# Showing something isn't a manifold

So I'm following some notes that are introducing manifolds with pretty minimal prerequisites. What I want to do is show where the image of $\phi: \mathbb{R}\rightarrow \mathbb{R^2}$ $t\mapsto (t-\sin(t),1-\cos(t))$ isn't a manifold. Since this is the standard cycloid, it's pretty clear that things go bad at the cusps, but how do I rigorously show that $\phi(\mathbb{R})$ isn't a manifold there? The derivative of $\phi$ isn't 1:1, but that's not enough is it? How do I know there's not some better parametrization of $\phi(\mathbb{R})$ around the cusps?

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It depends on what kind of "manifold" you're talking about. If you mean topological manifold (i.e. a space locally homeomorphic to $\mathbb R$), then the cycloid is in fact a manifold. One way to see this is to note that $f(t) = t-\sin t$ is a continuous, strictly increasing function from $\mathbb R$ to $\mathbb R$, and hence injective; it's easily seen to be surjective as well, so it's a homeomorphism. The map $\psi: \text{Im}(\phi) \to \mathbb R$ given by $\psi(x,y) = f^{-1}(x)$ is a continuous inverse for $\phi$ and thus a homeomorphism. – Jack Lee Nov 7 '12 at 22:17
@JackLee: Yeah, tis true! I'm given a pretty naive definition, essentially that it can be covered by smooth chart. I should have made the distinction that I wanted it locally diffeomorphic to $\mathbb{R}$. – AsinglePANCAKE Nov 8 '12 at 5:56
OK. You could try assuming there is a regular parametrization (i.e., a smooth parametrization whose velocity never vanishes) of the same set, $\psi:\mathbb R\to \mathbb R^2$, written $\psi(t)=(x(t),y(t))$. Then see what you can say about $x'(0)$ and $y'(0)$ to derive a contradiction. – Jack Lee Nov 8 '12 at 15:09
@JackLee I tried, and I'm failing. Anymore hints? Also, how do I check that this inverse is continuous? Finding it explicitly seems...messy. – AsinglePANCAKE Nov 8 '12 at 22:27
What I had in mind about $x'(0)$ and $y'(0)$ was something like this: first, since $y(t)$ takes a minimum at $t=0$, we have $y'(0)=0$. Then use the fact that $\psi'(t)$ has to be parallel to $\phi'(t)$ when $t\ne 0$ to conclude that $y'(t)/x'(t)\to\infty$ as $t\to 0$. This is only possible if $x'(t)\to 0$, which contradicts the assumption that $x'$ is continuous and $\psi'(0)\ne 0$. – Jack Lee Nov 8 '12 at 23:37

tl;dr version: skip the first two paragraphs.

Let's focus on the question first. We are told to prove that the set $X=\phi(\mathbb R)$ "is not a smooth manifold". What does this mean? The set already has a topological manifold structure. Are we to prove that it does not admit a differentiable structure? But that would be false. In fact one can put a differentiable structure on $X$ simply by declaring that $\phi$ is a diffeomorphism (i.e., smooth charts on $X$ would be defined as compositions of $\phi$ with smooth charts on $\mathbb R$). The set $X$ can be a smooth manifold. [Added] I'll elaborate: suppose $X$ be a topological space and $\phi:\mathbb R\to X$ is a homeomorphism. We can introduce differentiable structure on $X$ be declaring $\phi$ to be a smooth chart. The result is a smooth manifold with an atlas of one chart. (If your definition of manifold involves a maximal atlas, Zorn's lemma provides a maximal atlas containing $\phi$.)

But $X$ cannot be an submanifold of $\mathbb R^2$. Being a submanifold requires that the inclusion map $\imath :X\to\mathbb R^2$ be smooth with derivative of maximal rank (in this case, 1). Let's assume this and get a contradiction. Let $u : X\to \mathbb R$ be a local coordinate at a cusp such that $u(0)=0$. Then $\imath\circ u^{-1}$ is a smooth map from $\mathbb R$ to $\mathbb R^2$ with nonzero derivative at $0$. The differentible structure on $X$ is now out of the way. (It did not actually have to get in the way, if one uses a more direct definition of "submanifold of $\mathbb R^n$".)

It remains to prove that there is no smooth map $x=X(s),y=Y(s)$ that takes values in $X$, sends $0$ to cusp $(0,0)$, and has nonzero (rank 1) derivative at $0$. Since $Y$ has a minimum at $0$, it follows that $Y'(0)=0$. The derivative requirement implies $X'(0)\ne 0$. Invoking the limit definition of the derivative, we see that $\lim_{s\to 0}\frac{Y(s)}{X(s)}=0$. Our original parametrization $t\mapsto (x,y)$ also sends $0$ to $(0,0)$. Hence, when $t$ is small, the point $(x(t),y(t))$ is close to $(0,0)$. This implies $(x(t),y(t))=(X(s),Y(s))$ where $s=s(t)\to 0$ as $t\to 0$. Therefore, $\lim_{t\to 0}\frac{y(t)}{x(t)}=0$. But calculation shows that $\lim_{t\to 0}\frac{1-\cos t}{t-\sin t}$ does not exist.

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Awesome. Might you elaborate a bit on the first paragraph? I'm assuming you're making the distinction between embedding a manifold into a higher dimensional space and considering it more intrinsically. Also, would you have a handy reference for this limit invariance under reparametrizations? – AsinglePANCAKE Dec 20 '12 at 23:02
@AsinglePANCAKE 'limit invariance' was really a fancy way of saying that taking a limit can be exchanged with composition with a continuous map. Which is more or less the definition of continuity. $\lim_{t\to 0} f(t)=A$ implies $\lim_{t\to 0} f(g(t))=A$ whenever $g$ is continuous with $g(0)=0$. – user53153 Dec 21 '12 at 3:00

If a subset of $\mathbb{R}^{2}$ is a 1 dimensional manifold, then it is locally homeomorphic to $\mathbb{R}^{1}$. I do not think in a cusp you can find such a chart.

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Totally agree! But how do I prove this? – AsinglePANCAKE Nov 7 '12 at 3:53
at the point of intersection you maps the point connected with 4 path components, and if it homemomorphic to $\mathbb{R}^{1}$ you maps the point connected with 2 path components. Since path-connectivity is preserved by homeomorphism, this showed such a chart cannot exist. I think it might be cumbersome to write the formulas. – Bombyx mori Nov 7 '12 at 3:58
another option is to use implicit function theorem, which asserts a function's solution is a submanifold if and only if the function's derivative is non-singular at every point. However I do not see an elegant way to construct such an implicit function at here. One need to cancel out the sins on the left hand by the cos on the right hand, and so forth and so on, feel quite complicated. – Bombyx mori Nov 7 '12 at 4:01
@user32240: actually, the implicit function theorem is not an "if and only if" statement, just "if." For example, the function $f:\mathbb R^2\to \mathbb R$ given by $f(x,y) = x^2$ has zero derivative everywhere along its zero set (the $y$-axis), but its zero set is nonetheless a smooth submanifold. – Jack Lee Nov 7 '12 at 22:20
Are you sure Cliff Taubes wrote that? I just looked through the first few pages of his book, and didn't find any such "if and only if" statement. I would be surprised to find a mistake like that in his book. – Jack Lee Nov 8 '12 at 23:49