# Why does $S^n$ satisfy the local $n$-slice condition? (From Lee's Smooth Manifolds)

Example $5.9$ on page $103$ of John Lee's Smooth Manifolds says the following:

The intersection of $S^n$ with the open subset $\{x:x^i>0\}$ is the graph of the smooth function $$x^i=f(x^1,\dots,x^{i-1},x^{i+1},\dots,x^{n+1})$$ where $f\colon B^n\to\mathbb{R}$ is $f(u)=\sqrt{1-|u|^2}$. The intersection of $S^n$ with $\{x:x^i<0\}$ is the graph of $-f$. Since every point in $S^n$ is in one of these sets, $S^n$ satisfies the local $n$-slice condition, this is an embedded submanifold.

The terminology is that if $M$ is a smooth manifold, and $S\subset M$ a subset, then $S$ satisfies the local $k$-slice condition if each point of $S$ is contained in the domain of a smooth chart $(U,\varphi)$ for $M$ such that $S\cap U$ is a single $k$-slice in $U$.

I don't see how this makes $S^n$ satisfy the local $n$-slice condition. Presumably the chart on $\mathbb{R}^{m+1}$ is $(U=\{x:x^i>0\},\mathrm{id})$, so that $S^n\cap\{x:x^i>0\}$ is an $n$-slice of $U$? But this doesn't seem right since $\mathrm{id}(S^n\cap U)$ is a hemisphere of $S^n$, but that's not a $n$-slice in the corresponding half-place $\mathrm{id}(U)$?

• You can certainly do this by constructing explicit charts as @PhoemueX did. But what I had in mind here was to use Proposition 5.4, which says that a graph of a smooth function is an embedded submanifold, and Theorem 5.8, which says that an embedded submanifold satisfies the local slice condition. – Jack Lee Aug 8 '14 at 16:49
• @JackLee Thanks Professor Lee. – Clara Aug 9 '14 at 1:12
• But here $S^n$ is not the graph of one smooth function, it is considered as the union of the graphs of 2 smooth functions, why does proposition 5.4 still apply?@JackLee – koch Nov 9 '16 at 1:46

The chart is

$$\{(x_1, \dots , x_{n+1}) \mid (x_1, \dots x_{i-1}, x_{i+1}, \dots x_{n+1}) \in B^n \text{ and } x_i > 0\} \to \Bbb{R}^{n+1}, (x_1, \dots,x_{n+1}) \mapsto (x_1, \dots, x_{i-1},x_i - f(x_1, \dots x_{i-1}, x_{i+1}, \dots x_{n+1}), x_{i+1}, \dots x_{n+1}) .$$

• Thanks. If we're trying to show $S^n$ is an embedded submanifold of $\mathbb{R}^{n+1}$, wouldn't the chart have to be from $\mathbb{R}^{n+1}$? – Clara Aug 7 '14 at 5:37
• Oh, sure. Replace $n$ by $n+1$ everywhere. I will edit the post. – PhoemueX Aug 7 '14 at 5:40
• I'm sorry, but what is the domain of the chart? It's not $\{x:x^i>0\}$ is it? Because $f$ is not defined there, but it's not $B^{n+1}$ either, because $S^n\cap B^{n+1}=\emptyset$? – Clara Aug 7 '14 at 5:43
• I also edited that, sorry :) – PhoemueX Aug 7 '14 at 5:46
• After thinking about your question, I actually think that we have to add the condition $x_i >0$ to the domain. Otherwise, there are points on the sphere (with $x_i <0$ for example), which are not mapped into the "slice". I will edit that too. – PhoemueX Aug 7 '14 at 6:20

I'm also confused by that. But after read Prof. Lee and koch comments i realized that maybe what it means is this :

$\mathbb{S}^n$ covered by the graphs of smooth function. The graph of smooth function is embedded submanifold. But an embedded submanifold satisfy local k-slice condition. So each graph that cover $\mathbb{S}^n$ satisfy local slice condition. Because each point in $\mathbb{S}^n$ is in these graphs, $\mathbb{S}^n$ satisfy local slice condition. Therefore $\mathbb{S}^n$ is embedded submanifold.

• Yes, exactly right. – Jack Lee Sep 22 '17 at 21:19