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Homework Solutions for Math 4800 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 knot 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 unknot 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 trefoil knot 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 knots 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 link 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
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