Is it possible to prove that the circle is a manifold using open spheres?

[davidge]

The answer to the question of the thread title is yes, according to what I found on web. Now a manifold is by definition a topological space that (aside from other conditions) is locally Euclidean.

What does such condition means? Is it the same as saying that each of its points must have a homeomorphic neighborhood to $\mathbb{R}^n$ for a n-dimensional manifold?

If so, how can we prove that it's possible to find an open neighborhood around each point of the circle? Could we use open spheres to do that?

I have tried to prove this by considering the unit circle in $\mathbb{R}^2$ (with the usual metric) as
$$\mathbb{S}^1: \big[x \in \mathbb{R}^2 \ | \ d(x,0) = 1 \big]$$ where $0$ is the point where the circle is centered in. Let $P_0$ be the center of an open sphere on the circle.
We have $d(0,x) = 1 = d(0,P_0)$

Using the "triangle inequality" we have
$d(0,P_0) + d(P_0, x) \leq d(0,x)$

The solution is $d(P_0,x) = 0$. This says that $P_0$ and $x$ are the same point! That is, the open sphere contains only its center point. Let's call the open sphere $A$.

Now as this open sphere is in $\mathbb{R}^2$ a parametrization for its points could be $\gamma: [0,\delta) \times [0, 2 \pi) \longrightarrow A$ (the inverse mapping is the usual spherical polar coordinates). We would need both intervals in $[0,\delta) \times [0, 2 \pi)$ to be open in order to it to be homeomorphic to $\mathbb{R}^2$. But if we do that for our open sphere $A$, we will exclude its unique point, namely the central point $P_0$. The solution would be that it doesn't contain any points at all. But then our proof that $\mathbb{S}^1$ is a manifold fails. How do we solve the problem?

 

[fresh_42]

The answer to the question of the thread title is yes, according to what I found on web. Now a manifold is by definition a topological space that (aside from other conditions) is locally Euclidean.

Yes.

What does such condition means? Is it the same as saying that each of its points must have a homeomorphic neighborhood to $\mathbb{R}^n$ for a n-dimensional manifold?

Yes.

If so, how can we prove that it's possible to find an open neighborhood around each point of the circle? Could we use open spheres to do that?

Why open spheres? The usual charts are the two stereographic projections from the poles.

I have tried to prove this by considering the unit circle in $\mathbb{R}^2$ (with the usual metric) as
$$\mathbb{S}^1: \big[x \in \mathbb{R}^2 \ | \ d(x,0) = 1 \big]$$ where $0$ is the point where the circle is centered in. Let $P_0$ be the center of an open sphere on the circle. We have
$d(0,x) = 1 = d(0,P_0)$
Using the "triangle inequality" we have
$d(0,P_0) + d(P_0, x) \leq d(0,x)$
The solution is $d(P_0,x) = 0$. This says that $P_0$ and $x$ are the same point! That is, the open sphere contains only its center point. Let's call the open sphere $A$.

Now as this open sphere "lives" in $\mathbb{R}^2$ a parametrization for its points could be $\gamma: [0,\delta) \times [0, 2 \pi) \longrightarrow A$ (the inverse mapping is the usual spherical polar coordinates), with $\delta > 0; \delta \in \mathbb{R}$. We would need both intervals in $[0,\delta) \times [0, 2 \pi)$ to be open in order to it to be homeomorphic to $\mathbb{R}^2$. But if we do that for our open sphere $A$, we will exclude its unique point, namely the central point $P_0$. The solution would be that it doesn't contain any points at all. But then our proof that $\mathbb{S}^1$ is a manifold fails. How do we solve the problem?

I don't understand what you want in $\mathbb{R}^2$. The circle is one dimensional, so $\mathbb{R}^1$ is the Euclidean space of choice. Draw a line from the northpole through a point of the circle until you reach the $x-$axis ($\mathbb{R}^1$) and you get on the $x-$axis the homeomorphic Euclidean chart. For the points below the $x-$axis do the same from the south pole. This gives you the two charts you need.

 

[davidge]

The usual charts are the two stereographic projections from the poles

This is a very easy way to show that $\mathbb{S}^1$ minus one of its points is homeomorphic to $\mathbb{R}$. But is there a way of showing that by using open "1-dimensional spheres"?

Why open spheres?

Just to see how the proof works if we use open spheres.

I don't understand what you want in $\mathbb{R}^2$. The circle is one dimensional, so $\mathbb{R}^1$ is the Euclidean space of choice. Draw a line from the northpole through a point of the circle until you reach the $x-$axis ($\mathbb{R}^1$) and you get on the $x-$axis the homeomorphic Euclidean chart. For the points below the $x-$axis do the same from the south pole. This gives you the two charts you need.

I see

 

[fresh_42]

A sphere isn't flat, Euclidean spaces are. Additionally why replace one sphere by another? With which goal? Of course you can as well use many small sections of tangents as charts, similar to ordinary atlases in the real world. This decreases the error to the price of more charts.

 

[WWGD]

Just look at arcs about a point. They are twisted open intervals in the line. The twisting is " gentle-enough" to preserve (smooth, not just topological) equivalence with the Reals; no corners, nothing too nasty. EDIT: though the cricle is homeomorphic, though not diffeomorphic with a square (in standard embedding, differentiable structure)

 

[WWGD]

The answer to the question of the thread title is yes, according to what I found on web. Now a manifold is by definition a topological space that (aside from other conditions) is locally Euclidean.

What does such condition means? Is it the same as saying that each of its points must have a homeomorphic neighborhood to $\mathbb{R}^n$ for a n-dimensional manifold?

If so, how can we prove that it's possible to find an open neighborhood around each point of the circle? Could we use open spheres to do that?

I have tried to prove this by considering the unit circle in $\mathbb{R}^2$ (with the usual metric) as
$$\mathbb{S}^1: \big[x \in \mathbb{R}^2 \ | \ d(x,0) = 1 \big]$$ where $0$ is the point where the circle is centered in. Let $P_0$ be the center of an open sphere on the circle.
We have $d(0,x) = 1 = d(0,P_0)$

Using the "triangle inequality" we have
$d(0,P_0) + d(P_0, x) \leq d(0,x)$

The solution is $d(P_0,x) = 0$. This says that $P_0$ and $x$ are the same point! That is, the open sphere contains only its center point. Let's call the open sphere $A$.

Now as this open sphere is in $\mathbb{R}^2$ a parametrization for its points could be $\gamma: [0,\delta) \times [0, 2 \pi) \longrightarrow A$ (the inverse mapping is the usual spherical polar coordinates). We would need both intervals in $[0,\delta) \times [0, 2 \pi)$ to be open in order to it to be homeomorphic to $\mathbb{R}^2$. But if we do that for our open sphere $A$, we will exclude its unique point, namely the central point $P_0$. The solution would be that it doesn't contain any points at all. But then our proof that $\mathbb{S}^1$ is a manifold fails. How do we solve the problem?

On top of the local homeomorphism that must exist, this local homeomorphisms, must be "glued " to each other nicely-enough. You may say that a manifold is just a collection of Euclidean parts glued in/with "nice-enough" maps.

 

[lavinia]

Here is another way to show that spheres are manifolds. Take the standard $n$-sphere of radius 1 centered at the origin of $R^{n+1}$ Project the northern hemisphere onto the open unit disk in $R^{n}$ by dropping the last coordinate. (For the 2-sphere in $R^3$ one would drop the z coordinate.) This is a coordinate neighborhood of the north pole. Now take any other point on the sphere and rotate the sphere so that this point moves to the north pole. Follow this rotation by the same projection.

- For the 2-sphere considered as the Riemann sphere, the two charts may be chosen to be $z$ and $1/z$ .

- As you may know from calculus of several variables, the Implicit Function Theorem says that if $f$ is a smooth function defined on $R^{n+1}$ and its gradient is nowhere zero on the set $f(x) = c$ for some constant $c$, then this set can be locally parameterized by a open region in $R^{n}$. The inverses of these local parameterizations are coordinate charts.

Consider the function defined on $R^{n+1}$, $f(x) = ||x||^2$. For instance in $R^2$ this function is $f(x,y) = x^2+y^2$. A sphere of radius $k$ centered at the origin is the set of points $f(x) = k^2$. Notice that the gradient of $f$ is nowhere zero on this set. So by the Implicit Function Theorem, the sphere is a manifold (actually a smooth manifold).

More generally the Implicit Function Theorem works for a smooth function defined on any manifold not just Euclidean space. For instance stand a torus up on the plane and let $f$ be the height function of each point above the plane. $f$ is smooth and only has four critical points, the top and bottom of the torus - these are the absolute maximum and minimum - and the tops and bottom of the inner circle - these are the two saddle points. Everywhere else $df$ is non-singular. The Implicit Function Theorem guarantees that the points $f(x) = c$ is a 1 dimensional manifold whenever none of the four critical points are in $f^{-1}(c)$.

The Implicit Function Theorem is critically important. It would be useful to understand how it is used to find manifolds.

 

[davidge]

Just look at arcs about a point. They are twisted open intervals in the line. The twisting is " gentle-enough" to preserve (smooth, not just topological) equivalence with the Reals; no corners, nothing too nasty. EDIT: though the cricle is homeomorphic, though not diffeomorphic with a square (in standard embedding, differentiable structure)

Yes, this is a good way to see how it works, though it's a bit qualitative

On top of the local homeomorphism that must exist, this local homeomorphisms, must be "glued " to each other nicely-enough. You may say that a manifold is just a collection of Euclidean parts glued in/with "nice-enough" maps.

Cool

A sphere isn't flat, Euclidean spaces are. Additionally why replace one sphere by another? With which goal? Of course you can as well use many small sections of tangents as charts, similar to ordinary atlases in the real world. This decreases the error to the price of more charts.

I see

@lavinia it's interesting to know about this way of finding a manifold. Thanks for pointing out.

 

[WWGD]

Yes, this is a good way to see how it works, though it's a bit qualitative
.

Sorry, I can't think of a way of making this more formal without making use of e.g., tangent spaces and the tangent map. Have you seen them?

 

[davidge]

Sorry, I can't think of a way of making this more formal without making use of e.g., tangent spaces and the tangent map. Have you seen them?

Yes, I do. Go ahead.

 

[WWGD]

Yes, I do. Go ahead.

Ok, a diffeomorphism gives rise to a(n) linear isomorphism between the tangent spaces ( of/between standard circle, square) . But the tangent space at the corners/edges of the square is not well-defined, so there can be no tangent space isomorphism, therefore no diffeomorphism.

 

[davidge]

the tangent space at the corners/edges of the square is not well-defined

Is the Jacobian determinant equal to zero for those points?

 

[WWGD]

Yes, but you would have to parametrize both and find a map between them. Once the result holds, it is independent of the choice of parametrization. EDIT: You can, e.g., inscribe the square in the circle and then map a point in the circle to the square by drawing a line from the center of the circle until it hits the square.

 

[davidge]

Yes, but you would have to parametrize both and find a map between them. Once the result holds, it is independent of the choice of parametrization.

Ah, ok. Thanks.

 

[WWGD]

Maybe this is more precise: consider an arc in the circle mapping just left of the , say, (1,1) corner in the square, i.e., on (x(t),1) and another arc, mapping along (1, y(t)) and see what happens when $x(t),y(t) \rightarrow 1$ . You can use coordinates and compute the actual tangent map (i.e., the Jacobian) , which will not be invertible at that point.

 

[lavinia]

On top of the local homeomorphism that must exist, this local homeomorphisms, must be "glued " to each other nicely-enough. You may say that a manifold is just a collection of Euclidean parts glued in/with "nice-enough" maps.

You seem to be thinking of a smooth manifold where the composition of a chart with a local parameterization must be a diffeomorphism. For a general topological manifold there are no such requirements as far as I know.

 

[WWGD]

You seem to be thinking of a smooth manifold where the composition of a chart with a local parameterization must be a diffeomorphism. For a general topological manifold there are no such requirements as far as I know.

Yes, you're right, I was assuming a smooth manifold, still, the gluing maps must be at least nice-enough to be homeomorphisms.

 

[lavinia]

Yes, you're right, I was assuming a smooth manifold, still, the gluing maps must be at least nice-enough to be homeomorphisms.

The gluing maps are automatically homeomorphisms.

If one defines a manifold as a locally Euclidean space of a fixed dimension: That is there is a open neighborhood of any point that is homeomorphic to an open set in $R^{n}$ then the inverse of any such homeomorphism is by definition also a homeomorphism. So the composition of an inverse with another local homeomorphism must be a homeomorphism on the overlap of the two open sets.

 

[WWGD]

The gluing maps are automatically homeomorphisms.

If one defines a manifold as a locally Euclidean space of a fixed dimension: That is there is a open neighborhood of any point that is homeomorphic to an open set in $R^{n}$ then the inverse of any such homeomorphism is by definition also a homeomorphism. So the composition of an inverse with another local homeomorphism must be a homeomorphism on the overlap of the two open sets.

Well, yes, this is what I said, that if we start with a collection of Euclidean open sets, we patch them together into ( topological, C^k, smooth) manifolds by using maps in the respective categories.

 

[lavinia]

Well, yes, this is what I said, that if we start with a collection of Euclidean open sets, we patch them together into ( topological, C^k, smooth) manifolds by using maps in the respective categories.

Right. But that is a different question. In this thread one starts with a topological space and asks when is it an n dimensional manifold. For this one only needs to say that each point has an open neighborhood that is homeomorphic to and open set in $R^{n}$.Nothing needs to be said about overlapping domains. If one wants additional structure such as differentiability then one needs to say something about the coordinate transformations. But this is a restricted situation.

If one starts with a collection of sets that are homeomorphic to open sets in $R^{n}$ then one only needs to say how they overlap. Nothing else about gluing is required to make a topological manifold.

 

[WWGD]

Right. But that is a different question. In this thread one starts with a topological space and asks when is it an n dimensional manifold. For this one only needs to say that each point has an open neighborhood that is homeomorphic to and open set in $R^{n}$.Nothing needs to be said about overlapping domains. If one wants additional structure such as differentiability then one needs to say something about the coordinate transformations. But this is a restricted situation.

If one starts with a collection of sets that are homeomorphic to open sets in $R^{n}$ then one only needs to says how they overlap. Nothing else about gluing is required to make a topological manifold.

Well, yes, but I was following up on a comment I made in another post that I thought OP was referring to, and gluing just means to me putting the patches together according to some rule, here the rule being identifying a point with its image using a homeomorphism.

 

[WWGD]

Right. But that is a different question. In this thread one starts with a topological space and asks when is it an n dimensional manifold. For this one only needs to say that each point has an open neighborhood that is homeomorphic to and open set in $R^{n}$.Nothing needs to be said about overlapping domains. If one wants additional structure such as differentiability then one needs to say something about the coordinate transformations. But this is a restricted situation.

If one starts with a collection of sets that are homeomorphic to open sets in $R^{n}$ then one only needs to says how they overlap. Nothing else about gluing is required to make a topological manifold.

Well, yes, but I was following up on a comment I made in another post that I thought OP was referring to, and gluing just means to me putting the patches together according to some rule, here the rule being identifying a point with its image using a homeomorphism. Still, I will make an edit to make it explicit:

EDIT: My comment about patching together pieces of Euclidean open sets is a side note, not related directly to the matter of this post.

 

[WWGD]

This is, I think, another argument: we know the circle $\mathbb S^1$ is a smooth manifold, while the square S is not; a smooth manifold cannot be diffeomorphic with a non-smooth manifold ( let's show these two). ..

1) Square ( $\partial( [0,1] \times [0,1])$) is not a smooth manifold: Rotate the square by 45 degrees in either direction, so that we end up with an " inverted v " or a /\ . We can parametrize the /\ using two line segments. But the two line segments will have slopes of different signs. This means any diffeomorphism between any interval (a,b) and /\ will have both positive derivatives and negative derivatives, meaning it will have f'(x)=0 at some point, so it cannot be a diffeomorphism. Now, rotating back using R will not make a difference, since rotation is a diffeomorphism, i.e., if R(S) admits smooth charts, so does $R^{-1}R(S)=S$. So there is no differentiable map between ( a corner of ) the square and an interval (a,b) , so the square cannot be given smooth charts.

2) Now we need to show that a smooth manifold cannot be diffeomorphic with a non-smooth manifold : if there was a diffeomorphism, then we would be able to pullback smooth charts using the diffeomorphism.

Please check the argument.

EDIT: Of course we can give the square smooth charts using this: If M,N are homeomorphic and M is a smooth manifold, then N can be made into a smooth manifold by pulling back smooth charts. from M using the homeomorphism.

 

[lavinia]

This is, I think, another argument: we know the circle $\mathbb S^1$ is a smooth manifold, while the square S is not; a smooth manifold cannot be diffeomorphic with a non-smooth manifold ( let's show these two). ..

1) Square ( $\partial( [0,1] \times [0,1])$) is not a smooth manifold: Rotate the square by 45 degrees in either direction, so that we end up with an " inverted v " or a /\ . We can parametrize the /\ using two line segments. But the two line segments will have slopes of different signs. This means any diffeomorphism between any interval (a,b) and /\ will have both positive derivatives and negative derivatives, meaning it will have f'(x)=0 at some point, so it cannot be a diffeomorphism. Now, rotating back using R will not make a difference, since rotation is a diffeomorphism, i.e., if R(S) admits smooth charts, so does $R^{-1}R(S)=S$. So there is no differentiable map between ( a corner of ) the square and an interval (a,b) , so the square cannot be given smooth charts.

2) Now we need to show that a smooth manifold cannot be diffeomorphic with a non-smooth manifold : if there was a diffeomorphism, then we would be able to pullback smooth charts using the diffeomorphism.

What you are getting at seems to be that at the corners a square does not have a well defined tangent space. So it is not a smoothly embedded submanifold of the plane. It one takes as the definition of a differentiable manifold a subset of Euclidean space that has a well defined continuously varying tangent plane at each of its points then the square is not even a differentiable manifold.

It seems then that it makes no sense to talk about a diffeomorphism between it and a circle - since it is not a differentiable manifold. Yes?

To me, a smooth manifold is defined as a locally Euclidean topological space whose coordinate transformations are diffeomorphisms. The Whitney Embedding Theorem says that one can always smoothly embed a smooth manifold in some Euclidean space so in some sense it suffices to think of smooth manifolds as subsets of Euclidean space. From this point of view, the square is not a smooth manifold since it is not a smooth embedding of a smooth manifold into the plane.

 

[lavinia]

@WWGD There is a subtlety here which merits mentioning. For an abstract smooth manifold coordinate charts are defined as homeomorphisms and the differentiable structure arises by requiring that the coordinate transformations be diffeomorphisms. That is: No concept of smoothness is used in defining a chart. The smoothness arises from the coordinate transformations on overlapping coordinate domains.

For a smooth manifold embedded in Euclidean space however, a coordinate chart must be smooth. What does this mean? After all a submanifold has measure zero. One needs an idea of smoothness for a subset of Euclidean space. The general definition is that the map on the subset must be extendable to a smooth map on an open set in Euclidean space around of each point. This definition works for any subset of Euclidean space. One then requires the chart to be smooth and that its inverse should also be smooth.

In this way the manifold inherits its differentiable structure from Euclidean space(and also its geometry). Without these requirements the differentiable structure is not inherited and one could define the square as a smooth manifold.

 

[WWGD]

@WWGD There is a subtlety here which merits mentioning. For an abstract smooth manifold coordinate charts are defined as homeomorphisms and the differentiable structure arises by requiring that the coordinate transformations be diffeomorphisms. That is: No concept of smoothness is used in defining a chart. The smoothness arises from the coordinate transformation on overlapping coordinate domains.

For a smooth manifold embedded in Euclidean space however, a coordinate chart must be smooth. What does this mean? After all a submanifold has measure zero. One needs an idea of smoothness for a subset of Euclidean space. The general definition is that the map on the subset must be extendable to a smooth map on an open set in Euclidean around of each point. This definition works for any subset of Euclidean space. One then requires the chart to be smooth and that its inverse should also be smooth.

In this way the manifold inherits its differentiable structure from Euclidean space(and also its geometry). Without these requirements the differentiable structure is not inherited and one could define the square as a smooth manifold.

Right, good point; we can just pullback charts along the homeomorphism, to smooth out the square: Say $(U, \Phi), (V, \Psi)$ are smooth charts for the circle so that $\Phi \circ \Psi^{-1}$ and $\Psi \circ \Phi^{-1}$ are smooth, so that $h(x)$ in $U \cap V$ . Then we can send $x$ in the square to $h(x)$ in the circle, which is a smooth chart.at $h(x)$ so that $\Psi \circ Phi^{-1}(h(x))$ and $\Phi \circ \Psi^{-1}(h(x))$ are smooth charts for x .Is that what you were thinking? I think a continuous injection between the two is enough.

I don't like so much the heavy machinery, but it can come in handy:
This may be a case of just pullbacks , in a more general sense, with M,N smooth manifolds as objects and difeomorphisms as the morphisms in a category for smooth manifolds.

 

[mathwonk]

one nice way to recognize a set as a smooth manifold is to use the implicit function theorem. i.e. if we have a smooth map f:R^n-->R^m and a point p in the target R^m such that at every point of the preimage f^-1(p), the derivative map is surjective as a linear map, then that fiber is a manifold.

in this case if we consider the map R^2-->R define by (x,y)-->x^2+y^2, then the circle is the inverse image of the point 1 in R, and since the derivative as a linear map is the matrix [ 2x 2y], which is a surjection at every (x,y) with x^2+y^2 = 1, we are home free!

 

[ConfusedMonkey]

I think a continuous injection between the two is enough.

I think you need a homeomorphism. You can "transport the smooth structure" of the circle onto the square by taking smooth charts on the square of the form $(V, \varphi \circ h^{-1})$ where $h$ is a homeomorphism from the square onto the circle (in fact I think a local homeomorphism is enough). If $h$ was just a continuous injection then there is no guarantee that the maps $\varphi \circ h^{-1}$ are continuous.

 

[WWGD]

I think you need a homeomorphism. You can "transport the smooth structure" of the circle onto the square by taking smooth charts on the square of the form $(V, \varphi \circ h^{-1})$ where $h$ is a homeomorphism from the square onto the circle (in fact I think a local homeomorphism is enough). If $h$ was just a continuous injection then there is no guarantee that the maps $\varphi \circ h^{-1}$ are continuous.

Good point, but there is a result that a continuous bijection between compact and Hausdorff is a homeomorphism.. Should have said/meant to say: continuous bijection in previous.

 

[ConfusedMonkey]

Ah, yes. I always forget about that very useful result.