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  Kähler Cones in $\mathbb{C}^4$ and a foliation of $\mathbb{P}^3$

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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


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?

This post imported from StackExchange MathOverflow at 2015-03-02 12:58 (UTC), posted by SE-user Darius Alexander
asked Jan 29, 2012 in Mathematics by Darius Alexander (10 points) [ no revision ]
retagged Mar 2, 2015
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.

This post imported from StackExchange MathOverflow at 2015-03-02 12:59 (UTC), posted by SE-user Robert Bryant
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 $\mathbb{C}^4$.

This post imported from StackExchange MathOverflow at 2015-03-02 12:59 (UTC), posted by SE-user Darius Alexander
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.

This post imported from StackExchange MathOverflow at 2015-03-02 12:59 (UTC), posted by SE-user Robert Bryant

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