrm{\ell }-b-2$. The module $X=\mathrm{\Delta }(a\mathrm{\ell }-1)\otimes \mathrm{\Delta }(b+1)$ is tilting by step 2.1 and [F2]. Its Weyl labels are $n_j=m-2j$, $0\le j\le b+1$, each once. Use the central quantum Casimir $C=(q-q^{-1}{)}^{2}FE+qK+q^{-1}{K}^{-1}$. Its commutator with $E$ vanishes by substituting $[F,E]=-(K-K^{-1})\mathrm{/}(q-q^{-1})$ and $KE=q^{2}EK$. The involution $E\leftrightarrow F$, $K\leftrightarrow K^{-1}$ fixes $C$ (use the mixed relation), so its commutator with $F$ vanishes too. It commutes with every divided power and toral binomial: the generic commutators vanish in the integral subalgebra before specialization. Its scalar on $\mathrm{\Delta }(n)$ is $c_n=q^{n+1}+q^{-n-1}$, evaluated on the generating highest vector. Now $c_{n_j}=(-1{)}^a(z_j+z_j^{-1})$ with $z_j=q^{b+1-2j}$. Since $0<b+1<\mathrm{\ell }$ and $-(b+1)\le b+1-2j\le b+1$, equality with $c_m$ occurs exactly at $j=0,b+1$: equality of $z+z^{-1}$ means $z=q^{b+1}$ or $q^{-b-1}$. The generalized $c_m$-eigenspace $Y$ is a module summand. Primary-decomposition polynomial projectors preserve submodules and quotients, so applying this projector to a Weyl filtration retains precisely $\mathrm{\Delta }(m)$ and $\mathrm{\Delta }(m^{\mathrm{\prime }})$. Thus $Y$ is tilting with character $\chi (m)+\chi (m^{\mathrm{\prime }})$.

To prove $Y$ indecomposable, first show $\mathrm{\Delta }(m)\cong ̸\mathrm{\nabla }(m)$. The vector $v_{b+1}$ is highest of weight $m^{\mathrm{\prime }}$: for $1\le r\le b+1$, its $E^{(r)}$ coefficient is ${\left[\begin{array}{c}a\mathrm{\ell }-1+r\\ r\end{array}\right]}_q=0$, because its numerator contains $[a\mathrm{\ell }]$ and its denominator indices are below $\mathrm{\ell }$; larger $r$ give zero by the string convention. Its generated submodule is spanned by $F^{(j)}{v}_{b+1}$. Terms with $j>m^{\mathrm{\prime }}$ vanish. Indeed a possibly nonzero basis term requires $j\le a\mathrm{\ell }-1$; in the remaining interval write $j=(a-1)\mathrm{\ell }+d$ with $\mathrm{\ell }-b-1\le d<\mathrm{\ell }$. The upper residual digit of $b+1+j$ is $b+1+d-\mathrm{\ell }<d$, so step 1.1 gives ${\left[\begin{array}{c}b+1+j\\ j\end{array}\right]}_q=0$. The generated submodule is therefore supported on $[-m^{\mathrm{\prime }},m^{\mathrm{\prime }}]$. The highest-vector property [F2] gives a nonzero map $\mathrm{\Delta }(m^{\mathrm{\prime }})\to \mathrm{\Delta }(m)$. Hom orthogonality in [F2] forbids this if $\mathrm{\Delta }(m)\cong \mathrm{\nabla }(m)$, since $m^{\mathrm{\prime }}<m$.

Decompose $Y$ into indecomposable tiltings using [F2]. Its unique highest-weight-$m$ summand is $T(m)$. Its only two Weyl factors are those found in step 3.1, each once. If $Y$ were decomposable, the top summand would have the sole Weyl factor $\mathrm{\Delta }(m)$ and its complement the sole factor $\mathrm{\Delta }(m^{\mathrm{\prime }})$. Then $T(m)=\mathrm{\Delta }(m)$ would be tilting. Weyl characters are triangular and linearly independent, so a dual-Weyl filtration of character $\chi (m)$ has the sole factor $\mathrm{\nabla }(m)$, contradicting step 4.1. Hence $Y=T(m)$, proving the two-factor character formula. Step 2.1 gives the Steinberg formula.

Expanding coevaluation in a weight basis, [F4] gives ${\dim }_q\mathrm{\Delta }(n)={\sum }_{j=0}^n{q}^{n-2j}=[n+1{]}_q$. The two characters in step 5.1 have dimensions $[a\mathrm{\ell }+b+1{]}_q=(-1{)}^a[b+1{]}_q$ and $[a\mathrm{\ell }-b-1{]}_q=(-1{)}^{a+1}[b+1{]}_q$, which cancel. The Steinberg dimension is $[a\mathrm{\ell }{]}_q=0$. For retained labels [F3] gives $[m+1{]}_q>0$.

Let $f$ be an endomorphism of an indecomposable $T(m)$. Over $\mathbb{C}$ it has an eigenvalue $\lambda$. Then $f-\lambda I$ is not invertible and is nilpotent by the Fitting argument of [F2]. It preserves each integral weight space; on that space it has ordinary trace zero, so also weighted trace zero after multiplication by $q^{\text{weight}}$. Therefore $\mathrm{tr}(Kf)=\lambda {\dim }_{q}T(m)$. By step 6.1 this vanishes for every $m\ge \mathrm{\ell }-1$ and every $f$. In particular all test composites have zero trace, so the identity is negligible. Finite sums remain negligible because trace is the sum of diagonal-block traces. For the retained labels simplicity follows from step 2.1 and nonzero dimension from step 6.1. This proves all claims, without a linkage theorem or an oddness hypothesis.