Show that $\Bbb S^n$ contains an affine independent set with $n+2$ points

Question

Show that $\Bbb S^n$ contains an affine independent set with $n+2$ points. (Hint: For every $k\ge 0$, euclidean space $\Bbb R^n$ contains $k$ points in general position.)

I couldn't find any intuitive explanation for points being in general position. Apparrently in $\Bbb R^2$ the points $\{a_1,a_2,a_3\}$ are in general position if neither of them are collinear? So is the idea for the proof that in say $\Bbb S^1$ you cannot pick three points to lie on the same line etc?

Answer 1

I want to present two proofs of the Theorem. The first is very simple but not constructive. The second is decidedly more complicated (this is the first that came to mind: perhaps I have a contorted mind...), but it has the merit of being constructive: I present it here purely out of curiosity in the hope that it will be helpful.

$\raise1ex{\quad}$
Proof$\;$ 1

The points of $\;\mathbb R^{n+1}$:

$$ e_1=(1,0,\ldots,0), \;e_2=(0,1,0,\ldots,0),\ldots, \;e_{n+1}=(0,0,\ldots,1), \;\varepsilon=(-1,0,\ldots,0) $$

are $\;n+2\;$ points of $S^n$, and these points are affine independent, because the vectors

$$ e_1-\varepsilon =(2,0,\ldots,0), \;e_2-\varepsilon =(1,1,0,\ldots,0),\ldots, \;e_{n+1}-\varepsilon=(1,0,\ldots,0,1) $$

are $n+1$ vectors of $\;\mathbb R^{n+1}\;$ linearly independent, because the determinant of their representative matrix (in the standard basis) is

$$ \det\left[\matrix{ 2 & 0 & 0 & 0 &\ldots & 0 \\ 1 & 1 & 0 & 0 &\ldots & 0 \\ 1 & 0 & 1 & 0 &\ldots & 0 \\ 1 & 0 & 0 & 1 &\ldots & 0 \\ \ldots &&&&& \\ \ldots &&&&& \\ 1 & 0 & 0 & \ldots & 0 & 1} \right] = 2\det(I_n) = 2 \neq 0. $$

$\raise3ex{\quad}$
Proof $\;$ 2

Induction on $n$.

The case $\;\mathbf{n=0}\;$ is trivial: $\;S^0=\{-1,1\}\subseteq \mathbb R^1$, and the points $\;e_1=1, \;\varepsilon^1=-1\;$ are affine independent since $\;e_1-\varepsilon^1=2\neq 0$. [Little Note: in accordance with what we will say below, $e_1$ should be written $e_1^1$.]

Step $\;\mathbf{n\to n+1}.\;$ Suppose that there exist $n+2$ affine independent points in $S^n\subseteq \mathbb R^{n+1}$. And we suppose that these points are

$$ e_1^{n+1},\;\ldots,e_{n+1}^{n+1},\;\varepsilon^{n+1} $$

where the family $\;(e_1^{n+1},\ldots,e_{n+1}^{n+1})\;$ is the standard basis of $\;\mathbb R^{n+1}\;\;$[the upper index reminds us of this]. This means that the vectors

$$ e_1^{n+1}-\varepsilon^{n+1},\;\ldots, e_{n+1}^{n+1}-\varepsilon^{n+1} $$

are linearly independent. Consider then the following points in $S^{n+1}\subseteq\mathbb R^{n+2}=\mathbb R^{n+1}\times \mathbb R$:

$$ e_1^{n+2},\;\ldots, e_{n+1}^{n+2}, \;e_{n+2}^{n+2}, \;\varepsilon^{n+2}:=(\varepsilon^{n+1},0), $$

and prove that they are affine independent. Indeed, if

$$ \lambda_1(e_1^{n+2}-\varepsilon^{n+2})+\;\ldots+\lambda_{n+1}(e_{n+1}^{n+2}-\varepsilon^{n+2})+\lambda_{n+2}(e_{n+2}^{n+2}-\varepsilon^{n+2})=0, \tag{*} $$

by taking the standard scalar product by $e_{n+2}^{n+2}$, we obtain

\begin{align} \Big(\lambda_1(e_1^{n+2}&-\varepsilon^{n+2})+\;\ldots+\lambda_{n+1}(e_{n+1}^{n+2}-\varepsilon^{n+2})+\lambda_{n+2}(e_{n+2}^{n+2}-\varepsilon^{n+2})\;\Big|\;e_{n+2}^{n+2}\Big)=\\[1.5ex] &=\sum_{k=1}^{n+1}\Big(\lambda_k(e_k^{n+2}-(\varepsilon^{n+1},0))\;\Big|\;e_{n+2}^{n+2}\Big)+\Big(\lambda_{n+2}(e_{n+2}^{n+2}-(\varepsilon^{n+1},0))\;\Big|\;e_{n+2}^{n+2}\Big)=\\[1.5ex] &= \sum_{k=1}^{n+1} \;0 + \lambda_{n+2}\Big(e_{n+2}^{n+2}\;\Big|\;e_{n+2}^{n+2}\Big)-\Big((\varepsilon^{n+1},0)\;\Big|\;e_{n+2}^{n+2}\Big)=\lambda_{n+2}=0 \end{align}

because $e_k^{n+2}$ for $k=1,\;\ldots n+1$, and $\;(\varepsilon^{n+1},0)\;$ are all orthogonal to $\;e_{n+2}^{n+2}$.
The (*) then gives

\begin{align} &\hskip3ex\lambda_1(e_1^{n+2}-\varepsilon^{n+2})+\;\ldots+\lambda_{n+1}(e_{n+1}^{n+2}-\varepsilon^{n+2})=\\[1.5ex] &=\lambda_1\big((e_1^{n+1},0)-(\varepsilon^{n+1},0)\big)+\ldots+ \lambda_{n+1}\big((e_{n+1}^{n+1},0)-(\varepsilon^{n+1},0)\big)=0 \end{align}

which implies

$$ \lambda_1\big(e_1^{n+1}-\varepsilon^{n+1}\big)+\ldots+\lambda_{n+1}\big(e_{n+1}^{n+1}-\varepsilon^{n+1}\big)=0 $$

and this, for the inductive hypothesis, gives $\lambda_1=\ldots=\lambda_{n+1}=0$.

Notice that the preceeding costruction immediately gives $\;\varepsilon^{n+1} = (-1,0,\ldots,0)$ with $n$ zeros.

Answer 2

General position is even more general than affine independence, because it applies in much more generality than just finite sets. But let's not go there, because this is really simpler than it sounds.

You are right that three points in $\mathbb R^n$ (not just in $\mathbb R^2$) are affinely independent if and only if they are not collinear. More generally, $k+2$ points in $\mathbb R^n$ are affinely independent if and only if they do not line on any $k$-dimensional affine subspace. This has a simple reformulation in vector language: a subset $\{p_0,p_1,...,p_k,p_{k+1}\} \subset \mathbb R^n$ is affinely independent if and only if the the set of displacement vectors $$\vec v_1 = p_1-p_0 \qquad ... \qquad\vec v_k = p_k - p_0 \qquad \vec v_{k+1} = p_{k+1}-p_0 $$ is linearly independent.

The problem asks you to prove that $\mathbb S^n$ contains an affinely independent set with $n+2$ points. Keep in mind: to solve this problem, for each $n$ you just need one subset of $n+2$ points in $\mathbb S^n$ which you can verify is affinely independent. Let me suggest that in the case $n=1$ you can find a very concrete, affinely independent subset of three points in $\mathbb S^1$. And then you can do this again for a four point subset in $\mathbb S^2$. And then you should be do this with a pattern that easily generalizes to $\mathbb S^n$ for all $n$.