I admit up front i'm a complete newbie not even close to being a geometer/topologist, but i need some advice about what the product of spheres would look like. A 1 sphere times a 1 sphere is a two dimensional torus, which is a closed surface, but what happens when we up the ante to an n sphere times an n sphere? In particular a 3 or 4 sphere times a 3 or 4 sphere? Is that also called a torus and is it also a closed surface in some higher dimensional space? And how would we set up a coordinate system to describe that object and how would we project those coordinates into three dimensions so we could see what it looked like? I know we could never see the whole thing. How would we do this in mathematica so we could make a pretty picture out of it?

And a related question: what would the product of an n sphere by an m sphere look like? Perhaps this is the question I should have asked first.


To be selfcontained, note that the usual notation for the $n$-dimensional sphere is $S^n$. This is the surface of the $(n+1)$-dimensional ball and can be given as

$$S^n=\{(x_1,...,x_{n+1})\in\Bbb R^{n+1}\mid x_1^2+\cdots+x_{n+1}^2=1\}.$$

The product space $S^n\times S^m$ will give a surface, but no torus in general. Note that an $n$-dimensional torus is defined as

$$\underbrace{S^1\times\cdots\times S^1}_{n\text{ times}}.$$

Also, just because you can describe a space in some topological sense, this does not mean that there is the unique representation of it in a higher dimensional space. There are many ways how $S^n\times S^n$ can be embedded into higher dimensional Euclidean spaces.

Here is one example: the set

$$X:=\{(x_1,...,x_{n+1},y_1,...,y_{m+1})\in\Bbb R^{n+m+2}\mid x_1^2+\cdots+x_{n+1}^2=1\,\wedge\,x_{1}^2,...,x_{m+1}^2=1 \}$$

is topologically equivalent to $S^n\times S^m$ as a subspace of $\Bbb R^{n+m+2}$. This can be improved to lower dimensions for the embedding space. For example: $S^1\times S^1$ (the usual torus) is here embedded into $\Bbb R^4$. we know we can do better. One way to do this in general would be to see that $X$ is a (scaled) subset of $S^{n+m+1}$ and use stereographic projection to map this into $\Bbb R^{n+m+1}$.

For a visualization you could look at $\Bbb R^3$-slices. But I think this will give not much information on the shape for $n>3$.

  • $\begingroup$ @MWinter. Thank you very much for putting so much effort into your reply. It was very helpful and definitely opened my eyes! But I don't understand what you mean when you say that X is a "scaled" subset of $S^(n+m+1)$. I see that the dimension has dropped by one but what was scaled? With regard to the stereographic projection, I assume it would be from the sphere's north pole. How does this work in the case of $S^1\timesS^1$? Is the donut torus the result of a stereographic projection? How do you get a donut, which has an explicit equation, out of the product of S^1 with itself? $\endgroup$ – wojo Sep 30 '17 at 21:52
  • $\begingroup$ @wojo 1. The set $X$ is actually a subset of $2\cdot S^{n+m+1}$ because the squares of all coordinates sum up to two. This is what I meant with scaled. 2. You can use the northpole for the projection, or any other point which is not in $X$ itself. There are plenty of them by dimensional considerations. 3. I played a little bit with the $S^1\times S^1$ example and its projections into $\Bbb R^3$. Unfortunately I have the results not here at the moment. I found a projection which looked like an intuitive torus, but not perfectly round like a circle. No idea if a "perfect torus" is possible :/ $\endgroup$ – M. Winter Oct 1 '17 at 13:12

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