Just wanted to write up some solutions to problems in hyperbolic geometry as practice. Here is one.

Proposition. An isometry of the upper half-space model whose only fixed point is $\infty$ has the form $(x, t) \mapsto (Ax + b, t)$ for some $A \in O(n-1), b \in \mathbb{R}^{n-1}$. More generally, any isometry with a single fixed point on the boundary fixes each horosphere setwise, and acts via a Euclidean isometry.

Proof. Let $f$ be the given isometry of the upper-half space. Let $H_t = {(x, t) \mid x \in \mathbb{R}^{n-1}}$ be the horocycle of $\infty$ at height $t$. First of all we know that $f(H_t)$ is also some horocycle since it is a surface which is perpendicular to all of the vertical lines (since the set of geodesics emanating from $\infty$ is permuted by $f$.)

We will next show that each horosphere is fixed setwise by showing that $H_t$ can be defined geometrically with reference to $\infty$.

Let’s show $f(H_t) = H_t$. Pick $p \in H_t$. Let $q$ be any point in $H_t$ hyperbolic distance 1 away from $p$. Consider the triangle $(p, q, \infty)$. Its isometry class on $p$ or $q$ since the isometry group of $\mathbb{H}^n$ acts transitively on pairs of points in $H_t$ at distance 1 (e.g., by translating and then rotating around the vertical axis, or reflecting in the two dimensional case). Call this triangle (whose isometry class is the same for all such pairs $p, q \in H_t$) $\Delta_t$ and let $a_t$ be its area.

I claim that $a_t > a_{t’}$ for $t < t’$. It’s enough to show that some $\Delta_t$ contains some $\Delta_{t’}$. Pick any distance 1 pair $(x, t), (y, t)$. Let’s show the triangle $((x, t), (y, t), \infty)$ contains $((x, t’), (y, t’), \infty)$. If not, then the geodesic segment $(x, t’) \to (y, t’)$ must enter the region below the arc $(x, t) \to (y, t)$, and so intersects that other geodesic. If it enters that region, it also must leave, so the two geodesic segments cross each other twice. But then the two segments must lie on the same geodesic, which is obviously not the case.

This essentially defines each horocycle in intrinsic terms (allowing reference to $\infty$). More specifically: Let $p, q \in H_t$ at distance 1. We know $f$ sends $(p, q, \infty)$ to $(f(p), f(q), \infty)$. We know from above that $f(p)$ and $f(q)$ lie on some $H_{t’}$ together, and we know that $a_t$, the area of $(p, q, \infty)$ is the same as the area $a_{t’}$ of $(f(p), f(q), \infty)$. But this implies $t = t’$ since $a_t$ is monotonically decreasing in $t$. Therefore $f(p)$ lies on $H_t$ as well.

Each horocycle is a Euclidean $(n-1)$-space because the UHS metric depends only on $t$.

All of this so far shows that $f$ has the form $(x, t) \mapsto (\phi(x, t), t)$. Let’s show $\phi$ does not depend on $t$. Consider the vertical line $L_x = { (x, t) \mid t \in \mathbb{R}}$. $f(L_x)$ is also a vertical line (since $\infty$ is fixed), necessarily equal to $L_{\phi(x, 0)}$. $(x, t)$ is the intersection point of $H_t$ and $L_x$, so $f((x,t))$ is the intersection point of $f(H_t) = H_t$ and $f(L_x) = L_{\phi(x, 0)}$, which is $(\phi(x, 0), t)$. Therefore $\phi$ does not depend on $t$. Since each $H_t$ is a Euclidean $(n-1)$-space, $\phi$ is a Euclidean isometry which has the desired form $Ax + b$ for $A \in O(n-1)$.

The general statement is true by conjugating by an isometry swapping the single fixed point with $\infty$.t wanted to write up some solutions to problems in hyperbolic geometry as practice.

Proposition. An isometry of the upper half-space model whose only fixed point is $\infty$ has the form $(x, t) \mapsto (Ax + b, t)$ for some $A \in O(n-1), b \in \mathbb{R}^{n-1}$. More generally, any isometry with a single fixed point on the boundary fixes each horosphere setwise, and acts via a Euclidean isometry.

Proof. Let $f$ be the given isometry of the upper-half space. Let $H_t = {(x, t) \mid x \in \mathbb{R}^{n-1}}$ be the horocycle of $\infty$ at height $t$. First of all we know that $f(H_t)$ is also some horocycle since it is a surface which is perpendicular to all of the vertical lines (since the set of geodesics emanating from $\infty$ is permuted by $f$.)

We will next show that each horosphere is fixed setwise by showing that $H_t$ can be defined geometrically with reference to $\infty$.

Let’s show $f(H_t) = H_t$. Pick $p \in H_t$. Let $q$ be any point in $H_t$ hyperbolic distance 1 away from $p$. Consider the triangle $(p, q, \infty)$. Its isometry class on $p$ or $q$ since the isometry group of $\mathbb{H}^n$ acts transitively on pairs of points in $H_t$ at distance 1 (e.g., by translating and then rotating around the vertical axis, or reflecting in the two dimensional case). Call this triangle (whose isometry class is the same for all such pairs $p, q \in H_t$) $\Delta_t$ and let $a_t$ be its area.

I claim that $a_t > a_{t’}$ for $t < t’$. It’s enough to show that some $\Delta_t$ contains some $\Delta_{t’}$. Pick any distance 1 pair $(x, t), (y, t)$. Let’s show the triangle $((x, t), (y, t), \infty)$ contains $((x, t’), (y, t’), \infty)$. If not, then the geodesic segment $(x, t’) \to (y, t’)$ must enter the region below the arc $(x, t) \to (y, t)$, and so intersects that other geodesic. If it enters that region, it also must leave, so the two geodesic segments cross each other twice. But then the two segments must lie on the same geodesic, which is obviously not the case.

This essentially defines each horocycle in intrinsic terms (allowing reference to $\infty$). More specifically: Let $p, q \in H_t$ at distance 1. We know $f$ sends $(p, q, \infty)$ to $(f(p), f(q), \infty)$. We know from above that $f(p)$ and $f(q)$ lie on some $H_{t’}$ together, and we know that $a_t$, the area of $(p, q, \infty)$ is the same as the area $a_{t’}$ of $(f(p), f(q), \infty)$. But this implies $t = t’$ since $a_t$ is monotonically decreasing in $t$. Therefore $f(p)$ lies on $H_t$ as well.

Each horocycle is a Euclidean $(n-1)$-space because the UHS metric depends only on $t$.

All of this so far shows that $f$ has the form $(x, t) \mapsto (\phi(x, t), t)$. Let’s show $\phi$ does not depend on $t$. Consider the vertical line $L_x = { (x, t) \mid t \in \mathbb{R}}$. $f(L_x)$ is also a vertical line (since $\infty$ is fixed), necessarily equal to $L_{\phi(x, 0)}$. $(x, t)$ is the intersection point of $H_t$ and $L_x$, so $f((x,t))$ is the intersection point of $f(H_t) = H_t$ and $f(L_x) = L_{\phi(x, 0)}$, which is $(\phi(x, 0), t)$. Therefore $\phi$ does not depend on $t$. Since each $H_t$ is a Euclidean $(n-1)$-space, $\phi$ is a Euclidean isometry which has the desired form $Ax + b$ for $A \in O(n-1)$.

The general statement is true by conjugating by an isometry swapping the single fixed point with $\infty$.