ArthurStuart
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 Apr14 asked Total curvature of an ovaloid. Apr14 comment Triangulations of surface. I'm sorry, what is a gon? Apr14 asked Asymptotic geodesic on hyperboloid. Apr13 asked Frenet frame and tangent space. Apr7 comment Maximal circles and sphere. $p$ and $v$ are ortogonal, and maximum circle is a great circle. Apr6 asked Maximal circles and sphere. Mar31 comment Characterize the sphere using mean curvature. But the connected sum of two $T^2$ has hyperbolic points... right? (Ok $T^2$ has only a hole) Mar31 comment Characterize the sphere using mean curvature. @HeeKnowLee Only if $\Sigma$ is a sphere... true? Mar31 comment Characterize the sphere using mean curvature. $\int |K|= 2 \pi \chi(T^2)= -4 \pi$ Mar31 comment Characterize the sphere using mean curvature. I'm sorry. If $\Sigma$ is a compact generic surface and $\int_{\Sigma} |K|= 4 \pi$, can I say that $\Sigma$ has no hyperbolic points? Mar31 accepted Characterize the sphere using mean curvature. Mar31 comment Characterize the sphere using mean curvature. Ok thank you! I have a little curiosity. We know that $H^2 \ge K$. If $\int_{\Sigma}K= 4 \pi$ can I deduce that $\Sigma$ has not hyperbolic points? Mar31 comment Characterize the sphere using mean curvature. @aGer Let's suppose that $\int_{\Sigma}H^2= 4 \pi$ but $\Sigma$ not a sphere. We know that $H^2 \ge K$, so $\int_{\Sigma}H^2 \ge \int_{\Sigma}K = 2 \pi \chi(\Sigma)$... Gauss-Bonnet? Mar31 asked Characterize the sphere using mean curvature. Mar31 comment Geometric interpretation of Gaussian curvature. Ok, thank you! The last thing: can I argue something about the elliptic or hyperbolic nature of the points of $\Sigma$? Is it possible to say that $\Sigma$ has an elliptic point? Moreover, is it possible to say that if $\int_{\Sigma}|K|=4 \pi$ than $\Sigma$ has not hyperbolic points? Mar31 comment Geometric interpretation of Gaussian curvature. $4 \pi = \mathcal{A}(S^2) = \int_{S^2} 1 \, d\mathcal{A}_{S^2} = \int_{\Sigma}|Jac(N)(p)| .$ We have by definition $|Jac(N)(p)|= |(dN)_p(e_1) \wedge (dN)_p(e_2)| ,$ where $\{e_1, e_2\}$ is an o.n. basis of $T_p\Sigma$. We choose $e_1$ and $e_2$ as the principal direction $\Sigma$ in $p$. We obtain $$|Jac(N)(p)|=|k_1(p)k_2(p)||e_1 \wedge e_2|= K(p) .$$ So to finish: $$4 \pi = \int_{\Sigma} K \le \int_{\Sigma} |K| .$$ I have a problem with the module also in this case... Mar31 comment Geometric interpretation of Gaussian curvature. $\int_{\Sigma} |K| \ge \int_{\Sigma}K = \int_{\Sigma} dN_p dA_{\Sigma} = \int_{S^2} dA_{S^2} = 4 \pi$. Is it correct? Mar31 comment Geometric interpretation of Gaussian curvature. @user86418 I can prove that the Gauss map is surjective, but I don't know how to continue.... Mar31 asked Geometric interpretation of Gaussian curvature. Mar30 revised Pappus theorem and area of a revolution surface. added 33 characters in body