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I spent quite some time on this problem without finding a solution, I hope it is not trivial. Regardless, I was not able to find an answer anywhere else.

Suppose to have a sufficiently smooth curve $\gamma$, embedded in a generic two-dimensional Riemannian manifold $M$, parametrized by arc-length $s$. Now define a new curve, $\gamma'$, as the one obtained by the union of parallel-transported points from $\gamma$ along one of the two normal directions for an arc-length distance $z$. The question is: how are the geodesic curvatures of $\gamma$ and $\gamma'$ related?

More rigorously: if $P=P(s)$ is a point on $\gamma$, and $\mathbb{n}\in T_P M$ is a vector normal to $\gamma$ at $P$, trace the geodesic curve starting at $P$ with initial direction $\mathbb{n}$. Follow this geodesic for an arc-length distance $z$. You will end up in a point $P'(s,z)$, "normally displaced" w.r.t. the original curve. The new curve is the union of such points, $\gamma'= \{\cup_s P'(s,z)\}$. The question becomes: how is $\kappa'$, the geodesic curvature of $\gamma'$, related to $\kappa$, the geodesic curvature of $\gamma$, and to the curvature of the ambient manifold?

It is rather easy to work out flat the case, $M=\mathbb{R}^2$. I find the curvature $\kappa'$ of $\gamma'$ to be

$$ \kappa'(s,z)= \frac{\kappa(s)}{1+z\; \kappa(s)} \,. $$ The extra factor $(1+z\; \kappa(s))^{-1}$ follows from parametrizing $\gamma'$ with $s$, i.e. the tangent vector of the new curve is not unit-normalized ($s$ is the arc-length of $\gamma$, not of $\gamma'$).

I expect the general expression to depend on some geometric invariants of $M$ as well. Also, I implicitly assume that $z$ is small enough so that the map between points of $\gamma$ and $\gamma'$ is bijective (i.e. no two geodesics cross each other), and $\gamma'$ is smooth. In the likely case it is not possible to find an exact and/or closed answer to the question, I would like to be able to obtain an answer in the limit $z\to 0$ correct at least at $O(z^2)$, and clearly see how it deviates from the $\mathbb{R}^2$ answer.

PS: this is my first post so I apologize if I did not format the question correctly. Also, I am a physicist so please forgive my lack of exact definitions.

edit: minor typos corrections

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  • $\begingroup$ I think it is very well stated and it is a relevant question. I would add that it is necessary to make a continuous choice of normal vectors along the curve (although I think this is what you meant). $\endgroup$ – Eduardo Longa Sep 17 '18 at 18:15
  • $\begingroup$ yes you are perfectly right! Thanks for pointing that out $\endgroup$ – Pier Sep 18 '18 at 16:36

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