Homeomorphism between quotient space and S^1 x S^(n-1)

[Joolz]

Hi there,

Can anyone please help me and explain to me why the following is true:

Let r >1 and define an equivalence relation on Rⁿ - {0} by :

(x ~ y) if and only if (there exists an s in Z such that r^sx = y)

Then the quotient space is homeomorphic to S¹ x S^n-1

So by quotient space they mean Rⁿ - {0}/~ right? To understand it better I tried to figure out how it works for n=2.

Then the quotient space is R² - {0}/~ with elements that are equivalence classes which are actually lines through (0,0) Is that a bit correct?

But then I got stuck because of the condition that r > 1 So let say we want to check to which class a point (x, -y) belongs...

Then we got (by the lines) that (x, -y) ~ (x/-y , 1) cause we take r^s = -y but that doesn't work when r needs to be > 1

So do they then mean that the lower half of the R x R axes arent in the equivalence relation?

And how is that homeomorphic to S¹ x S¹ ?


I would appreciate any help or hints... I really want to understand this..

Kind regards,

Joolz

 

[daon2]

Hi there,

Can anyone please help me and explain to me why the following is true:

Let r >1 and define an equivalence relation on Rⁿ - {0} by :

(x ~ y) if and only if (there exists an s in Z such that r^sx = y)

Then the quotient space is homeomorphic to S¹ x S^n-1

So by quotient space they mean Rⁿ - {0}/~ right? To understand it better I tried to figure out how it works for n=2.

Then the quotient space is R² - {0}/~ with elements that are equivalence classes which are actually lines through (0,0) Is that a bit correct?

But then I got stuck because of the condition that r > 1 So let say we want to check to which class a point (x, -y) belongs...

Then we got (by the lines) that (x, -y) ~ (x/-y , 1) cause we take r^s = -y but that doesn't work when r needs to be > 1

So do they then mean that the lower half of the R x R axes arent in the equivalence relation?

And how is that homeomorphic to S¹ x S¹ ?


I would appreciate any help or hints... I really want to understand this..

Kind regards,

Joolz

For R^2-{0}/~, the relation of course depends on the chosen r, so perhaps denote it with a subscript. This isn't lines through the origin. In the R^2 case, consider the points on the graph (x,1/2x) with r=pi. Is there an integer s such that pi^s(1,1/2) ~ (2,1)?

In this case, what would a point (a,b) be equivalent to? The set of all (a*pi^s, b*pi^s) for every integer s. If you have studied algebra, this is en example of a group action on R^2 by Z with uncountably many orbits (though this may be hard to show), and each orbit is countable (hence it cannot be lines through the origin). First let's forget that 0 is being removed. {(0,0)} is its own orbit! No point is equivalent to the origin except itself.

I would first look at this in R^1. Then we are working with just reals. What numbers are equivalent to 1? The answer is the set {r^s, s in Z}. Hence in general, the equivalence class [a]={ar^s, s in Z}. So, for n=1, your job is to show this is homeomorphic to S^1, which honestly I'm not too sure how to do. What jumps out to me is that the circle is also an exponential: in polar we have (t, re^(it)), the circle of radius r.

edit: I see now. Let r > 1. In the case for n=1, for each a in (1,r], there is an equivalence class for [a]. Now suppose b belongs to (r,r^2], say b=rs where 1< s <= r. Then b ~ s, ie. = ans s belongs to (1,r]. The same idea holds for the interval (r^i, r^{i+1}]. Hence every equivalence class has a representative in (1,r]. Hence the quotient space is the closed interval [1,r]/{1 ~ r} which is indeed homeomorphic to S^1.

 

[Joolz]

thank you very much for helping...

This part: "edit: I see now. Let r > 1. In the case for n=1, for each a in (1,r], there is an equivalence class for [a]. Now suppose b belongs to (r,r^2], say b=rs where 1< s <= r. Then b ~ s, ie. = ans s belongs to (1,r]. The same idea holds for the interval (r^i, r^{i+1}]. Hence every equivalence class has a representative in (1,r]". I get now

But why is the quotient space then [1,r]/{1~r}? I thought the quotient space was R-{0}/~ or are they the same? cause what about all the points in (-infinity, 1] -{0} ? They are just equivalent to and only to itself??? My question is thus, where do all the other points go to?

And how will this work for n=2?

I appreciate your help..

Kind regards,

 

[daon2]

thank you very much for helping...

This part: "edit: I see now. Let r > 1. In the case for n=1, for each a in (1,r], there is an equivalence class for [a]. Now suppose b belongs to (r,r^2], say b=rs where 1< s <= r. Then b ~ s, ie. = ans s belongs to (1,r]. The same idea holds for the interval (r^i, r^{i+1}]. Hence every equivalence class has a representative in (1,r]". I get now

But why is the quotient space then [1,r]/{1~r}? I thought the quotient space was R-{0}/~ or are they the same? cause what about all the points in (-infinity, 1] -{0} ? They are just equivalent to and only to itself??? My question is thus, where do all the other points go to?

And how will this work for n=2?

I appreciate your help..

Kind regards,

Okay, I have been out of point-set for a while, so use my statements as guidance and not rigorous fact. I do see a big issue with what I gave.

\(\displaystyle \displaystyle \mathbb{R} - \{0\} = \bigcup_{k\in \mathbb{Z} }(r^k,r^{k+1}] \cup \bigcup_{k\in \mathbb{Z}}[-r^{k+1},-r^{k})\)

The first big union is taken care of with the explanation I have given. For k<0 then those intervals "take up" (0, 1]. A popular way to represent the circle is identifying 0 and 1 on the unit interval and the homeomorphism between it and, say, the graph of x^2+y^2=1 is \(\displaystyle t\to (\cos(2\pi t),\sin (2\pi t) )\), then as [0,1] and [r^k, r^{k+1}] are homeomorphic, that's it.

The problem comes now from the negative real numbers which seem to give an extra copy of S1. So for now I'm not sure what to do about that. For n=2, just attempt something similar to obtain something like: [a,b]x[c,d]/~, where ~ is the relation turning the unit square into a torus ( S^1 x S^1). A similar issue may come up there. This problem does seem difficult.

 

[SlipEternal]

Let \(\displaystyle q: \mathbb{R}^n - \{0\} \to S^1 \times S^{n-1}\) define a function where
\(\displaystyle q(v) = (\cos{\frac{2\pi \ln{|v|}}{\ln{r}},\sin{\frac{2\pi \ln{|v|}}{\ln{r}}) \times \left\{\frac{v}{|v|}\right\}\) where \(\displaystyle |v|\) is the magnitude of the vector.

Step 1: Show that \(\displaystyle q\) is a quotient map.
Step 2: Show that \(\displaystyle q\) is constant on \(\displaystyle [v]\).
Step 3: Find the appropriate lemma that proves that there exists a homeomorphism.

 

[daon2]

^ That is pretty, I likely never would have come up with that.

On another note your post helped me realize that there actually should be two copies of S1 in the n=1 case, as S^0 is the boundary of D^1, giving S1 x {-1, 1} which my setup seems to work for.

 

[Joolz]

Thank you both very much for helping...

I still struggle with how to see a product of spheres... Can you explain for example the products S¹ x S⁰ and S¹ x S² Do they just mean the cartesian products?

Kind regards

 

[daon2]

Thank you both very much for helping...

I still struggle with how to see a product of spheres... Can you explain for example the products S¹ x S⁰ and S¹ x S² Do they just mean the cartesian products?

Kind regards

S^1 x S^0 = S^1 x {-1,1} gives two disjoint copies of S^1.

S^1 x S^1 gives a copy of a circle for every point on the circle. Normally it is done "nicely" to give the nice looking torus.

S^1 x S^2 gives a circle for every point on the sphere or vise versa. It is "hard" to visualize.

S^2 x S^2 gives a sphere at every point on the sphere. Again, "hard" to visualize.

Recall S^1 is the unit interval (D^1) with the end points (S^0) being glued. S^2 is the disk (D^2) with the boundary (S^1) glued. S^3 can be thought of as D^3 (the solid 3d sphere) where the boundary (S^2) is all "glued" to a point. It is hard to visualize but some intuition can be gained from the analogy.

Maybe someone else can give better interpretations. I'm no geometer.

 

[Joolz]

Thank you... again!! So in the case n=1 we send the interval (1, r] to a circle and the interval [-r, -1) to the other??

I would like to know how SlipEternal got his function... how were you looking at this... cause I would've never come up with the 2pi*ln|v|/ln(r) part... Can you tell me in words, what that means... why would you use ln|v|/ln(r) ?

Sorry It has been a while for me...

kind regards

 

[SlipEternal]

The homeomorphism is supposed to be between equivalence classes on a subset of \(\displaystyle \mathbb{R}^n\). \(\displaystyle S^1\) is homeomorphic to a quotient space of \(\displaystyle D^1 \subset \mathbb{R}\) as daon2 mentioned. The closest we have to a quotient space of \(\displaystyle \mathbb{R}\) is magnitude of a vector in \(\displaystyle \mathbb{R}^n\). Since the point \(\displaystyle 0\) is removed, all magnitudes are positive. So, we want to turn magnitude into a circle. This means that we want to send a vector to \(\displaystyle (\cos{|v|},\sin{|v|})\) in order to map it to the circle. We now have a quotient map. It is the wrong quotient map, currently.

We want vectors to be constant on equivalence classes. So, we want \(\displaystyle |rv|\) to go to some multiple of \(\displaystyle 2\pi+|v|\). Well, anytime you want to turn multiplication into addition, you need logs. It is like the old joke, Noah is upset because the snakes wouldn't reproduce. All the other animals got off the ark and were fruitful, but not the snakes. He asked them what the problem was. They said, they are adders. They need logs to multiply. So, \(\displaystyle \ln{|r^sv|} = \ln{r^s|v|} = \ln{r^s}+\ln{|v|} = s\ln{r} + \ln{|v|}\). Now, we need a factor of \(\displaystyle 2\pi\), and we need to get rid of that \(\displaystyle \ln{r}\). Well, \(\displaystyle \log_r{r}=1=\frac{\ln{r}}{\ln{r}}\). In general, \(\displaystyle \log_a{x} = \frac{\ln{x}}{\ln{a}}\). So, we need \(\displaystyle \frac{2\pi \ln{|v|}}{\ln{r}}\). Now, take cos and sin of that, and you have a circle that is constant where we want it to be constant.

Next, \(\displaystyle S^{n-1} \subset \mathbb{R}^{n}\). If you want a continuous surjective map, you need a continuous map. So, you cannot change the vector much. Since all vectors have positive magnitude, \(\displaystyle \left\{\frac{v}{|v|}\right\}\) is the normal vector in the direction of the original vector. That means that this is a continuous map taking a vector in \(\displaystyle \mathbb{R}^n\) to the unit sphere in that dimension. I don't care how many points go to the same point on the sphere, as each point has a different magnitude, yet all points with magnitude that is a factor of \(\displaystyle r^s\) times your starting vector all go to the same point in \(\displaystyle S^1\). Now, we have a continuous surjection that is constant on the equivalence classes we wanted.

 

[Joolz]

aaaaarrrggghh... I can't stand it, but I just don't see how you got from s ln r + ln|v| to 2pi *ln|v|/ln r

the rest is pretty clear..

thank you so much for taking the time to explain it.

 

[SlipEternal]

No worries
Here it is step-by-step.
Let \(\displaystyle q:\mathbb{R}^n - \{0\} \to S^1 \times S^{n-1}\) be a quotient map that we are going to build. We don't yet know what its formula is (nor that it is even necessarily going to be a quotient map). It is simply a place-holder for a function we want.

A vector in \(\displaystyle \mathbb{R}^n\) has two properties that make it unique: magnitude and direction.
Let's look at vectors in \(\displaystyle \mathbb{R}^n - \{0\} / \) ~.
Hence, if we have two vectors \(\displaystyle v,w\) and \(\displaystyle v\)~\(\displaystyle w\), we know that for some \(\displaystyle s \in \mathbb{Z}\), \(\displaystyle r^sv = w\). This means that \(\displaystyle w\) is a scalar times a vector, so its direction is the same as that of \(\displaystyle v\), and its magnitude is the only thing that may differ.

A vector in \(\displaystyle S^1\) has one property that makes it unique: direction. A vector in \(\displaystyle S^{n-1}\) also has only one property that makes it unique: direction. So, since magnitude is a single dimension and \(\displaystyle S^1\) is essentially a single dimension, we can map magnitude to \(\displaystyle S^1\) and direction (in \(\displaystyle \mathbb{R}^n\)) to \(\displaystyle S^{n-1}\) and this will preserve all of the data about our vector. We already discussed that \(\displaystyle v \to \left\{\frac{v}{|v|}\right\}\) preserves direction, but not magnitude. So, we satisfied the projection of our map that goes to \(\displaystyle S^{n-1}\). Now, we want some function that will take \(\displaystyle |v| \to S^1\). Additionally, we want it to be so that if this function takes \(\displaystyle |v|\) to a point \(\displaystyle (x,y)\), we want \(\displaystyle |r^sv|\) to also go to \(\displaystyle (x,y)\). We know that the map \(\displaystyle \theta \to (\cos{\theta},\sin{\theta})\) maps to a circle. So, that will be our starting point.

Let \(\displaystyle f:\mathbb{R}_{> 0} \to \mathbb{R}\) define a function. We will again build it gradually.
Now, let's define our map \(\displaystyle q\) a bit.
So, let \(\displaystyle q(v) = (\cos{f(|v|)},\sin{f(|v|)}) \times \left\{\frac{v}{|v|}\right\}\).
Now, we want \(\displaystyle f(|r^sv|) = 2\pi s + f(|v|)\) because both sin and cos are \(\displaystyle 2\pi\) periodic.
Keep in mind that we know \(\displaystyle |r^sv| = r^s|v|\).
So, let's take a derivative of this function:
\(\displaystyle f'(|v|) = \lim_{a \to 0}{\frac{f(r^a|v|) - f(|v|)}{r^a|v| - |v|}} = \lim_{a \to 0}{\frac{2\pi a}{r^a|v| - |v|}\)
By L'Hôpital's rule, we can take derivatives of top and bottom with respect to \(\displaystyle a\) (since both tend towards zero) and apply the limit again.
\(\displaystyle f'(|v|) = \lim_{a \to 0}{\frac{2\pi}{r^a\ln{r}|v|}}\)
Notice the \(\displaystyle - |v|\) is gone since its derivative with respect to \(\displaystyle a\) is zero and we picked up a \(\displaystyle \ln{r}\) in the denominator because \(\displaystyle \frac{d(r^a)}{da} = r^a\ln{r}\).
Now, we have:
\(\displaystyle f'(|v|) = \frac{2\pi}{\ln{r}|v|}\)
Solving this differential equation, we get:
\(\displaystyle f(|v|) = \frac{2\pi \ln{|v|}}{\ln{r}} + c\)
We don't care about the arbitrary constant, so set it equal to zero.

And now we are done! We solved for a function that works (granted, we could have used any value for \(\displaystyle c\)).
\(\displaystyle q(v) = (\cos{\frac{2\pi \ln{|v|}}{\ln{r}}},\sin{\frac{2\pi \ln{|v|}}{\ln{r}}}) \times \left\{\frac{v}{|v|}\right\}\)
And we have a quotient map with the desired characteristics.

Keep in mind, when I solved this originally, I didn't go through all of these steps. Solving the differential equation was not necessary. I just knew that I wanted my formula to have certain properties, and I just generated the formula that worked. This is just the evidence that it is the correct formula.