# One-Forms in Functional Space?

I am presently reading this 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.

## 1 Answer

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.