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Given the Poincare Disc $D$ and its ideal boundary $S^{1}$, I want to construct a homeomorphism between $S^{1}$ and the gromov boundary of $D$, $\partial D$, of equivalence classes of geodesics given some base point p.

I can construct a bijection between $S^{1}$ and $\partial D$ by identifying equivalence classes of rays with unique points on $S^{1}$ by noting that each line approaching infinity in $D$ is the unique radius for a given angle (so these lines are unique and so their intercepts in $S^{1}$ must also be unique, hence a bijection). What I am struggling with is showing that this bijection is continuous (it is not necessary to show continuous inverse in this case). I would like to show that, given a base point p and 2 geodesics from p to $x, y \ \epsilon \ \partial D$ where $x \neq y$ that: if $u$ on the line px and $v$ on the line py are close together (i.e. $d_{D}(u, v)$ is small), then the geodesics from $x , y$ to p are close together (i.e. the gromov product between the line px and py is large); and vice versa. Ideally, a relationship between $\alpha$, the angle between the 2 geodesics at p, and the length of the geodesic (I say length as this triangle xyp will be isoscles), would be most helpful.

Any help would be much appreciated!

Edit: To clear things up: if x, y both lie on the unit circle (the boundary of the poincare disc) and x$\neq$ y, and two points u and v in D, then given a base point p in D, we can define a geodesic L1 passing through p, u and x and a second geodesic L2 passing through p, v and y. Let the angle between the two geodesics at p be $\alpha$, and both L1 and L2 have length a. I also define L3, the line from u to v of length b. I would like to show that the gromov product of u with v (=$a - \frac{1}{2} b$ here) is dependent only on alpha. Hopefully that clears things up, sorry for not being more specific earlier.

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My interpretation of your question was different from Igor's. (That two different people can interpret your question differently is a bad sign.) I understood you to want the hyperbolic distance from p to the infinite geodesic joining x and y, in terms of the angle between xp and yp. This is easily computed from the Euclidean distance. I might have made a slip, but I got the Euclidean distance to be $\cos \alpha/2+\tan\alpha/2(\sin\alpha/2-1)$. –  HJRW Jan 7 '11 at 22:34
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2 Answers

up vote 4 down vote accepted

Let me first reformulate the question once again. Given a basepoint $p$ on the hyperbolic plane and two boundary points $x,y$, there are two measures of $x$ and $y$ being close as seen from $p$: the angle $\alpha$ between the geodesic rays issued from $p$ in the direction of $x$ and $y$, respectively, and the Gromov product $G=(x|y)_p$. The author wants to know whether $\alpha\to 0$ iff $G\to\infty$.

Indeed, as suggested by Igor, one can deduce it from the hyperbolic cosine law. As FuriousDee has already noticed, if $u$ and $v$ are two points on the hyperbolic plane with $d(p,u)=d(p,v)=a$, then $(u|v)_p=a-b/2$, where $b$ is determined from the equation $$ \cosh(b)=\cosh^2(a)-\sinh^2(a)\cos(\alpha) \;, $$ and $\alpha$ is the angle between $u$ and $v$ as seen from $p$. Now, in order to obtain the relation between $\alpha$ and $G$ one has to let $a$ go to infinity.

However, it is much easier to obtain the limit formula by doing a bit more of elementary geometry. Namely, let $w$ be the point on the geodesic $(x,y)$ such that the geodesic segment $(p,w)$ is perpendicular to $(x,y)$. Then the Gromov product $G$ is $d(p,x)-d(w,x)=d(p,y)-d(w,y)$. Of course, the distances in this formula are all infinite; nonetheless their differences still make sense if one understands them as the values of the Busemann cocycles determined by the boundary points $x$ and $y$. Now, the best way to deal with Busemann cocycles explicitly is to consider the upper half-space model, where the Busemann cocycle with respect to the point at infinity is just the logarithm of the ratio of Euclidean heights of the corresponding points. So, let $x$ be the point at infinity, and $p$ be the point with the coordinates $(0,1)$. Then the geodesic ray $(p,x)$ is just the vertical ray issued from $(0,1)$, whereas $(p,w)$ is an arc of the half-circle perpendicular to the boundary line at the endpoints, passing through the point $p$, and perpendicular at the point $w$ to the vertical line $(x,y)$ - therefore, the center of this half-circle is precisely the point $y$ on the boundary line (at this point it would actually be much easier just to draw a picture). Denote by $R$ the radius of the latter half-circle, then the value of the Busemann cocycle in question, i.e., the Gromov product $G$, is $\log R$. On the other hand, the angle between the Euclidean line joining the points $y$ and $p$ and the horizontal line is precisely the angle between $(p,x)$ and $(p,w)$, i.e., $\alpha/2$. Therefore, $R=1/\sin(\alpha/2)$, whence $G=-\log(\sin(\alpha/2))$.

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Problem solved. Thanks guys –  FuriousDee Jan 8 '11 at 11:21
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If I understand correctly, you are looking for the hyperbolic Law of Cosines http://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines

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This seems to help: I have the gromov product of u with b at a - b/2, and the cosh law gives cosh(b) = (1- cos($\alpha$))($cosh^{2}(\alpha)$) + cos($\alpha$) Would I be on the right lines here? –  FuriousDee Jan 7 '11 at 23:13
    
that should read u with v –  FuriousDee Jan 7 '11 at 23:19
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