# 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.

• There is such a term. It is not typically written because it is assumed (and its sort of true in some important cases) that $[\delta_1,\delta_2]=0$. I'll toot my own horn a bit a cite my work arxiv.org/abs/2009.14334 in which we present a proper review of covariant phase space including all such terms. We also discuss many other issues (like boundary terms) which one wouldn't find in a standard discussion. Commented Feb 26, 2021 at 14:10
• I would like to also mention this paper as being potentially useful. @PraharMitra is this really what people mean by (2)? I always interpreted it as being an explicit writing out of the expression $i_{V_1}i_{V_2}\delta\theta$. Also Gold, note we need $\omega$ to be a $(D-1)$-form over spacetime. This is easy to make contact with in explicit expressions and no fancy machinery: in the SHO the total derivative term is $~d(p\delta q)$. Clearly it's the $p\delta q$ we want for $\theta$ to produce $\omega~\delta p\wedge\delta q$, not its derivative. Commented Feb 26, 2021 at 18:01
• @RichardMyers - The explicit writing out of the expression $i_{V_1} i_{V_2} \delta \theta$ is $V_1(\theta(V_2)) - V_2(\theta(V_1)) - \theta([V_1,V_2])$ and OP is asking why the last term is not present in standard presentations of covariant phase space. Commented Feb 27, 2021 at 0:45
• @PraharMitra Yes, I meant in the sense of $i_{V_1}i_{V_2}(\alpha\wedge\beta)=alpha(V_2)\beta(V_1)-\alpha(V_1)\beta(V_2)$ for 1-forms, so if you write out things in components, it looks roughly like the expression above (without the bracket) given the understanding the $\delta_1$ doesn't act on the $\delta_2$ and visa versa, which isn't then the assertion $[\delta_1,\delta_2]=0$, but just misleading notation. Commented Feb 27, 2021 at 1:21
• @RichardMyers - "given the understanding that $\delta_1$ doesn't act on $\delta_2$" - you are right here. This prescription gives us the correct formula, but the prescription is not needed. If you use the correct definition of the exterior derivative (including the $[\delta_1,\delta_2]$ term you would end up getting the same answer as the prescription. Anyway, I don't it really matters - as long as we understand what the equation means and how to use it, that's all one needs. Commented Feb 27, 2021 at 18:30

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:

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

2. 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.

3. D. Harlow & J. Wu, Covariant phase space with boundaries, arXiv:1906.08616; p. 13-14.

• I think that this is really a matter of notation/convention. For example in the paper "Local symmetries and constraints" by Wald & Lee (doi.org/10.1063/1.528801) the authors state in the paragraph above eq. (2.24) that "A field configuration $\phi$ on space-time is represented as a point of $\mathscr{F}$. Similarly, a field variation $\delta\phi^a$ on spacetime about the field configuration $\phi$ may be viewed as a vector in the tangent space to $\mathscr{F}$ at point $\phi$". In particular that's the reference I had in mind when writing the post.
– Gold
Commented Feb 27, 2021 at 17:17
• Most other texts use $\delta\phi$ as a vector, eg papers by Wald, Iyer, Lee, Zoupas, etc. More over, in the context that @Gold is interested in, the papers written by Strominger, Laddha, Barnich, Compere, etc. all use $\delta \phi$ as a vector. In any case, that "misunderstanding" is not the issue that OP was having in the question. Commented Feb 27, 2021 at 18:33
• I updated the answer. Commented Feb 27, 2021 at 19:16

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.