Noether's first Vs Noether's Second theorem

Question

I am reading about the first and second Noether's theorem from https://arxiv.org/abs/2112.05289. In the text, there is this piece, which I am not sure I entirely understand.

Let us reflect briefly on how this proof compares to that of Noether’s 1st theorem. In both proofs, one needs some sort of varying arbitrary function in order get the final result. In Noether’s 1st theorem, it was $\epsilon(x)$; in Noether’s 2nd theorem, it was $\lambda(x)$. The difference is that in Noether’s 1st theorem, the transformation that depended on $\epsilon(x)$ was not really a symmetry of the action (even though it reduced to one when $\epsilon = \text{const}$) and we only had $\delta S \approx 0$ on-shell, meaning our conservation equation only held on-shell. However, in Noether’s 2nd theorem, we had a gauge symmetry transformation which left $\delta S = 0$ off-shell and already depended on an arbitrary function $\lambda(x)$ without the need of $\epsilon(x)$. Because $\delta S = 0$ off-shell, we could find an equation which held off-shell.

1. I know this sounds a bit basic, but is there a deeper reason as to why the action varies by a total derivative term? In my understanding, if one Taylor expands the Lagrangian in terms of the infinitesimal transformation, then this is what one would find (up to equations of motion).

2. The author speaks for $\epsilon(x)$ as if it is not a symmetry of the action. However, the action with $\epsilon(x)$ yields the same classical equations of motion as if the action for $\epsilon=\text{const}$. Therefore, the two parameters are both symmetries of the action, no? Isn't it always that a symmetry of an action leaves the classical equations of motion invariant under the symmetry transformation under consideration?

3. At some point in the text (p.12-13), the author states, after having used the equations of motion, that the current $J_{\lambda}^{\mu}$ can be written as the divergence of an antisymmetric tensor $K^{\mu\nu}$, i.e. $$J_{\lambda}^{\mu}=\partial_{\nu} K^{\mu\nu},\ K^{\mu\nu}=-\lambda F^{\mu\nu}$$ and then he states that because of $K^{\mu\nu}$ being an antisymmetric tensor, the conservation of $J^{\mu}_{\lambda}$ follows from the trivial equation $\partial_{\mu}\partial_{\nu}K^{\mu\nu}=0$, which holds both on- and off-shell. My last question is this: how can $J^{\mu}_{\lambda}$ be conserved trivially (i.e. on- and off- shell) if the equations of motion have been used to write it in terms of the antisymmetric tensor $K^{\mu\nu}$?

Answer 1

1. Unfortunately I don't understand this part of the question.
2. An infinitesimal transformation is a symmetry of the action if the action changes by a boundary term under it. To give an example, consider the scalar field Lagrangian $\mathcal L=-\partial_\mu\phi\partial^\mu\phi^\ast,$ where $\phi$ is complex valued. The transformation $$ \delta\phi=i\alpha\phi,\quad\delta\phi^\ast=-i\alpha\phi^\ast $$ is an infinitesimal symmetry if $\alpha$ is a constant, since $$ \delta\mathcal L=-\partial_\mu(i\alpha\phi)\partial^\mu\phi^\ast+\partial_\mu\phi\partial^\mu(i\alpha\phi^\ast) =-i\alpha\partial_\mu\phi\partial^\mu\phi^\ast+i\alpha\partial_\mu\phi\partial^\mu\phi^\ast=0. $$ But if $\alpha$ is a function on spacetime, we have $$ \delta\mathcal L=i\partial_\mu\phi\phi^\ast\partial^\mu\alpha-i\phi\partial^\mu\phi^\ast\partial_\mu\alpha\neq 0, $$ and this isn't even a total derivative term. So the transformation with $\alpha$ a function is not in general a symmetry.
3. The text does not appear to be very systematic regarding this, but if $J^\mu_\lambda$ is a current that depends on a set of functions $\lambda^a$ linearly and differentially, e.g. $$ J^\mu=J^\mu_a\lambda^a+J^{\mu,\nu}_a\partial_\nu\lambda^a+\dots+J^{\mu,\nu_1...\nu_r}_a\partial_{\nu_1...\nu_r}\lambda^a, $$ and it satisfies $ \partial_\mu J^\mu_\lambda=0$ for all $\lambda$ as an off-shell relation, then we can always write also off-shell that $$ J^\mu_\lambda=\partial_\nu K^{\mu\nu}_\lambda, $$ where $K^{\mu\nu}_\lambda$ is antisymmetric and also depends on $\lambda$ linearly and differentially. Moreover, if one gives a proper global formulation for these types of objects, it turns out that this exactness result is also true globally, i.e. there is no nontrivial notion of "de Rham cohomology" for these types of objects. $$ \ $$When $J^\mu_\lambda=J^\mu_a\lambda^a+J^{\mu,\nu}_a\partial_\nu\lambda^a$, this can be worked out easily: The conservation law gives $$ 0=\partial_\mu J^\mu_\lambda=J^{\mu,\nu}_a\partial_{\mu\nu}\lambda^a+(J^\nu_a+\partial_\mu J^{\mu,\nu}_a)\partial_\nu\lambda^a+\partial_\mu J^\mu_a\lambda^a, $$ and since $\lambda^a$ is arbitrary, this must separately vanish order-by-order in $\lambda^a$. The second order part gives $$ J^{(\mu,\nu)}_a=0\Longleftrightarrow J^{\mu,\nu}_a=K^{\mu\nu}_a, $$ where $K^{\mu\nu}_a$ is antisymmetric. The first order part gives $$ J^\mu_a=\partial_\nu K^{\mu\nu}_a, $$ and the zeroth order part is trivially satisfied then. Then if we define $$ K^{\mu\nu}_\lambda=K^{\mu\nu}_a\lambda^a, $$ we get $$ \partial_\nu K^{\mu\nu}_\lambda=\partial_\nu K^{\mu\nu}_a\lambda^a+K^{\mu\nu}_a\partial_\nu\lambda^a=J^\mu_a\lambda^a+J^{\mu,\nu}_a\partial_\nu\lambda^a=J^\mu_a. $$ Exercise: Work out this equation for the electromagnetic example in OP without using the field equations (hint: the second order part of the coservation law is the field equations themselves).