p, ) = q + q,\
\[eq:clebschLagrange:hamiltonianDirect:constraint\] = ((q, p, ), ),

for all $ \zeta \in \LieA{g} $.

The proof is by direct inspection. Let the curve $ t \mapsto p(t) $ be defined by $ p(t) \defeq \difpFrac{L}{\dot{q}} \bigl(q, \dot q + \xi \ldot q, \xi \bigr)(t) $. Using the Clebsch–Euler–Lagrange equation , we have $$\label{eq:clebschLagrange:hamiltonian:proof:pTime}
        \DifFrac{}{t} p = \dualPair{p}{\nabla \xi_*} + \difpFrac{L}{q}(q, \dot q + \xi \ldot q, \xi).$$ On the other hand, using the identification  of $ H $ as the Legendre transform of $ L $ with $ v = \dot q + \xi \ldot q $, we get $$\difpFrac{H}{q}(q, p, \xi)
            = \dualPair*{p}{\difpFrac{v}{q}} - \dualPair*{\difpFrac{L}{\dot q}}{\difpFrac{v}{q}} - \difpFrac{L}{q}(q, v, \xi)
            = - \difpFrac{L}{q}(q, \dot q + \xi \ldot q, \xi).$$ Comparing with , we see that  is equivalent to $$\difpFrac{H}{q}(q, p, \xi) = - \DifFrac{}{t} p + \dualPair{p}{\nabla \xi_*}.$$ Let us rewrite the last term on the right-hand side of this equation. For $ \xi \in \LieA{g} $, choose a curve $ \varepsilon \mapsto g_\varepsilon $ in $ G $ with $ g_0 = e $ and $ \difFracAt{}{\varepsilon}{0} g_\varepsilon = \xi $. Let $ \Upsilon: \R \to Q $, $ \Upsilon(\varepsilon) = g_\varepsilon 