Undergrad Does ##\mathbb{S}^1## embed in ##\mathbb{R}##?

[kent davidge]

Whitney's theorem tell us that a m-dimensional manifold embedds in $\mathbb{R}^{2m}$, but does not tells us if it embedds in $\mathbb{R}^n, n < 2m$. In special, I wanted to know whether the circle embedds in $\mathbb{R}$. For this, I tried constructing a function $f: \mathbb{S}^1 \longrightarrow \mathbb{R}$ given by $f(\cos\theta, \sin\theta) = \theta$ for $(\cos\theta, \sin\theta) \in \mathbb{S}^1$. The image is $f(\mathbb{S}^1) = (0, 2 \pi]$ if $\mathbb{S}^1$ is defined as usual as a subset of $\mathbb{R}^2$ with $\theta$ on that interval.

Such function $f$ seems to be the more natural mapping I can think of, for its inverse is how $\mathbb{S}^1$ is typically defined. The problem is that $(0, 2 \pi]$ is not open in $\mathbb{R}$ (usual topology) whereas $\mathbb{S}^1$ is open in itself (subspace topology). So should I conclude that $\mathbb{S}^1$ does not embedd in $\mathbb{R}$?

 

[fresh_42]

It depends on what you call an embedding. It's certainly not continuous at $\theta =0$, for you have to cut the circle. This changes its topological properties like genus. All you have is a function of sets: $\,\mathbb{S}^1 \leftrightarrows (0,2\pi] \hookrightarrow \mathbb{R}$.

 

[kent davidge]

It's certainly not continuous at $\theta =0$, for you have to cut the circle

I'm considering the circle as being all $(\cos\theta, \sin\theta) \in \mathbb{R}^2$ with $\theta \in (0,2 \pi]$. So the very definition doesn't include $\theta = 0$. Do we still need to cut the circle?

It depends on what you call an embedding

I'm using the one here http://schapos.people.uic.edu/MATH549_Fall2015_files/Survey Paul.pdf
It's right below the figure of the Klein Bottle, "Topological Embedding".

 

[fresh_42]

How do you distinguish between $\mathbb{S}^1$ and $(0,2\pi]$, i.e. what makes it a circle? If you simply define the circle by its angle, without matching the boundaries, then you have the same topological space. If we identify the boundaries, then approaching from left gives a different result than approaching from right.

 

[kent davidge]

How do you distinguish between $\mathbb{S}^1$ and $(0,2\pi]$, i.e. what makes it a circle? If you simply define the circle by its angle

I'm not defining it as the interval $(0,2\pi]$, rather

I'm considering the circle as being all $(\cos\theta, \sin\theta) \in \mathbb{R}^2$ with $\theta \in (0,2 \pi]$

 

[fresh_42]

So it is already embedded in the plane equipped with the subspace topology?

Let me write open intervals as $(a,b) = ]\,a,b\,[$ to avoid confusion with the points in $\mathbb{R}^2$.

Now a function $f : \mathbb{S}^1 \longrightarrow ]\,0,2\pi\,]$ is continuous at $x=(1,0)=(\cos 0,\sin 0)$, if for every neighborhood $V$ of $f((1,0))$ there is a neighborhood $U$ of $(1,0)$ with $f(U)\subseteq V$. The neighborhoods $U$ are all of the form $U=]\,(\cos(- \varepsilon), \sin(-\varepsilon))\,,\,(\cos(\varepsilon), \sin(\varepsilon))\,[$. Let's take $V=]\,-\varepsilon,2\pi\,]=]\,2\pi -\varepsilon , 2\pi\,]$ as neighborhood of $f((1,0))=2\pi \in (0,2\pi]$. Then all $f(U)= ]\,0,\varepsilon\,[ \,\cup \,]\,2\pi-\varepsilon ,2\pi\,] \nsubseteq V\,.$

 

[kent davidge]

This was very clarifying.

I went through a similar reasoning before the opening post, but I would define the neighborhood of $(1,0)$
as $U = \ ](\cos\epsilon,\sin\epsilon), (1,0)]$, in which case $f(U) = \ ]\epsilon, 2\pi] \subseteq V$ and the conclusion would be of continuity of $f$ at $(1,0)$. Could you point to me where I'm going wrong with this reasoning?

 

[kent davidge]

Oh, never mind. I realized that there are points to the right of $(1,0)$, which means one neighborhood of it would be of the form you said earlier. Then $f(U)$ is not in $V$. So our conclusion is that there cannot be an embedding because the mapping is not continuous, correct?

 

[fresh_42]

Yes, it's not continuous, unless we change the topologies, but then the statement loses its meaning. Or we could compactify $]0,2\pi]$ by identifying its ends, but then we have a circle again, just with two different representations: points or angles.

$\mathbb{S}^1$ is the projective real line $\mathbb{P}(1,\mathbb{R})$ where the points at infinity are identified.

 

[mathwonk]

the circle is connected and compact so it image in R under any continuous function is a closed bounded interval. If the map were an embedding, or just injective, it would be a homeomorphism onto this interval, but an interval can be disconnected by removing one point, while the circle cannot. hence no continuous map S^1-->R is an embedding.

 

[lavinia]

Whitney's theorem tell us that a m-dimensional manifold embedds in $\mathbb{R}^{2m}$, but does not tells us if it embedds in $\mathbb{R}^n, n < 2m$. In special, I wanted to know whether the circle embedds in $\mathbb{R}$. For this, I tried constructing a function $f: \mathbb{S}^1 \longrightarrow \mathbb{R}$ given by $f(\cos\theta, \sin\theta) = \theta$ for $(\cos\theta, \sin\theta) \in \mathbb{S}^1$. The image is $f(\mathbb{S}^1) = (0, 2 \pi]$ if $\mathbb{S}^1$ is defined as usual as a subset of $\mathbb{R}^2$ with $\theta$ on that interval.

Such function $f$ seems to be the more natural mapping I can think of, for its inverse is how $\mathbb{S}^1$ is typically defined. The problem is that $(0, 2 \pi]$ is not open in $\mathbb{R}$ (usual topology) whereas $\mathbb{S}^1$ is open in itself (subspace topology). So should I conclude that $\mathbb{S}^1$ does not embedd in $\mathbb{R}$?

Whitney's Theorem is about smooth embeddings of smooth manifolds in Euclidean space. A smooth embedding by definition has no singularities. For an embedding into $R$ this means that the function's derivative is nowhere zero.

A function from the circle into $R$ must have a maximum and a minimum - because the circle is compact - and at these points the function's derivative is zero.

Also, at these points the derivative reverses sign so the function is not 1-1 in neighborhood of the critical points.

- More generally one can not embed a compact n-manifold in $R^{n}$. Any n- dimensional submanifold of $R^{n}$ must be an open set since each point has an open neighborhood that is homeomorphic to an open n dimensional ball.

 

[kent davidge]

Whitney's Theorem is about smooth embeddings of smooth manifolds in Euclidean space

Thanks for pointing this out

For an embedding into $R$ this means that the function's derivative is nowhere zero

...because otherwise the derivative map would not be injective?

 

[lavinia]

Thanks for pointing this out

...because otherwise the derivative map would not be injective?

Yes. An embedding is a difeomorphisrm so the derivative map must be an isomorphism on each tangent plane.