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

Continuing in this manner, let k =1, and we see that the number of intersections of W
is (β 1 − 1)(α 1 − 1), since we disregard the first a and the last b and the b’s must intersect
the a’s in this particular lattice, as defined above. More specifically, if we assume the worst
case, each b must intersect all the a’s prior to it (with the exception of the first a), and all
the a’s must intersect all b’s prior to them. Notice this number matches the bound listed
above.
Suppose k =2, and we see that the number of intersections is β 1 (α 1 − 1) + α 2 β 1 +(β 2 −
1)(α 1 + α 2 − 1) = (α 1 + α 2 − 1)(β 1 + β 2 − 1), thebound.
Checking for k =3, we see the number of intersections is
β 1 (α 1 − 1) + α 2 β 1 +β 2 (α 1 +α 2 −1) + α 3 (β 1 +β 2 )+(β 3 − 1)(α 1 + α 2 + α 3 − 1) = (α 1 + α 2 +
α 3 − 1)(β 1 + β 2 + β 3 − 1); again, the bound. In each case, we see that each exponent above
the a’s is multiplied by each exponent ³X
above the b’s, then we subtract each of these exponents
once and add 1; thisalwaysgivesus (a exponents) − 1´³X
(b exponents) − 1´
.
By induction, we’ll assume that it holds for u that the number of intersections of W =
a α 1
b β 1
a α 2
b β 2
...a α k−1 b
β k−1a α u
b β u
is (α 1 + α 2 + ... + α u − 1)(β 1 + β 2 + ... + β u − 1) = m u . Now
we’ll add an extra a q b p to the end of the word. So the number of intersections of W become
β 1 (α 1 − 1) + α 2 β 1 + β 2 (α 1 + α 2 )+... +(q − 1)(α 1 + α 2 + ... + α u + p − 1) =
m u +b(α 1 +α 2 +...+α u −1)+q(β 1 +β 2 +...+β u )+(p−1)(α 1 +α 2 +...+α u +q −1)+1 =
(α 1 + α 2 + ... + α u + q − 1)(β 1 + β 2 + ... + β u + p − 1); thebound.
Alternatively, we can count this directly by simply looking at the number of intersections
of the "squares" generated by the intersection of each band. We know that the c 1 band
intersects every other d band resulting in at most (α 1 − 1) intersections. Similarly, the d k
band intersects every other c band resulting in at most (β k − 1) intersections. So we can
add the intersections in the following way:
int. ≤ (α 1 −1)(β 1 +β 2 +...+β k −1)+α 2 (β 1 +β 2 +...+β k −1)+...+α k (β 1 +β 2 +...+β k −1)
=(α 1 + α 2 + ... + α k − 1)(β 1 + β 2 + ... + β k − 1), thebound.
Corollary 3.2 Let W beawordoflengthn. Let S W be the minimum number of necessary
self-intersections for a curve represented by W .
Then
½ (
n
2
max{S W } =
− 1)2 ,n even
¡ n+1
¢ 2
− 3 ¡ ¢
n+1
.
2
2 +2,n odd
Proof. By Theorem 3.1, the bound for W is (k − 1)(n − k − 1), withk defined above.
Using n calculus, we find the maximum of this polynomial to be:
¡ o 1 n − 1¢ 2
, at ©£ k = 1n¤ª , as noted above. The only restriction is that n be an
2 2 2
integer; in other words n must be even.
Let us look at the closest value that yields an odd word near the maxima. Since the bound
is a quadratic in k with n fixed, we can differentiate it twice to verify that it is concave down.
Since it is concave down, and has maximum at k = n , we can look at the closest integer
2
values to the maximum: namely n−1 and n+1
n+1
. Consequently, the maximum is at with
2 2 2
value ¡ ¢
n+1 2
− 3 ¡ n+1
2
2
¢
+2.
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