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Is there some kind of intuitive/waving hand argument to explain that $$SO(3)/SO(2) \simeq S^2 \; ?$$

Anne O'Nyme
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    Think of group action. $SO(3)$ acts on $\mathbb S^2$, and $SO(2)$ is a stabilizer (any $SO(3)$ element fixing $(0,0,1)$ must be the rotation of the $x, y$ plane). –  Apr 17 '14 at 09:09
  • The discussion at page 590 of Tony Zees' Einstein Gravity in a Nutshell: https://books.google.de/books?id=5Dy1hlKvmCYC&lpg=PA590&ots=BQWJTqWzCH&dq=SO(3)%2FSO(2)%20equivalence%20classes&hl=de&pg=PA590#v=onepage&q&f=false is quite helpful to undestand this pictorially – jak Jan 06 '17 at 16:57

2 Answers2

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(To remove this question from the unanswered list):

Think of group action. $SO(3)$ acts on $\mathbb S^2$, and $SO(2)$ is a stabilizer (any $SO(3)$ element fixing $(0,0,1)$ must be the rotation of the $x,y$ plane)

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A quite intuitive way to understand that the coset space $SO(3)/SO(2)$ is indeed the sphere $S^2$ consists in finding the manifold $SO(3)$ and then identifying the points connected by rotations in a plane.

First the $SO(3)$ manifold. The oriented segment from the origin to a given point belonging to the manifold gives the oriented axis of rotation, $\hat n$, while the distance to the origin gives the angle, $\phi$. We adopt the right-hand rule thus a rotation $(\phi,\hat n)$ is equivalent to a rotation $(-\phi,-\hat n)$ and the union $(0\leq\phi\leq\pi,\hat n)\cup(0\leq\phi\leq\pi,-\hat n)$ gives all possible rotations about the axis $n$. Moreover, the rotation $(\pi, \hat n)$ is identified to the rotation $(\pi,-\hat n)$. Thus we find that the $SO(3)$ manifold is a ball of radius $\pi$ whose antipodal points on its surface are are identified.

Now chose a point of that manifold, namely a point in the surface. This singles out an oriented axis $\hat n$ and then we identify all points in that axis since they are related by rotations in the plane perpendicular to the axis, i.e., elements of $SO(2)$. Repeat this to all different axis and we are let with a sphere $S^2$.

Diracology
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