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Suppose that $X$ and $Y$ are two loxodromic isometries of the hyperbolic space and that the product $XY$ is also a loxodromic element. We consider the axes of these three elements. I'd like to know if we can say something about the mutual position of these axes, for instance if the fixed points of $XY$ have to respect some special placement with respect to the ones of $X$ and $Y$, or if we can say something about the angles between axes and common perpendicular lines and so on. I couldn't find any book or paper covering this topic, but also a reference would be greatly appreciated. Thank you.

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Can you clarify your use of the term “loxodromic” in this context? I know loxodromic Möbius transformations, but I'd classify hyperbolic isometries using terms from Euclidean isometries. So I assume you refer to those with 2 ideal fixed points, i.e. translations and perhaps glide reflections. In general, the placement of the axis of $XY$ isn't fixed by the other two axes, since it also depends on the displacement lengths of the component transformations. The combined axis will be closer to the component axis with larger displacement. – MvG Feb 9 '13 at 3:55
What do you mean by 'the hyperbolic space'? Two-dimensional, three-dimensional, arbitrary-dimensional? – Tara B Feb 9 '13 at 13:05
I know of one case where you can say something nice. If you have two axes $A$ and $B$, and they intersect, then ou can take a point on $A$ half a translation distance away, and do the same for $B$. Their product then has axis through those two points. The proof is by writing a hyperbolic isometry as a product of two reflections; then when you take the product of these two isometries, the reflections "in the middle" cancel out. I haven't thought how to do this in the general case, but it might be extendable, and at the very least might give you something to try! – user641 Feb 14 '13 at 10:10
Oh, I should add that with the new axis you've constructed, the translation distance is twice the distance from those two "midpoints" you constructed on $A$ and $B$. So this really does determine the isometry. – user641 Feb 14 '13 at 10:13
Sorry, I only just saw your replies today. I would have thought this would be in "The Geometry of Discrete Groups", by Alan Beardon, but I haven't found it so far. The case for a product of two-dimensional hyperbolic isometries is covered on pp.182-184. Does that help at all? – Tara B Feb 20 '13 at 17:31

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