Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

I need to prove the following formula for Gaussian curvature $K$ of an open subset $V$ of a surface $S$ (and then use it to find the Gaussian curvature of an ellipsoid):

$$ K=\frac{\big<D(fN)(v_1) \times D(fN)(v_2),fN\big>}{f^3} $$

where $N$ is the Gauss map, $f:V\rightarrow \mathbb{R}$ is a smooth nowhere-vanishing function, and $v_1$, $v_2$ are smooth tangent vector fields on V such that $(v_1,v_2,N)$ is an orthonormal right-handed basis.

Can anybody find any holes in the following argument?

RHS $=\Big<\Big([DN+\frac{Df}{f}N](v_1) \times [DN+\frac{Df}{f}N](v_2)\Big),N\Big> = \big<DN(v_1) \times DN(v_2),N\big> = \big<DN(v_1)\times DN(v_2),v_1\times v_2\big> = \det(DN) = K$

Then for the ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$, use $f=\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}$, $N=\frac{1}{f}\Big(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2}\Big)$, and $v_1$, $v_2$ be the normalised versions of $\Big(\frac{y}{b^2},\frac{-x}{a^2},0\Big)$ and $\Big(\frac{xz}{a^2c^2},\frac{yz}{b^2c^2},-\frac{x^2}{a^2}-\frac{y^2}{b^2}\Big)$ to get $K=\frac{1}{a^2b^2c^2f^4}$.

Many thanks for any help with this!

share|cite|improve this question
Is $f$ arbitrary? If it is, then that surely doesn't bode well for your formula for $K$. – treble Jan 1 '13 at 20:23
Yes, in the first part (not with the ellipsoid) $f$ is an arbitrary smooth nowhere-vanishing function. But if my working is correct, it's quite easy to cancel out $f$ in the formula for $K$. – Harry Macpherson Jan 1 '13 at 20:39
a clean method is to consider the Shape Operator which is the Jacobian, $JN$ of the Gauss' map. The Gauss' curvature is the determinant of $JN$ viewed as $JN:T_p\Sigma\to T_{N(p)}S^2$ – janmarqz Jan 10 '14 at 2:38

Your Answer


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

Browse other questions tagged or ask your own question.