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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...

Trang Oul
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Andy Tam
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  • 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. – J.G May 15 '16 at 23:34

3 Answers3

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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.

asmeurer
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Tsemo Aristide
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    This is not quite a proof. – Mariano Suárez-Álvarez May 15 '16 at 23:27
  • why............? – Tsemo Aristide May 15 '16 at 23:28
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    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. – Mariano Suárez-Álvarez May 15 '16 at 23:29
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    Well, what exactly does "you do not have a tangent space" mean? – Mariano Suárez-Álvarez May 15 '16 at 23:29
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    @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? – zyx May 16 '16 at 00:37
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    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. – Mariano Suárez-Álvarez May 16 '16 at 01:25
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    @zyx, I usually incorporate the fact that a manifold has a single dimension in the definition of what a manifold is (an «$n$-dimensional manifold» is a topological space together with a set of functions into $\mathbb R^n$ such that…) In any case, it is an elementary exercise that if the space is connected then a set of charts satisfying the required conditions but with possibly with different $\mathbb R^n$ as codomains then in fact the $n$ is constant. Now indeed my argument shows that the set is not a $1$-dimensional manifold but leaves the possibility that the dimension is larger. – Mariano Suárez-Álvarez May 16 '16 at 05:01
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    ...and yes, I think oe should prove that the set is not a manifold of any other dimension. Even after that, I guess the amount of work involved is less than the one required to make precise what not having a tangent space means and proving it in this case! – Mariano Suárez-Álvarez May 16 '16 at 05:02
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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.

Mariano Suárez-Álvarez
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  • even more, you can classify every 1-manifold, right? – L.F. Cavenaghi May 15 '16 at 23:30
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    Yes, every connected 1-manifold without boundary is diffeomorphic to either $\mathbb{R}$ or $S^1$. – Moya May 15 '16 at 23:35
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    But that result is certainly more difficult than the question in the post! – Mariano Suárez-Álvarez May 15 '16 at 23:38
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    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." – Daniel McLaury May 16 '16 at 00:10
  • 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. – Mariano Suárez-Álvarez May 16 '16 at 01:23
  • Okay, I see where you are going, but I think this statement works for any connected manifold, right? For example, if $M = { , (x, y, z) : z^2 = x^2 + y^2, z > 0 , }$, we can write $z = \sqrt{x^2 + y^2}$. Thus, at most we're still removing one point $(x, y, f(x, y))$ at a time. That is because $f$ is a function. – Andy Tam May 16 '16 at 01:26
  • One invariant is the number of connected components which you get when you remove a point, another is the dimension of span of the speeds of curves, another is that the speeds of curves form a subpace (here one can show that the set of speeds of curves through the origin is not a subpace), or something. Brute force, on the other hand, I don't look now what that means in this context. – Mariano Suárez-Álvarez May 16 '16 at 01:29
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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.

leftaroundabout
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