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Tom LaGatta
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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 $\gamma$.

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 $\gamma(0)$)

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 $\gamma$.

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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Tom LaGatta
  • 8.5k
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  • 43
  • 82

Prescribing a Riemannian metric along a given geodesic

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 $\gamma(0)$)

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 $\gamma$.