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  Kähler Cones in C4 and a foliation of P3

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Take the 3-dimensional complex projective space P3. Consider the action of the group SU(2)×SU(2). I have read in physics related articles that these group gives a singular foliation of P3 in three types of orbits: one is the Segré submanifold P1×P1, another is the real projective space RP3SO(3) and then a family of 5-dimensional surfaces that are non-trivial SO(3) fiber bundles over S2. I am trying to recover this foliation by working dircetly in C4. The idea (which could be incorrect) is to use the homogenous polynomial P(z)=z1z4z2z3 in C4. 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 S2×S3 and is a U(1) fiber bundle over S2×S2, i.e. the segré orbit of P3 is recovered in the base of this particular cone. The Kähler metric of this cone is of the form

ds2=dr2+r2dΣ2

where r is the radial coordinate and dΣ2 is the metric of the base of the cone. My question is the following: can I recover the remaining leaves of P3 in a similar way? I am almost sure that they can be recovered by "deforming" the equation of the cone by P(z)=12ϵ in C4 (there is a nice paper of Candelas et al. "comments on conifolds" were this is explained very well). The surfaces obtained for fixed ϵR are everywhere smooth cones and I believe that the remaining orbits of SU(2)×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 C4 as the space of 2-by-2 complex matrices Z, and then the action of the two 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)=det(Z). It also preserves Q(Z)=tr(ZZ), 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 S7P3 to each level set of P I should end up with the orbits of the SU(2)'s in P3 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 C4.

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 diag(r1eiθ,r2eiθ) q, where p and q belong to SU(2) and r1r20 while 0θπ2. Obviously, it's the ratio of r1 to r2 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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