(i)の二個目の等号は、$S$が対称テンソルだからそう.
$Xg(Y,Z)= g(\nabla_X Y, Z) + g(\nabla_Y X, Z) $で、$\alpha + (-\alpha) = 0$だから, $\alpha S_p(X,Y,Z) + (-\alpha) S_p(X,Z,Y) = \alpha S_p(X,Y,Z) + (-\alpha) S_p(X,Y,Z) = 0$
$g(\nabla^{(\alpha)}_{\partial_i} \partial_j ,\partial_k)= g((\overline{\nabla}_{\partial_i} \partial_j ,\partial_k) – \frac{\alpha}{2} S(X,Y,Z)$
\[
\partial_j \log p(\omega) = \frac{\partial_j p(\omega)}{p(\omega)}
\]
と, $\log$の二回微分の計算
\[
\partial_i\partial_j \log p(\omega)
= \frac{(\partial_i\partial_j p(\omega)),p(\omega)-(\partial_i p(\omega))(\partial_j p(\omega))}{p(\omega)^2}.
\]、あるいは別表現
\[
\partial_i\partial_j \log p(\omega)
= \frac{\partial_i\partial_j p(\omega)}{p(\omega)}-\frac{\partial_i p(\omega),\partial_j p(\omega)}{p(\omega)^2} = \frac{\partial_i\partial_j p(\omega)}{p(\omega)} -\quad (\partial_i \log p(\omega)) (\partial_j \log p(\omega))
\]
を使う.
上の $\log$の二回微分の計算を使うと,
\begin{align}
\Gamma_{ij,k}^{(-1)} &= \sum_{\omega =1}^n p(\omega) \{ (\partial_i \log p(\omega))(\partial_j \log p(\omega)) + \partial_i\partial_j \log p(\omega) \} (\partial_k \log p(\omega)) \\
&= \sum_{\omega =1}^n (\partial_i \partial_j p(\omega) ) (\partial_k \log p(\omega))
\end{align}
\[ \partial_b p(\omega)= \delta_b(\omega) – \delta_n(\omega) \]だから,
$ \partial_a\partial_b p(\omega)= 0$である.
したがって, $\Gamma_{ij,k}^{(-1)} =0$がわかる.