Table Of ContentTHE 27 POSSIBLE INTRINSIC SYMMETRY GROUPS OF TWO-COMPONENT LINKS
JASONCANTARELLA,JAMESCORNISH,MATTMASTIN,ANDJASONPARSLEY
ABSTRACT. Weconsiderthe“intrinsic”symmetrygroupofatwo-componentlinkL, definedtobetheim-
age Σ(L) of the natural homomorphism from the standard symmetry group MCG(S3,L) to the product
MCG(S3)×MCG(L).Thisgroup,firstdefinedbyWhittenin1969,recordsdirectlywhetherLisisotopicto
alinkL(cid:48)obtainedfromLbypermutingcomponentsorreversingorientations;itisasubgroupofΓ ,thegroup
2
2 ofallsuchoperations.
1 Fortwo-componentlinks,wecatalogthe27possibleintrinsicsymmetrygroups,whichrepresentthesub-
0
groupsofΓ uptoconjugacy. Weareabletoprovideprime,nonsplitexamplesfor21ofthesegroups;some
2
2
areclassicallyknown,somearenew.Wecatalogthefrequencyatwhicheachgroupappearsamongall77,036
n of the hyperbolic two-component links of 14 or fewer crossings in Thistlethwaite’s table. We also provide
a somenewinformationaboutsymmetrygroupsofthe293non-hyperbolictwo-componentlinksof14orfewer
J
crossingsinthetable.
3
1
]
T 1. INTRODUCTION
G
1.1. Symmetries ofLinks. The symmetrygroup of a linkLis awell-studied construction inknot theory.
.
h Itisdefinedtobethemappingclassgroupofthepair(S3,L)andisfrequentlydenotedSym(L). Avariety
t
a ofapproachesexistforcomputingsymmetrygroupsforprimeknotsandlinks;seeforinstance[3,4,6,8].
m
Thispaperconsidersadifferentbutrelatedgroupoflinksymmetries. Theimageofthegrouphomomor-
[ phism
1 π:Sym(L) = MCG(S3,L) → MCG(S3)×MCG(L).
v
represents what we call intrinsic link symmetries, where we focus upon the action on L itself and only
2
2 recordtheorientationoftheambientS3.
7 Without using the language of mapping class groups Whitten [9], following ideas of Fox, defined the
2 groupstructureofMCG(S3)×MCG(L)in1969. TheydenotedthisgroupaseitherΓ(L)orΓ ,whereµ
. µ
1 isthenumberofcomponentsofL. Thisgroup(cf.Definition2.1)canbedescribedasasemidirectproduct
0
of Z groups recording the orientation of each component of the link L with the permutation group S
2 2 µ
1 representingcomponentexchanges,crossedwithanotherZ2 recordingtheorientationofS3:
:
v Γ = Γ(L) = Z ×(Zµ(cid:111)S ).
µ 2 2 µ
i
X An element γ = ((cid:15) ,(cid:15) ,...,(cid:15) ,p) in Γ acts on L to produce a new link Lγ. If (cid:15) = +1, then Lγ and
0 1 µ µ 0
r L are the same as sets; however the components of L have been renumbered and reoriented. If (cid:15) = −1,
a 0
thenewlinkLγ isthemirrorimageofL,againwithrenumberingandreorientation. Wecanthendefinethe
intrinsicsymmetrygroupΣ(L)by
γ ∈ Σ(L) ⇐⇒ thereisanisotopyfromLtoLγ preservingcomponentnumberingandorientation.
Werefertoourpaper[2]forathoroughdescriptionofthehistory,construction,andapplicationsofintrinsic
symmetry groups. In that paper, we find the intrinsic symmetry groups for all 48 prime links of eight or
fewercrossings.
The goal of this paper is to describe how frequently the various possible intrinsic symmetry groups of
two-componentlinksoccurinexamples. Thereare27possiblesymmetrygroups,representingthedifferent
Date:Revised:January16,2012.
Keywordsandphrases. two-componentlinks,symmetrygroupofknot,linksymmetry,Whittengroup.
1
subgroupsofΓ uptoconjugacy. Thesubgroupsareonlyconsidereduptoconjugacyasconjugationcorre-
2
sponds to merely relabeling and reorienting the components of the link; see section 2.3. Figure 1 displays
the subgroup lattice of Γ . In Table 2, we present prime, nonsplit examples for 21 of these groups; some
2
examplesarealreadyknown,somearenew. ManynewexampleswerefoundusingthesoftwareSnapPea;
see section 3 for details. We present in Table 1 a report on how frequently each symmetry group occurs
amongthe77,036two-componenthyperboliclinkswith14orfewercrossings.
Insection5,werestricttothecaseofalternating,nonsplittwo-componentlinks,forwhichonly12ofthe
27symmetrygroupsarepossible. Inparticular,noalternatinglinkwithanevennumberofcomponentsmay
havefullsymmetry. Werealizealternating,primeexamplesfor11ofthese12groups.
2. THE WHITTEN GROUPΓ2
WebeginbygivingthedetailsofourconstructionoftheWhittengroupΓ andthesymmetrygroupΣ(L).
µ
Consider operations on an oriented, labeled link L with µ components. We may reverse the orientation of
anyofthecomponentsofLorpermutethecomponentsofLbyanyelementofthepermutationgroupS .
µ
However, these operations must interact with each as well: if we reverse component 3 and exchange com-
ponents3and5,wemustdecidewhethertheorientationisreversedbeforeorafterthepermutation. Further,
we can reverse the orientation on the ambient S3 as well, a process which is clearly unaffected by the
permutation. Toformalizeourchoices,wefollow[9]tointroducetheWhittengroupofaµ-componentlink.
Definition2.1. Considerthehomomorphismgivenby
ω : S (cid:55)−→ Aut(Zµ+1), p (cid:55)−→ ω(p)
µ 2
whereω(p)isdefinedas
(cid:0) (cid:1)
ω(p)((cid:15) ,(cid:15) ,(cid:15) ...(cid:15) ) = (cid:15) ,(cid:15) ,(cid:15) ...(cid:15) .
0 1 2 µ 0 p(1) p(2) p(µ)
Forγ = ((cid:15) ,(cid:15) ,...(cid:15) ,p),andγ(cid:48) = (cid:0)(cid:15)(cid:48),(cid:15)(cid:48),...(cid:15)(cid:48) ,q(cid:1) ∈ Zµ+1 (cid:111) S , wedefinetheWhittengroupΓ asthe
0 1 µ 0 1 µ 2 ω µ µ
semidirectproductΓ = Zµ+1(cid:111) S withthegroupoperation
µ 2 ω µ
γ ∗γ(cid:48) = ((cid:15) ,(cid:15) ,(cid:15) ...(cid:15) ,p)∗(cid:0)(cid:15)(cid:48),(cid:15)(cid:48),(cid:15)(cid:48)...(cid:15)(cid:48) ,q(cid:1)
0 1 2 µ 0 1 2 µ
= (cid:0)((cid:15) ,(cid:15) ,(cid:15) ...(cid:15) )·ω(p)((cid:15)(cid:48),(cid:15)(cid:48),(cid:15)(cid:48)...(cid:15)(cid:48) ),qp(cid:1)
0 1 2 µ 0 1 2 µ
(cid:16) (cid:17)
= (cid:15) (cid:15)(cid:48),(cid:15) (cid:15)(cid:48) ,(cid:15) (cid:15)(cid:48) ...(cid:15) (cid:15)(cid:48) ,qp .
0 0 1 p(1) 2 p(2) µ p(µ)
2.1. Intrinsicsymmetriesofalink. GivenalinkLconsistingofµorientedknotsinS3,wemayorderthe
knotssothat
L = K ∪K ∪···∪K .
1 2 µ
ConsiderthefollowingoperationsonL:
(1) PermutingtheK .
i
(2) ReversingtheorientationofanysetofK ’s.
i
(3) ReversingtheorientationonS3 (mirroringL).
Letγ beacombinationofanyofthemoves(1),(2),or(3). Wethinkofγ = ((cid:15) ,(cid:15) ,...(cid:15) ,p)asanelement
0 1 µ
ofthesetZµ+1×S inthefollowingway. Let
2 µ
(cid:40)
−1, ifγ mirrorsL
(cid:15) =
0
+1, ifγ doesnotmirrorL
and
(cid:40)
−1, ifγ reversestheorientationofK
(cid:15) = p(i)
i
+1, ifγ doesnotreversetheorientationofK
p(i)
2
Lastly, let p ∈ S be the permutation of the K associated to γ. To be explicitly clear, permutation p
µ i
permutes the labels of the components; the component originally labeled i will be labeled p(i) after the
actionofγ.
Foreachelement,γ inZµ+1×S ,wedefine
2 µ
µ
(1) Lγ = γ(L) = (cid:15) K(∗) ∪(cid:15) K(∗) ∪···∪(cid:15) K(∗) = (cid:91)(cid:15) K(∗)
1 p(1) 2 p(2) µ p(µ) i p(i)
i=1
where−K isK withorientationreversed,K∗ isthemirrorimageofK andthe(∗)appearsaboveifand
i i i i
(∗)
only if (cid:15) = −1. Note that the ith component of γ(L) is (cid:15) K the possibly reversed or mirrored p(i)th
0 i p(i)
component of L. Since we are applying (cid:15) instead of (cid:15) to K we are taking the convention of first
i p(i) p(i)
permutingandthenreversingtheappropriatecomponents.
WecannowdefinethesubgroupofΓ(L)whichcorrespondstothesymmetriesofthelinkL.
Definition 2.2. Given a link, L and γ ∈ Γ(L), we say that L admits γ when there exists an isotopy tak-
ing each component of L to the corresponding component of Lγ which respects the orientations of the
components. Wedefineastheintrinsicsymmetrygroup(alsocalledtheWhittensymmetrygroup)ofL,
Σ(L) := {γ ∈ Γ(L)|Ladmitsγ}.
TheintrinsicsymmetrygroupΣ(L)isasubgroupofΓ ,anditsleftcosetsrepresentthedifferentisotopy
µ
classesoflinksLγ amongallsymmetriesγ. Bycountingthenumberofcosets,wedeterminethenumberof
(labeled,oriented)isotopyclassesofaparticularprimelink.
Next, we provide a few examples of symmetry subgroups. Recall that the first Whitten group Γ =
1
Z ×Z hasorderfourandthatΓ = Z ×(Z ×Z (cid:111)S )isanonabelian16elementgroup.
2 2 2 2 2 2 2
Example2.3. LetL = 3 ,atrefoilknot. ItiswellknownthatL ∼ −LandL∗ ∼ −L∗,butL (cid:28) L∗,sowe
1
haveΣ(3 ) = {(1,1,e),(1,−1,e)}. ThismeansthatthetwocosetsofΣ(3 )are{(1,1,e),(1,−1,e)}and
1 1
{(−1,−1,e),(−1,1,e)},andtherearetwoisotopyclassesof3 knots. Atrefoilknotisthusinvertible. ♦
1
Example 2.4. Let L = 82 , whose components are an unknot K and a trefoil K . Since the components
12 1 2
K andK areofdifferentknottypes,weconcludethatnosymmetryinΣ(82 )cancontainthepermutation
1 2 12
(12). Aswedescribein[2],certainsymmetriesareimpossibleforalternatinglinkswithnonzeroself-writhe,
suchas82 . Inparticular,suchalinkneveradmitsasymmetrythatreversestheambientorientation,i.e.,one
12
with (cid:15) = −1. Together these two obstructions rule out 12 of the 16 elements in Γ , so we may conclude
0 2
thatΣ(82 ))hasorderfourorless.
12
It is not difficult to find isotopies that show that L is purely invertible, meaning isotopic to −L =
−K ∪ −K , and individually invertible in the first component, meaning isotopic to −K ∪ K . Thus,
1 2 1 2
both (1,−1,−1,e) and (1,−1,1,e) are included in the Whitten group for L. Hence, Σ(L) is the four el-
ement group Σ = {(1,1,1,e),(1,−1,−1,e),(1,−1,1,e),(1,1,−1,e)}. There are four cosets of this
4,2
four element group in the 16 element group Γ , so there are four (labeled, oriented) isotopy classes of 82
2 12
links. ♦
2.2. Notation. We introduce some notation and names for commonly occurring symmetries. Let m =
(−1,1,1,e) describe a mirror of the link and ρ = (1,1,1,(12)) denote a pure exchange symmetry. In-
verting component k is denoted by i = (1,ε ,ε ,e) where ε = −1 and ε = 1. Let a = ρ ∗
k 1 2 k 3−k k
i = (1,ε ,ε ,(12)); observe that these elements have order 4. A link has pure invertible symmetry if
k 1 2
PI = (1,−1,−1,e) ∈ Σ(L). Finally,letb = PI∗ρ = ρ∗PI = (1,−1,−1,(12)). Weshallsometimesuse
thisnotationinaproductform,e.g.,ma = (−1,−1,1,(12)).
1
Henceforth, ‘symmetry group’ will always refer to the intrinsic symmetry group Σ(L). We use the
notationfrom[2]forsymmetrygroups: Σ isthejthsubgroupoforderk inoursubgrouplatticeforΓ .
k,j 2
3
2.3. SubgroupsofΓ . Inordertoclassifythepossiblesymmetrygroupsoftwo-componentlinks,wemust
2
find the subgroups of Γ . For any γ ∈ Γ , the symmetry groups Σ(L) and Σ(Lγ) are conjugate via γ.
2 2
ThereforeifalinkLhassymmetrygroupΣ,whichisconjugatetogroupΣ(cid:48),thenthereexistssomeγ ∈ Γ
2
suchthatLγ hassymmetrygroupΣ(cid:48);linkLγ ismerelyarelabeledandreorientedcopyofL. Thus,thetruly
differentsymmetrygroupsarerepresentedbytheconjugacyclassesofΓ subgroups.
2
Proposition 2.5. There are 35 subgroups of Γ and 27 mutually nonconjugate subgroups. A maximal set
2
of nonconjugate subgroups consists of 7 subgroups of order two denoted Σ through Σ , 11 subgroups
2,1 2,7
oforderfourdenotedΣ throughΣ andsevensubgroupsofordereightdenotedΣ throughΣ ,as
4,1 4,11 8,1 8,7
wellasthetrivialsubgroupandthefullgroup. Generatorsforthesegroupsandthelatticestructureofthe
subgroupsappearinFigure1.
Wenotethatthispropositioncorrectsamistakeintheliterature[7],inwhichitisreportedthat28possible
symmetrygroupsexist.
Proof. Fromthesemidirectproductstructurewecanseethatallbutfourofthe15non-identityelementsof
Γ are order two. The exceptions are a = (1,−1,1,(12)), a = (1,1,−1,(12)), ma , and ma , each of
2 1 2 1 2
whichhasorderfour. Thus,thereare11subgroupsofΓ ofordertwo. ViacomputationsinΓ ,weproceed
2 2
tofind15subgroupsoforderfourand7subgroupsofordereight. Combinedwiththetrivialsubgroupand
allofΓ ,wearriveatatotalof35subgroups.
2
However, some of these subgroups occur in conjugate pairs. For example, the subgroups (cid:104)m,ρ(cid:105) =
{e,m,ρ,mρ} and (cid:104)m,b(cid:105) = {e,m,b,mb} are conjugate four-element subgroups, since i (m)i−1 = m,
1 1
i (ρ)i−1 = b,andi (mρ)i−1 = mb.
1 1 1 1
Therearefourconjugatepairsoffour-elementsubgroupsandfourconjugatepairsofeight-elementsub-
groups. Therefore precisely 27 mutually nonconjugate subgroups of Γ exist. Our calculations (by hand)
2
oftheseΓ subgroupsandthecorrespondinglatticestructuregiveninFigure1havebeenverifiedusingthe
2
softwareMagma. (cid:3)
3. COMPUTATIONAL EXAMPLES
ForahyperboliclinkL,wemaycomputeitssymmetrygroupΣ(L)usingthesoftwareSnapPea. This
software can calculate the mapping class group MCG(S3 \ L) of the link complement. The elements
of MCG(S3,L) which extend through the boundary tori to automorphisms of all of S3 form a copy of
MCG(S3,L)insideMCG(S3\L). Wecandetectsuchmapsusingthefollowingstandardlemma.
Lemma 3.1. A map in Aut(S3 \L) extends to all of S3 if and only if it sends meridians of the boundary
toritomeridians. Moreover,anytwosuchextensionsareisotopic.
We utilize the Python front end SnapPy for SnapPea written by Marc Culler and Nathan Dunfield.
Toeachmapontheboundarytoriofthelinkcomplement,SnapPyassignsacollectionofµmatricesalong
with a permutation element which records how the components of the link were permuted. Each matrix is
2×2andrecordstheimagesofthemeridiansandlongitudesoftheappropriatecomponent,alongwiththe
orientationoftheambientspace. Theeffectofthemapontheorientation(cid:15) ofthegivencomponentandthe
i
orientation(cid:15) ofS3 isgivenbytherulesbelow. Notethatifthematrixforoneboundarytorusindicatesthat
0
the orientation on S3 is reversed, then so will the matrices for all other boundary tori, since these matrices
resultfromrestrictingasinglemaponS3.
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
1 n −1 n 1 n −1 n
0 1 0 1 0 −1 0 −1
⇓ ⇓ ⇓ ⇓
(cid:15) = 1,(cid:15) = 1 (cid:15) = −1,(cid:15) = 1 (cid:15) = −1,(cid:15) = −1 (cid:15) = 1,(cid:15) = −1
0 i 0 i 0 i 0 i
4
Γ
2
Σ Σ Σ Σ Σ Σ Σ
8,7 8,2 8,1 8,4 8,6 8,3 8,5
(cid:104)i1,ma(cid:105) ε0ε1ε2=1 Nom (cid:104)m,a(cid:105) (cid:104)mi1,a(cid:105) Noρ ε1 =ε2
* * * *
Σ Σ Σ Σ Σ Σ Σ Σ Σ Σ Σ
4,2 4,4 4,1 4,11 4,3 4,5 4,9 4,6 4,7 4,8 4,10
(cid:104)i1,i2(cid:105) (cid:104)a(cid:105) (cid:104)PI,ρ(cid:105) (cid:104)mb,mρ(cid:105) (cid:104)mi1,mi2(cid:105) (cid:104)ma(cid:105) (cid:104)m,PI(cid:105) (cid:104)m,i1(cid:105) (cid:104)mi1,i2(cid:105) (cid:104)m,ρ(cid:105) (cid:104)mPI,mb(cid:105)
* * * *
Σ Σ Σ Σ Σ Σ Σ
2,2 2,3 2,1 2,4 2,5 2,6 2,7
(cid:104)m(cid:105) (cid:104)mPI(cid:105) (cid:104)PI(cid:105) (cid:104)i (cid:105) (cid:104)ρ(cid:105) (cid:104)mρ(cid:105) (cid:104)mi (cid:105)
1 1
{id}
FIGURE 1. The 27 subgroups (up to conjugacy) of the Whitten group Γ2 are depicted
above as a subgroup lattice. There are 8 pairs of conjugate subgroups of Γ ; above, one
2
representativeisdisplayedforeachconjugatepair;thesearedenotedby∗. For21ofthese
subgroups, we provide an example realizing this symmetry group. The 6 subgroups for
whichnoexamplehasbeenfoundaredepictedwithareddashedborder.
Using these rules, it is easy to extract Σ(L) from SnapPy. We computed the symmetry group for all
77,036 two-component hyperbolic links of 14 and fewer crossings in SnapPy’s database. Table 1 shows
the census of symmetry groups of hyperbolic links found using SnapPy. The data file containing the
5
WhittengroupelementsfortheselinksisincludedintheArxivdatarepositoryanditisafutureprojectto
incorporateourcomputationaltechniquesintoSnapPy.
Group NumberofLinks Group NumberofLinks Group NumberofLinks
{id} 53484 Σ 1396 Σ 52
4,1 8,1
Σ 2167 Σ 25
4,2 8,2
Σ 24 Σ 0
4,3 8,3
Σ 17951 Σ 0 Σ 1
2,1 4,4 8,4
Σ 7 Σ 12 Σ 2
2,2 4,5 8,5
Σ 9 Σ 0 Σ 0
2,3 4,6 8,6
Σ 1336 Σ 0 Σ 8
2,4 4,7 8,7
Σ 418 Σ 2
2,5 4,8
Σ 3 Σ 11
2,6 4,9
Σ 123 Σ 1 Γ 0
2,7 4,10 2
Σ 4
4,11
TABLE 1. The number of links admitting each symmetry group among the 77,036 hyper-
bolic two-component links included in the Thistlethwaite table distributed with SnapPy.
Amongtheselinks,almosttwo-thirdshadnosymmetry(trivialsymmetrygroup)andthere
werenoexampleswithfullsymmetry.
Thereare293non-hyperbolictwo-componentlinkswithupto14crossingsinThistlethwaite’slinktable.
For these links, we were unable to compute the symmetry groups using SnapPea. We were able to get
partialinformation aboutthe symmetrygroups using theMathematicapackageKnotTheory[1]. For
each link L and each Whitten group element γ, we computed a number of invariants of L and Lγ in an
attempt to rule out γ as an element of Σ(L). To rule out “exchange” symmetries, we applied two tests.
First, we computed the Jones polynomials of each component of L to try to rule out exchanges between
componentsofdifferentknottypes. Ifthatfailed,weturnedtothe“satellitelemma”,whichweuseoftenin
[2].
Lemma 3.2. Suppose that L(K,i) is a satellite of L constructed by replacing component i with a knot or
linkK. ThenLcannothaveapureexchangesymmetryexchangingcomponentsiandj unlessL(K,i)and
L(K,j)areisotopic.
Proof. Such a pure exchange would carry an oriented solid tube around L to a corresponding oriented
i
solid tube around L . If we imagine K embedded in this tube, this generates an isotopy between L(K,i)
j
andL(K,j). (cid:3)
The point of this lemma is that we can often distinguish L(K,i) and L(K,j) using classical invariants
which are insensitive to the original labeling of the link. In our computations, we replaced component i of
LandLγ witha(2,2i)toruslink(fori > 0)andcomparedtheJonespolynomialsoftheresultinglinks. An
exampleisshowninFigure2.
Ruling out elements of the Whitten group using these methods gave us a subgroup of Γ guaranteed to
2
containΣ(L). Wefound9linkswithΣ(L) < Σ ,89linkswithΣ(L) < Σ ,194linkswithΣ(L) < Σ
4,1 4,2 2,1
and 1 link with Σ(L) < Σ . Consulting the subgroup lattice of Γ along with Table 1, we saw that only
8,1 2
thislastlink,the194th14-crossingtwocomponentnon-hyperboliclinkinThistlethwaite’stable,mighthave
hadanewsymmetrygroupwhichwasnotalreadyrepresentedbyahyperboliclink. Thepotentialgroupwas
Σ < Σ . However,rearrangingthediagramasinFigure3ruledoutΣ byrevealingapureexchange
4,4 8,1 4,4
symmetry.
6
FIGURE2. Thelink72(or7a1inThistlethwaite’snotation)hastwounknottedcomponents
6
with linking number zero. To decide whether this link admits a pure exchange symmetry,
we construct two satellite links. The center link has one component replaced by a Hopf
link while the other is unframed, while the right link has the other component replaced by
a Hopf link. If the original 72 admits a pure exchange symmetry, these satellites would be
6
isotopic. HowevertheJonespolynomialofthecenterlinkisa10−2a9+a8−a6+2a5−
1 1 1 1
a4 − +2a3 + + − +2 while the Jones polynomial of the right-hand link is
a4 a3 a2 a
1 2 2
a7−2a6+2a5−2a4+2a3− +a2+ − +3. Thisprovesthat72 doesnotadmit
a3 a2 a 6
apureexchangesymmetry.
FIGURE 3. The 194th non-hyperbolic two-component link with 14 crossings in Thistleth-
waite’s table has Σ(L) < Σ according to our polynomial and satellite lemma calcula-
8,1
tions. Theoriginaldiagramofthislinkisshownatleft. SinceΣ < Σ ,thislinkcould
4,4 8,1
have the “missing” symmetry group Σ . However, rearranging the diagram as shown on
4,4
therightrevealsthatthislinkhaspureexchangesymmetry. Sincepureexchangeisnotpart
ofΣ ,weseethatthislinkcannotbeanexampleofalinkwithΣ symmetry.
4,4 4,4
HavingobtainedsupergroupsoftheΣ(L)groupsforthesenon-hyperboliclinks, weturnedtoobtaining
subgroups of Σ(L). To do so, we viewed each diagram as a polyhedral decomposition of the 2-sphere and
enumerated the combinatorial symmetries of each polyhedron using Mathematica. Those symmetries
whichextendedtosymmetriesofthelinkaftercrossinginformationwastakenintoaccountwereclassified
accordingtotheWhittenelementtheyrepresented. Wecalledthesediagrammaticsymmetriesofeachlink.
Ofcourse,differentdiagramsareexpectedtorevealadditionalsymmetries,sothediagrammaticsymmetry
7
groupofadiagramofalinkisonlyasubgroupofΣ(L). Inmostcases,westillhaveanumberofpotential
symmetrieswhichmayormaynotbepresentforthelink. However,wefound70caseswherethediagram-
maticsymmetrygroupsrepresentedallofΣ(L)sincetheyagreedwiththe“supergroups”computedearlier:
6linkswithΣ(L) = Σ ,4linkswithΣ(L) = Σ ,and60linkswithΣ(L) = Σ .
4,1 4,2 2,1
4. EXAMPLES OF LINKS WITH PARTICULAR SYMMETRY GROUPS
Table2listsexamplesofprime,nonsplitlinksbytheirintrinsicsymmetrygroups. Wepresentexamples
for21ofthe27differentsymmetrygroups.
SymmetryGroup Example
{id} 11a164
Σ = (cid:104)m(cid:105) DT[8,10,-16:12,14,24,2,-20,-22,-4,-6,-18]
2,2
Σ = (cid:104)mPI(cid:105) DT[8,10,-14:12,22,24,-18,-20,-4,-6,-16,2]
2,3
Σ = (cid:104)PI(cid:105) 7a2
2,1
Σ = (cid:104)i (cid:105) 10a98
2,4 1
Σ = (cid:104)ρ(cid:105) DT[14,16,44,20,4,2,18,10,12,8,28:36,38,22,42,26,24,40,32,34,30,6]
2,5
Σ = (cid:104)mρ(cid:105) DT[44,-20,-16,-22,-14,-4,-6,-10,-12,-28,-8:2,34,-18,24,42,26,40,30,32,36,38]
2,6
Σ = (cid:104)mi (cid:105) 10a81
2,7 1
Σ 7a3
4,2
Σ Noexampleknown
4,4
Σ 4a1
4,1
Σ DT[16,-6,-12,-20,-22,-4,28:2,14,26,-10,-8,18,24]
4,11
Σ 10n46,10a56
4,3
Σ DT[14,6,10,16,4,20:22,8,2,24,12,18]
4,5
Σ 10n36
4,9
Σ Noexampleknown
4,6
Σ Noexampleknown
4,7
Σ DT[10,-14,-18,24:2,28,-4,-12,-20,-6,-16,26,8,22]
4,8
Σ DT[10,-14,-20,24:2,28,-16,-4,-12,-6,-18,26,8,22]
4,10
Σ 10n59
8,7
Σ 2a1
8,2
Σ 5a1
8,1
Σ DT[14,-16,22,-28,24,-18:12,-20,-10,-2,26,8,5,-6]
8,4
Σ Noexampleknown
8,6
Σ Noexampleknown
8,3
Σ DT[10,-14,-18,22:2,24,-4,-12,-6,-16,8,10]
8,5
Γ Noexampleknown
2
TABLE 2. Examplesofprime, nontrivialtwo-componentlinksarecurrentlyknownfor21
ofthepossible27intrinsicsymmetrygroups.
Theexamplelinksforthegroups{id}andΣ wereoriginallyfoundbyHillman[7]. Helistedexamples
8,5
formanyofthe12subgroupsthatdonotincludei ori inthatpaperaswell. Theexamplesforthegroups
1 2
Σ ,Σ ,Σ ,Σ ,andΣ weregenerallyknown;forexample,proofsthattheselinksexhibitthegiven
8,1 8,2 4,1 4,2 2,1
symmetrygroupsarefoundin[2].
8
FIGURE 4. Three new example links for symmetry groups, not found using SnapPea.
Fromlefttoright,theselinksexhibitthesymmetrygroups(cid:104)ρ(cid:105),(cid:104)mρ(cid:105)andΣ .
8,7
The examples for (cid:104)ρ(cid:105), (cid:104)mρ(cid:105) and Σ are found in Figure 4 and are new to the literature; we prove that
8,7
thesesymmetrygroupsarecorrectbelow. AllotherexampleshavebeenfoundusingSnapPeaandarenew
totheliterature.
Theorem4.1. ThethreelinksinFigure4possessthesymmetrygroups(cid:104)ρ(cid:105),(cid:104)mρ(cid:105)andΣ ,respectively.
8,7
Proof. ThelinkonthefarleftinFigure4iscomprisedoftwocomponentsof9 ,whilethelinkinthecenter
32
is comprised of knots 9 and 9∗ . The knot 9 has no symmetry, therefore the only possible nontrivial
32 32 32
symmetries for these links are ρ and mρ respectively. These two links achieve said symmetries through a
180degreerotationaboutalineperpendiculartothepage.
Thetwocomponentsofthelink10n59inthefarrightinFigure4arealeft-andaright-handedtrefoil(3
1
and 3∗). Since the trefoil is not mirrorable, we cannot mirror link 10n59 without permuting components,
1
and vice-versa. Therefore its symmetry group is a subgroup of Σ = (cid:104)i ,mρ(cid:105). The link exhibits the mρ
8,7 2
symmetrythrougha180degreerotationaboutalineperpendiculartothepage;Figure5showsitalsoadmits
thesymmetryi . Thus,Σ(10n59) = Σ . (cid:3)
2 8,7
5. INTRINSIC SYMMETRY GROUPS OF ALTERNATING LINKS
Alternating, nonsplit two-component links only have 12 possible intrinsic symmetry groups; 15 groups
will never occur when the link is alternating. To understand why, we will utilize a property of Conway
polynomials[5]. Thoughthisresultisclassicallyknown,weprovideaproofhereforcompleteness.
Lemma5.1. LetLbealinkwithccomponents. Considerasymmetryγ = (−1,ε,ε,...,ε,p)thatreverses
the orientation of either all or no components of L. Then, the Conway polynomial of L changes sign
accordingtotheformula∇(Lγ) = (−1)c+1∇(L).
Proof. Weshallproceedbyinductiononthenumberofcomponents. Forknots(c = 1),thisiswellknown.
Letγ bedefinedasabove. LetK beoneofthecomponentsofL. Resolveeachofthenovercrossingsthat
K haswiththeothercomponentsofLaccordingtotheskeinrelationoftheConwaypolynomial;similarly,
resolvethecorrespondingnundercrossingsinLγ. AsthiswillturneverycrossingK haswithLorLγ into
overorundercrossings,weeventuallygetdowntosplitlinksL−K andLγ −K. Weobtain
∇(L) = (−1)i1z∇(L )+(−1)i2z∇(L )+...+(−1)inz∇(L )+∇(L−K)
0,1 0,2 0,n
and
∇(Lγ) = (−1)i1+1z∇(Lγ )+(−1)i2+1z∇(Lγ )+...+(−1)in+1z∇(Lγ )+∇(Lγ −K)
0,1 0,2 0,n
wherei = 0or1ifthejthovercrossingofK inLwasa+1crossingora−1crossing,respectively.
j
9
FIGURE 5. An isotopy diagram showing that link 10n59 admits symmetry i2 – we may
reversetheorientationofthesecondcomponent.
AsL−K andLγ−K aresplit,∇(L−K) = ∇(Lγ−K) = 0. Moreover,everycrossingweresolvedwas
anintercomponentcrossing,soL isalinkwithc−1componentsforeveryj;thereforebytheinductive
0,j
hypothesis,∇(L ) = (−1)c∇(Lγ ).
0,j 0,j
Wethushave
∇(L) = (−1)i1+cz∇(Lγ )+(−1)i2+cz∇(Lγ )+...+(−1)in+cz∇(Lγ )
0,1 0,2 0,n
= (−1)c+1∇(Lγ) (cid:3)
This lemma rules out four possible symmetries for nonsplit, alternating two-component links L. Recall
thattheConwaypolynomialdoesnotvanishfornonsplit,alternatinglinks. ThussuchalinkLcannotadmit
a symmetry of the form (−1,ε,ε,p). Only 12 subgroups of Γ fail to contain one of the four elements of
2
this form; the other 15 groups cannot be symmetry groups for an alternating two-component link, which
provesthefollowingproposition.
Proposition 5.2. Only 12 intrinsic symmetry groups are possible for alternating, nonsplit two-component
links. These12groupsmaybeviewedasasubdiagramofFigure1: theyformtheorderidealgeneratedby
twogroups,Σ andΣ ,i.e.,thesetwogroupsandalloftheirsubgroups.
8,1 8,2
Alternating examples appear in Table 2 for all but one of these 12 groups; the exception is Σ , where
4,4
noexample,alternatingornot,isknown.
6. FUTURE DIRECTIONS
We have found prime, nontrivial examples for 21 of the 27 possible intrinsic symmetry groups of two-
component links.. Do examples exist for the 6 remaining subgroups? In considering links of 14 or fewer
crossings,ourexhaustiveapproachhasnotcapturedthefullcomplexityofpossiblelinkstructures. Itseems
possiblethatsomeofthemissinggroups,especiallythesmallerones,willappearamonglinksof20orfewer
crossings. For instance, examples for three groups (Σ ,Σ ,Σ ) appeared first for 14-crossing links;
4,8 4,10 4,11
perhaps examples of links with the three missing groups of order four will appear ‘soon’ in the tables. In
particular,weconjecturethatthereisanalternatinglinkwithsymmetrygroupΣ ,whichwouldcomplete
4,4
thesetofexamplesforalternating,nonsplittwo-componentlinks.
10
