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Let $\gamma : [0,1] \to \mathbb R^2$ be a finite $C^2$-curve in the plane which does not intersect itself. Let $p(z)$ be a second-degree polynomial in $z \in \mathbb R^2$. Can we construct a Riemannian metric $g$ along $\gamma$ such that

  • $\gamma$ is a geodesic of $g$,
  • $\gamma$ has length 1, and
  • $p(z)$ is the 2-jet of $g$ at $\gamma(0)$? (i.e. this prescribes $g$ and its first and second derivatives at $\gamma(0)$) Edit: As per Sergei's comment, assume that $p$ is chosen so that $\gamma$ does in fact solve the geodesic equation.

I think so, and my sketch of an argument follows the proof of existence of Fermi coordinates in reverse. I haven't worked through this in detail yet, though, because I'm more concerned about the next question:

Let $\gamma$ and $\eta$ both be finite curves $[0,1] \to \mathbb R^2$ which do not intersect themselves, and such that $\gamma(0) = \eta(0)$ and $\gamma(1) = \eta(1)$ with no other intersections (i.e. $\gamma \cup \eta$ is a piecewise, simple, closed $C^2$-curve in the plane). Can we construct a metric $g$ as above? Note that if so, $\gamma(0)$ and $\gamma(1)$ will be conjugate points along $\gamma$.

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If you prescribe the first derivatives of $g$ at $\gamma(0)$, the geodesic equation gives you a specific expression of $\gamma''(0)$ in terms of $\gamma'(0)$. So you cannot make any given curve a geodesic. –  Sergei Ivanov Apr 10 '10 at 8:35
    
1) Are you constructing the metric $g$ as $p$ times the Euclidean metric? 2) If you define $p$ along $\gamma$ so that $\gamma$ is a geodesic and has length 1, what else is there to do? –  Deane Yang Apr 10 '10 at 23:47
    
For the second question, do you again have some $p$-jet condition? If not, can you construct the metric you want just by diffeomorph-ing a connected open neighbourhood of the two curves onto an open subset of a suitably-metrized sphere? (There will be a few cases depending on whether the angles of the curves at their meeting-points are acute-acute, acute-straight, acute-obtuse, etc.) –  macbeth Apr 11 '10 at 0:25
    
I don't see any issues with constructing the metric in the second case, except that everything has to match up properly at the endpoints. –  Deane Yang Apr 11 '10 at 1:02
    
What regularity do you want from $g$? If $g$ is $C^2$, the geodesics must be $C^3$. –  Sergei Ivanov Apr 11 '10 at 7:48

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