No metric on S^2 having curvature bounded above or below by 0


by Relative0
Tags: bound, genus, manifolds, metric, surface
Relative0
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#1
Feb27-13, 01:46 PM
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So I ran in to a question;

Show that there is no metric on S^2 having curvature bounded above by 0 and no metric on surface of genus g which is bounded below by 0.

honestly I have no idea what is going on here. I know that a Genus is the number of holes in some manifold or the number of cuts performed that will not leave the manifold disconnected. Also a metric, such as the distance - although I don't understand how it is applied here. But most of all, I have no idea what the point is and how it is the bounding even happens.


Note: this is not homework but a question in a list to study for an exam - so I am hoping to understand the intuition and how to even generalize this to something else.

Thanks, any help is much appreciated.

Brian
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fzero
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#2
Feb27-13, 02:18 PM
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You don't mention what sort of course this is, but the Gauss-Bonnet theorem is an important ingredient in the most direct proof of these assertions. The remaining part of a proof would seem to be some analysis of the integral of the mean curvature in order to establish the bound.
Relative0
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#3
Feb27-13, 10:21 PM
P: 11
thanks fzero. This is a course in Differential Geometry. I read through the Gauss-Bonnet Theorem from the Wiki (very interesting) and see how curvature comes in - so does "curvature above bounded by 0" mean something like that it has to have negative curvature - or that it is concave?

Thanks again,

Brian

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#4
Feb27-13, 11:25 PM
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No metric on S^2 having curvature bounded above or below by 0


Quote Quote by Relative0 View Post
thanks fzero. This is a course in Differential Geometry. I read through the Gauss-Bonnet Theorem from the Wiki (very interesting) and see how curvature comes in - so does "curvature above bounded by 0" mean something like that it has to have negative curvature - or that it is concave?

Thanks again,

Brian
"Bounded above" means "[itex]\leq [/itex]" (so a semi-definite bound) and similarly for bounded below. In the question at hand, for most cases, there is a stronger, definite, bound.

Now, I strike the 2nd part of my previous reply. It seems clear that the question is referring to the total curvature, rather than the Gaussian or mean curvature at a point (a golf ball is topologically a sphere, but has negative Gaussian curvature at the edges of the dimples). Then the result follows more or less immediately from Gauss-Bonnet.
lavinia
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#5
Feb28-13, 02:30 PM
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The torus has genus 1 and can be given a metric of curvature zero. So it's curvature is bounded below by 0. For genus greater than 1 the curvature can not be everywhere non-negative.

You can contruct a flat torus in R^4 as the image of the map

(x,y) -> (cos(x),sin(x),cos(y),sin(y))
lavinia
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#6
Mar2-13, 08:57 AM
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Another way to see that a sphere can non have everywhere non-positive Gauss curvature is to to look at the unit length vector field that is tangent to great circles intersecting at the north and south poles and points from north to south.

It is easy to see that the sum of the indices of this vector field is 2.

The vector field is a section of the tangent circle bundle above the entire sphere except for the north and south pole. If one integrates the curvature 2 form in the tangent circle bundle over this section one gets the sum of the indices of the vector field. Since the sum of the indices is 2, the Gauss curvature can not be everywhere non-positive.

Similar arguments apply for surfaces of genus greater than 1 but in these cases the sum of the indices is negative.

A second theorem shows that the sum of the indices of a vector field with isolated singularities is the same for all vector fields and then one can construct a particular vector field whose index sum is easily seen to be the Euler characteristic.

If the Gauss curvature is identically zero, then the horizontal planes tangent to the unit circle bundle form an involutive distribution. This means that they are tangent to a surface. Such a surface is a vector field with no singularities.
lavinia
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#7
Mar3-13, 11:42 AM
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Another method, that does not use the Gauss Bonnet Theorem comes from the theory of characteristic classes. Certain polynomials in the matrix of curvature 2 forms have the same integrals for all connections. For Levi-Civita connections one can rephrase this to say that these polynomials have the same integral independent of the Riemannian metric.

For oriented surfaces, this polynomial is just the Gauss curvature multiplied by the volume element of the surface divided by 2pi.

One can then compute these integrals from a single example because they are the same for all. This idea is the same as the idea mentioned above where the sum of the indices of all vector fields can be computed from a single easy example.

In the case of the sphere of radius one in 3 space, the integral is 2 since the Gauss curvature is everywhere equal to 1 and the surface area of the sphere is 4pi. If the sphere had a metric of non-negative curvature then then the integral could not be positive and so could not equal 2.
lavinia
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#8
Mar5-13, 09:59 AM
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Here is an example that shows the invariance of the integral of the Gauss curvature on a surface.

Imagine a surface in 3 spaces that moves around and changes shape then stops. It flows from is beginning shape to its final shape through a 1 parameter family of surfaces. The result is a solid sheet of surfaces. The Gauss curvature 2 form on the beginning and end surfaces can be extended to a 2 form on the entire sheet. It equals the Gauss curvature 2 form on each intermediate surface. Stokes theorem then tells you that the difference of the integrals of the Gauss curvature on the first and last surface equals the integral of the exterior derivative of the curvature 2 form over the entire sheet. But this form is closed so the integral over the sheet is zero.
Relative0
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#9
Mar21-13, 12:14 AM
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Thanks Lavinia for the in depth description, it very much helps for the intuition.

Brian
lavinia
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#10
Mar21-13, 09:20 AM
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Quote Quote by Relative0 View Post
Thanks Lavinia for the in depth description, it very much helps for the intuition.

Brian
You are welcome.

BTW: there is a whole theory that generalizes the Gauss Bonnet theorem and identifies a collection of integrals of polynomials in the curvature 2 forms. This is the theory of characteristic classes.

It is known that one set of these classes, the Pontryagin classes, also have combinatorial formulas just as the Euler characteristic. But to my knowledge these formulas are hard to find and may still be an area of research.

Interestingly these polynomials also exist for connections on any smooth vector bundle, not just the tangent bundle, and also may not be compatible with a Riemannian metric.


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I was trying to come up with another proof that a sphere cannot have curvature everywhere bounded above by zero. Maybe you can give the approach a try.

On a Riemannian manifold, one can map a tangent plane at any point onto the manifold by the exponential mapping. This map takes a vector in the tangent plane and maps it to the end point of a geodesic segment whose length is the same as the length of the vector.

These geodesics fan out in all directions and create a polar coordinate system centered at the point.

On the sphere, geodesics starting at the north pole follow great circles and meet again at the south pole. One can show that this type of convergence of geodesics can only occur when there is positive Gauss curvature and intuitively one sees that the convergence is forced by a rounding of the surface.

Thus for manifolds with everywhere non positive Gauss curvature, polar geodesics never converge although some of them may cross at different points on the surface. Analytically, this means that the map of the exponential map has no singularities and so is a smooth map of the plane onto the surface without singularities.

the idea is to show that no such map can exist from the plane onto the sphere.


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