# A question on an exercise to show that unit sphere could not be covered by a single chart

The following is an exercise from Marsden et al, Manifolds, Tensor Analysis, and Applications in the first chapter on manifolds. First let me cite three essential definitions:

Definition 1: Let $S$ be a set. A chart on $S$ is a bijection $\varphi$ from a subset $U$ of $S$ to an open subset of a Banach space. We sometimes denote $\varphi$ by $(U,\varphi)$, to indicate the domain $U$ of $\varphi$. A $C^k$ atlas on $S$ is a family of charts $\mathcal A = \{ (U_i, \varphi_i) : i \in I \}$ such that:

(i) $S = \bigcup \{ U_i : i \in I \}$

(ii) Any two charts in $\mathcal A$ are compatible in the sense that the overlap maps between members of $\mathcal A$ are $C^k$ diffeomorphisms: for two charts $(U_i, \varphi_i)$ and $(U_j, \varphi_j)$ with $U_i \cap U_j \ne \emptyset$, we form the overlap map: $\varphi_{ji} = \varphi_j \circ \varphi_i^{-1}{|\varphi_i(U_i \cap U_j)}$, where $\varphi_i^{-1}{|\varphi_i(U_i \cap U_j)}$ means the restriction of $\varphi_i^{-1}$ to the set $\varphi_i(U_i \cap U_j)$. We require that $\varphi_i(U_i \cap U_j)$ is open and that $\varphi_{ij}$ be a $C^k$ diffeomorphism.

Definition 2: Two $C^k$ atlases $\mathcal A_1$ and $\mathcal A_2$ are equivalent if $\mathcal A_1 \cup \mathcal A_2$ is a $C^k$ atlas. A $C^k$ differentiable structure $\mathcal D$ on $S$ is an equivalence class of atlases on $S$. The union of the atlases in $\mathcal D$, $\mathcal A_D = \bigcup \{ \mathcal A : \mathcal A \in \mathcal D \}$ is the maximal atlas of $\mathcal D$, and a chart $(U,\varphi) \in \mathcal A_{\mathcal D}$ is an admisssible local chart. [...] A differentiable manifold $M$ is a pair $(S, \mathcal D)$ [...]

Definition 3: Let $M$ be a differentiable manifold. A subset $A \subseteq M$ is called open if for each $a \in A$ there is an admissible local chart $(U,\varphi)$ such that $a \in U$ and $U \subseteq A$.

So according to Definition 1, the manifold has no topological structure, this is in some way induced by all possible charts, see Definition 3. As Definition 1 ensures that all compatible charts have the some Banach space as an image (but guess this should also be added to the definition in case $U_i \cap U_j = \emptyset$), this in some way "transfers" the topological structure of this Banach space to the manifold.

Now in an exercise to the chapter introducing all these notions, it is asked to prove that $S^n$ can not be covered by a single chart. I was wondering, by the above definitions (as there could be no topological structure assumed on $S^n$ a priori, so it does not has to be the standard one) this essentially ask that there could be no single bijection from $S^n$ to any Banach space. This question I asked a few minutes ago, and yes this certainly is not true.

So is this exercise wrongly stated, should it read "no single chart contained in a maximal atlas inducing the standard topology on $S^n$", or have I overlooked something?? As I see it any bijection would give some topology on $S^n$, i.e. the initial topology (despite that might be not very useful). But maybe I have overseen some subtle consequence of the definitions restricting the possible topologies further?

• Yes, what you quote does assume that you consider the usual topology of the sphere. Otherwise, only cardinality would matter and of course any sphere is equipotent to any euclidian space. But then there would be no point calling it "the sphere", it would just be any set with the cardinality of $\mathbb{R}$. – Captain Lama Mar 20 '16 at 22:41
• Ok, so I got it right that the definitions did not restrict the topology on the set $S$ further then what is said by Definition 3 (which I guess just says that we impose the initial topology on the set as given by a maximal atlas and its charts)? – StefanH Mar 20 '16 at 22:44
• By the way, note in Definition 2 that it is not obvious without further proof that equivalence-of-atlases is an equivalence relation! Neither is it obvious that $\mathcal{A_D}$ is an atlas at all. – Henning Makholm Mar 20 '16 at 23:02
• @HenningMakholm That $\mathcal A_{\mathcal D}$ is an atlas follows as two charts from $\mathcal A_{\mathcal D}$ are contained in two atlases from $\mathcal D$, hence the are compatible as they are both equivalent, too simple of an argument as this seems obvious to me? – StefanH Mar 20 '16 at 23:11
• @Stefan: Well, both properties follow from equivalence-of-atlases being a transitive relation, but it is not quite trivial to see given the definition of "equivalent" that it would be. (Though of course it is true). – Henning Makholm Mar 20 '16 at 23:39

## 1 Answer

To some extent you're caught between different but equivalent ways of setting up the definitions.

The one you quote is one way to go, but it is also possible (and arguably more intuitive) to require that a manifold must be a topological space first before you start making charts and atlases for it. Going that way, one would define a chart to be a homeomorphism between an open subset of the manifold and an open subset of a Banach space, and adjust the definition of "compatible" charts accordingly. Your definition 3 would then instead become a (quite trivial) theorem.

Viewed in this light, the exercise asks for proof that there is no chart defined on all of $S_n$ which induces the usual subspace topology it inherits from $\mathbb R^{n+1}$.

Alternatively, the exercise may be asking for proof that $S^3$ viewed as a manifold, that is, with an atlas defining its usual differentiable structure already given, does not admit any chart defined on all of $S^3$.

(A proof of the former will imply the latter, of course).