Let $C=S^1\times[0,1]$ be a compact cylinder. Given positive numbers $l,\lambda>0$. Is it possible to construct a smooth Riemannian metric on $C$ of constant Gauss curvature -1 such that one connected component of the boundary $\partial C$ will be geodesic, while the second one will have length $l$ and constant geodesic curvature (second fundamental form) $\lambda$?
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1$\begingroup$ This reduces to an initial value problem for an ODE. Have you tried to write that down? $\endgroup$– Deane YangCommented Nov 3 at 17:50
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$\begingroup$ @DeaneYang: How do you get ODE (rather than PDE)? $\endgroup$– asvCommented Nov 3 at 18:05
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$\begingroup$ In order to define sign of the curvature of a boundary curve, you have to specify the normal direction. Do you want the 2nd boundary component to be convex or concave? $\endgroup$– Moishe KohanCommented Nov 3 at 18:33
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4$\begingroup$ Assume the metric is rotationally symmetric $\endgroup$– Deane YangCommented Nov 3 at 18:58
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$\begingroup$ Did you realize that the accepted answer is incomplete and does not work if $|\lambda|\ge 1$? $\endgroup$– Moishe KohanCommented Nov 14 at 4:54
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1 Answer
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An equidistant curve to a line in hyperbolic plane has constant geodesic curvature. So pick a line $L$ in $\mathbb H^2$ and pick $r>0$ such that the $r$-equidistant curves have the desired geodesic curvature. Let $L_r$ be one of them. After this pick a translation $f: \mathbb H^2 \rightarrow \mathbb H^2$ along $L$ such that the quotient of $L_r$ by $f$ has length $l$. The quotient by $f$ of the strip bounded by $L$ and $L_r$ produces the desired cylinder.
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$\begingroup$ Keep in mind that the geodesic curvature of your $L_r$ (sometimes called a hypercycle) is between $0$ and $1$ (or $-1$ depending on your orientation convention), this is a nice exercise in differential geometry. Thus, if $|\lambda|\ge 1$ you will never reach the "desired" curvature. $\endgroup$ Commented Nov 14 at 4:52