aa r X i v : . [ m a t h . G T ] M a r The Rasmussen invariant of a homogeneous knot
Tetsuya Abe
Department of Mathematics, Osaka City UniversitySugimoto, Sumiyoshi-ku Osaka 558-8585, JapanEmail: [email protected]
Abstract
A homogeneous knot is a generalization of alternating knots and positive knots. We deter-mine the Rasmussen invariant of a homogeneous knot. This is a new class of knots such that theRasmussen invariant is explicitly described in terms of its diagrams. As a corollary, we obtainsome characterizations of a positive knot. In particular, we recover Baader’s theorem whichstates that a knot is positive if and only if it is homogeneous and strongly quasipositive.
In [25], Rasmussen introduced a smooth concordance invariant of a knot K by using the Khovanov-Lee theory (see [15] and [16]), now called the Rasmussen invariant s ( K ). This gives a lower boundfor the four ball genus g ∗ ( K ) of a knot K as follows. | s ( K ) | ≤ g ∗ ( K ) . (1.1)This lower bound is very powerful and it enables us to give a combinatorial proof of the Milnorconjecture on the unknotting number of a torus knot. Our motivation for studying the Rasmusseninvariant is to describe s ( K ) in terms of a given diagram of a knot K to better understand g ∗ ( K ).From this point of view, some estimations of the Rasmussen invariant are known (Plamenevskaya[24], Shumakovitch [30] and Kawamura [12]. See also Stoimenow [32]).Let O + ( D ) and O − ( D ) be the numbers of connected components of the diagrams which isobtained from D by smoothing all negative and positive crossings of D , respectively. Recently,Kawamura [13] and Lobb [20] independently obtained a more sharper estimation for the Rasmusseninvariant as follows. Theorem 1.1 ([13] and [20]) . Let D be a diagram of a knot K . Then w ( D ) − O ( D ) + 2 O + ( D ) − ≤ s ( K ) , where ω ( D ) denotes the writhe of D (i.e. the number of positive crossings of D minus the numberof negative crossings of D ) and O ( D ) denotes the number of the Seifert circles of D . Let ∆( D ) = O ( D ) + 1 − O + ( D ) − O − ( D ) (a graph theoretical interpretation of ∆( D ) due toLobb is given in Section 3). In addition to Theorem 1.1, Lobb [20] showed that if ∆( D ) = 0, then s ( K ) = w ( D ) − O ( D ) + 2 O + ( D ) − D satisfy the condition ∆( D ) = 0.Lobb [20] showed that if D is positive, negative, alternating, or a certain braid diagram, then In [13] and [20], it was denoted by l ( D ) and T + ( D )) respectively. D ) = 0. Note that these diagrams are all homogeneous (the definition is given in Section 2). Inthis paper, we show that if D is a homogeneous diagram of a knot, then ∆( D ) = 0 (the converseis also true. See Theorem 3.4) and our main result is to determine the Rasmussen invariant of ahomogeneous knot. This is a new class of knots such that the Rasmussen invariant is explicitlydescribed in terms of its diagrams. Theorem 1.2.
Let D be a homogeneous diagram of a knot K . Then s ( K ) = w ( D ) − O ( D ) + 2 O + ( D ) − . Ozsv´ath and Szab´o [22] and Rasmussen [26] independently introduced another smooth concor-dance invariant of a knot K by using the Heegaard Floer homology theory, now widely known asthe tau invariant τ ( K ). The Rasmussen invariant and tau invariant share some formal propertiesand these are closely related to positivity of knots. There are many notions of positivity (e.g. braidpositive, positive, strongly quasipositive and quasipositive). We recall these notions of positivityin Section 3. Let D be a diagram of a knot K . Then Kawamura [13] also proved w ( D ) − O ( D ) + 2 O + ( D ) − ≤ τ ( K ) . Note that, if ∆( D ) = 0, 2 τ ( K ) = w ( D ) − O ( D ) + 2 O + ( D ) − Therefore the correspondingresult to Theorem 1.2 holds for the tau invariant. In particular, we obtain τ ( K ) = s ( K ) / K .On the other hand, the Rasmussen invariant and tau invariant sometimes behave differently. Ithas been conjectured that τ = s/
2, however, Hedden and Ording [11] proved that the Rasmusseninvariant and tau invariant are distinct (see also [19]). It may be worth remarking that the Ras-mussen invariant is sometimes stronger than the tau invariant as an obstruction to a knot beingsmoothly slice ([11] and [19], see also [7]). This is the reason why we are more interested in theRasmussen invariant rather than the tau invariant.One can easily see that a braid positive knot is strongly quasipositive, however, it is not obviouswhether a positive knot is strongly quasipositive. Nakamura [21] and Rudolph [28] independentlyproved that a positive knot is strongly quasipositive. Not all strongly quasipositive knots are pos-itive. For instance, such examples are given by divide knots [27]. Rudolph [28] asked whetherpositive knots could be characterized as strongly positive knots with some extra geometric condi-tions. Several years later, Baader found that the extra condition is homogeneity. To be precise,Baader [2] proved that a knot is positive if and only if it is homogeneous and strongly quasipositive.As a corollary of Theorem 1.2, we obtain some characterizations of a positive knot.
Theorem 1.3.
Let K be a knot. Then (1) – (4) are equivalent. (1) K is positive. (2) K is homogeneous and strongly quasipositive. (3) K is homogeneous, quasipositive and g ∗ ( K ) = g ( K ) . (4) K is homogeneous and τ ( K ) = s ( K ) / g ∗ ( K ) = g ( K ) . In particular, we recover Baader’s theorem. Note that our proof is 4-dimensional in the sensethat we use concordance invariants, whereas Baader [2] used the Homflypt polynomial. As animmediate corollary of Theorem 1.3, we obtain the following.
Corollary 1.4.
Let K be a homogeneous knot. Then the following are equivalent. (1) K is positive. by using the fact that − τ ( K ) = τ ( K ) for any knot K [22], where K denotes the mirror image of K . K is strongly quasipositive. (3) K is quasipositive and g ∗ ( K ) = g ( K ) . (4) τ ( K ) = s ( K ) / g ∗ ( K ) = g ( K ) . It may be interesting to compare Corollary 1.4 and the following proposition by Hedden.
Proposition 1.5 ([10]) . Let K be a fibered knot. Then the following are equivalent. (1) K is strongly quasipositive. (2) K is quasipositive and g ∗ ( K ) = g ( K ) . (3) τ ( K ) = g ∗ ( K ) = g ( K ) . At a first glance, we wonder why similar results hold for fibered knots and homogeneous knots.However, it is not surprising since homogeneous knots are related to fiberedness. For instance, aknot which admits a homogeneous braid diagram is fibered (see Section 2 or Proposition 1.4 in[20]).This paper is constructed as follows. In Section 2, we observe a geometric aspect of a homoge-neous knot. In Section 3, we give a new characterization of a homogeneous diagram of a knot anddetermine the Rasmussen invariant of a homogeneous knot (Theorem 1.2). In Section 4, we recallsome notions of positivity for knots and give some characterizations of a positive knot (Theorem1.3). In Section 5, we propose a new approach to estimate the Rasmussen invariant of a knot.
Acknowledgments
The author would like to thank the members of Friday Seminar on Knot Theory in Osaka CityUniversity, especially Masahide Iwakiri and In Dae Jong. The author would like to thank SeiichiKamada, Tomomi Kawamura and Kengo Kishimoto for helpful comments on an earlier draft ofthis paper and Mikami Hirasawa for explaining to him cut-and-paste arguments of Seifert surfacesin detail. This work was supported by Grant-in-Aid for JSPS Fellows.
Cromwell [4] introduced the notion of homogeneity for knots to generalize results on alternatingknots. The notion of homogeneity is also defined for signed graphs and diagrams. For graphtheoretical terminologies in this paper, we refer the reader the book of Cromwell [5]. A graph is signed if each edge of the graph is labeled + or − . A typical signed graph is theSeifert graph G ( D ) associated to a knot diagram D : for each Seifert circle of D , we associate avertex of G ( D ) and two vertices of G ( D ) are connected by an edge if there is a crossing of D whoseadjacent two Seifert circles are corresponding to the two vertices. Each edge of G ( D ) is labeled +or − depending on the sign of its associated crossing of D . For convenience, we say a + or − edgeinstead of an edge labeled + or − .A block of a (signed) graph is a maximal subgraph of the graph with no cut-vertices. A signedgraph is homogeneous if each block has the same signs. A diagram D of a knot is homogeneous if G ( D ) is homogeneous. Cromwell [4] showed that alternating diagrams and positive diagrams arehomogeneous. There are many homogeneous diagrams which are non-alternating and non-positive. Example 2.1.
Let D be the non-alternating and non-positive diagram as in Figure 1. Then G ( D ) is homogeneous (see Figure 2). Therefore D is a homogeneous diagram which is non-alternatingand non-positive. Note that D is not minimal crossing diagram (this is not used later). In this paper, we use the notation “cycle” instead of “circuit”. G ( D ) is homogeneous4et B n be the braid group on n strands with generators σ , σ , · · · , σ n − . Stallings [31] intro-duced the notion of a homogeneous braid. A braid β = σ ǫ i σ ǫ i · · · σ ǫ k i k , ǫ j = ± j = 1 , · · · , k ) is homogeneous if(1) every σ j occurs at least once,(2) for each j , the exponents of all occurrences of σ j are the same.For example, the braid σ σ − σ σ − is homogeneous, however, the braid σ σ σ σ − is not homo-geneous. Stallings [31] proved that the closure of a homogeneous braid is fibered. The followinglemma is origin of the name “homogeneous”. Lemma 2.2 ([4]) . Let β be a braid whose closure is a knot. Then β is homogeneous if and only ifthe braid diagram of the closure of β is homogeneous. A knot K is homogeneous if K has a homogeneous diagram. The class of homogeneous knotsincludes alternating knots and positive knots. There are homogeneous knots which are non-alternating and non-positive and Cromwell [4] showed that the knot 9 is the simplest one. Oneof the distinguished properties of a homogeneous diagram is the following. Theorem 2.3 ([4]) . Let D be a homogeneous diagram of a knot K . Then the genus of K is realizedby that of the Seifert surface obtained by applying Seifert’s algorithm to D . Cromwell proved the above theorem algebraically. There is a geometric proof. Here we give anoutline of the proof, which is suggested by M. Hirasawa.The Seifert circles of a diagram is divided into two types: a Seifert circle is of type 1 if it doesnot contain any other Seifert circles in R , otherwise it is of type 2 . Let D ⊂ R be a knot diagramand C a type 2 Seifert circle of D . Then C separates R into two components U and V such that U ∪ V = R and U ∩ V = ∂U = ∂V = C . Let D and D be the diagrams formed form D ∩ U and D ∩ V by adding suitable arcs from C respectively. If both ( U − C ) ∩ D = ∅ and ( V − C ) ∩ D = ∅ ,then C decomposes D into a ∗ -product of D and D , which is denoted by D = D ∗ D . Then theSeifert surface obtained by applying Seifert’s algorithm to D is a Murasugi sum of Seifert surfacesobtained by applying Seifert’s algorithm to D and D respectively (for the definition of a Murasugisum, see [14] or [8]). A diagram is special if D has no decomposing Seifert circles of type 2. Aspecial positive (or negative) diagram is alternating. Cromwell implicitly showed the following (seeTheorem 1 in [4]). Lemma 2.4 ([4]) . Let D be a homogeneous diagram of a knot K . Then(1) there are special diagrams D , · · · , D n such that D = D ∗ D ∗ · · · ∗ D n ,(2) each special diagram D i ( i = 1 , · · · , n ) is the connected sum of special alternating diagrams,(3) each special alternating diagram corresponds to a block of G ( D ) . Let D be homogeneous diagram of a knot K . Then, by Lemma 2.4, the Seifert surface S obtained by applying Seifert’s algorithm to D is Murasugi sums of the Seifert surfaces obtained byapplying Seifert’s algorithm to the special alternating diagrams. The following lemma is classicalresults of Crowell and Murasugi. Lemma 2.5.
Let D be a alternating diagram of a knot K . Then the genus of K is realized by thatof the Seifert surface obtained by applying Seifert’s algorithm to D . In [9], Gabai gave an elementary proof of Lemma 2.5 by using cut-and-past arguments. ByLemma 2.5, S is Murasugi sums of minimal Seifert surfaces. Let R and R be two minimalSeifert surfaces. Then a Murasugi sum of R and R is a minimal Seifert surface due to Gabai [8].Therefore we obtain a geometric proof of Theorem 2.3.5igure 3: the graph G ∆ is tree In this section, we give a new characterization of a homogeneous diagram (Theorem 3.4). Inparticular, we show that if D is a homogeneous diagram of a knot, then ∆( D ) = 0. One canprove this by induction on the number of cut-vertices of G ( D ), however, we prove this more graph-theoretically. For the Seifert Graph G ( D ) associated to a knot diagram D , we construct a graph(which is denoted by G ( D ) ∆ later) such that the number of cycles of the graph is equal to ∆( D )and we prove if D is homogeneous, then the graph has no cycles. By using Theorems 1.1 and 3.4,we determine the Rasmussen invariant of a homogeneous knot (Theorem 1.2).Let G + and G − be the graphs which are obtained from a signed graph G by removing all − and + edges, respectively. Here we note that, by definition, each vertex of G belongs to exactlyone connected component of G + and G − respectively. Let G ∆ be the graph whose vertices are theconnected components of G + and G − and two vertices of G ∆ are connected by an edge if a vertexof G belong to the two connected components (which correspond to the two vertices). We give twoexamples, which provide us the idea of the proof of Lemma 3.3. Example 3.1.
Let G ( D ) be the signed graph as in Figure 2. We label and the connectedcomponents of G + and , , and the connected components of G − . Then G ∆ is the graph as inFigure 3 and it is tree. Example 3.2.
Let G be the signed graph as in Figure 4. Then G has only one block and it is G itself. Since the block contains + and − edges, G is not homogeneous. Note that G has a cyclewhich contains + and − edges (in this case, the cycle is unique). We label and the connectedcomponents of G + and , , and the connected components of G − . Then G ∆ is the graph asin Figure 4 and G ∆ has a cycle which is denoted by (1 , , , , (in this case, the cycle isalso unique).Conversely, let e e , e e , e e and e e be the edges of G ∆ and e v , e v , e v , e v and e v the vertices of G ∆ asin Figure 5. Let v i ( i = 1 , · · · , be the vertex of G which corresponds to e e i and let v = v . Then e v i +1 as a connected component of G + or G − contains v i and v i +1 and there exists a simple path l i in e v i +1 from v i to v i +1 . Therefore we obtain a cycle l l l l from v to v (= v ) . For a signed graph G , we denote by sign( e ) the sign of an edge e of G . We show the followinglemma to prove Theorem 3.4. To prove (2) ⇐ = (3) is essential.6igure 4: the graph G is not homogeneous v = v v v v Figure 5: the graph G is not homogeneous7 emma 3.3. Let G be a signed graph. The following are equivalent.(1) G is not homogeneous.(2) G has a cycle which contains both + and − edges.(3) G ∆ has a cycle.Proof. (1) = ⇒ (2) Since G is not homogeneous, by definition, there exists a block which contains+ and − edges. Then there exist a vertex v and edges e and e of the block such that one of theendpoints of e and e is v respectively and sign( e ) = sign( e ). Now v is not a cut-vertex since v is a vertex of the block. There G has a cycle which contains both + and − edges.(1) ⇐ = (2) Let e · · · e n be the cycle of G which contains both + and − edges. Then there exists anatural number i such that sign( e i ) = sign( e i +1 ). Let v be the vertex such that one of the endpointsof e i and e i +1 is v respectively. Since v is not a cut-vertex, edges e i and e i +1 belong to the sameblock. Therefore G is not homogeneous.(2) = ⇒ (3) Let ( e · · · e i ) · · · ( e i k − +1 · · · e i k ) · · · ( e i j − +1 · · · e i j ) be the cycle which contains both +and − edges, where sign( e i k − +1 ) = · · · = sign( e i k ) and sign( e i k ) = sign( e i k +1 ). Then the path e i k − +1 · · · e i k is also one in G sign( e ik ) , which contracts to a vertex e v k of G ∆ ( k = 1 , · · · , j ). Let e e k be the edge of G ∆ whose endpoints are e v k and e v k +1 ( k = 1 , · · · , j − v k such that one of the endpoints of e i k and e i k +1 is v k respectively. Let e e j be the edgeof G ∆ whose endpoints are e v j and e v , which is corresponding to the vertex v j such that one of theendpoints of e i j and e is v j respectively.Therefore e e · · · e e j is a path from e v to e v , possibly not a cycle. If the path is not a cycle, wechoose a cycle (as a subsequence of edges of the path). Therefore G ∆ has a cycle.(2) ⇐ = (3) Let e e · · · e e n be the cycle G ∆ ( i = 1 , · · · , n ) and denote e e i = ( e v i , e v i +1 ). Then e v n +1 = e v .Let v i be the vertex of G which corresponds to e e i ( i = 1 , · · · , n ) and v n +1 = v . Recall that a vertexof G ∆ corresponds to a connected component of G + or G − . Then e v i +1 (as a connected componentof G + or G − ) contain v i and v i +1 ( i = 1 , · · · , n ). There exists a simple path l i from v i to v i +1 .There we obtain a path l l · · · l n from v to v n +1 (= v ), possibly not a cycle. If the path is nota cycle, we choose a cycle (as a subsequence of edges of the path). By the construction, the cyclealways contains both + and − edges.Note that O + ( D ) and O − ( D ) are equal to the numbers of connected components of G ( D ) + and G ( D ) − , respectively. Therefore the number of vertices of G ( D ) ∆ is equal to O + ( D ) + O − ( D )and, by definition, the number of edges of G ( D ) ∆ is equal to O ( D ). Lobb [20] showed that∆( D ) = b ( G ( D ) ∆ ) for any diagram D . For the completeness, we recall the proof here. b ( G ( D ) ∆ ) = b ( G ( D ) ∆ ) − χ ( G ( D ) ∆ )= 1 − ( O + ( D ) + O − ( D ) − O ( D ))= ∆( D ) , where b i denotes the i -th Betti number ( i = 0 ,
1) and χ denotes the Euler characteristic. Then weobtain the following. Theorem 3.4.
A diagram D of a knot is homogeneous if and only if ∆( D ) = 0 .Proof. By the above argument, ∆( D ) = 0 if and only if G ∆ is tree. Therefore the proof immediatelyfollows from Lemma 3.3.Now we prove Theorem 1.2. Proof of Theorem 1.2.
By Theorem 3.4, we obtain ∆( D ) = 0. As mentioned before, Lobb [20]showed that if ∆( D ) = 0, then s ( K ) = w ( D ) − O ( D ) + 2 O + ( D ) −
1. This completes the proof.8igure 6: nugatory crossings
In this section, we recall some notions of positivity and give some characterizations of a positiveknot (Theorem 1.3). In particular, we recover Baader’s theorem which states that a knot is positiveif and only if it is homogeneous and strongly quasipositive.Let D be a diagram of a knot. We denote by D p the diagram which is obtained from D bysmoothing (along the orientation of D ) at a crossing p . A crossing of D is nugatory if there existsa curve l such that the intersection of D and l is only a crossing of D (see also Figure 6). Then itis easy to see that the following lemma holds. Lemma 4.1.
Let p be a crossing of D . Then p is nugatory if and only if the number of theconnected components of D p is two. Here we recall some notions of positivity for knots. A knot is braid positive if it is the closure ofa braid of the form β = Q mk =1 σ i k . A knot is positive if it has a diagram without negative crossings.L. Rudolph introduced the concept of a (strongly) quasipositive knot (see [27]). Let σ i,j = ( σ i , · · · , σ j − )( σ j − )( σ i , · · · , σ j − ) − . A knot is strongly quasipositive if it is the closure of a braid of the form β = m Y k =1 σ i k ,j k . A knot is quasipositive if it is the closure of a braid of the form β = m Y k =1 ω k σ i k ω − k , where ω k is a word in B n . The following are known.(1) Let K be a torus knot. Then τ ( K ) = s ( K ) / g ∗ ( K ) = g ( K ), where g ( K ) denotes the(Seifert) genus of K . This is due to Rasmussen for s [25] and Ozsv´ath and Szab´o for τ [22].These equalities provide a proof of the Milnor conjecture.(2) Let K be a strongly quasipositive knot. Then τ ( K ) = s ( K ) / g ∗ ( K ) = g ( K ). This is dueto Livingston [17].(3) Let K be a quasipositive knot. Then τ ( K ) = s ( K ) / g ∗ ( K ). This is due to Plamenevskaya[23] and Hedden (with a detailed and constructive proof) [10] for τ , and Plamenevskaya [24]and Shumakovitch [30] for s . A torus knot is strongly quasipositive.
9y using Lemma 4.1, we prove Theorem 1.3.
Proof of Theorem 1.3. (1) = ⇒ (2) A positive knot is strongly quasipositive ([21] and [28]).(2) = ⇒ (3) A strongly quasipositive knot K is a quasipositive knot with g ∗ ( K ) = g ( K ) [27].(3) = ⇒ (4) Since K is a quasipositive knot, τ ( K ) = s ( K ) / g ∗ ( K ). By the assumption, g ∗ ( K ) = g ( K ) . Therefore τ ( K ) = s ( K ) / g ∗ ( K ) = g ( K ).(4) = ⇒ (1) Let D be a homogeneous diagram of K . Then the genus of K is realized by that ofthe surface constructed by applying Seifert’s algorithm to D (Theorem 2.3). Therefore 2 g ( K ) =1 + c ( D ) − O ( D ), where c ( D ) denotes the number of crossings of D . By Theorem 1.2, we have s ( K ) = w ( D ) − O ( D )+2 O + ( D ) −
1. By assumption, s ( K ) = 2 g ( K ). This implies that O + ( D ) − n − ( D ), where n − ( D ) denotes the number of negative crossings of D .If there exists a non-nugatory negative crossing p of D , then D p is connected by Lemma 4.1.Therefore O + ( D ) − < n − ( D ) (since, in general, the difference of the numbers of the connectedcomponents of two link diagrams D and D such that D is obtained from D by smoothing at acrossing of D is 0 or 1). This contradicts the fact that O + ( D ) − n − ( D ). Therefore all negativecrossings of D are nugatory and D represents a positive knot.Corollary 1.4 immediately follows from Theorem 1.3. Let D be a diagram of a knot with ∆( D ) = 0. Then Kawamura-Lobb’s inequality may not besharp as follow. Example 5.1.
Let K be the pretzel knot of type (3 , − , − and D the standard pretzel diagramof K . Then ω ( D ) = 9 , O ( D ) = 14 , O + ( D ) = 3 and O − ( D ) = 11 . Therefore ∆( D ) = 1 and w ( D ) − O ( D ) + 2 O + ( D ) − . On the other hand, since K is strongly quasipositive [29], weobtain s ( K ) = 2 g ∗ ( K ) = 2 g ( K ) = 2 . Remark that K is topologically slice but not smoothly slice. We need a more shaper estimation to describe the Rasmussen invariant of the pretzel knotof type (3 , − , −
7) in terms of its standard pretzel diagram. Roughly speaking, there are twoapproaches to estimate or determine the Rasmussen invariant. One of them is to compute theKhovanov homology by using a computer and to use the spectral sequence which converges toLee’s homology. The other is to use some formal properties of the Rasmussen invariant (and thetau invariant). We propose a new and direct approach to estimate or determine the Rasmusseninvariant. We briefly recall the definition of the Rasmussen invariant to explain this. For a fullexplanation, see [25].Let D be a diagram of a knot K and C ∗ Lee ( D ) Lee’s complex (see [25] for the definition). ThenLee [16] proved that the homology group of C ∗ Lee ( D ) is independent of the choice of diagrams of K . Lee’s homology of K , denoted by H ∗ Lee ( K ), is defined to be the homology group of C ∗ Lee ( D ). Inaddition, for a diagram D of a knot K , Lee [16] associated two (co)cycles of C ∗ Lee ( D ), denoted by f o and f ¯ o , and proved that [ f o ] and [ f ¯ o ] are a basis of H ∗ Lee ( K ), in particular, that the dimensionof H ∗ Lee ( K ) is equal to two, where [ · ] denotes its homology class. This basis is called canonical sincethe basis is determined up to multiple of 2 c for K [25], where c is an integer.Rasmussen [25] defined a filtration grading q on a non-zero element of C ∗ Lee ( D ) (which inducesa filtration on C ∗ Lee ( D )). Then a filtration grading s on a non-zero element [ x ] of H ∗ Lee ( K ) (which In [29], K is denoted by P ( − , , In [25], these are denoted by s o and s ¯ o respectively H ∗ Lee ( K )) is defined as follows. s ([ x ]) = max { q ( y ) | [ x ] = [ y ] } . Then the Rasmussen invariant of K , denoted by s ( K ), is defined to be s ([ f o ]) + 1(= s ([ f ¯ o ]) + 1).Since s ([ f o ]) ≥ q ( f o ) and q ( f o ) = ω ( D ) − O ( D ) (by the definition of q ), we obtain s ( K ) ≥ ω ( D ) − O ( D ) + 1. This is the slice-Bennequin inequality for the Rasmussen invariant for K (see [24] and [30]). Theorem 1.1 implies that there exists a cycle f such that [ f o ] = [ f ] and q ( f ) = ω ( D ) − O ( D ) + 2 O + ( D ) −
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