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## Hot answers tagged hyperbolic-geometry

3

It is more a matter of notation. Take $z\in H$ and $x,y\in\mathbb R^2$. Then $$\langle d_zw\, x,d_zw \,y\rangle_{w(z)}=:\frac{|dw|^2}{Im(w)^2}(x,y)=\frac{|dz|^2}{Im(z)^2}(x,y):=\langle x,y\rangle_z.$$ Added: More explicitly, write $z=z_1+iz_2$, $x=(x_1,x_2)\equiv x_1+ix_2$ and $y=(y_1,y_2)\equiv y_1+iy_2$. Then, by definition, $$\langle ... 1 My interpretation is that we should choose three different points$$\left(u,{c^2\over u}\right),\quad\left(v,{c^2\over v}\right),\quad\left(w,{c^2\over w}\right)$$on the hyperbola xy=c^2 such that the three lines determined by these points are all tangent to the parabola y^2=4ax, a>0. The line through the first two points has equation ... 1 You can multiply any left-invariant metric by a positive number and you will get a different left-invariant metric. 1 Do dome hyperbolic trigonometry: as \cosh2s=2\cosh^2s-1, we can rewrite u as$$u= \tau\cosh^2s-\frac12\tau+\frac12\tau=x\cosh s.$$On the other hand, \;\cosh^2s-\sinh^2s=1, whence, as \cosh s\ge 1>0 for all s,$$\cosh s=\sqrt{\sinh^2s+1}=\sqrt{y^2+1}.$$1$$ u= \tau*(1+\cosh(2s))/2 = \tau \cosh^2 s = \tau (1+\sinh^2 s) = \tau (1+y^2)  =\frac {x}{\cosh s}(1+y^2) = x \frac {(1+y^2) }{\sqrt{1 + \sinh^2 s}} = (1+y^2) \frac{x}{ \sqrt{1+y^2} } = x \sqrt{1+y^2}. $$1 \sinh(x) being strictly increasing on the reals, has a unique inverse function. Specific values for the inverse can be found using the definition of \sinh(x) and the quadratic equation, and a logarithm extraction. You already have \sinh(b/2)=\sinh(s/2)/\cosh(d) in terms of s,d. So after applying inverse sinh and doubling you have b in terms of ... 1 I don't know about "obvious", but see Section 6.4 of Dave Witte's book. 1 One quick method is to use symmetry: First prove (using your ODE, if you like) that the only geodesic with a vertical tangent vector is a vertical geodesic x(t)=constant, y(t) = e^{\pm t + C}. Next prove that the metric is invariant under the action of the group SL(2,\mathbb{R}) acting by fractional linear transformations. Finally, transform all ... 1 Suppose we have 3 points p, q, r\in\mathbb{H}^2. We want to show that$$d(p, q)+d(q, r)≥d(p, r). How hyperbolic circles are also Euclidean circles, then we can move the picture by a symmetry so that $p=i, q=\alpha+ai, r=bi$. Let $Q=ai$ be the point on the $Y$ axis which is on the same horizontal line as $q$. Here $a$ and $b$ and $\alpha$ are ...

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We observe that the quadrilateral result is actually one about tetrahedra. A Euclidean quadrilateral with diagonals $p$ and $q$ is the projection of a tetrahedron with opposite edges $p$ and $q$ into a plane parallel to those edges. If $h$ is the corresponding perpendicular distance from edge $p$ to edge $q$, then the quadrilateral edges $a$, $b$, $c$, $d$ ...

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