How can I explicitly calculate all the orbits of the action of $SO(3)$ on $\mathbb C\mathbb P\_2$?
For example I know that one of the orbits is the quadric $\{[z\_0:z\_1:z\_2]\in\mathbb C\mathbb P\_2: z\_0^2 + z\_1^2 + z\_2^2= 0\}$ but I dont know how to calculate it. Other orbits are copies of $\mathbb P\_1$ and $\mathbb S^3$.
Any reference would be appreciated too. Thanks.



Orbits of an action
$SO\_3$ acts with cohomogeneity one on $CP^2$, where we view $SO\_3$ inside $SU\_3$ (the identity component of the isometry group of $CP^2$, up to a $Z\_3$-kernel) by the standard inclusion. If the metric in $CP^2$ is normalized to be the quotient metric given by the Hopf fibration 
$S^5(1)\to CP^2$, then the orbit space of $CP^2$ under the $SO\_3$ action is a segment of length $\pi/4$. In fact, $\gamma(t)=[\cos t:i\sin t:0]$ is a unit speed geodesic everywhere orthogonal to orbits, and it meets each orbit exactly once if $t\in[0,\pi/4]$; here the endpoints correspond to the two singular orbits: the orbit through $\gamma(0)=[1:0:0]$ is a totally geodesic $RP^2$ (with corresponding isotropy group $S(O\_1\times O\_2)$), and the orbit through $\gamma(\pi/4)=\frac1{\sqrt2}=[1:i:0]$ is the quadric $CP^1=S^2$ with equation $z\_0^2+z\_1^2+z\_2^2=1$ (and corresponding isotropy group $SO\_2$, embedded in $SO\_3$ as rotation in the first two coordinates). The principal isotropy group is the $Z\_2$-subgroup of $SO\_3$ generated by $\mathrm{diag}(-1,-1,1)$, so the principal orbit is $SO\_3/Z\_2$. 


The $SO\_3$ action on $CP^2$ lifts to a $SO\_3\times SO\_2$ action on $S^5$
which is the restriction of the representation $R^3\otimes R^2$. This is the isotropy representation of the Grasmann manifold $G\_2(R^5)=SO\_5/(SO\_3\times SO\_2)$ of oriented $2$-planes in $R^5$. Such orbit structures are well studied and well described. 


Alternatively, we can write $C^3=R^3\oplus R^3$ and let $SO\_3$ act on
$S^5\subset R^3\oplus R^3$. Denoting the coordinates by $(x,y)\in R^3\oplus
R^3$, we have that a complete set of invariants is given by $a=||x||^2$,
$b=||y||^2$, $c=x\cdot y$. Cauchy-Schwartz says $c^2\leq ab$, which is the 
solid interior of a cone, and restricting to the unit sphere gives $a+b=1$,
so the orbit space $S^5/SO\_3$ topologically is a $2$-disc (but metrically is 
a hemisphere of radius $1/2$). Now the (right) action of $SO\_2$ on $R^3 \oplus R^3$ preserves $c$ and so is easily seeen to be the rotation around the pole.
We recover the orbit space $CP^2$ under $SO\_3$ as the segment of radius $\pi/4$ and can also figure out the isotropy groups this way. 


