The sphere $S^n$ admits a partition of unity consisting of two functions

Question

I am reading the book "Differentiable Manifolds" by Brickell and Clark.

Here is one of its problems: Show that the sphere $S^n$ admits a partition of unity consisting of two functions.

I'm not sure how to show this. I know that there is a atlas with only two charts. These are obtained by stereographic projection of $S^n \subset \mathbb{R}^{n+1}$ from the points $(0,0,...,1)$ and $(0,0,...,-1)$ onto the plane $z^{n+1} = 0$. Suppose $U$ and $U'$ are the punctured spheres obtained from $S^n$ by omitting these two points respectively. But I cannot continue. How should I proceed?

Thanks for any help.

Answer 1

Let $g:\mathbb R\to[0,1]$ be a smooth function function with $g(x)=0$ for $x\leq 0$ and $g(x)=1$ for $x\geq 1$. For a construction of such a $g$ see here.

Then for $U'$, $U$ there is a partition of unity given by $\phi_1(z_1,...,z_{n+1})= g(z_{n+1}+\frac 1 2 )$ and $\phi_2=1-\phi_1$.

Answer 2

The usual construction of partitions of unity can be found in the proof of the following theorem:

For any open covering ${\cal{U}} = \{ U_i\}_{i \in I}$ of a manifold $M$ there exists a partition of unity $\{\phi_i\}_{i \in I}$ that is subordinate to $\cal U$.

To say that $\{\phi_i\}_{i \in I}$ is a partition of unity subordinate to $\cal U$ means:

1. $\phi_i : M \to \mathbb R$ is smooth function for each $i \in I$, its image is contained in $[0,1]$, and the set $$\text{supp}(\phi_i) = \text{closure}\{x \mid \phi_i(x) > 0\} $$ is contained in $U_i$.
2. For each $x \in M$, there are only finitely many values of $i$ for which the value of $\phi_i(x)$ is nonzero, and $\sum_{i \in I} \phi_i(x)=1$.

From this theorem it follows that if you have a covering with two sets then you have a partition of unity with two elements.

Note: The requirement that $\text{supp}(\phi_i) \subset U_i$ is sometimes overlooked; see for example other answers on this page. But it is nonetheless an important condition in many applications.

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In the context of the book you are reading, it's not clear to me whether the authors have stated this theorem prior to this exercise. If they have not, perhaps the intent of the exercise is for you to discover the proof for yourself. Either way, if you like you can go read the proof of the theorem.

Answer 3

Let $\Phi$ be the first of those stereographic projections and let $F\colon\mathbb{R}^{n+1}\longrightarrow[0,1]$ be a continuous map such that $\lim_{\lVert x\rVert\to\infty}F(x)=0$. Now, consider the map$$\begin{array}{rccc}F_+\colon&S^n&\longrightarrow&[0,1]\\&x&\mapsto&\begin{cases}F\bigl(\Phi(x)\bigr)&\text{ if }x\neq(0,0,\ldots,1)\\0&\text{ otherwise,}\end{cases}\end{array}$$which is continuous. Finally, let $F_-(x)=1-F_+(x)$. Then $\{F_+,F_-\}$ is a partition of unity.

Answer 4

The sphere $S^n$ admits an open covering made of two open sets: $$ S^n=U\cup V,\qquad U=S^n\setminus\{N\},\quad V=S^n\setminus\{S\} $$ where $N$ and $S$ are the north and south pole respectively.

Now, choose isomorphisms of analytic varieties $$ V\simeq\Bbb R^n,\qquad U\setminus\{S\}\simeq\Bbb R^n\setminus\{0\} $$ in such a way $N\in V$ is identified to $0\in\Bbb R^n$.

Now choose any analytic function $\phi$ on $\Bbb R^n$ with values in $(0,1]$ such that $\phi(0)=1$, $\phi(x)\neq1$ for all $x\neq0$ and $\phi(x)\mapsto0$ as $\mid\mid x\mid\mid\to\infty$.

Pull back $\phi$ to $V$ under the above isomorphim and define on $U$ the function that is $1$ on $S$ and $1-\phi(x)$ for a point corresponding to $x$ under the above isomorphism. These two functions give a partition of unity as wanted.

Answer 5

There is a atlas with only two charts. These are obtained by stereographic projection of $S^n \subset \mathbb{R}^{n+1}$ from the points $(0,0,...,1)$ and $(0,0,...,-1)$ onto the plane $z^{n+1} = 0$. Suppose $U$ and $U'$ are the punctured spheres obtained from $S^n$ by omitting these two points respectively. The functions $$y:U \to \mathbb{R}^n , \qquad y':U' \to \mathbb{R}^n$$ defined by $$y^i (z)= \dfrac{z_i}{1 - z_{n+1}}, \qquad y'^i (z)= \dfrac{z_i}{1 + z_{n+1}}$$ are both injections onto $\mathbb{R}^n$ and they are charts for $S^n$.

Let $f:\mathbb{R}\to \mathbb{R}$ be a differentiable function such that $f>0$ on $s>0$ and $f=0$ on $s‎\leq‎‎‎‎‎ 0$. Put $g := f(r^2 - \|y(z)\|^2)$, $g' := f(r^2 - \|y'(z)\|^2)$, $h:= g \circ y$ and $h':= g' \circ y'$ where $r>1$. Then we get

$$h(z) = \begin{cases} ‎f(\frac{r^2 -1 - (r^2 + 1)z_{n+1}}{1 - z_{n+1}}) ‎‎‎& ‎\text{if -1 ‎$\leq‎‎‎‎‎‎ z_{n+1}‎<‎ \frac{r^2 - 1}{r^2 + 1}‎$}\\‎ ‎ ‎‎‎‎0‎‎‎‎ ‎‎‎‎‎& \text{if $\frac{r^2 - 1}{r^2 + 1}‎‎\leq‎‎‎‎‎‎ z_{n+1}$‎ < 1‎‎‎‎‎‎ } ‎ \end{cases},$$

and

$$h'(z) = \begin{cases} ‎f(\frac{r^2 -1 + (r^2 + 1)z_{n+1}}{1 + z_{n+1}}) ‎‎‎& ‎\text{if $-\frac{r^2 - 1}{r^2 + 1}‎< z_{n+1}‎\leq‎‎‎‎‎‎$ 1}‎\\‎ ‎ ‎‎‎‎0‎‎‎‎ & \text{if -1‎<‎‎‎‎‎‎ $z_{n+1}‎\leq‎‎‎‎‎‎ -\frac{r^2 - 1}{r^2 + 1}‎$} ‎ \end{cases}.$$

Now we define $\phi := \frac{h}{h + h'}$ and $\phi' := \frac{h'}{h + h'}$. Therefore $$\phi(z) = \begin{cases} ‎1 ‎‎‎& ‎\text{if -1 ‎$\leq‎‎‎‎‎‎ z_{n+1}‎\leq‎ -\frac{r^2 - 1}{r^2 + 1}‎$‎}\\ \frac{h(z)}{h(z) + h'(z)} ‎‎‎& \text{if $-\frac{r^2 - 1}{r^2 + 1}‎ ‎<‎‎‎ z_{n+1}‎ < \frac{r^2 - 1}{r^2 + 1}‎$}‎\\‎‎ ‎‎‎0‎‎‎‎ ‎‎‎‎‎& \text{if $\frac{r^2 - 1}{r^2 + 1}‎\leq‎‎‎‎‎‎ z_{n+1}‎ \leq$ 1‎‎‎‎‎‎} ‎ \end{cases},$$ and $$\phi'(z) = \begin{cases} ‎0 ‎‎‎& ‎\text{if -1 ‎$\leq‎‎‎‎‎‎ z_{n+1}‎\leq‎ -\frac{r^2 - 1}{r^2 + 1}‎‎$}\\ \frac{h'(z)}{h(z) + h'(z)} ‎‎‎& \text{‎if $-\frac{r^2 - 1}{r^2 + 1}‎ ‎<‎‎‎ z_{n+1}‎< \frac{r^2 - 1}{r^2 + 1}‎‎$}\\‎‎ ‎‎‎‎‎1‎‎‎‎ ‎‎‎‎‎& \text{if $\frac{r^2 - 1}{r^2 + 1}‎‎\leq‎‎‎‎‎‎ z_{n+1}‎ \leq$ 1‎‎‎‎‎‎ } ‎ \end{cases}.$$ Then we obtain the partition of unity $\{(U, \phi) , (U' , \phi')\}$ on $S^n$.