The title really say's it all, but once again is a curve's curvature invariant under rotation and uniform scaling?
Geometry – Is a Curve’s Curvature Invariant Under Rotation and Scaling?
differential-geometrygeometry
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Well, in many ways, it depends on what definition of isometry you're using. As I understand it, an isometry usually means a mapping that preserves the metric, i.e., some means of understanding distance between points and lengths of vectors on a surface - so, a place to start is to understand exactly what the metric is. In an elementary differential geometry course (that is, one where things are mostly done with $\mathbb{R}^{3}$ in mind), this is usually seen in the form of the quantities $E, F,$ and $G$, where:
$$E = x_{u} \cdot x_{u}$$ $$F = x_{u} \cdot x_{v}$$ $$G = x_{v} \cdot x_{v}$$
where $x(u, v)$ is a coordinate patch and $\cdot$ denotes the usual Euclidean dot product. A standard theorem one usually covers in a differential geometry course is that the Gaussian curvature $K$ can be expressed solely in terms of $E, F, G$ and partial derivatives of $E, F,$ and $G$ with respect to $u$ and $v$. Since isometries preserve the metric, i.e. $E, F, G$, it can therefore be seen that isometries preserve the Gauss curvature. This can be understood in more broad, conceptual terms, of course: if a mapping preserves the notion of distance between points and lengths of vectors between surfaces, it would seem natural that the Gaussian curvature, a quantity intimately tied to the notion of length, would also be preserved. This intuition, of course, is very much predicated on your definition of Gauss curvature, as there are many definitions and though all are equivalent, many are very different in presentation and give different insights into the meaning of Gauss curvature. However, that the Gaussian curvature is dependent only on $E, F, G$ and appropriate derivatives is not a trivial result. Any standard intro differential geometry text should cover this result and these ideas; I suggest you look for information in a good intro text accordingly.
What is true is that if $f : M \to M$ is an isometry, , $f^*g = g$, and then $f^*\nabla = \nabla$ and consequently this leads to $f^*R = R$. So an isometry of $M$ leaves unvariant the curvature.
But this has nothing to do with parallel transport: say $\gamma : I \to M$ is a smooth path and let $P_t$ be the parallel transport from $\gamma(0)$ to $\gamma(t)$. Then: $$ P_t : \left(T_{\gamma(0)}M,g_{\gamma(0)}\right) \to \left(T_{\gamma(t)}M,g_{\gamma(t)}\right) $$ is a linear isometry. It is not an isometry of $M$.
In fact, on a connected riemannian manifold, any two points can be joined by a smooth path and thus, there are always parallel transport between them. But they may not even exist an isometry of $M$ sending one point to another given one in the general case. This would imply that $M$ is two-point homogeneous and this has huge consequences on the curvature.
Best Answer
A curve's curvature is invariant under rotation. Intuitively, a curve turns just as much no matter how it is oriented. More formally, for a curve $\gamma(s)$ that is parametrized by arc length, the curvature is $\kappa(s) = ||\gamma''(s)||$. Rotation does not change the length of the $\gamma''(s)$ vector, only the direction; therefore, rotation does not affect curvature.
A curve's curvature is not invariant under uniform scaling, however. Consider the example of a circle. All circles are the same up to scaling, but they don't all have the same curvature; in general, a circle of radius r has curvature 1/r.