Is $\{(x,y)\in \mathbb{R}^2|x^3=y^2\}$ not a manifold?

Question

In "Introduction to Differential Geometry" (https://people.math.ethz.ch/~salamon/PREPRINTS/diffgeo.pdf), Example 2.3.8 (pg 37) says that the following set :

$$C:=\{(x,y)\in \mathbb{R}^2|x^3=y^2\}$$

is not a submanifold. Submanifold is defined in Definition 2.3.1 as just a subset that's a manifold. As the Example mentions, this set is generated by the function $f(t) := (t^2, t^3)$ over $\mathbb{R}$. I think this function is bijective from $\mathbb{R}$ to $C$, $f(t)$ and its inverse are continuous, so why is $C$ not a manifold? Intuitively, although the way it is embedded into $\mathbb{R}^2$ has a kink, isn't it still just a 1 dimensional curve?

Thanks in advance.

Answer 1

Your comments are correct and well-phrased. Allow me to give an alternate perspective, in the hopes that it is also helpful for you.

Another way to define smooth structure, which allows for more general objects than smooth manifolds, is as follows.

A smooth structure on a topological space $X$ is an assignment, to each open set $U \subset X$, of a subset $C^\infty(U; \Bbb R) \subset C^0(U;\Bbb R)$ of functions which we call smooth. This is required to satisfy the properties: if $f \in C^\infty(U)$ is smooth, and $V \subset U$, then $f|_V$ is smooth; and if $\mathcal U = \{U_i\}$ is a collection of open sets whose union is $U$, then $f \in C^0(U)$ if and only if $f|_{U_i}$ is smooth for all $U_i \in \mathcal U$.

Now any subset of $\Bbb R^n$ inherits a smooth structure in this sense. If $U \subset X$ is open and $X \subset \Bbb R^n$, we say that $f \in C^0(U)$ is smooth if and only if there exists some open $V \subset \Bbb R^n$ with $V \cap X = U$ and an extension $F: V \to \Bbb R$ with $F|_U = f$, so that $F$ is smooth in the traditional sense.

A 'space with smooth structure' is called a smooth manifold if there exists an open cover $\mathcal U$ of $X$, with for each $U \in \mathcal U$ a homeomorphism to an open set $\phi: U \to V \subset \Bbb R^n$ so that $f \in C^\infty(U)$ if and only if $f \phi^{-1}: V \to \Bbb R$ is smooth in the traditional sense. (You can check that this implies the $\phi$ define a smooth atlas on $\Bbb R^n$.)

Exercise. A subset $X \subset \Bbb R^n$, equipped with the smooth structure defined above, is a smooth manifold in my sense if and only if $X$ is what is usually called a submanifold of $\Bbb R^n$ --- for each $x \in X$, there exists an open set $U \subset \Bbb R^n$ and a diffeomorphism $\varphi: U \cong \Bbb R^n$ so that $\varphi(U \cap X) = \Bbb R^k \subset \Bbb R^n$.

The point is that your set has a natural "smooth structure" (notion of smooth function) in which it is not locally Euclidean around (0,0).

Answer 2

If you consider topological manifolds, this curve is one-dimensional manifold, since it 'looks' locally (even globally) as a real line. On the other hand, if you consider differential manifolds, it isn't one, since it doesn't 'look' locally as a real line, i.e. there are no diffeomorphism near $(0,0)$. The function $f$ is a homeomorphism on the image, but the inverse function isn't differentiable. Yes, the existence of a kink makes this problem.