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Unfortunately, not much to report mathematically. 4-d case is now totally solved (modulo my adviser checking it). H
Here's what I'm dealing with:
Consider an arbitrary homomorphism from S1 into S3xS3xS3xS3 = G is given by z -> (za, zb, zc, zd).
I can use this to define an action of S1 on S3xS3 via z*(p,q) = (za p z-c, zbqz-d)
Now, it's clear that the kernel of the map from S1 into G acts trivially. Thus, I may as well divide S^1 by the kernel (in otherwords, the action only depends on the IMAGE of the map, not the map itself, so I can assume, by dividing by the kernel, that the map is 1-1).
An easy exercise in GCD's implies that the kernel is generated by (a,b,c,d). Hence I may assume (a,b,c,d) =1.
Now, I'm only interested in free actions, because in this case I know the quotient is a manifold. It is again an exercise in GCD manipulation to prove that the action is semi free (i.e, an element either acts freely or trivially) iff 1<= (a - c, b - d) <= 2 for any combination of plus signs and minus signs.
Now, in the case of effective actions (which turns out to be equivalent to asking that one of the 4 GCDs above equals 1), Eschenburg, working off of a technique developed by Borel, has shown how to compute the cohomology of the quotient. Singhoff modified an approach of Borel and Hirzebruch to show how to compute characteristic classes of the quotient. We'll get back to this in a second.
Doing the cohomology crunching, and applying Barden/Smale's theorem of the classificatino of 1-connected 5 manifolds, and a not too hard application of the gysin sequence to the principal fiber bundle S1 -> G -> G//S1 it follows that for any allowable choice of a,b,c, and d, the quotient is diffeomorphic to an S3 bundle over S2, of which there are precisely 2.
How are these two distinguished? Well, it turns out they have the same cohomology rings, same pontryagin classes (all trivial), and the same stiefel whitney classes (all trivial), EXCEPT w_2 is nontrivial for the twisted bundle and nontrivial for the trivial bundle.
Great, so now the goal is for each allowable choice of a,b,c,d, to compute w_2 of the quotient.
This computation relies heavily on knowing the 2-roots of the groups involved. The 2-roots are certained linear functionals from a maximal Z2n subgroup of the groups involved to Z2 obtained by restricting the adjoint action of the group to this maximal 2-group.
Here's where life gets interesting: the 2-roots for both S1 and S3 are trivial. It follows immediately from Singhoff's work that the stiefel whitney class of any EFFECTIVE action is trivial.
Hence, for any choice of a,b,c,d with at least one (and therefore all) (a -c, b -d) = 1, the quotient is S3xS2.
Note in particular that all homogenous examples (i.e., c = d = 0) are trivial. This follows because if c=d=0, then the condition (a,b,c,d)=1 implies that (a -c, b -d) = (a,c) = 1. By the way, this result followed by the work of smale awhile ago, but it's still refreshing to see my results are consistent with the rest of mathematics?
Now, what about the ineffective action? Turns out that this arises iff up to ordering, a = 2 mod 4, b = 0 mod 4, c = d = 1 mod 2.
First thing to note: if I pick a = 2, b = 0, c=d=1 then the action is z*(p,q) = (z2p, zqz-1). In otherwords, this is a hopf action at twice speed plus a rotation. But if you convert everything to SU(2), you see that the zqz-1 acts as a rotation at twice speed!
Thus, if you divide by the ineffective kernel (a Z2), then the action is hopf on one factor and rotation on the other. Now, if you look at the induced action on S^3 x equatorial S^2, it turns out that the quotient is CP2 # -CP2, a.k.a. the (unique) nontrivial S2 bundle over S2. It is well known (and easy to show, using the Hirzebruch signature theorem), that w2(CP2 # -CP2) is nontrivial. Then a simple whitney sum argument shows that w2(G//S1) is nontrivial. In particular, the nontrivial bundle DOES arise (this is a modified proof of Pavlov's).
What about all the other choices?
Well, frankly, I have no idea.
Everything I've tried has failed miserably, possibly because the only real option I have is the find a nice biquotient of the form S3x SO(3)//S1 on which S1 acts freely which no ineffective kernel. I did this, but the methods Borel et al developed (which involve lots of Serre spectral sequences) really only work when the quotient is simply connected (and it's not, in this case). Hence I'm still stuck. Tomorrow I try something new, and then I think I'm totally out of ideas.
Wonderful.
And I still miss Sara, but hey, she's coming to see me next Tuesday!
Here's what I'm dealing with:
Consider an arbitrary homomorphism from S1 into S3xS3xS3xS3 = G is given by z -> (za, zb, zc, zd).
I can use this to define an action of S1 on S3xS3 via z*(p,q) = (za p z-c, zbqz-d)
Now, it's clear that the kernel of the map from S1 into G acts trivially. Thus, I may as well divide S^1 by the kernel (in otherwords, the action only depends on the IMAGE of the map, not the map itself, so I can assume, by dividing by the kernel, that the map is 1-1).
An easy exercise in GCD's implies that the kernel is generated by (a,b,c,d). Hence I may assume (a,b,c,d) =1.
Now, I'm only interested in free actions, because in this case I know the quotient is a manifold. It is again an exercise in GCD manipulation to prove that the action is semi free (i.e, an element either acts freely or trivially) iff 1<= (a - c, b - d) <= 2 for any combination of plus signs and minus signs.
Now, in the case of effective actions (which turns out to be equivalent to asking that one of the 4 GCDs above equals 1), Eschenburg, working off of a technique developed by Borel, has shown how to compute the cohomology of the quotient. Singhoff modified an approach of Borel and Hirzebruch to show how to compute characteristic classes of the quotient. We'll get back to this in a second.
Doing the cohomology crunching, and applying Barden/Smale's theorem of the classificatino of 1-connected 5 manifolds, and a not too hard application of the gysin sequence to the principal fiber bundle S1 -> G -> G//S1 it follows that for any allowable choice of a,b,c, and d, the quotient is diffeomorphic to an S3 bundle over S2, of which there are precisely 2.
How are these two distinguished? Well, it turns out they have the same cohomology rings, same pontryagin classes (all trivial), and the same stiefel whitney classes (all trivial), EXCEPT w_2 is nontrivial for the twisted bundle and nontrivial for the trivial bundle.
Great, so now the goal is for each allowable choice of a,b,c,d, to compute w_2 of the quotient.
This computation relies heavily on knowing the 2-roots of the groups involved. The 2-roots are certained linear functionals from a maximal Z2n subgroup of the groups involved to Z2 obtained by restricting the adjoint action of the group to this maximal 2-group.
Here's where life gets interesting: the 2-roots for both S1 and S3 are trivial. It follows immediately from Singhoff's work that the stiefel whitney class of any EFFECTIVE action is trivial.
Hence, for any choice of a,b,c,d with at least one (and therefore all) (a -c, b -d) = 1, the quotient is S3xS2.
Note in particular that all homogenous examples (i.e., c = d = 0) are trivial. This follows because if c=d=0, then the condition (a,b,c,d)=1 implies that (a -c, b -d) = (a,c) = 1. By the way, this result followed by the work of smale awhile ago, but it's still refreshing to see my results are consistent with the rest of mathematics?
Now, what about the ineffective action? Turns out that this arises iff up to ordering, a = 2 mod 4, b = 0 mod 4, c = d = 1 mod 2.
First thing to note: if I pick a = 2, b = 0, c=d=1 then the action is z*(p,q) = (z2p, zqz-1). In otherwords, this is a hopf action at twice speed plus a rotation. But if you convert everything to SU(2), you see that the zqz-1 acts as a rotation at twice speed!
Thus, if you divide by the ineffective kernel (a Z2), then the action is hopf on one factor and rotation on the other. Now, if you look at the induced action on S^3 x equatorial S^2, it turns out that the quotient is CP2 # -CP2, a.k.a. the (unique) nontrivial S2 bundle over S2. It is well known (and easy to show, using the Hirzebruch signature theorem), that w2(CP2 # -CP2) is nontrivial. Then a simple whitney sum argument shows that w2(G//S1) is nontrivial. In particular, the nontrivial bundle DOES arise (this is a modified proof of Pavlov's).
What about all the other choices?
Well, frankly, I have no idea.
Everything I've tried has failed miserably, possibly because the only real option I have is the find a nice biquotient of the form S3x SO(3)//S1 on which S1 acts freely which no ineffective kernel. I did this, but the methods Borel et al developed (which involve lots of Serre spectral sequences) really only work when the quotient is simply connected (and it's not, in this case). Hence I'm still stuck. Tomorrow I try something new, and then I think I'm totally out of ideas.
Wonderful.
And I still miss Sara, but hey, she's coming to see me next Tuesday!