The hyperbolic plane $ℍ$ is the upper half plane $\{(x, y): y>0\}$ endowed with the first fundamental form \[ \frac{\mathrm{d} x^2+\mathrm{d} y^2}{y^2} . \] State, without proof, the Gaussian curvature of $ℍ$. Let $γ(s)=(u(s), v(s))$ be a smooth curve in $ℍ$. State, without proof, the general differential equations which $u$ and $v$ must satisfy for $γ$ to be a geodesic in $ℍ$ parametrized by arc-length.

The Gaussian curvature of $ℍ$ is $-1$. The general differential equations are \begin{aligned} \frac{\mathrm{d}}{\mathrm{d} s}(E\dot{u}+F\dot{v})&=\frac12(E_u\dot{u}^2+2F_u\dot{u}\dot{v}+G_u\dot{v}^2),\\ \frac{\mathrm{d}}{\mathrm{d} s}(F\dot{u}+G\dot{v})&=\frac12(E_v\dot{u}^2+2F_v\dot{u}\dot{v}+G_v\dot{v}^2). \end{aligned} in $ℍ$, $E=1/v^2,F=0,G=1/v^2$, $E_u=F_u=G_u=0$, $E_v=-2/v^3,F_v=0,G_v=-2/v^3$, the equations become \begin{aligned} \frac{\mathrm{d}}{\mathrm{d} s}\frac{\dot{u}}{v^2}&=0,\\ \frac{\mathrm{d}}{\mathrm{d} s}\frac{\dot{v}}{v^2}&=-\frac{\dot{u}^2+\dot{v}^2}{v^3} \end{aligned}

Show that for each $k ∈ ℝ$ the half-line \[ γ(s)=(k, e^s) \] is a geodesic parametrized by arc-length.

To verify that $γ(s)$ satisfies the equations \begin{aligned} \frac{\mathrm{d}}{\mathrm{d} s}\frac{0}{(e^s)^2}&=0,\\ \frac{\mathrm{d}}{\mathrm{d} s}\frac{e^s}{(e^s)^2}&=-\frac{0^2+(e^s)^2}{(e^s)^3} \end{aligned}

Let $c, r ∈ ℝ$ with $c>0$. Show that the map \[ (x, y) ↦(c x+r, c y) \] is an isometry of $ℍ$. Let $f_c(r)$ be the geodesic curvature of the line $y=c$ in $ℍ$ (parametrized so that arc-length increases with $x$) at the point $(r, c)$. Show that $f_c(r)$ takes the same value for all $r ∈ ℝ$ and $c>0$.

\[ \frac{\mathrm{d} x'^2+\mathrm{d} y'^2}{y^2}=\frac{c^2\mathrm{d} x^2+c^2\mathrm{d} y^2}{c^2y^2}=\frac{\mathrm{d} x^2+\mathrm{d} y^2}{y^2} \] so the map is an isometry.
Since geodesic curvature is preserved by an isometry $(x, y) ↦(c x+r, c y)$, it takes the same value for all $c$ and $r$.
The line $y=c$ is a horocycle that is naturally parametrized by arclength, $α(t) = (t, c)$. The tangent vector is the unit vector that points to the right. This makes it appear as if the horocycle doesn't curve but we should use parallel transport to judge whether two vectors are parallel.Error! Click to view log. Take two points $A=(t, c)$ and $B=(t+h, c)$ on the horocycle and draw a geodesic between them: it's an arc of a circle with Euclidean radius $\sqrt{c^2+(\frac{h}{2})^2}$. The tangent vector to $α$ makes the angle $\sin^{-1}(\frac{h}{2c})$ with the geodesic at both $A$ and $B$, but it's in opposite directions. So, transporting the vector $α'(t)$ from $A$ to $B$ along the geodesic, we see that at the point $B$ it makes the angle $2\sin^{-1}(\frac{h}{2c})$ with $α'(t+h)$. Since the unit tangent rotates by $2\sin^{-1}(\frac{h}{2c})$ over the distance $d_H(A,B)=\frac{h}c$, the geodesic curvature is $$ \lim_{h→ 0} \frac{2\sin^{-1}(\frac{h}{2c})}{\frac{h}c} =1 $$

Now let $γ$ be a positively oriented, piecewise smooth, simple, closed curve in $ℍ$, bounding a region $R$, with external angles $α_1, …, α_n$. State, without proof, the Gauss-Bonnet Theorem as applied to $γ$.
Use this result to show that $f_c(r)=1$ for all $r ∈ ℝ$ and $c>0$.

Error! Click to view log. Let $γ$ be the boundary of the region $R$ bounded by the horocycle $y=1,y=2$ and the geodesics $x=-\frac12,x=\frac12$. The Gauss-Bonnet Theorem applied to $γ$ \[∫_γ k_g \mathrm{d} s+∫_R K \mathrm{d} A=2π-\sum_{i=1}^nα_i\] Since $K=−1$, \[∫_γ k_g \mathrm{d} s-A=2π-\sum_{i=1}^nα_i\] Since geodesic curvature takes the same value on the horocycles $y=1,y=2$ (but takes negative sign on $y=2$ traversed in the opposite direction), it is $1$ on the vertical lines $x=±\frac12$ \[k_g(∫_{\{y=1,x∈[-\frac12,\frac12]\}}\mathrm{d}s-∫_{\{y=2,x∈[-\frac12,\frac12]\}}\mathrm{d} s)-A=2π-\sum_{i=1}^nα_i\] The length of the horocycle $y=1$ from $(-\frac12,1)$ to $(\frac12,1)$ is $\frac11$, that of $y=2$ is $\frac12$, so \[k_g (\frac11-\frac12)-A=2π-\sum_{i=1}^nα_i\] Since $2π-\sum_{i=1}^nα_i=2π-\fracπ2×4=0$, \[\frac12k_g=A\] Calculating the area of $R$, \[A=\int_{-\frac12}^{\frac12}\int_1^2\frac1{y^2}\mathrm{d} y\mathrm{d} x=\int_{-\frac12}^{\frac12}\left[-\frac1y\right]_1^2\mathrm{d} x=\int_{-\frac12}^{\frac12}\frac12\mathrm{d} x=\frac12\] so \[k_g=1\]