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The moment map of the action of $\operatorname{SO}(3)$ on the sphere can be thought of as inclusion from $S^2$ into $\mathbb R^3$ by identifying $\mathfrak{so}(3)$ (the Lie algebra of $\operatorname{SO}(3)$) with $\mathbb R^3$.

I am just learning symplectic geometry and this fact came up without explanation in a paper that I'm reading. Can someone explain this, preferably in an intuitive way?

Thanks!

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1 Answer 1

If you've studied the moment map for coadjoint actions, the answer is easy to see but takes a few words to describe. If you are not familiar with this, see the end of this answer for some references.

The Lie algebra $$\mathfrak{so}(3) = \{A \in M_{3 \times 3}(\mathbb{R}) : A = -A^T\}$$ can be identified with $\mathbb{R}^3$ via the identification $$\mathbb{R}^3 \longrightarrow \mathfrak{so}(3),$$ $$\xi = (\xi_1, \xi_2, \xi_3) \mapsto \begin{pmatrix} 0 & -\xi_3 & \xi_2 \\ \xi_3 & 0 & -\xi_1 \\ -\xi_2 & \xi_1 & 0 \end{pmatrix} = A_\xi.$$ Under this identification, we have that $$[A_\xi, A_\eta] = A_{\xi \times \eta},$$ where $\xi \times \eta$ is the cross product of vectors in $\mathbb{R}^3$. Furthermore, $$\mathrm{Tr}(A_\xi^T A_\eta) = 2 \langle \xi, \eta \rangle,$$ so the standard inner product on $\mathbb{R}^3$ induces an invariant inner product on $\mathfrak{so}(3)$ that we can use to identify $\mathfrak{so}(3)$ and its dual $\mathfrak{so}(3)^\ast$. Under these identifications, we get that the adjoint action of $\mathrm{SO}(3)$ on $\mathfrak{so}(3)$, $$X \cdot A = XAX^{-1},$$ corresponds to the usual left action of $\mathrm{SO}(3)$ on $\mathbb{R}^3$, $$A \cdot \xi = A\xi.$$ Furthermore, the adjoint action is isomorphic to the coadjoint action, so the coadjoint action corresponds to the usual left action of $\mathrm{SO}(3)$ on $\mathbb{R}^3$ as well. Therefore the coadjoint orbits are just $2$-spheres, and the Kirillov-Kostant-Souriau symplectic form $$\omega_\Omega(\mathrm{ad}_A^\ast \Omega, \mathrm{ad}_{A'}^\ast \Omega) = \langle \Omega, [A,A'] \rangle$$ on the coadjoint orbits corresponds to the usual symplectic form $$\omega_x(u,v) = \langle x, u \times v)$$ on $S^2$. Now the moment map for the obvious action of $\mathrm{SO}(3)$ on one of its coadjoint orbits is just the inclusion map of the orbit into $\mathfrak{so}(3)$ (this is true for the action of any compact, connected Lie group on one of its coadjoint orbits, see the references below), which in our case corresponds to the inclusion $$S^2 \hookrightarrow \mathbb{R}^3.$$


Here are some references on moment maps for coadjoint orbits:

  • Introduction to Symplectic Topology, Dusa McDuff and Dietmar Salamon, Oxford Science Publications 1998, 2nd Edition. Example 5.24 on page 168 discusses this topic.
  • The Topology of Torus Actions on Symplectic Manifolds, Michèle Audin, Progress in Mathematics no. 93, Birkhäuser 1991. See Section 3.3, page 49.
  • Lectures on Symplectic Geometry, Ana Cannas da Silva, Lecture Notes in Mathematics no. 1764, Springer 2001. Homework 17 on page 139 guides you through the basic theory.
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