In Hamiltonian Mechanics on a cotangent bundle $T^*M$, we can define the Hamiltonian vector field by $$ \iota_{X_H} \omega = \mathrm{d}H $$ where $\omega$ is the canonical symplectic 2-form on $T^*M$.
However, when it comes to the Lagrangian vector field, all the materials I could fetch says that it is the unique vector field whose integral curve satisfies the Lagrange equation $$ \frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot q_i}=\frac{\partial L}{\partial q_i} $$ on each local coordinates, which is derived by variational principle. This is not a coordinate-free definition.
Marsden's Book introduced a kind of method by which we can define a the Lagrangian vector without referring to any local coordinates by performing the fiber transform $\mathbb{F}L$ from the cotangent bundle $T^*M$. Indeed, this book also talked about deriving Lagrangian vector field from variational principle, but they introduced Lagrange equation on local coordinates during their derivation.
What I want is to derive the definition of Lagrangian vector field from the variational principle, but without introducing Euler-Lagrangian equation on local coordinates. Or in other words, I want an entirely coordinate-free method to derive an entirely coordinate-free definition of Lagrangian vector field from the variational principle.
How can I do this?
Here are some trials I made: I defined the variation of a smooth curve $\gamma: [T_0,T_1]\to M$ as $\delta \gamma: [T_0, T_1] \to TM$, so that
- $\delta \gamma(t) \in T_{\gamma(t)} M, T_0\leq t\leq T_1$
- $\delta \gamma(T_0) = \delta \gamma(T_1) = 0$ (the zero vectors on the corresponding tangent spaces)
- $\delta \gamma$ is smooth (can be extrapolated into a smooth vector field on $M$),
then I want to derive the counterpart of the variational problem $$ \delta\int_{T_0}^{T_1} L(\gamma^\sharp(t)) \mathrm{d} t=\int_{T_0}^{T_1} \delta L(\gamma^\sharp(t)) \mathrm{d} t = 0, $$ where $$ \gamma^\sharp:[T_0, T_1]\to TM, t \mapsto (\gamma(t), \gamma'(t)) $$ is the lift of curve $\gamma$.
Now here comes the problem: How should I represent $\delta L(\gamma^\sharp(t))$ with respect to $\delta \gamma(t)$?
I can make my questions more specific:
Firstly, how should I lift $\delta\gamma(t)$ to $\delta\gamma^\sharp(t)$? Should $\delta \gamma^\sharp(t)$ be in $T_{\gamma^\sharp(t)}(TM)$ or $T_{\delta\gamma(t)}(TM)$?
Secondly, in local coordinate $(U;q^1,\cdots,q^n)$ of $M$, we can write a tangent vector $v\in T(TM)$ as $$ v = \sum_{i=1}^n \alpha^i\frac{\partial}{\partial q^i} + \sum_{i=1}^n\beta^i\frac{\partial}{\partial \dot q^i}, $$ which can be splitted into a "coordinate derivative part $\alpha^i\frac{\partial}{\partial q^i}$" and a "velocity derivative part $\beta^i\frac{\partial}{\partial \dot q^i}$". My question is, can we natually decompose a tangent space of a tangent bundle into these two parts? More specifically, for each $p\in M$ and $v \in T_pM$, can we divide the linear space $T_v(TM)$ into the direct sum of two linear subspaces $T_v(TM)_1$ and $T_v(TM)_2$, so that $$ T_v(TM) = T_v(TM)_1 \oplus T_v(TM)_2 $$ and for each local coordinate $(U;q^1,\cdots,q^n)$ s.t. $p\in U$, we have $$ T_v(TM)_1 = \mathrm{span}\left\{\frac{\partial}{\partial q^1},\frac{\partial}{\partial q^2},\cdots, \frac{\partial}{\partial q^n}\right\}, $$ and $$ T_v(TM)_2 = \mathrm{span}\left\{\frac{\partial}{\partial \dot q^1},\frac{\partial}{\partial \dot q^2},\cdots, \frac{\partial}{\partial \dot q^n}\right\}? $$ I think this is important because we splitted these two parts when we derived Euler-Lagrange Equation.
After all, I think the main problem is just that I have too little understanding about the structure of high order tangent bundle $T(TM)$. But I found little materials talking about it. So up to now, I am stuck at this point.