Boundary and homeomorphism

[davidge]

The 2-sphere $\mathbb{S}^2$ can be expressed as the product $\mathbb{S}^1 \times \mathbb{S}^1$
Now can we express $\mathbb{S}^1$ as $\mathbb{S}^1 \subset (-a,a)$, where $(-a,a)$ is some open interval of $\mathbb{R}$? If so, then (I think) $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}$. Also, it's homemomorphic to $\mathbb{R}^2$ up to four coordinate charts covering it in $\mathbb{R}^2$.

If so, by the same reasoning $\mathbb{S}^2 \subset \mathbb{R}^2$ is homeomorphic to $\mathbb{R}^2$. Also, it's homeomorphic to $\mathbb{R}^3$, for we can define an embedding from it to $\mathbb{R}^3$.

Finally, if the above is correct, $\mathbb{S}^2 \times \mathbb{R}$ is homeomorphic to $\mathbb{R}^3$, though it's not compact.

I'm trying to get this because I'm interested in knowing whether $\mathbb{S}^2 \times \mathbb{R}$ as a manifold has a boundary or not.

I'll appreciate any help.

 

[zwierz]

The 2-sphere S2\mathbb{S}^2 can be expressed as the product S1×S1

I was taught long ago that $S^1\times S^1$ is two dimensional torus not sphere.

 

[davidge]

I was taught long ago that $S^1\times S^1$ is two dimensional torus not sphere.

Oh, I thought the 2-sphere could be represented as $\mathbb{S}^1 \times \mathbb{S}^1$ because each one of the coordinates of a point $x \in \mathbb{S}^2$, say $(\theta, \phi)$, obeys a "circle equation".

 

[TeethWhitener]

Also, $\mathbb{S}^1$ is not homeomorphic to $\mathbb{R}$ (because it's not homeomorphic to an open interval of $\mathbb{R}$). Think about the values $\theta$ ranges over.

 

[davidge]

Also, $\mathbb{S}^1$ is not homeomorphic to $\mathbb{R}$ (because it's not homeomorphic to an open interval of $\mathbb{R}$). Think about the values $\theta$ ranges over.

So why can we map the entire circle using four coordinate charts from an open interval of $\mathbb{R}$? Can you check out my work below and say whether it's correct or not?

 

[fresh_42]

Oh, I thought the 2-sphere could be represented as $\mathbb{S}^1 \times \mathbb{S}^1$ because each one of the coordinates of a point $x \in \mathbb{S}^2$, say $(\theta, \phi)$, obeys a "circle equation".

This is locally (in a neighborhood of each point) true, but not globally (on the entire sphere), because you will have to leave out one of the poles.

 

[TeethWhitener]

So why can we map the entire circle using four coordinate charts from an open interval of $\mathbb{R}$? Can you check out my work below and say whether it's correct or not?

View attachment 204017

I can't really read what's going on here, but the easiest way to show that $\mathbb{S}^1$ is not homeomorphic to an open interval of $\mathbb{R}$ is to note that $\mathbb{S}^1$ is compact whereas an open interval of $\mathbb{R}$ is not. Since homeomorphisms preserve compactness, we immediately get that the two spaces are not homeomorphic.

 

[davidge]

I can't really read what's going on here, but the easiest way to show that $\mathbb{S}^1$ is not homeomorphic to an open interval of $\mathbb{R}$ is to note that $\mathbb{S}^1$ is compact whereas an open interval of $\mathbb{R}$ is not. Since homeomorphisms preserve compactness, we immediately get that the two spaces are not homeomorphic.

What if we choose a closed interval instead of a open one. A closed interval is compact.

This is locally (in a neighborhood of each point) true, but not globally (on the entire sphere), because you will have to leave out one of the poles.

Then letting the "end points", namely the poles out of consideration is forbidden?

 

[davidge]

closed interval has boundary, the circle does not

yes, but it's precisely what I was trying to say in post number 1. Pick a closed interval. There's no way of homeomorphicaly map the entire circle to $\mathbb{R}^2$ using an closed interval (at least in that way I did in post #5). So you have to "remove the end points" of your closed interval. Then now you have an open interval that is homeomorphic to $\mathbb{R}$.

Can you point out where I'm wrong on the above ?

 

[fresh_42]

Then letting the "end points", namely the poles out of consideration is forbidden?

It's not forbidden, it simply makes the difference between local and global. As entire set $S^1 \times S^1 \ncong S^2$ as a torus (on the left) isn't a sphere (on the right), because one has a hole and the other has not. But on a small patch you can always apply coordinates like on earth. You simply cannot find a single chart describing both. As manifolds they all have local flat charts patched to a whole atlas, but this doesn't make them equal.

 

[TeethWhitener]

What if we choose a closed interval instead of a open one. A closed interval is compact.

yes, but it's precisely what I was trying to say in post number 1. Pick a closed interval. There's no way of homeomorphicaly map the entire circle to R2R2\mathbb{R}^2 using an closed interval (at least in that way I did in post #5). So you have to "remove the end points" of your closed interval. Then now you have an open interval that is homeomorphic to RR\mathbb{R}.

Can you point out where I'm wrong on the above ?

I think you answered your own question. Look here:
https://www.physicsforums.com/threads/why-the-circle-cant-be-homeomorphic-to-a-real-interval.537897/
http://mathhelpforum.com/differenti...-not-homeomorphic-any-interval-real-line.html
Combining with what I said before: an open interval on $\mathbb{R}$ is not homeomorphic to $\mathbb{S}^1$ because compactness isn't preserved. A half-open interval won't work for the same reason. And as you said, there's no homeomorphism between a closed interval of $\mathbb{R}$ and $\mathbb{S}^1$. The links above answer why: choose a closed real interval $[a,b]$ and assume a (edit: bijection homeomorphism) exists $f: [a,b] \mapsto \mathbb{S}^1$. Then deleting any point $p$ (not equal to $a$ or $b$) from the real interval gives another homeomorphism: $g: [a,p) \cup (p,b] \mapsto \mathbb{S}^1 - \{f(p)\}$. But the new real interval $[a,p)\cup(p,b]$ is not connected, whereas $\mathbb{S}^1 - f(p)$ is connected. Since connectedness is a topological property, $g$ can't be a homeomorphism (and by extension, neither can $f$).

EDIT: one important note. The set $\mathbb{R} \cup \{\infty\}$ is homeomorphic to the circle. This is sometimes called the "point at infinity." It's a special case of one point compactification:
https://en.wikipedia.org/wiki/Alexandroff_extension

 

[davidge]

the new real interval $[a,p)\cup(p,b]$ is not connected

So can we apply the definition of connectness to intervals like the one above? I thought the definition could be applied only to open intervals. Also, doing the same as you did, but with an open interval $(a,b)$, the same reasoning would led us to conclude that $\mathbb{S}^1$ and $(a,b)$ are not homeomorphic. Cool!

 

[zwierz]

There is another funny proof that $[a,b]$ is not homeomorphic to $S^1$. Indeed, any continuous function $f:[a,b]\to[a,b]$ has a fixed point. But there is a continuous function of $S^1$ to itself that does not have a fixed point.

 

[WWGD]

The 2-sphere $\mathbb{S}^2$ can be expressed as the product $\mathbb{S}^1 \times \mathbb{S}^1$
Now can we express $\mathbb{S}^1$ as $\mathbb{S}^1 \subset (-a,a)$, where $(-a,a)$ is some open interval of $\mathbb{R}$? If so, then (I think) $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}$. Also, it's homemomorphic to $\mathbb{R}^2$ up to four coordinate charts covering it in $\mathbb{R}^2$.

If so, by the same reasoning $\mathbb{S}^2 \subset \mathbb{R}^2$ is homeomorphic to $\mathbb{R}^2$. Also, it's homeomorphic to $\mathbb{R}^3$, for we can define an embedding from it to $\mathbb{R}^3$.

Finally, if the above is correct, $\mathbb{S}^2 \times \mathbb{R}$ is homeomorphic to $\mathbb{R}^3$, though it's not compact.

I'm trying to get this because I'm interested in knowing whether $\mathbb{S}^2 \times \mathbb{R}$ as a manifold has a boundary or not.

I'll appreciate any help.

There is a result, I think by Ulam, that $\mathbb S^n$ cannot be embedded in $\mathbb R^n$ -or-lower. Still, it would be interesting to see whether $\mathbb S^n$ " Is a square root" for some n , meaning whether it is homeomorphic to the product $Y \times Y$ of some topological space $Y$ . There is an argument showing $\mathbb R^{2n+1}$ is not a square root in this sense.

 

[WWGD]

There is another funny proof that $[a,b]$ is not homeomorphic to $S^1$. Indeed, any continuous function $f:[a,b]\to[a,b]$ has a fixed point. But there is a continuous function of $S^1$ to itself that does not have a fixed point.

Nice. And there is also the connectivity one that any point p separates [a,b] , i.e., for any p in [a,b] , [a,b] -{p} is disconnected, while the same is not true for $\mathbb S^1$ - or higher-dimensions. I don't know the formal name for this, though ; it comes down to that if X is homeo to Y through h, then X-{p} is homeo to Y-{h(p)}. EDIT 2, besides, with heavier machinery, you also have contractibility, etc. which is preserved under homotopy equivalence alone. EDIT 3:

 

[fresh_42]

There is another funny proof that $[a,b]$ is not homeomorphic to $S^1$. Indeed, any continuous function $f:[a,b]\to[a,b]$ has a fixed point. But there is a continuous function of $S^1$ to itself that does not have a fixed point.

Nice idea. Or one can take $S^1 \times S^1 \cong \mathbb{P}^2(1,\mathbb{R}) \cong \mathbb{P}(2,\mathbb{R}) \ncong \mathbb{P}(1,\mathbb{C}) \cong S^2\;$, which indicates the problem with the poles.

 

[davidge]

there is a continuous function of $S^1$ to itself that does not have a fixed point.

what do you mean by having a fixed point?

 

[WWGD]

Yet another way :

what do you mean by having a fixed point?

It means there is a point that maps to itself, i.e., f(p)=p.

 

[davidge]

Yet another way :

It means there is a point that maps to itself, i.e., f(p)=p.

Thanks

 

[WWGD]

Thanks

No problem, consider re, Zwierz' post and fixed points the (continuous) map from $\mathbb S^1$ to itself that rotates every point by a fixed amount that is not an integer multiple of $2\pi$ .

 

[TeethWhitener]

So can we apply the definition of connectness to intervals like the one above? I thought the definition could be applied only to open intervals. Also, doing the same as you did, but with an open interval $(a,b)$, the same reasoning would led us to conclude that $\mathbb{S}^1$ and $(a,b)$ are not homeomorphic. Cool!

You have to be careful here, because the topology on $S=[a,p)\cup(p,b]$ is inherited from $\mathbb{R}$, meaning that open sets on $S$ are defined as the intersection of an open set in $\mathbb{R}$ with $S$. So in this case, for example, $[a,p)$ would be an open set in the space $S$ (since it's the intersection of, e.g., $(a-1,p) \cap S$, where $(a-1,p)$ is an open set in $\mathbb{R}$), as would $(p,b]$. But of course, since $S-[a,p)=(p,b]$, this means that $S$ contains subsets that are both open and closed, and therefore $S$ cannot be connected.

 

[davidge]

You have to be careful here, because the topology on $S=[a,p)\cup(p,b]$

Why are you calling $S$ that way? I thought $S$ would be $[f(a), \ ... \ , f(p)) \cup (f(p), \ ... \ , f(b)]$ instead.

But of course, since $S-[a,p)=(p,b]$, this means that $S$ contains subsets that are both open and closed, and therefore $S$ cannot be connected.

I see

 

[TeethWhitener]

Why are you calling $S$ that way? I thought $S$ would be $[f(a), \ ... \ , f(p)) \cup (f(p), \ ... \ , f(b)]$ instead.

I see

Bad notational choice on my part. I meant $S$ as a subspace of $\mathbb{R}$, not a subspace of $\mathbb{S}^1$.

 

[davidge]

Just one more question: can we use the same argument to show that $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}^2$?

 

[fresh_42]

Just one more question: can we use the same argument to show that $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}^2$?

This cannot be shown because it is wrong.

 

[davidge]

This cannot be shown because it is wrong.

Since an embedding from $\mathbb{S}^1$ to $\mathbb{R}^2$ can be defined, I thought the former was homeomorphic to the later.

 

[fresh_42]

Homeomorphy means bijective and continuous in both directions. $\mathbb{S}^1$ is the unit circle. One can apply the stereographic projection from the northpole onto the equatorial plane and get a line $\mathbb{R}^1$ as a result. However, whereas the southpole can be seen as the origin, the northpole is the infinite point, and thus the circle is homeomorphic to the projective line. But there is no plane around, unless you do the same thing for the $2-$sphere.

The fastest way is still what @TeethWhitener said in post #7.

 

[WWGD]

Since an embedding from $\mathbb{S}^1$ to $\mathbb{R}^2$ can be defined, I thought the former was homeomorphic to the later.

You would require an embedding onto , not just into.

 

[davidge]

@fresh_42 and @WWGD
I think I see what you are saying.

You would require an embedding onto , not just into.

Suppose a parametrization $g: (-r,r) \rightarrow \mathbb{S}^1 \\ t \mapsto x$ where $(-r,r)$ is an open interval of $\mathbb{R}$.
Now define $f_1: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, \sqrt{r²-x²} \bigg{)}$
This function does not cover the entire circle $\in \mathbb{R}^2$, because the points $(r,0)$ and $(-r,0)$ $\in \mathbb{R}^2$ are excluded. Also, we would need to define another function $f_2: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, - \sqrt{r²-x²} \bigg{)}$

Is my above understanding correct?

 

[WWGD]

@fresh_42 and @WWGD
I think I see what you are saying.

Suppose a parametrization $g: (-r,r) \rightarrow \mathbb{S}^1 \\ t \mapsto x$ where $(-r,r)$ is an open interval of $\mathbb{R}$.
Now define $f_1: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, \sqrt{r²-x²} \bigg{)}$
This function does not cover the entire circle $\in \mathbb{R}^2$, because the points $(r,0)$ and $(-r,0)$ $\in \mathbb{R}^2$ are excluded. Also, we would need to define another function $f_2: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, - \sqrt{r²-x²} \bigg{)}$

Is my above understanding correct?

Yes, this is correct, but still does not show that the map is not onto $\mathbb R^2$, though. In this case you can also use some topological invariants to show the non-existence of a homeomorphism.