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I am looking at the proof for this Lemma:

If $S \subset \mathbb{R}^3$ is a regular, compact orientable surface, then $S$ has an elliptic point.

The proof concludes with stating that since $II_p$ has a fixed sign, then that implies that the Gaussian Curvature of $p$ on $S$ is positive.

I am having trouble seeing why this is so. I know that both the 2nd fundamental form and Gaussian curvature involves $dN_p$, but one is an Inner product and the other is determinant. I also know that the prinicpal curvatures are the eigenvalues of $dN_p$, but I can't see how they are related more explicitly.

Would really appreciate some help is developing a better understanding of this. Thanks!

JJJ
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2 Answers2

1

Gaussian curvature is $$K(p):={\rm det}\ dN_p $$

If $x(u,v)$ is a parametrization, and if $$ N_u= dN x_u ,\ N_v=dN x_v $$ then we have first and second fundament form : $$ (x_u,x_u)=E,\ (x_v,x_v)=G,\ (x_u,x_v)=F $$ $$ (N_u ,x_u)=e,\ (N_v,x_v)=g,\ (N_u,x_v)= f $$ and $$ {\rm det} dN =\frac{eg-f^2}{EG-F^2} $$

Here recall a normal curvature : $$k_n(p)= -(dN c'(0),c'(0)) $$ for a curve on $S$ passing through $p$. In further $dN$ is self adjoint so that there exist eigenvalues $dN v_i=\lambda_i v_i $. That is, $-\lambda_i$ are normal curvatures. That is sign of $K$ is equal to $\lambda_1\lambda_2$.

If $S$ is a compact then we choose some inner point $o$ which is not in $S$. So the farthest point $x\in S$ from $o$ has normal curvatures of same sign. Hence we are done.

HK Lee
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0

Only for elliptic points the sign is positive and for saddle/hyperbolic points it is negative.

By Euler relation in the fiber tangent bundle normal curvature sign passes from positive via zero to negative if anti-clastic type $ k_1k_2< 0 $ of surface and for syn-clastic $ k_1k_2> 0 $ both are on the positive side.

$$ k_n = k_1 \cos^{2}\psi + k_2 \cos^{2}\psi $$

The second fundamental form determinant can be expressed in terms of the first fundamental form coefficients & derivatives thereby making scalar Gauss curvature entirely isometric mapping independent, demonstrated in the proof of Gauss Theorema Egregium.

Narasimham
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