On the equivalence of several definitions of the torus

Let $I$ be a closed interval. It is convenient to associate it with $[0,1] \subset \Bbb{R}$, especially as we will wish to treat it as a metric space as well as a topological space. It should be no surprise that $I$ is homeomorphic to any other closed interval of $\Bbb{R}$, so the choice of the endpoints $0$ and $1$ is not essential. In the following, I will distinguish between $\Bbb{R}$, a topological space without a metric, and $\Bbb{E}$, the Euclidean line, which is $\Bbb{R}$ endowed with the usual metric, $d(x,y) = |x-y|$. (This distinction is frequently viewed as superfluous, since remembering "the $\Bbb{R}$ I've been studying for years has always been a metric space and sometimes we just choose to forget about the metric when we are doing topology and then suddenly remember it when it is convenient to do so" seems to be easy for most of us.)

$2 \leftrightarrow 1$

In plane geometry, rectangle is a quadrilateral with right angles in each of its four corners and each pair of opposite sides parallel. "Right angles" and "parallel" are meaningless in a topological space. Also, on some 2-manifolds (a reasonable large set of spaces on which one might try to talk about a conformal (= angle preserving) embedding of a rectangle) a geometric figure made of (geodesic) line segments turning through three right angles can close (uniform spherical geometry -- two vertices on the equator and one vertex at the north pole). On others (uniform hyperbolic geometry) a geometric figure made of (geodesic) line segments and three right angles does not close at a right angle (the first and fourth segments may not meet or may meet at some smaller angle). So to get a rectangle, the space has to be flat where we try to put the rectangle, so we can require uniformly flat. This gives one option -- a rectangle always lies in (some connected subset) of the Euclidean plane, $\Bbb{E}^2$.

In a metric space, we can distinguish different rectangles by their orientation and edge lengths, so three numbers. In a topological space, a rectangle and a circle are indistinguishable and edge length is meaningless, so all three of our numbers are useless in distinguishing rectangles in $\Bbb{R}^2$. (This is a two step process : first embed the rectangle in $\Bbb{E}^2$, then forget about the metric and observe that you have a bunch of points in the topological space $\Bbb{R}^2$, labelled "rectangle".) There are then several alternative notions of equivalence that one might apply -- the strictest is usually ambient isotopy and any rectangle in the plane is ambient isotopic to any other rectangle. So in a topological space, we lose nothing by deciding that "rectangle" means $I \times I$.

$I \times I = [0,1] \times [0,1] \subset \Bbb{E}^2$, its product topology is equivalent to its subspace topology inherited from $\Bbb{R}^2$, and its product metric is equivalent to its subspace metric inherited from $\Bbb{E}^2$. The specified quotient is a disjunction of three terms. The first term makes the identification $(0,y) \sim (1,y)$ for $y \in I$. The second term makes the identification $(x,0) \sim (x,1)$ for $x \in I$. (The third term leaves all the interior points alone.) These choices set the orientation of the identification. One could alternatively make the identification $(x,0) \sim (1-x,1)$, which would reverse the orientation of that piece of the quotient.

The following pair of diagrams show by arrows how the identification in the quotient preserves the orientation and how the proposed alternative reverses the orientation using arrows from points on the lower edge to the equivalent point on the upper edge.

Note that $I \times I$ is "for each point in one copy of $I$ a copy of $I$" (with the product topology). We are free to decide which factor is which copy in that phrase -- in fact, we can reverse our point of view and get the same result. This means that we can treat $I \times I$ as a bundle of horizontal copies of $I$ or as a bundle of vertical copies of $I$.

When we apply the $(x,0) \sim (x,1)$ equivalence, each line in the vertical bundle has its lower endpoint identified with its upper endpoint, producing a circle. So temporarily pausing after performing that part of the quotient, we have "for each point in the horizontal $I$, we have a vertical $S^1$". If we start at $I \times I$ and instead apply the $(0,y) \sim (1,y)$ part of the equivalence, we obtain "for each point in the vertical $I$, we have a horizontal $S^1$". In both cases, we have a cylinder (just the curved part, not including the two flat disks that are normally used to cap off the ends), closed since it includes the two circles on its boundary. When we finish the quotient, all of the vertical line segments quotient into a circle and all of the horizontal line segments quotient into a circle and for each point in one family of circles, we have a circle in the other family. That is, we have an $S^1 \times S^1$.

Now suppose we have an $S^1 \times S^1$. This naturally lives embedded in $\Bbb{R}^4$ (or $\Bbb{C}^2$) as the unit circle in the first two coordinates cross the unit circle in the second two coordinates, $$ T^2 \simeq \{(a,b,c,d) \mid a^2 + b^2 = 1, c^2 + d^2 = 1\} \text{.} $$ When you embed a torus in $\Bbb{R}^3$, you have to make a decision about which $S^1$ factor corresponds to the longitude of the torus and which factor corresponds to the meridian. The two are not interchangeable -- the longitude bounds a disk exterior to the torus and the meridian bounds a disk interior to the torus. In the $\Bbb{R}^4$ version, we need make no such distinction -- there is a rigid rotation of $\Bbb{R}^4$ that implements $\{a \leftrightarrow c, b \leftrightarrow d\}$, making the distinction between the factors superficial.

As I said, we can also express this in $\Bbb{C}^2$: $$ T^2 \simeq \{(w,z) \mid ||w|| = 1, ||z|| = 1 \} \text{.} $$ But for our purpose, it is better to use the polar representation of complex numbers, $$ T^2 \simeq \{(\mathrm{e}^{2\pi \mathrm{i} \theta}, \mathrm{e}^{2\pi \mathrm{i} \phi}) \mid \theta \in [0,1), \phi \in [0,1) \} \text{.} $$ Notice that we are almost there -- we are close to having our two parameters only vary over $[0,1]$, we just need to undo the quotient. Start by deleting the circle $\theta = 0$. This gives us a cylinder excluding its boundary circles. Now make two copies of the circle we just deleted, glue one along the $\theta = 0$ boundary, and glue the other along the $\theta = 1$ boundary. This gives $I \times S^1$. Now repeat with $\phi$ replacing $\theta$, cutting out a line and gluing in two lines, producing $I \times I$. (Instead of deleting, we can cut to obtain $[0,1)$ for one factor, then duplicate the circle or line at $0$ to glue to the boundary at $1$, yielding an $I$ factor. Recall that when we do any cutting the two new boundary components are in no sense "close" to each other -- we have cleaved apart all the open sets that crossed the cut.)

(From here, going back to the torus, we start with $I \times I$, identify two lines, producing one line (obtaining $S^1 \times I$), then identify two circles, producing one circle (obtaining $S^1 \times S^1$), just as we described when talking about horizontal and vertical line bundles.)

$3 \leftrightarrow \{1,2\}$

In this case, you will be much better served seeing these as maps onto $[0,1) \times [0,1)$. Then looking at a small open disk neighborhood of a point in the middle of this "half-open square" (small enough to not meet an edge or corner), such a disk centered on an edge (small enough to not touch another edge or corner), and such a disk centered on a corner. You will discover that you have rediscovered the $\theta-\phi$ parametrization of the torus from above. That is, the open sets will reveal the quotients of the ends of the horizontal and vertical half-open line segment bundles.

$\Bbb{R}^2 /G \simeq \Bbb{R}^2 / \Bbb{Z}^2$

Happily, $\Bbb{R}^2$ is a vector space, so if you pick any two nonzero linearly independent elements of $\Bbb{R}^2$, say $\{\alpha, \beta\}$, then $$ \Bbb{R}^2 /G \simeq \Bbb{R}^2 / \langle \alpha, \beta \rangle \simeq \Bbb{R}^2 / \Bbb{Z}^2 \text{,} $$ where $\Bbb{R}^2$ is treated as an abelian group of displacement vectors, and $\langle \alpha, \beta \rangle$ is the free abelian subgroup of $\Bbb{R}^2$ (equivalently, a module over the PID $\Bbb{Z}$) generated by $\alpha$ and $\beta$. $G$ is given as the free abelian subgroup spanned by the two displacement vectors $(1,0)$ and $(0,1)$ (or their transposes, depending on how you are thinking of this object), which are linearly independent. It may help to notice that the block square matrix $\left( \alpha \ \beta \right)$ is an invertible ("two linearly independent elements") linear map taking the generators of $G$ to $\alpha$ and $\beta$, respectively. A linear map is a homeomorphism. The various parallelograms with opposite sides identified (preserving orientation) are all equivalent (to the special parallelogram, $I \times I$). One should check that the three types of neighborhoods meet the quotient requirements, as discussed in the previous section.

Firstly, if we look at $\mathbb{R}/\mathbb{Z}$ as a group we see that the cosets can all be represented by numbers in the range $[0,1)$, which we can identify as the fractional portion of $x$, or the stuff after the decimal point. Note that if we choose such a representative then all the other elements of that coset are of the form $x+n$ for some integer $n$ so if $x$ is in the coset then $x+1$ is in the coset.

We can map this range to the circle $S^1$ by $x \rightarrow e^{2\pi ix}$, the usual exponential map and this is a group isomorhpism from $\mathbb{R}/\mathbb{Z}$ to unit circle in $\mathbb{C}$. Multiplication on the unit circle becomes "clock addition" in $\mathbb{R}/\mathbb{Z}$ in a natural way.

So We have representatives in $[0,1)$ in $\mathbb{R}/\mathbb{Z}$ can be seen as points on the complex unit circle $S^1$, and that $x+1$ is always in the same coset as $x$. All we need to now is see what happens when we move to two dimensions.

So now we have points of the form $(x,y)$ in $\mathbb{R}^2/\mathbb{Z}^2 = (\mathbb{R}/\mathbb{Z})^2$ (why?) and we can again choose representatives from the fractional portions of $x$ and $y$ respectively. This means we can restrict our attention to the square $[0,1) \times [0,1)$ which entirely represents the group, while the cosets will be all translated copies of it of the form $(x + n, y +m)$ for $n,m \in \mathbb{Z}$. If we hold some $y_0$ fixed then we have $(x_1,y_0) + (x_2,y_0) = (x_1 + x_2, 2y_0) = (x_1 + x_2, y_0)$ (because $2y_0-y_0 \in \mathbb{Z})$ and so we can see a copy of $\mathbb{R}/\mathbb{Z}$ exists for each $y_0$, that is to say each vertical line on the square and each horizontal line on the square is its own "clock addition", which we can identify with a circle. Said another way we're taking each point on a circle and attaching a circle to it, using complex multiplication instead.

But now we're done. We have $\mathbb{R}^2/\mathbb{Z}$, the complex torus $S^1 \times S^1$ and the quotient group formed by the translations of the form $(x + n, y + m)$ all represent the same group. Exponential map that turns line segments into circles and then we attach a circle to each point on a circle to get the usual torus.