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Let $\Sigma$ be a convex surface with strictly positive curvature in $\mathbb{R}^3$ and $\Gamma \subset \Sigma$ be a closed simple space curve with parameterization $\gamma : \mathbb{S}^1 \rightarrow \mathbb{R}^3$. Assume that a functional $\phi: \mathbb{S}^1 \times \mathbb{S}^1 \rightarrow \mathbb{R}^+$ defined as $\phi(u_1, u_2) := \Vert \gamma(u_2) - \gamma(u_1) \Vert^2$ has a local minimum at $(u_1, u_2)$. I would like to show that tangent vectors $T(u_1)$ and $T(u_2)$ must then be collinear.

It is easy to prove that the statement holds true when $\Sigma$ is a sphere and that its higher dimensional analogy for curves lying on hyperspheres in spaces of dimension strictly larger that 3 is false.

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    $\begingroup$ I'm not sure I got the statement: Suppose $\Sigma$ contains two flat parallel areas, e.g the unit disk on the x,y plane centred on the origin $O:=(0,0,0)$, and its translate by $(0,0,1),$ centred at $P:=(0,0,1)$. Suppose $\gamma$ has a straight segment passing for the origin, say $\gamma(u_1)=O$ and another straight segment passing for $P$, say $\gamma(u_2)=P$. This gives a local minimum of $\phi$ in $(u_1,u_2)$, with non collinear tangent vectors. $\endgroup$
    – Pietro Majer
    Commented Jun 3, 2019 at 12:48
  • $\begingroup$ @PietroMajer Thank you for the observation. You understand the statement correctly. I added an additional assumption on the surface $\Sigma$ to reflect your comment. $\endgroup$
    – Jiří Minarčík
    Commented Jun 3, 2019 at 13:45
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    $\begingroup$ Replace straight lines by circles of sufficiently big radii. The effect will be the same. Qualitative assumptions ($>0$ without explicit bounds) are usually useless in such matters... $\endgroup$
    – fedja
    Commented Jun 3, 2019 at 16:04

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I consider the question negatively answered by the counter examples provided in comments of @PietroMajer and @fedja. Thank you.

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