# Is It More Precise to Define Manifolds as Topological Vector Spaces?

• TrickyDicky
In summary, 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.
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:
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.

$\mathbb{R}^{n}$ 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.

micromass said:
A real vector space doesn't have a topology. A vector space is a purely algebraic object.
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.

micromass said:
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.
Ok, my fault, I was identifying vector spaces over the topological field of the reals as topological vector spaces.
micromass said:
In any case, I fail to see what you dislike about the $\mathbb{R}^n$ definition.
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.

WannabeNewton said:
$\mathbb{R}^{n}$ 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.

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.

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.

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.

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

Ben Niehoff said:
To define manifolds, one only needs the topological space.

Ok, I guess this answers my question, thanks.

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.

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

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

In mathematics, a manifold is a topological space that locally resembles Euclidean space near each point. Manifolds are fundamental objects in various branches of mathematics, including differential geometry, algebraic geometry, and topology. They provide a way to study and understand curved spaces in a more tractable manner.

Formally, a topological manifold is defined as follows:

1. Topological Space: A manifold is a topological space, which means it has a collection of open sets that satisfy certain properties (e.g., they are closed under finite intersections and unions).
2. Locally Euclidean: For each point in the manifold, there exists a neighborhood around that point that is homeomorphic (topologically equivalent) to an open subset of Euclidean space (usually Euclidean n-space, denoted as R^n). This means that near every point on the manifold, it looks like flat space.
Manifolds can have different dimensions, and they can be finite-dimensional or infinite-dimensional. Common examples of manifolds include:

1. Euclidean Spaces: Euclidean space itself is a manifold, where every point locally looks like the familiar flat space.
2. Spheres: A 2-dimensional sphere (a surface of a ball) is a 2-dimensional manifold.
3. Tori: A torus (like a doughnut) is another example of a 2-dimensional manifold.
4. Real Projective Space: A 1-dimensional manifold formed by identifying antipodal points on a circle.
5. Smooth Manifolds: When you want to study differentiable functions on manifolds, you often work with smooth manifolds, which have additional structure allowing differentiation.
Manifolds serve as a crucial framework for understanding geometric and topological properties of spaces, and they have applications in various fields, including physics (e.g., general relativity), computer graphics, and data analysis. They provide a way to generalize and analyze spaces beyond simple Euclidean geometry.

Last edited by a moderator:
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.

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.

lavinia said:
Coordinates imply more structure than just the topology and for very small distances a coordinate system will appear to be a vector space structure.

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.

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

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

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:
jgens said:
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.
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.

TrickyDicky said:
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.

I see. Glad there is no confusion then

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.

mathwonk said:
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.

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.

jgens said:
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.
And I think this is an important point if only because I'd say this is a common source of confusion when going from Euclidean space to general manifolds.
As I see it this lack of a canonical basis for the local charts on the the manifold is precisely what makes the use of connections necessary to relate vectors from different tangent spaces linearizing open neighbourhoods at different points.

Last edited:
I've been pondering a bit on these replies, at first they seemed to me quite reasonable, right now I feel they were not so on target.

micromass said:
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.
Well it is true that a manifold is first of all a topological space with wichever axioms you consider that topological space to have (here I include being Hausdorff, second countable etc), that's understood. But the key property of manifolds seems to be that they can be given charts(coordinate functions) locally, and this can be seen as the property that they can be linearized at any point (they can be assigned tangent spaces at every point).
so it seems the algebraic structure of vector spaces, namely linearity, has some important role in manifolds.

Ben Niehoff said:
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.
Such operations make sense once you have the tangent space at each point and this is made possible thru the requirement of having the neighbourhoods of points homeomorphic to finite dimensional real vector space (with its natural topology of course but this should be understood from the use of the term homeomorphic).

Last edited:
TrickyDicky said:
I've been pondering a bit on these replies, at first they seemed to me quite reasonable, right now I feel they were not so on target.

Well it is true that a manifold is first of all a topological space with wichever axioms you consider that topological space to have (here I include being Hausdorff, second countable etc), that's understood. But the key property of manifolds seems to be that they can be given charts(coordinate functions) locally, and this can be seen as the property that they can be linearized at any point (they can be assigned tangent spaces at every point).
so it seems the algebraic structure of vector spaces, namely linearity, has some important role in manifolds.

Now you are talking about differentiable manifolds. The linear structure is important there because the linear structure is important to talk about derivatives and integrals.
If we talk about topological manifolds, then the linear structure seems much less important. There is no (useful) analogue for tangent spaces. Furthermore, I rarely need a linear structure when talking about topological manifolds.

Such operations make sense once you have the tangent space at each point and this is made possible thru the requirement of having the neighbourhoods of points homeomorphic to finite dimensional real vector space (with its natural topology of course but this should be understood from the use of the term homeomorphic).

I don't really know how much the neighborhoods of the points matter if you want to define a tangent space. In the philosophy of Noncommutative geometry, you just need a function ring and derivations on it. So the concept of tangent space can be greatly generalized. On the other hand, if you want to see some kind of relation between the tangent space and curves through a point, then you need smooth charts. And this relation is of course of great importance.

micromass said:
Now you are talking about differentiable manifolds. The linear structure is important there because the linear structure is important to talk about derivatives and integrals.
If we talk about topological manifolds, then the linear structure seems much less important. There is no (useful) analogue for tangent spaces. Furthermore, I rarely need a linear structure when talking about topological manifolds.

I don't really know how much the neighborhoods of the points matter if you want to define a tangent space. In the philosophy of Noncommutative geometry, you just need a function ring and derivations on it. So the concept of tangent space can be greatly generalized. On the other hand, if you want to see some kind of relation between the tangent space and curves through a point, then you need smooth charts. And this relation is of course of great importance.

Ok, you are right, I should have made that distinction clearer. Point taken, thanks micro.

I would like to add that even though what you are saying is true in general, in special cases that are heavily used in physics and engineering the distinction between topological and differentiable manifolds vanishes in the sense that for low dimensional manifolds(n<4) not only all diffeomorphisms are homeomorphisms like it's always the case, but also all homeomorphisms are diffeomorphisms, there is a unique differentiable structure so that all charts are smooth.

TrickyDicky said:
but also all homeomorphisms are diffeomorphisms

De function $f:\mathbb{R}\rightarrow\mathbb{R}:x\rightarrow x^3$ is a homeomorphism and not a diffeomorphism. So the statement is not even true for n=1.

TrickyDicky said:
but also all homeomorphisms are diffeomorphisms
This is not even remotely true. You can define homeomorphisms between two topological manifolds that have no smooth atlas in which case the notion of being a diffeomorphism doesn't even make sense for the homeomorphism.

micromass said:
De function $f:\mathbb{R}\rightarrow\mathbb{R}:x\rightarrow x^3$ is a homeomorphism and not a diffeomorphism. So the statement is not even true for n=1.

Sorry. I'm talking about homeomorphic manifolds. Thought it was clear by the context.

TrickyDicky said:
Sorry. I'm talking about homeomorphic manifolds. Thought it was clear by the context.

But $\mathbb{R}$ is homeomorphic (and even diffeomorphic) to $\mathbb{R}$...

micromass said:
But $\mathbb{R}$ is homeomorphic (and even diffeomorphic) to $\mathbb{R}$...

Yes, I realize now what I posted is a non-sequitur, the fact that homeomorphic smooth manifolds with dim. n<4 are also diffeomorphic doesn't imply that there is no distinction between topological and smooth manifolds in those dimensions.
Gee, this is tough and I'm too fast drawing conclusions. I'll try to refrain that.

TrickyDicky said:
Well it is true that a manifold is first of all a topological space with wichever axioms you consider that topological space to have (here I include being Hausdorff, second countable etc), that's understood. But the key property of manifolds seems to be that they can be given charts(coordinate functions) locally, and this can be seen as the property that they can be linearized at any point (they can be assigned tangent spaces at every point).
so it seems the algebraic structure of vector spaces, namely linearity, has some important role in manifolds.

You can have a chart on the point of a cone.

People are actually trying to help you Tricky, best listen and ask for clarification where you don't understand, rather than jumping down their throats.

Ben Niehoff said:
You can have a chart on the point of a cone.
So what?. As micromass noticed I was actually demanding smooth charts, can you have a smooth chart on the point of the cone?

Ben Niehoff said:
People are actually trying to help you Tricky, best listen and ask for clarification where you don't understand,
That's exactly what I'm doing, read the posts.
Ben Niehoff said:
rather than jumping down their throats.
Hmmm, this inaccurate remark can only be attributed off the top of my head to some unexplainable personal grudge you hold against me. If that is the case I pity you, but there isn't much I can do and I particularly don't care anyway, it does bother me a bit that it might spoil an otherwise peaceful(even cheerful) thread. Hopefully it won't.

Just be careful in what structures you pre - impose on your manifold. Things that hold for smooth manifolds don't necessarily have to hold for the more general topological manifold. The point is that in field theories like GR or classical field theory, you almost always only deal with smooth manifolds for obvious reasons so the mention of topological manifolds doesn't really come up but that certainly doesn't mean there is a vanishing of the distinction between topological manifolds and smooth manifolds.

TrickyDicky said:
I've been pondering a bit on these replies, at first they seemed to me quite reasonable, right now I feel they were not so on target.

I was referring to statements like this where you externalize your misunderstandings. Everyone else is "not so on target" rather than you having made a mistake. It makes it annoying to answer your questions.

I see in your subsequent responses you acted differently. In the past, you often got belligerent; maybe you're not anymore.

WannabeNewton said:
Just be careful in what structures you pre - impose on your manifold. Things that hold for smooth manifolds don't necessarily have to hold for the more general topological manifold. The point is that in field theories like GR or classical field theory, you almost always only deal with smooth manifolds for obvious reasons so the mention of topological manifolds doesn't really come up but that certainly doesn't mean there is a vanishing of the distinction between topological manifolds and smooth manifolds.

You are totally right, thanks WN.

Replies
20
Views
3K
Replies
44
Views
2K
Replies
3
Views
702
Replies
1
Views
2K
Replies
21
Views
1K
Replies
7
Views
3K
Replies
4
Views
2K
Replies
25
Views
2K
Replies
13
Views
2K
Replies
14
Views
2K