Difference between Noether theorem analysis on the spacetime and on jet bundle

Question

I'm studying the 'Advanced Lectures on GR' by G. Compère and got confused regarding one point. In Lecture 1 he studies surface charges in theories with local symmetries. In pages 15 and 16 he introduces the jet bundle. Informally we have a manifold $\mathfrak{J}$ with coordinates $(x^\mu,\Phi^i,\Phi^i_\mu,\Phi^i_{\mu\nu},\dots)$ where $x^\mu$ are spacetime coordinates.

We then define the vertical differential or variational operator to be $$\delta =\delta \Phi^i\dfrac{\partial}{\partial \Phi^i}+\delta \Phi^i_\mu\dfrac{\partial}{\partial \Phi^i_\mu}+\cdots\tag{1.23}$$

We also define the horizontal differential $d = dx^\mu \partial_\mu$ where the $\partial_\mu$ operator is defined by $$\partial_\mu\equiv \dfrac{\partial}{\partial x^\mu}+\Phi^i_\mu\dfrac{\partial}{\partial \Phi^i}+\Phi^i_{\mu\nu}\dfrac{\partial}{\partial \Phi^i_\nu}+\cdots\tag{1.24}$$

In this context we have the following discussion in pages 20 and 21 about the Noether-Wald surface charge:

Let us now take the variation of $\mathbf{L}$ along any infinitesimal diffeomorphism $\xi^\mu$: \begin{align}\delta_\xi \mathbf{L}=\mathcal{L}_\xi\mathbf{L} &= d(i_\xi \mathbf{L})+i_\xi d\mathbf{L} = d(i_\xi \mathbf{L})\\ &= \dfrac{\delta \mathbf{L}}{\delta \Phi}\mathcal{L}_\xi\Phi + d\Theta[\mathcal{L}_\xi\Phi;\Phi]\end{align} By virtue of Noether's second theorem (Result 5), we get: $$d(i_\xi \mathbf{L})=d\mathbf{S}_\xi\left[\dfrac{\delta L}{\delta \Phi};\Phi\right]+d\Theta[\mathcal{L}_\xi\Phi;\Phi]\Longrightarrow \partial_\mu \big(\xi^\mu L-\Theta^\mu[\mathcal{L}_\xi\Phi;\Phi]-S^\mu_\xi\left[\dfrac{\delta L}{\delta \Phi};\Phi\right]\big) = 0.\tag{1.58}$$ The standard Noether current of field theories is the Hodge dual of the conserved $n-1$ form $$\mathbf{J}_\xi \equiv i_\xi \mathbf{L}-\Theta[\mathcal{L}_\xi\Phi;\Phi],\quad \text{with}\quad d\mathbf{J}_\xi=d\mathbf{S}_\xi\Rightarrow d\mathbf{J}_\xi \approx 0.\tag{1.59}$$ Now, a fundamental property of the covariant phase space is that a close form that depends linearly on a vector $\xi^\mu$ and its derivatives is locally exact. Therefore, this Noether current can be written as $\mathbf{J}_\xi = \mathbf{S}_\xi + d\mathbf{Q}_\xi$.

Now in this analysis I understand the $d$ operator is the horizontal differential of $\mathfrak{J}$ defined by (1.24). Therefore the statement that "closed implies locally exact" follows from the algebraic Poincaré lemma which if I understand is the analogue of the usual Poincaré lemma for this operator $d$.

The point is that all of this seems to come out of usual differential geometry on spacetime itself, without the jet bundle. I mean, the Lagrangian density is the $n$-form $\mathbf{L}$ and a symmetry obeys $\delta \mathbf{L}=d\Xi$ by definition. A general variation is always of the form $$\delta \mathbf{L}=\dfrac{\delta L}{\delta \Phi}\delta \Phi + d\Theta[\delta \Phi;\Phi].$$ If the symmetry is further local Noether second theorem implies that the first term in the last equation is $d\mathbf{S}$ where $\mathbf{S}$ is one $(n-1)$ form depending homogeneously on the equations of motion and hence vanishing on-shell. Putting it all together we find $$d\big(\Xi-\Theta-\mathbf{S}\big)=0,$$ finally since $\Xi-\Theta-\mathbf{S}$ is just one $(n-1)$-form on spacetime, the equation above says it is closed, and the standard Poincaré lemma tells that it is exact.

In that case I really don't get it. Why do we need to resort to the jet bundle here in order to make this discussion? What is wrong with the above discussion relying purelly on the spacetime manifold? What is the reason we need to use the algebraic Poincaré lemma applied to the horizontal differential (1.24) on $\mathfrak{J}$ instead of the usual Poincaré lemma applied to the spacetime exterior derivative?

In summary my question here is basically why Compère do as quoted above instead of the simpler evaluation I mentioned?

Answer 1

One problem is that we want the forms to be local in spacetime. The homotopy operator from the standard Poincare lemma may introduce non-locality. The homotopy operator from the algebraic Poincare lemma avoids this.

Answer 2

I am not sure I understand OP's problem, but aside from what Qmechanic already said, the standard Poincaré's lemma does not always correspond to the variational Poincaré's lemma when it is applied to functional/variational forms. In particular, consider the horizontal row of the variational bicomplex $$ 0\rightarrow\mathbb R\rightarrow\Omega^{0,0}(J^\infty E)\rightarrow ... \rightarrow \Omega^{0,n}(J^\infty E), $$ where all arrows aside from the first two are the horizontal differentials $\mathrm d_h$ (denoted as simply $\mathrm d$ in the OP). Here $E\rightarrow^\pi M$ is a fibre bundle over which the variational bicomplex is constructed, $J^\infty E$ is the infinite jet manifold of $E$, and by default I consider $J^\infty E$ to be fibred over the base space $M$. The notation $\Omega^{k,l}(J^\infty E)$ refers to vertical and horizontal degrees respectively, namely if $\omega\in \Omega^{k,l}(J^\infty E)$, then the local expansion of $\omega$ always consists of linear combinations of basis terms of the form $\mathrm dx^{\mu_1}\wedge...\mathrm dx^{\mu_l}\wedge \mathrm dy^a_{\nu_1...}\wedge...$ where the remaining $k-1$ factors do not involve $\mathrm d x^\mu$ at all.

Since the maximum horizontal order is $n$, it follows that for any $L\in\Omega^{0,n}(J^\infty E)$, we have $\mathrm d_hL=0$ by default, however this absolutely does not imply (even locally) that there is some $\Theta\in \Omega^{0,n-1}(J^\infty E)$ such that $L=\mathrm d_h \Theta$. Remember that $\mathrm d_h$-exact Lagrangians are the variationally trivial ones. If $\mathrm d_hL=0$ implied $L=\mathrm d_h\Theta$, this would mean that all Lagrangians are variationally trivial, a clear nonsense.

On the other hand, since the horizontal differential's definition$^\ast$ is $$ (j^\infty\Phi)^\ast\mathrm d_h\lambda=\mathrm d((j^\infty\Phi)^\ast\lambda), $$ where $\lambda$ is a form on $J^\infty E$ and $\Phi:M\rightarrow E$ is a section of $E$ with $j^\infty\Phi$ being its infinite jet prolongation (the $^\ast$ disclaimer is that this definition might be valid only for purely horizontal $\lambda$s, my jet bundle-fu is kinda dusty) and on the RHS $\mathrm d$ is the ordinary de Rham differential on $M$, we have $$ 0=(j^\infty\Phi)^\ast\mathrm d_hL=\mathrm d((j^\infty\Phi)^\ast L)\Rightarrow \exists \Theta\in\Omega^{n-1}(M),(j^\infty\Phi)^\ast L=\mathrm d\Theta. $$ Here existence is understood locally. Using local coordinates, this means that if $L=L(x^\mu,y^a,y^a_\mu,...,y^a_{\mu_1..\mu_k},...)\mathrm dx^1\wedge...\mathrm dx^n$, then locally there exists an $n-1$-form on $M$ such that $$(j^\infty\Phi)^\ast L|_x=L(x^\mu,\Phi^a_\mu(x),...,\Phi^a_{\mu_1...\mu_k}(x),...)\mathrm dx^1\wedge...\wedge\mathrm dx^n=\mathrm d\Theta|_x$$, where $\Theta|_x=\frac{1}{(n-1)!}\Theta_{\mu_1...\mu_{n-1}}(x)\mathrm dx^{\mu_1}\wedge...\wedge\mathrm dx^{\mu_{n-1}}$.

The catch is that while $L$ is defined on $J^\infty E$ and thus depends on $x^\mu$ as well as the fibre coordinates $y^a,y^a_\mu,...,y^a_{\mu_1...\mu_k},...$, the pullback $(j^\infty\Phi)^\ast L$ now only depends on $x^\mu$, thus the ordinary Poincaré lemma only guarantees that there is an ordinary $n-1$ form on $M$, $\Theta$, such that $(j^\infty\Phi)^\ast L$ is its exterior derivative. Since for each section $\Phi$, the form of $(j^\infty\Phi)^\ast L$ is different when considered as a form on $M$, for each $\Phi$ there is a different $\Theta=\Theta_\Phi$ that is the primitive of $(j^\infty\Phi)^\ast L$, however the dependence $\Phi\mapsto\Theta_\Phi$ can be very complicated and it does not imply that there is a $\theta\in\Omega^{0,n-1}(J^\infty E)$ such that $\Theta_\Phi=(j^\infty\Phi)^\ast\theta$.

This is what Qmechanic referred to when he said that using the ordinary Poincaré's lemma would produce a "nonlocal" primitive.

Note that if such a $\theta\in\Omega^{0,n-1}(J^\infty E)$ did exist, that $\Theta_\Phi=(j^\infty \Phi)^\ast\theta$, then this would imply that $L=\mathrm d_h\theta$, and it would imply that $L$ is variationally trivial.

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Note:

In the OP, this question was not posed with regards to Lagrangians, but different elements of $\Omega^{k,l}(J^\infty E)$, however the reason I have chosen to use Lagrangians as an example is because the space $\Omega^{0,n}$ has the particular property that $\mathrm d_h\Omega^{0,n}=0$, but not all of its elements are locally horizontally exact. The reason for that is that this horizontal row of the variational bicomplex can be continued as $\Omega^{0,n}(J^\infty E)\rightarrow^{\mathscr E}\mathcal E_1(J^\infty E)$, where $\mathscr E$ is the Euler-Lagrange operator, and $\mathcal E_1(J^\infty E)$ is a space that is either difficult to construct explicitly or is the quotient space $\Omega^{1,n}(J^\infty E)/\mathrm d_h\Omega^{1,n-1}(J^\infty E)$ (if the quotient space construction is followed, then $\mathscr E$ is simply the image of the vertical differential $\delta$ under the quotient). Then it is $\mathscr EL=0$ that implies locally that $L=\mathrm d_h\theta$ (for any $L\in\Omega^{0,n}(J^\infty E)$).

The other portions of the horizontal row, say $...\rightarrow\Omega^{0,k}\rightarrow^{\mathrm {d}_h}\Omega^{0,k+1}\rightarrow ...$ are actually locally exact (for $k<n$), but this doesn't change the fact that using the homotopy operator from Poincaré's lemma on the base space to find the primitive of such an element when pulled back to the base space via a section will not in general produce the pullback of an element of the variational bicomplex.