Outside-alcove indecomposable tiltings are negligible

Every weight $\eta$ of $L(\gamma )$ has norm at most $\mathrm{\parallel }\gamma \mathrm{\parallel }$. Indeed its dominant Weyl conjugate ${\eta }^{+}$ is a weight by [F6], so $\beta =\gamma -{\eta }^{+}\in Q_{+}$. Then $\mathrm{\parallel }\gamma {\mathrm{\parallel }}^2-\mathrm{\parallel }{\eta }^{+}{\mathrm{\parallel }}^2=(\beta ,\gamma +{\eta }^{+})\ge 0$, by dominance of both labels. Every extremal weight $w\gamma$ has multiplicity one, by Weyl invariance and the one-dimensional highest line. Also, if $L(b)$ occurs in $L(a)\otimes L(\gamma )$, then $b-a$ is a weight of $L(\gamma )$. To prove this last assertion, project a highest vector to first-factor weight spaces, and take a maximal first weight $\eta$ in its support. For each simple raising operator its first-factor contribution at $\eta +{\alpha }_i$ cannot cancel a second-factor contribution, because that first weight is absent. Thus the component at $\eta$ lies in the simultaneous raising kernel in the first factor. Simplicity of $L(a)$ makes that kernel precisely its highest line: any other highest vector would generate a proper highest submodule. Consequently $\eta =a$, and the highest vector has nonzero projection to $L(a{)}_a\otimes L(\gamma {)}_{b-a}$. This proves the assertion and the tensor bound $\mathrm{\parallel }b-a\mathrm{\parallel }\le \mathrm{\parallel }\gamma \mathrm{\parallel }$.

Suppose $\mu$ is regular and outside the bottom alcove, and put $y=\mu +\rho$. Its alcove $C$ is inside the strict finite dominant chamber. Choose a point $p$ in the bottom alcove, sufficiently close to a small positive multiple of $\rho$, so that $(p,\alpha )<\mathrm{\ell }$ for every positive root. Perturb $p$ within that open set so that the segment from $p$ to the regular point $y$ crosses no codimension-two intersection of walls. Such a perturbation exists because only finitely many walls meet the bounded segment neighborhood, and each forbidden intersection imposes a proper affine condition on $p$. The segment stays in the strict dominant chamber by convexity and crosses at least one wall because $C$ is not the bottom alcove. Its last crossing enters $C$ through a facet $F$. A crossed wall has positive-root equation $(z,\alpha )=m\mathrm{\ell }$ with $m\ge 1$; the linear pairing on the segment crosses this level upwards, since its initial value is below $\mathrm{\ell }$. Thus $F$ is a lower facet of $C$. Its relative interior is internal to the strict dominant chamber: it is a facet of an alcove in that chamber and is not one of the finite chamber walls.

Apply [F4] to choose a dominant integral label $\lambda$ with $x=\lambda +\rho$ in the relative interior of $F$, and with $\gamma$ the ordinary dominant representative of $\mu -\lambda$ in the closed bottom alcove. Put $\delta =\mu -\lambda \in W\gamma$, so $\mathrm{\parallel }x-y\mathrm{\parallel }=\mathrm{\parallel }\gamma \mathrm{\parallel }$. By [F3], ${\mathrm{\Delta }}_{\zeta }(\gamma )$ is tilting. The tensor $M=T_{\zeta }(\lambda )\otimes {\mathrm{\Delta }}_{\zeta }(\gamma )$ is therefore tilting by [F1]. The label $\lambda$ is affine singular. Every Weyl section of $T_{\zeta }(\lambda )$ is singular by [F2] and has zero dimension by [F3]. Additivity gives $d_{T_{\zeta }(\lambda )}=0$, so its identity, and then $1_M$, are negligible by [F5].

The classical multiplicity of $L(\mu )$ in $L(\lambda )\otimes L(\gamma )$ is one. The Racah formula of [F6] is $c_{\lambda \gamma }^{\mu }={\sum }_{v\in W}\det (v)\dim L(\gamma {)}_{v(y)-x}\mathrm{.}$ The identity term equals one since $y-x=\delta \in W\gamma$, by step 1.1. For $v\mathrm{\ne }1$, strict finite dominance of both $x$ and $y$ gives $(x,y-vy)>0$: $y-vy$ is a nonzero nonnegative root combination, as follows by a reduced reflection word from the dominant $y$. Hence $\mathrm{\parallel }vy-x{\mathrm{\parallel }}^2-\mathrm{\parallel }y-x{\mathrm{\parallel }}^2=2(x,y-vy)>0$. Step 1.1 makes the other weight multiplicities zero. This proves the claimed coefficient without assuming that every extremal tensor label occurs.

Let ${\lambda }^{\mathrm{\prime }}$ be a Weyl section label of $T_{\zeta }(\lambda )$, and suppose a classical tensor constituent $L(\nu )$ occurs in $L({\lambda }^{\mathrm{\prime }})\otimes L(\gamma )$ with $\nu$ in the affine dot orbit of $\mu$. Put $x^{\mathrm{\prime }}={\lambda }^{\mathrm{\prime }}+\rho$, $y^{\mathrm{\prime }}=\nu +\rho$. Step 1.1 gives $\mathrm{\parallel }{y}^{\mathrm{\prime }}-x^{\mathrm{\prime }}\mathrm{\parallel }\le \mathrm{\parallel }\gamma \mathrm{\parallel }=\mathrm{\parallel }y-x\mathrm{\parallel }$. By [F2], $x^{\mathrm{\prime }}$ is obtained from $x$ by a downward chain. The lower-facet conclusion of [F4] therefore gives $\nu \le \mu$. If $\nu =\mu$, the closest-pair equality case provides a common affine isometry $\tau$ with $(x^{\mathrm{\prime }},y)=(\tau x,\tau y)$. Regularity of $y$ forces $\tau =1$, so ${\lambda }^{\mathrm{\prime }}=\lambda$. Weyl flags tensor to Weyl flags and their section multiplicities are exactly the classical coefficients by [F1] and [F6]. The top $\lambda$ occurs once in $T_{\zeta }(\lambda )$; step 3.1 and the exclusion just proved imply $(M:{\mathrm{\Delta }}_{\zeta }(\mu ))=1$, and no Weyl section in the $\mu$ orbit is strictly above $\mu$.

Decompose $M$ into indecomposable tiltings by [F1]. Its $\mu$ Weyl section belongs to some $T_{\zeta }(\tau )$. This forces $\tau \ge \mu$ by highest-weight dominance and places $\tau$ in the same affine orbit by [F2]. The last assertion of step 4.1 forces $\tau =\mu$. Thus $T_{\zeta }(\mu )$ is a direct summand of the negligible object $M$, and its identity is negligible by composing $1_M$ with the inclusion and projection. Its dimension is zero by [F5]. If $\mu$ is singular instead, [F2] and [F3] directly give dimension zero for all its Weyl sections, hence for $T_{\zeta }(\mu )$, and [F5] applies. This covers every label outside $A_{\mathrm{\ell }}$, including its upper boundary. For labels inside $A_{\mathrm{\ell }}$ the final assertion is exactly [F3].