[–]Applied Math 0 points1 point

sorry, this has been archived and can no longer be voted on

What about this way? We'll view the circle as an ellipse with equal length axes, and call it [; E_0 ;] Now squish the circle, so that the semi-minor axis (I'm visualizing this as the vertical axis, but squishing along a diagonal axis would give you your 'M' and offsetting the circle vertically would give you your 'b') is half the original radius, and the semi-major axis is twice the original radius, preserving the area of the circle. Call this [; E_1 ;]. We'll keep with this pattern, so that, letting [; r ;] be the radius of the original circle, [; E_n ;] is an elipse with semi-minor axis of length [; \frac{r}{2^n} ;] and semi-major axis of length [; r2^n ;]. In this way we keep stretching the circle out while preserving it's area.

Now let's take [; lim_{n\rightarrow \infty} E_n ;]. We find a sort of double-tailed Dirac Delta function with infinitessimal width, infinite length, and an area of [; \pi r^2 ;], which to my estimation is as much a line as anything else you could think of.

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

Do you want a "morph" kind of transform, like the Y axis in this animation?

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

To understand why this isn't as simple as it seems (because of the lack of a full homeomorphism between the two), try this lecture on Algebraic Topology.

Now, be warned: Professor Wildberger is rather famous for categorically denying the existence of infinite sets (and, with them, things like the Real numbers). However, this finitist "ultra-constructivism" has an advantage in pedagogy (in my opinion). He likes to show things very graphically and also uses projective geometry to create a compass and straightedge illustration of the group structure of the circle. So while I strongly disagree with his philosophical stance, it makes for a great way to learn algebraic topology and, in turn, why what you're asking for can't be done in a way that preserves structure.

[–] 11 points12 points

sorry, this has been archived and can no longer be voted on

I'm not sure if I totally understand your question, but you could use a stereographic projection to map (most of) the points of a circle to the points on a line. If you remove the north pole from your circle, this map is smooth and bijective, which is rather nice.

Alternatively if we write the equation for the circle in polar coordinates (r, θ), it's r = R, which technically is affine linear.

[–]Low-Dimensional Topology 8 points9 points

sorry, this has been archived and can no longer be voted on

Pick your favorite Möbius transformation z -> (az+b)/(cz+d), where you think of R2 as the complex plane and a,b,c,d are complex numbers such that ad-bc is nonzero. If the complex number -d/c lies on your circle, then this transformation will send the circle to a line. For example, if your circle is defined by |z|=r then the transformation z -> 1/(z-r) will suffice.

The reason for this is that Möbius transformations are really transformations of the Riemann sphere (a sphere composed of the complex plane plus a point at infinity) which send circles to circles. A line in the plane is the same thing as a circle through that point at infinity on the Riemann sphere, and the transformation z -> f(z) =(az+b)/(cz+d) will send any circle through -d/c to a circle through f(-d/c) = \infty, i.e. to a line.

[–] -1 points0 points

sorry, this has been archived and can no longer be voted on

Could you elaborate what you mean by "line" in this case? The full S1 cannot be mapped homeomorphically to a one-dimensional subvectorspace of R2 .

[–]Low-Dimensional Topology 5 points6 points

sorry, this has been archived and can no longer be voted on

The transformation is undefined at -d/c, so as a map from C to C it's really sending the complement of that point to the line. In the Riemann sphere the point -d/c would go to the point at infinity, sending the original S1 to the new circle formed by {line} u {infinity}.

I would also caution against using fancier language then necessary here (or in general), because it can either obfuscate what you're trying to say or make it wrong. A simple "R" or "the real line" would have sufficed instead of "a one-dimensional subvectorspace of R2 ". In fact, the subspace language makes it sound like you only want lines passing through the origin, when really this applies to any line at all in the plane, and for that matter the topology of the real line has nothing to do with how you might embed it in the plane. And if you were using that language to also imply that you want the image to be a straight line, then "homeomorphically" isn't the right choice of word because it applies just as well to, say, the graph of sin(x). This is why it's best to be concise, choose your definitions carefully, and even use active voice if you can: for example, "S1 is not homeomorphic to the real line."

[–] -1 points0 points

sorry, this has been archived and can no longer be voted on

I would also caution against using fancier language then necessary here (or in general), because it can either obfuscate what you're trying to say or make it wrong.

I intended to make myself more precise, as a "line" can be everything from a parametrisation of some compact intervall to the real line. Instead of going through all cases, I'd rather a fairly easy one. In any case, since you can identify any line parametrised over R with any one-dimensional vector space as a topological space, this discussion is really moot. As you seemed to know what you were talking about, I had hoped to make it more clear what I intended "line" to mean for the OP.

Homeomorphically is indeed the right choice of word in my view. It is the correct use both my statement and what I considered "making a circle into a line" could mean. Yes, the graph of sin(x) could also be a line, but since this is again homeomorphic to the real line, why should one bother to make it more complicated?

Of course, it is not the only one. But before one starts with isometries or anything more fancy, one should use more down-to-earth formulations, as you have told me yourself.

In this light, I'd like to make clear that I have no intentions to hide behind language, but I appreciate your input. I do not understand how this would apply to definitions or my wordings however, as I only use the very canonical ones everyone makes use of.

Now, I would like to make sure I understood what you stated, as I am not very familiar with Möbius transformations. The Möbius transformation sends an S1 to the compactification of R, which is, again S1?

And how is this less complicated than what I stated, or even a "line", as stated like this

If the complex number -d/c lies on your circle, then this transformation will send the circle to a line

by yourself?

[–]Low-Dimensional Topology 3 points4 points

sorry, this has been archived and can no longer be voted on

Let me address the math first: you're right about the image being S1 in the Riemann sphere. If you're just thinking about the Möbius transformation as acting on points in the usual complex plane than the image really is a line, because there's no point at infinity. (The Riemann sphere is a more natural setting -- in a complex analysis class or beyond you may see that the group formed by Möbius transformations is exactly its automorphism group, PGL(2,C) -- but we don't necessarily have to use it.) Also "the" compactification of R is misleading because S1 isn't the only compactification; in many other contexts it's natural to compactify it as a closed interval, by adding both a positive infinity and a negative infinity.

As for the wording, you need to consider your audience. The extra language about vector spaces could definitely confuse an OP who admits to having only taken courses up to differential equations, because they may not have ever seen linear algebra, much less homeomorphisms or any sort of topology. On the other hand, a lot of people learn about the complex plane in high school, so I hope (but can't guarantee) that my answer won't be lost on the OP.

If you had included the same language in a research paper I was refereeing anonymously, I would still ask you to change it in my report because it's less readable even to a reader who understands all of these terms than "the real line", which is precise and 100% unambiguous as far as I can tell. Why bring linear algebra into the picture when you're just talking about a particular topological space? If you only care about the homeomorphism type of that space, why mention an embedding of it into some other space? (Chekhov's gun applies to both of these questions.)

I don't mean to accuse you of having impure intentions or anything, and I don't claim to be a flawless writer myself, but when you spend lots of time reading other people's papers you really start to appreciate it when the author makes an effort to communicate things as clearly and as simply as possible. I think that a lot of mathematicians (and this is not a personal accusation directed at you) underestimate the importance of good writing skills in general.

[–] -1 points0 points

sorry, this has been archived and can no longer be voted on

Let me address the math first: you're right about the image being S1 in the Riemann sphere. If you're just thinking about the Möbius transformation as acting on points in the usual complex plane than the image really is a line, because there's no point at infinity. (The Riemann sphere is a more natural setting -- in a complex analysis class or beyond you may see that the group formed by Möbius transformations is exactly its automorphism group, PGL(2,C) -- but we don't necessarily have to use it.) Also "the" compactification of R is misleading because S1 isn't the only compactification; in many other contexts it's natural to compactify it as a closed interval, by adding both a positive infinity and a negative infinity.

Yes, you're right in all you say. I assumed I was talking to you, not the OP now, so I sloppily only called it "the compactification". Sorry about that.

I still need to know, however. I am not familiar with much complex analysis, but you can just about throw anything at me to explain it: is there anyway we can obtain a topological space to view the image of S1 under a Möbius transformation to be a line? That's what I read in your first post, and I am still confused.

I know I'm repeating myself here, but can you tell me what you mean by line, which space we are considering, and how the S1 is mapped to that?

As for the wording, you need to consider your audience. The extra language about vector spaces could definitely confuse an OP who admits to having only taken courses up to differential equations, because they may not have ever seen linear algebra, much less homeomorphisms or any sort of topology.

I might have overestimated what people do in differential equations courses then. How do you do ODEs or PDEs without knowing linear algebra and some metric spaces (of which topology is just the generalisation)?

Without this, you aren't even able to introduce the differential of maps between vector spaces of arbitary dimensions (or, if you like, just Rn and Rm ). How are you supposed to do differential equations without that?

I'm really confused now, as you can tell.

If you had included the same language in a research paper I was refereeing anonymously, I would still ask you to change it in my report because it's less readable even to a reader who understands all of these terms than "the real line", which is precise and 100% unambiguous as far as I can tell

And to me it isn't unambiguous, that's where our troubles start, I guess. If we first agree on calling it closed or open interval, then there's no problem. But just imagining a line .. what do you see? The real line, or maybe a closed interval?

As it wasn't clear what he meant by line, I didn't know if he knew that when I state "real line", I really mean it. It could just be "parts of the real line", or maybe he didn't really know the difference between taking different parts - I know, not a very common mistake, but I've seen it done once or twice already.

So I wanted to make sure this wouldn't occur again, and based my assumption on that you need linear algebra to do differential equations (which I now accept to be wrong, apparently).

I agree with you on the last part, but as you can see, I wasn't trying to make it more complicated. Quite to the contrary. My mistake then was that I was thinking too highly of the prerequisites of differential equations, or too much about how "real line" could be mistunderstood. How ironic.

[–]Low-Dimensional Topology 2 points3 points

sorry, this has been archived and can no longer be voted on

is there anyway we can obtain a topological space to view the image of S1 under a Möbius transformation to be a line?

If we view it as a transformation of the complex plane (no Riemann sphere, no point at infinity), then the answer is yes: the transformation z -> (az+b)/(cz+d) is simply undefined at z=-d/c, so if -d/c lies on the circle then the Möbius transformation when restricted to the circle is only defined on the complement of that point, and since S1\{*} is homeomorphic to the real line there's no problem. Otherwise, if you have a point at infinity then the answer is no, since the continuous image of a compact set (the circle) must be compact (not a line).

How do you do ODEs or PDEs without knowing linear algebra and some metric spaces (of which topology is just the generalisation)?

It's maybe not an ideal way to do it, but it does happen, especially in lower-level courses aimed at non-majors. See for example the basic ODE classes at MIT or at UT Austin, both highly ranked departments, where neither course has a linear algebra prerequisite. You just don't focus on things like systems of first-order equations (or you cover just enough basic material about eigenvalues and eigenvectors to solve them) or existence and uniqueness theorems, and you leave the more advanced material for later courses.

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

Thank you for your explanations. I think my problem here was that I'm learning in a different environment, where we begin with proof-based lectures from day one. I wasn't familiar with the American system.

[–] 4 points5 points

sorry, this has been archived and can no longer be voted on

You cannot. The "circle" you are thinking of is the n-sphere for n=1.

What you are trying to do is to obtain an homeomorphism between the S1 and a line (for example expressed as a one-dimensional subvectorspace of R2 ).

That is, however, impossible, due to the following: a property called connectedness is left untouched via homeomorphism. This means that if one of the two homeomorphic spaces is connected, the other one is too. In addition to this, if you take away one point away from each "space", they remain homeomorphic.

In this case, taking a point away from the S1 leaves it connected, but taking away one point from any line divides it into two connected components, while the line without a point itself is not connected. Ergo, the S1 and the line cannot be homeomorphic.

What others have told you is, canonically, the stereographic projection from the S1 without a point to the real line. As you can see, taking away two points of the S1 gives you two connected parts again .. just like with R. And indeed, this stereographic projection is a homeomorphism, even a smooth diffeomorphism!

[–] 2 points3 points

sorry, this has been archived and can no longer be voted on

I agree with the poster above that calling a line one dimensional subvectorspace is bizarre and not quite accurate, since if that line doesn't pass through the origin, it isn't even a vector space anymore, since it doesn't contain 0. Also, wouldn't it be more straightforward to say that S1 is compact and R is not? This avoids talking about removing a point.

[–] -1 points0 points

sorry, this has been archived and can no longer be voted on

We are talking about topological spaces. A "line" in Rn is nothing more than a subvectorspace translated. That's even a diffeomorphism .. where exactly is this confusing?* The OP had differential equations as prerequisite - how do you even do basic ODE, less alone PDE theory without knowing linear algebra?

You cannot use the compactness argument that easily for two reasons:

1. I'd have to tell you what compactness means in a topological sense. This is, in my view, more of a step than imagining taking two points away. If you only know about metric spaces, it might be a stretch to explain why it isn't always possible to have compactness by sequence or even Heine-Borel if you only dealt with R, but now you need that ugly definition about open covers.

2. I still wouldn't know if by line he meant a compact intervall or the real line. I cannot use the compactness argument if he meant a compact interval. Even if he meant an open intervall, what if the next question is about a compact one? I'd have to use the point-solution due to this natural extension of the question in any case. I assumed he meant an open interval now, but just in case I gave the idea for a proof that can be easily extended to a closed interval as well.

Nevertheless, yours is a good complaint. I thought of using compactness first as well, and then decided against it due to the above reasons.

*I now aknowledge that my assumptions about what you need for differential equations were wrong. I see how this can be confusing then.

[–]Mathematical Physics 3 points4 points

sorry, this has been archived and can no longer be voted on

Not exactly what you're looking for, but conformal mapping [see example 7 and pay attention to the colored dots] does a somewhat similar sort of thing. [Hint: it's helpful to already know complex analysis.]

To do exactly what what you mentioned, use polar coordinates.

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

Just for an idea of what your asking, describe a few circles and describe what lines you might want to get from them. Maybe a line through the center or tangent to the edge

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

Hmmm, I remember something about the W-plane where you can map lines into circles and vice versa, however I think it only works with complex numbers.

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

What does it mean? What do you want to accomplish? As nerkbot says, you can map all the points on a circle (or all but one) to points on a line, using a stereographic projection, or simply by giving the angle in radians between the point, the center of the circle, and a horizontal line passing through the center of the circle. There's no "mx+b" business here though, because a line in and of itself only has one coordinate; it's only when you place a line on a plane, with freedom to move it around and rotate it, that you need two.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}