# Tag Info

4

Things scale in a reasonable way: The metric $$ds^2 = \frac{4}{(1-x^2-y^2)^2}(dx^2 + dy^2)$$ has $K=-1$. The metric $$ds_c^2 = \frac{c^2}{(1-x^2-y^2)^2}(dx^2 + dy^2)$$ has $K=-4/c^2$.

2

Do you know how to construct a circle inversion? If so, draw any two hyperbolic lines $g_1$ and $g_2$ through $C$ but not through $P$. Invert $P$ in $g_1$ to obtain $P_1$ and also in $g_2$ to obtain $P_2$. The Euclidean circle through $P,P_1,P_2$ is the hyperbolic circle. This is because a point on the circle, reflected on a diameter of the circle, will ...

1

The synthetic viewpoint is more general! This is a benefit since stating things abstractly in terms of primitive notions is probably easier to grasp rather than delving immediately into coordinates. Compare: Two lines intersect in at most one point The system $\{ax+by+c=0;dx+ey+f=0\}$ has at most one solution. The first comes with less baggage than the ...

1

$X$ is dense in the 2-sphere boundary $S^2_\infty = \partial\mathbb{H}^3$, and the limit set equals $S^2_\infty$. An equivalent way of saying this is that closed orbits of the geodesic flow are dense in the unit tangent bundle of $\pi_1(M)$, and this is closely connected to E. Hopf's theorem that the geodesic flow is ergodic.

1

If the segments are geodesics in your surface, then the Gauss Bonnet Theorem will tell you that (letting $Q$ denote the quadrilateral and $K$ the (Gaussian) curvature of the surface) $$\iint_Q K\,dA = \sum_{j=1}^4 \iota_j - 2\pi,$$ where $\iota_j$, $j=1,\dots,4$, are the interior angles. In your case, you'll have $\iota_4-\pi/2$ for that right-hand side. ...

1

Hyperbolic isometries in the Poincaré disk preserve angles (both hyperbolic and Euclidean, since they are the same) and hyperbolic distances (i.e. distances measured using the hyperbolic metric), but in general not Euclidean distances. The only hyperbolic isometries which also preserve Euclidean distances are reflections in the diameters of the Poincaré ...

1

Here is an illustration of what's going on. Here I chose the common tangent to be orthogonal to the diameter through the given point. This ensures that the point lies on the symmetry axis of the crescent, so this is the most symmetric situation. For other tangents, the points where the hypercircle touches the boundary of the model would move at different ...

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I suggest you reflect the point in the hyperbolic line. This is just an inversion in a circle. Then you can compute the distance between the point and its image, and divide that by two. Moving one of the points which span the line into the origin may help as well, but it's not immediately obvious. And quite an unsymmetric approach. You might want to ...

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