How can I show that the cuspidal cubic $y^2 = x^3$ is not an embedded submanifold of $\Bbb{R}^2$? By embedded submanifold I mean a topological manifold in the subspace topology equipped with a smooth structure such that the inclusion of the curve into $\Bbb{R}^2$ is a smooth embedding. I don't even know where to start please help me. All the usual tricks I know of removing a point from a curve and see what happens don't work. How can I extract out information about the cusp to conclude it is not? Also can I put a smooth structure on it so it is an immersed submanifold? THankz.
[Math] $y^2 – x^3$ not an embedded submanifold
differential-geometry
Best Answer
Suppose the cuspidal cubic $A=\{(x,y)\in\Bbb R^2|y^2=x^3\}$ is an embedded submanifold of $\Bbb R^2$, and let the inclusion map $i:A\hookrightarrow \Bbb R^2$ denote the smooth embedding. For $A$, there is a smooth chart $(U,\varphi)$ containing $(0,0)$ such that $\varphi(U)=(-\epsilon,\epsilon)$ and $\varphi(0,0)=0$, then $\varphi^{-1}: (-\epsilon,\epsilon)\to U$ is a diffeomorphism, so $i\circ \varphi^{-1}$ is a smooth map. Let $\pi:\Bbb R^2\to \Bbb R$ denote the projection $(x,y)\mapsto y$, we define $f:=\pi\circ i\circ \varphi^{-1}$, then $(i\circ \varphi^{-1})(t)=(f^{\frac{2}{3}}(t),f(t))$, because $f(0)=0$, $f^{\frac{2}{3}}(t)$ is not differentiable at $t=0$, which is a contradiction.