# Gaussian Curvature

I am able to show that if a curve lies in a plane then it's curvature at a point $p$ is $$\kappa=\lim_{\mu\to 0}\frac{\sigma}{\mu}$$ where $\mu$ is the length of a segment of the curve containing p, and $\sigma$ is the length of the arc on $S^1$ formed by translating the velocity vectors of this segment to the origin.

Analogously, it can be shown that the Gaussian curvature at a point p on a surface, is given by $$\lim_{A\to 0}\frac{A^\prime}{A}$$ where $A$ is the area of a region containing $p$ and $A^\prime$ the area of the corresponding region on $S^2$ under the Gauss map. (Apparently this is how Gauss defined the curvature)

We also know that the Gaussian curvature is the product of the principal curvatures.

But the principal curvatures are the curvatures of plane curves by definition (curvatures of normal sections). Hence the principal curvatures are given by the first limit above.

So I guess this means that $$\lim_{A\to 0}\frac{A^\prime}{A}=\left(\lim_{\mu_1\to 0}\frac{\sigma_1}{\mu_1}\right)\left(\lim_{\mu_2\to 0}\frac{\sigma_2}{\mu_2}\right)$$ where $\mu_1,\mu_2,\sigma_1,\sigma_2$ correspond to the two normal sections.

How does this happen? Does this mean that the area $A^\prime$ is equal to the product of the two arc lengths $\sigma_1\sigma_2$? It's definitely not true if the region on the sphere is a cap!

• can you give sources? even the given formula's look incorrect to me (but i am only a beginner in this subject) – Willemien Nov 22 '14 at 10:19
• I learned all of this from Do Carmo's Differential Geometry of Curves and Surfaces. The formula for curves follows from exercise 3 on pages 22 - 23. The formula for the Gaussian curvature is Proposition 2 on page 167. He also gives the formula for curves on this page. – JonHerman Nov 22 '14 at 19:20
• I had a look at DoComo's book, I think you made a mistake: on page 167 $\sigma$ stands for the angle in radians, not for the arc length of the curve (the book says" the arc length of its image in the indicatrix of tangents") and the "indicatrix of tangents" is an unit circle so has a fixed radius of 1 see en.wikipedia.org/wiki/Tangent_indicatrix . The formula you gave has a fixed limit of 1. It is a variation of $\lim_{\sigma\to 0}\frac{\sigma}{\sin \sigma }$ GOOD LUCK – Willemien Nov 23 '14 at 9:08
• Thank you for looking into this. On 167 he is referring back to the exercise on page 23. In this exercise you show that, supposing $p=\gamma(s)$, then the curvature at $p$ is $\kappa=\lim_{\widetilde s\to s}\frac{\theta(\widetilde s)-\theta(s)}{\widetilde s-s}$. But $\widetilde s-s$ is the arc length of a segment containing $p$, and $\theta(\widetilde s)-\theta(s)$ is the angle between the two normals. On the unit circle, angle from $x$-axis is equal to arc length. – JonHerman Nov 23 '14 at 20:25
• I don't have doCarmo here with me, but I believe in your last comment you're referring to the angle $\theta$ that a plane curve (i.e., its unit tangent vector) makes with a given direction (traditionally, the $x$-axis). – Ted Shifrin Nov 23 '14 at 21:52

We're not saying that the area of a spherical cap, is per se, any such product. We're talking about an infinitesimal piece, say coming from a bit of arclength $\mu_1=\Delta\phi$ on the longitude circle and $\mu_2=\sin\phi\Delta\theta$ on a latitude circle. These get multiplied, respectively, by the principal curvatures, so that $\sigma_1 = k_1\mu_1$ and $\sigma_2=k_2\mu_2$. No contradictions that I can see, but ask further.
• Thank you for your answer and comment. I guess what is bothering me is that I fail to see what is special about the principal directions. Let $\tau_1$ and $\tau_2$ denote the normal sections, curves with initial velocities principal. Why does the formula above only hold for these normal sections? That is, why can't we take two different plane curves with principal initial velocities? Is there something special about how the normal vectors to $T_qS$ change along $\tau_1$ and $\tau_2$? – JonHerman Nov 24 '14 at 15:53
• the curvature of a plane =0 while the formula says that the curvature of a plane =1 ($A' = A$) so how can it be correct? – Willemien Nov 25 '14 at 10:09
• No, @Willemien, the Gauss map of a plane is constant, so $A'=0$. – Ted Shifrin Nov 25 '14 at 12:59