Since the special fibre is reduced, by equation (2.29)

$$c_{1}(L,\|\cdot\|)\land\delta_{X_{\Sigma}}=\frac{1}{e}\sum_{i=0}^{k}\left(\deg_{\mathcal{L}}E_{i}\delta_{\xi_{i}}+\sum_{j\in\Theta_{i}}\deg_{\mathcal{L}}F_{i,j}\delta_{\xi_{i,j}}\right).$$

Denote this measure temporarily by $\mu$. Then

	$\displaystyle(\theta_{\Sigma})_{\ast}(\rho_{\Sigma})_{\ast}\mu$	$\displaystyle=\frac{1}{e}\sum_{i=0}^{k}\left(\deg_{\mathcal{L}}E_{i}+\sum_{j\in\Theta_{i}}\deg_{\mathcal{L}}F_{i,j}\right)\delta_{\xi_{i}}$
		$\displaystyle=\frac{1}{e}\sum_{i=0}^{k}\left(D\cdot E_{i}+\sum_{j\in\Theta_{i}}D\cdot F_{i,j}\right)\delta_{\xi_{i}}$
		$\displaystyle=\frac{1}{e}\sum_{i=0}^{k}\sum_{l=0}^{k}\left(\alpha_{l}E_{l}+\sum_{s\in\Theta_{l}}\alpha_{l,s}F_{l,s}\right)\cdot\left(E_{i}+\sum_{j\in\Theta_{i}}F_{i,j}\right)\delta_{\xi_{i}}$
		$\displaystyle=\frac{1}{e}\sum_{i=0}^{k}\left(\alpha_{i-1}E_{i-1}+\alpha_{i}E_{i}+\alpha_{i+1}E_{i+1}\right)\cdot\left(E_{i}+\sum_{j\in\Theta_{i}}F_{i,j}\right)\delta_{\xi_{i}}$
		$\displaystyle=\frac{1}{e}\sum_{i=0}^{k}(\alpha_{i-1}-2\alpha_{i}+\alpha_{i+1})\delta_{\xi_{i}}.$

In the previous computation, we have used that, since $E_{l}\cdot\operatorname{div}(\varpi)=F_{l,s}\cdot\operatorname{div}(\varpi)=0$, then

	$\displaystyle F_{l,s}\cdot(E_{i}+\sum_{j\in\Theta_{i}}F_{i,j})$	$\displaystyle=0,\text{ for all }i,j,l,s,$
	$\displaystyle E_{l}\cdot(E_{i}+\sum_{j\in\Theta_{i}}F_{i,j})$	$\displaystyle=\begin{cases}0,&\text{ if }l\not=i-1,i,i+1,\\ 1,&\text{ if }l=i-1,i+1,\\ -2,&\text{ if }l=i.\end{cases}$

An analogous computation shows that

(5.64)

$$c_{1}(L,\|\cdot\|_{\mathbb{S}})\land\delta_{X_{\Sigma}}=\frac{1}{e}\sum_{i=0}^{k}(\alpha_{i-1}-2\alpha_{i}+\alpha_{i+1})\delta_{\xi_{i}}.$$

∎