(no subject)
Hooray!
This morning I finished the 5-d case. 2 down, 2 to go!
This is why mathematicians live - for that feeling that comes after toppling a problem after months or years of perpetual effort - from the miraculous insight that comes in the shower to the innumerable hours spent staring at scribbles on a page. We try 10,000 wrong things, each time learning how to approach the problem slightly differently. Then one day, the problem crumbles into fine sand. And in retrospect, we wonder - why was it so hard to begin with?
I went through approximately 200 pages of paper trying wrong things. In the end, the first messy draft of the solution fits on 3 pages.
Here's the theorem:
(Let = denote Lie group isomorphism or diffeomorphism where appropriate)
Let M be simply connected and assume M = G//H can be written as a biquotient (and assume G and H are written according to Totaro's convention). Then,
G = SU(3), H = SU(2) , M = S^5 and the action is homogeneous OR
G = Spin(6), H = Spin(5), M = S^5 and the action is homogeneous OR
G = SU(3), H = SO(3), M = Wu manifold and the action is homogeneous OR
G = S^3 x S^3, H = S^1.
In the final case, a general action is given by z*(p,q) = (z^a p z^-c, z^b q z^-d) with (a,b,c,d) = 1 and (a+-c, b+-d) = 1 (for all choices of sign) OR (a+-c, b+-d) =2 (for all choices of sign). The second case can arises iff, up to order, a is congruent to 0 mod 4, c is congruent to 2 mod 4, and both b and d are odd.
In the first case, the action is effective and the quotient is S^3 x S^2.
In the second case, the action has a Z_2 ineffective kernel and the quotient is the unique nontrivial S^3 bundle over S^2. More specifically, in the first case, the second stieffel whitney class is trivial while in the second case it's nontrivial.
Corollories:
1. If G//H = S^3 x S^3 / S^1 (homogenous), (i.e., c=d=0), then the quotient is S^3 x S^2 (already known, but it's cool that I have it as a corollary to my work).
2. Choosing a = 2, b = 1, c = 0, d = 1, one sees that that {a,b,c,d} satisfies the second case, in particular, the nontrivial S^3 bundle over S^2 DOES arise.
3. If M is 1-connected, dim <=5, and M = G//H where H acts effectively on G, then M is diffeomorphic to a homogeneous space.
The timing could not be better - Sara joins me in Rio TOMORROW for 2 weeks!!
This morning I finished the 5-d case. 2 down, 2 to go!
This is why mathematicians live - for that feeling that comes after toppling a problem after months or years of perpetual effort - from the miraculous insight that comes in the shower to the innumerable hours spent staring at scribbles on a page. We try 10,000 wrong things, each time learning how to approach the problem slightly differently. Then one day, the problem crumbles into fine sand. And in retrospect, we wonder - why was it so hard to begin with?
I went through approximately 200 pages of paper trying wrong things. In the end, the first messy draft of the solution fits on 3 pages.
Here's the theorem:
(Let = denote Lie group isomorphism or diffeomorphism where appropriate)
Let M be simply connected and assume M = G//H can be written as a biquotient (and assume G and H are written according to Totaro's convention). Then,
G = SU(3), H = SU(2) , M = S^5 and the action is homogeneous OR
G = Spin(6), H = Spin(5), M = S^5 and the action is homogeneous OR
G = SU(3), H = SO(3), M = Wu manifold and the action is homogeneous OR
G = S^3 x S^3, H = S^1.
In the final case, a general action is given by z*(p,q) = (z^a p z^-c, z^b q z^-d) with (a,b,c,d) = 1 and (a+-c, b+-d) = 1 (for all choices of sign) OR (a+-c, b+-d) =2 (for all choices of sign). The second case can arises iff, up to order, a is congruent to 0 mod 4, c is congruent to 2 mod 4, and both b and d are odd.
In the first case, the action is effective and the quotient is S^3 x S^2.
In the second case, the action has a Z_2 ineffective kernel and the quotient is the unique nontrivial S^3 bundle over S^2. More specifically, in the first case, the second stieffel whitney class is trivial while in the second case it's nontrivial.
Corollories:
1. If G//H = S^3 x S^3 / S^1 (homogenous), (i.e., c=d=0), then the quotient is S^3 x S^2 (already known, but it's cool that I have it as a corollary to my work).
2. Choosing a = 2, b = 1, c = 0, d = 1, one sees that that {a,b,c,d} satisfies the second case, in particular, the nontrivial S^3 bundle over S^2 DOES arise.
3. If M is 1-connected, dim <=5, and M = G//H where H acts effectively on G, then M is diffeomorphic to a homogeneous space.
The timing could not be better - Sara joins me in Rio TOMORROW for 2 weeks!!