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Set $M = \{ \, (x, y) : x^2 = y^2 \, \}$. If for every point $(a, c)$ in $M$, there exists a neighborhood $U$ containing $(a, c)$ and function $\phi(x, y)$ such that:

  1. $\phi(x, y) = 0$ on $M \cap U$;
  2. The Jacobian matrix associated with $\phi$ has rank $1$ on $U$. (In general, it does not have to be rank $1$. But here the only choice is $1$.)

Then, $M$ is a manifold. If the Jacobian matrix has ranks greater than $0$, then we have use $\phi$ to carry out the implicit function theorem, and construct a function such that $(x, y) = (x, f(x))$ on $M \cap U$. But I don't know how to go in reverse; what is the contradiction if $M$ is a manifold?

An educated guess says that $(0, 0)$ is our trouble spot. The function $\varphi(x, y) = x^2 - y^2$ equals $0$ on $M$. But the Jacobian matrix has zero rank at $(0, 0)$. So, we cannot use $\varphi$ to carry out the implicit function theorem...

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  • $\begingroup$ As an easy generalization, you might think about for what $n$ the set $M=\{ (x,y): x^n=y^n\}$ is a manifold. It should be precisely when $n$ is odd I believe. $\endgroup$ – J.G May 15 '16 at 23:34
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$x^2-y^2=0$ is equivalent to $(x-y)(x+y)=0$, it is equivalent to $x=y$ or $x=-y$. Thus it is the union of two lines of $R^2$ which intersects at $(0,0)$, it is not a manifold since you don't have a tangent space at $(0,0)$.

You can also say that if it was a manifold, it would have been a $1$-dimensional manifold, but a connected neighborhood of $(0,0)$ can't be diffeomorphic to an interval, since if you remove $(0,0)$ from it, you have at least four connected components.

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    $\begingroup$ This is not quite a proof. $\endgroup$ – Mariano Suárez-Álvarez May 15 '16 at 23:27
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    $\begingroup$ You can argue that the speed vectors of all curves through a point of a submanifold span a subpace of the same dimension as the submanifold, and then show that in this case this is not the case. $\endgroup$ – Mariano Suárez-Álvarez May 15 '16 at 23:29
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    $\begingroup$ Well, what exactly does "you do not have a tangent space" mean? $\endgroup$ – Mariano Suárez-Álvarez May 15 '16 at 23:29
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    $\begingroup$ @MarianoSuárez-Alvarez , these criticisms can get arbitrarily pedantic. In your own answer, are we supposed to know invariance of smooth dimension -- that (if the locus is a manifold) the dimension would be constant at all points, and if constant that the constant would equal 1? $\endgroup$ – zyx May 16 '16 at 0:37
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    $\begingroup$ I don't think observing that "you don't have a tangent space" is hand-waving is being pedantic, at all. I still don't know what it meana, exactly. $\endgroup$ – Mariano Suárez-Álvarez May 16 '16 at 1:25
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A 1-dimensional connected manifold has the property that when you remove any one of its points you get at most two connected components.

Prove this and then use it to show that your set is not a manifold.

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  • $\begingroup$ even more, you can classify every 1-manifold, right? $\endgroup$ – L.F. Cavenaghi May 15 '16 at 23:30
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    $\begingroup$ Yes, every connected 1-manifold without boundary is diffeomorphic to either $\mathbb{R}$ or $S^1$. $\endgroup$ – Moya May 15 '16 at 23:35
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    $\begingroup$ But that result is certainly more difficult than the question in the post! $\endgroup$ – Mariano Suárez-Álvarez May 15 '16 at 23:38
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    $\begingroup$ This is of course a perfectly good proof, but if the OP is struggling with the problem perhaps it's better for him to unpack the definitions and solve the problem by "brute force." $\endgroup$ – Daniel McLaury May 16 '16 at 0:10
  • $\begingroup$ And what exactly is brute force? One needs to find some invariant whosr value one knows for 1-dimensional manifolds and which is computable for this set. $\endgroup$ – Mariano Suárez-Álvarez May 16 '16 at 1:23
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$\phi$ has not just the purpose to indicate which points are in the manifold. It also gives a kind of “local compass”, telling you from each point in which unique direction the manifold continues, namely (one-dimensional case) in direction $\pm\nabla \phi$. Note that really only the direction is meaningful: the distance which you can go to stay in the manifold is in general just infinitesimal; so it's actually rather $\frac{\nabla\phi}{\|\nabla\phi\|}$ that's interesting. But this is not defined if $\nabla\phi=0$ at some point – in this case it can't point you to any one direction where the manifold continues. In your example, at $p=(0,0)$ there are actually two directions in which $M$ extends: $(1,1)$ and $(1,-1)$.

As you say, one way to grasp this problem is to say that it prevents you from applying the implicit function theorem.


By the way, the space of these “local directions” forms itself a manifold structure.

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