Why is the unit square a 2-manifold?

[JG89]

I've shown that the unit square, $[0,1]^2$, is a smooth 2-manifold in $\mathbb{R}^2$. But it has a corner at the point (1,1). I thought smooth manifolds weren't suppose to have corners, cusps, etc. So what exactly is special about the unit square?

 

[micromass]

I thought smooth manifolds weren't suppose to have corners, cusps, etc.

This is quite a common misconception. For a more dramatic example:

$$M=\{(x,y)\in \mathbb{R}^2~\vert~y=|x|\}$$

(this is just the graph of the absolute value function which has a corner/cusp in 0).

Now consider the atlas consisting of only the function $\varphi:\mathbb{R}\rightarrow M:x\rightarrow |x|$.

This puts a smooth structure on M since there is only one function in the atlas!

Being a smooth manifold depends on the transition functions!


(also note that $[0,1]^2$ is your question is a manifold with boundary)

 

[morphism]

I've shown that the unit square, $[0,1]^2$, is a smooth 2-manifold in $\mathbb{R}^2$.

(Emphasis mine.) Careful with your wording...

There's a subtle difference between something being a smooth manifold and something being a smooth submanifold of R^2 (or R^n).

As micromass mentioned, this is a pretty common pitfall. Basically what's going on in your example is that the unit square is homeomorphic to the unit disc and therefore can be given the smooth manifold (with boundary) structure of the latter. However, the unit square is not a smooth submanifold of R^2.

 

[JG89]

Morphism, could you please elaborate? My book always uses the word "k-manifold in $\mathbb{R}^n$. It never mentions the word "submanifold".

Since the unit square is a subset of $\mathbb{R}^2$, then how come it is not permissible to call it a manifold in $\mathbb{R}^2$

 

[JG89]

Also, could you please give me a quick definition of submanifold? I find the wiki page of it a bit confusing.

 

[lavinia]

I've shown that the unit square, $[0,1]^2$, is a smooth 2-manifold in $\mathbb{R}^2$. But it has a corner at the point (1,1). I thought smooth manifolds weren't suppose to have corners, cusps, etc. So what exactly is special about the unit square?

The unit square is not a smooth submanifold of the plane. This is because it does not have a tangent space at its corners.

Formally, I think that smoothness would require a smooth map of an open neighborhood of the corner - open in the plane - that maps the intersection of this neighborhood with the square to an open neighborhood of the x-axis is the upper half plane and maps the boundary diffeomorphically onto an open interval on the x-axis. This clearly can not be done.

As a topological manifold with boundary the square can be given a smoothness structure. However it does not inherit this structure from the plane.

 

[JG89]

The unit square is not a smooth submanifold of the plane. This is because it does not have a tangent space at its corners.

Formally, I think that smoothness would require a smooth map of an open neighborhood of the corner - open in the plane - that maps the intersection of this neighborhood with the square to an open neighborhood of the x-axis is the upper half plane and maps the boundary diffeomorphically onto an open interval on the x-axis. This clearly can not be done.

As a topological manifold with boundary the square can be given a smoothness structure. However it does not inherit this structure from the plane.

I also thought that it couldn't be a smooth manifold in $\mathbb{R}^2$ because I see that there can't be a tangent space at the corners. But what if each smooth map in the atlas consisted of the map $\alpha(x,y) = (x,y)$?. Take a corner in the unit square, say (0,0). Let V be any set containing (0,0) that's open in $I^2$. Clearly V is also open in $\mathbb{H}^k$. Note also that $\alpha$ is $C^{\infty}$, $\alpha^{-1}$ is continuous, and $D\alpha$ has rank 2. Thus $\alpha: V \rightarrow V$ is a coordinate patch on $I^2$ about (0,0).

But what the hell, it seems strikingly obvious to me that there CANNOT be a tangent space at (0,0)! What's wrong with this reasoning? By the way, by showing that $\alpha$ is C^r, its inverse is continuous and that its derivative has rank 2, I showed (using the definition in my book) that it satisfies the requirements for a manifold, just in case you didn't know what definition I was working with.

 

[JG89]

Let me give you guys my definition of a manifold, just so you know what I'm working with. I will type it word for word, because I have never heard of "sub-manifolds" and such.

Definition:

Let k > 0. A k-manifold in $\mathbb{R}^n$ of class $C^r$ is a subspace M of $\mathbb{R}^n$ having the following property: For each $p \in M$, there is an open set V of M containing p, a set U that is open in either $\mathbb{R}^k$ or $\mathbb{H}^k$, and a continuous map $\alpha: U \rightarrow V$ carrying U onto V in a one-to-one fashion, such that :

1) $\alpha$ is of class $C^r$
2) $\alpha^{-1}$ is continuous
3) $D\alpha(x)$ has rank k for each $x \in U$.



I think I showed in my above post that the map $\alpha(x,y) = (x,y)$ satisfies this? And furthermore, by my definition, would you call it a sub-manifold of $\mathbb{R}^2$?

 

[lavinia]

If you use only one chart in your atlas then the square is a manifold but it is not what is intended when speaking of smooth manifold. A smooth manifold is supposed to be a union of open sets that are diffeomorphic to open domains in Euclidean space - with the usual topology and smoothness structure.

But this I don't think is the real source of confusion here.

The square is homeomorphic to a manifold with boundary such as say the unit disc whose boundary is a circle. But as a subset of the plane, it is not diffeomorphic to the unit disc. If you think through the formal definition I posted above, you will see this.

As a subset of the plane, the square is what is called a manifold with corners.

 

[JG89]

If you use only one chart in your atlas then the square is a manifold but it is not what is intended when speaking of smooth manifold. A smooth manifold is supposed to be a union of open sets that are diffeomorphic to open domains in Euclidean space - with the usual topology and smoothness structure.

A smooth manifold is one whose transition functions are smooth? I.e. $C^{\infty}$, right? Then if I can cover my manifold with a single coordinate patch, then isn't it trivially smooth, since there really are no transition functions?

 

[JG89]

This is quite a common misconception. For a more dramatic example:

$$M=\{(x,y)\in \mathbb{R}^2~\vert~y=|x|\}$$

(this is just the graph of the absolute value function which has a corner/cusp in 0).

Now consider the atlas consisting of only the function $\varphi:\mathbb{R}\rightarrow M:x\rightarrow |x|$.

This puts a smooth structure on M since there is only one function in the atlas!

Being a smooth manifold depends on the transition functions!


(also note that $[0,1]^2$ is your question is a manifold with boundary)

Micromass, I have a question for you. In my definition of a manifold k-manifold, it requires that for any coordinate patch $\alpha$, we have the rank of $D\alpha$ to be k. It says in my book that this rules out the possibility of the manifold having any cusps or corners.

How do I reconcile this statement in my book with your statement that it is a misconception that a smooth manifold cannot have cusps or corners?

 

[lavinia]

A smooth manifold is one whose transition functions are smooth? I.e. $C^{\infty}$, right? Then if I can cover my manifold with a single coordinate patch, then isn't it trivially smooth, since there really are no transition functions?

yes.

 

[lavinia]

Let me give you guys my definition of a manifold, just so you know what I'm working with. I will type it word for word, because I have never heard of "sub-manifolds" and such.

Definition:

Let k > 0. A k-manifold in $\mathbb{R}^n$ of class $C^r$ is a subspace M of $\mathbb{R}^n$ having the following property: For each $p \in M$, there is an open set V of M containing p, a set U that is open in either $\mathbb{R}^k$ or $\mathbb{H}^k$, and a continuous map $\alpha: U \rightarrow V$ carrying U onto V in a one-to-one fashion, such that :

1) $\alpha$ is of class $C^r$
2) $\alpha^{-1}$ is continuous
3) $D\alpha(x)$ has rank k for each $x \in U$.



I think I showed in my above post that the map $\alpha(x,y) = (x,y)$ satisfies this? And furthermore, by my definition, would you call it a sub-manifold of $\mathbb{R}^2$?

there is no map of an open set in the plane that is mapped onto the closed square. There is no map, $\alpha$

 

[JG89]

For any point in the interior of the unit square there is of course a map $\alpha$ for it. For any point on the boundary of the unit square, there is a map $\alpha$ whose domain is $\mathbb{H}^k$ though. These points make up the boundary of the manifold. So isn't this collection of maps an atlas on the unit square

 

[Bacle2]

This may summarize some comments made here: you need to consider a choice of topology in order to state that something is a manifold. That something is a topological space. In your case, the unit square _with the subspace topology_ is not a smooth manifold. But , like others said, the square can be given a topology under which it is a smooth manifold. A general result is this: If (X,T), (Y,T') are homeomorphic topological spaces and either one of them is a manifold, so is the other. In this case, the square is homeomorphic to the unit circle (e.g., circumscribe the square in a circle and draw a line from the center of the circle; this gives you a homeo. between the two. )

 

[JG89]

The book I am studying from requires that we work in $\mathbb{R}^n$. It always uses the Euclidean metric, so let's just say that we give any subset of $\mathbb{R}^n$ the standard topology.

As for my unit square example, I now see that there are open sets (open in the square) containing the corner (1,1) that aren't diffeomorphic to any open set of $\mathbb{R}^2$ or $\mathbb{H}^2$, but diffeomorphic to certain open sets in $\mathbb{L}^2$ (the lower half plane). Is it permissible to have a coordinate patch whose domain is the upper half plane, and another coordinate patch whose domain is the lower half plane in the same atlas?

 

[JG89]

Also, could someone please give me a definition of "sub-manifold"? My book does not mention it and I find the wiki page confusing.

 

[JG89]

Formally, I think that smoothness would require a smooth map of an open neighborhood of the corner - open in the plane - that maps the intersection of this neighborhood with the square to an open neighborhood of the x-axis is the upper half plane and maps the boundary diffeomorphically onto an open interval on the x-axis. This clearly can not be done.

Lavinia, I now agree with what you have been telling me. I have a question though about what you posted above. I can clearly see that this is true. But if I wanted to go about proving this, rigorously, how would I start?

 

[Bacle2]

There aremany definitions of submanifold. One of them is that every point has a (subspace) 'hood (neighborhood) U that looks like the standard embedding of ℝⁿ in ℝ^n+k, i.e., you need a map ψ so that ψ(U)=(a₁,a₂,...,a_n,0,0,...,0). But the subspace 'hoods of the square that contain
(1,1) (with the corners having coordinates (0,0), (0,1),(1,0), (1,1) ) have a corner, unlike
the standard embedding (x₁,..,x_n)→(x₁,x₂,...,x_n,0,0,...,0)

 

[JG89]

Nevermind, I finally see how to prove it. If the unit square were a smooth 2-manifold in $\mathbb{R}^2$ then the interval [0, 0.5), for example, in the unit square would have to be diffeomorphic to an open interval (a,b) on the x-axis, which clearly isn't possible as they aren't even homeomorphic! This is what you were trying to tell me earlier, Lavinia, but I finally see it clearly.

Sorry for being such a pain with this guys, but it was necessary for my understanding

 

[Bacle2]

Don't mean to over-nitpick, but [0,0.5) is not open in the subspace topology, if this is what you meant. Still, the problem intervals must contain the corners.

 

[JG89]

^ I was just choosing a set that is open in the boundary of the square, which [0, 0.5) is.

 

[Bacle2]

My bad, sorry, you are correct.

My bad again: I don't think there is any open subset U of R^2 with U/\S=[0,0.5), where S is the unit square. I don't mean to just play games with you; this is a key
point in this argument, not an empty technicality.

 

[JG89]

Let me show me complete reasoning to make sure that everything is good. If $I^2$ is a 2-manifold in $\mathbb{R}^2$ then let $\alpha$ be a coordinate patch about the point (0,0). Since the interval $[0,0.5) \times 0$ (which is open in the boundary of the square) is part of the boundary of the square, it must be diffeomorphic to a set $(a,b) \times 0$ (which is open in the boundary of the upper half plane) where (a,b) is an interval on the x-axis. This clearly isn't possible as [0,0.5) isn't even homeomorphic to (a,b).

Is this correct?

 

[JG89]

@Bacle,

You're right, there isn't any such open set. But my point is that a set that is open in boundary of the unit square must diffeomorphically map to an open set on the boundary of the upper half plane (which is the cartesian product of the x-axis and {0}). But in the case of [0, 0.5) * {0} (open in the boundary of the square), this set cannot diffeomorphically map to a set that is open in the boundary of the upper half plane (such a set would look like (a,b) * {0}) where (a,b) is an open interval in R, and so the unit square is not a 2-manifold in 2-space

 

[Bacle2]

I'm not sure I get your layout; you are starting with the wrong premise that [0,0.5)x{0} is open in the subspace of the square. The potential problems would happen at the corners, where all the relative open sets are "L's" , i.e., lines forming a right angle. This is the open set you need to show that it cannot be diffeomorphic to a set (a,b).

 

[JG89]

Ah, you're completely right. I don't know how I missed that. [0,0.5) x {0} isn't open in the boundary of the square. Ok, now I see clearly what I have to do. Choose an open "L" interval containing a corner of the square and show that it cannot be diffeomorphically mapped to any open set in the boundary of the upper half plane. I'll get on this now :)

 

[JG89]

Would this be sufficient for a proof:

If $I^2$ is a smooth 2-manifold in $\mathbb{R}^2$, then $\partial I^2$ (the boundary of the square) is a smooth 1-manifold in $\mathbb{R}^2$. Now, obviously $\partial I^2$ is just the perimeter of the square. So if this is a 1-manifold, then there should exist a tangent space at each of its points. However, there are infinitely many lines that are tangent to the square at the corner (0,0), for example, and so there does not exist a tangent space at that point, which means that $\partial I^2$ cannot be a smooth 1-manifold, contradicting our original premise. Thus $I^2$ is NOT a smooth 2-manifold in $\mathbb{R}^2$.

How is that?

 

[Bacle2]

I don't know what theorem to use offhand to state that there must exist a tangent space at each point (tho this is a true fact), but I can think of stating that , if there was a diffeomorphism, it would give rise to ( or, if you want more formal obscurity, the diffeo would "induce" ) a vector space (meaning here tangent space) isomorphism between an open subset of R^n, and the "L" subset. But, as you said, the tangent space (which is a/the tangent line, by dimension arguments) is not defined at the corner ( you can make an argument for why it is not defined) , so the corner cannot be the image of a vector space isomorphism. I can't think offhand of a more clear argument --which of course does not imply one does not exist.

 

[JG89]

the tangent space (which is a/the tangent line, by dimension arguments) is not defined at the corner ( you can make an argument for why it is not defined)

For this I would just say that there are infinitely many lines passing through the corner (0,0) that do not intersect any other part of the square, and so there is no well-defined tangent space at (0,0) of the square.

 

[Bacle2]

Sorry for the delay; I've been kind of busy recently.
Yes; I can only think of variations of the fact that a function whose graph contains a corner is not differentiable at the corner, since the tangent space would be generated by the line y=f'(x)x+b.