Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to read a survey paper on the Willmore conjecture and I'm missing a lot of basic knowledge. In particular, let $u: \mathcal{M} \rightarrow S^3 \rightarrow \mathbb{R}^4$ be a smooth immersion of a compact orientable two dimensional surface into the standard 3-sphere, and let $\mathcal{M}$ take the metric induced by the ambient space. Writing the principal curvatures as $k_1$ and $k_2$, we have the Gaussian curvature given by $1+k_1k_2$.

I don't understand where the 1 in $K = 1+k_1k_2$ is coming from. The metric on $\mathbb{R}^4$ is the standard $g_{ij} = \delta_{ij}$, and restricting it to $S^3$ yields the round metric. $S^3$ is our ambient space, and restricting to $u$ gives us our induced metric.

I can think of two ways of showing this. The first would simply be to do everything out in local coordinates: for any $p \in \mathcal{M}$, we have a chart $\varphi: U \rightarrow V \subset \mathbb{R}^2$. Writing $u(p) = u(\varphi^{-1}(x^1,x^2)) = (y^1,y^2,y^3,y^4)$ and then projecting stereographically $\sigma: S^3 \rightarrow \mathbb{R}^3$, $\sigma(\vec{y}) = (\frac{y^1}{1-y^4},\dots,\frac{y^3}{1-y^4})$, we could recover $K$ via typical computations.

Alternatively, since $2K = \mathcal{R}$, the Ricci scalar, we could compute the curvature tensor and find it that way (unless there is some shortcut? I don't have much experience with these diffgeo objects).

I'm wondering if either these approaches would get me what I want, or if there is a naive reason for where the $1$ is coming from. Any help is appreciated. The relevant stuff is on p. 365 of the document, or 5th page from the start, section 2: the S^3 framework.

share|improve this question

1 Answer 1

For $3$- manifolds of constant curvature $K_0$ (your $S^3$ for which $K_0$ equals $1$) the Gauss curvature of a hypersurface is $ K=k_1*k_2+K_0 $, with $k_i$ being the principal curvatures - see, e.g., Volume 4 (chapter 7) of Spivak's 'Comprehensive Introduction to Differential Geometry', which contains a comprehensive (! - nomen est omen) discussion of the relevant equations, also including a discussion about higher dimensional manifolds. See in particular the proof of propostion 24. Alternatively Theorem 5.5 in Gallot, Hulin, Lafontaine, 'Riemannian Geometry'. I'd expect to find similar results in many books of differential geometry. Search for Gauss formulas and Gauss curvature.

(And it's 'principal curvature', not 'principle curvature').

share|improve this answer
    
Thanks for such precise references! I googled around a bit but couldn't find this, and couldn't find it in the few books on diffgeo I own. Nicely spotted on the principle/al, btw, I so often don't think when I write :) –  snarski Jun 22 '12 at 23:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.