MHB What is the relationship between Gaussian curvature and volume forms?

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Here's this week's problem!

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Problem. Let $X$ be a surface imbedded in $\Bbb R^3$. Show that if $K$ is the Gaussian curvature of $X$, then $$K(x) = \lim_{V\downarrow x} \frac{\text{vol}_{\Bbb S^2}(N(V))}{\text{vol}_X(V)}$$ where $N : X\to \Bbb S^2$ is the Gauss map, $\text{vol}_X$ and $\text{vol}_{\Bbb S^2}$ are volume forms on $X$ and $\Bbb S^2$, respectively, and the limit is taken over all neighborhoods $V$ of $x$ decreasing to $x$.

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No one answered this week's problem. You can find my solution below.

Spoiler

If $r(u,v)$ is a parametrization of $X$, then $dN(r_u) = N_u$ and $dN(r_v) = N_v$. Thus

$$\|dN(r_u) \times dN(r_v)\| = |\operatorname{det}(dN)| \|N_u \times N_v\| = K\|N_u \times N_v\|.$$

Since $dN(r_u) \times dN(r_v) = K(N_u \times N_v)$, we have $N^*(\sigma_{\Bbb S^2}) = K\operatorname{vol}_X$. Therefore,

$$\operatorname{Area}(N(V)) = \iint_{N(V)} \sigma_{\Bbb S^2} = \iint_V N^*(\sigma_{\Bbb S^2}) = \iint_V K\operatorname{vol}_X.$$ Therefore

$$K(x) = \lim_{V\downarrow x} \frac{\operatorname{Area}(N(V))}{\operatorname{vol}_X(V)}.$$