Office hours: Next few weeks: Tue 12:30–13:50 in 709 Evans.
2 (due 2/2) Chapter 1. [Euclid : Books I–IV.] Problems: 3.4, 3.5, 3.10, 4.7. These problems may be postponed until 2/9: 5.13, 5.15, 5.18.
Starting from homework 3, all the proofs must be complete: based on existing axioms, definitions, theorems etc, and not on "intuition as in high school".
3 (due 2/9) Sections 6, 7. Problems: 6.3, 6.7, 6.10. These problems may be postponed until 2/16: 7.2, 7.9, 7.10, 7.11.
4 (due 2/16) Definitions:
On a projective plane, given two lines l,l' and a point P not on these lines, the projection l --> l' with center at P is the mapping which sends a point L on l to the unique point L' on intersection of l' with line PL.
In what follows, a triangle is 3 points not on the same line. A configuration C on an incidence plane I is a collection of lines and points of I. An automorphism of C is a permutation f of points of C such that if points P,Q from C are on a line belonging to C, then f(P),f(Q) are also on a line belonging to C.
1. On a projective plane, consider two triangles: ABC and A'B'C' such that lines a=AA', b=BB', and c=CC' intersect at one point P, and 7 points P,A,B,C,A',B',C' are distinct. Then there is a unique point C'' such that projection pc with center at C' from a to b sends A,A' to B,B'; likewise, there is a unique point A'' such that projection pa with center at A'' from b to c sends B,B' to C,C'; and a unique point B'' such that projection pb with center at B'' from a to c sends A,A' to C,C'. Show that the following statements are equivalent:
1) for any choice of A,B,C,A',B',C' as above, the composition of pc and pa coincides with pb;
2) for any choice of A,B,C,A',B',C' as above, A'', B'', C'' are on the same line.
Update: when you show (1) ==> (2), you may assume that P, A'', C'' are not on the same line.
2. If A'',B'',C'' are on the same line, then the configuration of 10 points A,B,C,A',B',C',P,A'',B'',C'' and 10 lines a,b,c, ABC'', ACB'', BCA'', A'B'C'', A'C'B'', B'C'A'', A''B''C'' is called the Desargues configuration D. (Actually, "other coincidences" may happen; for example, P may be on the line A''B''C''; we assume they do not happen.) Show that for any two points X,Y of D, there is an automorphism of D which sends X to Y.
3. Calculate the number of automorphisms of the Desargues configuration D from the preceding problem. (Hint: given a point X of D, what is the number of automorphisms which send X to X?)
4. A "flat" on a Desargues configuration D is a set of 6 distinct points R,S,T,U,V,W of D such that V,R,S are on the same line of D, and same holds for subsets {V,T,U}, {W,S,T}, {W,R,U}. Show that the number of flats in D is 5. Show that for any permutation f of the set of flats, there is a unique automorphism of D which gives this permutation of flats.
Sketch (2) ==> (1): Look what changes when one replaces a point A' by a point A''', and chooses points B''', C''' so that A'''B''' passes through C'' (same as A'B' does), and B'''C''' passes through A'' (same as B'C'). (Essentially, we replaced A'B'C' by A'''B'''C''' without changing points A'' and C''). The point B'', would it change? (We must replace it by the intersection of AC and A'''C'''.) By (2), it must remain on the line A''C''; and it remains on the line AC. Unless the latter two lines coincide, the point B'' does not move (which proves (1)). If they coincide, the point A'' of intersection of BC and B'C' is on AC; so A''=C, so B'C' passes through C - contradiction.
Sketch (1) ==> (2): Consider the points of intersection of A''C'' with lines a,b,c; composition of pa and pc sends one of them to another (call them X and Y). By (1) X,Y,B'' are on the same line; unless X=Y, this finishes the proof. If X=Y, then they coincide with P; so P,A'',C'' are on the same line. Obviously, they are distinct, hence unless B'' is on this line, the line A''B'' does not pass through P. Now replace A''C'' in the argument above by A''B''.
2 and 3. Accumulate some examples. First, one can exchange A with B, A' with B', and A'' with B''. (Likewise for B and C; or for A and C). Second, one can exchange A,B,C with A',B',C' correspondingly. This still leaves P fixed, and A'', B'', C'' not mixing with A,B,C,A',B',C'; but it shows that: (i) all 3 lines passing through P may be freely exchanged, and (ii) on any of these 3 lines two points (those which are not P) may be exchanged simultaneously with exchanging corresponding pairs points on two other lines passing through P (and remaining 3 points being fixed). We must get a way to mix these 3 groups together.
Make a leap of faith in the fact that all the points are created equal: try to apply the same construction to A instead of P: there are 3 lines passing through A (a, AC'' and AB''), and there is a pair of points not equal to A on all of them; try exchanging these pairs: P with A', B with C'', C with B'', and leaving the remaining 3 points B',C',A'' fixed. This gives a permutation of 10 points; we MUST check that any of 10 "lines" (i.e., a triple of points of configuration) is permuted to another line. (Here is the proof: the "other" 3 points are on the same line; it is fixed; 3 lines through A are OK - fixed - by the construction; BCA'' <--> C''B''A''; A'B'C'' <--> PB'B; A'C'B'' <--> PC'C.
That's it: THIS example shows that P may be mixed into the subset A,B,C,A',B',C', and the same for C''. Hence any point may be sent to P. We saw that there are 6*2 automorphisms which send P |--> P (6 from permutation of a,b,c, and 2 from permutation of not-equal-to-P points on these lines). Since P can be send to any of 10 points, the total is 120 automorphisms.
4. Assume this checked: any point X on a flat has 2 lines passing through it; each of these lines has 3 points of the flat (denote other points on these lines Y,Y' and Z,Z'). For a flat which contains P, these lines are either a,b, or b,c, or a,c. If they are a,b, then {Y,Y'}={A,A'} and {Z,Z'}={B,B'}; hence the 6th point of the flat must be on AB and A'B', so it must be C''.
Hence there is a unique flat which contains P and has a,b as lines of the flat; hence there are 3 flats containing P. P can be replaced by any of 10 points, so there are 30 pairs a flat + a point on the flat; therefore there are 5 flats. Since flats are defined completely in terms of "a line contains a points", automorphisms permute flats; we obtain a homomorphism from the group of automorphisms into the group of permutations of flats.
Comparing 120 with 5!, it is enough to check that this homomorphism has no kernel. An element of kernel would send a flat to itself; in particular, it preserve 3 flats which contain P; but P is the only point of intersection (check!) hence it would preserve P. Likewise, it would preserve any point.
5 (due 2/23) Section 8. Problems: 8.3, 8.5, 8.7, 8.8, 8.9.
Problem 6: A 3-relation on a set S: it assigns to a triple A,B,C of elements of S one of two values: TRUE or FALSE. Consider a 3-relation denoted as A!B!C; call it symmetric provided A!B!C is TRUE if and only if C!B!A is TRUE. We say that it has a property On (with n > 2) if given n distinct elements of S, there are exactly 2 ways to enumerate them as X1, X2, ..., Xn so that Xr ! Xs ! Xt is TRUE if and only if the index s is between r and t. Show that for a symmetric 3-relation, properties O3 and O4 (taken together) imply that On holds for any n > 2. (Hint: Is it possible that A!X!B, B!X!C, and C!X!A?)