Quadruple crossing number of knots and links

0.31 MB·English

by  Colin Adams

QUADRUPLE CROSSING NUMBER OF KNOTS AND LINKS
COLINADAMS
ABSTRACT. Aquadruplecrossingisacrossinginaprojectionofaknotorlinkthathasfourstrandsoftheknotpassing
straightthroughit.Aquadruplecrossingprojectionisaprojectionsuchthatallofthecrossingsarequadruplecrossings.Ina
previouspaper,itwasprovedthateveryknotandlinkhasaquadruplecrossingprojectionandhence,everyknothasaminimal
quadruplecrossingnumberc4(K).Inthispaper,weinvestigatequadruplecrossingnumber,andinparticular,usethespanof
3 thebracketpolynomialtodeterminequadruplecrossingnumberforavarietyofknotsandlinks.
1
0
2
n 1. INTRODUCTION
a
In[1],amulti-crossing(alsocalledann-crossing)ofaknotisdefinedtobeasingularityinaprojectionsuchthat
J
nstrandsoftheknotcrossstraightthroughthesingularpoint. Inthatpaper,itwasprovedthatforeveryfixedn ≥ 2,
8
anygivenknotorlinkhasaprojectionsuchthatallsingularitiesaren-crossings. Hence,wecandefinec (K)tobethe
2 n
minimalnumberofn-crossingsinanysuchprojection. Inthatpaper,triplecrossingprojectionswereconsidered,and

] c (K)wasdeterminedforavarietyofknotsandlinks.
T 3
Inthispaper,wecontinuetheinvestigation,extendingresultstoquadruplecrossingprojectionsofknotsandlinks.
G
For simplicity, define q(K) = c (K). A quadruple crossing has strands at four heights, which we label top to bottom
4
h. as1,2,3,and4respectively. Readingclockwisearoundthecrossing,andalwaysstartingwiththetopstrand,thereare
t sixtypesofcrossings,denotedc ,c ,c ,c ,c ,andc . Notethatifwereflecttheprojectionofoneof
a 1234 1243 1324 1342 1423 1432
m thesecrossingsinalineperpendiculartothestrandlabelled1,wefindc1234 pairedwithc1432,c1243 pairedwithc1342
andc pairedwithc . Thispairingwillbehelpfullater.
[ 1324 1423

2 1 1 1
v
6 2 4 3
2 3 3 2
7 4 2 4
2
. (a)c (b)c (c)c
1 1234 1432 1324
1 1 1 1
2
1 4 3 2
v: 2 4 4
i 3 2 3
X
(d)c (e)c (f)c
r 1423 1342 1243
a
FIGURE1. Quadruplecrossings.
Notethatasingequadruplecrossingresolvesintosixdoublecrossings,andhence,q(K) ≥ c(K) wherec(K)isthe
6
traditionalcrossingnumber. Thisboundisrealizedbya(4k,4)-toruslinkL,whichhasfourcomponents,andisknown
tohavecrossingnumber12k. Eachmeridianalhalf-twistisrealizedbyaquadruplecrossing,soq(L)=2k.
InSection2,weconsiderfurthertherelationbetweenq(K)andc(K). In[1],itwasprovedthatc (K)≤c(K)–2for
3
allknotsandlinksexcept2-braidknots,andfor2-braidknots,c (K)≤c(K)–1.Onemighthopethatc(K)>c (K)>
3 3
c (K),however,thisisnotthecase. Forexample,thefigure-eightknotJ hasc (J) = c (J) = 2.Onemightalsohope
4 3 4
thatatleastc (K)≥c (K)forallk≥2. Atthispoint,wedonotknowthis. Wedoknowthatc (K)≥c (K)forall
n n+1 n n+2
k≥2. InSection2,weprovethatc (K)<c(K),thefirstproofofwhichwasgivenbyMichaelLandry. Thissectionis
4
independentofthesubsequentsections.
Date:January30,2013.
1
2 COLINADAMS
InSection3,werelatethequadruplecrossingnumberofaknotorlinktothespanofthebracketpolynomial. In[1],
itwasshownthatspan(<K >)≤8c (K). Here,weshowthatspan(<K >)≤16q(K). Fromthis,itfollowsthatifK
3
isanalternatingknotorlink,weobtainastrongerlowerboundq(K)≥ c(K).
4
InSection4,afterintroducingsomemovestoturnatraditionalprojectionintoaquadruplecrossingprojection,we
usethisresulttodeterminethequadruplecrossingnumberofavarietyofknots.Weincludeatableofspecificquadruple
crossingnumberformanyoftheprimeknotsof10orfewercrossings.
Further investigations into multi-crossings appear in [2], where it was proved that every knot and link has a pro-
jection with just one multi-crossing. The minimal n for which there is a single n-crossing is called the u¨bercrossing
numberoftheknot. Onecanfurtherprovethatforknots, thereissuchaprojectionthatresemblesaflower, calleda
petalprojection,sothattherearenonestedloopsleavingandre-enteringthesinglemulti-crossing. Hence,everyknot
hasapetalnumber,whichistheleastnforthesinglen-crossingofapetalprojection. U¨bercrossingnumberandpetal
numberandboundsonthemweredeterminedforavarietyofknotsandlinks.
I would like to thank B. DeMeo, M. Montee, S. Park, S. Venkatesh and F. Yhee for helpful conversations and
especiallyAlexLin,whodeterminedtheskeinrelationsforquadruplecrossingsthatappearbelowandMichaelLandry,
whofirstprovedTheorem2.1below.
2. UPPERBOUNDONQUADRUPLECROSSINGNUMBER
Inthissection,weprovethatq(K)<c(K)forallnontrivialknotsandlinks.Wefirstnotethatq(K)≤c(K).Thiswas
provedin[1], wheretheoperationasinFigure2allowsonetoturneverydoublecrossingintoaquadruplecrossing.
Theobviousgeneralizationimmediatelyimpliesc (K)≥c (K)forallk≥2.
n n+2
FIGURE2. Turningadoublecrossingintoaquadruplecrossing.
Wenotethefollowinggeneralizationoffolding,theoriginalversionofwhichwasalsointroducedin[1]. There,a
closednon-self-intersectingloopintheprojectionplanewascalledacrossingcoveringcircleifitintersectedtheknot
projection only in crossings, and at each crossing, there were two strands of the knot coming out to each side of the
loop. We call the length of the circle the number of crossings it passes through. Given such a circle that crosses an
evennumberofcrossings,wecanperformaquadruplefolding,asinFigure3. Firstwetaketheoverstrandatoneof
thecrossingshitbythecrossingcoveringcircleandwestretchitaroundthecircleasinFigure3(b),eliminatingone
crossing in the process and turning the other crossings on the circle into triple crossings. Then we take the second
strandmakinguptheoriginalcrossing,andwestretchitaroundthecirclealso. Butaswedosowepushitalternately
insideandoutsidethefirststretchedstrand,switchingaswepassthrougheachcrossing. Wethusobtainaprojection
wherethendoublecrossingsthatwereintersectedbythecrossingcoveringcirclehavenowbecomen–1quadruple
crossings.
(a) (b) (c)
FIGURE 3. Using an even length covering crossing circle to turn double crossings into quadruple
crossings.
QUADRUPLECROSSINGNUMBEROFKNOTSANDLINKS 3
Theorem2.1. q(K)<c(K)foreverynontrivialknotandlink.
Proof. Fromwhatwehavesaid,givenaminimalcrossingprojectionPofK,itsufficestofindasinglecrossingcovering
circleofevenlength,sayn.Thenbyquadruplefoldingit,weobtainn–1quadruplecrossings,andtheremainderofthe
doublecrossingscanbeconverteddirectlytoquadruplecrossingsasinFigure2, yieldingfewerquadruplecrossings
thandoublecrossings.
If the projection contains a complementary region with an even number of edges, then we can take our crossing
coveringcircletosurroundit,passingthrougheachofitscrossings,andwearedone. Sowemayassumethatallofthe
complementaryregionsofPhaveanoddnumberofedges.
Chooseacrossingx. Replacetheprojectionwiththecorrespondingplanargraph. Thefourcomplementaryregions
atxmustbealldistinct,sinceiftheywerenot,theoriginalprojectionwouldnothavebeenreduced. Calltheregions
A,B,CandD. LetR=A∪B∪C∪D. IfP–Risconnected,thenwecanformaloopthatshadowstheboundaryofR,
passingalternatelyinsideandoutsidetheouterboundaryofeachofthecomplementaryregions,asinFigure4. That
circlewillhaveevenlength.
A B
D C
FIGURE4. Findinganevenlengthcrossingcoveringcircle.
If P–R is not connected, then two of the regions must touch each other in more than one connected component.
Supposetwooppositeregionstoucheachotheratasecondcrossingy.Thenwecanformacircleoflength2thatpasses
throughthesetworegionsandthecrossingsxandy. Iftwooppositeregionsdonottoucheachother,sayAandC,we
cantakeourcoveringcrossingcircleofevenlengthasinFigure4,shadowingtheboundaryofR,totheoutsideonthe
boundariesofAandCandtotheinsideontheboundariesofBandD.
(cid:3)
3. QUADRUPLECROSSINGNUMBERANDTHEBRACKETPOLYNOMIAL
Inthissection,weconsiderwhatthespanofthebracketpolynomialcantellusaboutquadruplecrossingnumber.
Thebracketpolynomialwasintroducedin[3]andisdefinedbyconsideringallthewaystosplitcrossingsinafixed
projectionPofKandthentaking<P>=ΣAa(s)A–b(s)(–A2–A–2)|s|–1,wherethesumisoverallstatessofP,a(s)isthe
numberofA-splits,b(s)isthenumberofB-splits,and|s|isthenumberofcirclesinthestates. Althoughthebracket
polynomialdependsontheparticularprojection,itsspanisaninvariantforK.
Each quadruple crossing in a quadruple crossing projection will be one of the six possibilities mentioned above.
Whenweresolveasinglequadruplecrossingintosixdoublecrossingsandthensplitthosecrossingseachinthetwo
wayspossible,weobtain64possiblesplittings.Manyofthesecontainadditionalcomponents,butoncethecomponents
areremoved,weareleftwith14possiblesplittings,fourthatconsistoffourparallelstrands,calledparallelsplits,two
that consist of four u-shaped components, called U splits, and eight that have two parallel components and two U
components,calledmixedsplits. SothesecondlineofthefirstskeinrelationbelowconsistsofoneUsplit,twomixed
splitsandtwoparallelsplitsinthatorder.
4 COLINADAMS
Theorem3.1.
1
(3.1) < 2 >= A2(< > + < >)
3
4
+A0(< > + < > + < > + > + < >)
+A–2(< > + < > + < > + < > + < >)
+A–4(< > + < > + < >)+A–6(< >)
1
(3.2) < 4 >= A6 < > + A4(< > + < > + < >)
3
2
+A2(< > + < > + < > + < > + < >)
+A0(< > + < > + < > + < > + < >)
+A–2(< > + < >)
(3.3)
1
< 3 >= A4 < >
2
4
+A2(< > + < > + < > + < >)
+A0(< > + < > + < > + < > + < > + < >)
+A–2(2 < > + < > + < >)+A–4(< >)
(3.4)
1
< 4 >= A4 < > + A2(2 < > + < > + < >)
2
3
+A0(< > + < > + < > + < > + < > + < >)
+A–2(< > + < > + < > + < >)
+A–4(< >)
QUADRUPLECROSSINGNUMBEROFKNOTSANDLINKS 5
1
(3.5) < 3 >= A4 < >
4
2
+A2(< > + < > + < > + < >)
+A0(2 < > + < > + < > + < > + < >)
+A–2(< > + < > + < > + < >)
+A–4(< >)
1
(3.6) < 2 >= A4 < >
4
3
+A2(< > + < > + < > + < >)
+A0(2 < > + < > + < > + < > + < >)
+A–2(< > + < > + < > + < >)
+A–4(< >)
Proof. Thisfollowsimmediatelybyresolvingeachoftheparticularquadruplecrossingsintoitssixdoublecrossings,
andthenapplyingtheskeinrelationforthebracketpolynomialappliedtodoublecrossings(cf.[3])tospliteachofthe
crossings,eliminatetheclosedcomponentsandcleanuptheresult.TheserelationswerefirstdeterminedbyAlexTong
Lin. (cid:3)
Givenacrossingc,definethefirstlevelsplitstobethosesplitscorrespondingtothehighestpowerofAintheskein
relationforthatcrossinginTheorem3.1. Thesecondlevelsplitsaredefinedtobethosethatcorrespondtoapowerof
Atwolessthanthoseofthefirstlevelsplits,thethirdlevelsplitsarethosethatcorrespondtoapowerofAthatisfour
lessthanthoseofthefirstlevelsplits,etc.
Given a particular splitting of a crossing that occurs in a state s, we can perform a split move which replaces two
adjacentstrandsinthesplittingbythetwootherstrandsthatconnecttheirendpointsasinFigure5. Theresultingstate
s(cid:48)willsatisfy|s(cid:48)|=|s|±1.
FIGURE 5. Performingasplitmoveonasplittingchangestheresultantnumberofcomponentsby
1.
Notethateachsplitcanbeobtainedfromanyothersplitbyasequenceofsuchsplitmoves. Eachmovechangesthe
numberofcomponentsinastateby±1.
6 COLINADAMS
Lemma3.2. Letsbeaparticularstate.
(1) Changingasplitinsofagivenleveltooneofalowerlevelthatisonesplitmovedifferentcannotincreasethe
highestpowerofAinthepolynomialassociatedtos.
(2) Changingasplitinsofagivenleveltooneofahigherlevelthatisonesplitmovedifferentcannotdecrease
thelowestpowerofAinthepolynomialassociatedtos.
Proof. In the first case, by the skein relations, the power of A from the relation will go down by at least 2. But the
numberofcomponentsinthestateincreasesbyatmost1,andthereforetheadditional(–A2–A–2)thatcomesfromthe
potentialincreaseinthenumberofcomponentscanatmostoffsetthelossof2inthepowerofA.Hencethenewstate
hasgreatestpowernolargerthantheoriginal. Asimilarargumentholdsforthesecondstatement. (cid:3)
Theorem3.3. Foranynontrivialknotornon-splittablelink,span(<K >)≤16q(K).
Proof. Let P be a quadruple crossing projection of K with q triple crossings. Each crossing is one of our six types.
Defineastatetobes ifitisobtainedbysplittingeachcrossingotherthanc orc asafirstlevelsplit,suchthat
max 1342 1243
ifthereismorethanonefirstlevelsplit(asoccursforc ),wechooseacollectionoffirstlevelsplitsthatmaximizes
1234
thenumberofcomponentcirclesinthestate. Inthecaseofc orc ,weconsidertheirparallelsecondlevelsplit.
1342 1243
We use it rather than the first level split, if it intersects four components of the state or it intersects two components
and the parallel fourth level split intersects four components. We call each of the splittings used to form s a high
max
splitting.
Similarly,definethestates tobethestateobtainedbysplittingeachcrossingotherthanc orc asafifth
min 1342 1243
levelsplit,suchthatifthereisachoiceoffifthlevelsplits(asoccursforc ),wechooseacollectionthatmaximizes
1432
thenumberofcirclesins . Forc andc ,weconsidertheirparallelfourthlevelsplit. Weuseitratherthanthe
min 1342 1243
fifthlevelsplitifweusedthesecondlevelsplitfors . Wecalleachofthesplittingsusedtoforms alowsplitting.
max min
We first prove that there is no state with a higher power of A than s . Any other state is obtained from s by
max max
changingsomenumberofthefirstlevelsplitsonthecrossingsotherthanc orc . Bycheckingthefourskein
1342 1243
relationsforthefourothercrossingsinTheorem3.1,weseethateachofthesecondlevelsplitsisobtainedfromoneof
thefirstlevelsplitsbyasinglesplitmove. Eachthirdlevelsplitisobtainedfromasecondlevelsplitbyasplitmove.
Andsoon. Similarly,eachfourthlevelsplitisobtainedfromafifthlevelsplitbyasinglesplitmove,etc.
Thus, in the cases of c ,c and c , where there is only one first level splitting, Lemma 3.2 implies that
1432 1324 1423
changing the first level split to a second level split results in a new state that has greatest power no larger than the
original.
Inthecaseofc ,wheretherearetwofirstlevelsplittings,eachsecondlevelsplittingisonesplitmoveawayfrom
1234
oneofthefirstlevelsplittings. Ifwechangethefirstlevelsplittingins toasecondlevelsplittingandthatsplitting
max
differsfromthefirstonebyonesplitmove,thenthenumberofcomponentscangoupbyatmostone,andthegreatest
powerofAisunchanged. Ifwechangethefirstlevelsplittingtoasecondlevelsplittingandthenumberofcomponents
goesupbyatleast2,thentheotherfirstlevelsplitting,whichdiffersfromthesecondlevelsplittingbyonlyonesplit
move,wouldhavegeneratedmorecomponentsfors thandidtheonewehad,acontradictiontoourprocedurefor
max
choosingthesplittingsthatforms .
max
Finally,weconsiderthepairofcrossingsc andc . Exceptfortheparallelsecondlevelsplit,allothersecond
1243 1342
levelsplitsdifferfromthefirstlevelsplitbyonesplitmove,andhenceLemma3.2showsthatwecannotincreasethe
highestpowerofAbyswitchingtooneofthose.Thesameholdsforthethird,fourthandfifthlevelsplits.Supposenow
thattheparallelsecondlevelsplitisusedins . Thentherearetwocases. Itcouldbethattheparallelsecondlevel
max
split generates four components, which implies that the the first level split and the fifth level split each generate one
componentandtheparallelfourthlevelsplitgeneratestwo. Oritcouldbethattheparallelfourthlevelsplitgenerates
fourcomponents,whiletheparallelsecondlevelsplitgeneratestwocomponentsandthefirstlevelandfifthlevelsplits
each generate one component. In either case, changing this split cannot yield a higher power of A, since in the first
case,thehighestpowerofAinthepolynomialtermoftheparallelsecondlevelsplitbeatsthatofthefirstlevelsplitby
A4 andinthesecondcase,thehighestpowerofAinthepolynomialtermoftheparallelsecondlevelsplittiesthatof
thefirstlevelsplit,andnoothersplithasahigherpowerofAinitspolynomialtermthanthefirstlevelsplit.
Thesametypeofargumentsapplytoshowthatthepolynomialtermofs possessesthelowestpowerofA.
min
GivenaprojectionP,defineM tobethehighestexponentofAinthepolynomialtermassociatedtos andm to
P max P
bethelowestexponentofAinthepolynomialtermassociatedtos .
min
Let|c |bethenumberofcrossingsoftypec inP. SoΣ|c |=q. Accordingtotheskeinrelation,
ijk ijk ijk
M ≤6|c |+4(|c |+|c |+|c |+|c |)+2|c |+2|s |–2
P 1432 1342 1243 1324 1423 1234 max
QUADRUPLECROSSINGNUMBEROFKNOTSANDLINKS 7
m ≥–2|c |–4(|c |+|c |+|c |+|c |)–6|c |–(2|s |–2)
P 1432 1342 1243 1324 1423 1234 min
Hence,
span(<K >)≤M –m ≤8Σ|c |+2(|s |+|s |)–4
P P ijk max min
=8q+2(|s |+|s |)–4
max min
Thus,itsufficestoprovethat|s |+|s |≤4q+2foranyconnectedquadruplecrossingprojectionwithqquadruple
max min
crossings.
Weapplyinductiononq. Ifthereisasinglequadruplecrossing, thesinglesplittingfors isdeterminedbythe
max
typeofcrossing,butinallcasesthehighsplittingwilleitherbeaU-splittingasinFigure6(a)oraparallelsplittingas
inFigure6(b)and6(c),allowingforrotation. Ifthehighsplittingisoneortwoofthese,thelowsplittingorsplittings
comefromthosethatremain. NotethattherelativeorientationsofthesethreesplittingsmustappearasinFigure6but
allthreecanberotatedtogether.
(a) (b) (c)
FIGURE 6. Splittings that can correspond to high and low states, allowing rotation of all three to-
gether.
Itisthenstraightforwardtocheckthatifthehighsplittingyields4components, thelowsplittingmustyield2. If
the high splitting yields 3 components, the low splitting must yield 1 or 3 components. If the high splitting yields 2
components, thelowsplittingmustyield2or4. Andifthehighsplittingyields1component, thelowsplittingmust
yield1or3. Thus,|s |+|s |≤6.
max min
Supposenowthatwehaveprovedthatforaconnectedprojectionwithqquadruplecrossings,|s |+|s |≤4q+2.
max min
Given a connected quadruple crossing projection P with q+1 quadruple crossings, choose one and split it as a high
split. Intheonecaseofacrossingoftypec wheretherearetwopossiblehighsplits,choosetheonethattogether
1234
withalltheotherhighsplitswouldyieldthegreatestnumberofcomponents. Aftersplitting,wehaveaprojectionP(cid:48)of
alinkwithqquadruplecrossings. Thisprojectionhas1,2,3or4connectedcomponents. Supposefirstthatithasfour
connected components, labelled L ,L ,L and L . Each of the components contains some number of the quadruple
1 2 3 4
crossings,denotedq ,q ,q andq respectively. Welabeltheirmaxandminstatesass(cid:48) ands(cid:48) .
1 2 3 4 max,i min,i
Then by our induction hypothesis, |s(cid:48) |+|s(cid:48) | ≤ 4q +2 for each i. But |s | = (cid:80)|s(cid:48) | since we split the
max,i min,i i max max,i
first crossing as a high split. To obtain |s |, we take |s(cid:48) | for each component, but then we must change the first
min min,i
crossing we set from the high splitting to the low splitting. So the total number of components will go down by 2.
Hence,|s |=(cid:80)4 |s(cid:48) |–2.Thus,|s |+|s |≤(cid:80)4 (4q +2)–2=4q+6=4(q+1)+2aswewantedtoshow.
min i=1 min,i max min i=1 i
Inthecasethatthehighsplitatthefirstcrossingyieldsthreecomponents,thelowsplitatthesamecrossingyields
atmostthreecomponents. Hence,|s |+|s |≤(cid:80)3 (4q +2)=4q+6=4(q+1)+2aswewantedtoshow.
max min i=1 i
Ifthehighsplitatthefirstcrossingyields2components,thefactthelowsplitcanyieldatmost4componentsyields
|s |+|s | ≤ (cid:80)2 (4q +2)+2 = 4q+6 = 4(q+1)+2. Finally, ifthehighsplityieldsonlyonecomponent, the
max min i+1 i
lowsplitcanyieldatmostthreecomponentsso|s | = |s(cid:48) |and|s | ≤ |s(cid:48) |+2. Thus,wehave|s |+|s | ≤
max max min min max min
(4q+2)+2≤4(q+1)+2. Hence,inallcases,|s |+|s |≤4(q+1)+2. (cid:3)
max min
Corollary3.4. ForanynontrivialalternatingknotorlinkK,q(K)≥ c(K).
4
Proof. Resultsof[3],[4]and[7]yieldthefactthatforareducedalternatingknot,
span(<K >)=4c[K]. Hence,byTheorem3.3,4c(K)≤16q(K),yieldingtheresult. (cid:3)
Corollary3.5. LetT denotean(r,s)-torusknot. Thenq(T )≥ r+s–2.
r,s r,s 4
Proof. Thisfollowsimmediatelyfromthefactthatfollowsfromresultsin[3]thatthespanofthebracketpolynomial
forsuchaknotis4(r+s–2). (cid:3)
8 COLINADAMS
Aswewillseeinthenextsection,thisdoesyieldtheexactquadruplecrossingnumberforthe2-braidknots,although
wewillnotusethistoproveit.
4. MOVESTOOBTAINQUADRUPLECROSSINGNUMBER
In this section, we exhibit in Figure 7 a collection of moves that allows one to convert certain standard double
crossingprojectionsofaknotintoquadruplecrossingprojections.WiththesemovesandTheorem3.3wecandetermine
thequadruplecrossingnumberofavarietyofknots.
Moves III and IV apply to regions of a projection containing 3 crossings. We call a region as appears on the left
inthosediagrams a3-set. Notethat MoveIVisinfacta specialcaseofMoveIII. Similarly MovesVandVIapply
to regions of a projection containing 4 crossings. We call such a region a 4-set. A single crossing is a 1-set and an
alternating bigon is a 2-set. In order to turn a standard projection into a quadruple crossing projection, we will be
lookingtosubdivideitinto4-setsand3-sets,withthepossibilityofa1-setor2-setaswell.
Corollary4.1. LetK beanalternatingknotorlinkinareducedalternatingprojectionPwithccrossings.
(1) Ifc=4kforsomek≥0,andareducedalternatingprojectioncanbesubdividedinto4-sets,thenq(K)= c.
4
(2) If c = 4k+1, anda reduced alternating projectioncan be subdivided intoa disjoint collection ofall 4-sets,
anda3-setanda2-set,orall4-setsanda1-set,orall4-setsandthree3-sets,thenq(K)= c+3.
4
(3) Ifc=4k+2,andareducedalternatingprojectioncanbesubdividedintoadisjointcollectionofall4-setsand
a2-set,orall4-sets,andtwo3-sets,thenq(K)= c+2.
4
(4) Ifc=4k+3,andareducedalternatingprojectioncanbesubdividedintoadisjointcollectionofall4-setsand
one3-set,thenq(K)= c+1.
4
Proof. By Theorem 3.3, we obtain a lower bound on quadruple crossing number which exactly matches the upper
boundobtainedbythenumberofsetsinoursubdivisionoftheprojection,eachsetofwhichbecomesasinglequadruple
crossing. (cid:3)
This corollary is relatively effective at determining the quadruple crossing number of small crossing alternating
knots. (See Table 1.) For instance, for the 5,6,7 and 9-crossing prime alternating knots, it determines exactly the
quadruple crossing number for all except 7 ,7 ,9 ,9 ,9 ,9 ,9 , and 9 . It is less effective at determining the
6 7 26 31 32 33 34 40
quadruplecrossingnumberofthe8-crossingknots,onlydeterminingq=2for8 ,8 and8 . However,bydetermining
3 4 9
allknotsgeneratedbyprojectionswithq = 2,andusingMoveVIIon8 ,8 ad8 ,thequadruplecrossingnumber
19 20 21
of 7 and 7 and all 8-crossing knots except four is determined. Corollary 4.1 also allows us to determine that the
6 7
quadruplecrossingnumberis3formanyoftheprime10-crossingknots. Notethatfor10 and10 ,onemustgoto
33 34
non-alternating11-crossingdiagramstoobtainthenecessarydecompositionintosets.Anyprime10-crossingknotthat
doesnotappearinthetablehasunknownquadruplecrossingnumber.
Corollary4.1appliestovariousinfinitefamiliesofknotsandlinks,includingavarietyofpretzelknotsandrational
knots. Wementiontwoparticularlysimplefamilieshere.
Corollary4.2. (1) LetK bea2-braidknotorlinkofncrossings. Thenq(K )=(cid:100)n(cid:101).
n n 4
(2) LetJ bearationalknotwithalternatingConwaynotation3a. Thenq(J )=(cid:100)3+a(cid:101).
n n 4
Proof. InthecaseofK ,thefourcasesofCorollary4.1yieldtheresult. InthecaseofJ ,thereisa4-setcomingfrom
n n
the3andthefirstcrossinginthesequenceofacrossings. Theremaininga–1crossingscanbeeithersubdividedinto
all4-sets,if3+a=4k,orone1-setandtherest4-setsif3+a=4k+1,orone2-setandtherest4-setsif3+a=4k+2
orone3-setandtherest4-setsif3+a=4k+3. (cid:3)
REFERENCES
[1] C.Adams,Triplecrossingnumberofknotsandlinks,ArXiv:1207.7332,toappearintheJournalofKnotTheoryanditsRamifications(2012),
[2] C.Adams,T.Crawford,B.DeMeo,M.Landry,A.Lin,M.Montee,S.Park,S.Venkatesh,F.Yhee,Knotprojectionswithasinglemulti-
crossing,ArXiv:1208.5742(2012),
[3] V.F.R.Jones,Heckealgebrarepresentationsofbraidgroupsandlinkpolynomials,Ann.ofMath.126(1987),335388.
[3] L.Kauffman,Newinvariantsinthetheoryofknots,Amer.Math.Monthly95(1988),195242.
[4] K.Murasugi,Jonespolynomialsofalternatinglinks, Trans.Amer.Math.Soc.295(1986),147174.
[5] J.Pach,G.To´th,Degeneratingcrossingnumbers,DiscreteComput.Geom.41(2009),376384.
[6] H.Tanaka,M.Teragaito,Triplecrossingnumbersofgraphs,ArXiv:1002.4231(2010),
[7] M.Thistlethwaite,AspanningtreeexpansionoftheJonespolynomial,Topology26(1987),297–309.
QUADRUPLECROSSINGNUMBEROFKNOTSANDLINKS 9
(a)MoveI(holdsforany
choiceatcrossing)
(b)MoveII(holdsforany
choiceatcrossings)
(c)MoveIII(holdsforany
choiceatcrossing)
(d)MoveIV
(e)MoveV
(f)MoveVI
(g)MoveVII
FIGURE7. Movestoobtainquadruplecrossings.
DEPARTMENTOFMATHEMATICSANDSTATISTICS,WILLIAMSCOLLEGE,WILLIAMSTOWN,MA01267
E-mailaddress:[email protected]
10 COLINADAMS
Knot q(K) Knot q(K) Knot q(K) Knot q(K)
3 1 9 3 10 3 10 ?
1 8 1 43
4 2 9 3 10 3 10 ?
1 9 2 44
5 2 9 0 3 10 3 10 ?
1 1 3 45
5 2 9 3 10 3 10 3
2 11 4 46
6 2 9 3 10 3 10 3
1 12 5 47
6 2 9 3 10 3 10 3
2 13 6 48
6 2 9 3 10 3 10 3
3 14 7 49
7 2 9 3 10 3 10 3
1 15 8 50
7 2 9 3 10 3 10 3
2 16 9 51
7 2 9 3 10 ? 10 3
3 17 10 52
7 2 9 3 10 3 10 3
4 18 11 53
7 2 9 3 10 3 10 ?
5 19 12 54
7 3 9 3 10 3 10 ?
6 20 13 55
7 3 9 3 10 3 10 ?
7 21 14 56
8 3 9 3 10 3 10 ?
1 22 15 57
8 3 9 3 10 3 10 ?
2 23 16 58
8 2 9 3 10 ? 10 ?
3 24 17 59
8 2 9 3 10 3 10 ?
4 25 18 60
8 3 9 ? 10 3 10 3
5 26 19 61
8 3 9 3 10 3 10 3
6 27 20 62
8 3 9 3 10 3 10 3
7 28 21 63
8 3 9 3 10 3 10 3
8 29 22 64
8 2 9 3 10 3 10 3
9 30 23 65
8 3 9 ? 10 3 10 3
10 31 24 66
8 3 9 ? 10 3 10 3
11 32 25 74
8 3 9 ? 10 3 10 3
12 33 26 76
8 ? 9 ? 10 ? 10 3
13 34 27 77
8 3 9 3 10 3 10 3
14 35 28 124
8 3 9 3 10 3 10 3
15 36 29 125
8 ? 9 3 10 3 10 3
16 37 30 126
8 ? 9 3 10 3 10 3
17 38 31 127
8 ? 9 3 10 3 10 3
18 39 32 130
8 2 9 ? 10 3 10 3
19 40 33 131
8 2 9 3 10 3 10 3
20 41 34 134
8 2 9 3 10 3 10 3
21 42 35 135
9 3 9 3 10 3 10 3
1 43 36 139
9 3 9 3 10 ? 10 3
2 44 37 140
9 3 9 3 10 ? 10 3
3 45 38 142
9 3 9 2 10 ? 10 3
4 46 39 144
9 3 9 ? 10 ? 10 3
5 47 40 142
9 3 9 3 10 ? 10 3
6 48 41 148
9 3 9 3 10 ? 10 3
7 49 42 161
TABLE1. Knownquadruplecrossingnumbers.