Hot answers tagged

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 ...


1

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 $a$, $b$, $c$, $d$ ...



Only top voted, non community-wiki answers of a minimum length are eligible