We have the following result from Do Carmo book of differential Geometry: "Let $p$ be a point of a surface $\Sigma$ such that the Gaussian curvature $K(p) \neq 0$, and let $V$ be a connected neighborhood of $p$ where $K$ doesn't change sign. Then $$ K(p)= \lim_{A \rightarrow 0} \frac{A'}{A} ,$$ where $A$ is the area of a region $B \subset V$ containing $p$, $A'$ is the area on the image of $B$ by the Gauss map $N: \Sigma \rightarrow S^2$, and the limit is taken through a sequence of region $B_n$ that converges to $p$, in the sense that any sphere around $p$ contains all $B_n$, for $n$ large enought". How can I prove, using this result, that is $\Sigma$ is a compact surface $$ \int_{\Sigma}|K| \ge 4 \pi \,\,\,? $$

  • $\begingroup$ A couple of hints (whose usefulness may depend on how much "you know (at this stage of the book)" about closed surfaces in space: The $4\pi$ comes from the area of the unit sphere. Show $N$ is surjective (this may require a geometric argument relying on the fact that $\Sigma$ is the boundary of a compact region), then use the change of variables theorem to express the integral of $|K| = |\det DN|$ over $\Sigma$ in terms of the area of $S^{2}$. $\endgroup$ – Andrew D. Hwang Mar 31 '15 at 11:30
  • $\begingroup$ @user86418 I can prove that the Gauss map is surjective, but I don't know how to continue.... $\endgroup$ – ArthurStuart Mar 31 '15 at 11:47
  • $\begingroup$ This question, particularly the comment involving the change of variables theorem, should help finish the proof. :) $\endgroup$ – Andrew D. Hwang Mar 31 '15 at 12:15
  • $\begingroup$ $\int_{\Sigma} |K| \ge \int_{\Sigma}K = \int_{\Sigma} dN_p dA_{\Sigma} = \int_{S^2} dA_{S^2} = 4 \pi $. Is it correct? $\endgroup$ – ArthurStuart Mar 31 '15 at 13:13
  • $\begingroup$ "Is this correct?": That's the general idea, but it's not quite OK as stated. It turns out that the total curvature of a closed surface is $2\pi$ times its Euler characteristic, which can be non-positive. Rather than dropping the absolute value, use the fact that integrating $|\det DN|$ gives the unsigned area of the image of $N$ (i.e., area not counting orientation; if $N$ "folds back on itself", the multiply-covered area gets counted multiple times with a positive sign). $\endgroup$ – Andrew D. Hwang Mar 31 '15 at 13:57

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