The Khovanov width of twisted links and closed 3-braids

0.37 MB·English

by  Adam Lowrance

THE KHOVANOV WIDTH OF TWISTED LINKS AND CLOSED 3-BRAIDS
ADAMLOWRANCE
9
0
0 Abstract. Khovanov homology is a bigraded Z-module that categorifies the Jones polynomial.
2 The support of Khovanov homology lies on a finite number of slope two lines with respect to the
bigrading. The Khovanov width is essentially the largest horizontal distance between two such
n
lines. We show that it is possible to generate infinite families of links with the same Khovanov
a
J width from link diagrams satisfying certain conditions. Consequently, we compute the Khovanov
width for all closed 3-braids.
6
1

]
T 1. Introduction
G
Let L ⊂ S3 be an oriented link. The Khovanov homology of L, denoted Kh(L), was introduced
.
h byMikhailKhovanovin[13],andisabigradedZ-modulewithhomologicalgradingiandpolynomial
t
a (or Jones) grading j so that Kh(L) = Khi,j(L). The graded Euler characteristic of Kh(L) is
i,j
m
the unnormalized Jones polynomial:
[ L
(q+q−1)V (q2) = (−1)irank Khi,j(L)qj.
L
2
i,j
v X
6 The support of Kh(L) lies on a finite number of slope 2 lines with respect to the bigrading.
9 Therefore, it is convenient to definetheδ-grading byδ = j−2i sothat Kh(L)= Khδ(L). Also,
1 δ
2 either all theδ-gradings of Kh(L) areodd, or they all are even. Let δmin betheminimum δ-grading
L
. where Kh(L) is nontrivial and δ be the maximum δ-grading where Kh(L) is nontrivial. Then
1 max
0 Kh(L) is said to be [δmin,δmax]-thick, and the Khovanov width of L is defined as
9
1
0 w (L)= (δ −δ )+1.
Kh max min
: 2
v
In this paper, we show the following:
i
X
• If a crossing in a link diagram is width-preserving (defined in Section 3), then it can be
r
a replaced with an alternating rational tangle and the Khovanov width does not change
(Theorem 3.4).
• We compute the Khovanov width of all closed 3-braids (Theorem 4.10).
• We determine the Turaev genus of all closed 3-braids, up to an additive error of at most 1.
• We show that for closed 3-braids the Khovanov width and odd Khovanov width are equal
(Corollary 5.5).
Thepaperisorganizedasfollows. InSection2,wereviewsomepropertiesofKhovanovhomology.
In Section 3, we describe the behavior of Khovanov width when a crossing is replaced by an
alternating rational tangle. In Section 4, the Khovanov width of any closed 3-braid is computed.
Finally, in Section 5 we show that Khovanov width and odd Khovanov width for closed 3-braids
are equal.
Date: January 15, 2009.
1
2 A.LOWRANCE
Acknowledgements. I wish to thank Scott Baldridge, Oliver Dasbach, Mikhail Khovanov, and
Peter Ozsv´ath for many helpful conversations. A portion of the work for this article was done
while the author was visiting Columbia University in the Fall of 2008. He thanks the department
of mathematics for their hospitality.
2. Khovanov homology background
In this section, we give background material on Khovanov homology. If D is a diagram for L,
then denote the Khovanov homology of L by either Kh(L) or Kh(D). Similarly, let w (L) and
Kh
w (D) equivalently denote the Khovanov width of L. If F is a field, then let Kh(L;F) denote
Kh
Kh(L)⊗F and w (L;F) denote the width of Kh(L;F).
Kh
Let L and L be oriented links, and let C be a component of L . Denote by l the linking
1 2 1
number of C with its complement L −C. Let L′ bethe link L with the orientation of C reversed.
1 1 1
Denote the mirror image of L by L and the disjoint union of L and L by L ⊔L . Thefollowing
1 1 1 2 1 2
proposition was proved by Khovanov in [13].
Proposition 2.1 (Khovanov). For i,j ∈ Z there are isomorphisms
Khi,j(L′) ∼= Khi+2l,j+2l(L ),
1 1
Khi,j(L ;Q) ∼= Kh−i,−j(L ;Q)
1 1
Tor(Khi,j(L )) ∼= Tor(Kh1−i,−j(L )), and
1 1
Khi,j(L1⊔L2) ∼= ⊕k,m∈Z(Khk,m(L1)⊗Khi−k,j−m(L2))⊕
⊕k,m∈ZTor1Z(Khk,m(L1),Khi−k+1,j−m(L2))
Let D be a diagram for L and D′ be the diagram D with the component C reversed. Denote
1
the number of negative crossings in D by neg(D), where the sign of a crossing is as in Figure 1.
Set s= neg(D)−neg(D′). Then Proposition 2.1 implies
Khδ(D′) ∼= Khδ+s(D), and
Khδ(L ;Q) ∼= Kh−δ(L ;Q).
1 1
In [14], Khovanov introduced the reduced Khovanov homology. For a knot K, this theory
is denoted Kh(K). For links of more than one component, the reduced Khovanov homology
depends on a choice of a marked component, and hence is denoted Kh(L,C), where C is the
marked comgponent of L. Similar to the unreduced version, Kh(L,C) is a bigraded Z-module with
ig,j
homological grading i and Jones grading j so that Kh(L,C) = Kh (L,C). Thegraded Euler
i,j
g
characteristic of Kh(L,C) is the ordinary Jones polynomial:
L
g g
i,j
g VL(q2) = (−1)irank Kh (L,C)qj.
i,j
X
g
As with Khovanov homology, if δ is the minimum δ-grading where Kh(L,C) is nontrivial
min
and δ is the maximum δ-grading where Kh(L,C) is nontrivial, then we say that Kh(L,C)
max
is [δmin,δmax]-thick. The reduced Kehovanov width is defined as wKgh(L) = 12g(δmax−δmin)+1.
AseaedaandPrzytycki[2]showthatthereisgalongexactsequencerelatingreducedandugnreduced
Kheovanoev homology. e e
THE KHOVANOV WIDTH OF TWISTED LINKS AND CLOSED 3-BRAIDS 3
Theorem 2.2 (Asaeda-Przytycki). There is a long exact sequence relating the reduced and unre-
duced versions of Khovanov homology:
··· → Khi,j+1(L,C) → Khi,j(L) → Khi,j−1(L,C) → Khi+1,j+1(L,C) → ···
Corollary 2.3. Let L be a link with marked component C. Then Kh(L) is [δmin,δmax]-thick if and
g g g
only if Kh(L,C) is [δmin +1,δmax −1]-thick. Hence wKh(L)−1= wKgh(L).
Proof. The long exact sequence of Theorem 2.2 can be rewritten with respect to the δ-grading as
g
δ+1 δ−1 δ−1
··· → Kh (L,C) → Khδ(L) → Kh (L,C) → Kh (L,C) → ··· .
δ
SupposeKh(L) is [δ ,δ ]-thick. ThereforeKh (L,C) = 0for δ > δ +1 andfor δ < δ −1.
gmin max g g max min
δmax+1
Suppose Kh (L,C) is nontrivial. Then for some i and j where j−2i = δ +1, the group
max
i,j g
Kh (L,C) is nontrivial. By repeatedly applying the long exact sequence of Theorem 2.2, one sees
i+kg,j+2k δmax+1
that Kh (L,C) is nontrivial for all k ≥ 0. However, the group Kh (L,C) is finitely
g δmax+1 δmin−1
generated. Hence Kh (L,C) is trivial. Similarly, one can show that Kh (L,C) is also
g g
trivial.
The long exact gsequence also implies that Khδmax−1(L,C) and Khδmin+g1(L,C) are nontrivial.
Thus Kh(L,C) is [δ +1,δ −1]-thick.
min max
Suppose Kh(L,C) is [δ +1,δ −1]-thicgk. Similar to the casge above, if either Khδmin(L) or
min max
Khδmagx(L) are trivial, then one can show that Khδmin+1(L,C) or Khδmax−1(L,C) respectively are
infinitely gegnerated. Hence Kh(L) is [δ ,δ ]-thick. (cid:3)
min max
g g
Corollary 2.3 implies that if C and C′ are two components of L, then Kh(L,C) is [δ ,δ ]-
min max
thick if and only if Kh(L,C′) is [δmin,δmax]-thick. Hence, the notation wKgh(L) is unambiguous.
g e e
g e e
D D D D
+ − v h
Figure 1: The links in an oriented resolution. D is a positive crossing, and D is a negative
+ −
crossing.
Let D ,D ,D and D be planar diagrams of links that agree outside a neighborhood of a
+ − v h
distinguished crossing x as in Figure 1. Define e = neg(D ) − neg(D ). There are long exact
h +
sequencesrelatingtheKhovanovhomologyofeachoftheselinks. Khovanov[13]implicitlydescribes
these sequences, and Viro [25] explicitly states both sequences. The graded versions are taken from
Rasmussen [20] and Manolescu-Ozsva´th [16].
Theorem 2.4 (Khovanov). There are long exact sequences
··· → Khi−e−1,j−3e−2(D )−→ Khi,j(D ) −→ Khi,j−1(D ) −→ Khi−e,j−3e−2(D ) → ···
h + v h
and
··· → Khi,j+1(D ) −→ Khi,j(D )−→ Khi−e+1,j−3e+2(D ) −→ Khi+1,j+1(D ) → ··· .
v − h v
4 A.LOWRANCE
When only the δ = j −2i grading is considered, the long exact sequences become
fδ−e gδ hδ−1
··· → Khδ−e(D )−−+−→ Khδ(D )−→+ Khδ−1(D )−−+−→ Khδ−e−2(D )→ ···
h + v h
and
··· → Khδ+1(D ) −f−−δ−+→1 Khδ(D ) −g→−δ Khδ−e(D ) −h−δ−−−→e Khδ−1(D )→ ···
v − h v
There are versions of these long exact sequences where Khovanov homology is replaced with re-
duced Khovanov homology. In the reduced sequences, the gradings are identical to the unreduced
sequences.
D D D
0 1
Figure 2: The links in an unoriented resolution.
Let D, D and D be link diagrams differing only in a neighborhood of a crossing x of D (as in
0 1
Figure 2) with associated link types L, L and L respectively. The set Q of quasi-alternating links
0 1
is the smallest set of links such that
• The unknot is in Q.
• If the link L has a diagram with a crossing x such that
(1) both of the links, L and L are in Q,
0 1
(2) det(L) = det(L )+det(L ),
0 1
then L is in Q. We will say that D is quasi-alternating at x.
In[15], Lee showed that alternating links have reducedKhovanov width1. Thesetof alternating
links is a proper subset of the set of quasi-alternating links. Manolescu and Ozsv´ath [16] use the
long exact sequences for reduced Khovanov homology to show that the same result holds for quasi-
alternating links.
Theorem 2.5 (Manolescu-Ozsva´th). Let L be a quasi-alternating link. Then Kh(L) is supported
entirely in δ-grading −σ(L), where σ(L) denotes the signature of the link.
g
Theorem 2.5 together with Corollary 2.3 imply that if L is quasi-alternating, then Kh(L) is
[−σ(L)−1,−σ(L)+1]-thick and w (L) = 2.
Kh
Theorem 2.4 directly implies the following corollary:
Corollary 2.6. Let D+,D−,Dv and Dh be as in Figure 1. Suppose Kh(Dv) is [vmin,vmax]-thick
and Kh(Dh) is [hmin,hmax]-thick. Then Kh(D+) is [δm+in,δm+ax]-thick, and Kh(D−) is [δm−in,δm−ax]-
thick , where
min{vmin+1,hmin +e} if vmin 6= hmin+e+1
δm+in = vmin+1 if vmin = hmin+e+1 and hv+min is surjective
vmin−1 if vmin = hmin+e+1 and hv+min is not surjective,
max{vmax +1,hmax +e} if vmin 6= hmax +e+1
δm+ax = vmax −1 if vmax = hmax +e+1 and hv+max is injective
vmax +1 if vmax = hmax +e+1 and hvmax is not injectve,
+


THE KHOVANOV WIDTH OF TWISTED LINKS AND CLOSED 3-BRAIDS 5
min{vmin−1,hmin +e} if vmin 6= hmin+e−1
δm−in = vmin+1 if vmin = hmin+e−1 and hv−min is surjective
vmin−1 if vmin = hmin+e−1 and hv−min is not surjective,
and


max{vmax −1,hmax +e} if vmax 6= hmax +e−1
δm−ax = vmax −1 if vmax = hmax +e−1 and hv+max is injective
vmax +1 if vmax = hmax +e−1 and hv+max is not injective.
 3. Twisted Links
3.1. Khovanov width of twisted links. Let τ = C(a ,...,a ) be a rational tangle, and let
1 m
D be a link diagram with a distinguished crossing x. Suppose the slopes of the arcs near x are
±1. Define D twisted at x by τ to be the diagram obtained by removing x and inserting τ such
that a neighborhood of the rightmost crossing or topmost crossing of τ in D looks exactly like a
τ
neighborhood of x in D. The resulting link diagram is denoted D . See Figure 3.
τ
D C(2,3,4) C(−4)
D D
C(2,3,4) C(−4)
Figure 3: The diagram D twisted by C(2,3,4) and C(−4).
The main result of this section, Theorem 3.4, is a generalization of a proposition proved by
Champanerkar and Kofman in [8].
Proposition 3.1 (Champanerkar-Kofman). Let D be a link diagram with crossing x, and let τ be
an alternating rational tangle such that D is twisted at x by τ. If D is quasi-alternating at x, then
D is quasi-alternating at each crossing of τ.
τ
Let D be a diagram with crossing x. Resolve D at the crossing x to obtain diagrams D
v
and D . Suppose Kh(D ) is [v ,v ]-thick and Kh(D ) is [h ,h ]-thick. As before, set
h v min max h min max
e = neg(D ) − neg(D ), where D is the same diagram as D except if the crossing x in D is
h + +
negative, then it is changed to positive in D . The diagram D is said to be width-preserving at x
+
if either of the following conditions hold.
• If x is a positive crossing in D, then both v 6= h +e+1 and v 6= h +e+1.
min min max max
• If x is a negative crossing in D, then both v 6= h +e−1 and v 6= h +e−1.
min min max max
Proposition 3.2. Let D be a link diagram with crossing x. If D is quasi-alternating at x, then D
is width-preserving at x.
6 A.LOWRANCE
Proof. SupposeD is quasi-alternating at x. Let D and D be the two resolutions of D at x. Since
v h
D is quasi-alternating at x, it follows that D and D are also quasi-alternating. Theorem 2.5
v h
implies that Kh(D), Kh(D ) and Kh(D ) are each supported entirely in one δ-grading. Suppose
v h
Kh(D ) is supportedin δ-grading v and Kh(D ) is supportedin δ-grading h. Corollary 2.2 implies
v h
that Kh(D )gis [v−1g,v+1]-thick gand Kh(D ) is [h−1,h+1]-thick. Let e = neg(D )−neg(D )
v h h +
wghere D is the same diagram as D excgept if x is negative in D, then it is changed to positive in
+
D . Since det(D) = det(D )+det(D ), it follows that the nontrivial parts of Kh(D), Kh(D ) and
+ v h v
Kh(D ) lie in three consecutive spots in the long exact sequence of Theorem 2.4 such that Kh(D )
h v
and Kh(D ) are not adjacent. Therefore, if x is positive, then v = h+e−1,gand if xgis negative,
h
tghen v = h+e+1. The result follows directly. g (cid:3)
g
Lemma 3.3. Let D be an oriented link diagram with crossing x, and let τ be an alternating rational
tangle with exactly two crossings x and x . Let D be D twisted at x by τ. If D is width-preserving
0 1 τ
atx, thenfor anyorientation, D iswidth-preserving atx andx . Moreover, w (D) = w (D ).
τ 0 1 Kh Kh τ
Proof. There are two ways to twist D at c, either horizontally or vertically. Let τ = C(2) and
1
τ = C(−2). For each case, it is only necessary to prove the result for one choice of orientations on
2
D and D . Proposition 2.1 implies the result for all other choices of orientations on D and D .
τ τ
Let D and D be the diagrams obtained by resolving D at x, and let Di and Di be the
v h v h
diagrams obtained by resolving D at the crossing x for i = 0,1. Suppose Kh(D ) and Kh(D )
τ i v h
are [v ,v ]-thick and [h ,h ]-thick respectively. Let e = neg(D )−neg(D ) where D is
min max min max h + +
the same diagram as D except if the crossing x is negative in D, then it is changed to positive
in D . Similarly set e = neg(Di)−neg(Di ) where Di is the same diagram as D except if the
+ i h + + τ
crossing x is negative in D , then it is changed to positive in D .
i τ +
x
1
x
0
0 0
D D D D D D
v h C(2) v h
Figure 4: The resolutions for x positive and τ = C(2).
Suppose x is positive. Choose the orientation on D given in Figure 4. Also, Figure 4 shows
τ1
the resolutions D0 and D0.
v h
Observe that x is positive in D for i = 0,1. Corollary 2.6 implies that Kh(D) is [α,β]-thick
i τ1
where α = min{v +1,h +e} and β = max{v +1,h +e}. The diagrams Di and D
min min max max v
represent the same link, and the diagrams D and Di represent the same link. Therefore, Kh(Di)
h h v
is [α,β]-thick and Kh(Di) is [h ,h ]-thick. The diagram Di is the same as the diagram D
h min max h h
except Di has one additional negative Reidemeister I twist, and hence neg(Di) = neg(D )+1.
h h h
Since the diagrams D and Di are identical, neg(D) = neg(Di). Thus e = e + 1. Since D is
v v i
width-preserving, it follows that v 6= h +e+1 and v 6= h +e+1. Therefore,
min min max max
h +e +1= h +e+2 6= α,
min i min
THE KHOVANOV WIDTH OF TWISTED LINKS AND CLOSED 3-BRAIDS 7
and
h +e +1= h +e+26= β.
max i max
HenceD is width-preservingatx . Also, Corollary 2.6implies thatKh(D )is[α+1,β+1]-thick,
τ1 i τ1
and thus w (D) = w (D ).
Kh Kh τ1
x x
1 0
0 0
D D D D D D
v h C(−2) v h
Figure 5: The resolutions for x positive, τ = C(−2), and with the depicted strands of D in the
same component.
The possible orientations of D depend on whether the strands forming the crossing x are in
τ2
the same component of D or different components of D. Suppose they are in the same component.
Choose the orientation on D given in Figure 5. Also, Figure 5 shows the resolutions D0 and D0.
τ2 v h
Observe that x is positive in D for i= 0,1. With suitably chosen orientations, we have
i τ2
(3.1) neg(D ) = neg(D) = neg(Di),
v h
and
(3.2) neg(Di )= neg(D ).
+ h
ThediagramDi isthesameasD exceptDi hasonecomponentreversedandanadditionalpositive
v v v
Reidemeister I twist. Therefore, Proposition 2.1 implies that Kh(Di) is [v −e,v −e]-thick.
v min max
Also, equations 3.1 and 3.2 imply that e = −e. The diagram Di is identical to D. Therefore,
i h
Kh(Di) is [α,β]-thick where α= min{v +1,h +e} and β = max{v +1,h +e}. Since
h min min max max
D is width-preserving at x, we have v 6= h +e+1 and v 6= h +e+1. Therefore,
min min max max
α+e +1= min{v +1,h +e}−e+1= min{v −e+2,h +1} =6 v −e,
i min min min min min
and
β+e +1= max{v +1,h +e}−e+1= max{v −e+2,h +1} =6 v −e.
i max max max max max
Thus D is width-preserving at x . Moreover, Corollary 2.6 implies that Kh(D ) is
τ2 i τ2
[α−e,β −e]-thick, and hence w (D)= w (D ).
Kh Kh τ2
x x
1 0
0 0
D D D D D D
v h C(−2) v h
Figure 6: Theresolutions for xpositive, τ = C(−2), andwith the depicted strandsof D indifferent
components.
Suppose the strands forming the crossing x are in different components of the link. Choose the
orientation on D given in Figure 6. Also, Figure 6 shows the resolutions D0 and D0. Observe
τ2 v h
8 A.LOWRANCE
that x is a negative crossing in D for i= 0,1. Orient Di so that it represents the same oriented
i τ2 h
link as D . With a suitably chosen orientation on D , we have
v h
(3.3) neg(D) = neg(D ) = neg(Di),
v h
and
(3.4) neg(D )+1 = neg(Di ) = neg(Di).
h + v
Equations 3.3 and 3.4 imply that e = −e−1. The diagram Di is the same as D except Di has
i v v
one component reversed. Equations 3.3 and 3.4 along with Proposition 2.1 imply that Kh(Di) is
v
[α−e−1,β−e−1]-thick whereα = min{v +1,h +e}andβ = max{v +1,h +e}. Since
min min max max
Di and D represent the same oriented link, it follows that Kh(Di) is [v ,v ]-thick. Since D
h v h min max
is width-preserving at x, we have v 6= h +e+1 and v 6= h +e+1. Therefore,
min min max max
α−e−1 = min{v −e,h −1} =6 v −e−2 = v +e −1,
min min min min i
and
β−e−1= max{v −e,h −1} =6 v −e−2= v +e −1.
max max max max i
Thus D is width-preserving at x . Moreover, Corollary 2.6 implies that Kh(D ) is [α−e−1,
τ2 i τ2
β−e−1]-thick, and hence w (D)= w (D ),
Kh Kh τ2
The case where x is a negative crossing in D is proved similarly. (cid:3)
Theorem 3.4. Let D be a link diagram with crossing x, τ be an alternating rational tangle, and
D be the diagram D twisted at x by τ. If D is width-preserving at x, then w (D) = w (D ).
τ Kh Kh τ
Proof. Let τ = C(a ,...,a ). Since τ is alternating, either a > 0 for all i or a < 0 for all i.
1 m i i
Suppose a > 0 for all i. Beginning with the diagram D and the crossing x, one can alternate
i
twisting the diagram by C(2) and C(−2). Replacing the appropriate crossings m times results
in the diagram Dτ′ where τ′ = C(2,1,...,1). Lemma 3.3 implies that each crossing in Dτ′ is
width-preserving, and wKh(D)= wKh(Dτ′).
Replace crossings corresponding to the m-th term in τ′ by C(2) until the resulting diagram is
obtained by twisting D by C(2,1,...,1,a ) at x. Next, replace crossings corresponding to the
m
(m−1)-st term in C(2,1,...,1,a ) with C(−2) until the resulting diagram is obtained by twisting
m
D by C(2,1,...,1,a ,a ) at x. Continue replacing crossings in the tangle by either C(2) or
m−1 m
C(−2) until the resulting diagram is obtained by twisting D by C(a ,...,a ) at x. Since at each
1 m
step, the only tangles used are C(2) and C(−2), Lemma 3.3 implies that w (D) = w (D ). The
Kh Kh τ
case where each a < 0 is proved similarly. (cid:3)
i
C(2) C(−2) C(2) C(2)
C(1) C(2) C(2,1) C(2,1,1) C(2,1,2)
C(2) C(2) C(−2) C(−2)
C(2,1,3) C(2,1,4) C(2,2,4) C(2,3,4)
Figure 7: The inductive process of Theorem 3.4. At each step, the circled crossing is replaced with
either C(2) or C(−2).
THE KHOVANOV WIDTH OF TWISTED LINKS AND CLOSED 3-BRAIDS 9
Remark 3.5. Watson[26]provesthatw (D )isboundedbyw (D )andw (D ). Byassuming
Kh τ Kh v Kh h
that D is width-preserving at x, we are able to strengthen the result and calculate w (D ).
Kh τ
SupposeD is an oriented diagram with crossing x. If D is twisted at x byτ = C(n)as in Figure
n
8, then the assumptions of Theorem 3.4 can be relaxed and a slightly stronger result holds. The
following technical result is needed to compute the Khovanov width of closed 3-braids.
} }
n n
D D D D
+ C(n) − C(−n)
Figure8: Forn > 0,twistD byC(n)andtwistD byC(−n). Thenchoosetheaboveorientations
+ −
for D and D .
C(n) C(−n)
Proposition 3.6. Suppose D is an oriented diagram with crossing x. Suppose D is twisted at
x by τ = C(n) as in Figure 8. Let D and D be the two resolutions of D at x. Suppose
n v h
Kh(Dv) is [vmin,vmax]-thick and Kh(Dh) is [hmin,hmax]-thick. Let α± = min{vmin ±1,hmin +e}
and β± = max{vmax ±1,hmax +e}.
(1) Let n > 0. Suppose that vmin 6= hmin +e+1. If vmax = hmax +e+1, then suppose that
there exist integers i and j such that j−2i= vmax, Khi,j(Dv) is nontrivial, and Khk,l(Dh)
is trivial for all k whenever l ≤ j −3e−1. Then Kh(D ) is [n+α ,n+β ]-thick.
τn + +
(2) Let n < 0. Suppose that vmax 6= hmax +e−1. If vmin = hmin +e−1, then suppose that
there exist integers i and j such that j−2i = vmin, Khi,j(Dv) is nontrivial, and Khk,l(Dh)
is trivial for all k whenever l ≥ j −3e−1. Then Kh(D ) is [n+α ,n+β ]-thick.
τn − −
Proof. Let n > 0. Since D is twisted at x by τ as in Figure 8, it follows that x is a positive
n
crossing. If both v 6= h +e+1 and v 6= h +e+1, then D is width-preserving at x. It
min min max max
follows from the proof of Theorem 3.4 that Kh(D ) is [n+α ,n+β ]-thick.
τn + +
Supposev 6= h +e+1 and v = h +e+1. Thus there exist integers i and j such that
min min max max
j −2i = v , Khi,j(D ) is nontrivial, and Khk,l(D ) is trivial for all k and for all l ≤ j −3e−1.
max v h
Since v 6= h + e + 1, it follows that the minimum δ-grading where Kh(D ) is nontrivial
min min τn
is n+α . We show, by induction on n, that Khi,j+n(D ) ∼= Khi,j(D ). This implies that the
+ τn v
maximum δ-grading supporting Kh(D ) is n+β .
τn +
If n= 1, then the long exact sequence of Theorem 2.4 looks like
0→ Khi,j+1(D) → Khi,j(D )→ Khi−e,j−3e−1(D )→ ··· .
v h
By hypothesis, Khi−e,j−3e−1(D ) is trivial, and hence Khi,j+1(D) ∼=Khi,j(D ).
h v
Suppose, by way of induction, that Khi,j+n(D ) ∼= Khi,j(D ). Resolve D at any crossing
τn v τn+1
in τ to obtain diagrams D′ and D′. Let e = neg(D′) − neg(D ). Since neg(D′) =
n+1 v h n+1 h τn+1 h
neg(D )+n and neg(D ) = neg(D), it follows that e =e+n. Observe that D′ and D are
h τn+1 n+1 v τn
10 A.LOWRANCE
the same diagram, and D′ and D are diagrams for the same link. Hence the long exact sequence
h h
of Theorem 2.4 looks like
0 → Khi,j+n+1(D )→ Khi,j+n(D )→ Khi−e−n,j−3e−3n−1(D )→ ··· .
τn+1 τn h
Since j − 3e − 3n − 1 ≤ j − 3e − 1, it follows that Khi−e−n,j−3e−3n−1(D ) is trivial. Thus
h
Khi,j+n+1(D )∼= Khi,j+n(D )∼= Khi,j(D ). Therefore Kh(D ) is [n+α ,n+β ]-thick.
τn+1 τn v τn + +
The case where n < 0 is proved in a similar fashion using the second sequence from Theorem
2.4. (cid:3)
3.2. The Turaev genus of twisted links. Each linkdiagramD hasanassociated Turaevsurface
Σ . Let Γ be the plane graph associated to D. Regard Γ as embedded in R2 sitting inside R3.
D
Outside the neighborhoods of the vertices of Γ is a collection of arcs in the plane. Replace each arc
by a band that is perpendicular to the plane. In the neighborhoods of the vertices, place a saddle
so that the circles obtained from choosing a 0-resolution at each crossing lie above the plane and
so that the circles obtained from choosing a 1-resolution at each crossing lie below the plane (see
Figure9). Theresultingsurfacehasa boundaryof disjointcircles, with circles correspondingto the
0 0
1
1
Figure 9: In a neighborhood of each crossing, insert a saddle so that the boundary above the plane
corresponds to the 0 resolution and the boundary below the plane corresponds to the 1 resolution.
all 0-resolution above the plane and circles corresponding to the all 1-resolution below the plane.
For each boundary circle, insert a disk to obtain a closed surface Σ known as the Turaev surface
D
(cf. [23]). The genus of this surface is denoted g(Σ ), and can be calculated by the formula
D
2−s (D)−s (D)+c(D)
0 1
g(Σ ) = ,
D
2
wherec(D)isthenumberofcrossingsinDands (D)ands (D)arethenumberofcirclesappearing
0 1
in the all 0 and all 1 resolutions of D respectively. The Turaev genus of a link is defined as
g (L) = min{g(Σ ) | D is a diagram for L}.
T D
TheTuraevgenusofalinkLisameasureofhowfarLisawayfrombeingalternating. Specifically,
Dasbach et. al. [10] prove the following proposition.
Proposition 3.7 (Dasbach-Futer-Kalfagianni-Lin-Stoltzfus). A link has Turaev genus0 if and only
if it is alternating.
Also, the Turaev genus of L gives a bound on the Khovanov width of L. Manturov [17] and
Champanerkar-Kofman-Stoltzfus [9] prove the following inequality.