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Show that the cone given by $C = \{(x, y, z) \in \mathbb{R}^3 \mid z = \sqrt{x^2 + y^2}\}$ is not a smooth manifold

In the definition I'm using of a smooth manifold, each point $x \in M \subseteq \mathbb{R}^k$ (where $M$ inherits the subspace topology from $\mathbb{R}^k$ with the usual topology) has a neighborhood $U$ of $x$ in $M$ such that $U$ is diffeomorphic to some open set of $\mathbb{R}^n$ for some $n > 0$. (This is the definition I'm using from Guillemin and Pollack's Differential Topology book)

Now to prove that $C$ is not a smooth manifold I'd have to show that there exists a point $x \in C$ such that any neighborhood $U$ of $x$ in $C$ is not diffeomorphic to any open set of $\mathbb{R}^2$.

Now I recall reading that $C$ is a topological $2$-manifold, hence each point of $C$ would have a neighborhood in $C$ homeomorphic to an open subset of $\mathbb{R}^2$. So the smoothness of the homeomorphism must fail at some point in $C$.

It seems that smoothness will fail at $x = 0 \in C$. I want to find out how to rigorously prove this.

I'm guessing the proof outline will go something like this;


Proof Outline: Let $U$ be a neighbourhood of $0$ in $C$ and suppose that there existed a diffeomorphism $f : U \to \widehat{U}$ where $\widehat{U}$ is an open subset of $\mathbb{R}^2$. We show that this results in a contradiction, hence it will follow that no neighborhood of $0$ in $C$ is diffeomorphic to an open subset of $\mathbb{R}^2$ and hence $C$ consequently will not be a smooth manifold.


However since I've picked an arbitrary diffeomorphism $f$, I don't know of any way to go about finding a contradiction.

How can I go about proving this?

Also in my proof outline I wrote above, the proof really only shows that $C$ is not a smooth $2$-manifold, wouldn't I need to show that $C$ is not a smooth manifold for any $n > 0$? In that case I'm guessing that $C$ wouldn't even be a topological manifold for any $n \neq 2$.

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  • $\begingroup$ If you remove the origin the space is disconnected. No open subset of $\mathbb{R}^2$ has this property. $\endgroup$ Apr 9, 2018 at 0:26
  • $\begingroup$ @ReneSchipperus How do you know that $C \setminus \{0\}$ is disconnected? Can you give me two open sets $U, V$ in $C$ such that $U \cup V = C \setminus \{0\}$? $\endgroup$ Apr 9, 2018 at 0:37
  • $\begingroup$ Yeah $z<0$ and $z>0$. $\endgroup$ Apr 9, 2018 at 0:41
  • $\begingroup$ The partial derivatives in any neighborhood of the origin don't exist because the right and left hand limits disagree. $\endgroup$
    – John Douma
    Apr 9, 2018 at 0:42
  • $\begingroup$ @ReneSchipperus I think you're referring to the double cone, in this case $z \not<0$ in any point of $C$ $\endgroup$ Apr 9, 2018 at 0:50

1 Answer 1

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Assume $f:U\to C$ is a diffeomorphism of $U\subseteq \mathbb{R}^2$ open onto a nbd of $0$. Let $p:C\to \mathbb{R}^2$ the projection onto the $x,y$ plane then $p\circ f$ is a diffeomorphism and has an inverse by the inverse function theorem. Call it $g$. Then $f\circ g(x,y)=(x,y,\sqrt{x^2+y^2})$ is differentiable, a contradiction.

Although, I am not happy with this answer, as the real meaning of the question is to show that the cone in not the IMBEDDING of a manifold into three space. In fact the cone as a topological space is homeomorphic to $\mathbb{R}^2$ which is a smooth manifold.

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    $\begingroup$ why $p\circ f$ is a diffeomorphism and isnt $f\circ g(x,y)=p^{-1}$ $\endgroup$
    – Jam
    Sep 1, 2018 at 21:53

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