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First of all! I mean a cylinder with a circle at the top and a circle at the bottom. Not the volume but its surface.

Intuition tells me YES. But...

How can charts be defined for the points on the edge? What I thought first is that since a sphere is a differentiable manifold and I could find a bijection projecting the cylinder on a sphere placed in its center (of symmetry) I could extend the properties of the sphere to the cylinder.

Now some doubts arise since the metric that I would define on the cylinder would be spherical, while I would expect to get standard euclidean geometry on the flat surfaces. Anyway, no metric is yet defined on differentiable manifolds and I can imagine to define the metric that best fits after. So

  1. Wthout a metric, does the differential manifold structure allows me to distinguish between a cylinder and a sphere?
  2. With a metric can I be sure whether my manifold is a cylinder or a sphere or anything else?
  3. Where can I get a handy summary on the structure information given by topology, manifold, and metric when I only know some of the three? thank you in advance
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  • $\begingroup$ Your manifold isn't a differentiable manifold because of the sharp edges at the top and bottom. $\endgroup$ – Martin Aug 11 '16 at 11:21
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    $\begingroup$ Maybe you should be explicit about what you mean by a "cylinder". I'd have thought maybe $\{(x,y,z) : x^2+y^2 = 1\}$, but on the other hand one says that Archimedes discovered that the volume of a sphere is $2/3$ the volume of the smallest right circular cylinder into which the sphere fits, and there we might mean $\{(x,y,z): x^2+y^2\le r\text{ and } |z|\le r\}$. At any rate I think there is a definition of a "manifold with boundary" that is not, strictly speaking, a manifold. $\qquad$ $\endgroup$ – Michael Hardy Aug 11 '16 at 12:24
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    $\begingroup$ It's basically the same considerations as in this question: it is not a submanifold of $\mathbb R^3$, but viewed abstractly it can be given a smooth manifold structure. $\endgroup$ – Najib Idrissi Aug 11 '16 at 13:01
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If you want your cylinder to have an "edge" then intuitively, it should not be a differentiable manifold. If you take a point in a differentiable manifold, then there is a regularly parametrized curve through the point in each direction. (Just take a straight line in a chart in transport it onto the manifold.) This means that you can "pass through that point" in each direction on a smooth curve with non-zero velocity. If you apply this to a point on the edge, this means that you can get from the side of cylinder to the top of the cylinder by a smooth curve whose speed is non-vanishing at each point. But obviously this should not be possible at the edge, since the speed would have to jump from having non-zero vertical component to being horizontal. (What I am trying to explain in a non-technical way here is that the cylinder has no well defined tangent plane at points of the edge.)

The problem with a formal answer to your question is that in order to get a formal answer, you would first have to specify what it means to "make the cylinder into a manifold". As you said, you can choose a bijection to a sphere (which even is a homeomorphism) and use this to carry the manifold structure over to the cylinder. But in this way, you have "removed" or "flattened out" the edge. Certainly, the cylinder is not a submanifold in $\mathbb R^3$, which can be seen by making the above argument with tangent spaces precise.

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Assume the cylinder is not solid and does not have a top or a bottom then yes. Its a differentiable manifold. This cylinder in $\mathbb{R}^3$ could be defined by $\{x \in \mathbb{R}^3 \mid x_1^2 + x_2^2 = R^2 \text{ and } |x_3|\leq C\}$. Now this cylinder is different from a sphere by the curvature: the curvature on a sphere is everywhere and in every direction the same. The curvature on the cylinder differs. Its $1$ if you go parallel to the $xy$-plane, its $0$ parallel to the $z$-axis. In general curvature is a good way to differentiate manifolds "by themselves".

Edit/Added: Finding a bijection between two manifolds does not make them equivalent, the sphere is not equivalent to the cylinder just because you can relate them 1-to-1. The problem with a top/bottom is that you will have a kink on the edges. At those kinks the manifold is not differentiable, just like $|x|$ is not differentiable at $0$.

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  • $\begingroup$ 1. What's the problem with the top and bottom and what'wromg with my spherical construction? 2. Is C constant? In case that's just a circle embedded in R^3. 3. Any hint for the handy summary? $\endgroup$ – ThePunisher Aug 11 '16 at 12:21
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    $\begingroup$ The set you define, $\{x \in \mathbb{R}^3 \mid x_1^2 + x_2^2 = R^2 \text{ and } x_3=C\}$, is a circle, not a cylinder. If instead of $x_3=C$ you had $x_3$ in a specified interval, then it would be a cylinder. $\qquad$ $\endgroup$ – Michael Hardy Aug 11 '16 at 12:27
  • $\begingroup$ you're right - I corrected it in an edit! $\endgroup$ – don-joe Aug 11 '16 at 12:46
  • $\begingroup$ So the problem is that you would like any smooth path on the manifold also to be differentiable as a path in R^3, right? $\endgroup$ – ThePunisher Aug 11 '16 at 13:19
  • $\begingroup$ If you want it to be a differentiable manifold, then yes... to be exact take the definition of a differentiable manifold: you have to cover it with differentiable maps. however there is no differentiable map for a kink. I guess the transformation you are thinking of is that of a homeomorphism. This kind of transformation can flatten out kinks. You can say: A cylinder and a sphere are homeomorphic. $\endgroup$ – don-joe Aug 11 '16 at 13:39

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