Defining a manifold without reference to the reals

In summary, the conversation revolves around the definition of a manifold and the difficulties in defining it without invoking the real number system. The speaker presents a set of axioms for defining a manifold that avoids using the reals, but it is pointed out that some of the axioms still indirectly refer to the real line. The conversation ends with the speaker acknowledging the need to find another way to define a manifold without using the real number system.
  • #1
bcrowell
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Most definitions I've seen for a manifold are based on the idea that small neighborhoods are homeomorphic to [itex]\mathbb{R}^n[/itex]. To me this feels a little like defining a bicycle as a car that's missing the engine and both the wheels on one side. The real number system is this big, sophisticated piece of mathematical technology that includes all kinds of metrical apparatus, but the whole idea of a manifold is that it's nonmetrical. I've been trying to work out a definition of a manifold that avoids this unaesthetic feature. I'm not intent on reinventing the wheel, so if someone says "I know a definition that doesn't invoke the reals, it's on page x of book y," that would be great. But anyway, here's what I've come up with so far:

A manifold is a Hausdorff, first-countable topological space T such that:
(M1) Self-similarity: Given any two points in T, any sufficiently small open neighborhoods around these points are homeomorphic.
(M2) Density: T is completely metrizable.
(M3) Not fractal: T's Hausdorff dimension and Lebesgue covering dimension are equal.

([EDIT] This set of axioms has problems. #23 is an attempt at fixing them.)

I intend M1 to rule out things like a a manifold with boundary, or a line glued onto a plane, and M2 to rule out things like the rationals.

Does anyone see anything wrong with the general approach, any fundamental reason why this can't possibly work, even with tinkering to fix specific flaws? Can anyone think of something that satisfies these axioms but isn't a manifold under some standard definition?

The next step would be to see if I can prove that this definition is equivalent other definitions. I think I can basically do this in the case of one dimension, although I haven't rigorously filled in all the steps. My argument doesn't explicitly make use of M3, which makes me think that there may be some hidden flaw in it (unless M3 somehow is only needed in 2 or more dimensions...?). I could sketch my argument, but it might make more sense to get general comments first on whether I'm on the right track, or reinventing the wheel.

Thanks in advance for any help!

-Ben
 
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  • #2
You don't need to talk about the reals if you're only interested in topological manifolds, but a smooth n-manifold is locally like an n-dimensional real vector space (which need not be represented as n-tuples of real numbers). This real vector space structure of the tangent spaces is essential to define smoothness.
 
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  • #3
On second thought, I'm not really sure how to define a topological manifold without invoking the real line in some way...
 
  • #4
Do your axioms ensure that the dimension is a positive integer?
 
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  • #5
dx said:
Do your axioms ensure that the dimension is a non-negative integer?

I think so. If the Hausdorff dimension wasn't a nonnegative integer, then it wouldn't match the Lebesgue covering dimension, which is always a nonnegative integer, so it would fail M3.
 
  • #6
Ya I just realized that. But I think lebesgue dimension can be zero right? Is that allowed for topological manifolds?
 
  • #7
Or, what I should have said is, do your axioms ensure that a space defined as you have with dimension zero imply that the manifold is the trivial one: {}?
 
  • #8
A point is a manifold, and so is any discrete set. (Some cardinality condition may apply if you put countability conditions in your definition of manifold)

The empty set has a dimension more like -1 or -infinity, if it has dimension at all.
 
  • #9
This discussion confuses me a little bit. A manifold is a space that is locally homeomorphic to Euclidean space. It seems then that one way or another you will have to define what this topology is. No matter how you do it though you will end up with the same thing.

So to me the question is not what a manifold is but whether you can define the topology of euclidean space in another way.
 
  • #10
I think the Lebesgue covering dimension is defined so that, as Hurkyl says, the dimension of set of discrete points is zero. I think the Lebesgue covering dimension of an empty set is undefined. The inductive dimension of an empty set is -1.
 
  • #11
How can we describe the topology of a line segment without invoking the real line?
 
  • #12
lavinia said:
So to me the question is not what a manifold is but whether you can define the topology of euclidean space in another way.
I would basically agree with this, with the caveat that we only need to define the local topology of euclidean space in another way.
 
  • #13
dx said:
How can we describe the topology of a line segment without invoking the real line?
Well, I'm claiming that that's what I did in #1, although I'm not sure I got it right. Concretely, it would be interesting if anyone could come up with a counterexample: something that satisfies my definition, and is one-dimensional, but isn't locally topologically the same as an open interval of the real line.

I don't think it would be surprising if we could capture the topological properties of the reals without explicitly building up the whole real number system. After all, Euclidean geometry captures all the properties of the real line without ever mentioning concepts like sets, numbers, or arithmetic operations.

[EDIT] Oops, a couple of mistakes -- added "locally" above, and deleted "and connected".
 
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  • #14
bcrowell said:
I would basically agree with this, with the caveat that we only need to define the local topology of euclidean space in another way.

Euclidean space is homeomorphic to any open ball
 
  • #15
lavinia said:
bcrowell said:
I would basically agree with this, with the caveat that we only need to define the local topology of euclidean space in another way.
Euclidean space is homeomorphic to any open ball
By "in another way," I meant without referring to an already-defined euclidean space, the real number system, or any metrical notions.
 
  • #16
Doesn't "completely metrizable" refer to the real line, via its use of metrics? Hausdorff dimension too?
 
  • #17
Hurkyl said:
Doesn't "completely metrizable" refer to the real line, via its use of metrics?
Yep, you're right, thanks for pointing that out. So if I want to achieve my original goal, I need to find some other way of doing that. I think the basic idea that I need to capture is probably much weaker than complete metrizability, but I don't know enough topology to express it properly. Essentially we don't want curves to be able to cut through each other without having a point in common...?
 
  • #18
Hurkyl said:
Hausdorff dimension too?
Yep, you're right there as well.
 
  • #19
To capture completeness, I don't think it's necessary to have the reals defined already. Otherwise it would be a tautology to say that the reals are complete. I think the right notion is probably that it's a complete uniform space: http://en.wikipedia.org/wiki/Uniform_space#Completeness

I'm sure there must be some simple way to define the distinction between fractal dimensions and non-fractal dimensions, without referring to a metric or the real number system, but I don't know what it is.
 
  • #20
I have a question: does Lebesgue covering dimension = 1 imply that the space is topologically a line (locally)?
 
  • #21
bcrowell said:
By "in another way," I meant without referring to an already-defined euclidean space, the real number system, or any metrical notions.

Why would you not want to construct the 1 dimansional manifold then get the others through Cartesian products?
 
  • #22
dx said:
I have a question: does Lebesgue covering dimension = 1 imply that the space is topologically a line (locally)?

I don't think so. The real line is just one example of a one-dimensional space. There are lots of others, like the rationals, the long line, etc.

lavinia said:
Why would you not want to construct the 1 dimansional manifold then get the others through Cartesian products?

I guess locally that might work, although globally, e.g., the torus isn't just a product of two lines.
 
  • #23
OK, here's another shot at defining a manifold without referring to the reals.

An n-manifold is a normal, Hausdorff, second-countable topological space T such that:
(N1) All points alike: T is a homogeneous space under its own homeomorphism group.
(N2) Completeness: T is a complete uniform space.
(N3) Dimension: T's Lebesgue covering dimension is n.

I'm pretty sure that N1-N3, unlike M1-M1 from #1, is free of any hidden references to the reals.

I think N1 is stronger than M1, and I think it's strong enough to rule out fractals. Every fractal I can think of has some points that have different topological properties than other points. Urysohn's theorem ( http://en.wikipedia.org/wiki/Inductive_dimension#Relationship_between_dimensions ) applies, and tells us that several different measures of the dimension agree on n. This is also a good sign in terms of ruling out fractals; I'm not sure, but I think that the dimensions referred to in Urysohn's theorem would probably disagree in the case of a fractal.

There is something called the Nöbeling-Pontryagin theorem ( http://en.wikipedia.org/wiki/Inductive_dimension#Relationship_between_dimensions , http://eom.springer.de/d/d032500.htm ) that allows us to say that T is homeomophic to a subspace of a Euclidean space. That doesn't immediately show that it is homeomorphic to a Euclidean space, but it is a good sign. I think it rules out things like the long line. Although the rationals, for example, are a subspace of the real line, I think N2 rules those out.

I think under these axioms I can sketch a proof that a 1-manifold is locally homeomorphic to the real line. The idea is that N3 allows us to make a cover with only 2-fold overlaps. We can arbitrarily pick one point from each of these regions of overlap, and the cover then induces an ordering on these points, so we can identify the points with some integers. We then take the open set lying between two integers and repeat the process, identifying the new points with fractions. By N2, we can continue this process indefinitely, and we won't run out of points before we get infinite sequences of fractions, which can be identified with reals.
 
  • #24
It would be very interesting to see how the topological characteristics of a line can be captured using only the terminology of point set topology, though I'm in no position to attack this since I have little understanding of the various notions of topological dimension. If you figure it out, do let me know.
 
  • #25
dx said:
It would be very interesting to see how the topological characteristics of a line can be captured using only the terminology of point set topology, though I'm in no position to attack this since I have little understanding of the various notions of topological dimension. If you figure it out, do let me know.

Thanks, dx, lavinia, and Hurkyl, for your help with debugging my initial axiomatization! I think the one in #23 may be more free of obvious flaws, but I'm going to need to solidify my very casual knowledge of topology a lot more before I can make any further progress.
 
  • #26
I think I've made some progress on this, and since others here expressed interest, I thought I'd post an update.

My axiomatization in #23 was too weak, because (I think) it would have allowed the Cantor set as a 1-manifold. So because of this, I think I need to add a condition that the space be locally connected.

I've been worried about excluding fractals, but it occurs to me that not all fractals are necessarily bad. I could be wrong, but it seems to me that many fractals are fractal only because a certain metric is associated with them, not because of any inherent topological property. For example, I think the Koch curve is homeomorphic to the real line.

For the one-dimensional case, I found out that it is indeed possible to give a purely topological characterization of the real line. Franklin and Krishnarao proved ca. 1971 that "a connected and locally connected separable and regular space, in which every point is a strong cut point, is homeomorphic to the real line." Here a cut point means a point such that if you delete the point, you split the space into two parts. This is not exactly suitable for my purposes, since it is much too specific (rules out other 1-manifolds besides the line, doesn't deal with more than 1 dimension). Nevertheless, I think it shows that it's reasonable to hope to be able to characterize a manifold purely topologically.

My latest formulation is this:

An n-manifold is a completely normal, second-countable, locally connected topological space T such that:
(M1) All points alike: T is a homogeneous space under its own homeomorphism group.
(M2) Completeness: T is a complete uniform space.
(M3) Dimension: T's Lebesgue covering dimension is n.

What follows immediately from this:
Hausdorff, normal, and regular (because completely normal).
Metrizable (by Urysohn's thm, because 2nd countable, regular, and Hausdorff).
Separable, i.e., there's a countable dense set (because because 2nd countable).
Paracompact (because metrizable).
Lindelof.
Small and large inductive dimensions agree with Lebesgue covering dimension (by Urysohn's theorem, which requires normal and 2nd-countable, http://en.wikipedia.org/wiki/Inductive_dimension).
Can be embedded in a Euclidean space (Nöbeling-Pontryagin theorem).

I've looked through Steen and Seebach, "Counterexamples in Topology," for counterexamples. The following are not counterexamples:
29 - Cantor set: not locally connected
46,47 - long line: not separable
51 - right half-open interval topology: not second-countable
116-118 - topologist's sine curve, and variants; violate M1, because origin is special, and also not locally connected
119-122 - brooms; violate M1, because they have a special point
 

Related to Defining a manifold without reference to the reals

1. What is a manifold?

A manifold is a mathematical space that is locally similar to Euclidean space, meaning that it looks flat when examined closely. It can be described as a topological space that is locally homeomorphic to Euclidean space.

2. What does it mean to define a manifold without reference to the reals?

When defining a manifold without reference to the reals, we are essentially describing the structure and properties of the manifold without using any real numbers or coordinates. This approach allows for a more abstract and general definition of a manifold that can be applied to different mathematical contexts.

3. How is a manifold defined without reference to the reals?

A manifold can be defined without reference to the reals using the concept of charts and atlases. A chart is a function that maps a subset of the manifold onto an open subset of Euclidean space. An atlas is a collection of charts that cover the entire manifold and satisfy certain compatibility conditions.

4. What are some advantages of defining a manifold without reference to the reals?

Defining a manifold without reference to the reals allows for a more general and abstract understanding of the concept. It also allows for the possibility of studying manifolds in different mathematical contexts, such as in algebraic geometry or differential geometry.

5. Are there any limitations to defining a manifold without reference to the reals?

One limitation is that it may be more difficult to visualize or work with manifolds that are defined in this way, as they do not have a specific set of coordinates or numbers associated with them. Additionally, some concepts and theorems in manifold theory may rely on the use of real numbers and coordinates, making them more challenging to apply in this general setting.

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