This site is being phased out.

# Torus

It a tire. Don't confuse with the donut - no air (void) inside.

## Contents

## As a product space

Both copies of ${\bf S}^1$ lie in ${\bf R}^2$, so the product would lie in ${\bf R}^4$. One, however, might think of small circles attached, vertically, to each point of the large circle on the plane.

Building a torus this way is a stretch - the way they are constructed will result in "a bag of points". That's where product topology comes into play.

## As a quotient space and a cell complex

This is how it is constructed as a cell complex.

One can get the torus ${\bf T}^2$ from the cylinder by gluing the top to the bottom (the red arrows):

$$(x,0) \sim (x,1).$$

Or directly from the square:

See Quotient spaces.

Torus is a surface, i.e., it's locally homeomorphic to the plane.

Represent the torus as a simplicial complex (i.e., its triangulation):

The Euler characteristic is

$$\chi(T^2) = 1 - 2 + 1 = 0.$$

The $n$-*torus*, $T^n$, is an orientable $n$-dimensional manifold defined as the product of $n$ 1-spheres
$$T^n = \underbrace{ S^{1} \times \cdots \times S^{1}}_{n},$$
or as a quotient space (exercise).

## Homology

Compute its homology as a vector space:

$B_1(K)$ $= {\hspace{3pt} span}\{ \partial \tau \} = {\hspace{3pt} span}\{a + b - a - b \} = {\hspace{3pt} span}\{0 \} = 0,$ so

$H_1(K)$ $= {\hspace{3pt} span}\{a, b \} / 0 = {\hspace{3pt} span}\{a, b \}.$In particular, the the first Betti number is $2$.

The homology over ${\bf R}$

$${\bf R}, {\bf R}^2, {\bf R}, 0, \ldots$$

does not distinguish the torus from the Klein bottle, but the homology over ${\bf Z}$ does (see Homology as a group).

The torus is an orientable surface as it does not contain the Mobius band.

One can create the *double torus* by attaching two tori to each other in this fashion (connected sum).

## As a parametric surface

Torus is a surface in space given parametrically by the formula:

Domain: ${\theta} {\in} [ 0, 2 {\pi} ), {\varphi} {\in} [ 0, 2 {\pi} ]$.

We can think of it as a "parametric surface". Indeed we rotate a circle (off center) in the $yz$-plane around the $z$-axis:

Let's explain where this construction comes from. Consider the formula again:

$$f( {\theta}, {\varphi} ) = ( ( a + b \cos {\varphi} ) \cos {\theta}, ( a + b \cos {\varphi} ) \sin {\theta}, b \sin {\varphi} ).$$

1. Fix ${\theta} = 0$, find the image with one parameter (parametric curve).

$$\begin{array}{} f( 0, {\varphi} ) &= ( ( a + b \cos {\varphi} ) \cos 0, ( a + b \cos {\varphi} ) \sin 0, b \sin {\varphi} ) &= ( a + b \cos {\varphi}, 0, b \sin {\varphi} ) &= ( a, 0, 0 ) + b ( \cos {\varphi}, 0 \sin {\varphi} ). \end{array}$$

What is it? A circle in the $y_1y_3$-plane with center $( a, 0, 0 )$ and radius $b$. Fix ${\theta} = 2{\pi}$, find the image.

$$\begin{array}{} f( 0, {\varphi} ) &= ( ( a + b \cos {\varphi} ) \cos 2{\pi}, ( a + b \cos {\varphi} ) \sin 2{\pi}, b \sin {\varphi} ) \\ &= ( a + b \cos {\varphi}, 0, b \sin {\varphi} ), \end{array}$$

no change to previous calculation as $2{\pi}$ is the period of the $\sin$ and the $\cos$.

2. Fix ${\varphi} = 0$, find the image with one parameter.

$$\begin{array}{} f( {\theta}, 0 ) &= ( ( a + b \cos 0 ) \cos {\theta}, ( a + b \cos 0 ) \sin {\theta}, b \sin 0 ) \\ &= ( ( a + b ) \cos {\theta}, ( a + b ) \sin {\theta}, 0 ). \end{array}$$

What is it? A circle in the $y_1y_2$-plane with center $( 0, 0, 0 )$ and radius $( a + b )$.

To put this together: