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In the definition of a parametrised surface $S$, for every point in the surface, $p \in W \subseteq S$, where $W$ is open, there exists a coordinate chart or patch , $F :U\to \mathbb{R}^n$ that maps to $p$ from an open subset $U \in \mathbb{R}^n$

Is that right? If anyone knows of a more general definition, I'm willing to learn. It sounds a lot like a manifold, which I'm not entirely familiar with.

In this definition, the number of surface patches in the atlas is not stipulated. Given a parametrisable surface, is a finite number of charts sufficient to describe the surface? Can we find a surface that requires infinitely many patches to fully chart? If so, in which $\mathbb{R}^n$ does the first such surface occur? In which dimensions is it always possible to find a finite number of patches for any given surface?

EDIT: Added requirement that such a surface (manifold) be connected. One made from infinitely many disconnected subsets would have to be charted infinitely.

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You keep writing "surface" but then speak about different dimensions. The word "surface" usually indicates a 2-dimensional manifold. – Jason DeVito Oct 25 '11 at 0:54
@JasonDeVito. Perhaps I should talk about manifolds instead. The only manifolds I've encountered so far are surfaces; I'm not sure how to generalise. – Samuel Tan Oct 25 '11 at 2:16
This is just intuition working, but I'm thinking that the surface of infinite genus requires infinitely many charts to specify an atlas. – Jonathan Gleason Oct 25 '11 at 2:18
If you allow your coordinate charts to be arbitrary open subsets of $\mathbb R^n$, which is how the OP stated it, then you can cover a surface of infinite genus by four charts. You can think of such a surface as an infinitely long strip with infinitely many dual pairs of bands attached, to which an infinitely long strip is attached to the boundary (which consists of two lines). The four charts are a neighborhood of the first strip, a neighborhood of one half the bands (which is disconnected), a neighborhood of the other half, and finally the last strip. – Grumpy Parsnip Oct 25 '11 at 2:46

3 Answers 3

up vote 12 down vote accepted

I think for a compact connected surface (a 2-dimensional manifolds) you only need three charts to cover the manifold. The rough idea is like this: think of a minimal CW-decomposition of the surface. Your first chart is a regular neighbourhood of the maximal forest in the 1-skeleton. The 2nd chart will be the union of the regular neighbourhoods of the remaining 1-cells, plus little arcs that connect them all up. Your remaining chart will be the union of the interiors of the 2-cells, with little arcs connecting them together.

Similarly, I think all compact 3-manifolds can be covered by 4 or 5 charts.

You can get lower bounds on the number of charts needed for any manifold by cup product arguments -- see:

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Oh yeah, I guess you don't need four for surfaces like I thought. I didn't need to divide the 1-handles into two groups. – Grumpy Parsnip Oct 25 '11 at 2:53
A few questions... ...what is a CW-decomposition? What does it mean for it to be minimal? Forest and skeleton? I'm guessing that the CW-decomp is made of 1-cells and 2-cells joined with little arcs? Maybe I should ask another question. I wonder why they are called 'little' arcs? – Samuel Tan Oct 25 '11 at 5:57
@Samuel: I suggest trying out a Google search for "CW complex". It's an old theorem of Whitehead's that smooth manifolds admit triangulations -- these are a special type of CW-decomposition. CW-decompositions and handle decompositions are one of the more relevant tools to use in answering your question. – Ryan Budney Oct 25 '11 at 23:06
Your guess for threefolds does work: see my answer. – Mariano Suárez-Alvarez Dec 20 '11 at 7:28

If $M$ is an $n$-dimensional manifold, then it is in particular a normal space of covering dimension $n$, and Ostrand's theorem tells us that that if we start with a locally finite open covering $\mathcal U$ of $M$ consisting of coordinate charts (such a thing exists), there is a refinement $\mathcal V$ of $U$ such that $\mathcal V$ is the union of $n+1$ subfamilies $\mathcal V_1$, $\dots$, $\mathcal V_{n+1}$ such that each $\mathcal V_i$ is a disjoint family.

If we allow for non-connected coordinate charts, then by looking at the union of each $\mathcal V_i$ we obtain an atlas consisting of $n+1$ charts. (If we insist on connected charts, we should be able to connect the components with thing open tubes...)

This gives Ryan's bound for $n=3$, for example.

Later. One knows that the clique number of a graph is a lower bound for the chromatic number: Ostrand's theorem says that up to refinement, equality can be achieved if the graph comes from a covering of a normal space. This is a very nice result!

share|cite|improve this answer any connected n-dimensional manifold can be charted with $n+1$ connected charts, or $n$ not necessarily connected charts. The result you claimed to be nice is probably quite pretty, but I unfortunately lack the tools to comprehend it. Thanks anyway. – Samuel Tan Dec 20 '11 at 23:40

In face, any triangulated $n$-manifold (not necessarily compact) can be covered by $n$ coordinate charts! However, this is a bit of a cheat — the elements of your atlas will be disconnected (in particular, this does not conflict with the bounds that Ryan Budney discussed coming from the Lusternik-Schnirelmann category, which come from connected covers).

I don't know a good reference for this, but here's the idea. Choose a triangulation of your manifold. For each vertex $v$ of your triangulation, choose a neighborhood $U^0_v$ such that $U^0_v$ is homeomorphic to a ball and such that $U^0_v \cap U^0_w = \emptyset$ for distinct vertices $v$ and $w$. Let $U^0$ be the union of all the $U^0_v$. It is clear that $U^0$ is homeomorphic to the disjoint union of countably many open balls in Euclidean space.

Next, for each edge $e$ of your triangulation, choose a neighborhood $U^1_e$ of $e \setminus U^0$ such that $U^1_e$ is homeomorphic to a ball and such that $U^1_e \cap U^1_f = \emptyset$ for distinct edges $e$ and $f$ (draw yourself a picture to convince yourself this is possible!). Let $U^1$ be the union of all the $U^1_e$. Again, it is clear that $U^1$ is homeomorphic to the disjoint union of countable many open balls in Euclidean space.

The pattern is now clear — repeat this procedure on the the $2$-simplices, the $3$-simplices, etc. The result is that $M = U^0 \cup U^1 \cup \cdots \cup U^n$.

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That's a very neat result! So the minimum number of charts required for an $n$-manifold is just $n$!. I can understand that, and that only--the rest is a little beyond me. I know that a triangulation of the manifold is where you divide the manifold up using triangles, but I'm not sure how the disjoint neighbourhoods of each vertex homeomorphic to $\mathbb{R}^n$ and the neighbourhood of an edge (just points with distance $\epsilon$ away from the edge?) tie in with simplices. Your result looks stronger than Ryan's, who seems to say $n+1$ charts are needed for an $n$-manifold. – Samuel Tan Oct 25 '11 at 6:00
I wonder what happens for manifolds that don't have a triangulation? Explicit examples are known in dimension 4. – Grumpy Parsnip Oct 25 '11 at 13:00
@Jim : I have no idea what happens for manifolds that can't be triangulated, though if I had to make a guess I would expect that it could be done for them too (maybe using the Kirby-Siebenmann topological handle decomposition of such a manifold). – Adam Smith Oct 25 '11 at 14:46
@SamuelTan : Our results are different in that he requires his charts to be connected while I don't. – Adam Smith Oct 25 '11 at 14:47
@JimConant : Addendum : I guess the topological handle decomposition only works in dimension 5 and greater. I have no idea what happens in dimension 4. – Adam Smith Oct 25 '11 at 15:25

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