Why there is no commutator term in the pre-sympletic density? In this post I'm considering the Covariant Phase Space (CPS) formalism as presented by Lee & Wald in "Local symmetries and constraints
". In the CPS formalism we take the Lagrangian form $L$ and write its variation as $$\delta L=E_i[\phi]\delta \phi^i+d{\pmb\Theta}[\phi,\delta \phi]\tag{1}.$$
This ${\pmb \Theta}[\phi,\delta\phi]$ is linear in $\delta \phi$ and if we view $\delta\phi$ as a tangent vector to the space $\mathscr{F}$ of allowed field configurations at $\phi\in \mathscr{F}$ we can think of ${\pmb\Theta}$ as a one-form in $\mathscr{F}$. So this is a $(D-1)$-form in the $D$-dimensional spacetime and a $1$-form in $\mathscr{F}$ so we call this a $(D-1,1)$-form.
This is the pre-sympletic potential density. Now, since in a phase space the sympletic form comes from the sympletic potential by one exterior derivative, often defines the pre-sympletic density to be one anti-symmetrized variation $${\pmb\omega}[\phi,\delta_1\phi,\delta_2\phi]=\delta_1{\pmb\Theta}[\phi,\delta_2\phi]-\delta_2{\pmb\Theta}[\phi,\delta_1\phi]\tag{2}.$$
This should be $\mathbf{d}{\pmb\Theta}$ where $\mathbf{d}$ is the exterior derivative in $\mathscr{F}$. But if we look at the invariant formula for the exterior derivative in some generic manifold $M$ it reads $$d\theta(X,Y)=X(\theta(Y))-Y(\theta(X))-\theta([X,Y])\tag{3}$$
where $X$ and $Y$ are vector fields over $M$ and where $\theta$ is here viewed as a map $\theta : \Gamma(TM)\times\Gamma(TM)\to C^\infty(M)$.
Comparing (2) and (3) they are similar in the first two terms, but the usual exterior derivative formula has the third term which has to do with the fact that vector fields in general do not commute. Only for commuting vector fields does (2) and (3) agree.
So why is (2) used for the exterior derivative $\mathbf{d}{\pmb\Theta}$ in $\mathscr{F}$? Why there is no term associated to $[\delta_1,\delta_2]$ like we see in the usual formula (3)? I believe we are restricting to vector fields in $\mathscr{F}$ which commute, $[\delta_1,\delta_2]=0$, but I don't see why do such a thing.
 A: Ref. 1 is considering a 2-parameter family of solutions $\phi(\lambda_1,\lambda_2)$, so that the vector-fields $X_1$ and $X_2$ (corresponding to $\frac{d}{d\lambda_1}$ and $\frac{d}{d\lambda_2}$, respectively) commute on such solutions. Therefore
$$\delta_1\phi^j(x)~=~i_{X_1}\delta\phi^j(x)
~=~\frac{d\phi^j(x)}{d\lambda_1}$$
and
$$\delta_2\phi^j(x)~=~i_{X_2}\delta\phi^j(x)
~=~\frac{d\phi^j(x)}{d\lambda_2}$$
may be viewed as vector components.
In general vector-fields don't commute, and there will be a Lie-bracket term $[X_1,X_2]$ in the invariant formula for the exterior derivative of 1-forms.
Here $\delta$ is the exterior derivative on the (infinite-dimensional) space of classical $\phi$-solutions, while $\mathrm{d}$ is the exterior derivative on spacetime, cf. e.g. Refs. 2-3. In particular, $\delta\phi^j(x)$ is a basis co-vector/one-form, not a vector.
References:

*

*J. Lee & R.M. Wald, Local symmetries and constraints, J. Math. Phys. 31 (1990) 725 (pdf); p. 730.


*C. Crnkovic & E. Witten, Covariant description of canonical formalism in geometrical theories. Published in Three hundred years of gravitation (Eds. S. W. Hawking and W. Israel), (1987) 676.


*D. Harlow & J. Wu, Covariant phase space with boundaries, arXiv:1906.08616; p. 13-14.
A: If you have say a field $\phi_i$ where $i$ are the indices of the field, then each $\phi_i$ for some $i$ can be considered a function on the space of the field. More than that, they represent local coordinates.
If you take $d$ of a one form in local coordinates (basically the curl) then there's no Lie bracket term because the vector fields arising from local coordinates commute.
If I recall correctly, when I've seen stuff actually worked out in this formalism they always use field indices like $\phi_i$ when something must be worked out explicitly.
