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Since the plane and the cylinder have zero Gaussian curvature, I'm wondering, is there an "intrinsic" way of telling one from the other?

By "intrinsic" here I loosely mean a property that can be calculated and/or deduced by inhabitants of the manifold itself, without "seeing" it from a higher dimensional space.

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There are paths of finite length and fixed direction on the cylinder that return a traveler to their starting point. – RecklessReckoner Mar 24 at 18:15
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They are topologically different. But locally they are the same thing. – Masacroso Mar 24 at 18:15
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Every loop in the plane can be shrunk to a point, not so with the cylinder. Topologically, the plane is a punctured sphere, while the cylinder is a twice-punctured sphere. – MPW Mar 24 at 18:17
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"Practically" speaking, if the radius of the cylinder is very large, there may not be any empirical way to compare models that would make the two situations distinguishable. The inhabitants may not live long enough to ever travel a "closed loop", and even differences in quantum field behavior (such as Eric Towers suggests) may become difficult to detect. – RecklessReckoner Mar 24 at 18:25
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In a plane you cannot draw the complete graph with 5 vertices and 10 edges without having some edges cross each other at non-vertex points, but you can do it on a cylinder. – user254665 Mar 24 at 22:29
up vote 31 down vote accepted

Locally there is no difference, but globally there is a difference.

Pick a point on the cylinder; look at a small neighborhood of that point. One can unroll the neighborhood and lay it flat in a plane without stretching the surface, so all distances in the plane are the same as the corresponding intrinsic distances on the cylinder. So you can't tell the difference.

However, if you go far enough in a particular direction on the cylinder, you'll return to where you started. That doesn't happen in the plane.

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Assuming the cylinder has finite circumference, no? Though then you're stretching the definition of a cylinder. – Nate Diamond Mar 24 at 23:31
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@NateDiamond : I was assuming a right circular cylinder as the term is used in traditional Euclidean geometry. $\qquad$ – Michael Hardy Mar 24 at 23:44
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@NateDiamond can you provide a meaningful difference between a plane and the "curved" surface of a cylinder-of-infinite-radius? – scubbo Mar 25 at 21:29
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@NateDiamond I don't think so. My point is that (the surface of) a cylinder of infinite radius is a plane. The original question probably isn't asking about ways to tell the difference between them, because there (by definition) is no such difference. Thus any answers to the question must be discussing cylinders of finite radius. So the assumption that "the cylinder has finite circumference" is implicit. To put it another way - you say "assuming the cylinder has finite circumference, no?". I claim that, taking the inverse assumption, the original question is meaningless. – scubbo Mar 25 at 23:20
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@Nate: There's no such thing as "infinite radius". Radius is defined as the distance from all points to something else (in the case of a circle, a point; in the case of a rt. circ. cylinder, a line) - distances cannot be infinite. – Deusovi Mar 26 at 3:19

Algebraic topology is the answer to such questions. In short, the first homotopy groups of these spaces are different. Every closed loop in the plane can be continuously contracted to a point, but not so for some loops in the cylinder.

See my answer here for what it might be like to live in a cylindrical space.

The intrinsic property that the plane has that the cylinder does not is simple connectedness.

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I thought the question was about metric properties and not just topological properties. Do you disagree with that? $\qquad$ – Michael Hardy Mar 24 at 19:05
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@MichaelHardy The OP mentions curvature, but does not make an explicit request, merely an intrinsic reason why the two are different. The fundamental group is pretty intrinsic. – Alex S Mar 24 at 21:01
    
I associate the word "intrinsic" with the "intrinsic metric", according to which the distance between two points in the manifold is the length of the shortest path connecting them within the manifold (as opposed to distance along a chord in the ambient space). $\qquad$ – Michael Hardy Mar 24 at 22:55
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I see. I have heard the term intrinsic referring to properties of a space which are independent of any embedding use to place this space into some ambient space. – Alex S Mar 24 at 23:01
    
That's just what I had in mind, but I thought it meant metric properties of the embedded space. $\qquad$ – Michael Hardy Mar 24 at 23:04

A resident of cylinderland would find that at each point, there are two distinguished directions to shine a laser pointer where the pointer (eventually) illuminates itself. Equivalently, along this distinguished axis, this universe appears to be periodic.

Quantum fields on cylinderland would have discrete excitations associated with this compact dimension.

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Hint:

not the curvature, but the existence of closed paths (loops) that cannot be reduced to a point by a continuous transformation.

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very good point. – marco trevi Mar 24 at 18:19
    
Although, in the case of a parabolic-cross section cylinder this would not apply – marco trevi Mar 24 at 18:20
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@marcotrevi Do you mean like $\{ (x,y,z) \mid y=x^2 \}$. That surface is diffeomorphic to $\mathbb{R}^2$ and the diffeomorphism can be chosen as an isometric isomorphism ((global) isometry). So they are indistinguishable intrinsically, both locally and globally. – Jeppe Stig Nielsen Mar 25 at 16:06

$\pi :{\bf R}^2\rightarrow C:={\bf R}^2/{\bf Z}$ is a universal covering where $C$ is cylinder. Since $C$ is quotient of ${\bf Z}$-isometric action, so $\pi$ is local isometry. Hence locally same but different fundamental group

(In further (1) they have different volume growth $f(R):={\rm vol}\ B(x_0,R)$ and (2) In ${\bf R}^2$ any two points in any ball $B(x_0,R)$ can be connected by line segment in it but if we have suitable $B(x_0,R)$ in cylinder, this can not happen.)

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If you define lengths (and angles of curved surface that can be computed out of it) then there is no difference between a cone, a cylinder or other developable of zero $K$ Gauss double curvature (isometric equivalents). $K=\frac {LN-M^2}{EG-F^2}$, comprises of coefficients of first/second fundamental forms in Classical surface theory is entirely determinable by Gauss Egregium theorem from the first form alone. It remains the same in Bendings of surface without tearing apart points in the fabric / membrane of space-time

FlatLanders in relativity theory (only parts I remember from book Differential Geometry & Relativity Theory: An Introduction by R.L. Faber 1983 CRC Press) can get information about surrounding space even if they are third (normal) dimension "blind" , or unable to compute $( L,M,N)$ directly. Without using telescope, gyroscope GPS or other external inputs how that is possible I too wonder, would like to know.

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Whenever one considers vectors on a surface S based at the point p, one consider this vectors as elements of tangent space. As a matter of fact the tangent space to the surface S at p exists if $X^{-1}(p) = q$, where $X$ is a parametrization of S, the differential $dX_q$ exists, and $dX_q$ has maximal rank. For a cylinder a natural direct product on the tangent space is the same that for a plane. This is because a plane and a cylinder shares many properties and locallyyou cannot distinguish them. As a matter of fact they have the same first fundamental form. Of course, there is no doubt that a plane and a cylinder are different surfaces. When looking on how the normal vector evolves on a cylinder in comparison to how it evolves on a plane you can understand the difference. Cylinder and plane have different second fundamental forms.

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