# Completeness of Upper Half Plane

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.

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Your first sentence, stating what you want to prove, is incomplete both in that it is actually not complete (you do not really tell us what you want to prove!) and in that you should specify what metric you are considering on the half-plane. –  Mariano Suárez-Alvarez May 3 '11 at 2:23
You may want to more or less rewrite the whole thing into being more understandable, really… –  Mariano Suárez-Alvarez May 3 '11 at 2:24

$\def\eps{\varepsilon}$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.

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árez-Alvarez Please see my reply below as I can't type it here. –  fpqc May 3 '11 at 11:57

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.

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Let me try to finish the argument: $|x_n - x_m| < |a_n + b_n i - a_m - b_m i| < |a_n| + |b_ i| + |a_m| + |b_m i|$ which goes to infinitiy so $x_n$ is not cauchy in the hyperbolic metric? –  fpqc May 3 '11 at 14:55
@D Lim: You are trying to prove that the original Cauchy sequence converges to some point $x \in \mathcal H$. General metric space argument show that this will be true if it is true for some subsequence. Now general topological arguments (i.e. compactness) show that there is a subsequence that converges to a point $x \in \overline{\mathcal H}$. If we show that this subsequence in fact converges to a point of $\mathcal H$, we are done. Thus, we have to show $x \in \mathcal H$. –  Matt E May 3 '11 at 14:55
Note: the preceding comment is a response to a now deleted request for clarification. –  Matt E May 3 '11 at 15:01
@D Lim: I'm not sure what you are trying to argue. I am making the claim that a sequence of points with imaginary part approaching infinity cannot be Cauchy in the hyperbolic metric. If you want to prove this carefully, the formula for the hyperbolic metric should appear somewhere in your proof. After posting my answer I read over the exchange of comments between you and Theo Buehler and saw that the same issue came up in your discussion with him. So I suggest that you begin by focussing on proving this statement carefully. –  Matt E May 3 '11 at 15:05
@D Lim: One other thing: after reading your exchange with Theo it seems that you're a student, and this question may well be homework for a class. In such situations it's polite to put the [homework] tag on your question. (I am a professor and out of courtesy I prefer not to completely answer other professor's students' homework.) –  Matt E May 3 '11 at 15:09

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.

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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.

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It is a general fact that a Cauchy sequence $(x_n)_{n=1}^{\infty}$ in a metric space is bounded: Given $\varepsilon \gt 0$ we find $N \in \mathbb{N}$ such that $n,m \geq N$ implies $d(x_n,x_m) \lt \varepsilon$. In particular $d(x_N, x_m) \lt \varepsilon$ for all $m \geq N$. Putting $R = \varepsilon + \max{\{ d(x_N,x_i)\,:\,1 \leq i \leq N\}}$ we have that $x_{i} \in B_{R}(x_{N})$ for all $i \in \mathbb{N}$. –  t.b. May 3 '11 at 12:20
If I understand your first argument correctly, you're using that a closed and bounded subset of $\mathbb{H}$ wrt the hyperbolic metric also is closed and bounded wrt the Euclidean metric. Why exactly is that? (I see that Mariano has given this as a hint, but I see no argument). Also, why exactly is $\operatorname{Im}{p} = 0$ excluded? –  t.b. May 3 '11 at 12:27
@TheoBuehler No not really, what I said is assume that I have a closed and bounded subset of $\mathbb{H}$ and have a cauchy sequence $p_n \in (X,d)$. Then $\exists p_{n_k}$ such that it converges to a point $p$ in the set $X$, and so I use this fact to show that the hyperbolic metric goes to zero (i.e. that $d(p_{n_k},p)$ goes to zero in the hyperbolic metric. My problem is, firstly, is the proof above correct and secondly how can I argue that not just a compact subset of $\mathbb{H}$ is complete but also the whole of $\mathbb{H}$. –  fpqc May 3 '11 at 12:33
But why "Then $\exists p_{n_k}$ ..."? –  t.b. May 3 '11 at 12:34
@TheoBuehler, sorry should have said: "$\exists p_{n_k}$ such that it converges to $p$ in the Euclidean metric". –  fpqc May 3 '11 at 12:36