Siberian Mathematical Journal, Vol. 55, No. 4, pp. 687–695, 2014Original Russian Text Copyright c© 2014 Razumovskiı R.V.

SOME CLASSES OF FIBERED LINKS

R. V. Razumovskiı UDC 514.76

Abstract: We present two infinite classes of links and prove their fiberedness. The grid diagrams areused for combinatorial description. The first class generalizes the Lorentz links and is characterizedby the fact that every second vertex in the diagram of each representative of the family lies on thecoordinate diagonal of the grid diagram. The complements of the knots of the second class admit a freeaction of Zn.

DOI: 10.1134/S0037446614040107

Keywords: fibered links, grid diagrams, Heegaard–Floer homology, (n, p, q)-periodic knots

Introduction

The classical knot theory is a subarea of three-dimensional topology which studies smooth embeddingsof families of circles into the three-dimensional sphere up to ambient isotopy. An elementary introductionto the classical knot theory can be found in [1, 2]. There exist generalizations studying embeddings intoother three-dimensional manifolds and generalizations connected with dimension lifting both for theambient spaces and embedded spheres.One of the problems in knot theory consists in distinguishing classes with interesting properties. An

important example is given by fibered links.

Definition 1. An oriented link L is called fibered if the complement of L is fibered over the circleinto connected oriented surfaces having L as the boundary.

The fibered links appear naturally in the study of the singularities of planar algebraic curves [3]. Aswas shown in [4], the closures of strictly positive braids are fibered links. In particular, this applies totoric links, iterated toric (and hence all algebraic) and twisted links (or, equivalently, toric links [5]). Theknots obtained by A′Campo’s construction are also fibered [6].For describing the notions introduced in the article, we use grid diagrams.

Definition 2. A planar link diagram is called grid if it is piecewise linear and contains only verticaland horizontal edges. The edges are in general position; i.e., each vertical or horizontal straight linecontains at most one edge; the number of vertical edges is called the complexity of the diagram; in eachself-intersection, a vertical edge is over a horizontal edge.Each link has a grid diagram [7]. For oriented links, the vertices are colored in black and white

so that the vertical edges be oriented from black to white vertices. We may assume that, in a griddiagram of complexity n, the coordinates of all vertices take integer values from 1 to n. Therefore, itis convenient to represent these diagrams as square matrices of order n in which the nonempty entriesare those 2n corresponding to n black and n white vertices. The cyclic permutations of the rows of thematrix correspond to isotopies of the link taking the upper edge to the lower row over the remainingedges. Similarly, the cyclic permutations of the rows of the matrix correspond to isotopies of the linkstaking the leftmost edge to the rightmost column under the other edges. We may regard grid diagramsas depicted on the torus T 2 embedded canonically in S3 except for selfintersections of the diagram, wherewe will need small thickenings of the surface. We may assume that the coordinates of the vertices takevalues in Zn = Z/nZ. We will use n ∈ Zn instead of 0 ∈ Zn. Applying an appropriate number ofcyclic permutations of edges, we may assume that the point (1, 1) belongs to the set of black vertices ofthe diagram.

Moscow. Translated from Sibirskiı Matematicheskiı Zhurnal, Vol. 55, No. 4, pp. 840–850, July–August, 2014.

Original article submitted January 29, 2013. Revision submitted January 27, 2014.

0037-4466/14/5504–0687

c©

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Fig. 1. The knot T (4, 5).

The simplest class of fibered links is given by the family of toric links. Theycan all be represented by grid diagrams of a special kind: the link T (p, q) isdefined by the diagram of complexity p+ q whose all black vertices lie on thediagonal y = x and all white vertices lie on the shifted diagonal y = x + p.For example, Fig. 1 shows T (4, 5).It is natural to generalize these diagrams in the two ways: by perturb-

ing points outside the diagonal y = x (and so obtaining diagonal links) orpreserving the Zn-invariance of the sets of black and white vertices.As we will show, both classes consist of fibered links. In the former case we explicitly construct the

Seifert surface being the fiber of the link. In the later case for the proof we use results from the theoryof Heegaard–Floer homology.

1. Diagonal Links

Let us generalize the notion of diagonal link which was introduced in [8] by a wider class of permu-tations.

Definition 3. Let σ ∈ Sn be a fixed-point-free permutation. By a link of type diag(σ) we meana link having a rectangular diagram of complexity n with the set of vertices combining the set of blackvertices {(i, i) | i = 1, . . . , n}, and the set of white vertices {(i, σ(i)) | i = 1, . . . , n}.We will also denote σ ∈ Sn by the vector (σ(1), σ(2), . . . , σ(n)).Example 1. Fig. 2 shows the knot diag(3, 8, 7, 1, 14, 9, 15, 10, 2, 5, 13, 6, 4, 11, 12).

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Fig. 2

The number of connected components of some link diag(σ) is obviously equal to the number of thecycles in the permutation σ. The vertical and horizontal edges over the diagonal are oriented upwardand rightward respectively, whereas the vertical and horizontal edges under the diagonal are orienteddownward and leftward respectively. Therefore, all crossroads on the diagram diag(σ) are negative. If σis a reshuffle then diag(σ) is a Lorenz link [8]. The details on Lorenz links can be found in [4, 5].If we replace σ by tkσt−k, where t = (2, 3, . . . , n− 1, n, 1) then we will obtain the isotopic link with

the isotopy defined by a cyclic shift on the torus along the diagonal. Replacing σ by ΣσΣ−1, where

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Σ = (n, . . . , 1), we also obtain some isotopic link—it suffices to rotate the diagram around its centerby 1800. The links diag(σ) and diag(σ−1) are isotopic to each other up to reversing orientation: thediagram must be reflected with respect to the diagonal y = x, change the signs of all crossroads, andreverse the orientation of each component.Consider the case when a link L = diag(σ) consists of at least one component. We want to understand

when it is a boundary link. Since all crossroads on the diagram are negative, the presence of intersectionsfor two components gives a strictly negative linking number, which means linkedness. Otherwise, a Jordancurve corresponds to the projection of the separating sphere. Construct a graph whose vertices correspondto the components of the link and whose edges correspond to the presence of a nonzero linking numberbetween the corresponding components. The connected components of the graph are no longer boundarybut it is not hard to prove that a connected component is separated from its complement by a sphere.Consequently, the connected components give maximal nonboundary sublinks.

Definition 4. Call a permutation σ ∈ Sn nonsplit if σ has no proper invariant segments, i.e., thereis no pair (i, j) with the properties i < j, (i, j) �= (1, n), σ({i, . . . , j}) = {i, . . . , j}.Theorem 1. A link L = diag(σ) is fibered if and only if σ is a nonsplit permutation.

Fig. 3. Separation of intersections.

Proof. Apply the Seifert algorithm, i.e., separate all intersec-tions (Fig. 3), get the Seifert disks, and join them by short bandstwisted by half-turn corresponding to the intersections of the dia-gram. As we show below, the so-obtained surface S with ∂S = L isthe fiber of the fibration.

Example 2. Fig. 4 shows the structure of the surface corresponding to the knot of Example 1. Itpresents four levels of embeddedness for disks, where darker shades corresponds to higher levels. The diskson adjacent levels are joined by the bands twisted by half-turn at the common points of the projectionsof the boundary circles of the disks.Frame the surface S by the field of vectors completing bases in the tangent planes of the surface to

negative bases of the space. The collision mapping f : S → S3 \ S along the so-constructed field inducesthe homomorphism f∗ : π1(S)→ π1(S3 \S). If f∗ is an epimorphism then L is fibered [9]. Let us describethe properties of the surface which are sufficient for the sequel:• The surface must be connected.• All Seifert circles must be oriented in one common direction.• For every disk P , the selfintersections of the boundary links corresponding to the bands joining it

with the ambient disk have the common sign defined by P . Here the global selfintersections can havedifferent signs.Call surfaces with these properties good.If σ is a split permutation then the link is boundary and so it cannot be fibered [10]. If σ is a nonsplit

permutation then the surface is connected. Indeed, if it is nonconnected then the two possibilitiesare open:(1) There are more than one external Seifert circles; choose one of them. Then the set of the diagonal

vertices in the corresponding disk is a segment invariant under σ. Hence, the permutation splits.(2) There is an embedded Seifert circle not joined by bands with the circle containing it. Similarly,

the collection of the diagonal vertices in the corresponding disk is a segment invariant under σ.The two other sufficient properties are pointed out immediately after the definition of diagonal link.

The theorem ensues from the

Lemma. Let S be a good surface. Then the link ∂S is fibered.

Proof. Use the construction of [4] for proving the surjectivity of f∗. Let Sp be a good surfacecomposed of p Seifert disks. Induct on p. If p = 1 then π1S1 ∼= π1(S3 \ S1) ∼= {1}. Suppose that theassertion is proved for all p ≤ k and prove it for p = k + 1. Consider the surface Sk obtained from Sk+1by removing some disk D not containing the other disks on the projection of the diagram. Suppose that

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Fig. 4. Construction of a surface with four levels of embeddedness of disks.

D is joined with the disk containing it by r bands. Then

π1(Sk+1) ∼= π1(Sk) ∗ Fr, π1(S3 \ Sk+1) ∼= π1(S3 \ Sk) ∗ Fr.Here Fr is a free group of rank r. Let a be a simple basis cycle in π1(Sk+1) passing through two adjacentbands of the r bands under consideration. Fig. 5 shows that f∗a is a basis cycle in π1(S3\Sk+1) embracingone of the bands. This is guaranteed by the fact that the selfintersections of the link have a common sign.Hence, f∗(Fr) ∼= Fr, where, on the left-hand side, Fr corresponds to the subgroup in π1(Sk+1) generatedby the basis cycles passing through the adjacent bands of the r bands under consideration and, on theright-hand side, Fr corresponds to the subgroup in π1(S

3 \Sk+1) generated by the basis cycles embracingthe r bands. By the induction hypothesis, f∗π1(Sk) ∼= π1(S3 \ Sk). Consequently,

f∗π1(Sk+1) ∼= f∗π1(Sk) ∗ f∗Fr ∼= π1(S3 \ Sk) ∗ Fr ∼= π1(S3 \ Sk+1).The lemma and the theorem are proved.

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Fig. 5. Collision of loops.

Corollary to the Lemma. Let b be a strictly positive braid, i.e., a braid represented by a wordw ∈ Bn that includes all generators σ1, . . . , σn−1 and does not include their inverses. Consider anarbitrary collection ε = (ε1, . . . , εn−1) consisting of ±1. Denote by ε(w) ∈ Bn the word obtained from wby replacing each σi by σ

εii . Then the closure of ε(w) is a fibered link.

Proof. The surface given by applying the Seifert algorithm to the closure of the braid ε(w) possessesthe following properties: It consists of n Seifert disks, the boundary circles of the disks are oriented inone direction, the surface is connected due to the strict positivity of w; and the selfintersections of thelink joining the ith and (i+ 1)th Seifert circles (i = 1, . . . , n− 1) have a common sign εi. Consequently,the surface is good, and the link is fibered. �

2. (n, p, q)-Periodic Knots

Definition 5. Let n, p, q be a triple of naturals with (n, p) = (n, q) = 1, 1 ≤ p, q ≤ n − 1. An(n, p, q)-periodic knot is a knot having a grid diagram of complexity n with the following set of vertices:the set of black vertices {(1+ iq, 1+ ip) | i = 0, . . . , n−1}; the set of white vertices {(1+ iq, 1+(i+1)p) |i = 0, . . . , n− 1}.Example 3. Fig. 6 shows the grid diagram of a (13, 7, 3)-periodic knot.

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Fig. 6

Each toric knot T (p, q) is isotopic to a (p+ q, p, p)-periodic knot. Some iterated toric knots are also(n, p, q)-periodic.

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In the grid diagram of an (n, p, q)-periodic knot, of n vertical edges, p edges are directed downwardand n− p edges are directed upward; and of n horizontal edges, q edges are directed leftward and n− qedges are directed rightward. Indeed, a vertical edge has length p and is oriented upward when the initialpoint has vertical coordinate at most n−p; in the remaining p cases, a vertical edge has length n−p andis oriented downward; the situation with horizontal edges is similar. Therefore, an (n, p, q)-periodic knotcan be represented as the closure of braids on p, q, n− p and n− q strands. This is a general method forobtaining the closure of braids in grid diagrams, which consists in disconnecting the edges of a certaindirection and orienting and connecting them via infinity [7]. For min(p, q, n − p, n − q) = 1, we obtaina trivial knot since the braid group on one strand is trivial. For 1 < p < n−1, every (n, p, 2)-periodic knotis isotopic to a toric knot T (k, 2) for some k since the closures of braids on two strands give these links.The sets of black and white vertices in (n, p, q)-periodic knots are invariant under the shift by (q, p),

and the shift mapping has order n. This circumstance will be used in the proof of

Theorem 2. (n, p, q)-Periodic knots are fibered. The three-dimensional genus of an (n, p, q)-periodicknot is at most (min(p(n− p), q(n− q))− n+ 1)/2.Proof. For proving fiberedness, use some results from the theory of Heegaard–Floer homology.

In [11], there was constructed an invariant of oriented knots, the finite-dimensional bigraded vector space

HFK(K) =⊕

m,s

HFKm(K, s)

over Z2. Here K is an oriented knot, m and s are the values of the Maslov and Alexander gradings

respectively. The invariant HFK(K) is connected with the symmetrized Alexander polynomial by therelation

ΔK(t) =∑

m,s

(−1)m rank HFKm(K, s) · ts.

We will need the following assertions:By [12], the three-dimensional genus of a Seifert knotK is equal to the maximal value of the Alexander

grading that contains nontrivial homology HFK(K); i.e.,

g(K) = max{s ∈ Z | HFK∗(K, s) �= 0}.By [13], a knot K is fibered if and only if the homology HFK(K) in the higher Alexander grading isone-dimensional; i.e.,

dim HFK∗(K, g(K)) = 1.

The differential ∂ of a chain complex defining HFK(K) preserves the Alexander grading. Hence, inthe particular case of only one generator w of a maximal Alexander grading, we can prove the fiberedness

of the knot. Indeed, ∂w = 0, and the grading A(w) contains nothing else, and so dim HFK∗(K,A(x)) = 1,and the complement to K is fibered on a surface of genus g(K) = A(w).

The next construction of a chain complex defining HFK(K) was described in [14]: Let Γ be a griddiagram of complexity n for a knot K. Assign to it the chain complex (C(Γ), ∂). As generators forthe module C(Γ) over Z2, take the n-point sets X = {(i, σ(i)) | i = 1, . . . n, σ ∈ Sn}, whose numberis n! = |Sn|. To define the Alexander grading A : X → Z, associate with Γ a quadratic matrix a oforder n as follows: aij is minus the number of the turns of the projection of the knot around the point(i−1/2, j−1/2). Then, for each generator x, we have A(x) =∑p∈x ap up to a common additive summand.Example 4. Fig. 7 presents the diagram Γ of Example 3 with the associated matrix a.Next, in [14] there are defined the Maslov grading M : X → Z and the differential ∂ : C(Γ)→ C(Γ)

preserving the Alexander grading (we will not need the definitions). Let V be a bigraded two-dimensionalvector space with generators of the gradings (−1,−1) and (0, 0). Then the homology H∗(C, ∂) of theabove-described complex is isomorphic to the bigraded group HFK(K)⊗ V ⊗(n−1).692

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Fig. 7. The matrix a for the diagram of Example 3.

Define the family of matrices (that includes a) determining the Alexander grading up to an additivesummand. Put

U = {u ∈Mn(R) | ui,j + ui+1,j+1 = ui,j+1 + ui+1,j , i = 1, . . . , n− 1, j = 1, . . . , n− 1}.

It is easy to check that ui,j = ui,1 + u1,j − u1,1; hence, dimU = 2n − 1, and we can take the collection{ui,j | min(i, j) = 1} as a system of independent parameters in U . Let La = a + U be a (2n − 1)-dimensional affine plane in Mn(R). For each z ∈ La, we can define the grading Z : X → R by theformula Z(x) =

∑p∈x zp. Since, for every two of these gradings Z1 and Z2 and x ∈ X, we have

Z1(x)− Z2(x) =∑

i=2,...,n−1(z1 − z2)i,1 +

∑

j=2,...,n−1(z1 − z2)1,j + (2− n)(z1 − z2)1,1 = const(z1, z2)

(i.e., the difference does not depend on x) and A(x) =∑p∈x ap up to an additive summand, it suffices

to prove our assertion for every Z-grading. For each z ∈ La,

zi,j + zi+1,j+1 − zi,j+1 − zi+1,j =⎧⎨

⎩

0, if (i, j) is not a vertex of the diagram;

1, if (i, j) a black vertex;

−1, if (i, j) is a white vertex.Indeed, these equalities hold for a ∈ La, and the addition of u ∈ U changes nothing. On the other hand,these conditions characterize La fully.Now, use the Zn-invariance of the diagrams of (n, p, q)-periodic knots. The sets of white and black

vertices are stable under the shift by a vector (q, p); this transformation has order n. The shift inducesthe transformation of the matrix a consisting in the q-fold cyclic permutation of the columns and a p-fold cyclic permutation of the rows. After one shift, we obtain the matrix φa ∈ La. Suppose thata = 1

n

∑n−1k=0 φ

ka, i.e., that a is the average of a over the action of φ. Since La is an affine plane, a ∈ La(the sum of the weights is equal to 1). Denote by A the grading defined by a.

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Fig. 8. The average a of a over the action of Zn.

Example 5. Fig. 8 shows an example of a. The generator containing all elements 1013 has maximal

grading A, and there are no other generators with such a grading.Let r = pq−1 ∈ Zn (q−1 ∈ Z∗n). Then, for each i, the (i + 1)th column of a is obtained from the

ith column by an r-fold cyclic permutation of the elements. Since (n, r) = 1, for proving the existenceof w ∈ X with a maximal grading A, it suffices to demonstrate that the first column of a containsa unique maximum (in this case all maxima over columns will be situated in different rows). Assumethat si = ai+1,1 − ai,1, i = 1, . . . , n. Then

si−r − si = ai−r+1,1 − ai−r,1 − ai+1,1 + ai,1

= ai+1,2 − ai,2 − ai+1,1 + ai,1 =⎧⎨

⎩

1, i = 1,

−1, i = p+ 1,0 otherwise.

The last passage is explained by the fact that a ∈ La, and the cases of i = 1 and i = p+ 1 correspond toa black vertex and a white vertex. Therefore, for some d, under the condition 1− kr ≡ 1+ p(modn), wehave

(s1, s1−r, s1−2r, . . . , s1−kr, s1−(k+1)r, . . . , s1−nr) = (d, d+ 1, d+ 1, . . . , d+ 1, d, . . . , d).

Hence, k = n− q. The equality 0 =∑ni=1 si = nd+ k implies d = q−nn , d+ 1 =

qn .

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It is easy to show that the sequence (a1,1, a1−r,1, . . . , a1−(n−1)r,1) contains the unique maximal element.Indeed, had this fail the two maxima would be joined by u values q−nn and v values

qn with 0 < u+v < n.

We infer 0 = u q−nn + vqn =

(u+v)q−unn . The numerator of the fraction divides by n, and (q, v) = 1;

therefore, u+ v divides by n. The so-obtained contradiction proves fiberedness.It remains to prove the constraint on the genus. Since

H∗(C, ∂) ∼= HFK(K)⊗ V ⊗(n−1)

andΔK(t) =

∑

m,s

(−1)mrank HFKm(K, s) · ts

for some half-integer w, we have

ΔK(t) =det(tai,j ) ∗ tw(t− 1)n−1 .

Let us estimate the difference of the maximal and minimal degrees of the polynomial det(tai,j ) from above.The collection {si} contains n − q values qn > 0 and q values q−nn < 0. Hence, the difference between

the maximum and the minimum in each column of a is at most q(n−q)n . Consequently, the difference ofthe extremal degrees in det(tai,j ) is at most q(n − q). Therefore, the difference of the extremal degreesin ΔK(t) is at most q(n − q) − n + 1. Since the genus of the fibered knot is equal to the higher degreeof the symmetrized Alexander polynomial, we obtain the upper bound (q(n − q) − n + 1)/2. A similarargument for rows gives the bound (p(n − p) − n + 1)/2. Taking the minimum of the two bounds, weobtain the theorem. �Remark. The techniques of the proof of Theorem 2 could be applied to the case of diagonal knots.

The only generator with the higher Alexander grading is defined by the elements lying on the diagonal.Instead, we used a more transparent geometric argument by presenting the Seifert surface that is thefiber of the fibration.

The author expresses his gratitude to his supervisor Professor I. A. Dynnikov for posing the problem,numerous discussions, and invaluable help in work.

References

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R. V. RazumovskiıMoscow State University, Moscow, RussiaE-mail address: razumovskymail@mail.ru

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