In the book I'm reading ("Differential Geometry Curves-Surfaces-Manifolds by Wolfgang Kuhnel") two definitions of prinicpal curvatures directions are presented:

  1. The extramum values of $II(X,X)$ under the constraint $I(X,X)=1$
  2. The eigenvectors of the shape operator $L$

I'm having some difficulty proving this equivalence. It was also mentioned that the gradient of $II(X,X)$ is simply $LX$ ($L$ operating on $X$). So I thought if I prove that, it follows that (1) holds iff the gradient of $II(X,X)$ is proportional to the gradient of $I(X,X)$ (which is clearly $X$) iff $LX$ and $X$ are proportional iff $X$ is an eigenvector.

  • How do I compute the gradient of $II(X,X)$ and show it is equal to $LX$. It seems like it is a simple task that I'm missing some idea or technique.
  • I'll be glad to hear of any interesting insights regarding prinicpal curvatures in order to have a better intuition.
  • $\begingroup$ What's $LX$?${}{}$ $\endgroup$ – user99914 May 25 '15 at 8:20
  • $\begingroup$ The shape operator L operating on the vector X. I'll clarify this $\endgroup$ – Itamar Vigi May 25 '15 at 8:21
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    $\begingroup$ Do you know that $II(X, Y) = I(LX, Y)$? (might be up to a constant depending on the definition) $\endgroup$ – user99914 May 25 '15 at 8:44
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    $\begingroup$ Then the first two equivalence has nothing to do with differential geometry. It's result in linear algebra. $\endgroup$ – user99914 May 25 '15 at 8:46
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    $\begingroup$ math.stackexchange.com/questions/1199852/… $\endgroup$ – user99914 May 25 '15 at 8:48

I suppose you could basically identify $T_uf$ as $\mathbb{R}^2$, then you could carry out the calculation in the usual way.
To be more specific about the gradient w.r.t second fundamental form:
\begin{equation} \frac {\partial}{\partial x_1}<LX,X>=<L'(X)\frac {\partial X}{\partial x_1},X> + <LX,\frac {\partial X}{\partial x_1}>=2<LX,\frac {\partial X}{\partial x_1}>=2<LX,(1,0)^t> \end{equation} which is the first coordinate of $LX$. Similarly, $$ \frac {\partial}{\partial x_2}<LX,X>=2<LX,(0,1)^t> $$ which is the second coordinate of $LX$.
Combine this two equations, you can get the result.

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  • $\begingroup$ What is your definition of second fundamental form here? $\endgroup$ – user99914 Oct 12 '15 at 14:27
  • $\begingroup$ the second fundamental form is defined to be the dot product between LX and X on the tangent surface $\endgroup$ – Xiaoyu Xie Oct 13 '15 at 1:32
  • $\begingroup$ and L is defined to be $D\nu$ composite $(Df)^{-1}$ which is known as Weingarten map, here $\nu$ is the Gauss map along $f$ $\endgroup$ – Xiaoyu Xie Oct 13 '15 at 1:36

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