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I want to figure out why the set of zeros of the function $g:\mathbb{R}^{2} \to \mathbb{R}$ defined as $g(x,y) = x^2 - y^2$ is not a differentiable manifold. So what I want to use is the following result:

Let $A \subset \mathbb{R} $ open, $p<n$, $g:A \to \mathbb{R}^p$ of class $C^1$ such that $g'(x)$ has rank $p$ for all $x \in A$ such that $g(x)=0$ therefore $g^{-1}(\{0\})$ is a differentiable manifold in $\mathbb{R}^n$ of dimension $n-p$.

So I computed the Jacobian matrix

$$(2x \;\; -2y )$$

Then in the origin this does not have rank 1, is has rank $0$ and we note that this is the only point where this happens therefore it can't be a differentiable manifold.

The other thing I was trying to do is to give an example of a system of coordinates (commonly charts) where some condition fails, and I thought in $F:\mathbb{R}^2 \to \mathbb{R}^3$ as $F(x,y)=(x,y,x^2-y^2)$ and its Jacobian in zero has rank 3, but I don't know how to proceed to find the contradiction to my definition of differentiable manifold :

A subset $M \subset \mathbb{R}^n$ is a differentiable manifold of dimension $k$ if, for each $x \in M$, exists open sets $U \subset \mathbb{R}^k$ and $V \subset \mathbb{R}^n$ and a class $C^1$ function $f:U \to V$ such that:

1) $x \in V$

2) $f(U)=V \cap M$, $f$ is an homeomorfism

3) for each $y \in U$ the jacobian matrix has rank k

Can you help to proceed in this or tell if my approach are right? thanks a lot :)

Bernard
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user162343
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    The Jacobian of $F$ has rank $2$ everywhere, but now you're looking at the graph of $f(x,y)=x^2-y^2$, which is, in fact, globally a smooth surface. There is no contradiction to anything. For your original problem, the issue is that you cannot parametrize the curve $y^2=x^2$ in a neighborhood of the origin. – Ted Shifrin May 13 '15 at 19:23
  • Ok, so how can you prove that we can't parametrize the curve in the origin? – user162343 May 13 '15 at 19:26
  • It is not even a topological manifold. – Anne Bauval Mar 24 '24 at 17:33

2 Answers2

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Take any neighborhood of the origin, we want to show it isn't diffeomorphic to the open ball $B^n$. The standard arguments from topology are:

  1. If we delete the origin, we disconnect the neighborhood. No point in $B^n$ has this property for $n \geq 2$.
  2. If we delete the origin, there are at least 4 connected components. On the other hand deleting a point in the interval gives only two components.

Thus, the neighborhood is not even homoeomorphic to any open ball.

Sam Clearman
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  • Thanks a lot, but Can you explain this please (in a manner of multivariable analysis jajaja )? Because I don't know topology jet – user162343 May 13 '15 at 19:33
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Here is a "non-topological" proof.

The set $M$ is simply the union of two lines $M = L_1 \cup L_2$ where $L_1$ is the line $y=x$ and $L_2$ is the line $y=-x$.

Arguing by contradiction, let's assume that $M$ is a differentiable manifold. It's dimension must equal $1$, because $M-\{(0,0)\}$ clearly is a differentiable manifold of dimension $1$, and removing a single point from a differentiable manifold of dimension $k$ produces a differentiable manifold of dimension $k$.

Let $L_3$ be the tangent line of $M$ at the point $(0,0)$. For every $C^1$ curve $$\gamma : (-1,+1) \to M $$ such that $\gamma(0)=(0,0)$, the vector $\vec \gamma'(0)$ is tangent to the line $L_3$.

Since $L_1 \ne L_2$, the line $L_3$ is unequal to one of $L_1$ or $L_2$ (possibly both).

If $L_3 \ne L_1$ then we obtain a contradiction by using the curve $\gamma(t) = (t,t)$, whose tangent vector $\vec \gamma'(0)$ is tangent to $L_1$ but not to $L_3$.

If $L_3 \ne L_2$ then we obtain a similar contradiction using $\gamma(t) = (t,-t)$.

Lee Mosher
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