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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.2011-05-03
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    You may want to more or less rewrite the whole thing into being more understandable, really…2011-05-03

4 Answers 4

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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 $i \in \mathbb{N}$.2011-05-03
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    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?2011-05-03
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    @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 $\mathbb{H}$.2011-05-03
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    But why "Then $\exists p_{n_k}$ ..."?2011-05-03
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    @TheoBuehler, sorry should have said: "$\exists p_{n_k}$ such that it converges to $p$ in the Euclidean metric".2011-05-03
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    That's what I understood, but why exactly is that?2011-05-03
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    @TheoBuehler How can I generalise my argument to the whole upper half plane? I am really confused, I would appreciate if you could help. Thanks.2011-05-03
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    I'm trying my best to do so... But first we need to get your argument straight. Once again: how exactly do you conclude the existence of a convergent subsequence of $(p_n)$ wrt the Euclidean metric? My first comment should then answer your second question.2011-05-03
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    I live in a closed and bounded set, so it is compact, and so if I have a sequence that lives in that set, does it not follow that I have a subsequence that lives in the set and converges to a point in the set? I am just using what I know about $\mathbb{R}^n$ and trying to transfer that into the upper half plane.2011-05-03
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    Yes, I see that this is the punchline. So we've come full circle to my second comment, agreed?2011-05-03
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    Yeah ok. So I exclude the bit that $Im(p_n)$ being equal zero as this would contradict $p_n$ being a cauchy sequence in the hyperbolic metric.2011-05-03
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    Yes, that's what you need to show. In fact, that's precisely what Mariano was getting at with his hint, so you should try to come up with a good argument for that.2011-05-03
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    @Theo Buehler, I may have misunderstood you, you said "That's what I needed to show". Do you mean show that $Im(p_n)$ is never zero, or that I can generalise my argument from a compact subset to the whole upper half plane? If it is the former, $Im(p_n)$ cannot be zero as this would contradict $p_n$ being cauchy in $(X,d)$, $d$ the hyperbolic metric. If it were the latter, then can I say that as my choice of subset was arbitrary, I can generalise to the whole of $\mathbb{H}$? Thanks.2011-05-03
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    So far you have not given an argument establishing Mariano's hint, so I mean the former. You keep repeating that $\operatorname{Im}(p_n) \neq 0$. That's clearly true because $p_n \in \mathbb{H}$. But *why* does $\operatorname{Im}{p_n} \to 0$ (or $|p_{n}| \to \infty$ in the Euclidean metric) contradict the fact that $p_n$ is Cauchy wrt the hyperbolic metric? Now if you can give an argument for that, you've established Mariano's hint and you're essentially done. I'll explain that later.2011-05-03
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    I mean exactly what I wrote.2011-05-03
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    @TheoBuehler I think you mean that $|p_n|$ goes to infinity in the hyperbolic metric. In the euclidean metric everything is well and good. It's simply because I define hyperbolic distance $d_\mathbb{H}$ as the integral along the hyperbolic line joining two points with respect to the hyperbolic element of arc-length $\frac{|dz|}{Im(z)}$. So intuitively it tells me that if $Im(z) = 0$ something is going to go wrong and the integral blows up.2011-05-03
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    Ah, we're finally approaching an argument here! No, I really meant $|p_n| \to \infty$ wrt the Euclidean metric (maybe after passing to a subsequence). So now go and flesh out the details why a Cauchy sequence wrt hyperbolic metric must be contained in a set of the form as Mariano describes it (Mariano also speaks of the Euclidean norm, by the way).2011-05-03
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    @Theo Buehler I don't understand how can $|p_n|\rightarrow \infty$ in the euclidean metric??2011-05-03
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    I'm getting real confused now.2011-05-03
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    Once again (I'm sorry that I'm apparently confusing you more than I help): You're *assuming* that $(p_n)$ is a Cauchy sequence with respect to the hyperbolic metric. You *want to show* that $(p_n)$ is in a set as Mariano describes it. Assume by contradiction, it isn't. After passing to a subsequence, either $(\operatorname{Im}{p_{n_k}}) \to 0$ or $|p_{n_k}| \to \infty$ wrt the Euclidean metric.2011-05-03
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    @TheoBuehler Ok ok I can form a better picture in my mind now. Suppose $p_n$ does not stay in the set as Mariano has described. Then if $Im(p_{n_k})=0$, this would mean that $d(p_m, p_{n_k}) > \epsilon$ would approach infinity as $k \rightarrow \infty$ for some fixed $m$ in the hyperbolic metric contradicting $p_m$ being cauchy in the hyperbolic metric. On the other hand, I don't know how the second case could arise of $p_{n_k}$ approaching infinity in the euclidean metric.2011-05-03
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    Hmm... You *never* have $\operatorname{Im}{p_{n_k}} = 0$ because $p_{n_{k}} \in \mathbb{H}$. Here's the thing: If the sequence $(p_n)$ doesn't stay in any "Mariano-set" then either you can extract a subsequence such that $(\operatorname{Im}{p_{n_k}}) \to 0$ or you can extract a subsequence such that $|p_{n_\ell}| \to \infty$ by the definition of a "Mariano-set". Now show that both things cannot happen if $(p_n)$ is hyperbolic Cauchy, that is both lead to divergence wrt the hyperbolic metric. I guess it's only your intuition telling you that $|p_{n_\ell}| \to \infty$ cannot happen.2011-05-03
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    @Theo Buehler, sorry I think I wrote above that $Im(p_{n_k})$ is impossible as this would lead to divergence of the hyperbolic metric as my hyperbolic element of arclength is $|dz|/Im(z)$. Now in a closed and bounded subset of $\mathbb{H}$, is it not plain that $|p_{n_k}| \rightarrow \infty$ cannot happen in the euclidean metric (and hence cannot in the hyperbolic metric as $\log (x)$ is unbounded, and my numerator of the term inside $\log (x)$ is bigger than the denominator? As I said before, how do go from a closed and bounded subset to the whole of $\mathbb{H}$?2011-05-03
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    Well, that sounds good and can certainly be made into an argument! I didn't say you were far away from getting where you want... Now write this up in detail! *Hypothesis:* $(p_n)$ is Cauchy wrt hyp. metric. *Claim:* There exist $\varepsilon \gt 0$ and $R \lt \infty$ such that $p_n \in \{z\,:\,\operatorname{Im}{z} \geq \varepsilon, |z| \leq R\}$. *Proof:* suppose not. Then either... or ... and give your argument in the last comment with a bit more details.2011-05-03
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    @Theo Buehler at last: Suppose $p_n$ is cauchy in $h$, the hyp. metric. Claim: $\exists \epsilon>0, R < \infty$ such that $p_n \in \{z : Im z \geq \epsilon, |z|\leq R\}$. Assume not. Then passing to a subsequence if necessary, then either \exists $p_{n_k}$ such that $Im(p_{n_k}) = 0$ or $|z| > R$ in the euclidean metric. Now first case is impossible as this would mean that $d(p_m, p_{n_k}) > \epsilon$ $\forall n_k > m$, $m$ fixed contradicting $p_n$ being a cauchy sequence. See next comment.2011-05-03
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    @TheoBuehler The second case is impossible as well as this would mean that $|p_{n_k}|$ diverges in the euclidean metric and hence diverges as well in the euclidean metric (as $\log (x)$ is unbounded. So we are done. Now how do I generalise this to the whole upper half plane and not just some compact set?? I have tried hours on end to nothing.2011-05-03
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    Do you mean $\operatorname{Im}(p_{n_k}) \to 0$, above? The rest seems ok. But then you have what you want: You started with a Cauchy sequence in $\mathbb{H}$. The entire Cauchy sequence is in a "Mariano-set", which is closed and bounded in the Euclidean metric. Now extract a subsequence which converges in the Euclidean metric. Then you show that this subsequence converges in the hyperbolic metric as well. Since a Cauchy sequence has at most one accumulation point (in any metric space) and if it happens to have one then it converges, the sequence you started with converges as well. Done.2011-05-03
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    @Theo Buehler Yeah sorry it was meant to be that. I understand what you mean that the original sequence converges (That's pretty easy using the triangle inequality) but I have just shown it converges in the compact subset and not $\mathbb{H}$, no???? I am just getting real confused. By the way I am a first year student and we're just learning how about determinants!!2011-05-03
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    @TheoBuehler Oh wait I get it I get it I just assumed that $p_n$ was arbitrary anywhere in $\mathbb{H}$ so infact my argument is general and I'm done! No????2011-05-03
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    Yes, but let $p$ be the limit point in that compact set. Then this means that $d_{h}(p_n,p) \to 0$, no? But this is completely independent of whether $p_n,p$ belong to this compact set or not, isn't, it? So the sequence $p_n$ converges to $p$ in $\mathbb{H}$ as well. Finally, no worries, I know about your level, Ben :)2011-05-03
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    Ah, you've seen it yourself! **YES, EXACTLY!** (I wrote my previous comment without seeing your last one).2011-05-03
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    @TheoBuehler Ahhhh that feels soooo goood man woohoo!!2011-05-03
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    Ok, great :) I think you've earned a biiig loong break and a big pot of coffee! See you around!2011-05-03
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    @Theo Buehler I would like to use my answer somehow in this proof, can I just say (I mean before my answer) that because of the facts we have just discussed, $p_n \in \mathbb{H}$ implies it is a cauchy sequence in some closed and bounded set?2011-05-03
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    I'm not sure I understand your question. If $p_n$ is a Cauchy sequence then it lies inside a bounded set (that's my very first comment). Now our long discussion established that such a sequence lies in a "Mariano-set". This makes the second paragraph ("Now as...") precise. Inserting the main point of our discussion there yields a complete argument (when you insert "and since a Cauchy sequence has at most one accumulation point, the entire sequence $(p_n)$ must converge") before the paragraph "Another way...".2011-05-03
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    @Theo Buehler Thanks very much. I think I've got it now. Man that was exhausting!!2011-05-04
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    @Theo Buehler Oh I noticed a flaw in the proof. You know when I said that assuming its not bounded we have the numerator of the log function being bigger than the denominator, etc... It's not true, as taking one element, either $x$ or $y$ and letting it approach infinity in the formula for $d_\mathbb{H}$, I actually get that the terms inside the log function go to 1 and hence log of that goes to zero. But I think I'm rescued by the fact that every cauchy sequence is bounded. What do you think??2011-05-04
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    Ben: Yes, I think this is correct. Sorry for the late response, but I was quite busy the last few days. Are all your doubts resolved now (I figure that's the case by your acceptance of your own answer)?2011-05-07
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    Yeah, don't worry about it I used the reverse triangle inequality to get me out of jail.2011-05-07
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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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$\def\eps{\varepsilon}$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.2011-05-03
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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?2011-05-03
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    @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 $x \in \mathcal H$.2011-05-03
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    Note: the preceding comment is a response to a now deleted request for clarification.2011-05-03
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    @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.2011-05-03