I am presently reading [this](https://arxiv.org/abs/1906.08616) paper on covariant phase space and I have difficulty understanding the following formalism developed:


In the paper (section $2.2$, pg. $12$), the authors have introduced the notion of pre-phase space and go on to reinterpret differential forms by their functional counterpart. Instead of viewing $\delta$ as the variation of a functional, it is viewed as an exterior derivative living in the configuration space. Thus, the action of $\delta \phi^{a}$ is given by
$$\delta \phi^{a}\left(\int d^{d}x'f^{b}\left(\phi,x' \right)\frac{\delta}{\delta \phi^{b}(x')} \right)=f^{a}(\phi,x)$$  

They go on to derive a formula for the pre-symplectic current by making the assumption that $\delta^{2}=0$ (which holds since the functional is being viewed as an exterior derivative). Finally, in section $2.3$, they follow this formalism to define a vector field as follows
$$X\_{\xi}\equiv\int d^{d}x\mathcal{L}\_{\xi}\phi^{a}(x)\frac{\delta}{\delta \phi^{a}}$$
such that $\cdot$ in $X\_{\xi}\cdot \delta \phi^{a}(x)$ denotes the insertion of a vector into the first arguement of the differential form. 


I don't follow the formalism used, are they stating that the differential forms have the above-stated form in the functional space? If this is so then how does one prove this and that the assumption $\delta^{2}$ holds. 



One-Forms in Functional Space?
There is no problem in defining the exterior differential $\delta$ on infinite-dimensional manifolds such as the function space. In particular, $\delta^2 = 0$ follows from a similar calculation as in finite dimensions.


I guess the notation in the paper (as with almost every physics paper on this subject) should be understood in a somewhat formal way. They write the function $\phi \mapsto \phi^a(x) \in \mathbb{R}$ simply as $\phi^a(x)$. Thus the exterior differential $\delta \phi^a(x)$ is a $1$-form on the function space. Then $\frac{\delta}{\delta \phi^b(y)}$ is "defined" via duality by the formula 
$$\delta \phi^{a}\left(\int d^{d}y f^{b}\left(\phi,y \right)\frac{\delta}{\delta \phi^{b}(y)} \right)=f^{a}(\phi,x).$$
This is in analogy with the usual coordinate expression $d q^i (X^j \frac{\partial}{\partial q^j}) = X^i$, but $\phi \mapsto \phi^a(x)$ does not give a local chart on the function space (in contrast to $q \mapsto q^i$) so this should be understood more as a notation than as a definition.


