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Away from a zero of $\varphi$, the first Bogomolny vortex equation givesSubstitution into the second equation, with ${\nabla }^2=4{\partial }_z{\partial }_{\bar{z}}$, givesNow set $\tilde{\Omega }=\Omega |\varphi {|}^2$. The Gaussian curvature formula givesTherefore $(\tilde{K}+1/2)\tilde{\Omega }=(K+1/2)\Omega$ implieswhich is exactly the vortex equation.

The metricis the pullback of the $n=1$ Poincare disc model metric under $w=z^n$. It consequently has $K=-1/2$ away from the origin. Comparing it with $d{s}_1^2$ gives the Witten hyperbolic vortexPutting $n=N+1$ gives winding number $N$. A gauge choice with positive radial factor isand then

The Poincare disc model has metricRotations about the origin are hyperbolic isometries, so a Euclidean circle centred there is a hyperbolic circle. Its radius is the radial length

The Poincare disc model isIts geodesics through the origin are the Euclidean diameters. To see directly that the radial segment from $0$ to $r{e}^{i{\theta }_0}$ minimizes length, write an arbitrary joining curve as $z(t)=r(t)e^{i\theta (t)}$. Its hyperbolic length satisfiesThe radial segment has constant $\theta$ and monotone $r$, so equality holds. Henceand every radial diameter is length minimizing. Rotational symmetry and uniqueness for the geodesic equation show that these are all geodesics through $0$.

Given any geodesic and a point $p$ on it, a disc-preserving maptakes $p$ to the origin. By part (a), $T$ is a hyperbolic isometry and commutes with $J(z)=1/\overline{z}$. The transformed geodesic is a diameter, hence a generalized circle invariant under $J$. Its inverse image is therefore also a Generalized circle under a Möbius transformation invariant under $J$. Thus every hyperbolic geodesic is the part in $\mathbb{D}$ of a Euclidean line or circle preserved by reflection in the unit circle; equivalently, it is a diameter or a circle orthogonal to the unit circle. This is the Geodesics of the Poincare disc description.

After the $2n$ sides are identified in pairs, the polygonal-schema Euler count gives one face, $n$ edges and some number $V\ge 1$ of vertices. The Euler characteristic of a closed orientable genus-$g$ surface is $2-2g$, so

For genus two, take a regular octagon in the Poincare disc model and identify each side with its opposite side, with the orientation reversed along the boundary. One cyclic labelling isThe quotient has $V=1$, $E=4$, and $F=1$, hence Euler characteristic $-2$ and genus two. All eight vertices become one point, so smoothness requires their angles to sum to $2\pi$. Each interior angle is consequentlyEquivalently, the hyperbolic polygon area formula shows that the regular hyperbolic octagon fundamental polygon has areaas required by the Gauss-Bonnet theorem for a genus-two surface of curvature $-1$.

For $|a|<1$, the Möbius transformation $f_a(z)=(z-a)/(1-\overline{a}z)$ is an isometry of the Poincare disc model sending $a$ to zero. Rotations combined with these transformations describe its orientation-preserving isometries.