2012 Siemens Paper Sc Bx

Counting Absolute Maxima of a Sequence of Partial

Sums: Searching for Knots with Fibered Knot

Complements

Abstract

In this paper, we expand upon previous work by Brown and Stallings, whose separate

results together showed that in a lens space L(p, q), the knot complement of a simple knot

K(p, q, k) is fibered over the circle if and only if the corresponding sequence of partial sums

Ap,q,k = {Ap,q,k(i)}p1i=0 of a sequence of values, derived from a two-value function e(i) on

variables p, q, k, and i, has a unique absolute maximum, allowing us to translate the knot

theory question of finding simple knots with fibered knot complements into the number the-

ory problem of determining conditions on p, q, and k such that Ap,q,k has a unique maximum.

We expand the number of known simple knots with fibered knot complements by determin-

ing relationships among and conditions on the triples p, q, k that ensure that the sequence

of partial sums has a unique maximum, which we do by first forming conjectures with the

aid of a computer program and proving these conjectures using number theory techniques.

The knot theory of lens spaces has largely been unexplored but has begun to gain atten-

tion from low-dimensional topologists in the last decade. Fiberedness of the knot complement

of simple knots is an important piece of information about the basic structure of the knot,

and our work greatly expands the number of known simple knots with fibered knot comple-

ments. Therefore, our research provides invaluable information for knot theorists pursuing

this area of study and is likely to find extensive application in the near future.

1

AdminSticky NoteTurn OFF hyphenation to avoid words getting broken up. Makes the paper easier to read.

Executive Summary

The knot theory of lens spaces has largely been unexplored but has begun to gain atten-

tion from low-dimensional topologists in the last decade. The so-called simple knots have

been identified as fundamental in this theory. Fiberedness of the knot complement of simple

knots is an important piece of information about the basic structure of the knot, and our

work greatly expands the number of known simple knots with fibered knot complements.

Therefore, our research provides invaluable information for knot theorists pursuing this field

of study.

The separate results of Brown and Stallings imply a theorem which translates the question

of fiberedness of the knot complement from a knot theory question into the less complicated

number theory question of counting the number of unique absolute maxima in a sequence of

partial sums, thus simplifying our search for known simple knots with fibered knot exteriors

significantly. We further expand upon this theorem by developing relationships among and

characteristics of triples of p, q, k such that the sequence of partial sums, which we denote

in the paper by Ap,q,k, has a unique absolute maximum.

Knot theory is a rapidly expanding field of mathematics that has numerous applications

to everyday life in Physics, Computer Science, and Organic Biochemistry (especially in

mapping DNA structure). Because lens spaces are a concept that is relatively new within

knot theory, our results are likely to find application in the near future.

1

AdminSticky Notechange to "significantly simplifying".Delete the "significantly" at the end of the sentence.

AdminSticky NoteAdd one or two more sentences here to give the reader a more solid idea of how knots can be applied. Right now you're still talking in the theoretical realm. Bring it down to earth a bit more for the layperson.

1 Introduction

1.1 The Problem

In this paper, we examine a function ep,q,k(i) that takes an input i as well as parameters

p, q, and k and returns as output either a positive or a negative number. We then create a

sequence {Ap,q,k(j)}j=0, where Ap,q,k(j) =

ji=0

ep,q,k(i), a sequence of partial sums of ep,q,k(i).

The specific problem statement is given below:

Problem. Given integers p, q, k such that gcd(p, q) = 1, define the function e(i) =

ep,q,k(i) as follows:

e(i) = ep,q,k(i) =

(p k) if 0 iq < k

k if k iq < p

where a denotes the residue of a (mod p) in {0, 1, . . . p 1}.

Define {Ap,q,k(i)}i=0 to be an infinite sequence of partial sums such thatA(j) = Ap,q,k(j) =j

i=0

ep,q,k(i).

Characterize triples p, q, k such that Ap,q,k = {Ap,q,k(i)}p1i=0 has a unique absolute maxi-

mum.

*It is important to recognize that for our purposes Ap,q,k, which we use as shorthand for

a sequence, is not equivalent to Ap,q,k(i), which represents a value.

In this paper, we determine fundamental relationships among the three variables p, q,

and k and relationships among various triples that ensure the sequence of partial sums has

a unique maximum.

1.2 Background in Knot Theory and Motivation

A lens space is a certain type of 3-manifold (a shape in which each point has a neighborhood

that looks like the three dimensional space) created by using coprime variables p and q. To

form the lens space L(p, q), we take a square, identify or glue together opposite edges of

1

AdminSticky NoteLet's avoid equations being broken up over multiple lines.Make your paper easier to read, not harder.

the square to form a torus, and draw a line of slope 0 on this square and another of slope

p

q, continuing them such that they exit on one side and enter on the other. For example,

if p = 5 and q = 3, we have a line of slope 53and another of slope 1. Note that we have

labeled the sides S1, S2, S3, and S4. Next, we identify side S1 to side S3 to form a cylinder

(Figure 1).

Figure 1: Creating the Cylinder. To form the torus we identify S2 and S4.

We then identify sides S2 and S4, now looped in the shape of a circle, forming a torus:

we call this torus T. The original line of slope 0 now forms a curve that we call curve and

the original line of slope pqforms a curve that we call curve . We then create two new solid

tori, V and V, which we glue to each side of T so that the curves and are meridians for

V and V, respectively. A meridian for a solid torus is a curve in its boundary torus that

bounds a disk in the interior of the torus but does not bound a disk in the boundary torus.

Glued together along T, V and V form the lens space.

The parameters p, q, and k in the problem define a simple knot K(p, q, k) in the lens

space L(p, q). To form the knot, we label the p intersections between curve and the

curve from 0 to p 1. We know there exists an arc c that connects points 0 and k and

2

whose interior lies in the interior of the disk that is bounded by . By symmetry, there is an

analogous arc c. We then glue together c and c to form the K(p, q, k). Both lens spaces

and simple knots are very difficult to represent graphically.

It follows from Brown [1] that if Ap,q,k has a unique absolute maximum, the fundamental

group of the knot complement admits a surjection to the integers with a finitely generated

kernel. It has been shown that by Stallings [2], this implies that the knot complement is

fibered over the circle, meaning that it can be expressed as a twisted circle of homeomorphic

surfaces (like a Mobius strip):

Theorem 1.2.1. Ap,q,k has a unique absolute maximum if and only if the corresponding knot

K(p, q, k) in lens space L(p, q) has a knot complement that is fibered over the circle.

The significance and implications of K(p, q, k) having a fibered knot complement are

discussed in section 3 (applications and future work). The study of the knot theory of lens

spaces has only recently begun to receive more attention from low-dimensional topologists:

therefore, our research on the fiberedness of the knot complement of simple knots, an im-

portant fundamental question in the knot theory of lens spaces, is both relevant and crucial

for these topologists.

1.3 Geometric Symmetries

Because this paper deals with knot theory, we can look at the lens spaces and knots we

produce from a geometric point of view. Taking such a perspective allows us to exploit

certain symmetries, which we prove here. We also provide original number theory proofs

for these theorems in section 2.1. These symmetries can be applied to the specific cases we

prove in section 2.2 to increase the scope of the cases solved regarding the search for triples

p, q, k that give a unique maximum in the sequence Ap,q,k.

Theorem 1.3.1. The sequence Ap,q,k has a unique maximum if and only if Ap,q,k has a

unique maximum.

3

Proof. From the original definition of our lens space, q primarily defined the slope of one of

our lines and therefore our curve . Reversing the sign of q only changes the orientation of ,

so the lens space formed by the variables p and q is identical to that produced by p and q

up to symmetry. Therefore the partial sums sequences Ap,q,k and Ap,q,k are symmetrically

similar.

Theorem 1.3.2. The sequence Ap,q,k has a unique maximum if and only if Ap,q,k has a

unique maximum.

Proof. Because the knot Kp,q,k is congruent to Kp,q,k (they are related by a reflection), the

corresponding sequences Ap,q,k and Ap,q,k have the same number of absolute maxima.

2 Main Results

2.1 General Symmetries

In this section, we present various symmetries we have proven that are applicable to the

specific cases we find in the next section. This significantly increases the number of p, q, k

for which we know the uniqueness of the absolute maximum of the corresponding sequence

of partial sums.

Immediately from the problem statement the following is evident:

Remark. Ap,q,k has a unique absolute maximum in each interval of size p if and only if

Ap,q+mp,k+np has a unique absolute maximum in each interval of size p, because the function

A depends on q and k.

Theorem 2.1.1. Ap,q,k(i) = Ap,q,k(i+ np) for any integer n.

Proof. Notice that Ap,q,k(i+np) = Ap,q,k(i)+i+np

j=i+1 ep,q,k(j). Then, since every p consecutive

terms of ep,q,k consist of k instances of (p k) and p k instance of k,

x+p1j=x

ep,q,k(j) =

k(p k) + (p k)k = 0 for any integer x. Then,

4

Ap,q,k(i+ np) = Ap,q,k(i) +

i+npj=i+1

ep,q,k(j)

= Ap,q,k(i) +

i+pj=i+1

ep,q,k(j) +

i+2pj=i+p+1

ep,q,k(j) + +

i+npj=i+(n1)p+1

ep,q,k(j)

= Ap,q,k(i) + 0 + 0 + + 0 = Ap,q,k(i).

Now that we have shown that Ap,q,k is periodic with period p,the following corollary holds

by the periodicity of the absolute maxima:

Theorem 2.1.2. Ap,q,k has a unique maximum on [0, p) if and only if Ap,q,k has a unique

maximum on every interval of size p.

Proof. From the periodicity of Ap,q,k, we know that the absolute maxima recur periodically

with period p. The theorem follows.

*Note: From here on, in this paper, we write the term unique maximum, which is

equivalent to the term unique absolute maximum in each interval of size p by Theorem

2.1.2.

Notice that ep,q,k is a periodic integer function with period p. Ap,q,k is also periodic,

because the sum of p consecutive values of ep,q,k(i) is 0.

Following from the periodicity of the function Ap,q,k, if the graph of Ap,q,k has a unique

absolute maximum on the interval [0, p), then the graph has a unique absolute maximum on

any interval of size p. In general, the number of absolute maxima and absolute minima is

constant on any interval of size p, independent of the position of the interval.

We can visualize the function ep,q,k on a linear scale. Consider the infinite integer number

line. Then considering the valuations of the points on this number line in mod p, we define a

negative interval to be an interval such that for each n in the interval, n [0, k). Similarly,

define a positive interval to be an interval such that for each n in the interval, n [k, p).

Define this new number line as a positive-negative identified number line. Then if iq lies

in a negative interval, ep,q,k(i) = e(i) = (p k); on the other hand, if iq lies in a positive

interval, e(i) = k. Analogously, we can model consecutive values of e(i) (that is, e(i) and

5

e(i+ 1), etc.) by looking the type, positive or negative, of interval a bug, which hops q

numbers to the right each time i increments by 1, is in.

The following theorems were stated and proven geometrically in the previous section. We

now offer concrete number theory proofs of these theorems.

Theorem 2.1.3. The sequence Ap,q,k has a unique maximum if and only if Ap,q,k has a

unique maximum.

Proof. Define f(i) = e(i), where e(i) = ep,q,k(i). Then f , which is essentially ep,q,k, has

the same characteristics as e: f is periodic with period p and can be viewed with respect to

a positive-negative identified number line. Notice that e(i) represents a function in terms

of p, q, and k that is analogous to how f(i) represents a function in terms of p,q, and k.

Consider e(i). Using the bug analogy, consider a bug that starts at the left hand side

of a negative interval, at a number congruent to 0 mod p, moving toward the right with

steps of size q. Similarly, analyzing f(i), we consider another bug that is moving to the

left by q. However, because we can let the second bug start anywhere on our positive-

negative identified number line, a convenience made possible by the periodicity of ep,q,k, let

the starting position of the second bug be at the right hand side of a negative interval, at

a number congruent to k 1 (mod p). Notice that the two bugs are traveling in opposite

directions at the same speed. This means that they are always on the same type of interval

by symmetry; analogously, this implies that the two sequences e(i) and f(i) take on the

same values but are a phase shift apart e(i) = f(i+ k 1). Then Ap,q,k(i) = Ap,q,k(i+ ),

q k 1 mod p, by the definition of A(i). Hence Ap,q,k and Ap,q,k are a phase shift apart,

which by theorem 2.1.2 implies that they have the same number of absolute maxima in each

interval of size p.

More formally, choose i0 such that i0(q) k 1 (mod p). Then f(i0 + i) = e(i)

by symmetry, meaning that the sums [f(0) + + f(i0 1)] + f(i0) + + f(i0 + i) =

[f(0) + + f(i0 1)] + e(0) + e(1) + + e(i) have a unique maximum if and only if

Ap,q,k(i) = e(0) + + e(i) has a unique maximum. Lastly, we can apply theorem 2.1.1

6

because Ap,q,k has a unique maximum on [0, p) if and only if it has a unique maximum on

[i0, i0 + p), so Ap,q,k has a unique maximum if and only if Ap,q,k has a unique maximum, as

desired.

We now prove a lemma that will help us prove Theorem 1.3.2.

Lemma 2.1.4. The sequence Ap,q,k has a unique maximum if and only if it has a unique

minimum.

Proof. For the sake of contradiction, assume that on the interval [0, p) the sequence Ap,q,k

has a unique absolute maximum and multple absolute minima (let two of these absolute

minima have indices i and j). Use the same definition for f(i) as we used in the previous

proof. Then, if n is the number such that Ap,q,k has an absolute maximum at n and if i

and j are distinct numbers in (n p, n] such that Ap,q,k has absolute minima at i and j,

e(i+ 1) + e(i+ 2) + + e(n) = e(j + 1) + e(i+ 2) + . . . e(n) are both equal to the width of

the sequence Ap,q,k (define the width to be max(Ap,q,k)min(Ap,q,k)). Also, e(n) = f(n),

e(n 1) = f(n + 1), and so on. Therefore f(n) + f(n + 1) + + f(i 1) =

f(n) + f(n + 1) + + f(j 1). These are both equal to the width of the sequence

Ap,q,k, because each set of p consecutive terms of the sequence of f(i)s contains the same

elements as each set of p consecutive terms of the sequence of e(i)s. However, because i 6= j,

this implies that the sequence Ap,q,k has two absolute maxima in an interval of size p, a

contradiction.

The reverse (that given Ap,q,k has a unique minimum, Ap,q,k has a unique maximum) is

similarly proven.

Lemma 2.1.4 provides a nice proof to the next symmetry:

Theorem 2.1.5. The sequence Ap,q,k has a unique maximum if and only if Ap,q,k has a

unique maximum.

Proof. This is equivalent to showing that Ap,q,k has a unique maximum if and only if Ap,q,pk

has a unique maximum. Notice that ep,q,k(i) = ep,q,pk(i (p k)) by the definition of

7

e(i), so Ap,q,k(i) = Ap,q,pk(i ), where q p k (mod p). Hence Ap,q,k and Ap,q,pk are

related by a phase shift. Then since Ap,q,pk has a unique maximum, Ap,q,pk has a unique

minimum, which implies by lemma 2.1.3 that Ap,q,k has a unique maximum.

Most importantly, this gives us a useful and symmetric corollary:

Corollary 2.1.6. The sequence Ap,q,k has a unique maximum if and only if Ap,q,pk has a

unique maximum.

2.2 Characterizing p, q, k that give a unique maximum in Ap,q,k

In this section, we will present various conditions under which the sequence Ap,q,k has either

a unique maximum or multiple maxima. These cases were determined by analyzing the

output from the Mathematica code (Figure 2) we used. The code returns, as p, q and k run

from 1 to any number we choose (here, we chose 100), the p, q, k such that Ap,q,k has multiple

maxima (a slight edit would have allowed for the output of cases giving a unique maximum

in sequence) in the sequence Ap,q,k. In searching for patterns among the cases returned by

the code, we can identify general relationships among the three variables that seem always

to yield a unique maximum or multiple maxima in the sequence Ap,q,k. After making these

conjectures, we set out to prove them, using number theory techniques.

Figure 2: Mathematica Code

8

First, we present general definitions and observations that assist us in proving the theo-

rems we present later in this section.

We can write, by the division algorithm, p = bq + r and k = cq + s, where b, c 0 and

r, s {0, 1, . . . , q 1}. We look at the sequence consisting of the residues of iq in mod p as

i ranges from 0 to p 1. The sequence becomes

{0, q, 2q, . . . bq, qr, 2qr, . . . bqr, q 2r, 2q 2r, . . . , bq 2r, . . . , r, q+r, 2q+r, . . . , (b1)q+r}.

Then we can divide this sequence into q subsequences

{0, q, 2q, . . . , bq}, {q r, 2q r, . . . , bq r} . . . , {r, q + r, 2q + r, . . . (b 1)q + r}.

We number these subsequences, calling the first subsequence the 0th subsequence and the

last subsequence the q 1th subsequence, and denote this sequence of subsequences by

{Ii}q1i=0 where Ii denotes the ith subsequence. Notice that the first value of each Ii is less

than k, while the last value of each Ii is greater than or equal to k. This implies that our

sequence ep,q,k(i) consists of several (p k)s followed by several ks, and so on. Moreover,

the number of (p k)s in each string of (p k)s and the number of ks in each string of

ks can be determined. By analyzing the individual subsequences, we can write the sequence

of ep,q,k(i)s precisely.

Define a sequence {ti}q1i=0 such that ti is the value of the first term in Ii. Notice that

ti+1 ti p (mod q), i.e. ti+1 ti r. Furthermore, since t0 = 0, we have the general

formula ti = ir, where the residue is taken mod q. We then have four cases for each Ii,

which we call A1 and A2 (which share the characteristic that ti r 1), and B1 and B2

(which share the characteristic that ti r 1):

9

Types of Subsequences

Subsequence Type Condition Breakdown of Corresponding ep,q,k

A ti r 1 b+ 1 terms

A1 ti s 1 c+ 1 instances of (p k), b c instances of k

A2 ti s c instances of (p k), b c+ 1 instances of k

B ti r b terms

B1 ti s 1 c+ 1 instances of (p k), b c 1 instances of k

B2 ti s c instances of (p k), b c instances of k

In each type A subsequence, because ti r 1, we have b + 1 terms: ti, ti + q, . . . , ti +

bq in the subsequence. On the other hand, each type B sequence only contains b terms:

ti, ti + q, . . . , ti + (b 1)q. We determine the breakdown of the corresponding ep,q,k sequence

by looking at how many terms in the subsequence are less than k how many terms are

greater than or equal to k: for instance, in a type A1 subsequence, we have the terms

ti, ti + q, . . . ti + cq, . . . , ti + bq, of which c+ 1 are less than k = cq + r and b c are greater

than or equal to k, corresponding to c+ 1 instances of (p k) and b c instances of k in

the sequence of e(i)s.

In addition, notice that in comparing r and s, if r > s, we cannot have Ii be type B1;

similarly, if r < s, we cannot have Ii be of type A2.

In the sequence Ap,q,k, the terms corresponding to the last terms of the Ii correspond to

what could be local maxima in {Ap,q,k}, because in each subsequence, the values increase,

implying that in the corresponding values in ep,q,k, the (p k) terms precede the k terms.

We refer to these terms (in Ap,q,k) as candidate local maxima. (Notice while A(0) is a local

maximum, we do not add it to our list of candidate local maxima because A(0) = (pk) 1 = r, which means that Ii cannot be of type A2.

Because r = 1, we have ti = q i, where the residue is taken mod q, i.e. {ti}q1i=0 =

{0, q 1, q 2, . . . 1}.

Now we can characterize each Ii. Notice that since 0 r1 = 0 and 0 s, we have that

I0 is type A1. Then, since for 1 i q s, ti [s, q 1], then each Ii where 1 i q s

is type B2. Lastly, for q s + 1 i q 1, ti [1, s 1] = [r, s 1], each Ii with

q s+ 1 i q 1 is type B1.

Since our candidate local maxima occur at the ends of the subsequences, our candidate

local maxima are M0 = A(b),M1 = A(2b), . . . ,Mq1 = A(bq). Then, as i ranges from 0 to

12

q s 2, A((i+1)b)A(ib) = c(p k)+ (b c)(k) = cp+ bk = bcq rc+(bcq+ bs) =

bs rc = bs c. Notice that this must be positive because bs cr = pskrq

and p k and

s > 1 = k.

Also, as i ranges from qs1 to q1, A((i+1)b)A(ib) = (c+1)(pk)+(bc1)k =

cp p+ bk = (bcq rc p) + (bcq + bs) = bs c bq r.

Since bs c bq r < bs bq < 0, we have a unique absolute maximum in the sequence

Ap,q,k(i) at i = (q s 1)b.

Theorem 2.2.5. If p 1(mod q) (or equivalently, r=q-1) then the sequence Ap,q,k has

multiple maxima if bs = cr. Otherwise, the sequence has a unique maximum.

Proof. Since p 1(mod q), we know that r = q 1 > s, so we cannot have B1 occur in Ii.

Notice if r = s, then q|(p k) and the sequence Ap,q,k has a unique maximum.

In addition, since r 1(mod q), ti i (mod q). Hence, for i [0, s1], ti [0, s1], so

Ii is type A1. For i [s, r1], ti [s, r1] so Ii is type A2. Lastly, for i [r, q1] = {q1},

ti = r = q 1 so Iq1 is type B2.

Note that M0 through Ms1 form an arithmetic sequence and Ms1 through Mr1 form

an arithmetic sequence, so the values M1 through Ms2 and the values Ms through Mr2

cannot be unique absolute maxima. Therefore, we only need to compare the four values

M0,Ms1,Mr1, and Mq1.

Since I0 is type A1, M0 = bs cr (p k).

I0 through Is1 are type A1, so for i [0, s1), Mi+1Mi = (c+1)(pk)+(bc)k =

bk cp = bs cr + (k p). Hence Ms1 = s(bs cr) + s(k p).

Is through Ir1 are type A2, so Mr1 = (r s)[c(p k) + (b c + 1)k] + Ms1 =

(rs)[c(pk)+(bc+1)k]+s(bscr)+s(kp) = (rs)[(b+1)kcp]+s(bscr)+s(kp) =

(r s)[(bs cr) + k] + s(bs cr) + s(k p) = r(bs cr) + s(k p) + (r s)k.

Finally, since Iq1 is type B2, Mq1 = Mr1 c(p k) + (b c)k = Mr1 + bk cp =

Mr1 + bs cr = q(bs cr) + s(k p) + (r s)k = 0.

We analyze three cases: when bs cr is 0, positive, or negative. We write, for the sake

13

of simplicity, l = bs cr.

Case 1 (l = 0): Since l = 0, we know M0 = k p, Ms1 = s(k p), Mr1 = s(k

p) + (r s)k, and Mq1 = s(k p) + (r s)k = Mr1. Because r s 0 and k > 0,

Mq1 = Mr1 Ms1. Thus, if M0 Mq1, there are multiple maxima in Ap,q,k, and if

M0 > Mq1, there is a unique maximum.

In order for M0 Mq1, k p s(k p) + (r s)k (s 1)(p k) (r s)k

(s 1)p (r 1)k ps p rk k. But since bs = cr ps = kr, our final inequality

becomes k p, which is always true (we are considering k as a residue in mod p). So this

case always gives multiple maxima for Ap,q,k.

Case 2 (l > 0): Because r s and l > 0, we have Mq1 > Mr1 > Ms1. We recognize

that while M0 6= Mq1, Ap,q,k will have a unique maximum.

Next, notice that Xi can have a value of l (p k) if Ii is type A1, l+ k if Ii is type A2,

and l if Ii is type B2. Then since

q1i=0

Xi = CMq1 = 0, and l + k > 0 and l > 0, we need

l (p k) < 0. But because I0 is type A1, M0 = X0 = l (p k) < 0, so M0 6= Mq1 = 0,

implying that Ap,q,k has a unique maximum.

Case 3 (l < 0): Because k p < 0 and l < 0, M0 > Ms1 and Mr1 > Mq1. Hence if

M0 6= Mr1, Ap,q,k has a unique maximum.

For the sake of contradiction, assumeM0 = Mr1. Then l+kp = rl+s(kp)+k(rs)

(s 1)(p k) = (r 1)l + (r s)k (s 1)p = (r 1)l + (r 1)k (s 1)(bq + r) =

(r 1)l+ (r 1)(cq+ s) bqs+ rs bq r = (r 1)l+ crq+ rs s cq l(q r+1) =

(p k) 2l = p k. However, l < 0 and p k > 0, so we cannot have these values be

equal, so this case yields a unique maximum in Ap,q,k.

Theorem 2.2.6. Given p 2 (mod q) and k 1 (mod q), Ap,q,k has a unique maximum

unless b = 2c.

Proof. In this case, r = 2 and s = 1. Hence, since r > s, Ii cannot be type B1. Because

gcd(p, q) = 1, and q must be odd, otherwise p and q would both be even. We write q = 2l+1.

Then, since t0 = 0, I0 is type A1. Notice that when i = l, ti q (2lr) = 1 r 1 so Ii is

14

type A2. For all other i, Ii is type B2.

Then, we have M0 = X0 = (c+ 1)(k p) + (b c)k = bk cp+ k p = b 2c+ (k p),

since I0 is type A1.

Because for i [1, l1], Ii is type B2, we have Xi = c(kp)+(bc)k = bkcp = b2c.

Hence Ml1 = M0 + (l 1)(b 2c) = (l 1)(b 2c) + (k p).

Since Il is type A2, we have Xl = c(k p) + (b c + 1)k = b 2c + k, so Ml =

Ml1 + b 2c+ k = l(b 2c) + k + (k p).

Lastly, Mq1 = 0.

Notice that because the values M0,M1, . . .Ml1 and the values Ml,Ml+1, . . . ,Mq1 both

form arithmetic sequences, the values M1 through Ml2 and the values Ml+1 through Mq2

can not be absolute maxima. We therefore analyze the four values M0,Ml1,Ml, and Mq1.

Now we have three cases:

Case 1 (b 2c = 0): In this case, we have M0 = Ml1 and Ml = Mq1, so Ap,q,k must

have multiple maxima unless l = 1, in which q = 3. But if q = 3, we know M0 = k p 0,

so M0 Ml = Mq1, implying that Ap,q,k has multiple maxima.

Case 2 (b2c < 0): In this case, we know that the arithmetic sequences M0,M1, . . .Ml1

and Ml,Ml+1, . . .Mq1 both have negative constant differences, so it remains to compare M0

and Ml. If these values are equal, then there are multiple absolute maxima, and otherwise

there is a unique absolute maximum. Assume these two values are equal. Then k p =

k p+ l(b 2c) + k. But this implies that 2cl bl = c(2l + 1) + 1 bl + c+ 1 = 0, which

is impossible because all the terms on the left hand side are positive.

Case 3 (b 2c > 0): In this case, we have Xi > 0 for i > 0, so Mq1 is a unique absolute

maximum.

Hence, Ap,q,k has a unique maximum unless b 2c = 0.

Theorem 2.2.7. Given k2|p, Ap,q,k has a unique maximum.

Proof. For this case, we look at the set of local maxima (not candidate local maxima). Define

this set to be the set of all Ap,q,k(i)s where iq [k, p 1] (implying e(i) = k) and (i+ 1)q

15

[0, k 1] (implying e(i+1) = k p). Note that while i = 0 does not satisfy these conditions

and is a local maximum, we do not consider it because A(0) = (p k) < A(p 1) = 0, so

A(0) cannot be an absolute maximum.

Then, assume for the sake of contradiction that some two local maxima A(i) and A(j) are

equal, and that i 6= j. Then e(i+1)+e(i+2)+ +e(j) = 0. Among these e(i)s, suppose d1

have a value of (pk) and d2 have a value of k. Hence d1(pk) = d2k d1p = (d1+d2)k.

Note that d1+d2, the number of e(i)s in the sum, is equal to j i. Therefore d1p = (j i)k.

Because k2|p, we have k|(j i).

Since A(i) and A(j) are local maxima, e(i+1) = (pk) and e(j+1) < 0. By definition,

e(i + 1) < 0 only if (i+ 1)q [0, k 1]. Similarly, (j + 1)q [0, k 1]. Since (i+ 1)q and

(j + 1)q are different because i 6= j, |(j + 1)q (i+ 1)q| (0, k 1]. Most importantly, this

implies that |(j + 1)q (i+ 1)q| cannot be divisible by k.

Note that k|(ji) k|((j+1)q(i+1)q). Therefore, there exists an integer n such that

kn = (j+1)q (i+1)q. Since k|p2, we have k|p. Thus, (i+1)q = (i+ 1)q+x1p = (i+ 1)q+

kmx1, for some integer x1. Similarly, for some integer x2, we know (j+1)q = (j + 1)q+kmx2,

which implies that k|(j + 1)q (i+ 1)q, a contradiction because we showed k cannot divide

(j + 1)q (i+ 1)q.

Hence, no two local maxima are equal, so the absolute maximum is unique.

Applying Corollary 2.1.6 to Theorem 2.2.7 gives the following corollary:

Corollary 2.2.8. Given (p k)2|p, Ap,q,k has a unique maximum.

Theorem 2.2.9.. Given n = bc= r

ssuch that n > 1 is an integer, Ap,q,k has multiple

maxima.

Proof. Notice that since r = ns, r > s so B1 does not occur in Ii. Also, Ii is type A1 when

ti [0, s 1], type A2 when ti [s, r 1] = [s, ns 1], and type B2 when ti [ns, q 1].

Because ti (i 1)r (mod q), so ti+1 ti r ti ns (mod q). Hence, if Ii is type

A2, then ti [s, ns 1], which means that ti+1 [q (n 1)s, q 1]. Note that because

16

q r q (n 1)s s, ti+1 cannot be in the interval [0, s 1], which implies that Ii+1

is either type A2 or B2, so each string of type A2 subsequences is followed by a type B2

subsequence. Hence the local maxima of the sequence of candidate local maxima are all

repeated. Because the absolute maximum is taken from this set, the absolute maximum is

repeated.

3 Conclusion

In this paper, we built upon the previous work of Brown [1] and Stallings [2], whose work

together showed that the knot complement of a knot K(p, q, k) in lens space L(p, q) is fibered

if and only if the sequence {Ap,q,k(i)}p1i=0 we defined in section 1.1 has a unique maximum.

This opened an accessible mathematical approach to the question of fiberedness.

First, we proved some general symmetries that we noticed from analyzing the problem

statement and the knot theory background of the problem. These symmetries can be applied

to the more specific results we found in the second part of our results to increase the number

of known triples p, q, k such that Ap,q,k has a unique maximum. Because we aimed to try

to characterize all such p, q, k, these symmetries were a powerful tool that simplified the

question. We were able to find many of these desired p, q, k triples by computer analysis

and we proved the conjectures we derived from our program output using number theory

techniques.

3.1 Application and Future Work

Our work is useful in gaining a better understanding of the knot theory of lens spaces, a

topic that has not been well studied.

Dehn Surgery is a process that uses a knot in one three-manifold to produce another

three-manifold. It is known that if Dehn surgery on a knot in a lens space produces either

S3 (the three-sphere) or S1xS2 (the circle times the sphere), then the knot complement is

17

fibered. It is conjectured that such a knot must also be a simple knot. This, among other

conjectures and theorems, exemplifies the importance of simple knots in low dimensional

topology. In this paper we discovered many simple knots K(p, q, k), which greatly expands

the knowledge we have of the characteristics of simple knots.

Each lens space L(p, q) contains p free homotopy classes of oriented loops. Because

homotopy is an equivalence relation, in organizing the knots in L(p, q) we first partition

them according to their homotopy classes. Each homotopy class contains a simple knot that

helps reduce geometric and topological considerations. These simple knots are analagous to

the unknot in the three-sphere, an important comparison that can lead to minimization of

considerations in 3-space.

There is much to be discovered about the knot theory of lens spaces and therefore simple

knots that will likely find application in the future. Understanding which simple knots are

fibered and which are not is an open question that our research is relevant to and that we

would like to approach in the future. The exploration of the fiberedness of simple knots is

a question about their basic structure, an open line of research: those simple knots that are

fibered have been conjectured to demonstrate interesting behavior among contact structures

on lens spaces [3]; on the other hand, those that are not fibered could prove to be even more

curious, since computer experiments have suggested them to be quite rare.

One important step towards studying these open questions we could take is to completely

characterize all p, q, k such that Ap,q,k has a unique maximum. While we covered a large

number of these and even eliminated a considerable amount, it still remains to find all such

p, q, k. We pose this as an open question we would also like to continue to work on in the

future. Furthermore, wed like to investigate those knots whose corresponding sequence of

partial sums does not have a unique maximum by developing a method of counting the

number of maxima the sequence has. We believe that such an investigation would also lead

to interesting results.

18

4 References

[1] Kenneth S. Brown, Trees, valuations, and the Bieri-Neumann-Strebel invariant, Invent.

Math. 90 (1987), no. 3, 479504.

[2] John Stallings, On fibering certain 3-manifolds, Topology of 3-manifolds and related

topics (Proc. The Univ. of Georgia Institute, 1961), Prentice-Hall, Englewood Cliffs, N.J.,

1962, pp. 95100.

[3] http://www.nd.edu/ rhind/cont-lens2.pdf

[4]

19

AdminSticky NoteThis [4] is empty