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Let $\Sigma$ be a non-orientable surface possibly with boundary or punctures. Is it possible that a one-sided loop in $\Sigma$ is always realized as a geodesic?

In the orientable case, it is well-known that an essential loop (i.e., not homotopic to a puncture or a constant loop) can be realized as a geodesic.

Any comments or references are appreciated.

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    $\begingroup$ One-sided loops are never null-homotopic, so they are always realised by a (unique) geodesic. As another way to think about this: consider the preimage of the one-sided loop in the orientation double cover. Tighten there and push down. $\endgroup$ Commented Oct 2, 2024 at 9:20
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    $\begingroup$ @Sam Nead: Except that some geodesics can go through the punctures. $\endgroup$ Commented Oct 2, 2024 at 11:32
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    $\begingroup$ Do you assume your metric to be complete and have geodesic (or at least convex) boundary? Then the answer is positive. $\endgroup$ Commented Oct 2, 2024 at 15:24
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    $\begingroup$ There's metrics on the Moebius band that give no geodesic representatives. But you can ensure they're always "relative geodesic", i.e. when in the interior they are geodesic, and otherwise they are just segments of the boundary. It's like trying to take the shortest route around a race track with walls. $\endgroup$ Commented Oct 2, 2024 at 19:54

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