# $y^2 - x^3$ not an embedded submanifold

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.

• I think that using $t=x/y$ to parametrize the points will allow you to give a differentiable structure on the curve. As $t^2=y^2/x^2=x^3/x^2=x$ and, consequently, also $t^3=y$, you can probably do something. Don't know about the first part of your question. May 29, 2013 at 9:37
• By "manifold" do you really mean "differentiable manifold" or maybe "smooth manifold" rather than just "manifold"?
– user14972
May 29, 2013 at 10:34
• @Jesse: The meaning of Xmanifold (e.g. with X="embedded sub") depends on the meaning of manifold.
– user14972
May 29, 2013 at 10:43
• @JyrkiLahtonen Yes, that gives it a smooth structure, and makes it diffeomorphic to the real line. But then the map embedding it in $\mathbb{R}^2$ is not smooth! May 29, 2013 at 10:44
• @Hurkyl Smooth manifolds.
– user23086
May 29, 2013 at 10:58

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.

• Why $f^{\frac{2}{3}}(t)$ is not differentiable at $t=0$? For example, if (by some miracle) we have $f(t)=t^3$ then both $f$ and $f^{\frac{2}{3}}$ are smooth... Oct 22, 2019 at 18:48

It is better to view $y$ as the independent variable and $x=y^{2/3}$. Since $2/3<1$, this has infinite slope at the origin for positive $y$ and infinite negative slope for negative $y$. Hence the origin is not a smooth point of this graph, which is therefore not a submanifold.

• Thanks for your answer. I don't understand why this means though that it is not an embedded submanifold.
– user23086
May 31, 2013 at 0:08
• If you zoom in on an infinitesimal neighborhood of the origin, what you will see is two rays pointing in the same direction, namely the positive direction along the $x$-axis. Meanwhile, if you zoom in on an infinitesimal neighborhood of a point on a smooth curve, you will always see a straight line (namely, rays pointing in opposite directions). You should be able to turn this into a formal argument. If not, I can provide more hints. May 31, 2013 at 8:19
• I am still a beginner at these things, please help me I don't think I capable of make formal argument from this. Please provide more hints thank you.
– user23086
May 31, 2013 at 8:25
• Try using the implicit function theorem: if the curve were smooth at the origin, it would be the graph of a function over the $x$-axis in a neighborhood of the origin. But there are no points over the negative $x$-axis. Alternatively, there aere two points instead of one over the positive $x$-axis. May 31, 2013 at 8:31
• Another approach: for a smooth point on a curve, the intersection of the curve with a small circle centered at the point with give two points at an angle close to $\pi$. Meanwhile, for this curve, the intersection with a small circle centered at the origin will give two points forming an angle that tends to zero instead of $\pi$. Jun 2, 2013 at 7:34

Suppose $$C=\{(x,y):y^2=x^3\}$$ is an embedded submanifold of $$\mathbb R^2$$. Consider a smooth curve $$\gamma:(-\epsilon,\epsilon)\to C$$ such that $$\gamma(0)=(0,0)$$. Since the inclusion is smooth $$\iota:C\to\mathbb R^2$$, we have a smooth curve $$\iota\circ\gamma:(-\epsilon,\epsilon)\to\mathbb R^2, t\mapsto (\gamma_1(t),\gamma_2(t))$$. Note the relation $$\gamma_2(t)^2=\gamma_1(t)^3$$, which yields $$\gamma_1(t)=\gamma_2(t)^{2/3}$$. The function $$x\mapsto x^{2/3}$$ is smooth for $$x>0$$. Since $$\gamma$$ is smooth, its derivative is smooth too, and hence we have $$\dot\gamma_1(0)=\lim_{t\to 0}\dot\gamma_1(t)=\lim_{t\to 0}\frac 23\gamma_2(t)^{-1/3}\dot\gamma_2(t).$$ Since $$\lim_{t\to 0}\gamma_2(t)^{-1/3}=\infty$$, the above limit can only be finite if $$\lim_{t\to 0}\dot\gamma_2(t)=0$$, and hence $$\dot\gamma_2(0)=0$$. This means that $$\gamma(t)=(\gamma_1(t),0)$$. Since $$\gamma_1(t)^3=0$$, we have in fact $$\gamma\equiv 0$$. We've shown that the tangent space of $$C$$ at $$(0,0)$$ is zero-dimensional, which contradicts that $$C$$ is a curve.