# Manifold definition

1. Feb 6, 2013

### TrickyDicky

In general terms a manifold can be defined simply as a topological space locally resembling Euclidean space with the resemblance meaning homeomorphic to Euclidean space, plus a couple of point set axioms that avoid certain "patological" manifolds and that some authors reserve for the definition of differentiable manifolds. My doubt is that since the category of homeomorphisms doesn't include notions of distance and angles, that is, metric properties are not included, wouldn't it be more precise defining manifolds as topological spaces locally homeomorphic to real topological vector spaces (or complex topological vector spaces in the case of a complex manifold) rather than to Euclidean spaces?

Not that the usual definition is wrong, but IMO it also might be misleading wrt the often ignored difference between real topological vector spaces and Euclidean vector spaces (only the latter has the Euclidean inner product).Calling both entities R^n doesn't help either.
In the wikipedia page on manifolds one can see it defined both ways but there is much more insistence in the "locally resembling Euclidean space" definition.

Last edited: Feb 7, 2013
2. Feb 6, 2013

### micromass

A real vector space doesn't have a topology. A vector space is a purely algebraic object. So locally homeomorphic makes no sense.

Of course, it is entirely possible to give a real vector space a natural topology, but strictly speaking, we aren't dealing with vector spaces anymore, but topological vector spaces.

In any case, I fail to see what you dislike about the $\mathbb{R}^n$ definition.

3. Feb 6, 2013

### WannabeNewton

$\mathbb{R}^{n}$ is a real vector space. Any finite dimensional vector space can be given a norm. All norms on a finite dimensional vector space generate the same topology. Any finite dimensional normed vector space is homeomorpic to $\mathbb{R}^{n}$ for some $n$. Not sure what point you are trying to make.

4. Feb 6, 2013

### TrickyDicky

Hmmm, I had always in mind a real vector space with the natural topology. Isn't a vector space over the reals automatically acquire a topology? I guess not.

Ok, my fault, I was identifying vector spaces over the topological field of the reals as topological vector spaces.
I dislike the ambiguity of the symbol, for instance right now I don't know if you are talking about Euclidean space R^n or an algebraic real vector space R^n or the topological space R^n.

5. Feb 6, 2013

### TrickyDicky

As I was telling micromass I was thinking about topological vector spaces when I said vector spaces. Sorry about that. Maybe after correcting it you see my point.

6. Feb 6, 2013

### micromass

I guess I don't know why you specifically want a vector space structure to define manifolds.

All you need to define a manifold is the topological space $\mathbb{R}^n$, and not the addition and multiplication structure.

If you want to define a manifold as something locally homeomorphic to a real vector space (with some natural topology), then you are kind of implying that the vector space structure is important when defining manifolds. But in reality, there is no need for any algebraic structure when defining manifolds. So I'm not sure why you want to look at topological vector spaces to begin with.

7. Feb 6, 2013

### Ben Niehoff

Some authors use $\mathbb{R}^n$ to refer to the topological space, and $\mathbb{E}^n$ to refer to the topological space endowed with the Euclidean metric. To define manifolds, one only needs the topological space.

As for vector spaces, howerver...

I don't see any natural way that a manifold can be locally homeomorphic to a vector space. As Micromass points out, the sentence itself sounds "grammatically incorrect" to me. More to the point, what makes a vector space a vector space is its algebraic structure; that elements of the space can be added, subtracted, and mutiplied by scalars.

Such operations do not, however, make sense on a simply-connected open neighborhood of some manifold. Therefore a manifold cannot be locally compared to a vector space in any meaningful way. The tangent space at each point is a vector space, but points on the manifold itself have no algebraic operations defined on them.

8. Feb 6, 2013

### TrickyDicky

I corrected my post, does it make more sense now?

9. Feb 7, 2013

### TrickyDicky

Ok, I guess this answers my question, thanks.

10. Feb 7, 2013

### dextercioby

My opinion:

1.$\left(\mathbb{R}^{n},\mathbb{R},\langle ,\rangle\right) \equiv \mathbb{E}^{n}$ is an n-dimensional topological vector space (a Banach space actually) with the metric topology induced by the euclidean scalar product through the norm.

2. $\mathbb{R}^n$ is a topological space with the product topology, the open interval topology inherited from $\mathbb{R}$ + the cartesian product.

So when one says that a finite dimensional topological manifold is locally homeomorphic to $\mathbb{R}^{n}$, do they mean 1. or 2. ? My guess is 1. :uhh:

11. Feb 7, 2013

### jgens

Since the spaces have the same topology it does not matter.

12. Feb 7, 2013

### mathwonk

That is the point. The word "homeomorphic" already tells you that only the topology is being considered.

The situation with infinite dimensional real vector spaces is that they have more than one topology, so then you can't say simply "homeomorphic" since there is no canonical choice.

Moreover one does consider infinite dimensional manifolds modeled locally on Banach spaces. In those cases the norm is given on the Banach space, but merely the induced (complete) metric topology is relevant.

So if one wants to define a manifold as locally modeled on some real vector space, one has to give the topology somehow, and a norm is a convenient way.

Moreover one usually wants to employ local coordinates in discussing manifolds, and here it is helpful that one has modeled ones manifold on R^n, i.e. Euclidean coordinate space. So it seems to me that in discussing manifolds, one usually wants to use both the topology and the coordinates.

Indeed in classical physics texts, one does not even have much intrinsic grip on the manifold, but discusses tensors et al as if they were always given in coordinates, plus some rules for changing those coordinates.

I sympathize with the confusion however. I once was greatly puzzled when a professor showed me how to prove a certain function was continuous by choosing coordinates. I thought that was no good since originally the space had no coordinates, only a topology. I failed to see that a function on a finite dimensional real vector space is continuous in the usual topology if and only if it is continuous in terms of any choice at all o coordinates.

In fact this failure of mine demonstrated that, in spite of all the manifold stuff I had memorized, I did not grasp the whole point of continuously compatible local coordinate systems. Namely that the compatibility means that questions about continuity can be checked in any of them.

So if you like, a finite dimensional real vector space is itself a manifold with a huge number of global charts, one for each basis. Moreover each basis defines the same topology so anyone can be used to check continuity of functions.

13. Feb 8, 2013

### lavinia

While locally homeomorphic refers only to the topology, the idea of a manifold is that it can be locally coordinatized in the same way as Euclidean space. Coordinates imply more structure than just the topology and for very small distances a coordinate system will appear to be a vector space structure.

14. Feb 8, 2013

### jgens

True. But it is also worth noting that this local vector space structure is not canonical. Different coordinates around the same point will give you a different vector space structure.

15. Feb 8, 2013

### TrickyDicky

I don't think this is correct, it is coordinate independent.

16. Feb 8, 2013

### jgens

Nope. Manifolds generally have no coordinate independent vector space structure. Not even locally. The idea lavinia mentioned was to use charts to locally pull the vector space structure on $\mathbb{R}^n$ back to the manifold, but this structure depends heavily on which coordinates you choose. For example, consider the space $\mathbb{R}$ as a topological manifold. Then both $(\mathbb{R},\mathrm{id})$ and $(\mathbb{R},\mathrm{id}^3)$ give us global charts, but they define totally different vector space structures. The same problem occurs in the smooth case.

Edit: To help clear up confusion, regard the domains of $\mathrm{id}$ and $\mathrm{id}^3$ as purely topological spaces. Do not give them any algebraic structure. View the codomains of these maps as real vector spaces with the natural topology.

Edit II: For further clarification, all this argument shows is that charts cannot be used to give a canonical local vector space structure on a manifold, since each choice of chart will generally produce a different a different local vector space structure. So it does not matter that $\mathbb{R}$ has a natural vector space structure.

Last edited: Feb 8, 2013
17. Feb 9, 2013

### TrickyDicky

I think we are just talking about different things. You are dealing with the local vector space structure on a manifold while I was thinking about the manifold structure of a finite real vector space that is basis independent up to diffeomorphism.
I didn't realize it when I said you were not right.

18. Feb 9, 2013

### jgens

I see. Glad there is no confusion then

19. Feb 9, 2013

### mathwonk

i assumed lavinia was thinking of smooth manifolds and thinking of the infinitesimal vector space, i.e. tangent space, structure that gives at a point.

20. Feb 9, 2013

### jgens

lavinia specifically mentioned coordinate systems providing a local vector space structure in post 13 (the post that I responded to) which makes me think he/she meant pulling a vector space structure back using charts. He/she is of course right that you can do this, I just wanted to note that this structure will generally not be coordinate independent.