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I'm studying Classical Mechanics and there is one object that appeared recently on the book I'm not being able to get a physical intuition about it. The mathematical definition goes as follows:

Let $M$ be a smooth manifold together with a sympletic form $\omega$ and suppose $G$ acts on the left on $M$ such that the action preserves the sympletic form. This means that if $\delta_{g} : M\to M$ is the diffeomorphism associated to $g\in G$, then

$$\delta_g^\ast \omega=\omega$$

Now let $\mathfrak{g}$ be the Lie algebra of $G$ and $\langle,\rangle : \mathfrak{g}^\ast\times \mathfrak{g}\to \mathbb{R}$ the pairing

$$\langle \varphi, A\rangle = \varphi(A),$$

if we denote $X^A$ the vector field in $M$ associated to $A$ then one can see that $\eta = X^A\lrcorner \ \omega$ is closed because the action preserves $\omega$. In that case, we define a momentum map as a function $\mu:M\to \mathfrak{g}^\ast$ such that

$$d(\langle \mu,A\rangle) = X^A \lrcorner \ \omega.$$

Now, for Classical Mechanics we are interested in the case $M = T^\ast Q$ where $Q$ is the configuration manifold.

In that case I assume there should be some good intuition about what momentum maps really are and what they represent. In truth, even the name invites us to think there are some important implication in Physics from this definition above.

In that case, in Classical Mechanics, what momentum maps defined as above really are? What they represent and what is a good intuition about them?

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The equivariant moment map has several applications. Its meaning is that it provides an encoding of how the Lie group $G$ acts on the phase space, and it gives you a way to find the observables corresponding to the conserved quantities/generators of the symmetry $G$: It also defines the process of symplectic reduction to a reduced phase space.

Given that $$ \mathrm{d}(\langle\mu(\dot{}),g\rangle) = \omega(\rho(g),\dot{})$$ where $\rho(g)$ denotes the (Killing) vector field associated to the infinitesimal action of $\mathfrak{g}$, the 1-form $\omega(\rho(g),\dot{})$ is closed due to $\mathrm{d}^2 = 0$. If one assumes the group action to be Hamiltonian, then one assumes that the form is also exact and thus there is a smooth function $f_g$ with $\mathrm{d}f_g = \omega(\rho(g),\dot{})$. A Hamiltonian action is also assumed to give a Lie algebra homomorphism $$ \mathfrak{g} \to C^\infty(M), g \mapsto f_g$$ and its image are precisely the generators of the symmetry in the classical sense. In this way, the moment map provides a coordinate-free description of how the Lie algebra of the symmetry embeds into the full Poisson algebra of observables. For example, for the rotation group, the image is the Lie algebra of angular momentum (thus the name!).

One may now use the symmetry to reduce the phase space to a surface on which the conserved quantities are constant. This is done by picking any regular value of $\mu$ and taking its preimage, then diving out the group action. This surface is invariant under the Hamiltonian flow and gets its own dynamics, allowing us to discard the rest of the phase space if we are only interesting in this particular value for the charges generating the symmetry.

The moment map thus tells you how to find "surfaces with constant charges" in the total phase space.

Of particular importance is the case of a gauge group representing constrained dynamics, where the correct choice is naturally prescribed by the fact that the generators vanish on the constraint surface, so the correct symplectic reduction is $\mu^{-1}(0)/G$. This may be used to give a high-level account of how BRST cohomology obtains the correct algebra of gauge-invariant observables, see this answer of mine.

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  • $\begingroup$ Is it correct to say that the the moment map (when it exists) simultanously solves for all the conserved quantities? (conserved by the symmetries). So that in some sense it gets you half way through to solve the integrable system? $\endgroup$ – Saal Hardali Apr 19 '16 at 20:09

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