Invariance of the l.h.s. of Euler-Lagrange equation

Question

Let $M^n$ be a smooth manifold equipped with a nondegenerate Lagrangian $L:TM\to\mathbb R$, $L=L(x,y)$, $x\in M$, $y\in T_xM$. The stationary points of the corresponding integral functional on curves are the solutions of the Euler-Lagrange equation, which in coordinates reads $$ \frac d{dt} \frac{dL}{dy_i}(x(t),\dot x(t)) - \frac{\partial L}{\partial x_i}(x(t),\dot x(t)) = 0, \qquad i=1,\dots,n . $$ Consider a smooth curve $t\mapsto x(t)$ which is not stationary. Plugging it into the l.h.s. of the equation yields coordinates of a co-vector (from $T^*_{x(t)}M$) which depends on the curve but not on the coordinate system. The invariance of this co-vector can be seen e.g. from the first variation formula for the functional.

Question: Is there a coordinate-free definition of this co-vector?

Actually I am interested only in the case when $L$ is the Lagrangian associated to a Finsler metric (i.e. $L$ is quadratically homogeneous).

Notes

• The equation itself (i.e. the property that the co-vector is zero) has an invariant expression e.g. with Hamiltonian formalism.

• In the Riemannian case, the co-vector in question (for a unit-speed curve) corresponds to the geodesic curvature vector under the isomorphism between $TM$ and $T^*M$ defined by the metric. This can be defined invariantly via the Levi-Civita connection.

Answer 1

There is a coordinate-free description using only natural objects on $TM$. Here is one way to do it.

First, consider the basepoint submersion $\pi:TM\to M$. For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-fiber through $v$ is equal to $T_{\pi(v)}M$, a vector space. It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$. Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$, and let $\nu_v : T_v(TM) \to T_v(TM)$ be the nilpotent endomorphism $\nu_v = \iota_v\circ \pi'(v)$.

Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. Define a $1$-form $\omega_L$ on $TM$ by $$ \omega_L(w) = dL\bigl(\nu_v(w)\bigr) $$ for all $w\in T_v(TM)$. Let $R$ be the vector field on $TM$ that is tangent to the fibers of $\pi$ and that is the natural radial vector field on each such (vector space) fiber, and set $E_L = dL(R) - L$. (If $L$ is quadratic homogeneous on each $\pi$-fiber, then, by Euler's relation, $E_L = L$.)

Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then, for each $t\in[a,b]$, consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule $$ \beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w) $$ for $w\in T_{\gamma'(t)}TM$. Then $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so $\beta(t) = \pi'(\gamma'(t))^\ast(\delta\gamma(t))$ for a unique co-vector $\delta\gamma(t)\in T^\ast_{\gamma(t)}M$.

This assignment $t\mapsto \delta\gamma(t)$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.