ScalingStacks

Assume ζ∈L⁡(θ)\zeta\in L(\theta). Since ζ¯\bar{\zeta} is negative, it follows that ζ\zeta is positive, then (θζ)s=θs(\theta^{\zeta})_{s}=\theta_{s} for s∉{s1,s2}s{\not\in}\{s_{1},s_{2}\}, while (θζ)s1=βs2∘ζ(\theta^{\zeta})_{s_{1}}=\beta_{s_{2}}\circ\zeta and (θζ)s2=αs1∘ζ−1(\theta^{\zeta})_{s_{2}}=\alpha_{s_{1}}\circ\zeta^{-1}. We deduce that (β⋅u)⊠v=θζ(\beta\cdot u)\boxtimes v=\theta^{\zeta} if β⋅u≠0\beta\cdot u\neq 0. So, the assertion (8.1.3) is a consequence of the following:

(8.1.4) given ​ζ∈L⁡(θ)​ positive, we have ​β⋅u≠0​ if and only if ​ζ∈D⁡(θ).\text{given }\zeta\in L(\theta)\text{ positive, we have }\beta\cdot u\neq 0\text{ if and only if }\zeta\in D(\theta).

Original mathematics by the credited authors. Source collection and HTML conversion remain in progress.

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2