Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the simplest way to show that the cone $\{(x,y,z)\in\mathbb R^3\,|\,z=\sqrt{x^2 + y^2}, z\geq 0\}$ is not diffeomorphic to $\mathbb R^2$?

After some comments, I realize that this question The cone is not immersed in $\mathbb{R}^3$ is more in the the spirit of what I wanted to know.

What I'm really trying to ask in a physical sense is in what sense is the vertex of the cone a point at which the cone is not smooth, and how can one prove this?

share|improve this question
What is your smooth structure on the cone? –  Sanchez Mar 24 '13 at 19:01
With your $z > 0,$ it is not connected. –  Will Jagy Mar 24 '13 at 19:04
@WillJagy Thanks I altered the set. –  joshphysics Mar 24 '13 at 19:06
@Sanchez Yeah ok that's a good point. Is it even possible to put a smooth structure on the cone? My intuition says no and that your question is rhetorical. –  joshphysics Mar 24 '13 at 19:09
joshp, you can take a bijection of your set to any smooth manifold you like, of any dimension, they all have the cardinality of the continuum. As a subset of $\mathbb R^3,$ this thing is not smooth at the origin. –  Will Jagy Mar 24 '13 at 19:12
show 4 more comments

1 Answer

Your original question makes perfect sense. A diffeomorphism between a set $X \subset \mathbb R^n$ and a set $M \subset \mathbb R^k$ means a 1-1 continuous function $f : X \to M$ which is `smooth' in the sense that for every $p \in X$ there is an open neighbourhood $U$ of $p$ in $\mathbb R^n$ and a smooth map

$$ \tilde f : U \to \mathbb R^k$$

such that $\tilde f(q) = f(q)$ for all $q \in U \cap X$. You also demand that $f^{-1} : M \to X$ is smooth.

The reason why a cone isn't diffeomorphic to a euclidean space basically just boils down to an implicit function theorem argument. You ask yourself, what is the tangent space at the cone point? The implicit function theorem tells you that near any point in the manifold, the manifold is the graph of a smooth function from the tangent space to its orthogonal complement. So you just have to argue that regardless of what the tangent space is, there's no smooth function that it's the graph of. There's also a coordinate-axis version of the implicit function theorem which is more convenient in this case. If it was a smooth manifold, the function $z = \sqrt{x^2+y^2}$ would have to be a smooth function.

share|improve this answer
Hmm all right that's reasonable. Thanks for the response Ryan. DO you think this is the "best" way to capture the intuitive notion that the vertex of the cone is a location at which the manifold is "not smooth," especially in comparison to the other question I linked to about proving immersing the cone in $\mathbb R^3$? –  joshphysics Mar 24 '13 at 21:28
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.