I wonder "which quantity" is conserved in relation to a specific symmetry.

I guess it is in some meaning simply the generator (in the context of Lie theory) of the symmetry, as it is true for angular momentum (conserved) being generator of rotation (symmetry).

I want clear formulation of idea but I can't complete it:

Let $M$ be configuration manifold of a classical system. $L(q,\dot q,t)$ it's Lagrangian, $h^s: M\rightarrow M$ a one-parameter group of diffeomorphisms of $M$ which preserves $L$. then $$I(q,\dot q)= \frac{\partial L}{\partial \dot q}\frac{d h^s(q)}{ds}|_{s=0}$$ is a conserved quantity along trajectory $q$. on the other hand $h^s$ produce a flow on manifold and we may denote its vector field by $\frac{\partial}{\partial s}$ which we may call generator of diffeomorphism. so in particular the generator of diffeomorphism along a trajectory $q$ is a vector field along it not a scalar field to could be discussed a conserved quantity or not (for example conserved quantity in relation to rotation be its generator namely angular momentum)

you may say by formula of $I$ just act $\frac{\partial L}{\partial \dot q}$ on this vector field to get a scalar field! but why?!! (I mean by which principle?). and then in what meaning $I$ is genarator of $h^s$?

  • $\begingroup$ As I said I actually want formulation of idea (because there are subtleties). most of all is this: generator of symmetry is a vector field over configuration manifold not simply a number. so by which mean generator of symmetry is constant over trajectory of motion? (I'm confused because angular momentum of a system is a number and can be constant, but angular momentum as the generator of rotation is not a number!) @Qmechanic $\endgroup$
    – moshtaba
    Jun 1, 2018 at 17:33
  • $\begingroup$ If (in a representation) a state belongs to an eigenspace of a generator, then we may replace the generator by its eigenvalue. $\endgroup$
    – Qmechanic
    Jun 1, 2018 at 17:43
  • $\begingroup$ Sorry, do you discuss in context of QM? I'm discussing in context of classical mechanics considering configuration manifold. @Qmechanic $\endgroup$
    – moshtaba
    Jun 1, 2018 at 17:51
  • $\begingroup$ Consider to modify and clarify your post accordingly in a self-contained manner. Remember that the comment section does not count as explanation. $\endgroup$
    – Qmechanic
    Jun 1, 2018 at 18:23

2 Answers 2


Elements of your question appear a little vague, so I will review Noether's theorem and consider an example. Hopefully, we can address

I wonder "which quantity" is conserved in relation to a specific symmetry

We will utilise differential geometry since OP formulates their question using this terminology.

Let $\phi$ be the symplectic action of a Lie group $G$ on configuration manifold $Q$. We can lift this action to the phase space $T^*Q$ as $\phi^{T^*}$. The action has an $Ad^*$-equivariant momentum mapping given by:

$$ J:T^*Q\rightarrow \mathfrak g ^*; \ \hat J (\xi )(\alpha _q) = \alpha _q \cdot \xi _Q(q)\qquad \tag{1} $$ where $\xi_q$ is the infinitesimal generator of $\phi$ on $Q$ for $\xi \in \mathfrak g$, $J$ is the momentum map and $\hat J$ is a homomorphism of the Lie algebra. Equivariant means the cocycle is zero. Let $X$ be a vector field on $Q$, the momentum is then:

$$ P(X):T^*Q\rightarrow \mathbb R ; \ \alpha_q \mapsto \alpha _q\cdot X(q)\tag{2} $$

and thus $\hat J(\xi) = P(\xi_q)$.

As an example let $Q=\mathbb R^n`$, $G=\mathbb R^n`$,and let $G$ act on $Q$ by translation:

$$ \phi :G\times Q \rightarrow Q: (q',q)\mapsto q'+q $$

The infinitesimal generator is $\xi _Q (q) = \xi$ and the momentum map is:

$$ \hat J(\xi)\cdot (q,p) = p\cdot \xi $$

So $J$ is the linear momentum.

As another example suppose that $H$ is invariant under the action $\phi$: $$ H(x) = H(\phi_g(x)) \qquad \forall \ x\in (q,p), \ g \in G $$

then $J$ is an integral for $X_H$. In other words, $J$ is invariant under the flow of $X_H$.

To summarise:

  • we had a Lie group $G$ and a Lie algebra $\xi \in \mathfrak g$. We demanded that for each $g\in G$ the action is symplectic.

  • we introduced the momentum mapping for the action as a map $J$ where $J : T^*Q\rightarrow \mathfrak g^*$ provided that $\forall \ \xi \in \mathfrak g$ we have:

$$ d\hat J(\xi) = \iota_{\xi_{T^*Q}}\omega $$

  • in direct answer to your question, under the symplectic action of $G$, the momentum $J$ is an integral of the vector field associated to the invariant function.

I don't have the time now to exactly verify all subtleties, however, the following should nontheless answer your concerns.

We assume that a Lie group $G$ acts on the configuration manifold on the left, the action being $G\times M\rightarrow M$. Let this be denoted as $(g,x)\mapsto gx\equiv l_g(x)$.

It is more convenient to work with a right action, so let $\rho_g:M\rightarrow M,\ \rho_g(x)=l_{g^{-1}}(x)$, which is now a right action.

The finite group action also determines infinitesimal actions. If $A\in\mathfrak g$ is a Lie algebra element, then the infinitesimal transformation associated with $A$ is the vector field $X_A$ defined by $$ X_A|_x=\left.\frac{d}{d\epsilon}\rho_{\exp(\epsilon A)}(x)\right|_{\epsilon=0}. $$ It can be verified that the map $A\mapsto X_A$ is a Lie algebra homomorphism. Moreover, if the group action $\rho$ is "faithful" in the sense that the map $G\mapsto\text{Diff}(M),\ g\mapsto\rho_g$ is an injective group homomorphism, then $A\mapsto X_A$ is an injective Lie algebra homomorhpism, hence if $T_a$ ($a=1,...,\dim G=k$) are a set of generators for the Lie algebra $\mathfrak g$, and if $$ [T_a,T_b]=C^c_{ab}T_c, $$ then $$ [X_a,X_b]=C^c_{ab}X_c, $$ where $X_a:=X_{T_a}$.

Note that the vector field $X_A$ is what one would write as $\delta q^i$ in more... traditional notation.

We can make $G$ act on $TM$ instead of $M$ by tangent prolongation (I don't wanna do much details here), hence, there are also infinitesimal transformations on $TM$, given by a vector field $\bar{X}_A$ (which is defined on $TM$, as opposed to $M$).

The group $G$ is an infinitesimal symmetry group for the action, if under infinitesimal actions (eg. $\bar{X}_A$), the Lagrangian changes by a total derivative: $\delta L=\mathcal L_{\bar{X}_A}L=\frac{d}{dt}K_A$.

Then, the charge $$ Q_A=\frac{\partial L}{\partial\dot{q}^i}X_A^i-K_A $$ is a constant of motion.

Considering that any infinitesimal transformation (from $\mathfrak g$) is a symmetry, we can do this for every generator $T_a$, and get $k$ conserved charges $$ Q_a=\frac{\partial L}{\partial\dot{q}^i}X^i_a-K_a. $$

The charges are functions on the velocity phase space $TM$. However, you can use the Legendre transform to reinterpret them as functions on the momentum phase space.

Then, as it turns out, we have the followsing Poisson bracket relations: $$ \{Q_a,Q_b\}=C^c_{ab}Q_c+c_{ab}, $$ where the Poisson algebra of the charges $Q_a$ are allowed to have central charges $c_{ab}$ that commute with all generators.

In short, the Lie algebra of the symmetry group $G$ have three distinct realizations here.

  • An abstract one given by the abstract generators $T_a$.

  • The Lie algebra of the infinitesimal transformations $X_A$'s (these are the "variations" or "derivations" as often called).

  • The Lie-subalgebra of the Poisson algebra of phase-space functions given by the $k$ conserved charges $Q_a$. This one is allowed to have a center however.

In your case of the angular momentum, the symmetry group is $\text{SO}(3)$, the "abstract" algebra is given by the usual rotation generator matrices, the "derivation" algebra is given by three vector fields corresponding to the algebra's generators via the group action (and these are essentially the three rotational Killing vector fields of the euclidean metric ! ), and the "Poisson" algebra is given by the angular momentum components $L_i$, which are also the conserved charges.


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