In the context of classical Hamiltonian systems, Noether's theorem asserts that if a Hamiltonian system $(M,\omega,H)$ has a 1-parameter group of Hamiltonian symmetrys $\{\varphi_s\}$ generated by a Hamiltonian vector field $X_A$, then the quantity $A$ is conserved along the solutions of the Hamilton's equations (the flow of $X_H$), i.e., $A$ is a first integral. Conversely, if a smooth function $A$ is conserved along the solutions of the Hamilton's equations, then there exists a Hamiltonian symmetry generated by a Hamiltonian vector field $X_A$. This result shows a deep connection between the symmetries of a Hamiltonian system and its conserved quantities.
Proof
Indeed, since $\phi_s$ are Hamiltonian symmetries, $X_A(H)=0$. But we know that
$$ X_A(H)=dH(X_A)=\omega(X_H,X_A)= $$ $$ =-\omega(X_A,X_H)=-dA(X_H)=-X_H(A) $$from where we conclude the result. $\blacksquare$
In @olver86 we find the interesting Proposition 6.28
Proposition. In a Poisson manifold, a function $P(x,t)$ is a first integral of the Hamiltonian system of ODEs:
$$ \frac{dx}{dt}=J(x)\nabla H(x,t) $$if and only if
$$ \frac{\partial P}{\partial t}+\{P,H\}=0. $$$\blacksquare$
Poor's man version.
One of the great advantages of Lagrangian formulation is the application of this theorem. It lets us reduce variables if we know a symmetry of the system.
Given a Lie group acting over the configuration space, we can considerate the infinitesimal action $q\mapsto q+\delta X_V$ (see Lie algebra action) where $X_V=(f_i(q))$ in a particular coordinate system. If $p_i$ are the generalized momenta in this coordinates $q_i$, Noether's theorem says that if the infinitesimal action doesn't change the Lagrangian then the quantity
$$Q= \sum_i p_i\cdot f_i$$
is conserved (see The theoretical minimum 1, page 139).
Full version.
A vector field $\eta$ on a manifold $M$ is said to be an infinitesimal symmetry for a tensor field $Q$ (maybe a function) defined on $M$ if
$$ \mathcal{L}_{\eta} Q=0 $$This is equivalent to saying that $Q$ is invariant under the flow $g_{\eta}^t$ associated to $\eta$, i.e.,
$$ (g_{\eta}^t)^*(Q)=Q $$On the other hand, a vector field $\eta$ on $M$ can be lifted to a unique vector field $\tilde{\eta}$ on $TM$ in such a way that its flow is the corresponding lifted flow. I think it should have something to do with the prolongation of vector fields...
This means that for $(m,v)\in TM$ and $t\in \mathbb{R}$, the lifted flow takes $m$ to $g_{\eta}^t(m)$ and $v$ to the Lie transport of $v$ under $g_{\eta}^t$.
It is easy to show that, in coordinates, if $m=(q_i)$, $v=(\dot{q}_i)$ and $\eta=(a_i)$:
$$ \tilde{\eta}=\sum_{i=1}^{n}\left(a_{i} \frac{\partial}{\partial q_{i}}+\sum_{j=1}^{n} \dot{q}_{j} \frac{\partial a_{i}}{\partial q_{j}} \frac{\partial}{\partial \dot{q}_{i}}\right) $$Theorem:
Let $L:TM\mapsto \mathbb{R}$ be a Lagrangian function, and $\eta=(a_i)$ a vector field such that $\tilde{\eta}$ is an infinitesimal symmetry for $L$. Then for any critical path $\gamma$ for $L$, the function
$$ \sum_{i=1}^{n} a_{i} \frac{\partial L}{\partial \dot{q}_{i}}$$
is constant along the lift of $\gamma$ on $TM$.
Proof: A brief introduction to physics for mathematicians, page 14.
It is almost trivial. See formulation of QM#Symmetries.
See @olver86 Theorem 4.29.
Suppose $G$ is a one-parameter local group of transformations of variational symmetry of the variational problem
$$ J[u]=\int_{\Omega}L(x,u^{(n)})dx $$Let
$$ \mathbf{v}=\sum_{i=1}^p \xi^i(x, u) \frac{\partial}{\partial x^i}+\sum_{\alpha=1}^q \phi_\alpha(x, u) \frac{\partial}{\partial u^\alpha} $$be the infinitesimal generator of $G$, and $Q=(Q_1,\ldots,Q_q)$ its characteristic. Then $Q$ is also the characteristic of a conservation law for the Euler-Lagrange equations.
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Author of the notes: Antonio J. Pan-Collantes
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