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The set $N = \left\lbrace \left(x,y\right)\in\mathbb R^2\mid x^2-y^2-x^4=0\right\rbrace$ is given and I want to prove that $N$ is not a submanifold of $\mathbb R^2$. Our definition of a subset $N\subseteq \mathbb R^n$ being a $k$-dimensional submanifold is that for every $x\in N$ there exists an open neighborhood $U_x\subseteq\mathbb R^n$ and a diffeomorphism $\phi:U_x\to \phi(U_x)=:V_x$ with $V_x\subseteq \mathbb R^n$ also being open so that $$\phi(N\cap U_x)=V_x\cap\left(\mathbb R^k\times\{0\}^{n-k}\right)=\left\lbrace y\in V_x \mid \forall i>k:y_i=0\right\rbrace.$$ From this definition it follows readily that an $n$-dimensional submanifold of $\mathbb R^n$ is open and a $0$-dimensional one is a discrete set. Both of these are not true for my set $N$, so the only remaining option would be a one-dimensional submanifold.

Looking at the plot of this curve on WolframAlpha I think the "problematic" point is $(0,0)$, but I don't know how to prove that there exists no such diffeomorphism at this point. How should I do that?

Bonus question: How can I find these problematic points and get an idea of what the curve looks like just from the equation given and without WolframAlpha (i. e. in an exam)?

Mophotla
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Yes, if it were going to be a submanifold, it would have to be $1$-dimensional. To see locally what it looks like (without Wolfram), notice that near the origin it looks like $x^2-y^2=0$, i.e., an X. If it fit your definition, after some suitable change of coordinates, a neighborhood of the origin would look like $y=0$ sitting inside a neighborhood of the origin in $\Bbb R^2$. Can you see what that's not possible?

The Implicit Function Theorem guarantees you that you will have a submanifold locally around any point where the derivative of your defining function is nonzero. So the problem points come from critical points of that defining function.

For future reference, you might want to look at the answer to this question.

Ted Shifrin
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  • I can see why that isn't possible; no neighborhood in which two lines intersect can resemble just one line, regardless of how much you tilt or shift it. But how could I express this formally? Or is just noticing that there are two intersecting lines in the origin formal enough? You are right that the implicit function theorem is helpful in finding critical points, but unfortunately it does not work in the opposite direction, i. e. does not guarantee a critical point will not have a local submanifold around it. – Mophotla Apr 21 '17 at 22:06
  • @Mophotla: Yes, you're correct. The Implicit Function Theorem (working over $\Bbb R$) is not an if-and-only-if statement. But it will identify the potential "problem points." Nevertheless, the answer I linked you to gives you a powerful way to settle this question constructively. ... You need to use a basic notion of analysis/topology to see why an X can't be homemorphic to an interval. Can you think of the trick? – Ted Shifrin Apr 21 '17 at 22:26
  • Referring to the linked answer: Intuitively speaking, if this X were the graph of a real-valued function (maybe after change of coordinates), then every line parallel to the $y$-axis would only intersect this X in only one point, which obviously isn't possible. However, I can't think of the trick you are mentioning. – Mophotla Apr 21 '17 at 22:51
  • Well, you'd have to check that it's a graph over neither coordinate axis. ... Re trick: Consider removing a carefully chosen point. – Ted Shifrin Apr 21 '17 at 22:52
  • If we remove the origin, our X is not connected anymore and consists of four disjoint open sets, let's say $A,B,C,D$. Is it that if a homeomorphism $\phi$ existed, $\phi(A\sqcup B\sqcup C\sqcup D)$ would also have to be four open sets (which would be four open intervals in $\mathbb R$), but $\phi(A\sqcup B\sqcup C\sqcup D\sqcup (0,0))$ would have to be connected, because $A\sqcup B\sqcup C\sqcup D\sqcup (0,0)$ is connected? So $\phi(0,0)$ would have to "fill" three gaps between four intervals, which is impossible? – Mophotla Apr 21 '17 at 23:16
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    More succinctly, you have 4 components versus 2 components (when you remove $\phi(0)$ from the interval). Right. – Ted Shifrin Apr 21 '17 at 23:17
  • Just to be clear: So, a homeomorphism can never map 4 disjoint open sets onto 2 disjoint open sets? – Mophotla Apr 21 '17 at 23:20
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    Think about the inverse mapping. :) – Ted Shifrin Apr 21 '17 at 23:34