Math 333 Key to Homework # 2 October 5, 1998
1. Plane dual to Fano's geometry axioms. (5 pts)
a. There exists at least one point.
b. There are exactly three lines on every point.
c. Not all lines are on the same point.
d. There is exactly one point on any two lines.
e. There is at least one line on any two distinct points.
Representation of the model and justification. (5 pts).
Fano's geometry has the property of being self-dual; i.e., the plane dual axioms of Fano's geometry are all true in Fano's geometry. Therefore, a model for the plane dual axioms is any model of Fano's geometry.
2. Show that each axiom of Fano's Geometry is independent. (2 pts for each axiom).
a. Axiom 1: a single point and no lines. The model "no points and no lines" would work vacuously for any set of axioms. Therefore, I gave credit for it but I really didn't like it.
b. Axiom 2: Points are A, B, C, and there are 3 lines {A,B}, {A,C}, and {B,C}.
c. Axiom 3: Points are A, B, C, and one line consisting of all three points {A,B,C}.
d. Axiom 4: Points are A, B, C, D, and the lines are {A,B,C} and {A,B,D}.
e. Axiom 5: A model for Young's geometry works. Let the points be A B, C, D, E, F, G, H, I, and the lines be the columns in the following table. (Note: There are nine points and twelve lines.)
| A |
D |
A |
A |
B |
B |
B |
C |
C |
D |
G |
H |
| B |
E |
D |
E |
E |
D |
F |
F |
E |
H |
H |
F |
| C |
F |
G |
I |
H |
I |
G |
I |
G |
C |
I |
A |
3. Prove: Fano's geometry consists of exactly seven lines. (5 pts for proving at least seven, and another 5 pts for proving there can be no more than seven lines.)
Proof: First we show that there are at least seven lines. We know by Ax 3 there is a point P not on a line l. l has 3 points by Ax 2. By Ax 4 there is a line from P to each of the three points on l, call the points A, B, and C on l and the lines l1, l2, and l3 respectively. Note that all four of these lines are distinct. These lines each have 3 points and the third point on each of lines l1, l2, and l3 can not be A, B or C because of Ax 4. In particular, if not, there would be two distinct lines that had two points in common, P and one of A, B, or C. So let's call the third point on l1 D, the third point on l2 E and the third point on l3 F. D, E, and F are all distinct points. If not, Ax 4 is violated. By Ax 4, A must be joined to F and C must be joined to D by two additional distinct lines, say l4 and l5. Note that B is not on either of l4 and l5 . Finally B must also be joined to D and F, and hence we have at least seven lines.
Suppose there is an eighth line l8. By Ax 5, it must intersect l. Without loss of generality (wlog) we can assume it intersects l in A. L8 has two other points by Ax 2. However, neither of these two points can be B,C,D,E,F or P because of Ax 4. Note also that we already know that Fano's geometry has exactly seven points by a Theorem proved in class. Hence l8 has two additional points beyond the seven and that's a contradiction.
4. Prove: In Fano's geometry, each point is on exactly three lines. (5 points for at least 3 lines; 5 points for at most 3 lines. If you did not start with an arbitrary point, 2 points were deducted.)
Proof: Let P be any point. Then there is a line l with P not on it. This follows from a previous result proved in class about the existence of any point P and two other points in Fano's geometry not all collinear. This line l has three distinct points by Ax 2. By Ax 4 here are lines from P to each of these three points and these lines are distinct by Ax 4. Hence we have at least three lines on P.
Suppose there is a fourth distinct line l4 on P. Then by Ax 5, l4 and l have a point in common. However if it is any of the three points on l then l4 equals one of the other three lines already considered above by Ax 4. Hence there must be a fourth point on l but that contradicts Ax 2.