# Least number of charts to describe a given manifold

Hello, I'm wondering if there is a standard reference discussing the least number of charts in an atlas of a given manifold required to describe it.

E.g. a circle requires at least two charts, and so on (I couldn't manage to get anything relevant neither on wikipedia nor on google, so I guess I'm lacking the correct terminology).

Edit: in the case of an open covering of a topological space by n+1 contractible sets (in that space) then n is called the Lusternik-Schnirelman Category of the space, see Andy Putman's answer. The following book seems to be the standard reference http://books.google.fr/books?id=vMREfNN-L4gC&pg=PP1

Great, now I'm still interested by the initial question: does anybody know of another theory without this contractibility assumption (hoping that it allows more freedom)? e.g. would it lead to different numbers say for genus-g surfaces?

Final edit: yes different numbers for genus-g surfaces (see answers below), but not sure there is a theory without contractibility. Right, really lots of interesting literature on the LS category nevertheless, hence the accepted answer. For example there are estimates for non-simply connected compact simple Lie groups like PU(n) and SO(n) in Topology and its Applications, Volume 150, Issues 1-3, 14 May 2005, Pages 111-123.

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It's not quite the same thing, but a related object is the Lyusternik–Schnirelmann category of a topological space. See

http://en.wikipedia.org/wiki/Lyusternik-Schnirelmann_category

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It's pretty close, thanks! –  Thomas Sauvaget Oct 26 '09 at 14:25
Is there some reason this has the same name as the "category" that has objects and morphisms, or is this purely a case of the same English word getting used for two mathematical objects? –  Michael Lugo Oct 26 '09 at 14:34
The concepts are unrelated. The notion of the L-S category predated the "category-theoretic" notion of a category. I suspect that its inspiration comes from the Baire category theorem. –  Andy Putman Oct 26 '09 at 14:47

To answer your last question, the least number of charts needed to cover any orientable 2-manifold is 2. Consider the usual embedding of an orientable surface Σ in R3 which is symmetric across the plane z = 0 (as shown here), and let ε > 0 be sufficiently small. The open subsets Σ ∩ {z > -ε}, Σ ∩ {z < ε} form a covering of Σ by charts: by Morse theory Σ ∩ {z > -ε} is diffeomorphic to Σ ∩ {z > ε}, which is diffeomorphic to an open subset of R2 by projecting onto the xy-plane.

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Note that the two open sets in Reid's answer may not be made contractible in general. Indeed, for orientable 2-manifolds, 3 contractible sets are required, except for S^2, where 2 are required. –  j.c. Oct 26 '09 at 17:53
Yeah, I figured out after asking, so really the LS category measures something different, thanks! –  Thomas Sauvaget Oct 26 '09 at 18:06

I have found on the second page of Michor "Topics in Differential Geometry": "Note finally that any manifold $M$ admits a finite atlas consisting of $\dim{M}+1$ not connected charts. This is a consequence of topological dimension theory [cf. Nagata, Modern Dimension Theory]; a proof for manifolds may be found in [cf. Greub, Halperin, Vanstone, Connections, curvature and cohomology.I]."

I hope to have been useful.

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Yes, sounds like interesting references, and the result is intuitively satisfying, thanks! I'll try to find a library that has them... –  Thomas Sauvaget Feb 5 '11 at 21:19

After "dimension" this is the most basic numerical invariant of a manifold and the least explored. I found this reference some years ago: I. Bernstein, "On Imbedding Numbers of Differentiable Manifolds", Topology, Vol. 7, pp. 95-109.

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Wow, thanks for pointing out this reference! Mike Hopkins has also written on this concept: springerlink.com.ezproxy.webfeat.lib.ed.ac.uk/content/… In particular I think he answers the OP's question for (almost) all projective spaces. –  Mark Grant Jun 17 '11 at 12:52

Orthogonal question: Does the (minimum) number of charts needed to describe a manifold tell you anything about the manifold?

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Is this appropriate as an answer? I would have commented on the question if I had the rep. –  Sonia Balagopalan Oct 26 '09 at 16:15
For the L-S category, having an upper bound on the category can bound the complexity of your manifold (e.g. in terms of cohomology) from above –  j.c. Oct 26 '09 at 17:31
The L-S category gives a lower bound on the number of critical points of a functional on your manifold. That's the big abstract theorem that Lusternik and Schnirelmann proved. –  Jeff Strom Jul 29 '10 at 13:38

I believe that Cech cohomology could yield the sort of answer you're looking for (at least in the contractible case). The general idea is that it computes cohomology based on nothing but the so-called "incidence data" of a good cover (that is, which n-fold intersections of open sets in the cover are nonempty) -- in fact, a n-chain is nothing more than a formal sum of (nonempty) (n+1)-fold intersections, with R coefficients. Of course, this cohomology theory is usually isomorphic to singular cohomology, de Rham cohomology, et al. (in particular, they agree in the case of manifolds). So if your manifold has a high h1=rank(H1), for example, then there must be lots of different 2-fold intersections to generate H1, and also if your manifold has nonzero Hk then there must exist a (k+1)-fold intersection of sets in the open cover, which means that you must have at least that many sets in any good cover. (Note that if M is a k-dimensional orientable manifold, then Hk(M)=R by Poincare duality.)

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Cech cohomology is too weak to give a full answer, though, because it only gives you a bound on how to cover a space such that all nonempty intersections are contractible, rather than just the individual open sets themselves. For example, a sphere S^n can be covered by 2 contractible sets, even though it takes n+1 to give a good cover. Also, you could have an acyclic space where you see no cohomological obstructions to the space itself being contractible, but \pi_1 is nonzero so you certainly need at least two contractible sets to cover. –  Eric Wofsey Oct 26 '09 at 19:48
Right. (I also just realized that my last statement is clearly false; e.g., the unit disk has trivial cohomology.) But maybe there's a weaker Cech-ish cohomology that can address this anyways. What happens if we work with a complex where only the individual open sets must be contractible? It seems possible that there's a statement along the lines of "the cohomology of this complex injects into the cohomology of a good cover", and maybe we could get a lower bound on the difference in rank (based on higher-dimensional cohomology of M and/or the cohomology of the nonempty intersections...?). –  Aaron Mazel-Gee Oct 27 '09 at 5:03