Lemma 8.2.4 (0PCZ) · ScalingStacks

0PCZ

Lemma 8.2.4. Let $\alpha\in G_{n}-\{0\}$ .

Given $i\in(1,n-1)$ , the following assertions are equivalent

(1)

$\alpha(i-n-1)>\alpha(i-n)$
(2)

$L(\alpha_{|\{i-n-1,i-n\}})\neq\emptyset$
(3)

$[i-n-1\to i-n]\in D(\alpha)$
(4)

$\alpha\in G_{n}T_{i}$
(5)

$\alpha T_{i}=0$ .

There exists $i\in(1,n-1)$ such that $\alpha\in G_{n}T_{i}$ if and only if $L(\alpha_{|(-n,-1)})\neq\emptyset$ .

0PD0

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. ∎