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Let $$f:(0,\pi/2) \longrightarrow S^1 $$ be given by $f(t)=(\cos t,\sin t)$. Let us assume we have the induced metric on $S^1$ from the standard embedding into $\mathbb{R^2}$. Call the inclusion map i. Now we know that with respect to the induced metric, $S^1$ has constant positive curvature. Let us pull back the metric to $(0,\pi/2)$.

$$ f^*(i^*(dx\otimes dx+dy \otimes dy))=(i\circ f)^*(dx\otimes dx+dy \otimes dy)\\=(i\circ f)^*dx \otimes (i\circ f)^*dx+(i\circ f)^*dy \otimes (i\circ f)^*dy\\=(-\sin t)dt \otimes (-\sin t)dt + (\cos t)dt \otimes (\cos t)dt=dt\otimes dt$$

Now we have the standart metric on $(0,\pi/2)$. Moreover f is a diffeomorphism onto its image and hence an isometry. Isometries preserve curvature but $S^1$ has positive curvature and $\mathbb{R}$ has 0 curvature. What am i missing?

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  • $\begingroup$ $f$ is not an isometry. $\endgroup$ – uniquesolution Jul 25 '17 at 7:23
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    $\begingroup$ Ok, thank you. Now, why do you think isometries preserve curvature of curves? Gauss' theorem is about Gaussian curvature being preserved by isometries of surfaces. $\endgroup$ – uniquesolution Jul 25 '17 at 7:40
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    $\begingroup$ Yes, they do, but not curvature of curves. The curvature they preserve is the Gaussian curvature, which is actually a multiple of principal curvatures, or the determinant of the shape operator, if you are well versed with differential geometry. Obviously you are bending here a piece of a line into the plane. Obviously the line is flat, but so is the bended line when bended back! The curvature one-over-radius is NOT the Gaussian curvature. In particular, Gaussian curvature is valid for surfaces, not curves. $\endgroup$ – uniquesolution Jul 25 '17 at 7:48
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    $\begingroup$ If I recall correctly, isometries preserve geodesic distances, but I don't see why they should preserve curvatures of geodesics. In particular, note that some notions of curvature (e.g. mean curvature) are not intrinsic, whereas others (e.g. Gaussian curvature) are. $\endgroup$ – user3658307 Jul 25 '17 at 17:21
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    $\begingroup$ Isometries preserve Riemannian curvature. What is the Riemann curvature tensor of a $1$-dimensional manifold? Is not trivially zero? You're mixing different curvatures and thus the confussion. $\endgroup$ – mfl Jul 26 '17 at 12:32
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As I understand it, isometries preserve the metric in a certain sense, which means that intrinsic curvatures will not change, but other notions of curvature certainly can.

Note that isometry preserves the Riemann curvature tensor, which means that it also preserves the Ricci tensor and scalar curvature $R$. When you showed that there is an isometry from the circle to the uncurved line, you essentially showed that the circle has zero (Riemannian) curvature (as does any 1D Riemannian manifold).

However, this "intrinsic" curvature being zero does not mean there is no curvature. For instance, a space curve under arclength parameterization $\gamma(s)$ has curvature $\kappa(s)=||\ddot{\gamma}(s)||_2$. (Yet still has zero Riemannian curvature; perhaps calling $\kappa$ "curvature" is a disservice to students).

Another example is for 2D Riemannian manifolds in 3D, where the Theorema Egregium of Gauss applies. It essentially states that Gaussian curvature $K$ is invariant under isometry. However, the mean curvature $H$ is not. This is not so intuitive, since $H$ is simply the average of the principal curvatures, while $K$ is the product; why should one be invariant and the other not? However, in this case, $R=2K$, i.e. the Gaussian curvature is just half the Ricci curvature scalar, which we know must be preserved. The mean curvature is "extrinsic", on the other hand. (Indeed, the shape operator and thus the principal curvatures appear to depend on embedding, so one reasonably expects $H$ and $K$ to as well; the fact that $K$ is in fact an intrinsic invariant was actually rather surprising, hence the very remarkable name given to the theorem by Gauss.)

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