3

So, I would like to prove that the curve $\alpha :y^2 -x^3 =0$ is not a differential submanifold of $\mathbb{R}^2$.

My notes are quite messy about this, and at the time it was an argument that I really didn't get. Moreover, it's one of the first times I have to deal with manifolds. I know that I am supposed to use (I mean, the teacher used) the implicit function theorem, and see the curve as the locus of zeros of a differentiable function in $\mathbb{R}^2$, because I need a submanifold of $\mathbb{R}^2$. If you consider such a function, you can prove that in $(0,0)$ both partial derivatives are zero. Then you say you can't apply the implicit function theorem and so the curve is not a submanifold of $\mathbb{R}^2$.

There are a few things I am not sure about. However, the most troublesome is by far the application of the implicit function theorem. I mean, the theorem is great if you want to prove that some curve has a regular parametrization without bothering searching for an explicit one, which is great if I had wanted to prove that a curve is a differential submanifold. Here I cannot apply the theorem in $(0,0)$. How can you conclude then? Doesn't the implicit function theorem give only sufficient conditions? I mean, if you prove that every differentiable function in $\mathbb{R}^2$ with $\alpha$ as locus of zeros has partial derivatives equal to $0$ at the origin, why should it mean that no structure of differential submanifold is possible at all?

tommy1996q
  • 3,344
  • What is your definition of differential subvariety? Is this equivalent to differentiable manifold? – Severin Schraven Aug 20 '17 at 20:57
  • @SeverinSchraven yes, I've edited the question – tommy1996q Aug 20 '17 at 21:56
  • You might want to think about this answer to a very similar question. – Ted Shifrin Aug 20 '17 at 23:23
  • @TedShifrin I checked your other answer, but I don't understand why it should be that way. I mean, I've only studied the inverse function theorem with $\mathbb{R}^n$, but I think I get the definition of smooth function from a differential variety to $\mathbb{R}^n$ (it's that there must be a chart such that the transiction function is smooth , right?). It's just I don't get why every submanifold should locally be the graph of a smooth function. If it's a difficult theorem I am going to trust you, because I am not very used to these things, but if it's easy to understand I would like to – tommy1996q Aug 21 '17 at 08:01
  • 1
    A $1$-dim. manifold in $\Bbb R^2$ must in particular locally (around each point) be either a graph $y=\phi(x)$ or a graph $x=\psi(y)$. This follows because the tangent line at any point must project isomorphically onto (at least) one of the coordinate axes. The Inverse Function Theorem (applied in a chart at that point) tells you that the projection of a neighborhood of that point in the curve onto said coordinate axis is a diffeomorphism. The smooth inverse of that projection gives you the graph you wanted. – Ted Shifrin Aug 21 '17 at 16:02
  • @TedShifrin I think I got it now. Let's say we have an homemorphism between a nbhd of some point on the curve and $\mathbb{R}$. Then the inverse must be something like $(f(t),g(t))$ and changing variables you get either $(t,h_1 (t))$ or $(h_2 (t),t)$, and those $h_1$ and $h_2$ should be smooth, shouldn't they? However at this point how do you conclude? How do you show that curve is not a graph of a smooth function in $(0,0)$? The only think I can think about is that $y(x)=\pm x^{3/2}$ which should create problems in the origin – tommy1996q Aug 21 '17 at 17:36
  • You can't have a $\pm$ in a function. Having $y$ as a function if $x$ is impossible, and having $x$ as a function of $y$ violates differentiability. By the way, your chart needs to be a diffeomorphism. This curve is a topological submanifold. – Ted Shifrin Aug 21 '17 at 18:04
  • @TedShifrin yeah sorry, with that $\pm$ I meant that solutions are symmetrical in $y$ and the problem lies in $(0,0)$ (in other points there should be no problem, right?), where it seems no differentiable change of charts is possible. About it being a topological submanifold I agree with you, you could just project everything on the $y$ axis and have an homeomorphism, the problem is about giving it a differential structure, and it's this impossibility that I can't understand – tommy1996q Aug 21 '17 at 20:33
  • With a differentiable chart, you proceed to apply the Inverse Function Theorem to deduce it must (in a neighborhood of the origin) be one of the two graphs, as we said. But neither works. So it can't be a differentiable manifold in a neighborhood if the origin. – Ted Shifrin Aug 21 '17 at 22:48
  • @TedShifrin Ok I finally get it, thanks. So the key to solve the problem is to use the Inverse Function Theorem to state that around each point it must be the graph of a smooth function. But this happens only because we want $\alpha$ to be a submanifold, right? If we just wanted to put a structure of 1-manifold on $\alpha$ without bothering with the rest of $\mathbb{R}^2$ we could have done so, couldn't we? – tommy1996q Aug 22 '17 at 11:37
  • The question in the title does not make any sense — "blah is a manifold in R^2" does not mean anything really. Being a submanifold, on the other hand, does make sense. – Mariano Suárez-Álvarez Jan 24 '23 at 10:47
  • @MarianoSuárez-Álvarez edited – tommy1996q Jan 24 '23 at 23:26

1 Answers1

3

I have made @TedShifrin's argument rigorous here. Suppose for the sake of contradiction that the curve $y^{2}-x^{3} = 0$ is a smooth manifold. In this case we may choose $\eta\in(0, 1)$ such that $\Gamma:=\{\|x\|_{\infty} < \eta\}\cap\{y^{2}-x^{3} = 0\}$ is diffeomorphic to an open interval $I$ (since this set is connected). Let $g = (g_{1}, g_{2}): I\rightarrow\Gamma$ denote this diffeomorphism. Note that the function $g_{2}: I\rightarrow(-\eta^{3/2}, \eta^{3/2})$ is smooth, injective, and in particular, $g_{2}'\neq 0$ on $I$. Therefore, using the Inverse Function Theorem, it is a diffeomorphism and thus $g\circ g_{2}^{-1}:(-\eta^{3/2}, \eta^{3/2})\rightarrow\Gamma$ is a diffeomorhism. For any $t\in(-\eta^{3/2}, \eta^{3/2})$ we have $g\circ g_{2}^{-1}(t) = (g_{1}(g_{2}^{-1}(t)), t)\in\Gamma$ and thus we must have $g_{1}(g_{2}^{-1}(t)) = t^{2/3}$. This implies that $g\circ g_{2}^{-1}(t) = (t^{2/3}, t)$ is a smooth, which is clearly false.

Karthik Kannan
  • 1,094