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In 3-dimensional manifold theory, I have encountered the manifold $S^2 \times S^1$ many times. (The following story can be applied not only this manifold but also for any 3-dimensional manifold.)

But I don't have any geometrically or topologically image of the manifold in my head. How do you deal with this difficulties? Is there any good way to imagine the manifold in my head?

Since $S^1$ is a union of an interval and a point, I know it is a thick sphere identified the inner boundary with the outer boundary. But Still it is not that clear.

Or do you just deal the manifold algebraically without appearing any geometric intuition?

I appreciate any help or tip. Thank you in advance.

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I also like to think of $S^2 \times S^1$ as being fibered over $S^1$, i.e. "a trivial family of 2-spheres parametrized by the 1-sphere". (Or vice versa, of course...) –  Aaron Mazel-Gee Oct 15 '12 at 14:52

4 Answers 4

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Any product manifold $M \times N$ can be visualized as a configuration space for a pair of particles, one of which travels on $M$ and one of which travels on $N$. So $S^2 \times S^1$ can be visualized as the configuration space of a pair of particles, one of which travels on a sphere and one of which travels on a circle.

There is an alternate visualization as follows. First one thinks of $S^2 \times I$ as a thickened sphere (like a $3$-dimensional annulus), with an inner boundary sphere $S^2 \times \{ 0 \}$ and an outer boundary sphere $S^2 \times \{ 1 \}$. Then one identifies the two boundaries. (Edit: I did not notice that you had already talked about this visualization. I think it can be helpful.)

In general, probably different people will get different things out of different visualizations. Use whatever works for you.

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As Qiaochu Yuan and Mariano Suarez-Alvarez said, I find the easiest way of thinking of $\mathbb{S}^2\times \mathbb{S}^1$ is as a thickened sphere with two identified boundary components.

In this vein, you could also think of $\mathbb{S}^2\times\mathbb{S}^1$ as $(\mathbb{S}^2\times\mathbb{R}) / \mathbb{Z}$, where $\mathbb{Z}$ acts by translation on the second factor. It's easy to visualize $\mathbb{S}^2\times\mathbb{R}$ --- it's just a punctured $\mathbb{R}^3$.

Sometimes, though, I find it useful to think of $\mathbb{S}^2\times\mathbb{S}^1$ as a (degenerate) lens space, obtained by gluing two solid torii via the identity mapping class of $T^2$. Since a disk glued to a disk gives a sphere and the identity sends meridian to meridian, gluing compression disks along their boundaries, you can see that this gives a family of spheres parametrized by a circle.

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One could "imagine" it as a 4-dimensional "torus" formed by "revolving" a sphere in 4D space into a torus-like shape whose perpendicular (to the circle of revolution) cross-sections are spheres (or pairs of spheres). The surface of said "torus", to be precise, which has 3 dimensions. Though one can't really visualize 4D. Another way is to imagine a space that is "sphere-like" in 2 dimensions (that is, moving around in these two dimensions only is like moving around on the surface of a sphere), and "linear" (like 3D conventional/"flat" space) in the third, but such that after traveling a finite distance along the third, you end right back where you started -- though the other two dimensions have this property as well, but the geometry is such that traveling in them is like traveling in a sphere, but traveling on one of these, and traveling on the third, is like traveling on a torus.

Note that $S_1 \times S_1$ can be thought of as a normal torus via the same manner: think of a circle, revolved about an axis not crossing through it, and in its own plane.

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A thick sphere with its two boundary components identified.

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The OP mentions this in the question –  Jesse Madnick Oct 13 '12 at 5:52
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@JesseMadnick, but he asked how I imagined the space! –  Mariano Suárez-Alvarez Oct 13 '12 at 6:05

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