How the three-torus makes flat space finite

You need coordinates in three dimensions, the distance formula and vectors as arrows that can be added. The first article gives the picture without equations.

Start with a square whose edges join

Take a square of side $L$ in the plane, with coordinates $x$ and $y$ running from $0$ to $L$. Declare that the right edge is the left edge and the top edge is the bottom edge. In symbols, the points $(x, y)$, $(x + L, y)$ and $(x, y + L)$ are the same point. A walker who crosses the line $x = L$ reappears at $x = 0$ with the same $y$ and the same heading.

Nothing has been bent. In any small patch, the walker measures the angles of a triangle and finds a sum of 180 degrees. Parallel lines stay parallel. What changed is the global connection: a straight walk along $x$ returns to its start after a distance $L$.

The doughnut surface you can hold in your hand is a torus too, but it is curved. Its outer rim is longer than its inner rim. The flat torus is defined by the gluing rule alone.

The same rule in three dimensions

Now take a cube of side $L$ with coordinates $\mathbf{x} = (x, y, z)$. The gluing rule says that $\mathbf{x}$ and $\mathbf{x} + L\mathbf{n}$ are the same point for every vector of whole numbers $\mathbf{n} = (n_x, n_y, n_z)$:

This space is the three-torus. Its volume is $L^3$. A light ray that leaves through the face $x = L$ re-enters through $x = 0$ moving in the same direction, so straight lines stay straight.

There is a second way to picture the same space, and it is the one used for the rest of the course. Fill all of ordinary space with identical copies of the cube, like a wall of bricks that extends forever. Every object in the cube then has an image in every brick, at the positions $\mathbf{x} + L\mathbf{n}$. This filled-in picture is called the covering space. The images are not new objects. They are the same object seen along different straight paths that wrap around the box.

The choice of a cube is a convenience. One can cut the same space into tiles of other shapes, so the cube is a bookkeeping choice. The physics sits in the gluing rule, which says which points are the same.

Distance in a space that wraps

Between two points in the three-torus there are many straight paths: one for each image of the second point. The distance is the length of the shortest:

Worked example. Let $L = 1.2$ and take $\mathbf{x} = (0.1, 0, 0)$ and $\mathbf{y} = (1.1, 0, 0)$. The direct path has length $1.0$. The path through the face uses $\mathbf{n} = (-1, 0, 0)$: the difference is $0.1 - 1.1 + 1.2 = 0.2$. The two points are $0.2$ apart, even though their coordinates differ by $1.0$. In a space that wraps, compare coordinates only after removing whole box lengths.

Why local measurements cannot see the wrap

A triangle whose sides are much shorter than $L$ sits inside one brick. Its angles add to 180 degrees, exactly as in infinite flat space. Every local measurement of curvature gives the same answer in the three-torus and in infinite flat space. Telling them apart needs a path long enough to go around, which is why the question becomes one about the most distant light we can see.

The rule that links nearby points, the metric, is the same in both spaces. The rule that links distant regions, the topology, differs. General relativity relates matter to the local curvature, so it fixes the first and leaves the second open.

Copies of the observer

Place an observer at the origin. The gluing rule creates images at the lattice points

Group the images by $n^2 = n_x^2 + n_y^2 + n_z^2$, a shell. The first shell, $n^2 = 1$, holds the 6 images across the faces of the cube. The second, $n^2 = 2$, holds the 12 images across the edges. The third, $n^2 = 3$, holds the 8 images across the corners.

Worked example. Which images lie closer than 2 units for $L = 1.2$? The first shell sits at $1.2$, the second at $1.2\sqrt{2} \approx 1.70$, and the third at $1.2\sqrt{3} \approx 2.08$. So the first two shells qualify, $6 + 12 = 18$ images. They come in 9 pairs $\pm\mathbf{g}$, because an image on one side always has a partner on the opposite side. The next chapter shows why the number 2 matters: in units of the distance to the most distant light, images closer than 2 are the ones that draw circles on the sky.

Lab tasks

1. Predict how many images lie closer than 2 units when $L = 0.8$. Count shells up to $n^2 = 6$.
2. In the Lab, set the box edge to $0.8$ and read the number of matched circle pairs.
3. Drag the box corner slowly from $1.2$ to $1.5$. Watch which ring of copies moves out of reach first.
4. Explain why the readout drops from 9 to 3 pairs between $1.2$ and $1.5$.

Exercises

1. In a two-dimensional flat torus of side $L = 1$, find the distance between $(0.05, 0.9)$ and $(0.95, 0.1)$.
2. How many images of the observer have $n^2 = 4$, and how far away are they?
3. A triangle with sides of length $0.01$ is drawn in a three-torus of side $L = 1$. What is the sum of its angles?
4. Why does $L = 2.2$ give no image closer than 2 units?

For the task with $L = 0.8$: the shells $n^2 = 1$ to $6$ lie at $0.8$, $1.13$, $1.39$, $1.6$, $1.79$ and $1.96$, all closer than 2. They hold $6 + 12 + 8 + 6 + 24 + 24 = 80$ images, which form 40 pairs.

Sources: Cornish et al. 2004 on local geometry and topology, COMPACT collaboration preprint, section 2 on clones, tiles and fundamental domains, Cornish, Spergel and Starkman 1998.

Set the box edge to 1.2 and count the copies of you whose sphere crosses yours in the slice.