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I am trying to show that the set $$X = \{(x, y) \in \mathbb{R}^{2} \mid x^{2}=y^{3}\} \subset \mathbb{R}^{2} $$ is not a submanifold of $\mathbb{R}^{2}$. I understand that the curve isn't smooth at $(0,0)$ and this is why, but I just can't translate this into a proper proof. I know it has to do with the implicit function theorem (or inverse function theorem?) but I'm not sure how those theorems help here. The explanations for similar examples that I have seen make use of these tools but they are a little hand-wavey and I just don't see how to formalise the argument.

duelspace
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    $X$ is the union of two curves, the graph of $x = y^{3/2}$ and the graph of $x = -y^{3/2}$. These intersect at one point, the origin. So in any $\Bbb{R}^2$-neighborhood of the origin, the intersection of $X$ with that neighborhood is isotopic to "X", that is, two intersecting line segments... – Eric Towers Oct 29 '21 at 21:29
  • See the answer to this question. – Ted Shifrin Oct 29 '21 at 21:49
  • @TedShifrin Thanks for your reply. I have seen this but I don't understand how we can show the inverse function theorem implies that locally any submanifold of $\mathbb{R}^{n}$ is a smooth function. I've been re-reading through the statements of the theorem and definition of submanifold but i just dont see it. – duelspace Oct 30 '21 at 09:26
  • @EricTowers I don't understand the last bit - the intersection of $X$ with that neighbourhood is isotopic to $X$? What does isotopic mean here? What's the issue with a neighbourhood of the origin giving two intersecting line segments? If we take out a point the result is still homeomorphic to $\mathbb{R}$, no? – duelspace Oct 30 '21 at 09:29
  • The tangent line to the submanifold at any given point must project isomorphically to (at least) one of the coordinate axes. – Ted Shifrin Oct 30 '21 at 16:02
  • @TedShifrin I haven't learnt about tangents to manifolds yet :/ – duelspace Oct 30 '21 at 18:31
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    'the intersection of X with that neighborhood is isotopic to "X", that is, two intersecting line segments'... "X" is not $X$. "X" is a letter made by two linear strokes. The issue is that we either get a neighborhood of $\Bbb{R}$ or a neighborhood of $\Bbb{R}^2$ if this is a submanifold -- so which is is? If we take out a point, it isn't even connected, so cannot be homeomorphic to $\Bbb{R}$. – Eric Towers Oct 31 '21 at 15:49

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Let suppose that $X$ is a manifold, then locally in $(0,0)$ the set $X$ is the graph of a differentiable function $g$, which satisfies $g(0)=0$. Then $x^3=g(x)^2$. If $n$ is the order of $g$ at $0$ then $2n=3$. Contradiction.

Bernard
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Neo
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  • Thanks for your reply. Why must the set $X$ locally be the graph of a differentiable function? I don't see why this follows from the definition of a submanifold/inverse function thm. Also, what does the 'order' of g mean? – duelspace Oct 30 '21 at 09:30
  • I suggest this document for you (Proposition 3.2.1) https://maths-people.anu.edu.au/andrews/DG/DG_chap3.pdf – Neo Oct 31 '21 at 23:14