接続の変換則を導出する．
まず，共変微分 $\nabla_{\mu}A^{\nu}(x)$ が $(1,1)$ 型テンソルとして変換することを要請する：
\begin{align}
\nabla’_{\mu}A’^{\nu}(x’)
=\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\sigma}}\nabla_{\rho}A^{\sigma}(x).
\label{req}
\end{align}
\eqref{req} の右辺に \eqref{covariantderivative}を代入すると，
\begin{align}
\nabla’_{\mu}A’^{\nu}(x’)
&=\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\sigma}}\left(\partial_{\rho}A^{\sigma}(x)+\Gamma_{\rho\lambda}^{\sigma}(x)A^{\lambda}(x)\right)\notag\\
&=\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\sigma}}\partial_{\rho}A^{\sigma}(x)+\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\sigma}}\Gamma_{\rho\lambda}^{\sigma}(x)A^{\lambda}(x)
\label{nA1}
\end{align}
となる．
一方で，共変微分の定義の座標変換は，
\begin{align}
\nabla’_{\mu}A’^{\nu}(x’)
&=\partial’_{\mu}A’^{\nu}(x’)+\Gamma_{\mu\rho}’^{\nu}(x’)A’^{\rho}(x’)\notag\\
&=\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\sigma}}\partial_{\rho}A^{\sigma}(x)+\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial^2 x’^{\nu}}{\partial x^{\rho}\partial x^{\sigma}}A^{\sigma}(x)+\Gamma_{\mu\rho}’^{\nu}(x’)\frac{\partial x’^{\rho}}{\partial x^{\sigma}}A^{\sigma}(x)
\label{nA2}
\end{align}
となる．
ここで，\eqref{pA} を用いた．
\eqref{nA1} と \eqref{nA2} を比較すると，
\begin{align}
\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\sigma}}\Gamma_{\rho\lambda}^{\sigma}(x)A^{\lambda}(x)=\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial^2 x’^{\nu}}{\partial x^{\rho}\partial x^{\sigma}}A^{\sigma}(x)+\Gamma_{\mu\rho}’^{\nu}(x’)\frac{\partial x’^{\rho}}{\partial x^{\sigma}}A^{\sigma}(x)
\end{align}
が得られる．
左辺のダミー添字 $\sigma, \lambda$ を付け替えると，
\begin{align}
\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\lambda}}\Gamma_{\rho\sigma}^{\lambda}(x)A^{\sigma}(x)=\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial^2 x’^{\nu}}{\partial x^{\rho}\partial x^{\sigma}}A^{\sigma}(x)+\Gamma_{\mu\rho}’^{\nu}(x’)\frac{\partial x’^{\rho}}{\partial x^{\sigma}}A^{\sigma}(x)
\end{align}
となる．
$A^{\sigma}(x)$ は任意の反変ベクトルなので，
\begin{align}
\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\lambda}}\Gamma_{\rho\sigma}^{\lambda}(x)=\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial^2 x’^{\nu}}{\partial x^{\rho}\partial x^{\sigma}}+\Gamma_{\mu\rho}’^{\nu}(x’)\frac{\partial x’^{\rho}}{\partial x^{\sigma}}
\end{align}
となり，$\Gamma_{\mu\rho}’^{\nu}(x’)$ を含む項について解くと，
\begin{align}
\Gamma_{\mu\rho}’^{\nu}(x’)\frac{\partial x’^{\rho}}{\partial x^{\sigma}}=\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\lambda}}\Gamma_{\rho\sigma}^{\lambda}(x)-\frac{\partial x^{\rho}}{\partial x’^{\mu}}\frac{\partial^2 x’^{\nu}}{\partial x^{\rho}\partial x^{\sigma}}.
\end{align}
最後に，$\frac{\partial x’^{\rho}}{\partial x^{\sigma}}$ の逆変換を両辺に作用させることで，
\begin{align}
\Gamma_{\mu\rho}’^{\nu}(x’)
=\frac{\partial x^{\tau}}{\partial x’^{\rho}}\frac{\partial x^{\sigma}}{\partial x’^{\mu}}\frac{\partial x’^{\nu}}{\partial x^{\lambda}}\Gamma_{\sigma\tau}^{\lambda}(x)-\frac{\partial x^{\tau}}{\partial x’^{\rho}}\frac{\partial x^{\sigma}}{\partial x’^{\mu}}\frac{\partial^2 x’^{\nu}}{\partial x^{\sigma}\partial x^{\tau}}
\end{align}
が得られる．
これが接続の変換則である．