The Maximal Number of Transverse Self-Intersections of Geodesics ...

Outline of Proof: Now a similar situation occurs if the word is of odd length. Since the
word is odd, not all of the cyclic permutations of it have symmetrical counterparts (assuming
the above lemma holds). For example, if W = a 4 b 3 ,weseethatba 4 b has no symmetrical
partner. This means there is one less geodesic with roots near −φ +1and φ. Therefore,
the same lattice structure will occur in the fundamental region, except this time we count
n+1
− 1 intersections from the positive portion of H, and n+1 − 2 intersections from the
2 2
negative portion. These intersect in ¡ n+1
− 1 ¢¡ n+1
− 2 ¢ = ¡ ¢
n+1 2
− 3 ¡ ¢
n+1
2 2 2
2 +2 ways. The
graph below demonstrates this.
2
1.5
1
0.5
–1 –0.6 0 0.2 0.4 0.6 0.8 1
x
4 Miscellaneous Program Data and Observations
When this project began, it was my intent to use my program to check previous work dealing
with classes of curves. In particular, I wanted to verify that specific classes of curves were
the unique classes that generated a specific number of necessary self intersections. To do
this, I would find the possible combinations of k +1 simple loops, make these words minimal
(with the help of [CR]), and run them through my Maple script to ensure only specific classes
of words had k necessary self-intersections. Because of time constraints, I was only able to
verify this for k =2. The following is the theorem we wished to verify.
Theorem 4.1 (DIW, Theorem 3.2) The conjugacy class in π 1 (T) ofalooponT with
two non-trivial self-intersections is one of
(a) [(abAB) 3 ] or [(baBA) 3 ]
(b) [g(aabABabAB)]
(c) [g(abAbaB)]
(d) [g(aaabAB)]
(e) [g(abaBabAB)]
(f) [g(aabAAB)]
16