Euler-Lagrange's equations for a Lagrangian $L$ read $$\frac{d}{dt}\frac{\partial L}{\partial \bf v} = \frac{\partial L}{\partial \bf x} .$$ More precisely, the statement is that a trajectory $\gamma:\mathbb R\to M$ that minimises the corresponding action $\gamma\mapsto \int L\circ d\gamma$, satisfies the equation $$A_L[\gamma] \equiv (\partial_{\bf x}L)\circ d\gamma - (\partial_{\bf v}L\circ d\gamma)' = 0.\tag1$$ Here, $M$ is the configuration space; I'm thinking of the Lagrangian as a map $L:TM\to\mathbb R$ defined on the tangent bundle of $M$ (though I suppose this is strictly the case only when $L$ does not depend explicitly on the time), and the differential is understood as the mapping $$d\gamma:\mathbb R\to TM:t\mapsto (\gamma(t),d\gamma_t(\partial_t))\simeq (\gamma(t),\gamma'(t)).$$

In (1), I defined the operator $A_L$ mapping curves $\gamma:[a,b]\to M$ into functions $[a,b]\to\mathbb R$. Euler-Lagrange's equations, in terms of this operator, thus say that if $\gamma$ is stationary for the action, then $A_L[\gamma]$ is the zero function.

Is there a way to understand this operator $A_L$ on more general/abstract/geometric grounds? More specifically, does the specific structure of the differentials in the expression $\frac{\partial}{\partial\bf x}-\frac{d}{dt}\frac{\partial}{\partial \bf v}$ somehow arise geometrically?

  • 2
    $\begingroup$ You've already written down an essentially geometric interpretation - the zero locus $A_L^{-1}(0)\subset C^1([a,b],M)$ is the space of solutions for the equations of motion. Can you be a bit more specific what you're looking for? $\endgroup$
    – ACuriousMind
    Apr 25, 2021 at 11:17
  • $\begingroup$ @ACuriousMind I'm asking whether the specific structure of the differential operators in this expression can be seen as arising from some geometric principle. Whether $A_L$ comes "naturally" from some more general idea/operation. An answer like "$A_L$ arises from the exterior derivative of $L\circ d\gamma$" would be the ideal one (I know that is not the case and is nonsense, I'm just giving an example of the kind of answer I'm hoping for) $\endgroup$
    – glS
    Apr 25, 2021 at 11:23

1 Answer 1


I am going to use the $\infty$-jet formalism (one could also work with finite order jets but that's actually more complicated), and consider higher order variational problems as well as the case of multiple independent variables.

Jet spaces:

Let's fix some notation. I will consider a fibred manifold $\pi:Y\rightarrow X$, and I will suppose that $\pi$ is the only fibration on $Y$ thus will commit the slight abuse of notation of identifying the fibred manifold with the total space $Y$. A generic fibred chart is denoted $(W,x,y)$ where $W\subseteq Y$ is an open set and $x=(x^i)_{i=1}^n=(x^1,...,x^n)$ are the base coordinates and $y=(y^\kappa)_{\kappa=1}^m=(y^1,...,y^m)$ are the fibre coordinates.

The dynamical field in the variation problem is identified with a (smooth) section $\phi\in\Gamma_X(Y)$, $\phi:X\rightarrow Y$, $\pi\circ\phi=\mathrm{id}_X$. One could work with local sections as well, but I don't want the added notational shenanigans involved.

In a fibred chart, the section is described by the local functions $\phi^\kappa=y^\kappa\circ\phi$ (components).

For a nonnegative integer $r\ge 0$, the $r$th jet prolongation $J^rY$ of the fibred manifold is defined as follows. Two local sections $\phi,\phi^\prime\in\Gamma_p(Y)$ defined about $p\in X$ are $r$-equivalent at $p$ if in any (and thus all) fibred charts whose domain contains everything relevant we have $$ \frac{\partial^k\phi}{\partial x^{i_1}...\partial x^{i_k}}(p)=\frac{\partial^k\phi^\prime}{\partial x^{i_1}...\partial x^{i_k}}(p),\quad k=0,1,...,r, $$ that is the two local sections have the same partial derivatives up to and including order $r$. This is an equivalence relation and the class of $\phi$ is denoted $j^r_p\phi$, is said to be the $r$-jet of $\phi$ at $p$. The set of all $r$-jets at $p$ is $J^r_pY$ and we define $$ J^rY=\bigsqcup_{p\in X}J_p^rY. $$ This $r$th jet space $J^rY$ naturally has the structure of a smooth manifold, and we have a bunch of smooth surjective submersions (i..e fibrations), namely

  • Source projection: $\pi^r:J^rY\rightarrow X$, $j^r_p\phi\mapsto p$.
  • Target projection: $\pi^r_0:J^rY\rightarrow Y$, $j^r_p\phi\mapsto\phi(p)$.
  • Other projections: For $s< r$ we have $\pi^r_s:J^rY\rightarrow J^sY$, $j^r_p\phi\mapsto j^s_p\phi$.

All of these maps define fibred manifolds, for any $r\ge0$ the projection $\pi^{r+1}_r:J^{r+1}Y\rightarrow J^rY$ is an affine bundle ($J^0Y=Y$), thus for any $r>s$, $r,s\ge 0$ the projection $\pi^r_s$ is a locally trivial fibre bundle with $\mathbb R^N$ (for suitable $N$) as a fibre and $J^rY$ deformation retracts onto $J^sY$. Moreover, if the original fibration $\pi:Y\rightarrow X$ is a locally trivial fibre bundle then so is the source projection $\pi^r:J^rY\rightarrow X$.

If $(W,x,y)$ is a fibred chart of $Y$ then it induces natural fibred charts on $J^rY$ as follows. We have the chart $(W^r,x,y,y^{(1)},...,y^{(r)})$, where for $0\le s\le r$ we have $$ y^{(s)}=(y^\kappa_{i_1...i_{s}})_{i_1\le...\le i_s}, $$ defined by $$ y^\kappa_{i_1...i_s}(j^s_p\phi)=\frac{\partial^s\phi^\kappa}{\partial x^{i_1}...\partial x^{i_s}}(p). $$ The restriction $i_1\le...\le i_s$ is necessary because the derivatives are symmetric, thus unless one orders the indices, one doesn't get independent coordinate functions. Also $W^r=(\pi^r_0)^{-1}(W)$.

Given a local section $\phi\in\Gamma(Y)$, there is a corresponding local section of the source projection $\pi^r$, written as $j^r\phi$, defined by $j^r\phi(p)=j^r_p\phi$, and in coordinates given by $p\mapsto(x^i(p),\phi^\kappa(p),...,\phi^\kappa_{i_1...i_r}(p))$, where the subscripts are derivatives.

Finally, there is a mathematically rigorous way to take the limit $J^\infty Y$ as the jet order $r$ goes to infinity. This space is an infinite dimensional manifold, but it works almost exactly like a finite dimensional manifold. According to the conventions I use, all functions and differential forms on $J^\infty Y$ are globally finite order (i.e. they are induced from some finite order jet space).

Variational bicomplex:

Let $\Omega_\infty(Y)=\bigoplus_{k\in\mathbb Z}\Omega^k_\infty(Y)$ be the exterior algebra over $J^\infty Y$, considered as sheaf over $Y$ (direct image sheaf).

One can construct a natural bigrading on this algebra as $$ \Omega_\infty(Y)=\bigoplus_{(r,s)\in\mathbb Z^2}\Omega^{r,s}(Y). $$

There is an invariant way to define the bigrading but probably the most intuitive is to define (in a given fibred chart) the local $1$-forms $$ \theta^\kappa_I=dy^\kappa_I-y^\kappa_{Ii}dx^i, $$ where $I=(i_1,...,i_r)$ is a multiindex of length $r$, and the length $r$ ranges over all possible lengths. Then the forms $$ dx^i,\quad \theta^\kappa_I $$ form a local frame for differential forms on $J^\infty Y$. A differential form $\omega$ is an element of $\Omega^{r,s}(Y)$ if each term in the basis expansion has precisely $r$ $dx^i$s and $s$ $\theta^\kappa_I$s. We then say that $\omega$ has $r$ horizontal degree and $s$ vertical or contact degree. If $\omega\in\Omega^k_\infty(Y)$ has total degree $k$, then define $p_q$ ($0\le q\le k$) to be the projection $p_q:\Omega^k_\infty(Y)\rightarrow\Omega^{k-q,q}(Y)$, and also use the notation $h=p_0$ and $p=\sum_{q=1}^k p_q$.

The full de Rham differential then splits as $d=d_H+\delta$, where on a form of pure bigrade $(r,s)$, $d_H=h\circ d$ and $\delta=p\circ d$.

These differentials satisfy $d_H^2=0$, $\delta^2=0$, $d_H\delta=-\delta d_H$. Intuitively, $d_H$ takes total derivatives, and $\delta$ is akin to taking a variation in classical calculus of variations.

The differential double complex $(\Omega^{\bullet,\bullet}(Y),d_H,\delta)$ is the "bare" variational bicomplex associated with $Y$.

This bicomplex may be extended naturally as follows. For each $s>0$ define $$ \mathcal Q^s(Y)=\Omega^{n,s}(Y)/d_H\Omega^{n-1,s}(Y), $$i.e. $\mathcal Q^s(Y)$ is the cokernel of the horizontal differential $d_H$ on $\Omega^{n,s-1}(Y)$. Because $d_H$ and $\delta$ anticommute, there is an induced differential $$ \delta_\ast:\mathcal Q^s(Y)\rightarrow\mathcal Q^{s+1}(Y), $$ and this additional column is appended to the bicomplex asenter image description here.

I have denoted the quotient projection $\int:\Omega^{n,s}(Y)\rightarrow \mathcal Q^s(Y)$ as an integral symbol.

We may identify the space $\Omega^{n,0}(Y)$ as the space of all finite order Lagrangians, since in local coordinates we may write any element as $$ L=\mathcal L(x,y,...,y^{(r)})\nu, $$ where $\nu=dx^1\wedge...\wedge dx^n$, and $r$ is the order of the Lagrangian.

The operator $E:\Omega^{n,0}(Y)\rightarrow \mathcal Q^1(Y)$, $E=\int\circ \delta$ is the Euler-Lagrange operator. Since a Lagrangian has maximal horzontal degree, it follows that $dL=(d_H+\delta)L=\delta L$, thus we also have $E=\int\circ d$, i.e. the Euler-Lagrange operator is the exterior derivative followed by a quotient projection.

To understand why I have denoted the quotient projection with an integral sign, the following is a useful heuristic. A vertical vector field $\Xi=\Xi^\kappa\frac{\partial}{\partial y^\kappa}$ on $Y$ (we may also allow $\Xi$ to be a "generalized vertical vector field", which means that the components $\Xi^\kappa$ may also depend on the jet coordinates) has $\infty$-prolongation $$ j^\infty\Xi=\sum_{|I|<\infty}\Xi^\kappa_I\frac{\partial}{\partial y^\kappa_I}, $$ which is a $\pi^\infty$-vertical vector field on $J^\infty Y$, and $\Xi^\kappa_I=d_{i_1}...d_{i_r}...\Xi^\kappa=d_I\Xi^\kappa$, where $d_i$ is the total derivative with respect to $x^i$ (i.e. $d_i y^\kappa_{i_1...i_s}=y^\kappa_{i_1...i_si}$).

These vector fields represent field variations. That is if $\phi:U\rightarrow Y$ is a local section ($U\subseteq X$ open), then $\delta\phi=\Xi\circ\phi$ is a field variation in the usual sense.

It follows that if one is given an element $\omega\in\Omega^{n,s}(Y)$, and $s$ vertical vector fields $\Xi_1,...,\Xi_s$, then the expression $$ F_D[\phi;\delta_1\phi,...,\delta_s\phi]=\int_D(j^\infty\phi)\left(i_{j^\infty\Xi_1}...i_{j^\infty\Xi_s}\omega\right) $$ ($i$ is the interior product and $D$ is a compact $n$ dimensional domain in $X$) is an integral functional that is skew-symmetric and linear in the field variations $\delta_k\phi=\Xi_k\circ\phi$, so it can be seen as an $s$-form on the function space $\Gamma_D(Y)$ of sections over $D$.

It turns out (via the usual integration by parts argument) that as long as the vertical vector fields $\Xi_1,...,\Xi_s$ have support within $D$, then this functional is insensitive to adding a $d_H$-exact term to $\omega$ (remember that $d_H$ is basically taking total divergences!)

Such integral functionals are in one-to-one correspondance with elements of $\mathcal Q^s(Y)=\Omega^{n,s}(Y)/d_H\Omega^{n-1,s}(Y)$ for this reason.

This explains why $E=\int\circ d$ - the "formal integration", which is just taking the quotient essentially performs an abstract integration by parts.

For a Lagrangian $L=\mathcal L(x,y,...,y^{(r)})\nu$, we have $$ dL=\delta L=\sum_{|I|\le r}\frac{\partial^S\mathcal L}{\partial y^\kappa_I}\theta^\kappa_I\wedge\nu, $$ where $\partial^S/\partial y^\kappa_I$ is an appropriately normalized derivative so that we can sum over all multiindices without ordering the sums. Taking the formal integral essentially cuts off any $d_H$-exact parts.

By manually performing the integrations by parts, we arrive at the formula $$ E(L)=\int\sum\frac{\partial^S\mathcal L}{\partial y^\kappa_{I}}\theta^\kappa_I\wedge\nu=\int\sum(-1)^{|I|}d_I\frac{\partial^S\mathcal L}{\partial y^\kappa_I}\theta^\kappa\wedge\nu. $$ The latter expression is essentially a "canonical form" of $E$. Remember that the integrals here are purely symbolic and they just denote the quotient projection.

It is then common to use - instead of the quotient spaces $\mathcal Q^s(Y)$ - another construction, which I am going to call $\mathcal F^s(Y)$. Each of these spaces are subspaces $\mathcal F^s(Y)\le\Omega^{n,s}(Y)$ constructed in a way that $$ \Omega^{n,s}(Y)=\mathcal F^s(Y)\oplus d_H\Omega^{n-1,s}(Y), $$ and there is a projection $I:\Omega^{n,s}(Y)\rightarrow\mathcal F^s(Y)$ which is idempotent ($I^2=I$) and satisfies $I\circ d_H=0$. They are usually called interior Euler operators. The interior Euler operators essentially perform the integration by parts necessary to represent the quotient spaces $\mathcal Q^s$ as subspaces of $\Omega^{n,s}(Y)$. For the case of $s=1$, the space $\mathcal F^s(Y)$ consists of $n+1$-forms that look like $$ \varepsilon=\varepsilon_\kappa\ \theta^\kappa\wedge\nu, $$(these are called source forms) and then the Euler-lagrange operator is $$ E=I\circ d. $$

Lepage forms:

There is yet another approach which is probably closer to what OP considers geometric and is often favoured in the finite-order approach. Nontheless, I can define them in the $\infty$-jet formalism as well, and it is actually simpler to do so.

An $n$-form $\rho\in\Omega^n_\infty(Y)$ (usually not of pure bigrade) is called a Lepage form if it satisfies the fact that $p_1d\rho$ is $\pi^\infty_0$-horizontal, that is in a local chart it's $1$-vertical component has the form $$ p_1d\rho=\varepsilon_\kappa\ \theta^\kappa\wedge\nu, $$ i.e. this is basically a source form. This condition is completely invariant. Such forms have some rather salient properties. For example if two Lepage forms $\rho$ and $\rho^\prime$ have the same horizontal part, i.e. $h\rho=h\rho^\prime$, then we also have $p_1d\rho=p_1d\rho^\prime$, i.e. the $1$-vertical part of their exterior derivatives are also the same.

If one has a Lagrangian $L\in\Omega^{n,0}(Y)$, then a Lepage form $\rho$ is a Lepage equivalent of $L$ if it satisfies $L=h\rho$, i.e. the Lagrangian is the horizontal part of the Lepage form.

One can then prove that every Lagrangian of order $r$ has a globally defined Lepage equivalent of at most order $2r-1$.

Then the Euler-Lagrange form associated to the Lagrangian $L$ is given by $$ E(L)=p_1d\rho, $$ where $\rho$ is any Lepage equivalent. This expression is a lot nicer than the previous $E=I\circ d$, since the interior Euler-operator $I$ is quite complicated and somewhat "unnatural" in the sense that it is defined purely in local coordinates as a specialized integration-by-parts operator, it just turns out to be globallly well defined. By contrast, for Lepage equivalents, the definition $E=p_1\circ d$ is more natural, since the projection $p_1$ is easy to calculate and comes naturally from the structure of the jet bundles.

Lepage forms are closely related to the concept of the Poincaré-Cartan form, which is probably better known. In a local chart, every Lepage equivalent of $L$ can be written in the form $$ \rho=\Theta_L+d\eta+\mu, $$ where $\eta$ is an $1$-vertical form, $\mu$ is a form whose degree of verticalness is $\ge 2$, and $$ \Theta_L=\mathcal L\nu+\sum_{|I|\le r}\left(\sum_{|J|=0}^{r-|I|-1} (-1)^{|J|}d_J\frac{\partial^S\mathcal L}{\partial_{y^\kappa_{IJi}}}\right)\theta^\kappa_I\wedge\nu_i, $$ where $r$ is the order of $\mathcal L$, and $\nu_i=i_{\partial_i}\nu$.

The $\Theta_L$ is called the principal Lepage equivalent of $L$. The problem with it is that it is usually not well-defined globally, thus the correction terms $d\eta$ and $\mu$ are usually needed. However when the order of the Lagrangian is $1$ or $2$, then $\Theta_L$ is globally well-defined and "covariant". In this case it is usually called the Poincaré-Cartan form associated with $L$.

To give an example, for an order $1$ Lagrangian with $n=1$ $L=\mathcal L(t,q,\dot q)dt$, we have $$ \Theta_L=\mathcal Ldt+\frac{\partial\mathcal L}{\partial \dot q^\kappa}\theta^\kappa, $$and the exterior derivative is $$ d\Theta_L=\left(\frac{\partial L}{\partial q^\kappa}-\frac{d}{dt}\frac{\partial\mathcal L}{\partial\dot q^\kappa}\right)\theta^\kappa\wedge dt+\frac{\partial^2\mathcal L}{\partial q^\lambda\partial\dot q^\kappa}\theta^\lambda\wedge\theta^\kappa+\frac{\partial^2\mathcal L}{\partial\dot q^\lambda\partial\dot q^\kappa}\dot\theta^\lambda\wedge\theta^\kappa, $$ where now the dot is total differentiation, $$ \theta^\kappa=dq^\kappa-\dot q^\kappa dt, \\ \dot\theta^\kappa=d\dot q^\kappa-\ddot q^\kappa dt. $$ The $1$-vertical part of this expression is clearly the Euler-Lagrange form of $L$, and no integrations by parts had to be performed here.

Edit: As per OP's request, here is a simpler (less technically demanding) formulation for $n=1$ and $r=1$, that is for Lagrangians of order 1 with only one independent variable.

To fix notation, let M be an n dimensional manifold with local coordinates $q^{i}$, let $Q=\mathbb{R}\times M$ be the extended configuration space with coordinates $\left(t,q\right)$, let $\dot{Q}=\mathbb{R}\times TM$ with coordinates $\left(t,q,\dot{q}\right)$ be the extended velocity phase space, and $\dot{Q}^{\ast}=\mathbb{R}\times T^{\ast}M$ with coordinates $\left(t,q,p\right)$ is the extended momentum phase space. The Lagrangian is $L:\dot{Q}\rightarrow\mathbb{R}, L=L\left(t,q,\dot{q}\right)$, and its fibre derivative is $$ \mathbb{F}L:\dot{Q}\rightarrow\dot{Q}^{\ast},\quad\mathbb{F}L\left(t,q,\dot{q}\right)=\left(t,d\left(\left.L\right|_{\left\{ t\right\} \times T_{q}M}\right)\right), $$ i.e. in local coordinates it is $\mathbb{F}L\left(t,q,\dot{q}\right)=\left(t,q,p\right)=\left(t,q,\frac{\partial L}{\partial\dot{q}}\right)$. The energy function is $$ E\left(t,q,\dot{q}\right)=\mathbb{F}L\left(t,q,\dot{q}\right)\left(\dot{q}\right)-L\left(t,q,\dot{q}\right)=\frac{\partial L}{\partial\dot{q}^{i}}\left(t,q,\dot{q}\right)\dot{q}^{i}-L\left(t,q,\dot{q}\right), $$ it's basically the Hamiltonian from the Lagrangian point of view. Finally, let $$ \theta\in\Omega^{1}\left(\dot{Q}^{\ast}\right)=p_{i}dq^{i} $$ be the canonical $1$-form on $T^{\ast}M$ (canonically transferred to $\dot{Q}^{\ast}=\mathbb{R}\times T^{\ast}M$), also known as the tautological $1$-form or symplectic potential. Define $$ \Theta_{L}=\mathbb{F}L^{\ast}\theta-Edt $$ to be the Poincaré-Cartan form associated with $L$. In coordinates, this is $$ \Theta_{L}=L\left(t,q,\dot{q}\right)dt+\frac{\partial L}{\partial\dot{q}^{i}}\left(t,q,\dot{q}\right)\left(dq^{i}-\dot{q}dt\right). $$ We now discuss trajectories. Let $\gamma:\mathbb{R}\rightarrow Q$, $\gamma\left(t\right)=\left(t,q\left(t\right)\right)$, it's prolongation is $\dot{\gamma}:\mathbb{R}\rightarrow\dot{Q}$, $\dot{\gamma}\left(t\right)=\left(t,q\left(t\right),\dot{q}\left(t\right)\right)$. We also need to define variations of trajectories. Let $$ \Xi=\Xi^{i}\left(t,q\right)\frac{\partial}{\partial q^{i}} $$ be a vector field on $Q$ that is vertical in the sense that it does not involve the $\partial/\partial t$. It's flow is $F_{\epsilon}^{\Xi}$. We also define its prolongation to be $$ \dot{\Xi}=\Xi^{i}\left(t,q\right)\frac{\partial}{\partial q^{i}}+\dot{\Xi}^{i}\left(t,q,\dot{q}\right)\frac{\partial}{\partial\dot{q}^{i}}, $$ where $\dot{\Xi}^{i}\left(t,q,\dot{q}\right)=\frac{\partial\Xi^{i}}{\partial t}+\frac{\partial\Xi^{i}}{\partial q^{j}}\dot{q}^{j}$. This is a vector field on $\dot{Q}$. It's flow is $F_{\epsilon}^{\dot{\Xi}}$. Given a trajectory $\gamma$, we have the deformation $$ \gamma_{\epsilon}=F_{\epsilon}^{\Xi}\circ\gamma,\quad\dot{\gamma}_{\epsilon}=F_{\epsilon}^{\dot{\Xi}}\circ\dot{\gamma}, $$ and the derivatives at $\epsilon=0$ (variations) are $$ \delta\gamma=\Xi\circ\gamma,\quad\delta\dot{\gamma}=\dot{\Xi}\circ\dot{\gamma}. $$ The readers should check these manually. It is also useful to define the local coordinate forms $$ \delta q^{i}\left(t\right)=\Xi^{i}\left(t,q\left(t\right)\right),\quad\delta\dot{q}^{i}\left(t\right)=\dot{\Xi}^{i}\left(t,q\left(t\right),\dot{q}\left(t\right)\right). $$ The action functional is then $$ S\left[\gamma\right]=\int_{t_{0}}^{t_{1}}\left(L\circ\dot{\gamma}\right)\left(t\right)dt=\int_{t_{0}}^{t_{1}}\dot{\gamma}^{\ast}\Theta_{L}, $$ where we were allowed to replace the Lagrangian with the Poincaré-Cartan form since it differs from $Ldt$ by terms that vanish when pulled back along $\dot{\gamma}$. Then replacing $\gamma$ with $\gamma_{\epsilon}$ and taking the $\epsilon$-derivative at $0$ gives $$ \delta S\left[\gamma\right]=\int_{t_{0}}^{t_{1}}\gamma^{\ast}\mathscr{L}_{\dot{\Xi}}\Theta_{L}=\int_{t_{0}}^{t_{1}}\gamma^{\ast}\left(i_{\dot{\Xi}}d\Theta_{L}+di_{\dot{\Xi}}\Theta_{L}\right), $$ where we have expressed the variation as the pullback of the Lie derivative along $\dot{\gamma}$, and use Cartan's formula. Since the $d$ commutes with pullbacks, the last term is a total derivative and thus if the variation vanishes at the endpoints, we can throw it out. The first term we analyze in coordinates. We have $$ d\Theta_{L}=\left(\frac{\partial L}{\partial q^{i}}dq^{i}+\frac{\partial L}{\partial\dot{q}^{i}}d\dot{q}^{i}\right)\wedge dt+d\frac{\partial L}{\partial\dot{q}^{i}}\wedge\left(dq^{i}-\dot{q}dt\right)-\frac{\partial L}{\partial\dot{q}^{i}}d\dot{q}^{i}\wedge dt, $$ and then the interior product with $\dot{\Xi}$ is $$ i_{\dot{\Xi}}d\Theta_{L} =\left(\frac{\partial L}{\partial q^{i}}\Xi^{i}+\frac{\partial L}{\partial\dot{q}^{i}}\dot{\Xi}^{i}\right)dt+i_{\dot{\Xi}}d\frac{\partial L}{\partial\dot{q}^{i}}\left(dq^{i}-\dot{q}dt\right)-d\frac{\partial L}{\partial\dot{q}^{i}}\Xi^{i}-\frac{\partial L}{\partial\dot{q}^{i}}\dot{\Xi}^{i}dt\\ =\left(\frac{\partial L}{\partial q^{i}}dt-d\frac{\partial L}{\partial\dot{q}^{i}}\right)\Xi^{i}+i_{\dot{\Xi}}d\frac{\partial L}{\partial\dot{q}^{i}}\left(dq^{i}-\dot{q}dt\right). $$ Now we calculate the pullback along the trajectory: $$ \dot{\gamma}^{\ast}i_{\dot{\Xi}}d\Theta_{L} =\left(\frac{\partial L}{\partial q^{i}}\left(t,q\left(t\right),\dot{q}\left(t\right)\right)dt-d\left(\frac{\partial L}{\partial\dot{q}^{i}}\left(t,q\left(t\right),\dot{q}\left(t\right)\right)\right)\right)\delta q^{i}\left(t\right)\\ =\left(\frac{\partial L}{\partial q^{i}}\left(t,q\left(t\right),\dot{q}\left(t\right)\right)-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}^{i}}\left(t,q\left(t\right),\dot{q}\left(t\right)\right)\right)\right)dt\delta q^{i}\left(t\right). $$ The last term has vanished because $dq^{i}-\dot{q}^{i}dt$ vanishes on pullback with $\dot{\gamma}$. We have thus arrived at the result that for a given variation $\delta q^{i}\left(t\right)$ generated by the vector field $\Xi$, and a given trajectory $\gamma$, the Euler-Lagrange expression along the trajectory is given naturally as $$ E_{i}\left(L\right)\left(t,q\left(t\right),\dot{q}\left(t\right),\ddot{q}\left(t\right)\right)\delta q^{i}\left(t\right)dt=\dot{\gamma}^{\ast}i_{\dot{\Xi}}d\Theta_{L}. $$ The things I said about Lepage forms in the preceding section has been a grand generalization of this.


  1. Anderson: The Variational Bicomplex (Variational bicomplex, interior Euler operators)
  2. Olver: Applications of Lie Groups to Differential Equations (The quotient sequence formulation, formal integrals)
  3. Krupka - Introduction to Global Variational Geometry (Lepage forms)
  4. Bensoam, Baugé - Multisymplectic geometry and covariant formalism for mechanical systems with a Lie group as configuration space: application to the Reissner beam (The "classical" Poincaré-Cartan form; only a review in section II, couldn't think of a more relevant treatment)
  • $\begingroup$ bloody hell.. every time I think I got some kind of grip on the differential geometry lingo, I come across stuff like this. I mean, I know I asked for "abstract", so I guess I got what I deserved =). Jokes aside, this sounds great, thanks, but it unfortunately flies way over my head; it will take me some time to parse it. Thanks a lot for the sources, they are much appreciated. Any hope to dumb this down a little bit for someone that, say, understands what you'll find in "introduction to differential geometry" kind of books? $\endgroup$
    – glS
    Apr 26, 2021 at 8:31
  • 1
    $\begingroup$ @glS Sure, I'll add another section later when I get the chance. I'll also correct some typos. $\endgroup$ Apr 26, 2021 at 8:52
  • 1
    $\begingroup$ @glS I have expanded the answer with a section that involves only tangent and cotangent bundles. $\endgroup$ Apr 26, 2021 at 13:30

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