$. We stress that the latter symbol will always stand for the mapping given by , $$\dualPair*{\difpFrac{L}{\dot{q}}(q, v)}{w}
        = \difFracAt{}{\varepsilon}{0} L(q, v + \varepsilon w) \,, 
        \qquad w \in \TBundle_q Q.$$ In order to define the partial derivative $ \difpFrac{L}{q} $ in an intrinsic way, we need a linear connection in $ \TBundle Q $. By , every connection $ K: \TBundle (\TBundle Q) \to \TBundle Q $ in the bundle $ \TBundle Q \to Q $ yields a vector bundle isomorphism $$\label{eq:clebschLagrange:decompositionTTwo}
    \TBundle (\TBundle Q) \isomorph \TBundle Q \times_Q \TBundle Q \times_Q \TBundle Q.$$ Accordingly, we will write elements of $ \TBundle (\TBundle Q) $ as tuples $ (q, \dot{q}, \diF q, \diF \dot{q}) $, where $ \dot{q}, \diF q $ and $ \diF \dot{q} $ are elements of $ \TBundle_q Q $. We emphasize that the isomorphism  and, in particular, the component $ \diF \dot{q} $ depend on the connection $ K $. Now, for $ L: \TBundle Q \to \R $, the derivative of $ L $ at $ (q, \dot{q}) \in \TBundle Q $ with respect to the second component in the decomposition  will be written as $ \difpFrac{L}{q}(q, \dot{q}): \TBundle Q \to \R $. In 