Is there a compass and straightedge construction of parallel lines in hyperbolic geometry?
That is, given a line and a point not on the line, construct a line parallel to the given line.
[Math] compass and straightedge construction of parallel lines in hyperbolic geometry
geometric-constructionshyperbolic-geometrymg.metric-geometry
Related Solutions
I think they're equivalent. Clearly 3 ==> 2. To see 2 ==> 3, say an extension E/L of fields is 2-filtered if it has a tower of subextensions each of degree 2 over the next. Then note firstly that if E/L and E'/L are 2-filtered, then so is the compositum of E and E' in any extension of L containing both (induct on the length of the tower of, say, E); and secondly for any map f:E-->E' of extensions of L, if E is 2-filtered then so is f(E). Together these two facts imply that in 2 we may assume that K is Galois over Q, say with group G. Then Q(z)/Q corresponds to a subgroup H, and by Galois theory we just need to prove the following group-theoretic lemma:
Lemma: Let G be a 2-group and H a subgroup. Then there is a chain of subgroups connecting H to G where each is index 2 in the next.
Proof: Recall that we can find a central order two subgroup Z of G. If H\cap Z=1, then HZ gets us one up in the filtration. Otherwise H\cap Z=Z, i.e. Z is in H, and we can use induction on |G|, replacing G by G/Z.
A bounded-length straightedge can emulate an arbitrarily large straightedge (even without requiring any compass), so the rusty compass theorem is sufficient.
Note that, in particular, it suffices to show that there exists an $\varepsilon > 0$ such that a straightedge of length $1$ is capable of joining two points separated by any distance $\leq 1 + \varepsilon$ (and therefore emulates a straightedge of length $1 + \varepsilon$, and therefore arbitrarily long straightedges by iterating this process).
We can use Pappus's theorem to establish this result for any $\varepsilon < 1$:
https://en.wikipedia.org/wiki/Pappus%27s_hexagon_theorem
In particular, given two points $A$ and $c$ (separated by a distance slightly greater than 1) which we wish to join, draw a line $g$ through $A$ and a line $h$ through $c$ which approach relatively close to each other. Then add arbitrary points $B, C$ on $g$ and $a, b$ on $h$ such that the four new points are within distance $1$ of each other and the two original points. We assume wlog $b$ is between $a, c$ and $B$ is between $A, C$.
Then we can construct $X$ by intersecting the short (length $< 1$) lines $Ab, Ba$, and construct $Z$ by intersecting the short lines $Bc, Cb$. Then $Y$ can be constructed by intersecting the short lines $XZ$ and $Ca$.
Now, $Y$ is positioned collinear with $A$ and $c$ and between them, so we can use it as a 'stepping stone' to draw a straight line between $A$ and $c$.
The result follows.
EDIT: I decided to do this with the edge of a coaster and two points slightly too far apart to be joined directly:
Best Answer
The quickest way to get you started is to refer you to my article, reference [5] (a pdf) on
http://en.wikipedia.org/wiki/Squaring_the_circle
and then to the fourth edition (2008) of Marvin Jay Greenberg's book, which is reference [6].
I'm guessing what you want is Bolyai's construction, given a line and a point off the line, of the two rays through the point that are asymptotic to the line, one in each direction. When I wrote the article, I relied on an earlier edition of Marvin's book, along with The Foundations of Geometry and the Non-Euclidean Plane by George E. Martin, which has a nice little section at the very end. There is also, now, Geometry: Euclid and Beyond by Robin Hartshorne.
The most complete reference I know on constructions is in Russian, by Smogorshevski, other very helpful books by Kagan and by Nestorovich. Of course, at this point I have my own versions of it all.