### Solutions for First Test, Sep 19

1
The first example here is hyperbolic; the rest are neither hyperbolic nor affine nor projective.

In the first case, given any line l and point P, consider the two lines in the ambient plane which would intersect l on the unit circle. In other words, if l intersects that boundary circle at X, Y (not points in our model S), then PX and PY are two parallels to l through P.

In the other three cases, there are sometimes lines and points with unique parallels, and sometimes there are many parallels. To see the first case, take a line l for which the entire Euclidean line is in S, and in fact the Euclidean halfplane on one side of l is entirely in S, and take a point P in this halfplane. To see the second case, take a line l that passes through the "hole" missing from S. Given a point P, there is one line through P that is Euclidean-parallel to l, and another that would intersect l in this hole. Both are "parallel" in the model S.

2
This was problem 4.10 from the homework; we repeat the solution:

Starting from any projective plane (P,L,F), the claim is that (L,P,F') is another projective plane, whose points are the lines of (P,L,F) and vice versa.

We check the three axioms: Two points l,m of (L,P,F') determine a unique line p because the lines l,m of (P,L,F) intersect in the unique point p.

Similarly, two lines p and q of (L,P,F') intersect in the unique point l (which is the line of (P,L,F) through p and q).

Finally, let p,q,r,s be the four points of (P,L,F) guaranteed by axiom 3. Then lines pq, qr, rs, sp are four points of (L,P,F') which satisfy the same axiom.

3
The Poincare halfplane is a hyperbolic plane. For axiom 3, take for instace the four points (0,1), (1,1), (2,1), (3,1). No three lie on a line, since neither a vertical ray nor a semicircle can intersect a horizontal Euclidean line more than twice.

Two points determine a unique line: if they are vertically separated, this is a vertical ray; otherwise it is a semicircle. (To construct that semicircle, working in Euclidean geometry, we could note that any circle through P and Q has center on the perpendicular bisector of PQ. Since PQ is not vertical, this bisector has a unique intersection with the x-axis, which is the center point for our semicircle.)

4
Note that this model is isomorphic to the real projective plane whose points we defined as the set of all lines through the origin in R3, because a line through the origin intersects the sphere in two antipodal points, and two antipodal points determine a unique line through the origin in space. This observation is not necessary to solve this problem, but is useful to keep in mind. (We could think also of great circles as planes through the origin in space.)

Any great circle through P also passes through the antipodal point -P. If {P,-P} is distinct from {Q,-Q}, then P and Q are not antipodal, so there is a unique great circle through P and Q, which does also pass through -P and -Q. Thus it is the unique line through these two points of the Riemann projective plane.

Any two great circles on the sphere intersect; if they intersect at P, then they also intersect at -P, but at no other points. But on the Riemann projective plane, {P,-P} is a single point, the unique intersection of the two lines.