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January 8

Cardinality

I'm not sure if I've asked this question before. If I have, then forgive me. Given two differentiable manifolds X and Y, I want to know the cardinality of the space of smooth maps from X to Y, in terms of the cardinalities of X and Y. Forgive me if my question doesn't make perfect sense; I'm a novice in set theory. I have an idea that I'm trying to make sense of. I'm trying to express that the space of smooth maps from two-space to three-space is "much bigger" than the space of planes in three-space (which is diffeomorphic to the real projective plane RP²). I'm trying to make sense of "how many more" smooth surfaces there are than planes. Cardinality is an obvious way of doing it, but it seems very crude. Maybe there is another way of doing it? Hopefully you understand what I'm trying to get at. — Fly by Night (talk) 18:21, 8 January 2012 (UTC)[reply]

Cardinality is rarely a useful measure of size, except in discrete mathematics and naive set theory. In the case of smooth functions from one manifold to another, this always has the cardinality of the continuum, as do both X and Y, unless Y is zero dimensional (i.e., discrete points). Indeed, one can see that the cardinality of the set of smooth maps is not less than the continuum by considering just the constant maps into Y. For the opposite inequality, Y can be embedded into a ball in Rⁿ for sufficiently large n (by the Whitney embedding theorem). Smooth mappings from X to Y are thus contained in the separable Hilbert space

L^{2}(X)^{n}

, whose cardinality is the continuum.

There are many more useful ways to measure the "size" of a space than cardinality. Usually these involve introducing some topology on the space of interest, or taking some kind of quotient by an equivalence relation, or both. For instance, rather than counting mappings from X to Y, it might make more sense to count the homotopy classes of mappings. In the specific problem you are interested in, you can define a topology on the space of mappings in terms of which it becomes an infinite dimensional Frechet manifold. You probably also want to quotient by the group of automorphisms of X to get a meaningful comparison to the set of planes in R³, and that will complicate things. But, at any rate, one will be an infinite dimensional space, and the other a finite dimensional one. Sławomir Biały (talk) 11:52, 9 January 2012 (UTC)[reply]

Sławomir Biały has already given a nice answer, but let me expand a bit. I'd like to explain how you can see that the cardinality of smooth maps is at most continuum. The trick is that a manifold X is a separable space, which means that it has a dense subset of countable cardinality. Choose such a dense subset, say K. Now if you have even a map f from X to Y, then consider g = f|_K (the restriction of f to K) uniquely determines f. As this g has to be chosen as a function from a countable set (K) to a set of at most continuum cardinality (Y), the cardinality of such functions is also at most continuum. Note that all we are using here about the maps is that they're continuous, that is, we don't need to restrict to smooth maps. – b_jonas 15:37, 9 January 2012 (UTC)[reply]

Veering slightly off-topic here — whether a manifold has to be separable depends on the details of your definition of "manifold". An example of a non-separable manifold is the long line. Which by the way is one of my favorite topological spaces — I love the way that the construction itself can be pushed as far as you like through the ordinals, but if you go past ω₁, it's no longer a manifold --Trovatore (talk) 10:15, 10 January 2012 (UTC)[reply]

Thanks Sławomir and b_jonas. I've worked out what I need to do. I use the Whitney topologies on the space of map germs from X to Y. Assuming dim(X) < dim(Y), at a point p of X, I can write the image of X locally as the graph of a function. I get a map into a jet space, where any plane gets mapped to a point. — Fly by Night (talk) 17:39, 10 January 2012 (UTC)[reply]

Four color theorem

help me understand this theorem, does it mean that if you use less than four color on a map, two adjacent section will have the same color? If so, next question is does checkered pattern be accepted as a 'map'? becase you can do three colors on checkered pattern. MahAdik _usap 19:20, 8 January 2012 (UTC)[reply]

No it doesn't. It says that every separation of the plane can be coloured (i.e. assigned colour so that no two adjacent sections have the same colour) with at most four colours. The trivial separation, i.e. no boundaries, can be coloured with a single colour. If the split the plane into two or three pieces then you can colour it using two or three colours resp. The contrapositive of the theorem says that if you have less than four colours, say n, then there exists a separation which cannot be coloured with n colours. — Fly by Night (talk) 20:57, 8 January 2012 (UTC)[reply]

The contrapositive is that if a graph is not 4-colorable then it's not planar, or if a map requires more than 4 colors then its regions aren't contiguous. The statement you said, while true, is not equivalent to the four color theorem. Rckrone (talk) 01:45, 9 January 2012 (UTC)[reply]

Actually, that is the wrong contrapositive in this case, Fly By Night is saying that given a separation of the plane, then "If you have four, or more colours, you can colour it." The contrapositive would be, "If you cannot colour it, then you have less than four colours.", which is the same thing, basically, as his sentence (you would certainly be right if he were taking graphs as the basic object." Sorry for being contentious...Phoenixia1177 (talk) 16:44, 9 January 2012 (UTC)[reply]

I agree that the contrapositive depends on exactly how you state the theorem. However, what you said is also not the same as what Fly By Night said. You said that if a map is not n-colorable, then n < 4. Fly By Night's statement is a partial converse to that: if n < 4, then there exists a map that is not n-colorable. The 4 color theorem gives an upper bound on the number of colors needed, while his/her statement is that this bound is tight. Rckrone (talk) 02:09, 10 January 2012 (UTC)[reply]

I, suppose, that you are technically correct, but that the bound would need to be tight is rather obvious. More over, your version of the contrapositive is still off for the reasons I said. That said, I'm starting to feel really nit-picky and am going to stop:-) Phoenixia1177 (talk) 04:40, 10 January 2012 (UTC)[reply]

The issue isn't what statements are obvious, it's about what statements are logically equivalent. The statement "if you have less than four colours, say n, then there exists a separation which cannot be coloured with n colours," is not equivalent to the 4-color theorem. As a result it's not the contrapositive of any formulation of the 4-color theorem. Rckrone (talk) 05:29, 10 January 2012 (UTC)[reply]

...Wow, I'm feeling embarrassed, the entire time I've been reading that as something different. For some reason, my brain kept interpreting it as "If you have less than four colours, only then, and definitely then, do you have a separation that cannot be coloured.", I'm really not sure why (probably not paying attention closely...) At any rate, it seemed like you were complaining about the fact that this would not only claim you could always get by with four colours, but that sometimes they were required...which seemed unreasonable; obviously, you weren't being unreasonable, my apologies. Though, I still stand by my whining about whether you are talking about graphs or maps, etc. Sorry again:-) Phoenixia1177 (talk) 10:22, 10 January 2012 (UTC)[reply]