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Homotopical Khovanov homology 5 1 V.O. Manturov∗, I.M. Nikonov† 0 2 n a Abstract J WemodifythedefinitionoftheKhovanovcomplexfororientedlinksin 0 a thickening of an oriented surface to obtain a triply graded homological 2 link invariant with a new homotopical grading. ] T 1 Introduction G h. VirtualknottheorywasdiscoveredbyKauffmanin[5]. Itturnedoutthat t virtual links are isotopy classes of links in thickened surfaces considered a up to stabilizations. m Thus,virtualknotsandlinkspossesspropertiescomingfromknotsas [ well as those coming from curves on 2-surfaces. It turned out [8, 2, 6] that this homotopical information can be con- 1 verted into coefficients of polynomials, or, in other words, polynomials v 5 can be thought of to havehomotopy classes as coefficients. 0 Homotopy classes of curvesin a closed 2-surface can be thought of as 8 conjugacyclassesofthefundamentalgroupofthesurface. Therecognition 4 problem for such curvesand groups is solved, see, e.g. [4]. 0 Some numerical information coming from thesehomotopy classes can 1. be converted into additional gradings of the Khovanovhomology [3, 7]. 0 Inthepresentpaper,weshowhowthehomotopicalinformationabout 5 curves in Kauffman states can beconverted into additional gradings. 1 Thus,curveswhichappearassummands(withsomeothercoefficients) : are raised to the other level. Though some factorizations in the grading v space are needed for thecomplex tobe well defined,thehomotopy infor- i X mation whichis convertedintogradings, doesnotreducetosomehomol- r ogy classes or numerical grading; highly non-commutative group theory a comes into play in thenew form of gradings of theKhovanov complex. 2 Homotopical Khovanov homology LetSbeaconnectedclosedorientedtwodimensionalsurface. Weconsider linksinthethickeningS×[0,1]ofthesurface. Suchlinkscanbedescribed bydiagramslooklike4-valentgraphsembeddedintoS withthestructure 1Bauman Moscow State Technical University, Russia; Laboratory of Quantum Topology, ChelyabinskStateUniversity,Chelyabinsk,Russia 2DepartmentofMechanicsandMathematics,MoscowStateUniversity,Russia;Facultyof Management, NationalResearchUniversityHigherSchool ofEconomics,Russia 1 ofover-andundercrossingintheverticesofthegraphs. Differentdiagrams of thesame link differ bydiagram isotopies and Reidemeister moves. Definition 1. Let D be a link diagram in S. A source-sink structure on D is an orientation of edges of D (considered as a 4-valent graph) such thatateachcrossingofD therearetwooppositeincomingedgesandtwo opposite outcoming edges (see Fig. 1). Figure 1: Source-sink structure Manyinvariantsofclassicalknotsandlinkscanbeextendedtothecase of links in the surface S with minor changes of construction or without changes at all. Let us recall the construction of Khovanov homology of links. For simplicity we work overZ . 2 Let D be a diagram of some oriented link L in the surface S. Let X(D) be the set of crossing of the diagram D. At each crossing of D there are two possible resolutions 0 and 1 (see Fig. 2). A state is a map s: X(D) → {0,1} (or equivalently a sequence of 0s and 1s indexed by crossings of D). Any state s determines resolutions at all crossings of the diagram D. The diagram Ds which appears after resolution at each crossing according to the state s, has no crossings, and therefore it is a set of nonintersecting circles (closed simple curves) in S. Let γ(s) be the numberof circles in Ds and β(s) bethe numberof 1s in s. 0 1 Figure 2: Resolutions of a crossing Thus, D has 2n states, where n is the numberof crossings in D. The set of states {0,1}X(D) can be interpreted as the set of vertices of a n- dimensional cube (the state cube). The edges of the cube are oriented from thestate (0,0,...,0) tothe state (1,1,...,1). Let V be the two dimensional graded vector space with basis v and − v . Set the grading of the basis vectors to be deg(v ) = ±1. Let the + ± maps m:V ⊗V →V and ∆: V →V ⊗V be given by formulas m(v ⊗v )=v , m(v ⊗v )=v , ∆(v )=v ⊗v + + + + − − − − − m(v−⊗v+)=v−, m(v−⊗v−)=0, ∆(v+)=v+⊗v−−v−⊗v. (1) 2 With each state s, assign the state a space V(s)=V⊗γ(s). Now, the chain space of the Khovanov complex, we are going to construct for D, will be thesum [[D]]= V(s). M s∈{0,1}X(D) The space [[D]] has two gradings: the homological grading β and the quantum gradingq. Ifx=x1⊗x2⊗···⊗xγ(s) ∈V(s)⊂[[D]]wherexi= v±,i = 1,...,γ(s), then we set β(x) = β(s) and q(x) = Piγ=(s1)deg(xi)+ β(s). Anyedgee=s→s′ connectstwostatessands′. TheresolutionsDs and Ds′ differ at one crossing. If D admits a source-sink structure then oneofthefollowingpicturescanhappen: eitheracirclecansplitintotwo circles or two circles can merge intoone circle (see Fig. 3). γ1 γ2 γ m Δ Figure 3: Transformations of resolution In thefirst case, one should consider themap ∆⊗id⊗γ(s)−1: V(s)→ V(s′)where∆actsonthefactorV whichcorrespondstothesplittingcir- cle. Inthesecondcase,oneshouldconsiderthemapm⊗id⊗γ(s)−2: V(s)→ V(s′)wheremactsonthefactorsV ⊗V whichcorrespondtothemerging circles. TheconsideredmapfromV(s)toV(s′)iscalledthepartialdiffer- ential corresponding totheedgeeoftheKhovanovcomplex anddenoted as ∂e. The sum of partial differentials for all the edges of the cube of states d=Pe=s→s′∂e is the differential of Khovanov complex. The differential does not change quantumgrading and increases homology grading by 1. In order to construct an invariant homology theory,we should ensure amongotherthinginvarianceunderfirstReidemeistermoves. Tothisend, let usshift thehomological gradingby−n andthequantumgradingby − n −2n where n and n are the numbers of positive and negative + − + − crossings in the diagram D. Let C(D) be the shifted complex. Then homology Kh(D)=H(C(D),d) is the Khovanov homology of the link L. Definition 2. Let S be a closed two-dimensional surface. We consider theset L=[S1;S]ofall thehomotopyclasses offree orientedloops inS. Let (cid:13)∈L bethe homotopy class of contractible loops. Foranyclosedcurveγ onecanconsiderthecurve−γ obtainedfromγ bytheorientationchange. LetHbethequotientgroupofthefreeabelian group with generator set L modulo the relations (cid:13) = 0 and [γ] = [−γ] for all free loops γ. We consider a new homotopical grading in the complex [[L]] valued in the group H. Let s be a state and C ,C ,...,C be the circles of 1 2 γ(s) the resolution Ls. For each element x = x1 ⊗x2 ⊗···⊗xγ(s) ∈ V(s) 3 where xi =v± corresponds to the circle Ci,i= 1,...,γ(s), we define its homotopical grading as γ(s) h(x)=Xdeg(xi)[Ci]∈H. i=1 Let d be the ”part” of the differential d that does not change the h homotopical grading(i.e. thecomposition ofd withtheprojection tothe subspace with the correspondent homotopical grading). The map d is h presented as a sum dh =Pe∈E(∂e)h overall the edges of the state cube. Foranyedgee=s→s′thepartialmap(∂e)hhastheformmh⊗id⊗γ(s)−2 ifthepartialdifferentialisequaltom⊗id⊗γ(s)−2 and(∂e)h hastheform ∆ ⊗id⊗γ(s)−1ifthepartialdifferentialisequalto∆⊗id⊗γ(s)−1. Herem h h and∆ arethepartsofthemapsmand∆thatpreservethehomotopical h grading. The formula for the map m : V ⊗V →V, which corresponds to the h bifurcactionoftwocirclesγ andγ toacircleγ,dependsonthehomotopy 1 2 classes of the circles: m, [γ ]=[γ ]=[γ]=(cid:13); 1 2  m1, [γ ]=(cid:13),[γ ]=[γ]6=(cid:13); mh= m2hh, [γ12]=(cid:13),[γ21]=[γ]6=(cid:13); (2) m0, [γ]=(cid:13),[γ ],[γ ]6=(cid:13); h 1 2  0, [γ1],[γ2],[γ]6=(cid:13). The maps m0,m1,m0 are defined bythe formulas h h h m0(v ⊗v )=0, m1(v ⊗v )=v , m2(v ⊗v )=v , h + + h + + + h + + + m0(v ⊗v )=v , m1(v ⊗v )=0, m2(v ⊗v )=v , h + − − h + − h + − − (3) m0(v ⊗v )=v , m1(v ⊗v )=v , m2(v ⊗v )=0, h − + − h − + − h − + m0(v ⊗v )=0, m1(v ⊗v )=0, m2(v ⊗v )=0. h − − h − − h − − Analogously, the formula for the map ∆ : V → V ⊗V, which corre- h sponds to transformations of a circle γ to two circles γ and γ , depends 1 2 on thehomotopical classes of the circles as follows: ∆, [γ ]=[γ ]=[γ]=(cid:13); 1 2  ∆1, [γ ]=(cid:13),[γ ]=[γ]6=(cid:13); ∆v = ∆2hh, [γ12]=(cid:13),[γ21]=[γ]6=(cid:13); (4) ∆0, [γ]=(cid:13),[γ ],[γ ]6=(cid:13); h 1 2  0, [γ1],[γ2],[γ]6=(cid:13). The maps ∆0,∆1,∆0 are defined bythe formulas h h h ∆0(v )=v ⊗v +v ⊗v , ∆1(v )=v ⊗v , ∆2(v )=v ⊗v , h + + − − + h + + − h + − + ∆0(v )=0, ∆1(v )=v ⊗v , ∆2(v )=v ⊗v . h − h − − − h − − − (5) Remark1. Theformulas(2),(4)showthatapartialmapisnotzeroonly if one of thetransformed circles is not trivial. Theorem 1. The map d is a differential. h 4 Proof. Inordertoshowthatthemapd isadifferentialweneedtocheck h thatevery2-faceinthestatecubeofthediagramDiscommutative. Any 2-face corresponds to a resolution switch of two crossings of the diagram whereastheresolutionoftheothercrossingsisfixed. Sinceanyresolution of all thecrossings of the diagram except the chosen two crossings yields a diagram D′ with two crossings and source–sink structure on it, there threepossibleconfigurationsofthetworemainingcrossings’position (see Fig. 4). Note that any (partial) resolution of a diagram with source-sink structure inherits source-sink structure from the diagram. Here we consider onlythecomponentsofthediagram D′ which contain thecrossings anddon’tpointoutunder-andovercrossingsstructurebecauseitdoesn’t change thestates of the 2-face (but determines themaps of theface). Case I Case II Case III Figure 4: Diagrams with two crossings If the crossings belong to different components (case III) then the 2- face is obviously commutative. Case I. Among the resolutions of the face there are two states with twocircles andtwostateswith onecircle (seeFig.5). Thus,wehavetwo subcasesdependingonhowmanycirclesarethereintheinitialstate: with onecircle(forexample,e)orwithtwocircles(aandb). Thecorresponding diagramsareshowninFig.6. Nowweshouldcheckallthevariants,which homotopyclasses ofthecirclesa,b,c,d,e,f aretrivialandwhicharenot. a b e f c d Figure 5: Case I: Resolutions Case I.A. The initial state of the face has one circle. Let k be the number of homotopically trivial circles in the pair a,b, andl bethenumberofhomotopically trivialcirclesinthepairc,d. Then 0≤k,l≤2. Without loss of generality assume that k≤l. 5 V ∆h // V⊗2 V⊗2 mh //V ∆h (cid:15)(cid:15) (cid:15)(cid:15) mh mh (cid:15)(cid:15) (cid:15)(cid:15)∆h V⊗2 mh //V V ∆h // V⊗2 Case A Case B Figure 6: Case I: Face diagrams If k=l then the2-face is commutativedue tosymmetry. Letk=0,l=1. Assumethatthecirclecistrivialanddisnottrivial. Then [e]=[f]=[d] so e and f are not trivial. Hence, thecommutativity relation reduces to the equality m2∆2 = 0. But m2∆2(v ) = m2(v ⊗ h h h h ± h − v )=0, so theface is commutative. ± Let k=0,l=2. Theneandf aretrivialasconnectedsumsoftrivial circlescandd. Hence,weneedchecktheequalitym0∆0 =m∆. Butthe h h both maps are zero (overZ ), so theface is commutative. 2 The case k =1,l =2 is impossible. Indeed, let [a]=(cid:13) and [b]6=(cid:13). Then [e]=[b] is not trivial. On the other hand e must be trivial as a as a connected sum of trivial circles c and d. Case I.B. The initial state of theface has two circles. If e and f are either both trivial or both nontrivial then the face is commutative dueto symmetry. Let [e] = (cid:13) and [f] 6= (cid:13). Then the circles a,b,c,d must be all nontrivial. Commutativity of the face follows from the equality ∆0m0 = h h 0. Case II. Among the resolutions of the face there are one state with one circle, two states with two circles and two states with one circle and onestatewiththreecircles(seeFig.7). Therearethreesubcases(A,B,C) depending on how many circles are there in the initial state. The corresponding diagrams are shown in Fig. 8. a b c ab c a bc abc Figure 7: Case II: resolutions Letusenumeratepossiblevariantswhensomeofthecirclesa,b,c,ab,bc,abc are homotopically trivial: 1. The circles a,b,c are trivial. Thenthecirclesab,bc,abcaretrivialandwehaveafaceoftheusual Khovanovcomplex. 6 V ∆h // V⊗2 V⊗2 id⊗∆h//V⊗3 V⊗3 mh⊗id// V⊗2 ∆h (cid:15)(cid:15) (cid:15)(cid:15)∆h⊗id mh (cid:15)(cid:15) (cid:15)(cid:15)mh⊗id id⊗mh (cid:15)(cid:15) (cid:15)(cid:15) mh V⊗2 id⊗∆h// V⊗3 V ∆h //V⊗2 V⊗2 mh //V Case A Case B Case C Figure 8: Case II: face diagrams 2. Among a,b,c there is only onenontrivial circle. (a) [a]6=(cid:13) (b) [b]6=(cid:13) (c) [c]6=(cid:13) In this case [abc] 6= (cid:13), the circle ab is not trivial if [a] 6= (cid:13) or [b]6=(cid:13), and thecircle bc is not trivial if [b]6=(cid:13) or [c]6=(cid:13). 3. Among a,b,c only one circle is trivial. (a) [a]=(cid:13). Then [ab]=[b] is not trivial and [abc]=[bc]. i. [bc]=(cid:13) ii. [bc]6=(cid:13) (b) [b]=(cid:13). Then [ab]=[a], [bc]=[c] are not trivial. i. [abc]=(cid:13) ii. [abc]6=(cid:13) (c) [c]=(cid:13). Then [bc]=[b] is not trivial and [abc]=[ab]. i. [ab]=(cid:13) ii. [ab]6=(cid:13) 4. Noneof a,b,c is trivial. (a) [ab]=[bc]=(cid:13). Then [abc]=[a]=[c] is not trivial. (b) [ab]=(cid:13),[bc]6=(cid:13). Then [abc]=[c] is not trivial. (c) [ab]6=(cid:13),[bc]=(cid:13). Then [abc]=[a] is not trivial. (d) [ab],[bc]6=(cid:13). i. [abc]=(cid:13) ii. [abc]6=(cid:13) Any combination of cases A,B,C and 1–4.d.ii determines the maps of the face. Thus, we can write down commutativity relation of the face in table form (see Table 1). It is easy to check that all the equalities of the table hold, so thecase II is also commutative. The proof is finished. Remark 2. 1. Thedifferentials dandd don’tcommute. Indeed,ifthis h wereacasethenthedifferencedv =d−dh wouldhavebeenadifferential. But in the case I.B considered above in Theorem 1 with trivial e and nontrivial f we have v ⊗v 7→07→0 for the path of the face that goes + − through e, and v ⊗v 7→ v 7→ v ⊗v for the path of the face that + − − − − goes through f. So, the face is not commutative for the map dv. 7 Table 1. Commutativity relation. A B C 1 asinusualKhovanovcomplex 2.a (∆1h⊗id)∆1h=(id⊗∆)∆1h (m1h⊗id)(id⊗∆)=∆1hm1h m1h(m1h⊗id)=m1h(id⊗m) 2.b (∆2h⊗id)∆1h=(id⊗∆1h)∆2h (m2h⊗id)(id⊗∆1h)=∆1hm2h m1h(m2h⊗id)=m2h(id⊗m1h) 2.c (∆⊗id)∆2h=(id⊗∆2h)∆2h (m⊗id)(id⊗∆2h)=∆2hm2h m2h(m⊗id)=m2h(id⊗m2h) 3.a.i (∆2h⊗id)∆0h=(id⊗∆0h)∆ (m2h⊗id)(id⊗∆0h)=∆0hm m0h(m2h⊗id)=m(id⊗m0h) 3.a.ii 0=0 0=0 0=0 3.b.i (∆1h⊗id)∆0h=(id⊗∆2h)∆0h (m1h⊗id)(id⊗∆2h)=∆0hm0h m0h(m1h⊗id)=m0h(id⊗m2h) 3.b.ii 0=0 (m1h⊗id)(id⊗∆2h)=0 0=0 3.c.i (∆0h⊗id)∆=(id⊗∆1h)∆0h (m0h⊗id)(id⊗∆1h)=∆m0h m(m0h⊗id)=m0h(id⊗m1h) 3.c.ii 0=0 0=0 0=0 4.a (∆0h⊗id)∆2h=(id⊗∆0h)∆1h (m0h⊗id)(id⊗∆0h)=∆2hm1h m2h(m0h⊗id)=m1h(id⊗m0h) 4.b (∆0h⊗id)∆2h=0 0=0 m2h(m0h⊗id)=0 4.c 0=(id⊗∆0h)∆1h 0=0 0=m1h(id⊗m0h) 4.d.i 0=0 0=∆0hm0h 0=0 4.d.ii 0=0 2. If we set m0 = m1 = m2 = 0 and ∆0 = ∆1 = ∆2 = 0 in h h h h h h formulas (2), (4) the modified map d will also be a differential but the h complex will not beinvariant underReidemeister moves. Thus, ([[D]],d ) is a chain complex with triple gradings. We use the h conventional shift by −n of homological grading and the shift by n − − + 2n of quantum grading (where n and n are the numbers of positive − + − and negative crossings in the diagram) to obtain a chain complex C(D) whose homology Kh (D) = H(C(D),d ) we call homotopical Khovanov h h homology. It is an abelian group with threegradings Kh (D)= Kh (D) h X h i,j,h i∈Z,j∈Z,h∈H (here i is the homology grading, j is the quantim grading and h is the homotopical grading) whichturnsout tobealinkinvariant likeordinary Khovanov homology. Theorem 2. Homology Kh is an invariant of oriented links which ad- h mits source-sink structure. Proof. Let D beadiagram ofan oriented linkwith source-sink structure and let D′ be the diagram that obtained from D by a first Reidemeister move. Then the complex [[D′]] looks like [[D]] −∆→h [[D⊔(cid:13)]] (if the new crossingisnegative)or[[D⊔(cid:13)]]−m→h [[D]](ifthenewcrossingispositive). Thecomplex[[D⊔(cid:13)]]isisomorphicto[[D]]⊗V. Itcontainsacubcomplex [[D⊔(cid:13)]]v+ =[[D]]⊗v+ andwedenote[[D⊔(cid:13)]]v+=0=[[D⊔(cid:13)]]/[[D]]v+ the quotient complex by this subcomplex. Lemma 1. The maps [[D⊔(cid:13)]]v+ −m→h [[D]] and [[D]]−∆→h [[D⊔(cid:13)]]v+=0 are isomorphisms of complexes. Proof. Sincethesmallcircleistrivial,m =morm =m1. Inbothcases h h h m establishesanisomorphismbetweenV⊗v andV thatextendstothe h + desired isomorphism of complexes. For the second map we have∆ =∆ h or ∆1 which definean isomorphism between V and V ⊗V/(v =0). h + Withtheabovelemmawecanusetheproofofinvarianceoftheusual Khovanov homology (see, for example, [1]) without any changes. 8 Remark 3. Infact, homotopical Khovanovhomology can bedefinedfor linkswithout source-sink structure. Indeed,in general case an additional (nonorientable) bifurcation type appears where one circle transforms to another circle (see Fig. 9). 0 Figure 9: Nonorientable bifurcation of a cirle ThecircledcrossingsinFig.9,10areconsideredasvirtualbutnotreal crossings, i.e. as additional intersections which appear by projection of the resolution into theplane. Inthestatecubethisbifurcationcorrespondstozeromap. So,inorder toshowthathomotopicalKhovanovcomplexiswelldefinedwemustcheck 2-faces which include thenew typeof state transformation (see Fig. 10). Case IV Case V Case VI Case VII Case VIII Figure 10: Nonorientable diagrams with two crossings The commutativity relation for these faces is reduced to the equality 0=0forallcasesexceptone(seeFig.11)whereittakestheformm ∆ = h h 0. But thisequality holds because m∆=m0∆0 =m1∆1 =m2∆2 =0. h h h h h h V ∆h // V⊗2 0 mh (cid:15)(cid:15) (cid:15)(cid:15) V 0 // V Figure 11: Case VIII: nontrivial face diagram The proof of invariance of homotopical Khovanov homology needs no corrections since it does not encounterthe newbifurcation. Thus, the following theorem holds. Theorem 3. Homology Kh is an invariant of oriented links. h 9 Let us mention some simple properties of homotopical Khovanov homology. Proposition 1. If L is a classical link in S, i.e. there is a 3-ball B ⊂ S ×[0,1] which contains L, then homotopical Khovanov homology of L coincides with the usual Khovanov homology of L. Proof. IfLisaclassicallinkthenthehomotopicalgradingofallelements in the Khovanov complex is zero, so maps m and ∆ coincide with m h h and ∆. Hence, in this case we havetheusual Khovanovcomplex. Proposition 2. Let L be an oriented link in S. Let −L differ from L by orientation change and Lˇ be the link with the opposite source-sink structure. Then Kh (−L) = Kh (L) , Kh (L¯) =Kh (L) h i,j,h h i,j,h h i,j,h h −i,−j,−h and Kh (Lˇ) = Kh (L) for all i ∈ Z,j ∈ Z,h ∈ H. In other h i,j,h h i,j,h words, homotopical Khovanov homology does not depend on the choice of source-sink structure and does not detect invertibility of knots. The proof of the statement does not differ from the the proof for the usual Khovanov homology. Example 1 (Simple closed curves in the surface). Let γ be a simple closed curve in S. It can be considered as a knot diagram without crossings. UsualKhovanovhomologyofγisKh(γ)=Z ⊕Z withhomological 2 2 grading equal to 0 and quantum gradings equal to ±1. This homology does not depend on how the curve lie in the surface. Homotopical Kho- vanov homology is more sensitive. Kh (γ) = Z ⊕Z where the first h 2 2 summand has homological grading 0, quantum grading −1 and homotopical grading −[γ], and the second summand has homological grading 0, quantum grading 1 and homotopical grading [γ]. Thus, the homotopy typeof γ is kept in thehomotopical grading. Theorem 4. Let knots K and K′ in S have complexity 0, i.e. admit diagrams without crossings. Then K isotopic to K′ (up to orientation reversion) if and only if Kh (K)=Kh (K′). h h Proof. Let K and K′ are presented by simple closed curve in S. If Kh (K) = Kh (K′) then [K] = [K′] ∈ H and [K] = [±K′] ∈ L. Hence h h there is a homotopy between simple closed curves K and K′. Then by Baer–Zieschang theorem thereisanisotopy ofS thatsendsK toK′. Example 2 (Knotsintorus). LetK beaknotinthethickeningoftorus T2. For torus we have L = H (T2,Z). Let D be a diagram of K in the 1 torus. For any state s the resolution Ds is a set of nonintersecting circles in T2. If γ1 and γ2 are nontrivial circles in Ds then their homology classes coincideuptosign. ThenhomotopygradingsofelementsinV(s)equalto k[γ ],k∈Z. If thehomology class [D]=[K] of thesource-sink structure 1 of the knot is not zero the gradings will be proportional to [K]. If the homology class [D] is trivial then gradings can be nonproportional but must beeven. ConsideralinkLinthetoruswithtwocrossings(seeFig.12). Homo- topical Khovanov homology of the link L are isomorphic to (Z )8. The 2 10