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I am currently trying to understand the paper by Krioukov et. al. on hyperbolic networks, but since I do not have a background in hyperbolic geometry (or, in that sense, in geometry at all) I struggle to understand some points they are making.

For instance, they aim to compute the distance $x$ between two points $(r, \theta)$ and $(r', \theta')$ in the hyperbolic plane by using the hyperbolic law of cosines:

\begin{equation} \mathsf{cosh}(kx) = \mathsf{cosh}(kr) \mathsf{cosh}(kr') - \mathsf{sinh}(kr)\mathsf{sinh}(kr')\mathsf{cos}(\Delta\theta) \end{equation}

Here, $K=-k^2<0$ refers to the curvature, and $\Delta \theta = \pi - |\pi - |\theta - \theta'||$. I can intuitively follow this equation since it is similar to the regular law of cosines in the euclidean plane. However, in the next step they approximate it by assuming that $kr, kr'$ is large, and $\Delta \theta > 2 \sqrt{e^{-2kr}+e^{-2kr'}}$ as follows: \begin{equation} x = r + r' + \frac{2}{k} \ln \sin \frac{\Delta\theta}{2} \approx r + r' + \frac{2}{k} \ln \frac{\Delta\theta}{2}. \end{equation}

I am completely lost there, especially their assumption on $\Delta \theta$. Why does it make sense and how does it help simplify the fomula?

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up vote 3 down vote accepted

The derivation is a little long, and I admit to being mystified by their assumption on $\Delta \theta$ - it merely needs to be that $\Delta \theta$ is small compared to something $O(1)$. But anyway...write

$$\cosh{k x} = \cosh{k r} \cosh{k r'} (1-\tanh{k r} \tanh{k r'} \cos{\Delta \theta})$$

and note that $\lim_{t \rightarrow \infty} \tanh{t} = 1$. We can then write

$$\cosh{k x} \approx 2 \cosh{k r} \cosh{k r'} \sin^2{\left ( \frac{\Delta \theta}{2} \right )}$$

Use the definition of cosh and that $k r$ and $k r'$ are both large to derive the following equation:

$$\exp{(2 k x)} - \exp{(k (r + r'))} \sin^2{\left ( \frac{\Delta \theta}{2} \right )} \exp{(k x)} + 1 \approx 0 $$

This gives

$$ \exp{(k x)} \approx \exp{(k (r + r'))} \sin^2{\left ( \frac{\Delta \theta}{2} \right )} \frac{1}{2} \left [ 1 \pm \sqrt{1- 4 \exp{(-k (r + r'))} \csc^2{\left ( \frac{\Delta \theta}{2} \right )}} \right ] $$

The sqrt term on the right is very small. Assume the +ve solution, so the term in brackets is about 2. Take logs of both sides and get

$$x \approx r + r' + \frac{2}{k} \log{\sin{\left ( \frac{\Delta \theta}{2} \right )}} $$

I hope this helps.

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In fact, now that I look at it, I think the stated restriction on $\Delta \theta$ is there to keep that factor in the sqrt small. –  Ron Gordon Jan 4 '13 at 16:15
Indeed it does. How on earth did you see that? Anyway, if I am not mistaken there are 2 minor mistakes that cancel each other out: First, in your 3rd equation, the 2 should disappear, since you have 2 cosh on the right side, which both multiply a factor 1/2 when converting to exponentials. Second, I think you forgot to square the right term under the squareroot. If one corrects those steps, the $\log 2/k$ summand disappears as well. –  HdM Jan 4 '13 at 16:46
I believe you are right! Thanks - what teamwork! –  Ron Gordon Jan 4 '13 at 16:50
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