ScalingStacks

Proof. The equivalence between (1) and (2) follows from Lemma 7.4.20.

Assume (2). We deduce that $[i-n-1\to i-n]\in L(\alpha)$, hence $[i-n-1\to i-n]\in D(\alpha)$. So (3) holds.

Assume (3). Writing $\alpha=\alpha\cdot 1$, we deduce from Lemma 7.4.35 that (4) holds.

The implication (4)$\Rightarrow$(5) is immediate.

Asssume (5). We have $\alpha_{|\{i-n-1,i-n\}}\cdot([i-n-1\to i-n]\boxtimes[i-n\to i-n-1])=0$ by Remark 7.4.11. Lemma 7.4.9 shows that $i(\alpha_{|\{i-n-1,i-n\}})\neq 0$, hence (2) holds.

Assume now $L(\alpha_{|(-n,-1)})\neq\emptyset$. It follows from Lemma 7.4.20 that there is $i\in(1,n-1)$ with $\alpha(i-n-1)>\alpha(i-n)$, hence $\alpha\in G_{n}T_{i}$. This shows the last statement of the lemma. ∎