Total curvature

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odyssey

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Compute the total curvature of the the unit sphere \(\displaystyle S^2\) in \(\displaystyle \mathbb{R}^3\), that is compute \(\displaystyle \int_{S^2}K(p)dA.\) where \(\displaystyle K(p)\) is the Gauss Curvature.

How can I compute this?

I know if I use stereographic coordinates I can cover \(\displaystyle S^2\) minus a point with one coordinate chart, and therefore do all our computation in this coordinate chart but I don't know how to apply it?
 
The curvature of a sphere of radius R is 1/R^2. So you're integrating a constant which can be factored out. And a surface integral of just 1 is the are of the surface. So 1/R^2*4pi*R^2 = 4pi. Notice the R's cancel, they aren't relevant for the total curvature.

You can also use the Euler characteristic to obtain the result more quickly.
 
Daon - I am not sure I understand. I know that the curvature of a sphere of radius R is just \(\displaystyle 1/R^2.\) But are you saying the total curvature is just \(\displaystyle 4\pi\)?
 
odyssey said:
Daon - I am not sure I understand. I know that the curvature of a sphere of radius R is just \(\displaystyle 1/R^2.\) But are you saying the total curvature is just \(\displaystyle 4\pi\)?
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Yes. You might enjoy the Gauss-Bonnet theorem ;). Even if you take the sphere squish it and stretch it, make any ballon animal you please the total curvature is still 4pi!
 
Ah, that is a neat theorem!

How can I do it without using that theorem? I know the first fundamental form in stereographic coordinates is given by the spherical metric:

\(\displaystyle \frac{4}{(1+u^2+v^2)^2}\pmatrix{ 1 & 0 \\ 0 & 1 \\},\) but I don't know how to apply it to the given total curvature and I don't know what \(\displaystyle dA\) will be?
 
The normal definition of total curvature is a surface integral

\(\displaystyle \int_S K dS\)

Maybe it's your teacher's notation?

I haven't done the fundamental form stuff in a long while, but I think you're overcomplicating it:

Parameterize the sphere in the normal way.

\(\displaystyle \int_{S^3} K dS = \int_{0}^{2\pi} \int_{0}^{\pi} \underbrace{ 1 }_K \ \sin{\phi}\ \underbrace{d\phi\ d\theta}_{dA} = 4\pi\)
 
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If you know that the surface area of a sphere of radius R is \(\displaystyle 4\pi R\) then this is easy. K(p) is the constant 1/R and the integral of a constant over that surface is the constant times the area of the surface: \(\displaystyle (1/R)(4\pi R)= 4\pi\).
 
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