# Functional Poincaré's lemma and the inverse Lagrangian problem

I have only encountered the inverse Lagrangian problem in mathematics books that treat Lagrangian field theory using jet bundles and homological algebra, and while I am studying this approach, I still find the language somewhat difficult to understand/use.

In a more familiar, however much less precise formalism, I thought about the following.

If one's given a field $\psi$ and an action $$S[\psi]=\int d^nx\ \mathcal L(\psi,\partial\psi)$$ for this field, the Euler-Lagrange equations are given by $$E\mathcal L(x)=\frac{\delta S[\psi]}{\delta\psi(x)}=0.$$ The functional derivative here is an analogue of the gradient in finite dimensional calculus.

Naively, using finite dimensional intuition, one can argue that $$\frac{\delta^2S[\psi]}{\delta\psi(x)\delta\psi(y)}$$ is symmetric in $x,y$, and if an analogue of the Poincaré's lemma holds for "functionals" as well, then for a "functional" (more like a nonlinear operator, really) $F[\psi](x)$ it holds that $$\frac{\delta}{\delta\psi(y)}F[\psi](x)=\frac{\delta}{\delta\psi(x)}F[\psi](y),$$ then "locally" there should be a functional $$S[\psi]$$ such that $$F[\psi](x)=\frac{\delta}{\delta\psi(x)}S[\psi].$$

However, then, the inverse Lagrangian problem can be attacked as such:

Assume that a field $\psi$ is given with field equations $E[\psi](x)=0$, where $E[\psi](x)$ is some local function/field that depends on $\psi$ and its derivatives. If the equations of motion come from a variational principle, then $E[\psi]$ is "functionally exact", so the "functional exterior derivative" $$\mathcal{D}E[\psi](x,y)=\frac{\delta}{\delta\psi(x)}E[\psi](y)-\frac{\delta}{\delta\psi(y)}E[\psi](x)$$ should vanish, and by Poincaré's lemma, there is at least a "locally defined" action functional for $E$.

Questions:

• Is this approach in any way tenable? As in is there a functional version of Poincaré's lemma? If so it seems to me that determining whether a field equation is variational or not is reducible to a mechanical calculation.

• If there is a functional Poincaré's lemma, is there any explicitly calculable homotopy operator for it? The usual proof of Poincaré's lemma in finite dimensions also constructs the homotopy operator explicitly, and it can be used to calculate primitive forms.

Now, as a note, I'll say that probably this approach is not well defined in the rigorous sense, but physicists have a way of doing infinite-dimensional calculations via very very unrigorously used distributions and such pretty effectively, even if things are ill-defined mathematically, so I am only expecting answers on that level of rigour.

• If you're fine with the intuitive but imprecise physicist approach, why don't you discretise, use the discrete version, and take the continuum limit? This should give you a solid candidate for the homotopy operator you're after! – AccidentalFourierTransform Sep 17 '18 at 14:14
• @AccidentalFourierTransform That's one approach, I guess. I am mainly asking this question, because I have always got the impression that the inverse variational problem is difficult. However if this heuristic approach is workable then it should be nearly trivial to decide if an EoM is variational - though I would not be surprised if the homotopy operator is uncalculable. I might try to attack this in my free time but nontheless I'll wait for possible answers from people. – Bence Racskó Sep 17 '18 at 14:16
• Well, best case scenario, the operator will be given by a functional integral, so it will most likely be uncalculable! – AccidentalFourierTransform Sep 17 '18 at 14:19
• @AccidentalFourierTransform Unsure. For the finite dimensional homotopy operator you are integrating from the origin to $x$ via the curve $t\mapsto xt$ ($0\le t\le 1$), so I would imagine you have to integrate along a similar path in field space. – Bence Racskó Sep 17 '18 at 14:27

$$\text{Given a set of EOMs }F_i (x)\approx 0,\text{ find an action principle }S$$ is from a physics perspective not so much the technical issues with functional/variational differentiation. Rather it is the problems associated with accounting for the possibilities that we might have to, say, take linear combinations (with non-constant coefficients) and perhaps differentiate some of the EOM $F_i(x) \approx 0$, i.e rearrange the EOMs into $\widetilde{F}_i(x) \approx 0$ before we can solve$$\frac{\delta S}{\delta \phi^i(x)}~=~\widetilde{F}_i(x).$$ See also this related Phys.SE post.