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Let M be a surface in $R^3$ with principal curvatures $k_1=0, k_2\neq 0$.

$k_1$ is always zero and $k_2$ is never zero.

Suppose that $E_1, E_2$ are corresponding principal unit vector fields and $E_3$ is the unit normal vector field of M. In other words, $\{E_1, E_2, E_3\}$ constitutes a principal frame field on M.

Let $\alpha(t)$ be a principal curve for $k_1$ on M. How to prove that $\alpha''(t)=0$?

The hint says that $\alpha''=\nabla_{E_1}E_1$.

For a principal curve $\alpha$ of $k_1$ that is not unit speed, does one still have

$\alpha''=\nabla_{E_1}E_1$? Does the hint assume that the principal curve is unit speed?

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  • $\begingroup$ Yes unit speed is assumed here. $\endgroup$ – Xipan Xiao Oct 28 '13 at 0:04
  • $\begingroup$ What is the notation $\nabla_{E_1}$ defined as? $\endgroup$ – Lays Oct 29 '13 at 0:36
  • $\begingroup$ $\nabla_v W$ means the covariant derivative of a vector field W with respect to a vector v. $\endgroup$ – noot Oct 29 '13 at 2:01

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