Homework Solutions for Math

Tim Anderton Charlotte Cannon Miles Fore Brian Hunt

Jason Meakin Onyebuchi Okoro and Alex Pruss

University of Utah Spring

Chapter

Miles The ordering of the vertices of a is very imp ortant If the the order

of the vertices is changed you might not only change the knot but you might

not even get a knot Take for example the following

The rst picture shows an unknot but in the second where the order of the

vertices is changed we no longer have a knot b ecause it isnt simple

Another simple example shows how that can b e turned into the

unknot if the ordering of the vertices is changed Take the following trefoil

The rst picture shows the trefoil the second has the exact same vertices but

the order is again changed Notice in the second picture the unknot is the result

of changing the order of the vertices

Thus if the order of the vertices is changed you may get a dierent knot or you

may not even get a knot at all

Charlotte Prove that the vertices of a knot form a welldened set

3

A set of p oints fp g in a dening set for a knot K if the knot given by R is

i

fp g is K

i

A set of vertices for a knot K is a dening set fp g for K such that no prop er

i

subset of fp g is a dening set for K

i

A corner of a knot K is a p oint p on K so that the p oints on K near p do not

lie in a single line

The set of corners of a knot is welldened

We want to show that a p oint p of the knot K is a corner if and only if p is in

every set of vertices for K

Supp ose a p oint p of the knot K is a corner Then p must b e in every dening

set for K and so it must also b e in any set of vertices for K

Next we show that if p is in every set of vertices then p must b e a corner by

showing that if p is not a corner then p is not in the set of vertices

For a knot K dened by the p oints fp p p g if p is a p oint that is not

1 2 n 1

a corner then K can also b e dened by the subset fp p g b ecause a p oint

2 n

that is not a corner is a p oint lying on a straight line so removing such a p oint

do es not aect K So p is not in the set of vertices b ecause without p the

1 1

subset fp p g is still a dening set for K

2 n

So any set of vertices is exactly the set of corners

Alex

Alex

Alex Problem Show there is only one planar knot

This is equivalent to taking any planar knot K v v v and deforming it

1 2 n

into the unknot For any vertice v dene v the angle of the vertice as the

i i

angle b etween the line segments v v and v v Notice that v

i1 i i i+1 i

but j v j i K is convex

i

Weve already shown in that all convex planar are equivalent to the

unknot so these knots arent interesting Instead lets lo ok at knots that are

not convex Note that it suces to show that any nonconvex knot of n vertices

can b e reduced to a nonconvex knot of n vertices and the desired result

follows from induction

Knots that are not convex are knots K where j v j Since v

i i

v where v has opp osite sign from v Let all such v b e called switches

i i i1 i

Using any switch v we will b e able to nd a p oint we can remove with an

i

elementary deformation The reason is that to prevent v itself from b eing

i

removed we require another switch v somewhere in the knot See Figure

j

A With one exception this v will also require another switch to prevent itself

j

from b eing removed Since any knot has only a nite numb er of vertices there

can only b e a nite numb er of switches so at least one switch can b e removed

via an elementary deformation reducing K to a knot with n vertices

There is one exception to this rule It is p ossible that the switch v cannot b e

j

removed b ecause it is blo cked by the switch v But then the only way to blo ck

i

v will b e with a spiral See Figure B To escap e from the spiral will require

i

another switch v which in turn will require a switch to keep itself from b eing

k

removed By the same logic as b efore then there is at least one switch that

can b e removed simplyfying K into an n knot

Brian Thrm Any knot formed by vertices is the unknot We have three

basic cases

Case A square is formed

Case A overcrossing as seen going from P to P

Case A undercrossing as seen going from P to P

For case A we have a simple deformation that takes p oint to the center of

the square This move is seen as taking p oint to the dotted line This

automatically gives us the unknot

Figure Start as a square b ox and then take either p oint or and bring it to the

midp oint and you will form a triangle This is just the unknot

For case B and C lo ok at the following page for the gures

Since this contains all p ossible cases we have shown that all vertice knots are

the unknot

Tim Let K b e a knot determined by p oints p p p Show that there

1 2 n

is a numb er z such that if the distance from p to p is less than z then K

0

is equivalent to the knot determined by p p p Similarly show there is

2 n

1

a z such that every vertex can b e moved a distance z without changing the

equivalence class of the knot

for the region around p consider the two closest connected line segments

p p p p Because the knot is a simple p olygonal curve there must exist

1 2 n 1

an op en tubular region centered on each of these line segments of radii a and b

with a b such that no line segment other than p p p p p p p p

1 2 2 3 n n1 1 n

pass through these regions Because p p and p are centered in their tubular

1 2 n

regions if we move p by an amount z a Then we will have moved the p oint

1

by less than the radius of the tub e Therefore the resulting line segment will

also still b e in the op en tubular region and so the knot could not have p ossi

bly passed through itself Since we said b efore that no other segments passed

through this region Thus there always exists a z by which we may shift p to

1

0

p without changing the knot

1

a

if instead of cho osing z such that z a if for each p oint we cho ose z then

2

even if b oth end p oints are moved simultaneously the resulting line segment can

move less than half the distance towards the closest line and that line can move

less than half the distance also which means that no line segments can cross

and so there also must exist a z which preserves the knot when every p oint is

moved simultaneously

Figure Here we have

taken p oint P and taken it

to the midp oint of PP

After p erforming this el

ementary deformation we

have made the knot into a

Figure Here we start with

triangle This is again the

an overcrossing of our knot

unknot

Figure Here we start with Figure After an elemen

an undercrossing tary deformation we have a

triangle In this step we

have taken p oint P to the

midp oint of PP to get

rid of the p oint This shows

we get the unknot as ex p ected

Jason

Theorem Let K be a knot determined by p p p Then there is a

1 2 n

0

number z such that if the distance from p to p is less than z then K is equivalent

1

1

0 0

to the knot K determined by p p p

2 n

1

Let K b e a knot dened by p p p then by the stated theorem there exists

1 2 n

0

an equivalent knot K and this knot can b e created using simple deformations

0

to move p to the p osition p

1

1

By denition of a knot the cyclic p ermutation of p oints p p dene equiv

1 n

0

alent knots We then apply a cyclic p ermutation of p oints to the knot K to get

0

the knot dened by the p oints p p p p and then transform it into

i i+1 j

1

00 0 0

an equivalent knot K of p oints p p p p This pro cess continues an

i+1 j

i 1

arbitrary numb er of times for all knots equivalent to our general knot K

Tim Generalize the denition of elementary deformation and equivalence

to apply to links your denition should not p ermit one comp onent to pass

through another

A L and a link L consisting of sets of non intersecting knots are elementary

deformations of each other when L contains only one knot K whose p oints

dier from the p oints of the knots in L and K is an elementary deformation of

a knot K in L and The triangle spanned by the dierence of the knot in L

and L do es not intersect any other knot

Onye Supp ose K is a knot dened by p p p and J is a knot dened

1 2 n

by q q q If K and J have regular pro jections then the numb er of

1 2 n

2

for b oth of them in crossings R is the same Since K and J have the same

numb er of crossings and vertices and their diagrams are identical we can easily

undertake an elementary deformation such that p p p is equivalent to

1 2 n

q q q

1 2 n

Miles

Problem Sketch a pro of of theorem

Theorem Let K b e a knot determined by the ordered set of p oints p

1

0

p p For ever numb er t there is a knot K determined by an ordered

2 n

set q q q such that the distance from q to p is less than t for all i

1 2 n i i

0 0

K is equivalent to K and the pro jection of K is regular

A regular pro jection satises the following conditions

No line joining two vertices is parallel to the vertical axis

No vertices span a plane containing a line parallel to the vertical axis

There are no triple p oints in the pro jection

First lab el each arc a a a where a connects p to p a connects p

1 2 n 1 1 2 2 2

to p and so on ending with a connecting p to p Then for any t

3 n n 1

construct an op en ball at every vertex with radius t If a p oint p is moved to

i

any other p oint q in the op en ball around it the distance b etween p and q will

i i i

b e less than t

Fix p that is let q p Now if a causes part of the pro jection to not b e

1 1 1 1

regular it can b e moved via moving p so that the irregularities are gone and

2

the knot is still equivalent There is a single line running through p which is

1

parallel to the vertical axis If this line runs through the op en ball around p

2

then remove it from the op en ball Also there will b e a nite numb er crossings

in the diagram At each of these crossings create a plane that is parallel to the

vertical axis and runs through the pro jection of p and the crossing p oint If

1

any of these planes intersect the ball around p then remove that region from

2

the ball We know the deleted ball will b e nonempty b ecause a ball with a nite

numb er of slices taken out will always b e nonempty Now if p is moved to any

2

p oint in the ball that we have created it will satisfy the conditions of making

the pro jection regular and will b e within distane t of its old p osition However

we may still need to limit further the places p can b e moved to ensure that

2

0

K is equivalent to K Do not let a b e moved anywhere that would cause it

1

to cross another arc in the knot We know we can do this from theorem two

This still leaves innitely many places in the ball around p that we can move

2

p Pick one and lab el it q

2 2

Follow this same pro cedure for a only this time we need to conscider condition

2

ab ove for a regular pro jection If the plane that contains a and is parallel

1

to the vertical axis runs through the ball around p then remove that p ortion

3

of the ball also Continue doing this with the rest of the vertices When you

0

are nished there will b e a new knot K that is equivalent to K made up of

vertices q q q has a regular pro jection and every q is within distance t

1 2 n i

of p

i

Onye We are given that K is determined by the sequence p p p and

1 2 n

3 2

a regular pro jection This means that when K is pro jected from has R R to

no three vertices of K are collinear It also means that no singular p oint p of

i

0

K is on the same p oint as any other p oint p of K We are also given that K

l

has equal numb er of vertices as K such that t q i pi t for all i

3 2 0

it is going to b e a regular pro jection R to R Then when K is pro jected from

Having equal numb er of vertices and crossings as K it is therefore equivalent

0

to K b ecause we can undertake some elementary deformations to transform K

into K

Charlotte

Show that the trefoil knot can b e deformed so that its nonregular

pro jection has exactly one multiple p oint

Jason HW Denition of an oriented link Two denitions came to mind

when considering this what an oriented knot is The rst is the simplest an

oriented link is the union of oriented knots But this seems to lack a feel for the

word oriented and more restrictive denition might b e interesting for example

a link is oriented if at every crossing of knots k k in our diagramed link the

1 n

orientations of the overlapping strands is opp osite ie

or

Brian What is the largest p ossible numb er of distinct oriented n comp onent

links which can determine the same unoriented link up to equivalence

bShow that any two oriented links which determine the as an unoriented

link are oriented equivalent

Part a

Each link can b e oriented two ways which we will call FB Given two links we

can have FFFBBFBB This is just the binomial theorem of probability For

N

choices and N links we have p ossibilities that gives a set unoriented link as

a maximum One p ossibility of reaching this upp er b ound would b e to nd n

nonreverablegiven that there are an innite numb er of distinct nonreversable

knots knots and link them up into a long chain This would give us our upp er

b ound Part bBy lo oking at the b ottom of the following gure we see that the

unlink is just the union of two By the upp er four pictures it is shown

that each unknot is oriented equivalent So the Union of Unknots is oriented

equivalent as well

Miles Explain why if an oriented knot is reversible then any choice of orien

tation is reversible

Let K b e an oriented knot Then it can b e b e determined by an ordered set

r

of vertices p p The reverse of this knot is the knot K with the same

1 n

r

vertices but their order reversed A knot is reversible if K and K are oriented

equivalent If you chose an orientation of a reversible knot then K is orientation

r

However since there are only two ways to orient a knot and equivalent to K

Figure The top four gures show that the unknot is orientable equivalent The

b ottom two show that the unlink is just two unknots

r

if you chose the other orientation it is the knot K it follows that if a knot is

reversible then any choice of orientation is reversible

Tim Show the ppq pretzel knot is reversible

The ppq pretzel knot is equivalent to the p q p pretzel knot If you

consider the symmetric diagrams of the ppq pretzel knot it is clear that a

rotation by degrees changes the orientation of the diagram but not the

diagram itself Thus the p p q pretzel knot is reversible

The knot is the rst knot in the app endix that is not reversible Below is

17

a series of diagrams to show that the trefoil knot is reversible by Reidemeister

moves The second set of diagrams shows knot and then how it app ears if

1

you ip the knot in dimensional space Because it lo oks the same but oriented

in the opp osite direction this knot is reversible With great diculty this could

also b e shown by Reidemeister moves

Chapter

Onye

            

                   

Brian Show that the given knot is equivalent to the unknot by p erforming a

series of unknots

Figure Here we p erform steps of Reidemeister moves to show that this is equivalent

to the unknot

Miles I counted a total of four knots with with seven or fewer crossings that

were colorable They were the and the knots Here are the coloring

1 1 4 7

I found of them

1

1

4

7

Tim For which integers n is the n colorable For which values

of n is the n twisted double of the unknot colorable n is the numb er of

crossings in the vertical band of the ntwisted double of the unknot

if the diagram is colorable then the left strand of the torus knot must b e a

dierent color from the right one since if they were b oth the same color then

the entire knot would b e forced to b e the same color If we let the left strand

b e red and the right strand b e green then the rst crossing will leave us with a

green strand and then a blue strand from left to right then a blue strand and

a red strand and then a red strand and a green strand again Clearly after this

p oint the pattern rep eats We had to go through crossings to get to a p oint

with red on the left and green on the right which is the coloring we started with

so we have found a colorable knot and if we add three more crossings then we

will also have a colorable knot therefore any ntorus knot is colorable

Again the n twisted double of the unknot must start with two dierent color

strands on top or the whole thing would b e a single color Let the left strand

starting in to the twists b e red and the right strand b e green In order for a

valid coloring to b e found the vertical band of twists would need to end with

blue on the left and red on the right the color changes as b efore leaving us

with a sequence rg gb br

Onye I shall attempt to explain the colorability of a pqr pretzel knot with

out reference to the determinant or mo dularity of the knot since these ideas had

not b een intro duced in section of the chapter at the time of this exercise

My trial of all the p ossible combinations of colorings did not seem to work In

general a pqr pretzel knot where pq and r are dierent pro duce a pretzel

knot that is always not colorable However when p is o dd and is set to b e equal

to r and q is an o dd numb er less than p and r the system b ecomes colorable

Some examples of such a combination are the and

pretzel knots This concludes the discussion

Brian Part a For the we obviously dont change linking

numb ers since this is just an untwisting Even if the twisting crosses over

another link by p erforming a R move to sep erate them we get a link with no

crossing and we have the same linking numb er if R works So we must lo ok at

R and R These are p erformed on the next page as Fig diagram

Part b Show that the has linking numb er See gure on

following page Fig

Part c Two examples that have linking numb ers of and See last pageFig

Charlotte The sum used to compute linking numb ers can b e split into the

sum of the signs of the crossings where K passes over J which we will write

as crKJ and the sum of the signs of the crossings where J passes over K

Figure Reidemeister moves and to show that the linking numb er is invarient

to any R move that we can make on any links Obviously this also applies to more

than links by making sure each triangle only passes through link at a time

Figure Simply by putting the appropriate linking numb er of we show that it is

Figure Here are two examples with dierent linking numb ers

J K

JK +1

+1

−1

−1

crJK where each righthanded crossing is assigned a and each lefthanded

crossing is assigned a

a Use Reidemeister moves to show that each sum is unchanged by a deforma

tion

The rst typ e of move do es not aect crKJ or crJK b ecause it only involves

one of the knots either K or J passing over itself

The second typ e of move will always involve one righthanded and one

lefthanded crossing The sum of these is zero so this move will not aect

crKJ or crJK

The third typ e of move involves three strands and three crossings Regardless

of orientation and to which knot each strand b elongs this move will not aect

the linking numb er b ecause the three crossings that exist b efore the move is

p erformed are the same as the three crossings which result they simply o ccur

at a dierent place in the diagram Therefore crKJ and crJK are unaltered

b The value of crKJ crJK is unchanged if a crossing is changed in a

diagram

At any crossing either K passes over J or J passes over K There are also two

p ossible orientations

In the rst case if K passes over J and J passes under from the right we get

crKJ crJK If switched so that J passes over K then

K will pass under J from the left giving

In the second case if K passes over J and J passes under from the left we get

crKJ crJK If switched so that J passes over K then

K will pass under J from the right giving

So switching the crossing do es not aect crKJ crJK

c If the crossings are changed so that K always passes over J then crKJ

crJK is zero

Since deforming the diagram with Reidemeister moves do es not change either

crKJ or crJK and switching crossings so that K passes over J or J passes

over K do es not change the dierence of the sums crKJ crJK the diagram

of a link can b e deformed by these two metho ds until the pro jections are disjoint

At this p oint it is clear that crKJ crJK crKJ crJK

d The linking numb er is always an integer given by either of the two sums

crKJ or crJK

The text dened the linking numb er to b e an integer given by crKJ

crJK Since we have shown that crKJ crJK if all the

crossings are changed and changing a crossing do es not alter the value of

crKJ crJK then crKJ crJK must always b e zero Thus crKJ

and crJK must b e equal and by their denition integers So instead of

adding them together and dividing by we can take the linking numb er to b e

either value

Alex

Problem Show that a the presence of two colors on a colorable

knot diagram forces the presence of a third color but b there is a

diagram of the unlink of two comp onents that can b e colored with

two colors and one with three Why is this the case c Prove

Reidemeister moves dont change the colorability of links

a This follows from the observation that if a knot has two colors then there

must b e a vertex where b oth colors are present To color the knot this vertex

would require the presence of a third color as well

b Figure Ca can b e colored with two colors and Figure Cb requires three

c The pro of that works for knot diagrams works for diagrams of links as well

The reason for this is that nowhere in the pro of is it assumed that the strands

aected by any reidemeister move ever join together

Charlotte The Whitehead link is nontrivial

The unlink of two comp onents is trivial and has a diagram which is colorable

with three colors Since colorability is preserved under Reidemeister moves

every diagram of the unlink is colorable

To check for colorability color arc black Cho osing to make arcs and

black would force the whole diagram to b e black which is not a valid coloring

Coloring arc red forces arcs and to b e blue Arcs and force arc to b e

red Arc would then have to b e b oth blue and black which is also not valid

The Whitehead link is not colorable therefore it is not trivial

3

1

5

6

4

2

Tim If a knot is colorable there are many dierent ways to color it For in

stance arcs that were colored red can b e changed to yellow yellow arcs changed

to blue and blue arcs to red The requirements of the denition of colorability

will still hold There are six p ermutations of the set of three colors so any col

oring yields a total of six colorings For some knots there are more p ossibilities

a Show that the standard diagram for the trefoil knot has exactly six colorings

Since the standard diagram of the trefoil knot has only three arcs since all three

arcs are the same we can pick one of them at random We can cho ose any of

the colors to color this arc and we can pick two ways to color the second two

arcs Therefore we have choices of p ossible colorings which gives us the

colorings we get from p ermuting the colors

bHow many colorings do es the square knot have

Since the square knot is the connected sum of two colorable knots clearly we

can either make one of the two knots all the same color the color of the strand

connecting it to the other knot or we can color it in exactly the way we would

if the strands connecting it to the other knot were simply linked together and

we just had a single knot Since the rst trefoil has dierent colorings and for

each coloring the strands linking it to the other trefoil are the same color the

color of the linking strand makes the other knot b e colorable in only two ways

that arent mono chromatic The total numb er of colorings is then

c The numb er of colorings of a knot pro jection dep ends only on the knot that

is all diagrams of a knot will have the same numb er of colorings Outline a

pro of of this

two knot diagrams of the same knot dier only by Reidemeister moves and

b ecause the coloring of all of the diagrams of the Reidemeister moves are com

pletely determined by the color of the incoming strands they cannot p ossibly

change the numb er of colorings of the knot as a whole

dUse the connected sum of n trefoils to show that there are an innite numb er

of distinct knots

The rst knot in the chain can b e colored freely giving colorings for it and

each connected knot will have p ossible colorings that will b e consistent with

the color of the incoming strands Thus for a connected sum of n trefoil knots

there will b e n total colorings for n greater than Since the numb er of

colorings is dierent for each n then each n connected sum is a dierent knot

Therefore we have found an innite class of dierent knots

Brian For what values of p do es the trefoil knot have a mo d p solution Here

is a lab eling of a trefoil knot

Figure Lab eling for a trefoil knot

To show which mo d p solutions exist for the trefoil knot we must calculate the

determinate of the matrix given b elow minus one row and column

D etA

The only mo d p solution of this matrix is So the trefoil knot can b e lab eled

mo d

Onye Question

Supp ose a knot is lab elled mo d then the following relationship is correct

xyz mo d for any crossing p oint where x is the over crossing and y and z

are the undercrossings We had shown in a previous pro of that this relationship

holds true for all colorable knots

If the lab elling is therefore multiplied by then the new relationship b ecomes

xyz mo d This new relationship holds true for whatever values we

cho ose for x y and z

Supp ose x y z Then the original relationship implies that

mo d This is a true statement On the other hand the new relationship will

imply that mo d This is also a true statement

We may therefore conclude that a knot lab elled mo d could b e multiplied by

any prime numb er which will still preserve the colorability and lab elling of the

knot This concludes the pro of

Miles Problem If p is other diculties come up Explain why no knot can

b e lab eled mo d

If we think of trying to color a knot it makes sense that we cannot color it

mo d two as it would take at least three colors to color However if we lo ok at

the crossing relationship that x y z we see that it also do esnt make

sense Since there are three variables and only two colors it must b e that ei

ther x y or y z If x y then we have x x z x z x z

but then every arc would b e lab eled the same and the condition that there needs

to b e at least two distinct lab els wouldnt b e satised Also if y z then we

have x y z x z x z Again the condition of at least two

lab els b eing distinct do esnt hold Thus no knot can b e lab eled mo d

Tim For each knot with or fewer crossings nd the asso ciated matrix and

its determinant In each case for what p is there a mo d p lab eling

using the notation of the table in App endix

has an n n matrix 1

A

the determinant of is which implies a mo d lab eling

1

has matrix

1

C B

C B

A

the determinant of is which implies a mo d lab eling

1

has a matrix

1

C B

C B

C B

C B

A

the determinant of is which implies a mo d lab eling

1

has a matrix

2

C B

C B

C B

C B

A

the determinant of is which implies a mo d lab eling

2

has a matrix

1

C B

C B

C B

C B

C B

C B

A

the determinant of is which implies a mo d lab eling

1

has a matrix 2

C B

C B

C B

C B

C B

C B

A

the Determinant of is which implies a mo d lab eling

2

has a matrix

3

C B

C B

C B

C B

C B

C B

A

The determinant of is which implies a mo d lab eling

3

Brian Complete the rst part of the pro of that the determinate is the same

regardless of which row and column you remove

Since we are free to cho ose our lab eling we can set the last arc a This

n

gives us a matrix where the last column do es not matter so we can make all the

last values When a column b ecomes it gives us a nullity of Then when

a matrix has nullity is row reduced we get a row on the b ottom that contains

only s So we can remove b oth the row and column that are lled with s

Because the lab eling is arbitrary we could have picked any arc to equal An

arc is just a column in our matrix and so it do es not matter which column gets

removed Likewise b ecause we can do a row swap it do esnt matter which row

gets removed All we need to do b efore row reducing is swap one row for the

last row and it would make the new last row have s

Charlotte The diagram of the unknot which has no crossings and the

diagram of the unknot which has only one crossing do not have typical cor

resp onding matrices An n n matrix corresp onds to a diagram with n

arcs and the dete rminant of a knot is the absolute value of the asso ciated

n n matrix The problem with these two diagrams of the unknot

is that unlike any others they have only one arc each so the corresp onding

matrix for either diagram is a matrix and the matrix from which we are

supp osed to nd the determinant is a matrix We manage this problem

by dening the determinant and nullity of a matrix to b e A nullity of

means there is one solution Since the determinant is not the only existing

solution for the system of equations is trivial

Miles Problem Prove that the determinant of a knot is always o dd

We know from that if a knot has a determinant that is divisible by p then

a mo d p solution exists that satises the crossing equation at each crossing

That is to say if the determinant of a knot is divisible by p then the knot can

b e lab eled mo d p

Assume the determinant of a knot is l where l is an even integer That means

that l k for some integer k Since the determinant of the knot is k we know

that a mo d solution exists b ecause k is denitely divisible by However

we know this cant happ en from exercise Thus we have a contradiction

and it must b e that the determinant of a knot cant b e even

Alex Problem Show that if a knot has mo d p rank n then the

n

numb er of mo d p lab elings is pp

This follows from a basic understanding of what the rank of a knot signies

Let us lo ok at any n n matrix of a knot with mo d p rank n Then since

the rank of the knot is n this matrix has nullity n meaning that its kernel has

rank n This means that the space of vectors we can multiply by this matrix

that return have dimension n

But this is the space were interested in b ecause all lab elings of this knot when

written as a vector and multiplied by this matrix should return Notice then

n

that the amount of elements in an ndimensional mo d p space is p We toss

one of these elements out namely the trivial solution of the vector b ecause

although this vector multiplied by our matrix returns it corresp onds to a

lab eling of the knot using only either one or two colors and so is an illegitimate

n

solution So there are p solutions of the n n matrix

Now notice that to get down to the n n matrix we rst cho ose an arc

and x a lab eling for it As there are p lab els there are p ways to go from the

n n matrix to the n n matrix

But if there are p ways to cho ose an n n matrix and each of these

n

matrices has p solutions then the total numb er of solutions of the matrix

n

or equivalently the total numb er of lab eling of the knot is pp

Charlotte Relate the value of the Alexander p olynomial of a knot eval

uated at to the determinant of the knot dened in section

The rst metho d we learned for lab eling a knot making a matrix and nding

the determinant was in section In this matrix as in the Alexander matrix

each row represents a crossing and each column represents an arc

In the rst metho d we put a in every row in the column of the overcrossing

arc In the Alexander matrix we put t in for every overcrossing Evaluating

this at t we get

In the rst metho d we lab eled the two undercrossing arcs that approach and

leave each crossing with the value In the Alexander matrix we lab el the

undercrossing arc that approaches the crossing with a and the under

crossing arc that leaves the crossing with t Evaluating this at t we get

for b oth undercrossing arcs as we did in the rst metho d

Finally in b oth metho ds we put in the column of every arc that is not

involved in the crossing represented by any given row Since all entries of the

matrix are the same by either metho d when we evaluate the entries of the

Alexander matrix at t the value of the determinant will b e the same

by either metho d Therefore since the Alexander p olynomial of a knot K is

the determinant of the Alexander matrix for K the value of the Alexander

p olynomial when evaluated at t will b e the same as the determinant we

nd by the metho d in section

Miles

Miles Problem Compute the p olynomials of and to check they are

1 2

8 4

identical

Using the following lab elings and orientation for the knot

1

8

If we set up the matrix for the knot using the diagram shown and remove the

last row and column we get the determinant

t t

t

t t

t t

t t

t t

t

8 7 6 5 4 3 2

This determinant equals via TI t t t t t t t

knot Now use the following lab elings and orientation for the

2

4

If we set up the matrix for the knot using the diagram shown and remove the

last row and column we get the determinant

t t

t t

t

t t

t t

t t

t t

t t

8 7 6 5 4 3 2

Calculating this determinant we get t t t t t t t

Which is the exact same p olynomial that we got for the knot Thus the

1

8

Alexander p olynomial fails to distinguish the knot from the knot

2 1

4 8

Tim Prove that a knot and its mirror image have the same Alexander p oly

nomial

If the Knot has left and right reected then the right handed crossings will

b ecome left handed and vice versa but if we also reverse the orientation of the

knot then the handedness of the crossings remains unchanged If we lab el the

knot arcs in the same way as b efore then we will get precisely the same matrix

and so we must get the same determinant

Alex Problem Show the of a knot K

with its orientation reversed is obtained from the p olynomial of K by

1

substituting t for t multiplying by the appropriate p ower of t and

p erhaps changing sign

This is equivalent to asking to show that the Alexander Polynomial is invariant

under changes of orientation This can b e seen in the following way The eect

of a change of orientation on the matrix from which the Alexander Polynomial

is computed is to change all s to ts and vice versa It is merely necessary

then to show that this new matrix will have the same determinant as the old

up to changes of sign and p owers of t

Consider the columns of this matrix They will either i Have some amount

of t and t terms iia Will only have or iib will only have t terms

or iii will have some and t terms but no t terms Considering what

happ ens when nding the determinant by expanding down the column in each

of these cases along with some inductive reasoning will show the determinant

will only change by p ossibly a negative and a p ower of t

Case i Use row op erations to add a row containing t in this column to

all rows containing or t in this column The resulting column will lo ok like

your old column and the determinant of your matrix has not b een changed

This means that if b efore the determinant of the matrix from expanding down

this column was

detA detA tdetB detB tdetC detC

1 n 1 m 1 p

where A B C are the n n matrices obtained by removing the row

i i i

and column of each resp ective t and t term then now your determinant

is

0 0 0 0 0 0

detA detA tdetB detB tdetC detC

1 n 1 m 1 p

0 0 0

where A B C are obtained from A B C by replacing s with ts and vice

i i i

i i i

versa Clearly if we can demonstrate that something similar can b e done with

every single column then we have shown that the matrix of the reverseoriented

knot has the same determinant as the matrix of the original orientation of the

knot To show this we simply have to examine the other two cases

Case ii Simply expand down this column no changes necessary If b efore

expanding down this column gave determinant At expanding down it now

1 0 0

will either give t A t in case iia or will give tA t in case iib Note that

0

A t will dier from At by no more than a p ower of t and p ossibly a negative

sign by the same inductive reasoning used in i Since we are allowed to

multiply an Alexander p olynomial by a p ower of t this do esnt actually change

the p olynomial

Case iii If there is only one and t term simply switch the two rows con

taining these terms This will result in the original column and while a row

swap changes the sign of the determinant this isnt something we care ab out If

there is more than one and t term subtract a row with a in this column

from every other row with a in this column and do the same for rows which

have ts in this column This do es not change the derivative and will result in

a column with only one and one t term which was just discussed

This covers every case Weve now demonstrated that there is a way to ma

nipulate the matrix obtained from the reverse oriented knot in such a way so

that nding the determinant by expanding down any column will yield the same

determinant we had originally ignoring changes of sign and p owers of t This

shows that the Alexander p olynomial obtained from the reverse oriented knot

is identical to the one obtained from the originally oriented knot

Chapter

The surface in gure is a disk with two twisted bands attached This

surface is homeomorphic to the same surface with the bands untwisted This

is b ecause there exists a onetoone and onto map that cuts untwists and reat

taches the bands in the rst surface so that it lo oks in space like the second

surface This mapping is continuous b ecause p oints that are close to each other

on the rst surface map to p oints that are close to each other on the second

surface

Although the two surfaces are homeomorphic they cannot b e deformed into

each other in space We show this by demonstrating that their b oundaries are

dierent In exercise we saw that the b oundary of the surface with twisted

bands is a trefoil knot In the following diagram we see that the b oundary of

the surface with untwisted bands is the unknot

Miles

Miles Problem Use Theorem to nd the genus of the surface illustrated

b elow note the b oundaries for the knot have distinguished by coloring them

dierent colors

Theorem The genus of a connected orientable surface which is formed by at

taching bands to a collection of disks is given by disksbandsb oundary

comp onents

Our surface as four disks seven bands and three b oundary comp onents Thus

the genus

Tim Prove corollary

Corollary if two connected orientable surfaces intersect in a single arc con

tained in each of their b oundaries the genus of the union of the two surfaces is

the sum of the genus of each

2B

Since we are connecting the surfaces along the genus of a surface is g

2

the b oundary of each the union of the surfaces has a numb er of b oundary

comp onents B B Since the b oundary that we connect the two along

1 2

b ecomes the same b oundary in b oth surfaces and so would b e counted twice

By theorem in section we have that the euler characteristic of the union

of the two surfaces is the sum of b oth minus since they meet along a single arc

2 B B +2

1 2 1 2

Thus for the genus of the union of the two surfaces we have g

2

=2 B

1 1

which is clearly the sum of the genuses of the two surfaces alone g

1

2

2 B

1 1

and g

1

2

Problem Use Theorem to prove that the only genus surface

with a single b oundary comp onent is the disk

Theorem Every connected surface with boundary is homeomorphic to a sur

face constructed by attaching bands to a disk

Theorem If a connected orientable surface is formed by attaching bands to a

col lection of disks the genus of the resulting surface is given by

G disk s bands boundar y

Supp ose there were a surface S with one b oundary comp onenet S not homeo

morphic to the disk and GS By Thm we know S is homemorphic to

a disk with bands attached Since S is not homemorphic to the disk there is at

least one band attached But if we substitute anything but now let us compute

the genus of S using Thm We nd that GS r r where

r is the numb er of arcs But then the genus of S is not as we had assumed

so we have a contradictin

A punctured torus can b e deformed into a disk with two bands attached

Begin by stretching the hole around the surface until the torus lo oks like the

subsurface in the picture

Miles Problem Prove the genus of a surface is always nonnegative

We know from theorem that the genus of a knot can b e found using the

following equation g K disksbandsb oundary comp onents

We also know from theorem that every connected surface with b oundary is

homeomorphic to a surface constructed by attaching bands to a disk So it is

sucient to conscider surfaces that consist of only a disk with bands attached

Lets rst lo ok at the simplest surface just a disk The genus of this surface is

Now lets consider what happ ens when we add disks or bands to it What we

want to do is lo ok at the worst case scenarios and make the genus as small as

p ossible mayb e even negativeLo oking at the formula we see that this is done

by adding as many disks as p ossible adding as few of bands as p ossible and

making as many b oundary comp onents as p ossible

Notice that anytime another disk is added at least one band is required oth

erwise you would have more than one surface Also notice that to add more

b oundary comp onents requires adding bands or bands with disks If it is done

with just bands then the most b oundary comp onents can intro duce is but

this is cancelled out by the band added So what if we increase the numb er of

b oundary comp onents by adding a collection of bands with disks We run into

the same problems we had earlier Consider adding b oundary comp onents by

adding another disk with two bands running to the rst disk This will increase

the disk and b oundary numb er by one but also the band numb er by one The

same thing happ ens no matter how many disks and bands we may add to try

and increase the numb er of b oundary comp onents

Thus the biggest numb er that we can reduce the genus numb er by is It

follows that the smallest make the genus is

Miles Problem Why do es Seiferts algorithm always pro duce an orientable

surface

In short the reason this o ccurs is b ecause Seiferts algorithm pro duces a surface

which b oundary is an orientable knot

Supp ose Seiferts algorithm pro duced a nonorientable surface We already know

that the b oundary of the surface created is a knot However we reach a con

tradiction b ecause all knots are orientable and this would imply that the knot

wasnt orientable Thus it must b e that the is in fact orientable

Problem In applying Seiferts Algorithm a collection of Seifert

circles is drawn Express the genus of the resulting surface in terms

of the numb er of these Seifert circles and the numb er of crossings in

the knot diagram

This is simply an elementary applicatino of Thm shown ab ove Denoting

the numb er of seifert circles by s and the numb er of crossings by x we have

G s x This follows by noting that each seifert circle is a

disk each crossing is a band joining a disk to another disk and the resulting

surface has one b oundary comp onent namely the knot from which the surface

was obtained

Charlotte

Charlotte If K is nontrivial there do es not exist a knot J such that K J is

trivial If K J is trivial it is the unknot with genus The additivity of knot

genus states that the genus of the connected sum of two knots is the sum of the

genus of the two knots So for the connected sum of two knots to b e b oth

knots would have to b e trivial

Tim Use the connected sum of distinct knots to nd an example of a knot

which can b e decomp osed as a connected sum in two dierent ways

Since the knot is made up of three knots it follows that you can break it up

as a connected sum of one and the other two together or you can

break it up with that prime knot connected to one of the other two and a third

prime knot left out Of course the ab ove only holds if we are dealing with

dierent prime knots since otherwise the knots we are separating into would b e

indistinguishable

Use the genus to prove that there are an innite numb er of

distinct knots As a harder problem can you nd an innite numb er

of prime knots

Two knots are distinct if they have dierent genus But we can take a knot K

with genus g and construct K K with genus g Then K K is distinct from

K We can do this an innite amount of times nding an innite numb er of

distinct knots K K K K

Will think ab out the prime bit some more

Chapter

Tim Show that S is not commutative

6

Two elements of S are and just quickly checking we get

6

so clearly and do not p ermute and so S is not commutative

6

Miles

a Verify that the inverse to is

To do this we simply lo ok at the pro duct of these p ermutations and see if we get

the identity p ermutation Consider the pro duct In

the rst cycle go es to then in the third go es to Thus go es to Notice

go es to then ve go es to If you do this you will get the p ermutation

which is the identity p ermutation

bFind the inverses to and

The inverse to is the inverse to is

and the inverse to is Each of these can b e

veried why taking the pro duct of the p ermutation and the p ermutation that

is inverse to it and noticing that that identity is the result

c In general how do es one write down the inverse of a p ermutation given

in cyclic notation

First write the p ermutation down as a pro duct cycles such that no two of the

cycles will have an element in common we know we can do this by theorem

The inverse of the p ermutation is simply the set of cycles that were just found

but with all the elements in each cycle written in reverse order

1 1

d Let g b e the p ermutation Compute g g and g g

1

What is a short cut for computing g f g in general

1

Doing some simple computation we see that g g is the p ermutation

1

and also g g is the p ermutation

I dont know if this is what they were lo oking for but the following is a shortcut

1

kind of Conscider g g or Instead

1

of working with this huge mess we can just lo ok at g f and when we get to the

61

cycles in f just work backwards over g Lo oking at only

we see that go es to then three go es to then go es to Thus go es

to Now lo oking at go es to then go es to and then go es to

Therefore go es to Now lo ok at go es to then go es to then go es

to So go es to Then the rst cycle in the p ermutation is This can

b e continued to get the rest

Charlotte

Charlotte It is not p ossible in S for the pro duct of two cycles to b e an

8

cycle

Either the two cycles are disjoint or they have at least one common element

If they are disjoint then their pro duct is the same two disjoint cycles If they

have one element in common then their pro duct can have at most distinct

elements If they have more than one element in common their pro duct will

have fewer than distinct elements so it is not p ossible for an cycle to b e the

pro duct of two cycles

Problem In S a what is the order of b Verify

n

that the order of is c What is the largest order of an

element in S S S

7 10 20

a is a cycle and so has order

b and are disjoint has order and has order

n

So any n with jn jn will result in e The smallest such n is

c The largest order of S is which results from a cycle and a cycle

7

p ermutation In S the largest order is which results from a cycle cycle

10

and cycle In S the largest order is obtained from a cycle cycle

20

cycle and cycle

Charlotte a Check that the cycle is equal to the pro duct of

transp ositions

It is straightforward to check that In each

transp osition n n go es to and in the next transp osition go es to n so

the result is

b Write as a pro duct of three transp ositions

Write as a pro duct of transp ositions

c Argue that every p ermutation is the pro duct of transp ositions and more

sp ecically is a pro duct of transp ositions of the form n

Given a set of elements a a a some element a can b e moved to any

1 2 3 i

p osition in the p ermutation of the set by a series of transp ositions Similarly

each element of the set can b e moved to a dierent p osition by moving it one

p osition at a time trading places with a neighb oring element each time By

such a series of transp ositions we can make any p ermutation of the set Thus

all p ermutations are the pro duct of transp ositions

Show that every p ermutation is the pro duct of transp ositions of the form n

We know that the set of transp ositions generates S so now we need to show

K

that every transp osition can b e written as a pro duct of transp ositions of the

form n

The transp osition ab may b e written as the pro duct of transp ositions aba

So from transp ositions of the form n we can build transp ositions from which

we can build any p ermutation

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

d Show that every p ermutation can b e written as the pro duct of transp ositions

taken from the set f n ng To show this we will

consider an example Write as the pro duct of transp ositions To nd

a solution we draw a diagram showing each element from to and where it

is sent by the p ermutation The rst row of numb ers represents the original

elements The second row is the image under the p ermutation The lines match

each element to its image under the p ermutation

The second diagram is drawn to show more clearly that each move is simply

made up of transp ositions Each crossing represents a transp osition of two

elements The six transp ositions and o ccur

The result of these transp ositions is the desired p ermutation We can draw

such a diagram for any p ermutation thus showing that any p ermutation can

b e written as a pro duct of transp ositions

Miles

a Show that the set of cycles generates S

4

We know that the symmetric group can b e generated by the set of transp osi

tions thus if we can nd a way to write any transp osition as the pro duct of

cycles then we know that the set of cycles will generate S

4

Let A b e a set in S with A an element in the set Supp ose you want to

4 i

transp ose the i m element with the i n element Then let B b e the set

with B A if i m n If i m let B A likewise if i n let B A

i i m n n m

Then AB is the transp osition of the i m element of A with the i n element

of A

b Show that the cycles and generate S

4

Notice that

2 2

2

2

2

2 2

Thus and generate S

4

Tim Check that the consistency condition is satised at all the crossings in

the lab eled knot diagrams illustrated by gures and

For the lab elings in we have the conditions

all of which hold

and the conditions obtained from are

which also all hold so the consistency condition is satised at the crossings

Problem In gure two of the lab elings satisfy the consis

tency condition while one do es not Find the inconsisten lab eling

Figure b is inconsistent The b ottom intersection read o the diagram im

plies This equation actually equals

however

(12) (13) (23)

(23) (12)

(23)

(13) Figure A

(13)(245)

(35)(124) (12)(345)

Figure B

Charlotte Find a lab eling of the pretzel knot illustrated in Figure

using transp ositions from S Your lab eling should have every transp osition

4

app ear so it is clear that the lab els generate the group

Miles Problem Show that in some of the previous examples the lab eling b e

comes inconsistent if the orientation of the knot is reversed Show however

that is the orientation of the knot is reversed and each lab el is replaced with its

inverse then the lab eling will again b ecome consistent

Conscider the diagram in Figure Here it is with the orientation reversed

(1,2)

(1,3) (3,2)

(1,3)

(2,3) (1,2)

(1,3) (3,2)

(1,2)

Lo oking at the following vertices we see that the lab elings are not consistent

Now if we also replace each lab el with its inverse we get the following equa

tions at each crossing

Thus when the orientation was reversed and the lab elings were replaced with

their inverses the lab elings b ecame consistent

Problem Prove it is imp ossible to lab el the trefoil with trans

p ositions from S

4

The trefoil has only three arcs Once the lab elings on two of these arcs are

determined the third is forced But to b e consistent this third lab eling will

have to have the form xy where the rst lab eling takes the form xa and the

second takes the form y b But then the third lab eling is generated by the rst

two and it takes more than two transp ositions to generate S

4

Charlotte Check the claims ab out lab elings for diagrams and

1 46

Diagram cannot b e lab eled with transp ositions from S If we lab el arcs a and

1 4

b with the same lab eling for example b oth all arcs in the diagram are

forced to b e lab eled If the lab els for a and b overlap such as and

this forces a lab eling in transp ositions from S If the lab els for a and b

3

are disjoint we reach a contradiction For example if a is and b is

we will nd that arc d must b e b oth and Therefore cannot b e

1

lab eled with transp ositions from S

4

Diagram can b e lab eled with transp ositions from S as follows

46 4

Diagrams and can b e lab eled with cycles from S as follows

1 46 4

1 1 1 1 1 1

Miles Problem Why is xy x y x z xy xy x z xy equivalent to

1 1 1 1 1 1 1

y x z xy x y x z xy xy x

1 1 1 1 1

We know that y x z xy xy x z xy has an inverse in particular

1 1 1 1 1 1

y x z xy x y x z xy

1 1 1 1 1 1 1

Multiplying b oth sides of the rst equation on the left by this we get y x z xy x y x z xy xy x

1 1 1 1 1 1 1 1 1 1 1

y x z xy x y x z xy y x z xy xy x z xy

1 1 1 1 1 1 1

y x z xy x y x z xy xy x

Questions

Can we nd the cases that causes a certain invariant to not b e able to distin

guish two knots Put another way can we combine a couple of invariants we

know to obtain a true invariant

Do any of the invariants weve covered so far apply to linking numb er

Is there a simplest diagram for a knot How ab out a link

Can prove what the least numb er of crossings in a diagram equivalent to some

knot is

(1,2) (1,2) A E

(1,3) (2,3) (1,3) D (2,3)

(2,3) (2,3) (2,3)

C B (1,3) F

(1,2,3,4) (1,2,4,3) (1,2,3,4) (1,4,2,3) (1,4,2,3)

(1,2,4,3) ((1,3,2,4) (1,4,3,2) (1,2,4,3)

(1,3,4,2) ((1,3,2,4) (1,2,3,4)

(1,3,4,2) (1,4,2,3) (1,4,3,2)

Chapter

Problem Find the Seifert matrix of the knot in gure

Charlotte Find the Seifert matrix corresp onding to pqrpretzel knots with

p q and r o dd

As in gure in the text page a pretzel knot can b e deformed to lo ok

like a disk with bands attached by pushing the b oundary down through each

twist of the center band and leaving a space at its base b etween the two newly

formed sides of the center band Call the bands p q and r where the name

indicates the numb er of twists on the band In the new surface we have only

two bands which we will call P and R It will b e useful to refer to the twists

that were originally on p as P and the twists that were originally part of q but

p

are now part of P as P We will do the same for R What was formerly q is

q

now made up of the q twists of P and q twists of R

q q

Smo othing the bands in the center will either increase or decrease the numb er

of twists on p and r dep ending on the direction of the twists We will describ e

the case of q and p only realizing that the same is true for q and r If p and

q are twisted in the same direction then for every twist we push o of P we

q

automatically lo ose a twist on P That is each twist on P undo es a twist on

p q

P

p

If p and q are twisted in opp osite directions then for every twist we push o

of P we will add a twist to P In this case each twist on P simply b ecomes

q p q

instead a twist on P

p

We said that the numb er of twists on P and R is q So now we can say that the

q q

numb er of twists on P is the numb er of twists on P plus the numb er of twists on

p

P or simply p q Similarly the numb er of twists on R is r q Rememb er to

q

designate one direction of twists as negative and the other direction as p ositive

Because p q and r are o dd P and R will necessarily b e even This ensures that

our Seifert surface will b e orientable We draw our surface in the standard way

with curves P P R and R

P

Since P is counted in half twists the linking numb er of P and P is Similarly

2

R

the linking numb er of R and R is These are integers since P and R are

2

even

For q o dd the linking numb er of P and R will b e either q or q From

lo oking at gure you can see that this dep ends on whether P is ab ove or

b elow R at the top of band q For the purp ose of the matrix we are trying to

describ e we will say we have arranged the knot so that P is ab ove R at the top

which will mean that the linking numb er of P and R is q and the linking

numb er of R and P is q

This gives us the Seifert matrix

P

q

2

R

q

2

Miles Problem What would b e the eect of changing the orientation of the

Seifert surface on the Seifert matrix



We knot that the i j entry in the Seifert matrix is given by l k x x Clearly

i j

the magnitude of the linking numb er is going to stay the same What is going

to change is the sign If a crossing is left handed and the orientation is reversed

it will b ecome right hand and vice versa This corresp onds to a changing sign

Thus thus if a Seifert matrix is computed for a surface and then computed again

with the same surface with the orientation reversed the matrices will b e the

same except the signs on each entry will b e opp osite

Charlotte Check the calculation of the determinant that gives the Alexander

p olynomial of the knot in Figure

t

The matrix asso ciated with the knot in Figure is V tV where V is the

Seifert matrix of the knot

t

C B

t t t

C B

A

t t t

t t

Calculating the determinant by expanding on the rst column we get

t t t t

t t t t t t

t t t t t

4 3 2

t t t t

which is the Alexander Polynomial of the knot

Miles Problem Use the result of exercise to show that the

Alexander p olynomial of the conneted sum of knots is the pro duct of their

individual Alexander p olynomials

Result of exercise Let M and M me Seifert matrices for knots J and

J K

K resp ectively Then the Seifert matrix for J K is the matrix

M

J

M

K

The Alexander p olynomial of J K is then equal to

M tM

J J

M tM

K K

T T

which equals the detM tM Which is equal to detM tM

K K J J

Alexander p olynomial of J multiplied by the Alexander p olynomial of K

Questions

Brian

For this I think an interesting one would b e the one the b o ok mentions What is the

least numb er of Reidemeister moves that is required to transition the knot for Prob

into the unknot and can we prove that this is the least numb er required I see

a metho d of moves so far

QuestionHow many knots in the App endix can taken from link to links or

even n links Of these new links which ones are oriented equivalence

QuestionBy cutting a knot at a p oint A can you then make more crossings and

reconnect that knot and get all knots with crossings

Question Is there a p edigree tree such that by creating one knot you can get to only

some of the higher crossing knots

Question If there is a tree do es this help distinguish more knots apart

N

Question Can you get distinct orienations in other ways Do es making a chain

of n orientable equivalent knots such that it is not symmetric ab out the middle link

give us an upp er b ound

Question Are there knots that have a mo d p solution and a mo d q solution

Miles

Questions

Why are coloring numb ers restricted to just the o dd primes

If two knots can b oth lab eled mo d p will they share any other characteristics

If two knots are added together how will its coloring numb er relate to the two knots

coloring numb ers

To used the equation x y z mo d p to nd an invariant for knots Could we

p erhaps use some other equation that would give us a stronger invariant Questions

Is there a easier way to check that a certain set generates a group

How do you know if a set that generates a group is the smallest set least numb er

of elements that generates that set

In the previous chapter we learned that we could write down all p ossible surfaces

up to a homeomorphism Can we do the same sort of thing with groups

Is it necessarily the case that all surfaces which have that same knot K as their

b oundary are homeomorphic to each other

Is the least sticks question for a knot essentially the same as the least triangles for a

surface with the knot as its b oundary question

Is there a way to dene a prime knot such that prime factorizations is unique and a

two prime knots determine a single knot

Can we nd how many diagrams there are with n crossings using stereotypical pro

jection

Can we show there are innitely many distinct knots

Can any diagram of the unknot b e all the way unknotted using only the typ e Rei

demeister move

Can a knot ever b e lab eled the way we dened it in class with an ab elian group If

so what are some examples

Can we nd the cases that causes a certain invariant to not b e able to distinguish two

knots Put another way can we combine a couple of invariants we know to obtain a

true invariant

Do any of the invariants weve covered so far apply to linking numb er

Is there a simplest diagram for a knot How ab out a link

Can prove what the least numb er of crossings in a diagram equivalent to some knot

is

Tim

Are there arbitrarily many dierent knot invariants

Would the existence of an algorithm to distinguish any two knots in a nite

numb er of steps imply the existence of a nite numb er of invariants which would

distinguish any two knots

What is a simple way to enumerate the set of all knots allowing for duplicate

knots

What is the simplest algorithm to generate all p ossible knots without duplica

tion do es such an algorithm exist

The torus knots cover some p ortion of the set of all knots as do the p q r

pretzel knots Can you construct a nite set of simple classes of knots like these that

would cover the set of all knots these classes of knots would b e allowed to overlap

Do es the set of all pretzel knots of any numb er of strands cover the set of all

knots

Connecting the strands of a braid will give you some sub class of knots what

knots are not covered by this class of knots

Do es the class of all knots which are generated by connecting the strands of a

braid cover the set of all n stranded pretzel knots

Is there a closed form for the numb er of knots which can b e drawn with a

minimum numb er of crossings n is it related to the numb er of partitions of n

A knot diagram creates a series of distinct regions cut o from one another

by the arcs of the diagram Clearly every knot diagram must have at least regions

and every non trivial diagram must have at least regions If we take into account

the numb er of crossings on the b oundary of each region can we use this information

to determine what the minimum p ossible numb er of crossings a knot with a diagram

with these characteristics could have

Traversing the regions of a knot diagram moving around them in a counter

clo ckwise order and noting the order of the crossings gives a p ermutation of the

crossings Is there some natural way to translate cycles generated like this into a Sn

lab eling of the knot

Do es the Alexander p olynomial suce to distinguish all alternating knots

knots which go from over crossing to undercrossing to over crossing to

The alternating knots can all b e drawn such that every crossing is on the

outermost region of the knot Is there any other class of knots which can b e drawn

with all p oints exterior

Consider a knot to b e constructed of a material that has a certain exibility

but which wants lo cally to b e a straight line segment This puts a certain tension into

the knot Such a knot will have a minimum energy conguration where the amount

of curvature is least in some sense Given a knot made of such an elastic material

in any starting conguration will the knot always uncoil into this minimum energy

conguration frictionless environment of course and the knot would b e considered

to have a xed total length it cannot stretch Can such a knot have more than one

p ossible minimum energy conguration

What is the set of groups which are the fundamental groups of some knot

Treating knots as closed p olygonal curves with integer vertices how many

distinct knots can t into a cub e of side n for that matter how many dierent

representations are there

Knots can b e represented as a language a language is a set of strings and

a string is just a series of symb ols Given some way of representing a knot with

symb ols the language of a knot would b e the set of all p ossible representations using

that system of representation which represent equivalent knots The language of a

knot with a particular representation system is then a Is it p ossible to

have b oth a means of representing a knot which unambiguously determines the knot

and generates a language of all p ossible representations which is nite

What is the computational equivalent of the op eration of nding knot equiva

lence In other words what computational problems can we map onto the question of

equivalence of knots One such computational framework might b e to take a prop osi

tion and translate it into two knots or rather two knot representations in such a way

that the prop osition is true if and only if the two knots are equivalent Sp ecically

can we do this with the op eration of addition so can we asso ciate knots in such a

way to the prop osition a b c so that the question can b e answered by equivalence

of knots

Where on the Chomsky hierarchy is the language of knots

The language of at least some sub class of knot representations is decidable is

it p ossible to nd a class of knot representations such that the language of all knots

in that representation is a context sensitive language or a context free language or

even a regular language

How many essentially dierent diagrams of a knot are there which share the

same numb er of crossings Is the set of minimum crossing diagrams of a knot dierent

from the set of minimum arc diagrams of a knot If so in what cases

Onye

When are two diagrams of the same knot describ ed as isotopic

Why are knots not colorable mo d

When do two diagrams represent the same link

What is an alternating pretzelknot

What is knot addition and how do es it work

What is the mirror image of the gure knot

Do es a nontrivial knot in R necessarily have four collinear p oints

Are there any such things as ideal knots

When is a knot or link describ ed as prime

What is the unknotting numb er for the gure knot

What do es the term family of knots mean

Are there links that are colorable mo d

Is there only a unique way by which a knot or link may b e colorable