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I have a question concerning an exercises from a text call Topology and Groupoid authored by Ronald Brown

The question is as follows:

Let $E^2 = \{(x, y) \in \mathbb R^2 : x^2 + y^2 \leq 1\}$. The space $S^1 \times E^2$ is called the solid torus. Prove that the 3-sphere $S^3 = \{(x_1, x_2, x_3 , x_4) \in \mathbb R^4 : (x_1)^2+(x_2)^2+(x_3)^2+(x_4)^2 = 1\}$ is the union of two spaces each homeomorphic to a solid torus and with intersection homeomorphic to a torus [Consider the subspaces of $S^3$ given by $(x_1)^2 + (x_2)^2 \leq (x_3)^2 + (x_4)^2$ and by $(x_1)^2 + (x_2)^2 \geq (x_3)^2 + (x_4)^2$]

I am not certain I understand the hint from the square bracket in geometric terms.

From what I understand of how the 3-sphere can be constructed, one takes two 2-spheres and superimposes the boundary of one on top of the other and then glues both boundaries together.

The two 2-sphere can be represented as $S^3_+ = \{(x_1, x_2, x_3 , x_4) \in\mathbb R^4 : (x_1)^2+(x_2)^2+(x_3)^2 = 1, (x_4)^2 \geq 0\}$ and $S^3_- = \{(x_1, x_2, x_3 , x_4) \in\mathbb R^4 : (x_1)^2+(x_2)^2+(x_3)^2 = 1, (x_4)^2 \leq 0\}$

Is the question asking me to show that both $S^3_+$ and $S^3_-$ are individually homeomorphic to the solid torus and $S^3_+ \cap S^3$ is homeomorphic to the torus? If so how does the hint become relevant?

Thanks in advance

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You are talking about the usual construction of $S^{3}$ the one asked is bit unusual. In general there are many ways to triangulate a surface (and here break up a solid into polyherdral parts: essentially a CW complex). If you keep track of the smaller parts and their intersections you can decompose the solid into more manageable parts. The question is showing the sphere is two solid tori. –  s.b Sep 14 '12 at 19:19
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@s.b. it is actually quite usual in the correct circles! :-) Just picture the usual solid torus in $\mathbb R^3$ and look at its complement: it is also a solid torus minus a point, which is the one you would add in compactifying $\mathbb R^3$ to get a sphere. –  Mariano Suárez-Alvarez Sep 14 '12 at 19:22
    
@MarianoSuárez-Alvarez, I recall when I was first learning this stuff, this question stumped me for some time. Where as when you look at any sphere (after learning some basic general topology) the idea of breaking it up into two hem-spheres seems much more natural. –  s.b Sep 15 '12 at 16:50

2 Answers 2

Complex numbers make the description a little nicer. The 3-sphere can be represented by the set $\{(z,w) \in \mathbb C^2 : |z|^2+ |w|^2 =2\}$. This contains the subset $|z|=|w|=1$, which is the product of two circles, that is, a torus. This torus is the common boundary of the subsets where $|z|\le 1\le |w|$ and $|w|\le 1\le|z|$. For each of these subsets there is an explicit homeomorphism onto the product of closed disk with a circle (i.e. the solid torus). Namely, $(z,w)\mapsto (z, w/|w|)$ for the first subset, and similarly for the second.

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The question is not asking you to prove that the $S^3_\pm$ are homeomorphic to the solid torus. In fact, they are not.

The Hint describes precisely two subspaces of $S^3$: each of them is homeomorphic to a solid torus.

How to prove it? Well, first you have to see it! I honestly do not know any other way of doing this than by trying to figure out how those subspaces look like. It is not that hard if you try enough. Similarly, the two subspaces intersect in a torus: it should be easy to find an equation for the intersection, and then —again— you have to spend some time trying to visualize the resulting surface.

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Then according to the hint, the subsapce of S^3 are 1 spheres? Does the hint have to do with how a 3-sphere is constructed as a union of two 2-sphere? –  Seth Mai Sep 14 '12 at 19:22
    
Hm? No. As I wrote above, the two subspaces described in the hint are homeomorphic to a solid torus. –  Mariano Suárez-Alvarez Sep 14 '12 at 19:26
    
(Also: it is not true that the 3-sphere is constructued as a union of two 2-spheres...) –  Mariano Suárez-Alvarez Sep 14 '12 at 19:28
    
Then from the hint, am I suppose to interpret (x1)2+(x2)2≤(x3)2+(x4)2 as one single geometric object? –  Seth Mai Sep 14 '12 at 19:35
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You should note also that the intersection of the two subspaces is given by the set of points $(x_1,x_2,x_3,x_4)$ such that $x_1^2+x_2^2=x_3^2+x_4^2 =1/2$ and that is clearly (homeomorphic to) $S^1 \times S^1$. –  Ronnie Brown Sep 15 '12 at 14:15

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