The standard definition I've seen for a manifold is basically that it's something that's locally the same as $\mathbb{R}^n$, without the metric structure normally associated with $\mathbb{R}^n$. Aesthetically, this seems kind of ugly to me. The real line is a high-tech piece of mathematical machinery. We build up all that structure, then build the definition of a manifold out of it, then throw away most of the structure. It seems kind of like building an airplane by taking a tank, adding wings, and getting rid of the armor and the gun turret.

I've spent some time trying to figure out a definition that would better suit my delicate sensibilities, and have come up with the following sketch: An $n$-dimensional manifold is a completely normal, second-countable, locally connected topological space that has Lebesgue covering dimension $n$, is a homogeneous space under its own homeomorphism group, and is a complete uniform space.

Does this work? I should reveal at this point that I'm a physicist, and no more than a pathetic dilettante at math. I have never had a formal course in topology. My check on my proposed definition consisted of buying a copy of Steen's Counterexamples in Topology and flipping through it to try to find examples that would invalidate my proposed definition.

Since I'm not competent as a mathematician, what would probably be the best outcome of this question would be if someone could point me to a book or paper in which my idea is carried out by someone competent.

Clarification: I mean a topological manifold, not a smooth manifold.

Also, I should have mentioned in my original post that I had located some literature on the $n=1$ case in terms of characterizing the real line (which is not, of course, the same as characterizing a 1-dimensional manifold, but is a related idea):

P.M. Rice, "A topological characterization of the real numbers," 1969

S.P. Franklin and G.V. Krishnarao, "On the Topological Characterization of the Real Line: An Addendum," J. London Math Soc (2) 3 (1971) 392.

Brouwer, "On the topological characterization of the real line," http://repository.cwi.nl/search/fullrecord.php?publnr=7215

Kleiber, "A topological characterization of the real numbers," J. London Math Soc, (2) 7 (1973) 199

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  • Lang informally referred to (differentiable) manifolds as open subsets of Banach spaces being glued together by $C^p$-differentiable mappings (which identify parts of these subsets). So I think it would make more sense to improve this definition (using sheaves maybe, I don't know) rather than trying to come up with something that effectively obscures local homeomorphism to a Banach space. – Alexei Averchenko Jul 22 '11 at 04:09
  • @Alexei: But when you say "Banach space", you implicitly mean "Banach space over $\mathbb{R}$", don't you? I'm pretty sure a space that is locally homeomorphic to a Banach space over $\mathbb{Q}_p$ cannot be a topological manifold. – Zev Chonoles Jul 22 '11 at 04:28
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    Your delicate sensibility is going to have to prove that (assuming that list of properties does actually characterize manifolds) that "your manifolds" are the same as "ours" at some point or another... In a similar vein, even though we *know* that a Lie group is simply a topological group whose identity element has a neighborhood which does not contain a subgroup, no one would sanely use that as a definition, because to go from there to, well, Lie groups takes enormous effort. – Mariano Suárez-Álvarez Jul 22 '11 at 04:39
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    The long line is a perfectly good 1-dimensional manifold that isn’t second countable. – Brian M. Scott Jul 22 '11 at 06:20
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    To be honest and to play with your analogy (heehee!), I think you're trying to get a tank, strip it off, rebuild it without using the screws, and then still call it a tank in some manner. – Patrick Da Silva Jul 22 '11 at 06:56
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    @Brian M. Scott: is it? I believe that in general one defines a manifold $M$ to be Hausdorff, second countable and loc. hom. to $\mathbb{R}^n.$ If there is some reason why people do not demand second countability, I'd love to hear about it. – Gerben Jul 22 '11 at 08:45
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    @Zev: at first I thought you were nitpicking, but actually this may be a good idea to consider p-adic manifolds (defined as you put it) in their own right :) – Alexei Averchenko Jul 22 '11 at 08:51
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    What structure of $\mathbb R$ do we throw away by making the common definition of a manifold? – Eric O. Korman Jul 22 '11 at 10:02
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    I think this is nice question. I myself wondered about similar things in many different contexts. E.g. being path-connected requires $\mathbb R$ too and this also seems like an overkill (for finite spaces, say). Also, there is another question lurking close: what makes $\mathbb R$ so special that we use to build so many structures when (naively) many other objects could stand in its place. It seems to me that the answer will be that it's just what people had done and they found it's ok for their purposes. Yet, one can surely ponder replacing reals with something else. – Marek Jul 22 '11 at 10:17
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    This is completely irrelevant to the actual topic at hand, but I just really, _really_ wanted to link to [http://en.wikipedia.org/wiki/Winged_tank](http://en.wikipedia.org/wiki/Winged_tank). – Ilmari Karonen Jul 22 '11 at 12:18
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    What I don't like about this approach is that it might even lead to a purely topological definition of a topological manifold (and *that* may be considered as interesting), but I don't see how you're eventually going to differentiate, speak of tangent bundles etc in this framework. – t.b. Jul 22 '11 at 12:59
  • @Alexei: Actually, I meant to be nitpicking (sorry :)). The OP is looking for a definition of manifold without reference to the reals; I don't think your proposal meets that criterion. – Zev Chonoles Jul 22 '11 at 14:15
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    @Gerben: There is a considerable literature on non-metrizable manifolds; see the survey by Peter Nyikos in the *Handbook of Set-Theoretic Topology*, K. Kunen and J.E. Vaughen, eds. (A manifold is second countable iff it is Lindelöf iff it is paracompact iff it is metrizable.) – Brian M. Scott Jul 22 '11 at 18:45
  • @Brian: thanks, I'll have a look. – Gerben Jul 23 '11 at 06:39
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    "The real line is a high-tech piece of mathematical machinery." - and a manifold isn't ? – Mark Aug 07 '11 at 16:10
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    I like this question very much. At the same time, I think that the the "extra structure of the real line" may be more relevant than the "winged tank" metaphor suggests. Think of building a manifold using the "real line plus extra structure" like building a skyscraper using "steel girders plus scaffolding": the scaffolding may be removed from the final structure, but it makes it much easier to get all the other components in place. – Charles Staats Aug 08 '11 at 21:35
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    If one could give a strictly topological definition / characterization of manifolds, this in turn would give us a strictly topological definition / characterization of $\mathbb{R}$ as the unique non-compact 1-manifold. Would it assuage your sensibilities to find such a characterization? Then the definition of manifolds that we have now --- nice (2nd countable, Hausdorff) and locally an $n$-fold product of $\mathbb{R}$---would carry through, except that we never had the field structure on $\mathbb{R}$ to start with. – Robert Haraway Aug 11 '11 at 15:54

5 Answers5


I'm going to risk an answer to this one. It's a long answer, so I'll give a short summary first. One thing I'm not completely clear on is whether you mean topological manifolds or smooth manifolds. If you were a mathematician, I'd infer from your question the former, but as you're a physicist then I'm not confident of which.

There's a big difference between the two cases and the answers are very different. Here's the short version:

  • Topological Manifolds I have considerable sympathy for your point of view, but have to say, "Get used to it.". The point is that being a topological manifold is a property of a topological space and so is there whether you use it or not. We don't study topological manifolds because it makes us look good, but because many of the "usual" spaces that one encounters happen to be topological manifolds. That they are topological manifolds means that we have a great toolbox to use to study them, but if we ignored that toolbox then the spaces would still be topological manifolds.

  • Smooth Manifolds Here I have less sympathy with your point of view simply because the real line is so integral to calculus. The real line might be an incredibly complicated gadget, but then it needs to be to support calculus. Of course, there are variants of calculus (holomorphic, $p$-adic) but if the real line looks complicated, then I would be amazed to hear that the complex plane or the $p$-adics looked any simpler. Nonetheless, because being a smooth manifold is about structure, it is actually more feasible to entertain different definitions.

Okay, that was the short version. Now for the long version. First, I need to say something about definitions.

Mathematical Definitions

You say that you are a physicist, so it's possible that you haven't been let in on the secret about mathematical definitions. If you have, skip this bit. If not, I'll tell you. (But, hush! It's a secret. Don't tell anyone else.)

I'll illustrate the point I wish to make with an example that I hope is familiar to you. What is the definition of a continuous map between metric spaces? I teach this, and I teach three definitions:

  1. A function $f \colon M \to N$ is continuous if whenever $(x_n) \to x$ in $M$ then $(f(x_n)) \to f(x)$ in $N$. (Note: I chose metric spaces here, so sequences are sufficient.)

  2. A function $f \colon M \to N$ is continuous if for every $x \in M$ and $\epsilon \gt 0$ then there is a $\delta \gt 0$ such that whenever $d_M(x,y) \lt \delta$, $d_N(f(x),f(y)) \lt \epsilon$.

  3. A function $f \colon M \to N$ is continuous if whenever $U \subseteq N$ is open then $f^{-1}(U)$ is open in $M$.

These definitions are equivalent: they all agree which functions are continuous and which are not. So any statement made using one definition can be reformulated using another. But they have different uses, since they emphasise different aspects of what it means to be continuous. If you're interested in metric spaces because you use approximations then the first definition captures the idea of what you want to use: If I have an approximation of something, then after I hit it with a continuous function, it is still an approximation (of the image of the "something"). The second definition is actually the most practical when testing an explicit function for continuity: it's amenable to finding estimates and the like. Third is the most theoretically powerful: it's the first one we reach for when trying to prove theorems about continuous functions.

So just because a definition seems to be the "established" definition doesn't mean that that is the right way to think about it. Definitions are malleable, and we often use a different definition to the one that we truly believe is "right" simply because it is easier.

Topological Manifolds

Let me start with topological manifolds. I said at the beginning that the key here is that being a topological manifold is a property. That is, if I have a topological space then it either is or isn't a topological manifold. If it is, then I can use that fact when I study it; if it isn't, then I can't. If it is a topological manifold then I don't have to use that fact, but I'm likely to be making life difficult for myself if I don't. The key thing about a property is that if I ignore it, it is still there.

The people who study topological manifolds do so because many spaces of interest happen to be topological manifolds. If you change the definition, then those spaces will go on being locally Euclidean, and the people studying them will continue to use the fact that they are locally Euclidean, and all that will have changed is the language that they use. This is why I don't have much sympathy for your desire to change the definition.

Although most of the Euclidean structure doesn't have much topological influence on a topological manifold, it does provide a lot of useful tools in the analysis. For better or worse, Euclidean spaces are things that we simply know a lot about. So saying that a space is locally Euclidean means that we can use all our intuition and skills from the theory of Euclidean spaces to the study of the space. That's worth a lot, and you'd need to be very convincing to persuade people to give that up.

Now when thinking how to define a topological manifold, one encounters the question I alluded to in the above on definitions. Definitions come in all shapes and sizes. It's not clear from your question as to which definition you would like best. On the one hand, your dislike of the real line makes me think that you want the "pure" definition: the one that captures the soul of a topological manifold. I'll readily agree that the current definition is not that, it's more of the "body" type where it's easy to see how to use it. But the proto-definition that you give isn't that either: it's a mish-mash of topological concepts, each of which excludes a range of spaces, with the hope that in the end all you have left are the topological manifolds. I don't like that sort of definition, it's more of the $\epsilon$-$\delta$ type: has its place, but is neither the "soul" nor the "body".

However, what it feels most like is that you are playing that children's game where you have to explain what is an aardvark without using the words "aardvark", "anteater", "dictionary", or "pink panther".

An alternative is to come up with a definition that is actually different in that it doesn't completely agree with the current definition. In that case, your work is harder. You have to show why the new definition is better than the old one. The most convincing arguments would be either that your definition allows you to do more, or that it allows you to consider more spaces. But these are unlikely to both hold. If you allow more spaces, you probably lose out on abilities; if you find new tools, then you probably can't apply them to all the current topological manifolds. If you really want to do this then your best bet is not to mention topological manifolds at all, but to invent a wholly new concept, say "Topological foldimen" and simply say, "Topological manifolds that are X are foldimen, and foldimen that are Y are manifolds.". Then hope that there are plenty of interesting foldimen out there.

Smooth Manifolds

Smooth manifolds, on the other hand, are much more malleable. This is because being a smooth manifold is something a little bit extra. The standard definition of a smooth manifold starts with a topological manifold and then adds a little extra on top. Now, forgetting that extra does mean something. If you forget it, it goes away, and you can't be sure what it was.

As an illustration, if I have two topological manifolds, $X$ and $Y$, then the question "Is $X$ homemorphic to $Y$?" has the same answer if I remember that they are manifolds or not. But if I say that they are smooth manifolds, then the question "Is $X$ diffeomorphic to $Y$?" depends completely on their being smooth manifolds.

This actually gives us some room to manoeuvre. Because we need to construct the extra structure, we can consider different constructions. However, this is where your dislike of the reals counts against us. We cannot construct something from nothing: we need to start with something. There are many possible answers as to what that "something" is, but they all boil down to identifying certain spaces as "known" meaning that we decide what the structure for those spaces should be. Since they are "known", and everything else will be defined relative to them, they should be spaces that we really do know about. It's hard to get spaces that are more well-known than the Euclidean spaces. Certainly when calculus is concerned. All of the examples of this that I've seen have used Euclidean spaces, or "nice" subsets thereof.

So these are our "known" spaces, which we will use to define what a "smooth structure" means for "unknown" spaces. This is where we have some flexibility in the definition, and here is where we can get rid of that annoying "local stuff". Maybe, just maybe, we don't need to have actual charts and can get away with something weaker.

Actually, we can. No "maybe" about it. But the problem is that by weakening the definition, we end up with more things than we might like to admit to the hallowed halls of manifolddom. Nonetheless, there is some merit in pursuing this line as it separates out the construction aspect (of what "smooth" means) from the property aspect (of what a "manifold" is).

As I said, there are many approaches at this point. I'm going to outline one just so that you have one in mind. There are others, and this is not the place to evaluate them. If you don't like this one, the rest still carries through. I just want to be sure that you have one picture in your mind.

Here it is: it rests on the slogan that manifolds are all about smooth curves. If we both look at a manifold, we should agree on which curves in it are smooth and which are not. So one way to specify a "smooth space" is to give a list of all its smooth curves. One probably wants some conditions, but this can all be made precise. Thus a "smooth space" is a space in which we all agree on smooth curves.

As I said, there are other approaches, but whatever they are they still give us a list of the smooth curves. This will be important in a minute.

Now it turns out that this admits far more than just manifolds. In fact, too much. There are really weird spaces in our list now, and we'd like to get rid of them.

Remember the curves? Good. We're going to use them. Using the curves, one can define the tangent space of a "smooth space". Just the same as for a smooth manifold: derivatives of curves. Unlike smooth manifolds, this needn't be locally trivial, nor even the fibres vector spaces. There's also a notion of a topological tangent bundle, which is related to neighbourhoods of the diagonal.

Here's the definition:

A smooth manifold is a smooth space whose smooth and topological tangent bundles agree.

Now, I'm not 100% sure that this is exactly the same as "smooth manifold", but certainly all smooth manifolds fit this description, and it excludes smooth manifolds-with-boundary, but there might be the odd pathological case that this doesn't exclude. Nonetheless, it is a very powerful description: it implies that the tangent spaces are actual vector spaces - we didn't assume that, if you remember.


If you want to mess with definitions, go ahead: it's fun. But be careful that you know what type of definition you are aiming for. There are reasons that the current definitions are what they are and the reasons tend to be pragmatic: over the years, we've found certain definitions useful and others not. There is a lot of dead wood, and there are always new insights that shed new light on old concepts, but if you want to replace an old definition, then remember that it's been there for a while, working hard, and will strongly resist all attempts to "retire" it from service!

Andrew Stacey
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    Thanks for your answer. I had in mind the notion of a topological manifold, not a smooth manifold. Yes, I am aware of the fact that different definitions can be equivalent. I'm simply asking whether my proposed definition is or is not equivalent to the standard definition of a topological manifold. It's a yes/no question. An example of a helpful answer would be to provide a counterexample, or to provide a reference to a book or paper in which a similar approach is carried out. Although I appreciate your taking the time to write such a long answer, it doesn't address the specifics at all. –  Aug 12 '11 at 14:39
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    @Ben: I will admit that I paid considerable attention to the _comments_ in deciding what to answer and not so much to the actual question. Your question was very vague and the wording is such that it is not at all obvious that the "yes/no" part is the key part. Not that that matters to me - the "yes/no" part isn't all that interesting to me, for reasons that I hope I did make clear in the above. I would also be amazed to learn that the people who voted for this question are also _really_ interested in that exact definition that you propose! – Andrew Stacey Aug 12 '11 at 14:53
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    @Ben: regarding your "yes/no" question, my honest answer is: "I don't know, and I don't care.". It's a messy definition that is completely unmotivated: why _those_ conditions? In what way are they "better" than "locally Euclidean" which is concise and eminently usable? – Andrew Stacey Aug 12 '11 at 15:10
  • @Andrew: Are you sure that selecting a class of smooth curves is sufficient? In $\mathbb{R}^n$ we have functions which are not differentiable but have all directional derivatives, see http://www.math.tamu.edu/~tvogel/gallery/node17.html – Alexei Averchenko Aug 12 '11 at 15:51
  • @Alexei: For **smoothness** then smooth curves are sufficient (a theorem of Jan Boman). For differentiability of order $k$, then the appropriate curves work if you say "differentiable with locally Lipschitz derivative" rather than "differentiable with continuous derivative". Take a look in the early sections of "A Convenient Setting for Global Analysis" by Kriegl and Michor for more details (available free online from the AMS). – Andrew Stacey Aug 12 '11 at 18:53
  • *that children's game where you have to explain what is an aardvark without using the words "aardvark", "anteater", "dictionary", or "pink panther".* -- You mean https://en.wikipedia.org/wiki/Taboo_(game) ? I've had more fun playing that with adults ... – Torsten Schoeneberg Aug 24 '21 at 19:21

In [Harrold, O. G., Jr. A characterization of locally euclidean spaces. Trans. Amer. Math. Soc. 118 1965 1--16. MR0205240 (34 #5073)] there is a purely topological characterization of the $n$-dimensional sphere $S^n$ among metric spaces.

We can therefore characterize the $n$-sphere as the unique compact Hausdorff second-countable topological space (these conditions imply that the space is metrizable, in essentially one way in view of compactness) satisfying Harrold's conditions. This is a purely topological characterization which does not involve $\mathbb R$ at all.

Now define an $n$-manifold as a Hausdorff second countable space locally homeomorphic to $S^n$. This does what you want, I think.

Mariano Suárez-Álvarez
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    Here's the link to the paper (since dan232's answer doesn't spell it out): http://dx.doi.org/10.1090/S0002-9947-1965-0205240-6 – t.b. Aug 13 '11 at 00:17
  • Thanks, Mariano and Theo -- that's extremely helpful! –  Aug 14 '11 at 22:08
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    Techically shouldn't the condition be "Hausdorff second countable space locally homeomorphic to a proper open subset of $S^n$" or something similar? Otherwise e.g. $\mathbb R$ won't be a $1$-manifold – kahen Feb 13 '12 at 05:51
  • Any chance you could edit your answer to include the actual characterization? – ziggurism Nov 22 '20 at 18:00

Risking to receive a lot of downvotes, I'm writing this as an answer.

First of all, manifolds are generally not homogeneous with respect to their homeomorphisms group (even putting dimension aside, consider $S^2 \sqcup T^2$). Also, I have big doubts that all manifolds can be endowed with a uniform structure that is complete.

Now if we drop this requirement, all finite CW complexes fit in with the rest of your requirements (except maybe uniform completeness, but I think there must be a nice uniform structure on them since they are just different balls glued together, and I have already expressed my doubts about completeness).

So what it boils down to is: your proposed axioms do not capture the intuitive 'local sameness' of a manifold, let alone its local homeomorphism to a real Banach space. You attempted to capture it with the requirement of homogeneity with respect to the homeomorphisms group, but this requirement is simply not true.

Alexei Averchenko
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    Uniform completeness is easy to achieve. Every manifold can be embedded as a closed subset of some $\mathbb{R}^N$ by Whitney. (didn't downvote btw). – t.b. Jul 22 '11 at 12:15
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    @Theo: It is rare that Whitney's theorems be *easy*! :) – Mariano Suárez-Álvarez Jul 22 '11 at 15:46
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    But an abstract topological manifold does not come equipped with a uniform structure, and every noncompact manifold can be equipped both with a complete uniformity and an incomplete uniformity. So instead of "complete uniform space" the requirement should be "completely uniformizable space". I guess the only reason we are not saying "completely metrizable" is because of the cooties $\mathbb{R}$ has been endowed with for the purposes of this question? – Pete L. Clark Jul 22 '11 at 21:32
  • @Pete: I guess so, but that's following Bourbaki's magistral footsteps :) – t.b. Jul 23 '11 at 00:06
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    "Completely uniformizable" is a very weak condition -- any paracompact (hence any metric) space is completely uniformizable. (I'm assuming Hausdorffness, haven't bothered to determine exactly how to modify that statement without Hausdorff condition.) – Harry Altman Aug 11 '11 at 13:48
  • Re the requirement of a complete uniform space, I'm not sure if the dust has settled on the comments...? Does the notation $S^2 \sqcup T^2$ mean the disjoint union of a sphere and a torus? If so, then I guess this is a valid counterexample, but it seems relatively trivial to me. Maybe my proposed definition should say "A connected manifold is..." and add connectedness to the axioms. –  Aug 12 '11 at 15:19
  • Due to poor planning on my part, the bounty will expire this weekend while I'm on a backpacking trip. It's possible that this answer is exactly what I asked for, but due to my lack of fluency I'm going to need time to study. I'll award the bounty to this answer, since I'm out of time before my trip. –  Aug 12 '11 at 15:25
  • Jackpot! 8) I'll see if I can find an example of a connected manifold that's not homogeneous with respect to its autohomeomorphisms group. – Alexei Averchenko Aug 12 '11 at 15:45
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    @Alexei: Don't waste your time... Rather: prove as an exercise that every connected manifold is homogeneous :) – t.b. Aug 12 '11 at 16:48
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    @Theo: Preliminary sketch: (1) Using Brower's show that such homeomorphism exists for a closed ball, and it leaves the boundary in place; (2) For compact manifolds, select a finite cover and builds a 'path' of covering images of balls from point $x$ to point $y$, apply the previous construction recursively, use identity on the boundary to extend; (3) For non-compact case, pass to the compact case by selecting a closed functional neighborhood of a path connecting the two points. I'm not sure if step (2) can be removed by proving that such neighborhood is homeomorphic to a closed ball. – Alexei Averchenko Aug 12 '11 at 18:15
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    I'm not quite sure what you mean by "Using Brower's" but yes, that's good as a start. Now in any coordinate patch $\phi: U \to \mathbb{R}^n$ fix a disk $D_{2r} \subset \phi(U)$ and let $V = \phi^{-1}(D_r)$. For any $p,q \in V$ find a homeomorphism $\psi_{V}^{p,q}: M \to M$ such that $\psi_{V}^{p,q}(p) = q$ fixing the complement of $V$. Now given $x,y \in M$ choose a path connecting $x$ to $y$ and cover it by finitely many patches $V_1,\ldots,V_k$ as chosen before such that $V_{i} \cap V_{i+1} \neq \emptyset$... – t.b. Aug 12 '11 at 20:49
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    @Alexei: Alternatively, define the binary relation on $X$ by $x\sim y$ if there is a homeomorphism taking $x$ to $y$. Show that this is an equivalence relation, whose equivalence classes are open. If $X$ is connected, there can only be one equivalence class. – George Lowther Aug 13 '11 at 00:04
  • @Theo: at first I though Brower's fixed point theorem could be used instead of an explicit construction, but on the second thought conjugating some translation by $\mathrm{arctg}$ and then extending in an obvious way is the way to go. Now we finish the proof by selecting a cover consisting of balls at each point, selecting a path from $x$ to $y$, and then using compactness of its image to find a sequence of balls connecting $x$ and $y$, then applying the above construction. – Alexei Averchenko Aug 13 '11 at 06:07
  • @George: Each point of a manifold has a neighborhood homeomorphic to the open ball, and is thus transitive in a way shown above (conjugate translation with $\mathrm{arctg}$, extend) - brilliant! ^_^ – Alexei Averchenko Aug 13 '11 at 06:08
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    @Theo: UPD: My previous construction was wrong, here's another attempt. Consider $B^n = \lbrace x\in \mathbb{R}^n\mid\|x\|_2<1\rbrace$ with usual smooth and Riemannian structures, and two points $x,y\in B^n$. Let's denote $r=\frac{1}{2}(1+\max\lbrace\|x\|_2,\|y\|_2\rbrace)$, $X \in T_x B^n$ - the tangent vector to the line segment connecting $x$ and $y$, and $f(\frac{x}{r}) = \exp(\frac{\|x\|_2}{\|x\|_2-1})$. Then $Y=f(x)\hat X$, where $\hat X$ is LC-constant and equal to $X$ in $x$, is compactly supported, and the corresponding 1-parameter group has an autodiffeo in it that takes $x$ to $y$. – Alexei Averchenko Aug 13 '11 at 08:55
  • @Alexei: Yes, this looks good now (didn't check the formula). Note that it suffices to map $0$ to any point $x \in B_r$. For this choose an appropriate bump function $f$ and look at the vector field $X(p) = f(\|p\|) \cdot x$. The associated flow will transport $0$ to $x$ at some point in time. – t.b. Aug 13 '11 at 09:04
  • @Theo: To make the proof easier, we can pick a smooth function separating $\overline{B}_{\max\lbrace \|x\|_2, \|y\|_2\rbrace}$ and $B^n - B_r$ as the bump function in question, then it is trivial to prove that we get our autodiffeomorphism when the parameter of the flow is equal to $1$. – Alexei Averchenko Aug 13 '11 at 11:19
  • Right. That may be nicer. You know, it's been a while that I actually wrote that argument down in detail :) – t.b. Aug 13 '11 at 11:27

I'm sorry - this is to long for a comment...

You make me feel impelled to defend a very mighty definition. You said "We build up all that structure, then build the definition of a manifold out of it, then throw away most of the structure." but the spirit of this construction is that we do not loose the most of the structure! It keeps the quality and extend it to the possibility of building spaces with interesting e.g. geometrical properties.

First i have to declare: Its not hard to show that a manifold satisfies all the conditions you listed (except homogeneous space under its own homeomorphism group). I've seen an proof of equivalence in an very similar definition but i can't remember where, what i do remember is that is was hard to show that you really achieve a manifold by those requirements.

Another thing you should be told is that you don't loose metrisation in the sencefull (e.g. differentiable) examples. Each paracompact connected manifold is metrizable. Well, i have to admit - in viciously constructed examples this metric might not be intuitive, but at least in the case of an connected differentiable manifold there is a global metric, locally coinciding with the euclidean metric.

Addition: For complete uniformizability consider a theorem by Shirota that states: A completely regular Hausdorff space that is realcompact is completely uniformizable. A manifold is as well completely regular Hausdorff as realcompact since its second countable and so is countably compact.

P.S. @Brian M. Scott the real line is second countable...did you mean something else by "long line"?

Edit: Its time for an apology. The details might be not that easy i thought of and some of them are just wrong.

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    The long line is *not* the real numbers, but rather [this topological space](http://en.wikipedia.org/wiki/Long_line_%28topology%29). – Asaf Karagila Jul 22 '11 at 10:36
  • Okay, this is in fact a beautiful example for a local euclidean separable space which is not second countable and hence one of the reason for second countabillity to be appended to the axioms/requirements – Ben Jul 22 '11 at 10:56
  • Is every manifold a complete uniform space? How is such (unique? Corresponding to some metric?) uniformity defined on manifolds? How is every manifold homogeneous with respect to its homeo group? – Alexei Averchenko Jul 22 '11 at 10:57
  • I have to edit my post, an topol. manifold is metrizable if and only if it is paracompact (most of them will be). But i am quite sure it is uniformizable, going to look for a reference as far as i can. I have to admit - i missed the 'completeness'.. – Ben Jul 22 '11 at 11:05
  • @Alexei: paracompact manifolds are metrizable. In most of the references I've seen, people include paracompactness directly into the definition of a manifold (e.g. the long line would not be a manifold since it's not second-countable). Otherwise, one has only local metrizability. – Marek Jul 22 '11 at 11:07
  • @Marek: But how is that uniform structure unique (on non-compact spaces?) The real line and the interval $(0,1)$ are diffeomorphic but with respect to their usual metrics the uniform structure on $(0,1)$ is finer than the one on the real line. – t.b. Jul 22 '11 at 11:24
  • uniformizability: A manifold a is locally compact Hausdorffspace and so is $T_{3a}$ but this is implies the existence of an uniform structure. – Ben Jul 22 '11 at 11:25
  • @Ben: yes, but as I tried to point out, not intrinsically so. – t.b. Jul 22 '11 at 11:30
  • @Ben: this is all good (and every metric induces a uniform structure, too), but the OP said 'a complete uniform space', and it is this completeness that I asked about :) – Alexei Averchenko Jul 22 '11 at 11:31
  • @Alexei: you don't have completeness, of course. Look at my first comment here. – t.b. Jul 22 '11 at 11:32
  • So where is the error? By second countability it is Lindelöf and so is realcompact. Additional it is completely regular and this implies complete uniformizability. – Ben Jul 22 '11 at 11:43
  • @Theo: If I understood correctly, you spoke only of uniform structures induced by metrics, but it seems not entirely implausible that on every manifold there is a complete uniform structure. I don't know much more about uniform structure than its Wiki definition, though <_ – Alexei Averchenko Jul 22 '11 at 11:49
  • @Alexei: If a uniform structure is induced by a metric it is complete if and only if it the metric is complete. – t.b. Jul 22 '11 at 12:09
  • @Ben Anthes: The long line is not separable. – Nate Eldredge Jul 22 '11 at 17:50
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    @Alexei, Theo: In fact, any paracompact Hausdorff space (yes, any metrizable space, completely metrizable or not!) can be completely uniformized; complete uniformizability is a pretty weak condition. – Harry Altman Aug 12 '11 at 01:41

A topological space $X$ is called locally Euclidean if there is a non-negative integer $n$ such that every point in $X$ has a neighborhood which is homeomorphic to an open subset of Euclidean space $\mathbb{R}^n$. If can find an equivalent definition for a topological space to be "locally Euclidean" without the use of the real numbers, that's your answer. (paper on the subject)

Also realize that we need the real numbers to define de differential structure of a differentiable manifolds, for example. There are differentiable manifolds which are homeomorphic, but not diffeomorphic; so the structure doesn't depend only in the topology, but also in the way that the "locally Euclidean" property is carried out (the homeomorphisms defined between the manifold and the euclidean space, what is called the "atlas").

What I'm trying to say is that a topological manifold is much more than just a topological space when it has a surname like "differentiable" for instance.

Isn't it right?

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    If I'm wrong I would like to be corrected in addition to that downvote, thank you. – dan232 Aug 11 '11 at 13:14
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    I didn't down-vote. However, I don't think this answers the question. –  Aug 11 '11 at 14:58
  • I cannot comment, there's no bottom "add comment". I guessed it was because I had few reputation points since I'm new. – dan232 Aug 11 '11 at 16:02
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    I don't think defining manifolds by saying "locally homeomorphic to $\mathbb R^n$" counts as defining them without reference to the reals... – Mariano Suárez-Álvarez Aug 11 '11 at 17:06
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    @Mariano : The formal definition of manifolds is "topological space locally Euclidean" in many books. What I said is that if there's an equivalente definition to "being locally Euclidean" without the use of $\mathbb{R}^n$ there's the answer. Sorry if my English isn't that clear. – dan232 Aug 11 '11 at 17:42
  • @dan232: how can you define $\mathbb R^n$ without reference to the reals? – Mariano Suárez-Álvarez Aug 11 '11 at 17:59
  • @Mariano : not $\mathbb{R}^n$ itself, but if the are some topological properties which all together are equivalent to being "locally euclidean", such that this properties don't mention $\mathbb{R}^n$. May be it is imposible, I don't know. If it is imposible, then topological manifolds cannot be defined without $\mathbb{R}$. – dan232 Aug 11 '11 at 19:30
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    @dan232: that is my point, precisely: you have not answered the question :) – Mariano Suárez-Álvarez Aug 11 '11 at 21:37
  • @Mariano: I described how to get the solution. I strongly believe it is a still open problem which hasn't been solved, for what I have read. The thing is I pointed in the direction of where to look for the last theory about the subject. – dan232 Aug 12 '11 at 23:19
  • It is tautological that the problem of topologically characterizing spaces «locally homeomorphic to the space $X$ without reference to $\mathbb R$» can be reduced to the problem if «characterizing topologically the space $X$ without reference to $\mathbb R$». That is rather obvious, and that is what you have done. I do no think it is useful to continue this conversation, by the way. – Mariano Suárez-Álvarez Aug 12 '11 at 23:51
  • @Mariano: How can you be this way? You write all this things because at first you obviously didn't understood me. Now you take the paper I link in my answer and rephrase their ideas; wrongly by the way, if you're interested because "locally homeomorphic to $S^n$" is not the same to "locally homeomorphic to $\mathbb{R}^n$". On implies the other, but they are nor equivalent. – dan232 Aug 13 '11 at 14:31
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    I don't know if this provides any clarity to the above discussion or not, but I think that the OP of this answer may have meant "Euclidean space" in the sense of "arbitrary complete metric space satisfying the axioms of Euclidean geometry" and Mariano may have meant "Euclidean space" in the more commonly used sense of "literally $\mathbb{R}^n$". By various representation theorems, it turns out that any model of Euclidean space (in the first sense) is isomorphic in model theoretic terms to $\mathbb{R}^n$ (Euclidean space in the second sense), so the distinction is very difficult to understand. – Chill2Macht Jul 04 '17 at 10:22
  • But technically I think that the OP of this answer is correct inasmuch as Euclidean space (in the first sense, _not_ the second sense) can be defined without reference to the real numbers. This is a similar observation to that made by Robert Haraway in one of the comments below the question itself, where the $\mathbb{R}$ itself can be characterized purely topologically, without reference to a field structure -- and certainly structures isomorphic topologically/metrically to $\mathbb{R}$, as 1-dimensional Euclidean space (in the first sense) can also be so characterized. – Chill2Macht Jul 04 '17 at 10:25
  • Or at least the OP of this answer would have been correct had they defined Euclidean space as being something else then $\mathbb{R}^n$ (which re-reading this answer it seems like they failed to do), so inasmuch as the answerer fails to distinguish Euclidean space from $\mathbb{R}^n$, Mariano is definitely correct. So basically my above two comments are completely non-germane to the actual answer being discussed - not good. – Chill2Macht Jul 04 '17 at 10:27