# Isn't there a better way to put the canonical smooth structure on the $n$-sphere?

My differential geometry notes put a smooth structure on the $n$-sphere (denoted $S_n$) as follows. Firstly, $S_n$ is taken as a subset of $\mathbb{R}^{n+1}.$ Secondly, we define $U_0 = S_n \setminus \{(1,0,\ldots,0)\}$ and $U_1 = S_n \setminus \{(-1,0,\ldots,0).$ Charts with domain $U_0$ and $U_1$ ere then defined by stereographic projection.

I have a couple of issues with this:

• Its non-systematic; stereographic projection is "pulled out of hat" arbitrarily.

• Its non-explicit; it remains to figure out which charts on $S_n$ are actually compatible with this smooth structure.

Isn't there a better way of doing this?

• I don't understand what you mean by "it's non-explicit". It's not often that you can define a manifold using only two charts with completely explicit formulas and transition maps (you can literally write down the projections in terms of coordinates). Do you have an example of an "explicit" definition of a manifold? And I'd say that the stereographic projections are as "arbitrary" as deciding to put a manifold structure on a sphere... $S^n \cong \mathbb{R}^n \cup \{\infty\}$ is a very fundamental fact... – Najib Idrissi Jun 11 '15 at 14:01
• I agree with Najib. Also, why do you say that the stereographic projection is "non-systematic"? It has a very nice geometric interpretation. One could even say that it is the most natural projection of (part of) $S_n$ to $\Bbb{R}^n$, because $S_n$ is the one-point compactification of $\Bbb{R}^n$, where the inclusion is given by the inverse of the stereographic projection. – A.P. Jun 11 '15 at 14:10
• I am still not sure that I understand what you mean. Isn't that the definition of a chart? Maybe you're looking for something like diffeological spaces (where smooth structures are described by plots instead of charts)...? (Sorry if I come off as confrontative, I'm not being purposefully obtuse...) Another thought: you can view $S^n$ as the quotient of $\mathbb{R}^{n+1} \setminus 0$ (open subset of an Euclidean space) under the action of $\mathbb{R}_{>0}$. It's not a properly discontinuous action, but it's well-behaved enough... – Najib Idrissi Jun 11 '15 at 14:11
• How is $U\to\varphi(U)$ supposed to be smooth before the manifold has any smooth structure on it in the first place? The charts are being used to define the smooth structure, so you can't define the charts in terms of the smooth structure. The correct definition is that the transition maps are smooth, right? (Of course, after one finishes putting the smooth structure on the manifold, the charts are indeed smooth. But that's an automatic consequence.) Also, both of the bullet points in your question are plain wrong. – anon Jun 11 '15 at 14:19
• When defining a smooth structure on $S_n$ you shouldn't care about the structure of the space you are viewing it in. I believe that what you mean by an atlas of "charts compatible with the smooth structure of $\Bbb{R}^n$" is a submanifold structure on $S_n$. Sure, this structure does indeed make $S_n$ a smooth submanifold of $\Bbb{R}^n$ and we do have to prove it. We would still need to prove it for any other structure, though. – A.P. Jun 11 '15 at 14:19

You can apply to $f(x)=\sum x_i^2-1$ the theorem that the zero set $S$ of a smooth function $f$ defined on $\mathbb R^N$ is a smooth manifold as soon as $grad (f)$ is never zero on $S$.

In general the corresponding charts are deduced from the implicit function theorem and are thus not very explicit :-) But:
In the case of the sphere $S_n$ you have an atlas consisting of $2n+2$ very explicit charts.
Namely define (for $i=1,\cdots,n+1$): $$U_i^+=\{x \in S_n|x_i> 0\} \quad (\operatorname {resp.} U_i^{-\vphantom{|}}=\{x \in S_n|x_i\lt 0\})$$ and put $$\phi_i^+(x)=(x_1,\cdots,\hat {x_i},\cdots, x_n) \quad (\operatorname {resp.} \phi_i^{-\vphantom{|}}(x)=(x_1,\cdots,\hat {x_i},\cdots, x_n))$$ .

A subtle point
Given a subset $A\subset \mathbb R^N$ the question "Is $A$ a smooth submanifold of $\mathbb R^N ?$" is perfectly well posed and has an unambigous answer, which is (of course) "yes!" or "no!"
Let me emphasize that the answer has nothing to do with charts and atlases and is perhaps paradoxically easier to explain to someone who has never heard of abstract manifolds, defined as topological spaces endowed with a family of charts blah blah blah.
This point of view is explained on page 1 (!) of Topology from the Differentiable Viewpoint, a 64 page pamphlet by John Milnor, one of the most remarkable differential/algebraic topologists of all times.
That said, if the answer to the above question is "yes" (and the first two lines of my answer describe an example where this is the case) there is of course a canonical way to endow the smooth submanifold $A\subset \mathbb R^N$ with the structure of an abstract manifold with its required charts.
Still I find it healthy to be aware that there is a very low-tech notion of differentiable submanifold of $\mathbb R^n$, which logically and historically predates the abstract concept of manifold.

Edit: just watch!
Here is a wonderfully nostalgic video where Milnor defines the concept of submanifold at 13:53.

• Ah thanks, I had a feeling there was a theorem like that. What are the charts though? – goblin Jun 11 '15 at 14:26
• If a submanifold has a smooth structure such that the map $S \hookrightarrow M$ is a smooth embedding, then $S$ has a unique smooth structure, defined by slice charts: around every point in $S$ there is a chart $(U,\varphi)$ on $M$ such that $\varphi(S)$ is a level set in $\Bbb R^n$ (contained in some 'vertical' hyperplane $\mathbb R^{n-q} \times \{pt\}$. Restricting gives you your charts on $S$. – user98602 Jun 11 '15 at 14:32
• Could some expert please explain how to put a minus sign as an exponent to a symbol : my attempts $U_i^- , \phi_i^-$ are quite ugly. – Georges Elencwajg Jun 11 '15 at 15:02
• @Georges: This appears to be a problem with MathJax; U^{-} and U_i^{-} have their exponents at (almost) the same heights when used in LaTeX (picture link for reference), but in MathJax we get the abnormally low exponent on the latter: $U^{-}$ vs $U_i^{-}$ (picture link for reference). One workaround would be to use a \vphantom to give the exponent a better height, with maybe x or | as the argument; U_i^{-\vphantom{|}} produces $U_i^{-\vphantom{|}}$. A \strut would work also, but it is a little too large. – Zev Chonoles Jun 11 '15 at 15:43
• Hey that video is great! – goblin Jun 12 '15 at 5:48

Given a set of charts $\mathcal{U} = \{(U_{\alpha}, \phi_{\alpha})\}_{\alpha \in A}$ that covers a topological manifold $M^m$ (that is, any $x \in M$ is in some $U_{\alpha}$) and such that each transition map $\phi_{\alpha} \circ \phi_{\beta}^{-1} : \phi_{\beta}\left( U_{\alpha} \cap U_{\beta} \right) \to \phi_{\alpha}\left( U_{\alpha} \cap U_{\beta} \right)$ is smooth (as a function from an open set of $\mathbb{R}^m$ to another), we know from general principles that there exists a unique maximal atlas $\mathcal{A}$ for $M$ such that $\mathcal{U} \subseteq \mathcal{A}$. It is in fact the set of all pairs $(U, \phi)$ which are compatible with the charts in $\mathcal{U}$ (in the sense that all of the transition functions $\phi_{\alpha} \circ \phi^{-1} : \phi\left( U_{\alpha} \cap U \right) \to \phi_{\alpha}\left( U_{\alpha} \cap U \right)$ are diffeomorphisms). Indeed, compatibility between all of these added pairs is automatic. Depending on our knowledge and on the simplicity of $\mathcal{U}$, this might give an effective criterion to determine whether a given pair $(U, \phi)$ is in $\mathcal{A}$ or not.

In the context of your question, things are somewhat even better.

Let's think of $S^n$ as the unit sphere in $\mathbb{R}^{n+1}$. For $P \in S^n$, define $U_P = S^n \backslash \{ P \}$ and write $\sigma_P : U_P \to \mathbb{R}^n$ for the stereographic projection from the "P-pole" onto the tangent plane to $S^n$ at $-P$ (some smooth and standarized identification between this tangent plane and $\mathbb{R}^n$ is chosen).

You took $\mathcal{U} = \{(U_{N}, \sigma_{N}), (U_{S}, \sigma_{S}) \}$ where $N$ and $S$ denote the north pole and the south pole respectively. We want to identify what is the maximal atlas $\mathcal{A}$ associated to $\mathcal{U}$. Some computation shows that the set $\mathcal{V} = \{(U_{P}, \sigma_{P})\}_{P \in S^n}$ is in $\mathcal{A}$.

Now, any $(U_{\alpha}, \phi_{\alpha}) \in \mathcal{A}$ is such that $U_{\alpha} \subset S^n$ is an open but not closed subset of $S^n$ ; In particular, it is not all of $S^n$. Hence, there exists $P_{\alpha} \in S^n \backslash U_{\alpha}$ ; This means that $U_{\alpha} \subseteq U_{P_{\alpha}}$. We conclude that the transition map

$$\phi_{\alpha} \circ \sigma_{P_{\alpha}}^{-1} : \sigma_{P_{\alpha}} \left( U_{\alpha} \right) \to \phi_{\alpha}\left( U_{\alpha} \right)$$

is a diffeomorphism between open sets of $\mathbb{R}^n$ (equipped with its standard smooth structure). It is not hard to see that this transition map determines completely $(U_{\alpha}, \phi_{\alpha})$ as a function ; What's more is that any such diffeomorphism between open sets of $\mathbb{R}^n$ determines an element $(U_{\alpha}, \phi_{\alpha}) \in \mathcal{A}$ (here, we use the fact that $\mathcal{V} \subset \mathcal{A}$). This yields the following criterion :

"A function $\phi : U (\subset S^n) \to \mathbb{R}^n$ is an element of $\mathcal{A}$ if and only if there exists $P \in S^n \backslash U$ such that $\phi \circ \sigma_P^{-1} : \sigma_P(U) \to \phi(U)$ is a diffeomorphism with respect to the standard smooth structure on $\mathbb{R}^n$."

This shows that the knowledge of a "good" atlas such as $\mathcal{V}$ gives an even simpler criterion for compatibility.