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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 $\mu: M \cong \mathbb{R}^2 \to \mathfrak{g}^* \cong \mathbb{R}$?

ACuriousMind
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Yuhang Chen
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1 Answers1

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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)

ACuriousMind
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