In the paper that can be found in this link: https://arxiv.org/pdf/1712.00082.pdf
there is a definition of the geodesic curvature that I can't quite understand.
Here is how this article presents it:

Let $\vec{X}(s)$ be a curve on a 2D manifold in $R^3$ which is parametrized by the arc length $s$ as shown in the following figure.
Choose a vector function $\vec{V}(s)$ living in the tangent space at the position $\vec{X}(s)$ and is paralely transported along the curve. Then $k_g=\frac{d\theta}{ds}$, where $\theta$ is the angle between the velocity vector $\vec{T}=\frac{d\vec{X}(s)}{ds}$ and $\vec{V}(s)$, is defined as the geodesic curvature.
The paper states that the geodesic curvature $k_g$ reflects the deviation of the curve from the local geodesics.

enter image description here

What I don't understand are the following:
$1)$ Since $\vec{T}=\frac{d\vec{X}(s)}{ds}$ and $\vec{V}(s)$ are both parallely transported along the curve(former is a velocity vector and later is given), then shouldn't the angle between them remain constant? In Riemannian geometry, when we parallel transport a vector along a curve, its angle with the velocity vector of the curve remain constant.
Is it that the author might not use an affine connection that is compatible with the metric, which is the property that an affine connection needs in order to preserve lengths of vectors and angles between vectors during parallel transport? (see for example Do Carmo's Riemannian geometry, p.53 on Riemannian connections)

$2)$ How does the geodesic curvature relate to the Riemannian curvature found in Riemannian geometry? Also, is it an intrinsic quantity or an extrinsic one? Finally, how does it generalize to higher-dimensional manifolds that are not embedded in higher-dimensional manifolds?


$1)$ When we parallel transport a vector along a geodesic, its angle with the velocity vector remains constant. This is not true along an arbitrary curve $X:$ since the velocity $\dot X$ is not necessarily parallel, there's not much we can say about $\angle (\dot X, V).$ Think about simple examples in the Euclidean plane, where parallel just means constant.

One thing you should note from this is that the geodesic curvature $k_g$ of $X$ is zero if and only $X$ is a geodesic of the surface; so it is a sensible generalization of the curvature of a curve in Euclidean space, where curvature measures deviation from a straight line.

$2)$ The geodesic curvature can be defined for a curve in an arbitrary Riemannian manifold in exactly the same way, since parallel transport and angle measurements are intrinsic to the Riemannian structure. Alternatively, so long as $X$ is unit-speed we can simply define the geodesic curvature as the norm of the covariant acceleration of $X$.

I don't think throwing around the words extrinsic and intrinsic without more context is very useful. You could say the geodesic curvature of the curve $X$ in the surface $M \subset \mathbb R^3$ is intrinsic to the surface $M$, since it does not depend on the choice of (isometric) embedding of $M$ in $\mathbb R^3.$ On the other hand, when thinking about the curve as an embedding $X : I \to M$, the geodesic curvature is extrinsic: it tells you about how the embedding map bends inside $M$, not about any geometry of the interval $I$. Since the geodesic curvature is typically thought of more as a property of the curve than of the manifold, if I had to label it I would say it is an extrinsic curvature.

Regarding the relationship to the Riemannian curvature, you shouldn't expect something direct: $k_g$ measures extrinsic curvature of a 1-dimensional object while $R$ measures intrinsic curvature of a higher-dimensional object. The closest link is probably the Gauss-Codazzi relations: for $M \subset N$ a Riemannian submanifold, the second fundamental form is just a way of packaging together the geodesic curvatures (with respect to $N$) of curves in $M$. The Gauss equation then tells us that the Riemannian curvature $R^M$ can be written in terms of $R^N$ and the second fundamental form of the inclusion.

  • $\begingroup$ Nice answer. Could you please enlighten this "Think about simple examples in the Euclidean plane, where parallel just means constant" to me? I consider a circle in plane and velocity vectors are tangent to it. Now consider normal vectors (perpendicular to tangent and I think any vector works as well as normal vector!). it remain perpendicular during transport. What is the wrong here? $\endgroup$ – C.F.G Nov 5 '20 at 7:06
  • $\begingroup$ @C.F.G: Neither of these examples (the velocity of a circle or the normal to a circle) are parallel fields! In $\mathbb R^2$ in the standard coordinates the Christoffel symbols vanish, so the parallel transport equation $\dot V^k + \Gamma^k_{ij} \dot \gamma^i V^j = 0$ simply becomes $\dot V^k = 0,$ i.e. the components of the vector field should be constant. You should be sliding the same vector along your curve while holding it fixed, not turning it along with the curve. $\endgroup$ – Anthony Carapetis Nov 5 '20 at 9:44
  • $\begingroup$ Thank you so much. I know it algebraically. Since you wrote that "its angle with the velocity vector remains constant" so I fixed the velocity vector field and comparing the angle of other vector with it. $\endgroup$ – C.F.G Nov 5 '20 at 9:48

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