Timeline for Explanation for phenomenon in hyperbolic geometry
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 26, 2019 at 15:27 | comment | added | Holonomia | This is true in any Riemannian manifold. Experts known that the projection of a Killing vector field on a totally geodesic submanifold gives a Killing vector field of the totally geodesic submanifold. In your case the totally geodesic submanifold is the geodesic and the projection is a traslation (the minus sign in your equation is due to the way you took the tangent to the geodesic at x and y). | |
| Aug 26, 2019 at 4:49 | history | edited | Ivan Izmestiev |
removed a tag
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| Aug 26, 2019 at 4:48 | comment | added | Ivan Izmestiev | This is a special case of the first variation of arc length. But in can be proved in a more elementary way by differentiating $\langle x, y \rangle = \mathrm{const}$, (Minkowski scalar product) in the hyperbolic case and $\|x - y\|^2 = \mathrm{const}$ in the Euclidean case. | |
| Aug 26, 2019 at 2:37 | comment | added | burtonpeterj | In the Euclidean case when $\mathcal{F}$ is a rotation about a point $c$ with $|c-x| > |c-y|$ and such that $\angle cyx$ is right then $|d_x| > |d_y|$ but $d_y$ is parallel to $v_y$ while $d_x$ is at a nonzero angle to $v_x$. However, the formula I asked about still holds. | |
| Aug 26, 2019 at 2:24 | comment | added | burtonpeterj | In the Euclidean plane this is visually obvious for translations but not so obvious for rotations when $x$ and $y$ are at distinct distances from the center of the rotation. | |
| Aug 26, 2019 at 2:23 | comment | added | burtonpeterj | This is also true in the Euclidean plane, where I have verified it with explicit computation. | |
| Aug 26, 2019 at 0:41 | history | asked | burtonpeterj | CC BY-SA 4.0 |