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Let $S \subset \mathbb{R}^3 $ be a smooth surface, and let $S^"$ be the unit sphere, and let $n: S \to S^2$ be a given Gauss map.

I want to prove that the Gaussian curvature $K(p)$ at a point $p \in S$ is the limit of the ratio of areas we get when a neighbourhood $U$ shrinks around $p$ : $$ K(p) = \lim_{U \to p}\frac{\text{Area}(n(U))}{\text{Area}(U)} $$

I have shown already that $$\lim_{U \to p} \frac{\text{Area}(n(U))}{\text{Area}(U)} \le K(p) $$ letting $U = F(V)$ for some local parameterization $F$ from $V \subset \mathbb{R}^2$ to $U \subset S$, and using my definition: $$ \text{Area}(U) := \int_V ||\partial_x F \times \partial_y F || dx dy. $$ and the triangle inequality.

Is a lower bound obvious? or do we need to use the (reverse) triangle inequality or something?

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    $\begingroup$ As it stands, this isn't right, as Gaussian curvature could be negative. So it's really "signed area." The result follows from the meaning of the determinant as the factor by which a linear map scales area. And the Gaussian curvature $K(p)$ is defined to be the determinant of the "shape operator" $-dn_p\colon T_pS\to T_pS$. $\endgroup$ Mar 20 '15 at 14:44
  • $\begingroup$ Could you please say how you define Gaussian curvature? The desired equality should follow more or less immediately from the change of variables theorem, which says the "area distortion factor" is the determinant of the differential of the Gauss map. $\endgroup$ Mar 20 '15 at 14:44
  • $\begingroup$ I define Gaussian Curvature to be the determinant of the shape operator. Can you show me in an answer how it follows immediately? I can kinda see what you mean actually $\endgroup$
    – JC574
    Mar 20 '15 at 14:58
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    $\begingroup$ @JC574: $\newcommand{\Reals}{\mathbf{R}}$Let $U$ be a bounded open set, and $\phi:U \to \Reals^{2}$ a smooth map, written in coordinates as $\phi(x_{1}, x_{2}) = (y_{1}, y_{2})$. One relevant version of the change of variables theorem is $$ \operatorname{signed area}\bigl(\phi(U)\bigr) = \int_{U} \phi^{*}(dy_{1} \wedge dy_{2}) = \int_{U} (\det D\phi)\, dx_{1} \wedge dx_{2}. $$ Does that give you enough to finish the proof? $\endgroup$ Mar 21 '15 at 14:06
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Fist of all letme recast your original formulation: \begin{equation} |K(p)| = \lim_{U \to p}\frac{\text{Area}(n(U))}{\text{Area}(U)} \end{equation} I added an absolute value since the ratio of areas is non negative. Now let $X:V \rightarrow \mathbb{R}^{3}$ be a regular parametrization of a neighborhood of $p=X(u_0,v_0)$, where V is an open subset of $\mathbb{R}^2$ . Since X(V) is an open set, $U_ε \subset X(V)$ for all ε. Since we assumed that K(p) is non null, when considering the continuity of the Gaussian curvature function $K:V \rightarrow \mathbb{R}$, we can conclude that the Gaussian curvature does not change sign for ε chosen small enough. We can define the preimage $V_{\epsilon}=X^{-1}(U_{\epsilon})$. Since X is a regular parametrization, $(u_0,v_0)$ is the unique point in all $V_{\epsilon}$ for all positive $\epsilon$.Then by using the formula of the surface are, it is possible to write: \begin{equation} Area(U_{\epsilon})=\int \int_{V_{\epsilon}} ||X_u \times X_v|| du dv \end{equation} In a similar way on the unit sphere we have \begin{equation} Area(n(U_{\epsilon}))=\int \int_{V_{\epsilon}} ||N_u \times N_v|| du dv = \int \int_{V_{\epsilon}} |K(u,v)| ||X_u \times X_v|| du dv \end{equation} By using the Mean Theorem or double integrals we can write \begin{equation} \int \int_{V_{\epsilon}}||X_u \times X_v|| du dv = ||X_u(u_{\epsilon},v_{\epsilon}) \times X_v(u_{\epsilon},v_{\epsilon})|| \end{equation} And \begin{equation} \int \int_{V_{\epsilon}}||N_u \times N_v|| du dv = |K(u^{'}_{\epsilon},v^{'}_{\epsilon})|||X_u(u^{'}_{\epsilon},v^{'}_{\epsilon}) \times X_v(u^{'}_{\epsilon},v^{'}_{\epsilon})|| \end{equation} Thus, since when $\epsilon \rightarrow 0$ both $(u_{\epsilon},v_{\epsilon})$ and $(u^{'}_{\epsilon},v^{'}_{\epsilon})$ tend to $(u_0, v_0)$ we have: \begin{equation} \lim_{U \to p} \frac{\text{Area}(n(U))}{\text{Area}(U)} = \lim_{\epsilon \to 0} \frac{\text{Area}(n(U_{epsilon}))}{\text{Area}(U_{epsilon})}= \lim_{\epsilon \to 0} \frac{|K(u_{0},v_{0})|||X_u(u_{0},v_{0}) \times X_v(u_{0},v_{0})||}{||X_u(u_{0},v_{0}) \times X_v(u_{0},v_{0})||}=K(p) \end{equation}

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  • $\begingroup$ Could you explain how you got the area for A’. You’re taking the area over V_epsilon? What if there’s overlap? $\endgroup$ Apr 11 '20 at 18:28
  • $\begingroup$ What do you mean by A'? $\endgroup$
    – Upax
    Apr 12 '20 at 10:33
  • $\begingroup$ Apologies, I was using notation from another proof. Let me rephrase. You wrote $\int \int V_\epsilon |N_u \times N_v | dudv = A(n(U_\epsilon))$. Can you explain how you got that? Why does integrating over $ V_\epsilon $ give you the surface area? $\endgroup$ Apr 12 '20 at 23:31

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