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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.

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  • $\begingroup$ At first, I assumed that the action is by taking $SO(3)$ as a subgroup of $GL(3,\mathbb{C})$ in the standard embedding of $\mathbb{R}^3 \subset \mathbb{C}^3$, and then the standard action of $GL(3,\mathbb{C})$ on $\mathbb{CP}^2$. But that preserves the real points. What is the action? Do you mean $SO(3,\mathbb{C})$, not the compact form? $\endgroup$
    – Ben McKay
    Aug 7, 2013 at 17:21
  • $\begingroup$ The orbit that you wrote is not an orbit of the real form. So it seems you want the the $SO(3,\mathbb C)$-orbits. Also, shouldn't you write $[z_0:z_1:z_2]$ for $\mathbb{CP}^2$? On $\mathbb C^3$ your orbits are $\lbrace z_0^2+z_1^2+z_2^2 = const\rbrace$, so on $\mathbb{CP}^2$ the orbits are the projections of these, so how many other orbits do you have? $\endgroup$ Aug 7, 2013 at 17:48
  • $\begingroup$ Peter: What you are saying is wrong. There are many real Lie groups with complex orbits in a complex space!! $\endgroup$
    – user13559
    Aug 7, 2013 at 18:20
  • $\begingroup$ Ben: I mean $SO(3)$ as a compact subgroup of $SL(3,\mathbb R)$. So I am interested in the $SO(3)$ orbits in $X:= \mathbb C\mathbb P_2\setminus \mathbb R\mathbb P_2$. Please note that $SL(3,\mathbb R)$ is acting transitively on $X$ $\endgroup$
    – user13559
    Aug 7, 2013 at 18:25

1 Answer 1

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$SO_3$ 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.

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