I will be interested to know how to tell a symmetry is gauge symmetry from the Action without going to Hamiltonian theory, some says it has to do with Noether's 2nd theorem. Someone pls show me.
I will state without proof the 2nd Noether theorem and its connection with constraints.
The Lagrangian \mathcal{L}(\phi_{A},\partial_{a}\phi_{A}) with \phi_{A} (A=1,2,...,N) being generic fields is invariant under the infinitesimal transformations;
\phi^{A} = G^{A}_{\alpha}(\phi) \Delta^{\alpha}(x) + T^{Aa}{}_{\alpha}\partial_{a}\Delta^{\alpha}(x)
\alpha = 1,2,..,r
with arbitrary x-dependent functions \Delta^{\alpha} , i.e.,
\delta\mathcal{L}(x) = 0
if and only if the following relations hold
identically (i.e. irrespective of whether or not \phi_{A} are solutions of the field equations):
\partial_{a}(T^{Aa}{}_{\alpha}\frac{\delta}{\delta\phi^{A}}\mathcal{L}) = G^{A}{}_{\alpha}\frac{\delta}{\delta\phi^{A}}\mathcal{L} \ \ (1)
\partial_{a}K^{ab}{}_{\alpha} + J^{b}{}_{\alpha}=0 \ \ (2)
K^{ab}{}_{\alpha}+K^{ba}{}_{\alpha}=0
where
J^{a}_{\alpha}=G^{A}_{\alpha} \frac{\partial\mathcal{L}}{\partial\partial_{a}\phi^{A}} + T^{Aa}{}_{\alpha}\frac{\delta\mathcal{L}}{\delta\phi^{A}}
K^{ab}{}_{\alpha} = T^{Ab}{}_{\alpha}\frac{\partial\mathcal{L}}{\partial\partial_{a}\phi^{A}}
and
\frac{\delta\mathcal{L}}{\delta\phi^{A}}=\frac{\partial\mathcal{L}}{\partial\phi^{A}} - \partial_{a}(\frac{\partial\mathcal{L}}{\partial\partial_{a}\phi^{A}})
denotes Euler derivative.
Now, Eq(1) means that N-Euler derivatives are interrelated by r equations ( \alpha = 1,2,...r ), and hence, that the number of independent quantities among Euler derivatives is (N-r) in general. This means that the field equations:
\frac{\delta\mathcal{L}}{\delta\phi^{A}} = 0
for N unknown quantities \phi_{A} are not independent of each other, but that only N-r equations are independent owing to the constraints Eq(1).
regards
sam
