mathematical physics - Is the change in the Lagrangian always a total derivative for symmetry transformations of the action?

Let $\Omega\subset\mathbb R^n$ and consider an arbitrary functional $$ S\colon C^k(\Omega)\to\mathbb R $$ that is local, $$ \phi\overset S\mapsto \int_\Omega L(\phi,D\phi,\dots,D^k\phi) $$ for $L$ some $n$-form taking values in the $k$-jet space $J^k\phi$. We may call $S$ the action functional, and $L$ its associated Lagrangian (although we do not assume that $\delta S[\phi]=0$, i.e., we work off-shell).

There are two notions of a symmetry that are typically discussed:

1. Lagrangian symmetries, that is, transformations of the form $\phi\to\phi'$ that satisfy $$ L-L'=\mathrm d\Theta $$ for some $(n-1)$-form $\Theta$, and where $L'(J^k\phi)=L(J^k\phi')$.


2. Action symmetries, that is, transformations of the form $\phi\to\phi'$ that satisfy $$ S-S'=0 $$ where $S'[\phi]=S[\phi']$.

If we assume that the boundary conditions are such that $\Theta|_{\partial\Omega}=0$, then it is clear that Lagrangian symmetries are also action symmetries.

If we assume that $\Omega$ is "nice-enough" (e.g. star-shaped) then I believe that any action symmetry is also a Lagrangian symmetry, although I would like to have a more precise statement about this fact. In other words, my question is

Are these two notions of symmetry equivalent? Under what conditions?

Moreover, how much can we weaken the assumptions? For example, can we take $\Omega$ to be simply-connected instead of star-shaped? Can we take $\phi$ to only have weak derivatives instead of strong derivatives?

Answer

1. 
   

   OP's definition 1 is often called a quasi-symmetry$^1$ (QS) of the Lagrangian density ${\cal L}$, or equivalently, a QS of the Lagrangian $n$-form $$\mathbb{L}~=~{\cal L}~\mathrm{d}x^0 \wedge \ldots \wedge \mathrm{d}x^{n-1}.$$

   
   



2. 
   

   OP's definition 2 should be relaxed to a quasi-symmetry (QS) of the action $$S_{\Omega}~=~\int_{\Omega} \mathbb{L} ,$$ where the action is allowed to change with boundary terms. [This is partly because the Noether variations do not necessarily satisfy the boundary conditions that we usually impose when deriving the Euler-Lagrange (EL) eqs.]

   
   


3. 
   

   The following implication is trivial:

   
   

   
   

   A QS of the Lagrangian density ${\cal L}$ is also a QS of the action $S_{\Omega}$.

   
   

   
   

   OP's main question is

   
   

   
   
   

   What about the opposite direction?

   
   

   
   

   That's a good question! Assume that we know a transformation is a QS for $S_{\Omega}$ for at least a single fixed integration region $\Omega$. Then it is still possible to conclude that it is a QS of the Lagrangian density ${\cal L}$ if

   
   
   
   
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     (i) the Lagrangian density ${\cal L}$ is local and

     
     
   
   
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     (ii) if the field configuration space is contractible.

     
     
   
   
   
   
   

   This is due to an algebraic Poincare lemma, cf. Ref. 1. Note in particular that possible topological obstructions live in the field configuration space, not in the spacetime region $\Omega$.

   
   


4. 
   

   Example. Let $$\mathbb{L}~=~L~\mathrm{d}t, \qquad L~=~\frac{1}{2}\left(-ig^{-1}\dot{g}\right)^2 ~=~\frac{1}{2}\dot{\theta}^2 ,\qquad \Omega~=~[t_i,t_f],$$ where $$g~=~e^{i\theta}~\in~ U(1)~\cong~ S^1$$ is $U(1)$-valued with $|g|=1$. Here $\theta\in\mathbb{R}$ is a multi-valued angle variable $\theta\sim \theta+2\pi$, and therefore not a globally well-defined coordinate per se. Note in particular that the field configuration space $S^1$ is not contractible.

   
   

   Consider an infinitesimal transformation $$\delta g ~=~i\varepsilon t~ g, \qquad \delta\theta ~=~-ig^{-1} \delta g~=~\varepsilon t, $$ where $\varepsilon$ is a $t$-independent infinitesimal parameter. The transformation is not a QS of the Lagrangian $$ \delta L ~=~\varepsilon\dot{\theta} ~=~ \varepsilon\left(-ig^{-1}\dot{g}\right), $$ because it cannot be written as total time derivative using only good coordinates. But the change of the action is $$\delta S_{\Omega}~=~\varepsilon(\theta(t_f)-\theta(t_i))$$ which is a well-defined boundary integral, which is independent of the $\theta$-branch. So the transformation is a QS of the action. The corresponding Noether charge $$ Q~=~t\dot{\theta}-\theta $$ lacks good coordinates.

   
   

References:

1. G. Barnich, F. Brandt and M. Henneaux, Local BRST cohomology in gauge theories, Phys. Rep. 338 (2000) 439, arXiv:hep-th/0002245.

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$^1$ In this answer we will for simplicity only consider infinitesimal variations/transformations.