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I am trying to understand (simplify) the following space

$Z = S^2 \times S^2 / E$

Let $(a_1 , a_2 , a_3) \in S^2$ and $(n_1 , n_2 , n_3) \in S^2$. Here ${a_1}^2 + {a_2}^2 + {a_3}^2 = 1$ and ${n_1}^2 + {n_2}^2 + {n_3}^2 = 1$.

The equivalence relations $E$ are the following

1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$

2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)]$

where $(\pi; a_1,a_2,a_3)$ is a rotation of angle $\pi$ along the axis $(a_1, a_2, a_3)$. So the point $(\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)$ is the mirror image of $(n_1, n_2, n_3)$ in the plane whose normal is $(a_1, a_2, a_3)$.

The objective is to simplify the equivalence relation (2) by a homeomorphism of $S^2 \times S^2$. For example, the approach I am trying is to rotate the second $S^2$ such that the mirror plane becomes horizontal (XY plane). Since the normal of the mirror plane is $(a_1, a_2, a_3)$, the amount by which the second $S^2$ has to be rotated depends on the $(a_1, a_2, a_3)$ it corresponds to. And for this transformation to be a homeomorphism the rotation operations should vary continuously. The transformation should not complicate the first equivalence relation; one way this condition could be imposed is by ensuring that the rotation operation corresponding to $(a_1, a_2, a_3)$ and $(-a_1, -a_2, -a_3)$ are be the same. I am not able to come up with such a transformation. All my attempts to simplify (2) end up complicating (1).

To be more specific, I want to get a homeomorphism between the space $Z$ and the following space:

$Z^' = S^2 \times S^2 / E^' $

The equivalence relations $E^'$ are the following

1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$

2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (n_1, n_2, -n_3)]$

Any ideas are appreciated. Please let me know if you have any questions.

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Is this a Homework? – Djaian Nov 11 '10 at 15:38
No, I am graduate student in Materials Science department and it is a part of my project. Just to give a brief background: The space belongs to a product space of rotations and normal vectors $(SO(3) \times S^2)$. The space $S^2 \times S^2$ I mentioned above is a subspace of this space and my final objective is to be able to do statistical analysis on this space. – Srikanth Nov 11 '10 at 15:58
up vote 1 down vote accepted

This isn't quite what you asked for, but notice that (1) simply gives us $\mathbb{RP}^2 \times S^2$. I'm not sure I understand your notation for (2); it seems like it doesn't depend at all on $\pi$. If it's true that you're just reflecting $\vec{n}$ through the plane perpendicular to $\vec{a}$, then you're just getting a copy of $D^2$ over every point of $\mathbb{RP}^2$. This is a fiber bundle, and if you think through it I think you'll find that it's the disk bundle of the tangent bundle $T\mathbb{RP}^2$.

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You are right. It is just reflecting $\vec{n}$ through the plane perpendicular to $\vec{a}$. When you rotate a vector along an axis by $\pi$ and invert it, you end with a reflection through a plane that is perpendicular to that axis of rotation. I should have just said it is a reflection but I put it that way because that is how my equivalence relation looks after certain simplifications that I performed and I wanted to present the complete problem. – Srikanth Nov 11 '10 at 20:24
Anyways, I understand what you are saying but I am do not have enough experience with algebraic topology (am self-learning). I understand that (1) implies $Z$ is $\mathbb{RP}^2 \times S^2$. And yes, there is just a copy of $D^2$ over every point of $\mathbb{RP}^2$ and hence a fiber bundle. But what is the difference between this space and $Z^'$ that I mentioned above ? – Srikanth Nov 11 '10 at 20:25
$Z$ isn't $\mathbb{RP}^2\times S^2$, that's just what you get from $S^2\times S^2$ and apply only the first equivalence relation. However, now that I'm looking at it closely, the second is definitely not equivalent to the first. There's a fundamental difference between moving your reflecting plane and not. $Z'$ will be a trivial disk bundle, because over each point (= equivalence class $\{\pm \vec{a}\}$) in the first factor you'll have "the same" disk -- i.e. $Z'=\mathbb{RP}^2\times D^2$. This is different from the disk bundle of $T\mathbb{RP}^2$! Does that make sense? – Aaron Mazel-Gee Nov 12 '10 at 5:22
Thanks, Aaron. I totally get $Z^' = \mathbb{RP}^2 \times D^2$. I need to read up on tangent bundles and disk bundles to understand it better. Can you please guide me to some introductory books that introduce these concepts. Thanks, your help is much appreciated. – Srikanth Nov 12 '10 at 15:19
Sure. First of all, disk bundles are really pretty similar to vector bundles -- you can throw away the sphere bundle (i.e. in each fiber throw away the boundary of the disk) and recover something homeomorphic to the original vector bundle. The proof that $T\mathbb{RP}^2$ is nontrivial (i.e. $T\mathbb{RP}^2\not= \mathbb{RP}^2 \times \mathbb{R}^2$) uses something called Stiefel-Whitney classes, the best resource for which is probably Milnor & Stasheff. However, this requires some cohomology theory, so you should make sure that's not totally foreign to you. On the other hand, there's a very – Aaron Mazel-Gee Nov 12 '10 at 17:17

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