Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:V \to S^{2} \subset \mathbb{R}^{3}$ be a surface whose image is an open subset of the usual unit sphere $S^2$. Prove that there is NOT a change of variables $\phi: U \to V$ such that in the new variables the components of the first fundamental form are $g_{ij}=\delta_{ij}$ where $g_{ij}= \langle f_{v^{1}} ,f_{v^{2}} \rangle$ and $\delta _{ij}$ is the Kronecker delta.

Note: There is a result states that the Gauss curvature of $f$ at $v$ is zero if and only if there exists local coordinates in which the matrix rep of the first fundamental form of $f$ at $v$ is the identity matrix under the coordinates (for instance one may take the Fermi local coordinates).

Based on this result, is it true that if there is a global change of variables which makes $g_{ij}=\delta_{ij}$ hold then this change of varibles can also be considered as locally at each point on the surface, thus by above result the Gauss curvature of each point on the surface is zero?

share|cite|improve this question
Short answer: yes. Long answer: yes indeed. Even longer answer: form the Gaussian curvature formula, you can see that if locally both $g_{ij}=0$ and $\partial g_{ij}/\partial x_k=0$ then the local curvature vanishes. – yohBS Sep 23 '12 at 12:41
up vote 7 down vote accepted

Writing $u,v$ for parameters and $E,F,G$ for the coefficients of the fundamental form, we see on Wikipedia that when $F\equiv 0$, $$ K=-\frac{1}{2\sqrt{EG}} \left(\frac{\partial }{\partial u}\frac{G_u}{\sqrt{EG}}+\frac{\partial }{\partial v}\frac{G_v}{\sqrt{EG}} \right) $$ Which is kind of old-fashioned notation (the subscripts on the $G$s denote partial differentiation), but clear enough to conclude $K\equiv 0$; hence this cannot be the sphere.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.