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I am trying to learn differential geometry (i.e., teach myself!) So here is a question that came up.

For some $h > 0$, consider the cone

$C_h = \{ (x,y,z) \; : \; 0 \le z = \sqrt{x^2 + y^2} < h \} \subset \mathbb{R}^3$

endowed with subspace topology. It seems that we can cover this with a single chart $(U,\phi)$ where $U = C_h$ and $\phi$ is the projection $\phi(x,y,z) = (x,y)$. So it seems that this defines a differentiable structure and we get a smooth ($C^\infty$) 2-dimensional manifold. (Is it correct?)

Now consider the inclusion map $i : C_h \to \mathbb{R}^3$, is this maps smooth? It doesn't seem to me that it is. The expression of $i$ in the chart above is not smooth at $(0,0)$ and I don't seem to be able to find any other compatible chart around zero which has a smooth representation. (Haven't given it much thought though). If this is true how one shows that this map is not smooth. (Also, if this is true, a vague question is whether removing the origin is the only way to fix this problem)

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With regards to your last paragraph, it's worth noting that a function is smooth at p in one chart iff it's smooth at p in all charts. This follows from the fact that the coordinate interchange maps are smooth diffeomorphisms. – Jason DeVito Feb 17 '11 at 23:27
Thanks. I realized that after someelse's hint. – passerby51 Feb 19 '11 at 18:33

You've endowed $C_h$ with the structure of an abstract manifold but $C_h$ is not a submanifold of $\mathbb R^3$. The fact that your set isn't a submanifold boils down to two observations:

(1) The fact your map $i$ is not differentiable at the origin


(2) An application of the implicit function theorem gives the proof by contradiction. The implicit function theorem says that if your set was a submanifold, $i$ would have to be smooth -- technically you have to consider the two other coordinate projections $(x,y,z) \to (y,z)$, $(x,y,z) \to (x,z)$ but $C_h$ does not satisfy the "vertical line rule" so it can't be a graph of a function of $(y,z)$ or $(x,z)$.

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Thanks for the response. – passerby51 Feb 19 '11 at 18:34
@passerby51 That $C_h$ is not a differentiable submanifold of $\mathbb R^3$ is of course completely intuitively obvious since there is no well-defined tangent plane at $(0,0,0)$ (corner! cusp! singularity!). The manifold structure you give $C_h$, however, is not the one you see because what you've really done is declared $C_h$ to be diffeomorphic to the open disk. Hence, what the formal theory tells you with the failure of $i$ to be differentiable at $(0,0,0)$ is that whatever structure or geometry $C_h$ inherits as a subset of $\mathbb R^3$, it is not the geometry of the open disk. – Vladimir Sotirov Feb 27 '11 at 5:53
@Vladimir: Thanks for the illuminating and thoughtful comment. I see it better now. Here are some more questions! How do you formalize "whatever structure or geometry $C_h$ inherits as a subset of $\mathbb{R}^3$ ...", it seems a bit vague. Also, the differentiable structure I put on $C_h$ fails to make the inclusion map differentiable. Is there a way to prove that there is no (other) differentiable structure on $C_h$ that leads to the inclusion map being differentiable? – passerby51 Mar 5 '11 at 20:00
Regarding your latter question: if the inclusion had zero derivative at the singularity that would allow for the image to be $C_h$. So you could make your chart for $C_h$ the map $(x,y) \longmapsto ( \sqrt{(x^2+y^2)}x, \sqrt{(x^2+y^2)}y, (x^2+y^2) )$ (domain the unit disc in the plane), and then inclusion of $C_h$ in $\mathbb R^3$ would be differentiable and a topological embedding, but it would have zero derivative at $(0,0,0)$. Regarding your 1st question, yes, see the beginning of Guillemin and Pollack's "Differential Topology". – Ryan Budney Mar 5 '11 at 22:16

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