ScalingStacks

Proof. Given $s^{\prime}\in I^{\prime}$ and $s=\theta^{\prime}_{s^{\prime}}(1)$, the class $\theta_{s}\circ\theta^{\prime}_{s^{\prime}}$ is admissible, hence $\theta_{s}(0+)\cup\iota(\theta_{s}(0+))=\theta^{\prime}_{s^{\prime}}(1-)\cup\iota(\theta^{\prime}_{s^{\prime}}(1-))$ unless $s\in Z_{exc}$ and one of $\theta_{s}$ and $\theta_{s^{\prime}}$ is the identity, but not the other.

Given $c\in T(Z)$, we put

$$v_{c}=(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket)e_{c}=m_{c}(\llbracket\theta\rrbracket)e_{c}+m_{\iota(c)}(\llbracket\theta\rrbracket)e_{\iota(c)}=v_{\iota(c)}.$$

Let

$$a=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\cap Z_{exc}\\ \theta^{\prime}_{s^{\prime}}=\operatorname{id}\nolimits,\ \theta_{s^{\prime}}\neq\operatorname{id}\nolimits\\ c^{\prime}\in C(s^{\prime})\setminus\bigl((\theta_{s^{\prime}}(0+)\cup\iota(\theta_{s^{\prime}}(0+))\bigr)\end{subarray}}m_{c^{\prime}}(\llbracket\theta^{\prime}\rrbracket)e_{c^{\prime}}=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\cap Z_{exc}\\ \theta^{\prime}_{s^{\prime}}=\operatorname{id}\nolimits,\ \theta_{s^{\prime}}\neq\operatorname{id}\nolimits\\ c^{\prime}\in C(s^{\prime})^{+}\setminus\theta_{s^{\prime}}(0+)\end{subarray}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\llbracket\theta^{\prime}\rrbracket)e_{c^{\prime}}.$$

We have

$$m(\theta\circ\theta^{\prime})-m(\theta)-m(\theta^{\prime})=$$

$$=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ c^{\prime}\in(\theta\circ\theta^{\prime})_{s^{\prime}}(0+)\cup\iota((\theta\circ\theta^{\prime})_{s^{\prime}}(0+))\end{subarray}}m_{c^{\prime}}(\llbracket\theta\rrbracket)e_{c^{\prime}}-\sum_{\begin{subarray}{c}s\in I\\ c\in\theta_{s}(0+)\cup\iota(\theta_{s}(0+))\end{subarray}}m_{c}(\llbracket\theta\rrbracket)e_{c}-a$$

$$=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(0+)}-\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(1-)}-\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}=\operatorname{id}\nolimits\\ c\in C(\theta^{\prime}_{s^{\prime}}(1))\end{subarray}}v_{c}-a$$

Using (7.3.1), we find

	$\displaystyle\langle\llbracket\theta\rrbracket,\llbracket\theta^{\prime}\rrbracket\rangle$	$\displaystyle=-\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ c^{\prime}\in\theta^{\prime}_{s^{\prime}}(0+)\cup\iota(\theta^{\prime}_{s^{\prime}}(0+))\end{subarray}}v_{c^{\prime}}+\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ c\in\theta^{\prime}_{s^{\prime}}(1-)\cup\iota(\theta^{\prime}_{s^{\prime}}(1-))\end{subarray}}v_{c}$
		$\displaystyle=-\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(0+)}+\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(1-)}.$

We deduce that

$$\langle\llbracket\theta\rrbracket,\llbracket\theta^{\prime}\rrbracket\rangle+m(\theta\circ\theta^{\prime})-m(\theta)-m(\theta^{\prime})=-\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}=\operatorname{id}\nolimits\\ c\in C(\theta^{\prime}_{s^{\prime}}(1))^{+}\setminus\iota(\theta^{\prime}_{s^{\prime}}(1-))\\ \theta^{\prime}_{s^{\prime}}(1)\in Z_{exc}\end{subarray}}(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket)e_{c}-a$$

and the first equality of the lemma follows.

Consider $s^{\prime}_{1}\neq s^{\prime}_{2}$ in $I^{\prime}$.

If $s^{\prime}_{1}\in Z_{exc}$, $\theta^{\prime}_{s^{\prime}_{1}}=\operatorname{id}\nolimits_{s^{\prime}_{1}}$ and $\theta_{s_{1}}\neq\operatorname{id}\nolimits_{s_{1}}$, it follows from Lemma 7.3.21 that

$$\sum_{\begin{subarray}{c}s^{\prime}_{2}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}_{2}}\neq\operatorname{id}\nolimits\end{subarray}}i(\operatorname{id}\nolimits_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})=\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}_{2}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}_{2}}\neq\operatorname{id}\nolimits_{s^{\prime}_{2}}\\ c^{\prime}\in C(s^{\prime}_{1})^{+}\end{subarray}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\theta^{\prime}_{s^{\prime}_{2}})=\frac{1}{2}\sum_{c^{\prime}\in C(s^{\prime}_{1})^{+}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\llbracket\theta^{\prime}\rrbracket).$$

Similarly, if $s_{1}\in Z_{exc}$, $\theta^{\prime}_{s^{\prime}_{1}}\neq\operatorname{id}\nolimits$ and $\theta_{s_{1}}=\operatorname{id}\nolimits$, we have

$$\sum_{\begin{subarray}{c}s^{\prime}_{2}\in I^{\prime}\\ \theta_{s_{2}}\neq\operatorname{id}\nolimits\end{subarray}}i(\operatorname{id}\nolimits_{s_{1}},\theta_{s_{2}})=\frac{1}{2}\sum_{c\in C(s_{1})^{+}}(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket).$$

The second equality of the lemma follows. ∎