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Take the 3-dimensional complex projective space $\mathbb{P}^3$. Consider the action of the group $SU(2)\times SU(2)$. I have read in physics related articles that these group gives a singular foliation of $\mathbb{P}^3$ in three types of orbits: one is the Segré submanifold $\mathbb{P}^1\times\mathbb{P}^1$, another is the real projective space $\mathbb{RP}^3\cong SO(3)$ and then a family of 5-dimensional surfaces that are non-trivial $SO(3)$ fiber bundles over $S^2$. I am trying to recover this foliation by working dircetly in $\mathbb{C}^4$. The idea (which could be incorrect) is to use the homogenous polynomial $P(z)=z_1z_4-z_2z_3$ in $\mathbb{C}^4$. For $P(z)=0$ this is the equation of a well known 6-dimensional singular cone (it is fact a Conifold), it is easy to see that the base (the angular part) of the cone is topologically $S^2\times S^3$ and is a U(1) fiber bundle over $S^2\times S^2$, i.e. the segré orbit of $\mathbb{P}^3$ is recovered in the base of this particular cone. The Kähler metric of this cone is of the form

$ds^2=dr^2+r^2d\Sigma^2$

where $r$ is the radial coordinate and $d\Sigma^2$ is the metric of the base of the cone. My question is the following: can I recover the remaining leaves of $\mathbb{P}^3$ in a similar way? I am almost sure that they can be recovered by "deforming" the equation of the cone by $P(z)=\frac{1}{2}\epsilon$ in $\mathbb{C}^4$ (there is a nice paper of Candelas et al. "comments on conifolds" were this is explained very well). The surfaces obtained for fixed $\epsilon\in\mathbb{R}^*$ are everywhere smooth cones and I believe that the remaining orbits of $SU(2)\times SU(2)$ appear in the bases of these cones. Nevertheless, I am having some trouble to recover the 5-dimensional orbits. Is all my approach wrong??! This is kind of new for me. Any known literature or article that can help me with this?

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I guess the action you mean (there are two of them) is to regard $\mathbb{C}^4$ as the space of $2$-by-$2$ complex matrices $Z$, and then the action of the two $\mathrm{SU}(2)$s is by multiplication before and after. (This particular action is not a free action of the product group, btw.) This action does preserve the level sets of $P$, where $P(Z) = \mathrm{det}(Z)$. It also preserves $Q(Z) = \mathrm{tr}(ZZ^\ast)$, and these give you the level sets you want. –  Robert Bryant Jan 29 '12 at 16:23
    
Thank you for your comment. So if I restrict $Q(z)=1$ and restrict the Hopf projection $S^7\rightarrow\mathbb{P}^3$ to each level set of $P$ I should end up with the orbits of the $SU(2)$'s in $\mathbb{P}^3$ right? or am I missing something\everything? I know that the case $P(Z)=Q(z)$ is of interest, these level sets should be 3-sphere in $\mathbb{C}^4$. –  Darius Alexander Jan 29 '12 at 17:02
    
Yes, the usual polar decomposition says that you can write each $2$-by-$2$ complex matrix $Z$ in the form $Z = p\ \textrm{diag}(r_1e^{i\theta}, r_2e^{i\theta})\ q^\ast$, where $p$ and $q$ belong to $\textrm{SU}(2)$ and $r_1\ge r_2\ge 0$ while $0\le\theta\le \tfrac\pi2$. Obviously, it's the ratio of $r_1$ to $r_2$ that determines the orbit structure. –  Robert Bryant Jan 29 '12 at 19:10

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