1

I am looking for the most straightforward way of showing that the Gaussian curvature of the unit sphere is 1. How could I go about this?

I found a way for this which involves showing that $${R^{\theta}}_{\phi} = \sin(\theta) \; e^{\theta} \wedge e^{\phi}$$ however how do I relate the Ricci tensor to the Gaussian curvature?

John
  • 375

1 Answers1

2

The most straightforward way is using that $K(p) = \det(-{\rm d}N_p)$, where $N$ is a unit normal field to $\Bbb S^2$. For the unit sphere, the position vector $N(p) = p$ is normal to the sphere, so the result follows from noting that $\det({\rm Id}_2)=1$. Which allows you to compute the curvature like this is Gauss' Theorema Egregium.

Ivo Terek
  • 77,665
  • $N(p) = (\sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta)$, then what is $\mathrm{d}N$? – John Apr 29 '19 at 20:59
  • You do not need to use coordinates for this. And you have wrong notation/abuse of notation there, you would write $N(X(\theta,\phi))$ instead of $N(p)$ there. But if $N: \Bbb S^2 \to \Bbb S^2$ is the identity map, so is ${\rm d}N_p: T_p\Bbb S^2\to T_p\Bbb S^2$, for all $p\in \Bbb S^2$. – Ivo Terek Apr 29 '19 at 21:02
  • Hmm, thanks. But I don't understand this well enough. – John Apr 29 '19 at 21:06
  • Do you know what is the differential of a smooth map between manifolds? – Ivo Terek Apr 29 '19 at 21:08
  • I do, or at least I think I do. But I was hoping to figure the question from more basic principles. – John Apr 29 '19 at 21:11
  • Definitely computing the differential of a smooth function is more basic than actually computing components of the Riemann tensor. Weird approach. But ok, then use that for surfaces, $s = 2K$, where $s$ is the scalar curvature. – Ivo Terek Apr 29 '19 at 21:13
  • And to respect the index balance required by Einstein's convention, probably you should write $R_{\theta\phi}$ instead of $R^\theta_{\phantom{\theta}\phi}$. Moreover, this $R_{\theta\phi}$ should be a function, not a $2$-form. – Ivo Terek Apr 29 '19 at 21:15