# Proving a space is a manifold

Given a topological space defined as $A=A_1 \cup A_2$ with

$A_1=\{(x,y) \in R^2 \space \space|\space \space x^2+y^2=1, x<0\}$,

$A_2=\{(x,y) \in R^2 \space \space|\space \space |x|+|y|=1, x\geq0\}$.

Show that $A$ is a one-dimensional manifold.

So if we omit all the tiny details (e.g. showing $A$ is Hausdorff, connected etc.) and just get to the point of finding a differentiable structure on $A$, we need to find a collection of charts $(U_i, \phi_i)$ s.t. $A=\cup_iA_i$ and that the transition functions $\phi_j \circ \phi_i^{-1}$ are $C^{\infty}$.

So if I start by defining the subsets

$U_1=\{(x,y) \in A : y>0\}$

$U_2=\{(x,y) \in A : y<0\}$

$U_3=\{(x,y) \in A : x>0\}$

$U_4=\{(x,y) \in A : x<0\}$

it's clear $A=\cup_iA_i$. But what about the maps? Let's say I take:

$\phi_1=x$

$\phi_2=x$

$\phi_3=y$

$\phi_4=y$

However, I've just thought of these off the top my head without any justification other than the fact the variable for $\phi_i$ is different from the restricted variable in the subset $U_i$. (I don't even know why I'm doing that!) There must be some small trick that tells me how to chose my maps once I've defined my subsets?

I'm quite sure $\phi_j \circ \phi_i^{-1}$ are $C^{\infty}$, but how can I show this? In words I could probably explain it, but writing something mathematically correct is a bit different. What about e.g.:

$\phi_1(x,y)=x \implies \phi_1^{-1}(x)=(x,y) \implies \phi_3 \circ \phi_1^{-1}=\phi_3 ( \phi_1^{-1}(x)) = \phi_3 (x,y)=y$?

Which is obviously $C^{\infty}$. But I'm not convinced I've done that right. Any help?

• I don't think the space is a submanifold of $\mathbb{R}^{2}$ because there should be a singularity at $(1, 0)$, although I'm not certain about this point. However, there is definitely a smooth structure on $A$ because it is homeomorphic to a circle and so you can just define the smooth structure to be that of the circle. Apr 20, 2014 at 10:52

So, taking your coordinate chart maps, let's compute the transition maps.

On $U_{1} \cap U_{3}$: For $t \in \phi_{3}(U_{1} \cap U_{3}) \subseteq \mathbb{R}$, we have $$\phi_{1} \circ \phi_{3}^{-1}(t) = \phi_{1}(1 - t, t) = 1 - t$$ which is nice and $C^{\infty}$. Also, $$\phi_{3} \circ \phi_{1}^{-1}(t) = \phi_3(t, \sqrt{1 - t^{2}}) = \sqrt{1 - t^{2}}$$ which is not $C^{\infty}$ at $t = 1$ but that value is not in $\phi_{1}(U_{1} \cap U_{3}).$ So, these two transition maps are $C^{\infty}$ and the checks in the other charts are almost the same.

Edit: I made a pretty big mistake in the second computation (which would hold for $U_{2} \cap U_{3}$ not for the other direction of the map).

For $t \in \phi_{1}(U_{1} \cap U_{3})$, $\phi_{1}^{-1}$ maps into the first quadrant. Hence, we should have \begin{align*} \phi_{3} \circ \phi_{1}^{-1}(t) &= \phi_{3}(t, 1- t) \\ &\text{because t must be the x coordinate and }\\ &\text{1- t is the y coordinate for such points in the first quadrant}\\ &= 1 - t \end{align*}

This is still smooth. The computation I had previously done by mistake should have been done for $U_{1} \cap U_{4}.$

• Can you define $t$ in your answer please? Apr 20, 2014 at 11:08
• So, $\phi_{1}^{-1}$ is a map from $\mathbb{R} \rightarrow A$ and hence, $t$ is some arbitrary real number. I'll edit the answer. Apr 20, 2014 at 13:06
• Could you also tell me if there is a specific way to choose my maps for each subset? As I said in the question, I was just making these maps up because it seems like they work. Is there not some sort of trick I can use to choose maps for each subset? e.g. would $\phi_1=y$, $\phi_2=y$, $\phi_3=x$, $\phi_4=x$ still work? Apr 20, 2014 at 13:57
• The charts you chose are the natural choices: projection maps onto the $y$ and $x$ axes, on connected domains where these projections are one to one. It is a theorem that the charts of a submanifold of $\mathbb R^N$ can always be defined in a similar manner. It's confusing me that it seems to work here, though, because, as Christian Blatter points out, $A$ is certainly not a smooth submanifold of $\mathbb R^2$. Apr 20, 2014 at 14:14
• I think it works here because all of the kinks are not in the intersection of any two charts. Apr 20, 2014 at 16:58

Your space $A\subset{\mathbb R}^2$ is homeomorphic to $S^1\subset{\mathbb R}^2$ via $$f:\quad A\to S^1,\qquad {\bf z}\mapsto{{\bf z}\over |{\bf z}|}\ .$$ As $|{\bf z}|\geq{1\over\sqrt{2}}$ on $A$ this $f$ is continuous, and furthermore it is obviously a bijection. It follows that $f$ is a homeomorphism.

Now you could transport the differentiable structure on $S^1$ via $f\>$ to $A$, and then $A$ would become a smooth manifold. In addition the $C^\infty$ structure on $A$ would be compatible with the topological structure $A$ has inherited from ${\mathbb R}^2$.

But your space $A$ is not a smooth submanifold of ${\mathbb R}^2$ because of the kinks at $(0,\pm1)$. There is no window $W:=[-h,h]\times[1-h,1+h]$ such that $A\cap W$ can be written as a graph $$y=\phi(x)\qquad(-h<x<h)$$ with $\phi\in C^\infty$.

• You've shown me that we don't have a smooth manifold, however Siddharth's answer says we do have a smooth manifold and I can't see a problem with the solution presented. Apr 20, 2014 at 19:28
• Siddharth's answer corresponds to my second paragraph. What I'm saying in the third paragraph is that this manifold structure on $A$ is not directly inherited from the embedding of $A$ in ${\mathbb R}^2$. Apr 20, 2014 at 20:21