From group-theory perspective, what is Action?
I'm trying to self-study group theory in relation to physics, so I apologize in advance if I'm missing something obvious here or if the question is not clear.
We know there is a symmetry group, which can be either <PERSON> group or <PERSON> group, acting on physical space: a set of transformations that change $(t,x,y,z) \to (t',x',y',z')$ in a certain way.
However, what we need study is how they act on the space of States, and States (in classical mechanics) include momenta. The relation between, say, $p^x$ and $(t,x,y,z)$ comes from the Action $A(t,x,y,z)$:
$$ p^x = -\partial A(t,x,y,z)/\partial x $$ where $A(t,x,y,z)$ is the Action obeying Euler-Lagrange equations.
So there's this additional thing called Action which comes into play. 669
However, and it seems to me to be the key point, the Action cannot be chosen arbitrarily: it has to be constructed in such a way that a transformation from our group turns a solution of E-L equations into a solution of E-L equations with the same Action.
If we represent $A$ in the functional form: $$ A[x(t), y(t), z(t)] = \int L(t,x(t),y(t),z(t),\dot{x}(t), \dot{y}(t), \dot{z}(t)) dt $$ this condition implies that there exists a function $F(t,x,y,z)$, s.t. $$ L(t,x(t),y(t),z(t),\dot{x}(t), \dot{y}(t), \dot{z}(t)) dt - L(t', x'(t'),y'(t'),z'(t'),\dot{x}'(t'), \dot{y}'(t'), \dot{z}'(t')) dt' = dF $$ where primed values are related to unprimed values by our transformation.
One can check that given a particular group, it is possible to recover standard forms of galilean or relativistic actions (up to constants). For example, assuming homogeneity and isotropy of space and time, we can assume that $$ L(t,\vec{x}, \vec{v}) = \mathscr{L}(\vec{v}^2) $$ and then taking an infinitesimal transformation from Poincare group: $$ t' = t - (\vec{u},\vec{x})/c^2 $$ $$ \vec{x}' = \vec{x} - \vec{u}t $$ we conclude that this condition can only be satisfied if $$ \frac{\mathscr{L}}{c^2} + 2\left(1-\frac{\vec{v}^2}{c^2}\right)\mathscr{L}' = \textrm{const} $$ Solving it, we get $$ \mathscr{L} = \textrm{const}\times c^2 + \textrm{const}_2\times\sqrt{1-\frac{\vec{v}^2}{c^2}} $$ To recover the standard form just set $\textrm{const} = 0$, $\textrm{const}_2 = -mc^2$.
So, we have this full-differential condition which is some sort of condition on $L$ and hence on $A$, arising from the group we study.
My question is: from the group-theory perspective, what is this condition? What can we learn about $L$ (or $A$) given just the group, without explicitly solving the equation above? 804