Place your mouse cursor over the square grid on the left. Click and hold while moving the mouse to draw. The "corresponding" path will also be drawn in the torus canvas.

Click on the palette to select the pen color. The arrow keys
rotate the view in the torus canvas.
The `z`

-`x`

keys translate the domain
horizontally. The `c`

-`v`

keys translate the
domain vertically.

Familiarly, a *torus* is a mathematical model of the
surface of a doughnut. Alternatively, a torus is a two-dimensional
universe where every point has infinitely many avatars placed at
the grid points of a square lattice. In this "video-game" view,
the torus is constructed from a unit square by *gluing* the
left edge to the right edge by rolling the square into a cylinder,
and then similarly gluing the top edge to the bottom edge.
To achieve this gluing physically in Euclidean three-space we can
cheat by flattening the cylinder into a "double rectangle" with
attached (long) sides and open (short) ends and rolling up this
rectangle into a "quadruple square." In Euclidean four-space,
however, the gluing is accomplished honestly by the mapping
\[
\mathbf{x}(u, v) = \tfrac{1}{\sqrt{2}}
(\cos(2\pi u), \sin(2\pi u),
\cos(2\pi v), \sin(2\pi v)).
\]
(The components are scaled so the image lies in the unit sphere in
four-space.) The component functions are $2\pi$-periodic, and two
pairs $(u, v)$ and $(u', v')$ map to the same value if and only if
their difference $(u - u', v - v')$ is a pair of integers. This
occurs if and only if $(u, v)$ and $(u', v')$ are avatars of the
same point.

To visualize the flat unit-square torus we map back to
Euclidean three-space using *stereographic projection* from
the point $(0, 0, 0, 1)$ of the unit sphere. This mapping is
\[
\Pi(x, y, z, w) = \frac{(x, y, z)}{1 - w}.
\]
The torus in the right-hand canvas is the composition. After
canceling factors of $\sqrt{2}$, we have
\[
(\Pi \circ \mathbf{x})(u, v)
= \frac{(\cos(2\pi u), \sin(2\pi u), \cos(2\pi v))}{\sqrt{2} - \sin(2\pi v)}.
\]
This parametrization, which differs from that usually seen in
multivariable calculus, is *conformal*: The angle between
two crossing curves at a point $p$ in the Euclidean plane is equal
to the angle between the image curves in the torus at the point
$\Pi(\mathbf{x}(p))$.