Completeness of Upper Half Plane

Question

I am trying to prove that the upper half plane, defined as $\mathbb{H} = \{z \in \mathbb{C} : \Im(z)>0 \}$, is complete with respect to the hyperbolic metric.

First I note that if I have some closed and bounded subset $X$ of $\mathbb{H}$, it is complete. However, when dealing with $\mathbb{H}$, I would like to use the nice property that if I am in $\mathbb{R}^n$, and have some Cauchy sequence $x_n$ in a closed and bounded subset of $\mathbb{R}^n$, then it has a subsequence say $x_{n_k}$ which converges in my closed and bounded in the euclidean metric, viz. if I am in $\mathbb{R}^2$ it is just $|\mathbf{x} - \mathbf{y}|$, where $\mathbf{x}, \mathbf{y}$ some points in my set.

How do I deal with the fact that at the boundary, my euclidean metric remains bounded but the hyperbolic metric defined as

$$d(z_1,z_2) = \ln \left[ \frac{|z_1 - \bar{z_2}| + |z_1 - z_2 | }{|z_1 - \bar{z_2}| - |z_1 - z_2 | }\right]$$

goes to infinity?

In addition, the upper half of the complex plane is not closed, so how can I use nice properties like convergence of subsequences and stuff to prove that it is complete?

Thanks.

Answer 1

Let $p_n$ be a cauchy sequence in $(X, d)$, where $X$ is a closed and bounded subset of $H$ and $d$ is the hyperbolic metric in the upper half plane, defined as

$d(x,y) = \log\frac{|x-\overline{y}|+|x-y|}{|x-\overline{y}|-|x-y|}$

Now as $p_n$ is cauchy, there exists a subsequence $p_{n_k}$ that converges to $p \in X$ in the standard euclidean metric. We note that the possibility that the point $p$ be such that $Im (p) = 0$ is excluded as this would contradict $p_n$ being a cauchy sequence in $(X,d)$.

Now since $p_{n_k}$ converges to $p$ in the euclidean metric, it follows that $p_{n_k}$ converges to $p$ in the hyperbolic metric as well, using the formula above as $p_{n_k} \rightarrow p$ in the euclidean metric implies that $d(p, p_{n_k}) \rightarrow \log \frac{|p-\overline{p_{n_k}}|}{|p-\overline{p_{n_k}}|} = 0$.

Another way to think about it would be that the function $\frac{ \log\frac{|x-\overline{y}|+|x-y|}{|x-\overline{y}|-|x-y|}}{|x-y|}$ is a continuous function and hence must achieve its maximum and minimum values on a compact set, so that the function

$\log\frac{|x-\overline{y}|+|x-y|}{|x-\overline{y}|-|x-y|}$

is bounded above and below by $C|x-y|$ and $P|x-y|$, $C$ and $P$ some constants so by the squeeze theorem as $|x-y|$ goes to zero, $\log\frac{|x-\overline{y}|+|x-y|}{|x-\overline{y}|-|x-y|} \rightarrow 0$ as well.

So since a subsequence $p_{n_k}$ converges to $p$ in the hyperbolic metric, it follows that the original sequence $p_n$ converges to the $p$ as well in the hyperbolic metric, by a simple application of the triangle inequality.

My problem now is, I have proved this for a compact subset of $\mathbb{H}$. How do I prove this for the whole of the upper half of the complex plane?

Thanks.

Answer 2

I think it likely that you want this: every unit speed geodesic, every one, is given by one of two formulas: with real constant $A,$ real constant $B > 0,$ and real parameter $t,$ either $$ z(t) = A + i e^t, $$ or $$ z(t) = A + B \tanh t + i B\,\mathrm{sech}\,t. $$ Given any point and tangent direction, you can place one of these passing through that point in the desired direction, by taking appropriate values for $A,B.$ Furthermore the geodesic takes the variable $t$ from $-\infty$ to $\infty.$ The rest is called the Hopf-Rinow Theorem, you may need to read up on that.

Answer 3

$\def\eps{\varepsilon}$Suppose $(p_i)_{i\geq1}$ is a Cauchy sequence in $\mathbb H$. Show first there exist $\eps>0$ and $R>0$ such that $\operatorname{Im}p_i\geq\eps$ and $|p_i|\leq R$ for all $i\geq1$. It follows that the sequence does not leave the compact subset $$K=\{z\in\mathbb H:\operatorname{Im}z\geq\eps, |z|\leq R\}\subset\mathbb H.$$ Therefore you can use your initial observation to conclude in the general situation.

Answer 4

Here is a variant approach, which might be of interest. (It is a little different to Mariano's: instead of immediately trying to find a closed and bounded subset inside $\mathcal H$ which contains the Cauchy sequence, I will instead use the fact that $\mathcal H$ sits inside a certain closed and bound subset of the Riemann sphere. In short, instead of trying to work only "from the inside" of $\mathcal H$, I will let myself work "from the outside" instead.)

The upper half-plane $\mathcal H$ sits in the extended upper half-plane $\overline{\mathcal H},$ which is the closure of $\mathcal H$ in the Riemann sphere. So $\mathcal H$ is an open disk in the Riemann sphere (the open upper hemisphere, if you like), while $\overline{\mathcal H}$ is a closed disk (the closed upper hemisphere); it is the union of $\mathcal H$, the real line $\mathbb R$, and the point at infinty.

Let $x_n$ be a Cauchy sequence in the upper half-plane. We want to show that it has a limit $x$. By general metric space arguments, it is enough to show that some subsequence of $x_n$ has a limit.

Now $\overline{\mathcal H}$ is compact (i.e. closed and bounded in the Riemann sphere, if you like), and so any sequence in $\mathcal H$ has a subsequence which converges in $\overline{\mathcal H}$. By the remark of the preceding paragraph, it is suffices to show that this subsequence actually converges to a point of $\mathcal H$.

So, we have reduced to the following situation: If $x_n$ is a Cauchy sequence in $\mathcal H$ converging to a point $x \in \overline{\mathcal H}$, then $x$ in fact lies in $\mathcal H$.

Suppose (with the goal of getting a contradiction) that $x \not\in \mathcal H$. By applying an isometry (i.e. a point of $SL_2(\mathbb R)$) to the whole set-up, we may assume that $x$ is the point at infinity. (This is not necessary, but simplifies the computation that follows.)

So now we have a $x_n$ in $\mathcal H$ converging to the point at infinity, i.e. $x_n = a_n + b_n i$ with $b_n \to \infty$.

To get the desired contradiction, it suffices to show that this sequence is not Cauchy. This is an elementary computation using the definition of the hyperbolic metric.