Unit vectors

June 15, 20244 min read

Previously, we learned to calculate the length of vectors in $\VecR n,$ and thus the distance between points in $\AffR n.$ It's often useful to express a vector in terms of its length and its direction.

Let's start by defining direction.

If $v\In\VecR n$ is any nonzero vector, its direction is

\Dir v \Defeq \frac v{\Norm v}

This is also called the normalization of $v.$ By definition, we have

v = \Norm v \Dir v,

which expresses $v$ as its direction times its length. The $0$ vector has no direction.

Spheres

We can now give a formal definition of one of the most important spaces in geometry: the $n$ -sphere. This is defined to be the set of unit vectors in $\VecR{n+1}:$

\begin{aligned} \S n &\Defeq \{v\In\VecR{n+1} \mid \Norm v = 1\}\\ &= \left\{\ivec{x_0,\dotsc,x_n}\In\VecR{n+1} \mid \sqrt{\sum_{i=0}^n x_i^2} = 1\right\} \end{aligned}

Since $1^2 = 1,$ this is equivalent to

\S n = \left\{\ivec{x_0,\dotsc,x_n} \In \VecR{n+1} \mid \sum_{i=0}^n x_i^2 = 1\right\}

Note the shift in indexing: $\S1\subset\VecR 2$ is the usual unit circle in 2d, while $\S2\subset\VecR3$ is the usual unit sphere in 3d.

It's worth looking at the edge case

\begin{aligned} \S 0 &= \{v\In\VecR1 \mid \Norm v = 1\}\\ &= \{\pm1\} \subset \R \end{aligned}

That is, the $0$ -sphere consists of exactly two points! While this may seem strange, it fits nicely into a pattern about spheres in general:

every "slice" or "cross-section" of $\S2$ is a $1$ -sphere, except for the "north and south poles" $\ivec{0,0,\pm1}.$
every "slice" or "cross-section" of $\S1$ is a $0$ -sphere, except for the "north and south poles" $\ivec{0,\pm1};$

Let's prove this in general! This is a nice example of how we can use mathematical formalism to prove statements in $n$ dimensions based on our intuition in $\le 3$ dimensions.

Proposition

Consider the cross section

S^n_t \Defeq \{\ivec{x_0,\dotsc,x_n}\In S^n \mid x_n = t \}

Then $S^n_t$ consists of a single point when $t=\pm1,$ and is isomorphic to $S^{n-1}$ when $|t|<1.$

Proof

By definition,

S^n_t = \{\ivec{x_0,\dotsc,x_{n-1},t} \In \VecR{n+1} \mid \sum_{i=0}^{n-1} x_i^2 + t^2 = 1\}

Let $v=\ivec{x_0,\dotsc,x_{n-1}} \In \VecR n.$ We can rearrange the above condition to

\begin{aligned} \sum_{i=0}^{n-1} x_i^2 &= 1-t^2\\ \sqrt{\sum_{i=0}^{n-1} x_i^2} &= \sqrt{1-t^2}\\ \Norm v &= \sqrt{1-t^2} \end{aligned}

When $t=\pm1,$ this becomes $\Norm v=0,$ which implies $v=0.$ This means that $S^n_t$ consists of the single point

\ivec{0,\dotsc,0,t}.

When $|t|<1,$ we have that $S^n_t$ is isomorphic to the sphere of radius $\sqrt{1-t^2}$ in $\VecR n.$

Parametrization

We can use the defining equation of $\S n$ to find a parametrization of it. Let's start with the case of $\S1.$ Its defining equation is

x^2 + y^2 = 1

Let's rearrange this to solve for $y$ in terms of $x:$

\begin{aligned} x^2 + y^2 &= 1\\ y^2 &= 1 - x^2\\ y^2 &= \pm\sqrt{1 - x^2} \end{aligned}

So we can take $x=t$ as our parameter variable. We also need to find the domain of $t$ (equivalently $x).$ Remember that we can only take square roots of non-negative numbers. So we need to have

\begin{aligned} 0 &\le 1-x^2\\ x^2 &\le 1\\ |x| &\le 1 \end{aligned}

or equivalently

-1 \le x \le 1.

Let's try this in Desmos:

Unit vectors

Spheres #

Parametrization #

Spheres

Parametrization