Homotopical Khovanov homology

0.16 MB·English

by  Vassily Olegovich Manturov

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 con-
sider 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 corre-
sponding 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 ho-
mology.
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 struc-
ture. 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 cross-
ings. 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 homo-
topical 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
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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
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