I would just like to ensure that I have fully understood the content of Noether's theorem and a few of its details. The generic statement of Noether's theorem is relatively straight forward however there are subtleties associated with what exactly constitutes a symmetry and the consequences of working on- and off-shell.
Symmetry
I discuss two notions of symmetry here, they are:
Quasi-symmetry: In which, to first order, the action changes by a boundary term and/or the Lagrangian changes by a total derivative: $$\delta S=[B(q)]_{t_0}^{t_1},\quad \delta L=\frac{dF(q)}{dt} \tag{1.a}.$$
Symmetry: In which, to first order, both quantities are invariant: $$\delta S=0,\quad \delta L=0. \tag{1.b}$$
On-shell vs. off shell
By on-shell we mean the subset of curves through configuration space, $Q\cong R^N$, that solve the Euler-Lagrange equations.
By off-shell we mean the more general set of curves that do not necessarily solve the Euler-Lagrange equations.
Noether's Theorem
Now we discuss the actual content of Nother's theorem, that an off-shell symmetry (or more generally quasi-symmetry) of the action implies the existence of a conserved quantity on-shell. In other words, a generic transformation on the domain of the action functional implies a conserved quantity along the subset of the domain that solve the Euler-Lagrange equations.
A generic infinitesimal change to the action can be written:
$$\delta S[q(t)]=\int_{t_0}^{t_1} dt\left( \frac{\partial L}{\partial q}-\frac{d}{dt}\left( \frac{\partial L}{\partial \dot q} \right)\right)\delta q+\left[\frac{\partial L}{\partial \dot q}\delta q\right]_{t_0}^{t_1}, \tag{2}$$
If we now work on-shell, in which $q(t)$ solves the E-L equations and the integrand vanishes, we are left with a few possibilities:
$\delta q$ satisfies the boundary condition $\delta q(t_0)=\delta q(t_1)=0$, in which case the boundary term vanishes and this is simply the statement that all first-order transformations of the action along the equations of motion vanish.
$\delta q$ does not satisfy the boundary conditions, in which case we are left with two further possibilities, either $\delta q$ is a symmetry or $\delta q$ is a quasi-symmetry (assuming of course it is a symmetry at all).
If $\delta q$ is a symmetry, then the following is true:
$$\left[\frac{\partial L}{\partial \dot q}\delta q\right]_{t_0}^{t_1}=0\quad \Rightarrow \quad \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\delta q\right)=0, \tag{3}$$ and we obtain our conserved "Noether charge".
If, however $\delta q$ is a quasi-symmetry then we find by (1.a):
$$\left[\frac{\partial L}{\partial \dot q}\delta q\right]_{t_0}^{t_1}=\left[B(q(t))\right]_{t_0}^{t_1}\quad \Rightarrow \quad \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\delta q-B(q(t))\right)=0, \tag{4}$$
and we obtain a slightly different Noether charge. We can then finally (cf. this post) relate the quasi-symmetry of the action and the quasi-symmetry of the Lagrangian by:
$$B(q(t))=F(q(t)) \tag{5},$$
which closes the possible outcomes.
My question is are the above statements correct if we confine our attention to quasi-symmetries, and if not, where in my definitions and/or derivation have I made mistakes?
I understand that this is a bit of an open ended question, but this topic is discussed at length on this site, and after fairly extensive reading of related questions I think this summary question serves an ok role. But apologies if it is against the rules.
