# Why is $S^1 \times S^1$ a Torus?

Call a "torus" that geometric shape that "looks like" a doughnut. Frequently, one encounters the assertion that $S^1 \times S^1$ is a "torus" where $S^1$ is the unit circle. Now, if I think about this, I can understand the justification for calling this a torus, but I'm trying to understand how one would go about actually proving this. Indeed, there exist analytical descriptions of the torus such as this one provided by Wikipedia. So one could theoretically, with enough inspiration, find a homemorphism $h: S^1 \times S^1 \rightarrow G(T)$ where $G(T)$ denotes the graph of the torus as realized by the analytical description. This approach, assuming it works, uses coordinates and in any event wouldn't be very enlightening.

So, my question is, is there a coordinate-free way to prove that $S^1 \times S^1$ is homeomorphic to this thing we call a doughnut?

My thoughts on this are: I believe what is key is how one chooses to define a torus. I am familiar with constructing a torus by identifying opposite sides of a rectangle and this seems like a pretty natural definition. It is intuitively clear that if the rectangle is configured as \begin{align*} A---- & B \\ |\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;&| \\ C---- & D \end{align*} then one can map one circle homeomorphically onto the identification of AB and CD and then the other circle onto the identification of AC and BD. However, this really isn't a very precise argument and it seems to me that making it precise would eventually involve coordinates. Am I onto the right track with this approach or is there a better way of looking at the problem?

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In a very informal sense (which is why I'm making this a comment, not an answer), I think much of the difficulty comes from the fact that, while the torus can be embedded in $\mathbb R^3$, its "natural habitat" is really $\mathbb R^4$. In particular, the canonical embedding of $S^1$ into $\mathbb R^2$ induces a natural embedding of $S^1\times S^1$ into $\mathbb R^4$, the "flat torus". Any projection of the torus down into 3 dimensions necessarily involves some symmetry breaking, which complicates the description of the resulting surface. – Ilmari Karonen Jan 29 '12 at 22:21
Showing that wikipedia's parameterization of the torus is a parameterization by $S^1\times S^1$ doesn't require a lot of work: The parameterization $x(u,v)=(R+r\cos(v))\cos(u), y(u,v)=(R+r\cos(v))\sin(u),$ and $z(u,v)=r\sin(v)$ is $2\pi$ periodic in both $u$ and $v$. Since $S^1$ can be viewed as $\mathbb{R}/2\pi \mathbb{Z}$, the identification follows. – Adam Jan 30 '12 at 2:41

I think you're thinking of it in the right way, but you're also trying to have your cake and eat it here. If you want a completely rigorous argument, you can't get around having a rigorous description of the 3d shape that you're trying to prove homeomorphic to $S^1\times S^1$ in the first place. But that means, in particular, that you need a well-defined way of speaking on particular points on the geometrical torus, and how would you do that if not by coordinates?

In fact, one doesn't need that much inspiration to find a concrete coordinate-based homeomorphism. Let's define $S^1$ as the set $\{(x,y)\mid x^2+y^2=1\}\subset \mathbb R^2$ with the subspace topology. Then, $$h((x,y),(z,w)) = z(0,0,1)+(w+2)(x,y,0)$$ is your desired homeomorphism, into the set $$\{(x,y,z)\mid z^2 + \Bigl(2-\sqrt{x^2+y^2}\Bigr)^2 = 1\}\subset \mathbb R^3$$ which one ought to recognize as our geometrical torus.

To describe this as "not very enlightening" would be fair if it was about purely formal algebraic manipulation or black-box recipes for computation -- but in fact it has a clear geometric significance. One might say, instead:

For each $(x,y)$ we draw the $(z,w)$ circle on a vertical plane through the origin and oriented such that it passes through $(x,y)$, with the center of the $(z,w)$ circle being offset by $2$ units in the direction of $(x,y)$.

However, such a verbal description risks getting complex and hard to grasp, and it is all too easy for the writer to introduce ambiguities where it can be understood only if the reader already understands which construction is meant. The coordinate formulas, on the other hand, are unambiguous, and with a bit of practice one can see the geometric construction directly in them.

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Well, one way you might find fruitful is to consider $S^1$ as $\mathbb{R}/\mathbb{Z}$. Then if we identify the "square with sides identified" model with $\mathbb{R}^2/\mathbb{Z}^2$, we want to show that $\mathbb{R}/\mathbb{Z}×\mathbb{R}/\mathbb{Z}$ is homeomorphic to $\mathbb{R}^2/\mathbb{Z}^2$, which I find is much more natural since we don't have to treat boundaries as "special cases" when we define the homeomorphism.

Cheers,

Rofler

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If you like to think about the torus as the rectangle $[a,b]\times [c,d]$ with the opposite sides identified, then think what happens when you identify the opposite sides.

When you identify $[a,b]\times \{ c \}$ with $[a,b] \times \{ d \}$, every "vertical" segment $\{ x \} \times [c,d]$ becomes a circle $\{ x \} \times S^1$.

Thus, you get the cylinder $[a,b] \times S^1$.

Now, you identify the ends $a \times S^1$ with $b \times S^1$, every "horisontal" segment $[a,b] \times y$ becomes a circle, $S^1 \times y$, where $y$ is a point on $S^1$. Thus you get $S^1 \times S^1$.

P.S. This can also be seen easely by a cutting argument. If you cut $S^1$ at a point you get an interval. So cut $S^1 \times S^1$ twice, first at $\{a \} \times S^1$ to get $[a,b] \times S^1$ and cut it again. Now glue it together backwards and you get your definition of the torus.

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