Consider a particle with mass $m$ moving in a constant speed $v$ along the real line $\mathbb{R}$, with coordinate $q$. Then its linear momentum is $p = mv$.

Now let's translate the physics into math, namely a Lie group action on a symplectic manifold. The phase space is the symplectic manifold $M := T^*\mathbb{R} \cong \mathbb{R}^2$, with coordinates $(q,p)$ and a canonical symplectic form $\omega := dq \wedge dp$. The Lie group $\mathbb{R}$ acts on $\mathbb{R}^2$ by translation along $q$ direction. This is a symplectic and Hamiltonian action. Note that the Lie algebra $\mathfrak{g}$ of $\mathbb{R}$ is still $\mathbb{R}$. According to Wikipedia, the moment map generalizes the linear and angular momentum. So a linear momentum can be viewed as a special case of a moment map.

My question is:
In the above example, why is the linear momentum $mv$ a moment map $\mu: M \cong \mathbb{R}^2 \to \mathfrak{g}^* \cong \mathbb{R}$?

  • $\begingroup$ That would be the equation of motion. $\endgroup$ – puppetsock Dec 10 '19 at 18:22
  • $\begingroup$ @puppetsockreinstateMonica Care to explain more? What's $\mu$? $\endgroup$ – Yuhang Chen Dec 10 '19 at 18:54
  • $\begingroup$ Please do not edit your question to add a different question (since that invalidates existing answers). If you have a new question, ask a new question. $\endgroup$ – ACuriousMind Dec 11 '19 at 17:43
  • $\begingroup$ I'm sorry, but please also do not edit to follow-up questions into your questions. Questions are supposed to contain nothing but information relevant to the question itself. Also, there is no need to mark your edits with "Edit X", the revision history is accessible to all users if they are interested in the history of a post. $\endgroup$ – ACuriousMind Dec 11 '19 at 18:19
  • $\begingroup$ @ACuriousMind Thank you for your efforts in this question. Do you have an answer to the follow-up question, by any chance? Thanks again for your time. $\endgroup$ – Yuhang Chen Dec 11 '19 at 18:26

I explain in general how the moment map leads to an actual map from $\mathfrak{g}$ to the algebra of classical observables $C^\infty(M)$ in this answer of mine, but I will exemplify the reverse direction for 1D translations in the following:

  1. The translation algebra $\mathfrak{g}\cong \mathbb{R}$ acts on $M\cong \mathbb{R}^2$ as $f_c : (q,p)\mapsto (q + c, p)$ for some $c\in\mathfrak{g}$. The vector field that is the infinitesimal version of this action is of course $X_c = c\partial_q$.

  2. The contraction of the symplectic form with this vector field is $\iota_{X_c}\omega = (\mathrm{d}q\wedge \mathrm{d}p)(c\partial_q) = c\mathrm{d}p$, i.e. $$\mathrm{d}(\langle \mu, c\rangle) = c\mathrm{d}p\tag{1}$$ by definition of the moment map.

  3. The pairing $\langle \cdot, \cdot \rangle$ is just multiplication on $\mathbb{R}$ and so the momentum map for translation is $\mu : M\to\mathfrak{g}^\ast\cong \mathbb{R}, (q,p)\mapsto p$ since then $\langle \mu,c\rangle(q,p) = cp$ and the defining equation $(1)$ is true. So the moment map of translation is "the momentum" in the sense that it maps $(q,p)$ to the momentum part $p$.

(If you run this logic in reverse, you start from $\mu$ and get the map $\mathfrak{g}\to C^\infty(m), c\mapsto c\partial_q$, i.e. the algebra homomorphism mapping abstract infinitesimal translations to their concrete generators, which is the viewpoint I took in my post linked at the beginning)

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  • $\begingroup$ This looks reasonable although I haven't understood everything. But I'm confused after looking at an example from Page 74 in Audin's "Torus Actions on Symplectic Manifolds" (2nd Edition), which said: " If $H:W \to \mathbb{R}$ is any function and if the Hamiltonian vector field $X_H$ is complete, its flow defines a Hamiltonian $\mathbb{R}$-action, the momentum mapping of which is $H$. " Here $W$ is any symplectic manifold. We can take $W$ to be the manifold $M$ in our example. Then the moment map should be $H=p^2/2m$? $\endgroup$ – Yuhang Chen Dec 11 '19 at 16:25
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    $\begingroup$ @YuhangChen Why should the moment map be $H = p^2$? Don't be confused by the terminology - the function $H$ in your quote is not the Hamiltonian, the "Hamiltonian vector field" just means that $X_H$ is defined by $\mathrm{d}H = \iota_{X_H}\omega$. $\endgroup$ – ACuriousMind Dec 11 '19 at 17:06
  • $\begingroup$ So the function $H$ should be just $H: (q,p) \mapsto p$ instead of the total energy? $\endgroup$ – Yuhang Chen Dec 11 '19 at 17:16

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