Let $S=\{(x,y,z)\in \mathbf R^3 | x^4+y^4+z^4=1 \}$ .
To compute the Gaussian curvature $k$ of $S$, I tried an elementary method to find $dN_p$. Let $\alpha (t) = (x(t),y(t),z(t))$ be an parametried curve on $S$. Since $4x^3x'+4y^3y'+4z^3z'=0$, we obtain the normal map $N(t)=\dfrac{1}{n(t)}(x^3(t),y^3(t),z^3(t))$ where $n(t)$ is the absolute value of the vector $(x^3(t),y^3(t),z^3(t))$. Then we can find the explicit form of $dN_P$ from the relation $dN_p(\alpha ' (0))=N'(0)$.
However $N'(t)$ is much complicated by $n(t)$. So I doubt whether my approach is right. Is there an easy method of computing $k$?
My ultimate goal is to compute $\displaystyle \int_S k$. Should I try another approach?
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$\begingroup$ There is a way to calculate directly from level surface I think, see near page 210 in archive.org/details/ElementaryDifferentialGeometry Otherwise, we'll need to parametrize $S$ and go from there... $\endgroup$– James S. CookCommented Jun 25, 2015 at 23:26
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$\begingroup$ alternatively, and probably more to the point, there is a fairly nice formula for calculating curvature of $F(x,y,z)=0$ in en.wikipedia.org/wiki/Gaussian_curvature $\endgroup$– James S. CookCommented Jun 26, 2015 at 4:19
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2 Answers
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Using the nice formula at wikipedia. I calculate: $$ K = \frac{9x^2y^2z^2}{x^6+y^6+z^6}. $$
answered Jun 26, 2015 at 4:30
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$\begingroup$ Glad to help, but, do keep in mind, Leonardo's answer is a standard technique. It depends what you read, some folks use the classical formulas of Gauss with the $L,M,N$ etc.... but, the shape operator approach is very nice. That said, the unmotivated wiki formula is a convenient dark magic which yielded this answer with not too much effort. $\endgroup$ Commented Jun 26, 2015 at 4:48
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You can use the shape operator. You know that $U = \frac{1}{4\sqrt{x^6 + y^6 + z^6}}(4x^3,4y^3,4z^3)$ is the unit normal vector to this object. Now the shape operator is defined by $$S_p(v) = - \nabla_vU.$$ The gaussian curvature is defined by $$k_g := \det S_p.$$
answered Jun 25, 2015 at 23:24
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$\begingroup$ My book does not treat the shape operator but it seems very useful. So I'd like to study it. Thank you for your answer. $\endgroup$– JeongCommented Jun 26, 2015 at 4:36
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$\begingroup$ The shape operator ? I knew it as the operator of Weingarten! It is good to know all the names given to this really important object. $\endgroup$– NicolasCommented Jun 26, 2015 at 8:36
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$\begingroup$ Can you give me some hint for computing covariant derivative of U? I don't know how to compute it.. $\endgroup$ Commented May 10, 2016 at 3:21
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$\begingroup$ @quicksilver, of course! Let $U = U^ie_i$ (Here I am omitting the sum). Then $\nabla_vU = \nabla_v[U^ie_i] = v[U^i]e_i,$ where $v[U^i]$ means the directional derivative of the function $U^i$ on the $v$ direction. I hope that is is cleaner. Any questions comment. $\endgroup$ Commented May 10, 2016 at 17:11