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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.

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  • $\begingroup$ 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. $\endgroup$ May 29, 2013 at 9:37
  • $\begingroup$ By "manifold" do you really mean "differentiable manifold" or maybe "smooth manifold" rather than just "manifold"? $\endgroup$
    – user14972
    May 29, 2013 at 10:34
  • $\begingroup$ @Jesse: The meaning of Xmanifold (e.g. with X="embedded sub") depends on the meaning of manifold. $\endgroup$
    – user14972
    May 29, 2013 at 10:43
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    $\begingroup$ @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! $\endgroup$
    – Zhen Lin
    May 29, 2013 at 10:44
  • $\begingroup$ @Hurkyl Smooth manifolds. $\endgroup$
    – user23086
    May 29, 2013 at 10:58

3 Answers 3

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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.

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    $\begingroup$ 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... $\endgroup$
    – Arimakat
    Oct 22, 2019 at 18:48
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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.

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    $\begingroup$ Thanks for your answer. I don't understand why this means though that it is not an embedded submanifold. $\endgroup$
    – user23086
    May 31, 2013 at 0:08
  • $\begingroup$ 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. $\endgroup$ May 31, 2013 at 8:19
  • $\begingroup$ 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. $\endgroup$
    – user23086
    May 31, 2013 at 8:25
  • $\begingroup$ 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. $\endgroup$ May 31, 2013 at 8:31
  • $\begingroup$ 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 $\pi$. $\endgroup$ Jun 2, 2013 at 7:34
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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.

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