Is $ S = \{(x,y) \in \mathbb{R}^2 \colon x^2 = y^3\} $ an immersed submanifold of $\mathbb{R}^2$?

Question

I know that it is not an embedded submanifold because it is not locally Euclidean with the subspace topology, and I feel that it is not an immersed submanifold, because its tangent space at $(0,0)$ has dimension $2$ which would mean that it has to be a smooth manifold of dimension $2$. But if that happens, then since we have that the inclusion map is a smooth immersion, it ends up being both a smooth immersion and a smooth submersion (since the dimensions of $S$ and $\mathbb{R}^2$ are equal) which means that it's a local diffeomorphism. But, it is also bijective onto its image meaning that it is a diffeomorphism onto $S$ with the subspace topology, which is a contradiction. Kindly let me know if my reasoning and conclusion are correct or not.

Answer 1

To find out the dimension of $S$ you have to consider the tangent space at a smooth point. Since $S$ is the zero set of the function $f(x,y)=x^2-y^3$, a point $(x,y)$ is smooth (i.e. admits a neighbourhood diffeomorphic to euclidean space) iff $Df(x,y)\neq 0$.

This shows that $0$ is not a smooth point and it also shows (by using the implicit function theorem) that $S\backslash \{0\}$ is indeed a smooth 1-dimensional submanifold of $\mathbb{R}^2$.

On the other hand you can see that the map $\gamma:\mathbb{R}\rightarrow \mathbb{R}^2, \gamma(t) = ( t^3,t^2)$ is smooth map onto $S$. This is a topological embedding but not an immersion (since $\gamma'(0)=0$).

Edit: $S$ is not an immersed submanifold. Assume otherwise, then there is an immersion $I=(I_x,I_y):\mathbb{R}\rightarrow \mathbb{R}^2$ onto $S$. We may assume that $I(0)=(0,0)$. Again by the implicit function theorem we find a neighbourhood $(-\epsilon,\epsilon)$ of $0$, such that $I((-\epsilon,\epsilon))$ is an embedded submanifold of $\mathbb{R}^2$. $I$ has to be injective, because otherwise its image $S$ would contain a loop, which is impossible since we already know that $S$ is homeomorphic to $\mathbb{R}$. Injectivity of $I$ implies that $I_x$ is strictly monotic (assume increasing), thus $I((-\epsilon,\epsilon))=S\cap\{I_x(-\epsilon)<x<I(\epsilon)\}$. (We have just assured that the singularity at $(0,0)$ does not come from a self intersection nor from approaching itself) But this implies that $S\cap\{I_x(-\epsilon)<x<I(\epsilon)\}$ has a one dimensional tangent space at $(0,0)$ which is not true as you've already found out.