Let $M$ be a smooth $m$ dimensional manifold, let $\pi:E\rightarrow M$ be a smooth fibred manifold over $M$. Let us write generic fibred coordinates as $(x^i,y^\sigma)$ with $x^i$ being the base coordinates and $y^\sigma$ being the fibre coordinates. Einstein summation convention is in use unless otherwise stated. Let $J^rE$ denote the $r$th jet prolongation manifold of $E\rightarrow M$, the induced fibred coordinates are $$ (x^i,y^\sigma,y^\sigma_i,...,y^\sigma_{i_1...i_r}). $$For symmetric arrays the following conventions are in effect. If symmetric multi-index notation (like $I=(i_1,...,i_r)$) is used, then $$ A_I B^I=\sum_{i_1\le...\le i_r}A_{i_1...i_r}B^{i_1...i_r}, $$i.e. the summation is automatically nondecreasing. For derivatives this is inconvenient however, so we have $$y^\sigma_I\partial^I_\sigma f:=\sum_{i_1\le...\le i_r}y^\sigma_{i_1...i_r}\frac{\partial f}{\partial y^\sigma_{i_1...i_r}}, $$BUT if I write $\partial^{i_1...i_r}_\sigma$ instead of the multi-indices, this means symmetric derivative by default, which satisfies $$ \partial^{i_1...i_r}_\sigma y^\tau_{j_1...j_r}=\delta^\tau_\sigma \delta^{i_1}_{(j_1}...\delta^{i_r}_{j_r)}, $$ where the round brackets are the classic tensor calculus symmetrization brackets (including the factor of $(r!)^{-1}$).

I think that with these conventions we should have $$ y^\sigma_I\partial^I_\sigma f\equiv y^\sigma_{i_1...i_r}\partial^{i_1...i_r}_\sigma f, $$ i.e. we may switch from multi-index notation to unrestricted sums for derivatives with no additional factors.

Let $\lambda\in\Omega^{m,0}(J^\infty E)$ be a second-order Lagrangian of the local form $$ \lambda=L(x^i,y^\sigma,y^\sigma_i,y^\sigma_{ij})\nu, $$ where $\nu=dx^1\wedge...\wedge dx^m$.

I want to know two things:

  1. What are the (necessary and sufficient) conditions on $\lambda$ such that the Euler-Lagrange equations of $\lambda$ are second-order, rather than fourth order differential equations?
  2. What are the (necessary and sufficient) conditions on $\lambda$ such that there is (at least locally) an equivalent first-order Lagrangian, i.e. $\lambda=\lambda^\prime+D\Theta$, where $\lambda^\prime$ is a first-order Lagrangian and $\Theta\in\Omega^{m-1,0}(J^\infty E)$ is a first-order element of the variational bicomplex with $D:\Omega^{r,s}\rightarrow\Omega^{r+1,s}$ being the horizontal differential.

The only source I have seen treating this is Advanced Classical Field Theory by Giachetta, Mangiarotti and Sardanashvily (GMS from now on), specifically Section 2.4.5 "Reduced second order Lagrangians". However their treatment is not very detailed and I am having doubts about the correctness of the result presented within.

Specifically GMS states that the Euler Lagrange operator $E(\lambda)=E_\sigma(\lambda)\theta^\sigma\wedge \nu$ ($\theta^\sigma=dy^\sigma-y^\sigma_i dx^i$ is a basic contact form) acts on $\lambda$ as $$ E_\sigma(\lambda)=(\partial_\sigma-d_i\partial^i_\sigma +d_id_j\partial^{ij}_\sigma)L, $$ which is of course clear but then they simply state without proof that the two conditions for $E$ being a second-order operator are $$ \partial^{ij}_\sigma\partial^{kl}_\tau L=0\quad\quad\quad\quad\quad\quad(1) \\ \left(\partial^i_\sigma\partial^{jk}_\tau-\partial^j_\tau\partial^{ik}_\sigma\right)L=0.\quad\quad (2) $$The first condition is that the Lagrangian is affine in the second derivatives, i.e. $$L=F^{ij}_\sigma y^\sigma_{ij}+G, $$ where both $F$ and $G$ are first-order, which is coordinate-invariant because $J^2E\rightarrow J^1E$ is an affine bundle.

GMS now constructs the local differential form $$ F=F^{ij}_\sigma dy^\sigma_i\wedge \nu_j, $$ and if $\bar d$ is the "fibrewise" exterior derivative on $J^1E\rightarrow E$, they claim that condition (2) is equivalent to $ \bar dF=0 $, which is easy to see, then claim that $F$ is globally defined (also easy to see) and that since the affine bundle has contractible fibres, $\bar dF=0$ implies the global existence of a fibrewise $m-1$-form $\phi$ such that $F=\bar d\phi$.

They then show that $ L-D\phi $ is a first-order equivalent to $L$. It would thus seem that the answer to question 1. is that $L$ must be affine in the second derivatives and the fibrewise form $F$ that follows from the affinity of $L$ must be fibrewise closed, and question 2. is equivalent to question 1, i.e. if a second-order Lagrangian has first-order equations of motion, then there is an equivalent first-order Lagrangian. Cool.

However I have problems deriving (1) and (2), which, as I have said, GNS just puts there without explanation. It is not difficult to derive that the Euler-Lagrange operator on a second-order Lagrangian is

$$E_{\sigma}\left(\lambda\right) =Q_{\sigma}\left(\lambda\right)-y_{ijk}^{\kappa}\partial_{\kappa}^{jk}\partial_{\sigma}^{i}L+y_{ikl}^{\kappa}\partial_{j}\partial_{\kappa}^{kl}\partial_{\sigma}^{ij}L+y_{jkl}^{\kappa}\partial_{\kappa}^{kl}\partial_{i}\partial_{\sigma}^{ij}L+y_{ijk}^{\kappa}\partial_{\kappa}^{k}\partial_{\sigma}^{ij}L+y_{ijkl}^{\kappa}\partial_{\kappa}^{kl}\partial_{\sigma}^{ij}L +y_{ikl}^{\kappa}\phi_{jJ}^{\lambda}\partial_{\kappa}^{kl}\partial_{\lambda}^{J}\partial_{\sigma}^{ij}L+y_{iI}^{\kappa}y_{jkl}^{\lambda}\partial_{\kappa}^{I}\partial_{\lambda}^{kl}\partial_{\sigma}^{ij}L,$$ where $Q_\sigma(\lambda)$ is the collection of all terms that are at most second order.

There is only one fourth order term in this expansion, and since the derivative coordinates are all independent, it must vanish independently of the rest of the terms, thus $$ 0=y^\kappa_{ijkl}\partial^{kl}_\kappa\partial^{ij}_\sigma L, $$and if we could "cancel", we would obtain condition (1). However due to the symmetry of partial derivatives, the $y^\kappa_{ijkl}$ are symmetric in the lower indices and the best we can do is to choose $$ y^\kappa_{ijkl}=\delta^\kappa_\alpha\delta^a_{(i}\delta^b_j\delta^c_k\delta^d_{l)}, $$ then after renaming the indices we get $$ \partial^{(kl}_\kappa\partial^{ij)}_\sigma L=0. $$ I did not write out all the 24 terms in this symmetrization, but this must be nontrivially different from $\partial^{kl}_\kappa\partial^{ij}_\sigma L$ because this expression symmetric only under the exchange of $k$ and $l$, the exchange of $i$ and $j$ and the simultaneous exchange $(k,l,\kappa)\leftrightarrow (i,j,\sigma)$. At any rate I see that $\partial^{kl}_\kappa\partial^{ij}_\sigma L\Rightarrow \partial^{(kl}_\kappa\partial^{ij)}_\sigma L$, but I don't see the implication going backwards.

Going further, suppose that somehow (1) is true. Then reinserting (1) into the form of $E_\sigma(\lambda)$ one gets $$ E_\sigma(\lambda)=Q_\sigma(\lambda)-y^\kappa_{ijk}\partial^i_\sigma\partial^{jk}_\kappa L+y^\kappa_{ijk}\partial^k_\kappa\partial^{ij}_\sigma L, $$ thus the condition for the vanishing of the third-order parts becomes $$ y^\kappa_{ijk}\left( \partial^k_\kappa\partial^{ij}_\sigma L-\partial^i_\sigma\partial^{jk}_\kappa L \right)=0. $$

Once again this is only equivalent to (2) if a total "cancellation" is possible, but since $y^\kappa_{ijk}$ is symmetric in $ijk$ I get (I copied this from my document, so the indices here have a bit different names, shouldn't matter) $$0=\partial_{\sigma}^{k}\partial_{\lambda}^{ij}L+\partial_{\sigma}^{i}\partial_{\lambda}^{jk}L+\partial_{\sigma}^{j}\partial_{\lambda}^{ki}L+\partial_{\sigma}^{k}\partial_{\lambda}^{ji}L-\partial_{\lambda}^{i}\partial_{\sigma}^{kj}L-\partial_{\lambda}^{j}\partial_{\sigma}^{ik}L-\partial_{\lambda}^{k}\partial_{\sigma}^{ji}L. $$

Once again, this is a consequence of (2) but I do not see it being equivalent to (2).

However to this relation I cannot associate any meaning of a vanishing exterior derivative, therefore I cannot continue to the arguments of GMS from this condition about the existence of a first-order equivalent.

In fact, if the Lagrangian had a first-order equivalent, then the Euler-Lagrange equations must be affine in the second derivatives, however I have also calculated the explicit form of $Q_\sigma(\lambda)$, which is $$ Q_{\sigma}\left(\lambda\right) =\partial_{\sigma}L-\partial_{i}\partial_{\sigma}^{i}L-y_{i}^{\kappa}\partial_{\kappa}\partial_{\sigma}^{i}L-y_{ij}^{\kappa}\partial_{\kappa}^{j}\partial_{\sigma}^{i}L+\partial_{i}\partial_{j}\partial_{\sigma}^{ij}L +y_{i}^{\kappa}\partial_{j}\partial_{\kappa}\partial_{\sigma}^{ij}L+y_{ik}^{\kappa}\partial_{j}\partial_{\kappa}^{k}\partial_{\sigma}^{ij}L+y_{j}^{\kappa}\partial_{\kappa}\partial_{i}\partial_{\sigma}^{ij}L+y_{jk}^{\kappa}\partial_{\kappa}^{k}\partial_{i}\partial_{\sigma}^{ij}L +y_{ij}^{\kappa}\partial_{\kappa}\partial_{\sigma}^{ij}L+y_{i}^{\kappa}y_{j}^{\lambda}\partial_{\kappa}\partial_{\lambda}\partial_{\sigma}^{ij}L+y_{ik}^{\kappa}y_{j}^{\lambda}\partial_{\kappa}^{k}\partial_{\lambda}\partial_{\sigma}^{ij}L+y_{i}^{\kappa}y_{jk}^{\lambda}\partial_{\kappa}\partial_{\lambda}^{k}\partial_{\sigma}^{ij}L +y_{ik}^{\kappa}y_{jl}^{\lambda}\partial_{\kappa}^{k}\partial_{\lambda}^{l}\partial_{\sigma}^{ij}L, $$ and there is a term quadratic in $y^\sigma_{ij}$ here, which I don't see vanishing even if (2) is taken into account.

So in short

  • Are GMS correct? If so am I missing some trivial combinatorial shenanigans that would make my conditions reduce to that of (1) and (2)?
  • If GMS are incorrect and my conditions are correct, then is there any less complicated way of assessing them? GMS' $\bar dF=0 $ is quite elegant and easily calculatable as opposed to my debauch of indices. What about first-order equivalent Lagrangians? If I am correct and GMS are not then it would seem to me that not all 2nd order Lagrangians with 2nd order EoMs have 1st order equivalents. What is the condition to have a first-order equivalent?
  • What are some other sources that treat the question of second-order Lagrangians with second-order equations of motion? I would love to get my hands on a second (or third, and so forth) account, since GMS are not very detailed.
  • 3
    $\begingroup$ You are correct that (1) cannot be right. A famous example (when $m=2$ and the number of fiber variables is $1$) is the Gauss curvature 'null Lagrangian' $$ \lambda = \frac{y_{11}y_{22}-{y_{12}}^2}{(1+{y_1}^2+{y_2}^2)^{3/2}}\,\mathrm{d}x^1\wedge\mathrm{d}x^2 = \frac{\mathrm{d}y_1\wedge\mathrm{d}y_2}{(1+{y_1}^2+{y_2}^2)^{3/2}}.$$ In this case $E(\lambda)\equiv0$, but $\lambda$ is not affine in the second derivatives. For a serious treatment of this subject, you might want to look at the 2-volume treatise Calculus of Variations by Giaquinta and Hildebrandt. $\endgroup$ Commented Dec 13, 2020 at 11:34

1 Answer 1


I think you might find interesting §4.B of Anderson's The Variational Bicomplex [1] (Theorems 4.23, 4.29 and Corollary 4.30, to be more precise). However, these results are in a sense converse to your questions. For instance, that Theorem 4.29 computes, for a given variational equation, an upper bound on the order of a corresponding Lagrangian density (which can be achieved in the class of Lagrangian densities having the same Euler-Lagrange equation).

Closer to the direction of your question is Corollary 4.44 in Anderson's §4.C, which characterizes the dependence of null Lagrangian densities of a given order on jet coordinates of the highest order. So, for instance, if $\lambda$ is your second order Lagrangian, but $\lambda_1$ is a first order Lagrangian giving the same variational equations, then $\lambda-\lambda_1$ is a null second order Lagrangian, whose dependence on $y^\kappa_{ij}$ is then characterized by Anderson's result. I've not carefully read through those sections of Anderson's monograph before, so I will not attempt to decode them more explicitly here.

[1] Unpublished, but a copy is easily located with Google.


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