Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From page 2 in Lee's Introduction to topological manifolds:

enter image description here

Question 1: What does "describe parametrically" exactly mean? Is it a synonym for "global coordinate chart"? (that is, an atlas consisting of only one element, $(M,f)$ where $M$ is the entire manifold and $f$ is a homeomorphism $M \to \mathbb R^n$?)

Question 2: Can you give me an example of a $1$-manifold that does not admit a global coordinate chart? (that is, of a non-orientable curve?) (Is the answer that there cannot be such a manifold since non-orientable means that we can embed a Moebius band in it which we can't do in dimension $1$?)

Question 3: Is this definition compatible with this one? What's the domain of the map mentioned in the definition on Wolfram Alpha? Any one dimensional space? I doubt it since we also want $[0,1]$ to be the domain sometimes. Is this Wolfram entry incorrect?

Thanks for help!

share|cite|improve this question
I'd suggest not trying to read too much detail into the discussions in the first chapter of my book. What I was trying to do there was connect the idea of a manifold with things the reader has probably seen before, without introducing any rigorous definitions or proofs. When I wrote that space curves "are often described parametrically," I was just reminding the reader of the advanced-calculus notion of a parametric curve in $\mathbb R^3$. Most of your questions will be answered if you go on and read the next few chapters of the book. – Jack Lee Oct 22 '12 at 17:35
As an aside, there is a notion (used primarily for smooth manifolds) of a local parametrization, which is just a map whose inverse is a coordinate chart. But this isn't quite the same as a parametrized curve, because the latter need not be injective, as @KevinCarlson pointed out. – Jack Lee Oct 22 '12 at 17:37
@JackLee I am overjoyed to receive a comment from the master himself! Thank you very much, professor Lee. Also thank you for pinging me, I see you commented on one of the answers. – Rudy the Reindeer Oct 22 '12 at 17:37
up vote 1 down vote accepted

Question 1: A parameterization is not quite as strong as a(n inverse of a) global coordinate chart. For example, the curve in the line given by $|t|, -1< t < 1$ does not induce a coordinate chart, since it's not $1$-to-$1$. The image of this curve does, however, admit a global coordinate chart. The circle, on the other hand, does not, though it can be parameterized by $(\cos t,\sin t)$ with $0\leq t < 2\pi$.

Question 2: The classification of smooth $1$-manifolds already discussed does extend to topological $1$-manifolds. Specifically, the connected $1$-dimensional topological manifolds, up to homeomorphism:

  • The circle $S^1$
  • $\mathbb{R}$, or $(0,1)$
  • The half-open interval, e.g. $[0,1)$
  • The closed interval, e.g. $[0,1]$

I don't know of a detailed, published proof of this classification, but if you have access to JSTOR, here's an outline with copious hints.

Question 3: I don't think Lee means to equate curves with $1$-manifolds. One common definition is as follows:

A curve is a continuous map $f:I\to X$, where $I\subset \mathbb{R}$ is an interval and $X$ is any topological space.

The difference from $1$-manifolds is that curves may self-intersect, e.g. as the cuspidal cubic below. This is a manifold everywhere but the origin, but there it's homeomorphic to the cross.Cuspidal cubic

There are also curves that aren't $1$-manifolds anywhere. Consider the space-filling curves of Hilbert and Peano sending $[0,1]$ onto $[0,1]^2$: the images of these maps gives curves, according to the above definition, that are $2$-manifolds!

The sort of curve that is a $1$-manifold is usually called an arc-the injective image of an interval.

Wolfram's definition is strange. Since they define their context as analytic geometry, I doubt they meant the domain should be an arbitrary $1$-dimensional topological space. That would permit curves to be disconnected, too, which they normally aren't.

share|cite|improve this answer
Thank you! Regarding your definition of curve: It excludes the real line $\mathbb R$. But Wolfram seems to want the line to be also a curve. What do you think of this? – Rudy the Reindeer Oct 22 '12 at 10:10
Well, very often one is interested in curves defined on bounded intervals, but I didn't actually make that restriction: in my definition, we could have $I=\mathbb{R}$. – Kevin Carlson Oct 22 '12 at 10:12
Haha, : ) Of course, $I \neq [0,1]$! – Rudy the Reindeer Oct 22 '12 at 10:14
$[0,1]$ is the most common case! But, when we're not studying the fundamental group, it helps to let $(0,1)$ and $[0,1)$ in as well, those being enough to generate every possible $I$. – Kevin Carlson Oct 22 '12 at 10:20
You make a good point: I should have made sure to separate the curve, which is a function, from its image, which is what I meant admits a global chart. I was pointing out that this first example didn't really show that parameterized curves are a bigger class than manifolds with a global chart, while the following example of the circle does. – Kevin Carlson Oct 24 '12 at 12:22

It is an fact that every 1-dimensional Manifold is a circle or a piece of line. For instance, in the smooth case, you can take a look on the appendix of the book of Milnor - Topology from the differentiable viewpoint.

From this we can conclude that every 1-dimensional can be covered by one char (in the line case) and two charts (in the circle case). I dont know if the term "describe parametrically" is synonym for "global coordinate chart", but there is no problem in interpret it like this, in the case of line. Another interpretation is just that the author is showing us how a parametrization of an 1-manifold looks like. Probably the author will answer you question better than me and this is possible because he does participate on this forum.

Edit: I have changed my anwer a litlle bit, because Kevin Carlson pointed an error. (The circle dont have just one chart).

share|cite|improve this answer
Thank you for the reference to Milnor! The theorem in the appendix is about smooth manifolds but my question is about topological manifolds. Does the theorem in Milnor also apply to non-smooth manifolds? How should I think about it? – Rudy the Reindeer Oct 22 '12 at 8:16
@Tomas, what's a global coordinate chart for the circle? – Kevin Carlson Oct 22 '12 at 10:07
@KevinCarlson. You are right, doesnt exist. It is two charts. I will change it – Tomás Oct 22 '12 at 13:39
@MattN. The answer is no. The demonstration of this fact for the continuous case is much more elaborate. I will try to find some reference and then i post here. – Tomás Oct 22 '12 at 18:59
@MattN. Try this reference: <>;. I didnt read it, but i think it might be a good deal to verify if the demonstration is right. – Tomás Oct 22 '12 at 19:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.