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Here the cone surface is $C:=\{z^2=x^2+y^2,\, z\ge 0\}$ in $\Bbb R^3$. Sometimes it is said $C$ is not a 2-dimensional smooth manifold. Talking about manifolds without specifying the atlas is meaningless, of course. However, it's pretty easy to see that under many atlases, for example any atlas that's compatible with the atlas containing the projection to the $xy$-plane as the only chart, $C$ will indeed be a smooth manifold. So I wonder, in what "usual" context (topology, atlas etc), would anyone possibly regard $C$ not as a smooth manifold?

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  • $\begingroup$ Uhm, it's not differentiable at the peak, I guess? $\endgroup$ – mathreadler Dec 19 '17 at 11:33
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    $\begingroup$ @mathreadler that's a reasonable guess. However, if one equips it with a nice chart like projection, there'll be nothing special about the peak. $\endgroup$ – Vim Dec 19 '17 at 11:34
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    $\begingroup$ If one accepts the premise of your question, would there be any manifold at all that is not smooth? For any point or region of non-smoothness one could simply pick a chart in which it looks smooth. $\endgroup$ – Rahul Dec 19 '17 at 13:06
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    $\begingroup$ I see. The proper term should be "an embedded submanifold in $\Bbb R^3$". Thanks. @Rahul $\endgroup$ – Vim Dec 19 '17 at 13:58
  • $\begingroup$ I have not heard about these charts. They sound fishy. I could imagine you approximate the surface by some sequence of functions getting arbitrary close in some sense and still being differentiable there for all manifolds in that sequence. But I don't know if that is what charting means. $\endgroup$ – mathreadler Dec 19 '17 at 21:05
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It is not a smooth manifold in the sense of the differentiable structure it inherited as a subspace of $\Bbb R^3$. For instance, the smooth function $f:\Bbb R^3\to\Bbb R$ given by $(x,y,z)\mapsto z$ is not differentiable as a function on the cone.

To elaborate a bit, a subset $S$ is an embedded $n$-dimensional submanifold of an $m$-dimensional manifold $M$ iff there for every $s\in S$ exists an open $U\subseteq M$ with $x\in U$ and a chart $\phi:U\to \Bbb R^m$ such that $\phi(U\cap S)$ is an $n$-dimensional hyperplane in $\Bbb R^m$. In that case, the different charts $(\phi|_{U\cap S}, U\cap S)$ may be extended to a maximal atlas on $S$, and this is the differential structure that $S$ inherits from $M$. There is no such thing for the tip of the cone in $\Bbb R^3$.

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  • $\begingroup$ Thanks. Could you elaborate a bit on the differentianle structure inherited from $\Bbb R^3$. Can it be easily defined? $\endgroup$ – Vim Dec 19 '17 at 11:36
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An elementary property (that is often needed) of smooth manifolds is the tangent space at each point, which is obtained by taking derivatives of curves. Consider any curve on $C$ that passes through the peak. This curve is not differentiable at the peak (you can consider it as a curve in $\mathbb{R}^3$), hence we are missing a tangent space at this point.

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    $\begingroup$ why the curve is not differentiable at the peak? $\endgroup$ – Manolis Lyviakis Sep 1 '18 at 20:36
  • $\begingroup$ What should be the tangent space at the peak? What are tangents at the peak? They are not defined there. $\endgroup$ – T'x Sep 1 '18 at 20:41
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    $\begingroup$ Just consider a 1-d curve which has a similar peak. It‘s not possible to put a well defined tangent at that point, the limits for the differential quotient from right and left are different. $\endgroup$ – T'x Sep 1 '18 at 20:47
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    $\begingroup$ $C:(x,y,z):z=\sqrt{x^2+y^2}$ suppose it is a smooth manifold meaning there exists a smooth parametrization$ φ(u,v):U⊂R2→C$ such that $φ(u,v)=(φ_1(u,v),φ_2(u,v),φ_3(u,v))$ such that $φ_3=\sqrt{φ_1^2+φ_2^2}$ being continuously differentiable at every point but it fails at the peak since the limit at the peak of the derivative fails to be unique because if i approximate it through different paths(curves) it will be different. (So which is the the derivative and what limiti to take to continue my proof.? $\endgroup$ – Manolis Lyviakis Sep 1 '18 at 21:17
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    $\begingroup$ I give you a hint: consider a curve on there being like $x \mapsto -\vert x \vert$ (intersect a plane with the cone). This is clearly not differentiable at 0. $\endgroup$ – T'x Sep 2 '18 at 14:04
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For surfaces in $\mathbb{R}^3$ there is also an equivalent geometrical property that some set is a surface (smooth submanifold of dimension 2, without border) if and only if locally it can be represented by a graph of a smooth function. It means, locally your surface has a form: $$ x = f(y,z) \text{ or } y = f(x,z) \text{ or } z = f(x,y), $$ where $f$ is some infinitely smooth function. This comes from the fact that if you have general parametrization $p(u,v) = (x(u,v), y(u,v), z(u,v)), \, (u,v)\in U,$ where $(u,v)$ is the parametrization, then the differential of $p$ is injective (by definition of a surface). Injectivity of $p$ will imply that matrix of the differential will have rank 2, which says that locally you parametrize your surfrace by some pair $(x,y)$ or $(y,z)$ or $(x,z)$. Then you need to work a little with the implicit function theorem to obtain the formula above.

For the case of the cone it becomes obvious. The only way to choose coordinate system so that surface near the peak is a function, is to choose that peak "looks up". But in this case the obtained function is necessarily not smooth, hence it cannot be a surface.

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  • $\begingroup$ peak looks up? and why it is not smooth? $\endgroup$ – Manolis Lyviakis Sep 1 '18 at 20:36
  • $\begingroup$ "looks up" it means the tip of the cone looks in direction z and the function is a function of (x,y). in this case the derivative (or gradient) will have a jump at tip. Again, it is important that (x,y) is not just a parametrization of a surface of a cone, but the cone is embedded in R^3 $\endgroup$ – Fedor Goncharov Sep 1 '18 at 22:23
  • $\begingroup$ I can see the image but i cannot write the math $\endgroup$ – Manolis Lyviakis Sep 1 '18 at 22:24
  • $\begingroup$ math.stackexchange.com/questions/2902120/… $\endgroup$ – Manolis Lyviakis Sep 1 '18 at 22:25
  • $\begingroup$ If you assume that the statement above is true (which you need to prove of course), then at the vicinity of the tip your function can be represented only and only by sqrt(x^2 + y^2), (x,y) near 0 (because the statement above is for global coordinates (x,y,z)). $\endgroup$ – Fedor Goncharov Sep 1 '18 at 22:34

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