$S^2=SO(3)/SO(2)$. Does this mean that $S^2 = SU(2)/U(1) $?

Solution 1:

In general, to identify, say, a smooth manifold $X$ with a quotient $G/H$ of a Lie group $G$ by a closed Lie subgroup $H$, you need to produce a smooth transitive action of $G$ on $X$ with stabilizer $H$. For the classical compact Lie groups there are three standard series of transitive actions on spheres which go like this.

The special orthogonal group $SO(n)$ acts on $\mathbb{R}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{n-1}$ with stabilizer $SO(n-1)$, and hence

$$S^{n-1} \cong SO(n) / SO(n-1).$$

Furthermore, the natural map $Spin(n) \to SO(n)$ means we can also think about this action as an action of $Spin(n)$; then its stabilizer turns out to be $Spin(n-1)$, and so we can also identify

$$S^{n-1} \cong Spin(n) / Spin(n-1).$$

When $n = 3$ there is an exceptional isomorphism $Spin(3) \cong SU(2)$, and $Spin(2)$ inside $Spin(3)$ can be identified with the diagonal copy of $U(1)$ inside $SU(2)$. Note that the map $Spin(2) \to SO(2)$ is multiplication by $2$, as opposed to the standard map $U(1) \to SO(2)$ which is an isomorphism.

Similarly, the special unitary group $SU(n)$ acts on $\mathbb{C}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{2n-1}$ with stabilizer $SU(n-1)$, and hence

$$S^{2n-1} \cong SU(n) / SU(n-1).$$

Finally, the compact symplectic group $Sp(n)$ acts on $\mathbb{H}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{4n-1}$ with stabilizer $Sp(n-1)$, and hence

$$S^{4n-1} \cong Sp(n) / Sp(n-1).$$