# Definition of a circle in point set topology.

1. Jan 5, 2014

### center o bass

The circle seems to be of great importance in topology where it forms the basis for many other surfaces (the cylinder $\mathbb{R}\times S^1$, torus $S^1 \times S^1$ etc.). But how does one define the circle in point set topology? Is it any set homeomorphic to the set $\left\{(x,y) \in \mathbb{R}^2: x^2 + y^2 = 1\right\}$?

2. Jan 5, 2014

### tiny-tim

hi center o bass! happy new year!

it is, but as a definition of a 1D space, using a 2D subset with an induced topology seems over-elaborate

would you define a (2D) mobius strip using a subset of a 3D or 4D space?

the simplest definition would be the unit interval [0,1] (with the usual topology) with 0 and 1 identified

(or $\mathbb{R}/\mathbb{N}$)

3. Jan 6, 2014

### center o bass

Happy new year Tim! :)
I fully agree with you, and that was part of the reason why i posed the question. And thanks for the answer! Btw: Could you prehaps elaborate a bit on how
$\mathbb{R}/\mathbb{N}$ might serve as a definition of the circle? Is it somehow intuitive?

4. Jan 6, 2014

### tiny-tim

it's infinitely many copies of [0,1], with all the endpoints identfied

5. Jan 6, 2014

### center o bass

But is it homeomorphic to [0,1]? Intuitively it seems like [0,1] does not have a chance to be onto $\mathbb{R}/\mathbb{N}$.

6. Jan 6, 2014

### spamiam

Don't you mean $\mathbb{R}/\mathbb{Z}$ tiny-tim? $\mathbb{N}$ is not a group, so I'm not sure I know what it means for it to act on $\mathbb{R}$...

By the way, the notation $\mathbb{R}/\mathbb{Z}$ is somewhat ambiguous: we might mean that we are identifying all points in the subspace $\mathbb{Z} \subseteq \mathbb{R}$, which would result in an infinite bouquet of circles. Instead in this case we mean that the group $\mathbb{Z}$ acts on the topological space $\mathbb{R}$ by translations: given $n \in \mathbb{Z}$, the map $\varphi_n : \mathbb{R} \to \mathbb{R}$ given by $\varphi_n(x) = x + n$ is a homeomorphism. By identifying points that lie in the same orbit (i.e., real numbers that differ by an integer), we obtain a circle!

7. Jan 6, 2014

### lavinia

It means the same thing as when a group acts on a space. Each element of N acts as a translation of the real line and the action satifies the same conditions

0.x = x
(n+m).x = n.(m.x)

8. Jan 8, 2014

### center o bass

Btw Tim: would you define the circle similarly as a topological manifold? If so, what charts would you use to define the differential structure? :)

9. Jan 8, 2014

### WannabeNewton

Sure it's homeomorphic. Have you studied quotient spaces in topology and cosets in group theory? If so the fact that the two are homeomorphic should be a routine check for you. Keep in mind that $\mathbb{R}/\mathbb{Z}$ is the coset space, not the quotient space in the topological sense. In fact the quotient space, which is obtained by collapsing $\mathbb{Z}$ to a point, would be homeomorphic to a wedge sum of countably infinite many circles (c.f. exercise 3-18 in "Introduction to Topological Manifolds"-Lee).

10. Jan 8, 2014

### WannabeNewton

A smooth atlas (differentiable structure) is not the same thing as a topological atlas because the former requires smooth compatibility of charts so keep that in mind. The example of $S^n$ is an extremely standard one that you will find in literally every book on topological and smooth manifolds. See, for example, p.4 of "Differential Topology"-Guillemin and Pollack.

11. Jan 8, 2014

### center o bass

Yes and in all books that I've seen they define it as a set of points in R^(n+1). But like in the case of a circle in topology, isn't that superfluous? Could one not use the unit interval with identification as above and in addition specify an atlas? (I.e. a differential structure.)

12. Jan 8, 2014

### WannabeNewton

Why is that superfluous? That's the most natural definition of the unit circle $S^1$ i.e. as the set of all points a unit distance from an arbitrarily chosen origin in the affine space $\mathbb{R}^2$. The topological space that you get by considering $I/\sim$, where $\sim$ is the equivalence relation $0\sim 1$, is homeomorphic to $S^1$ (as defined in the previous sentence) so the two spaces are the exact same as far as topological structure is concerned. You aren't going to lose anything in showing $S^1$ is a topological manifold using charts in $\mathbb{R}^2$; it's the easiest way to do it and since $S^1$ is homeomorphic to $I/\sim$ there's nothing wrong in making life easier.

13. Jan 8, 2014

### economicsnerd

Silly question:
If $Y\subseteq X$ (say for topological spaces with no other structure), is it standard notation to write $X/Y$ for the space $X/\sim$, where $\sim:=\{(x,y)\in X^2:\enspace x=y \text{ or } x,y\in Y\}$? I'm not used to it, so the unit circle is the only guess I'd've had for $\mathbb R/\mathbb Z$.

14. Jan 8, 2014

### jgens

Yes.

15. Jan 8, 2014

### lavinia

A circle in geometry is the set of points in a plane that are equidistant to a single point. Circles can be centered anywhere in the plane and can have any radius.

The set of points of distance 1 to the origin is one example of a circle.

If one identifies the end points of the unit interval, one obtains a topological space. It is not a circle. Rather it is homeomorphic to a circle, in fact to any circle at all.

A one dimensional closed connected manifold without boundary is another topological space. It is not a circle either. One can prove that any such manifold is homeomorphic to a circle and also to the unit interval with its end point identified.

R/Z usually denotes the quotient group of the real numbers under addition by the integers. Since R is s topological space and since the integers act properly discontinuously on it, R/Z is also a topological space and it is homeomorphic to a circle. If one takes the homomorphism x -> exp(ix) the one gets a covering of the unit complex numbers by the real line and it is easy to use this to show that R/Z is homeomorphic to the unit complex numbers which in turn are just the unit circle in the plane.

In Topology, one often thinks not of specific spaces but of all spaces that are homeomorphic to a specific space. This leads to language that calls spaces such as R/Z or the interval with is end points identified as circles. A topologist would also call an ellipse a circle.

Last edited: Jan 8, 2014
16. Jan 9, 2014

### center o bass

A circle is equivalent with an ellipse as a manifold; I.e. a differential topologists would see these as different. It's only when curvature is taken into account (when one specifies a metric tensor), that the circle and the ellipse is equivalent.

17. Jan 9, 2014

### jgens

You have this backwards. From the perspective of differential topology a circle and an ellipse are equivalent. From the perspective of differential geometry the two are different since a circle has constant curvature while an ellipse does not.

18. Jan 9, 2014

### center o bass

I can think of some reasons: Since a manifold is a topological space with a differential structure, it has no apriori notion of distance, only the open sets and implied neighborhoods in the topology. Thus 'unit distance' is not a well defined notion as in the case with the topological circle. There is neither a notion of curvature, so the circle is equivalent to any ellipse as a manifold. Lastly, the circle is a one-dimensional manifold, and one should be able to make an 'intrinsic' definition without reference to a higher dimensional space.

I do however agree that there is nothing wrong with making life easier, but I worry that one might miss the above points by using the unit circle definition. (As I did.) I'm seeking a definition which put emphasis on the circle as a general manifold. (Not as a Riemannian one.)

I guess I could rephrase my question as: Can one define the circle, as a manifold, as the unit interval with points identified? In that case what charts should one use to specify it's differential structure?

19. Jan 9, 2014

### jgens

This is just like complaining that the standard definition of Rn comes with too much extra structure. Sure our natural choice of Sn comes with some additional structure but this is hardly a bad thing.

The standard definition as unit vectors does this. Whether the emphasis is as a differentiable or Riemannian manifold depends on the application and not the definition.

Yes this can be done. The charts are pretty easy to find.

20. Jan 9, 2014

### center o bass

It might be a bad thing if there is a risk that one might think there is something special about the unit circle from the point of view of general manifolds. I think that a definition, which does not have any additional structure, has the benefit that it emphasizes what properties a circle have, and do not have, as a general manifold.

If you look back at how this thread started with my question and TinyTim's answer we came to the agreement that a definition of a circle as the points in R^2 a unit distance away from the origin is superfluous from the point of view of topology. If one agrees with that, I do not think it is a far stretch to also agree that the same definition from the point of view of manifolds is also superfluous.

I do not think however that this definition, from the point of view of differential geometry (Riemannian manifolds), is superfluous, since then the 'canonical metric tensor' of the surface is the induced metric tensor the circle inherits from being a subspace of R^n with it's Euclidean metric tensor.