\cdot q $ be the orbit curve through $ q \in Q $. Considering $ \varepsilon \mapsto g_\varepsilon \ldot \dot{q} $ as a section of $ \Upsilon^* \TBundle Q $ for $ \dot{q} \in \TBundle_q Q $, we have $$\nabla_{\difp_\varepsilon}^\Upsilon (g_\varepsilon \ldot \dot{q})|_{\varepsilon = 0}
            = K \left(\difFracAt{}{\varepsilon}{0} g_\varepsilon \ldot \dot{q}\right)
            = K (\xi \ldot \dot{q})
            = K \bigl(\tangent \xi_* (\dot{q})\bigr)
            = \nabla_{\dot{q}} \xi_* \, .$$ A similar calculation yields $ \bar{\nabla}_{\difp_\varepsilon}^\Upsilon (g_\varepsilon \ldot p)|_{\varepsilon = 0} = \bar{K}(\xi \ldot p) $ for $ p \in \CotBundle_q Q $. Thus, taking the derivative of the identity $ \dualPair{p}{\dot{q}} = \dualPair{g_\varepsilon \cdot p}{g_\varepsilon \ldot \dot{q}} $ with respect to $ \varepsilon $ gives $$\label{eq:clebschLagrange:hamiltonian:proof:nablaXi}\begin{split}
        0 
        &= \difFracAt{}{\varepsilon}{0} \dualPair{g_\varepsilon \cdot p}{g_\varepsilon \ldot \dot{q}}
        \\
        &= \dualPair*{p}{\nabla_{\difp_\varepsilon}^\Upsilon (g_\varepsilon \ldot \dot{q})|_{\varepsilon = 0}} + \dualPair*{\bar{\nabla}_{\difp_\varepsilon}^\Upsilon (g_\varepsilon \ldot p)|_{\varepsilon = 0}}{\dot{q}}
        \\
        &= \dualPair{p}{\nabla_{\dot{q}} \xi_*} + \dualPair{\bar{K}(\xi \ldot p)}{\dot{q}}.
    \end{split}$$ That is $ \dualPair{p}{\nabla \xi_*} = - \bar{K}(\xi \ldot p) $. Hence, in summary, the first equation in  is equivalent to .

Moreover,  implies $$\difpFrac{H}{p}(q, p, \xi)
            = v + \dualPair*{p}{\difpFrac{v}{p}} - \dualPair*{\difpFrac{L}{\dot q}(q, v, \xi)}{\difpFrac{v}{p}}
            = \dot q + \xi \ldot q,$$ which yields the second equation in .

Similarly, the derivative of $ 