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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.