Width of the Whitehead double of a nontrivial knot
aa r X i v : . [ m a t h . G T ] J u l Width of the Whitehead double of anontrivial knot
Qilong Guo, Zhenkun LiJuly 16, 2019
Abstract
In this paper, we prove that w ( K ) = 4 w ( J ), where w ( . ) is thewidth of a knot and K is the Whitehead double of a nontrivial knot J . Width is a knot invariant introduced in Gabai [1]. It has been studiedintensively since then. Zupan in [4] conjectured that w ( K ) ≥ n w ( J ) , where K is a satellite knot with companion J and wrapping number n .This conjecture is still open. The authors of this paper proved in [2] aweaker version of the conjecture with wrapping number being replacedby winding number. In this paper we prove a special case of K havingwrapping number 2 and winding number 0, which is not covered by thediscussion in [2]. Theorem . Suppose J is a nontrivial knot and K is a Whitehead doubleof J . Then we have w ( K ) = 4 · w ( J ) . The gap between wrapping number and winding number in the proof inauthor’s previous paper [2] lies in lemma 4.4, which says that any properlyembedded surface representing a generator of H ( V, ∂V ) has the numberof intersection points with the satellite knot K being no smaller than the1 WIDTH OF A WHITEHEAD DOUBLE absolute value of the winding number of K . Here V is the tubular neighbor-hood of J containing K . This is in general not true for wrapping numberbut for the special case that K is a Whitehead double, we can somehowovercome this difficulty. Acknowledgement.
The first author was supported by NSFC (No.11601519)and Science Foundation of China University of Petroleum, Beijing (No.2462015YJRC034and No.2462015YQ0604). The second author was supported by his advisorTom Mrowkas NSF grant funding No. 1808794.
In what follows, we will use capital letters
K, J, L to denote the knot orlink classes while use lower case letters k, j, l to denote particular knots orlinks within the corresponding classes.
Definition 2.1.
Suppose l w = ˆ k ∪ ˆ j is a Whitehead link in S . Let ˆ V = S \ int( N (ˆ j )) be the exterior of ˆ j containing ˆ k in its interior. Since ˆ j is theunknot in S , ˆ V is a solid torus and ˆ k ∈ int( ˆ V ) can be thought as in figure1. Let j ⊂ S be a non-trivial knot and let V = N ( j ) be the closure of atubular neighborhood of j in S . Let f : ˆ V → S be an embedding suchthat f ( ˆ V ) = V , and let k = f (ˆ k ). Then k is called a Whitehead double of j and j is called its companion . ˆ k ˆ V Figure 1: The pattern of a Whitehead double.2
WIDTH OF A WHITEHEAD DOUBLE
Lemma 2.2.
Suppose k, j, V are defined as in definition 2.1, then anymeridian disk D of V would intersect k at least two times.Proof. See Rolfsen [3].
Remark . The above lemma actually means that the wrapping numberof k is 2. On the other hand, [ k ] = 0 ∈ H ( V ) so the winding number of k is 0. Definition 2.4.
Suppose h ′ : R → R is the projection onto the forthcoordinate, S ⊂ R is the unit 3-sphere, and h = h ′ | S . Fix any knot class K , let K be the set of knots k : S ֒ → S within the class K , so that k doesnot contain the critical points of h = h ′ | S , the function h ◦ k : S → R isMorse, and any two critical points have different critical values.For any k ∈ K , let c < ... < c n be all the critical values of h ◦ k. Pickregular values r , ..., r n − so that c < r < c < r < ... < c n − < r n − < c n , and define w ( k ) = n − X i =1 | h − ( r i ) ∩ k | , w ( K ) = min k ∈K w ( k ) . Here | · | means the number of intersection points. Then w ( K ) is called the width of K .As discussed in the introduction, the key lemma is the following. Lemma 2.5. (Key lemma.) Suppose S is a connected properly embeddedplanar surface inside V which represents a generator of H ( V, ∂V ) , then S intersects k at least 2 times. We will prove this lemma in next section while first use this lemma toprove our main result.
Theorem 2.6.
Suppose J is a nontrivial knot (class) and K is a Whiteheaddouble of J , then we have w ( K ) = 4 w ( J ) . THE PROOF OF THE KEY LEMMA
Proof.
It is easy to construct examples to show that w ( K ) ≤ w ( J ) . In order to prove the reversed inequality, we repeat the whole argument asin the authors’ previous paper [2] except for applying the key lemma 2.5.We only sketch the proof as follows. For details, readers are referred tothat paper.First pick a knot k ∈ K so that w ( k ) = w ( K ). There are correspondingcompanion knot j and its tubular neighborhood V so that k ⊂ V . Let T = ∂V . We can perturb V (while fixing k, j ) so that V does not containthe critical points of h = h ′ | S , h | T is Morse and all critical points of h | T have different critical values.As in [2], there is a Reeb graph Γ R ( V ) associated to the pair ( V, h | V ).There is a natural way to embed Γ R ( V ) into V and there is a unique loop l in Γ V so that(1). We have that l represents a generator of H ( V ).(2). We have that w ( L ) ≥ w ( J ).(3). We can isotope l into a new position l ′ with the following signifi-cance. If r is a regular level of h ◦ k , and m = | h − ( r ) ∩ k | , then there exists P , P , ..., P m as pair-wise disjoint connected components of h − ( r ) ∩ V , sothat for each j with 1 ≤ j ≤ m , P j intersects l ′ transversely at one point.Now let r , ..., r n − be all regular values picked as in definition 2.4. Foreach i with 1 ≤ i ≤ n −
1, we pick connected components P i, , ..., P i,m i of h − ( r i ) ∩ V as in (3). For each j with 1 ≤ j ≤ m i , we know that P i,j iscontained in h − ( r i ) ∼ = S so it is a planar surface. From (1) and the factthat P i,j intersects l ′ transversely at one point, we know that P i,j representsa generator of H ( V, ∂V ). Hence lemma 2.5 applies and | P i,j ∩ k | ≥ . Now we can apply the identical combinatorial arguments as in the proofof lemma 4.6 in [2] to finish the proof here.
In order to prove the key lemma, we need some preparations first.4
THE PROOF OF THE KEY LEMMA
Definition 3.1.
Suppose M = B or S is either a 3-ball or a 3-sphere.Suppose S is a properly embedded surface in M . This means when M = B , ∂S ⊂ ∂M and when M = S , S is closed. Then we can define a map C M,S : ( M \ S ) × ( M \ S ) → {± } as C M,S ( x, y ) = ( − | γ ∩ S | , where x, y ∈ M \ S are two points and γ is an arc connecting two points x and y that is transverse to S . The notation | · | means the number ofintersection points.The function C M,S ( x, y ) is independent of the choice of the arc γ . Sup-pose γ and γ ′ are two such arcs, then β = γ ∪ γ ′ is a closed curve. We knowthat | β ∩ S | ≡ S, ∂S ] ∈ H ( M, ∂M ) = 0. Hence we conclude | γ ∩ S | ≡ | γ ′ ∩ S | mod 2 . From the definition, we know that C M,S ( x, y ) = − x and y are not in the same component of M \ S .Another good property of this function is the following equality: forany 3 points x, y, z ∈ M \ S , we have C M,S ( x, y ) · C M,S ( y, z ) = C M,S ( x, z ) . (1) Remark . This type of functions are called potentials in Gabai [1].
Lemma 3.3.
Suppose S is a connected, properly embedded planar surfaceinside a 3-ball B . Let B and B be the two components of B \ S . Thenthere is no Hopf link l h = l ∪ l inside B such that l i ⊂ B i f or i = 1 , . Proof.
Assume, on the contrary, that there is a Hopf link l h = l ∪ l suchthat l i ⊂ B i . THE PROOF OF THE KEY LEMMA
Note B \ S has two components. To show this, let N ( S ) be a productneighborhood of S , and B ′ = B \ int( N ( S )). We know that int( B ′ ) and B \ S are homotopic. From the connectedness of S and Mayer-Vietoris sequence H ( B ) / / H ( ∂N ( S )) / / H ( N ( S )) ⊕ H ( B ′ ) / / H ( B ) / / / / Z ⊕ Z / / Z ⊕ H ( B ′ ) / / Z / / H ( B ′ ) = Z ⊕ Z . Let x ∈ l and x ∈ l be two points, thenfrom the hypothesis of the lemma, x and x lie in two different componentsof B \ S . Since there are only two components of B \ S , we know that C B,S ( x , x ) = − , and hence there exists a curve γ connecting x with x and having an oddnumber of intersections with S .We can pick another 3-ball B ′′ and glue it to B along their commonspherical boundary to get an S . Inside this S , l h = l ∪ l is still a Hopflink. Inside B ′′ , we can pick disks D , ..., D t so that S = S ∪ D ∪ , ..., D t ⊂ S is a 2-sphere. Since γ ⊂ B , we know that γ ∩ S = γ ∩ S and conclude that C S ,S ( x , x ) = − , so the two components of the Hopf link l h lie in two different componentsof S \ S , which is absurd since it is well-known that the Hopf link is non-separating.Now we are ready to prove the key lemma. proof of lemma 2.5. Suppose k is a Whitehead double of j , V is the tubularneighborhood of j containing k and S is a connected, properly embeddedplanar surface in V , representing a generator of H ( V, ∂V ). Assume that S and k have less than 2 intersections, then since [ k ] = 0 ∈ H ( V ), we knowthat S and k must be disjoint.It is straightforward to see that there is a meridian disk D of V , suchthat in B = V \ int( N ( D )), D intersects S transversely, and that after6 THE PROOF OF THE KEY LEMMA ❅❅❅
D k lVV
Figure 2: After cutting and re-guling we will get a Hopf link l .adding two small arcs to ( k − int( N ( D ))) near ∂N ( D ), we will get a Hopflink l out of k . Here N ( D ) is a neighborhood of D in V . See figure 2.If D ∩ S = ∅ , then we can apply lemma 3.3 directly to conclude acontradiction. Note the two components of the Hopf link l are in twodifferent components of B \ S . This argument for this is exactly the sameas the following more general case (when D ∩ S = ∅ ): in figure 3, we have C B,S ( u, v ) = − C B,S ( x, u ) = C B,S ( y, v ) = 0 asthere is no need for doing any surgeries.If D ∩ S = ∅ , we can assume that D ∩ S = β ∪ β ... ∪ β m , where β , ..., β n are intersection circles of the two surfaces and are in theorder such that β j bounds a disk D j ⊂ D disjoint from any β i for i < j .If all D i are disjoint from k , then we can do a series surgeries on S withrespect to D n , D n − , ..., D one by one to get a surface S ′′ so that S ′′ isdisjoint from D and k . We can pick some connected component of S ′′ , andit will also have such properties that we can apply the argument above toget a contradiction.Now we are in the most complicated case where some D i intersects k .Suppose j is the greatest index such that D j ∩ k = ∅ . We first claimthat D j cannot have a unique intersection point with k . Suppose thecontrary, then a sequence of surgeries on S with respect to D n , ..., D j +1 THE PROOF OF THE KEY LEMMA would generate a surface S ′ such that S ′ represent a generator of H ( V, ∂V ), S ′ is disjoint from k and D j ∩ S ′ = ∂D j = β j . Suppose S ′ is the component of S ′ containing β j , then S ′ is still a planarsurface. A surgery on S ′ with respect to D j would result in two surfaces S ′ and S ′ , each of which contains one copy of D j and hence has a uniqueintersection point with k . This is impossible since [ k ] = 0 ∈ H ( V ).Hence we conclude that D j has two intersection points with k , whichare all the intersection points of D with k . Now, as before, we can doa sequence of surgeries on S with respect to D n , ..., D j +1 , D j to get S ′ .When doing last surgery with respect to D j , we shall modify k at the sametime: cut k along D and glue two small arcs to the newly born boundarypoints near D to get a Hopf link l disjoint from S ′ and D . See figure 3.Now we can apply the remaining sequence of surgeries on S ′ , with respect to D j − , ..., D , to get a surface S ′′ which represents a generator of H ( V, ∂V )and is disjoint from D and l . D PPPPPPPPPPP D j − D j DD j − kS SS S lS ′ S ′ S ′ S ′ k lV V Figure 3: Surgery on S along D n , ..., D j − .Now we can cut off a neighborhood N ( D ) of D from V , which is disjointfrom l and S ′′ , to get a 3-ball B = V \ int( N ( D )). In order to see that thetwo components of l are in different components of B \ S ′′ , we can pickpoints x, y on two arcs of l which do not belong to k, and pick two points u, v near the boundary ∂N ( D ) but on different sides of D . See figure 4.8 THE PROOF OF THE KEY LEMMA llS ′′ S ′′ S ′′ S ′′ xuvy B = V \ int( N ( D ))Figure 4: Surgery on S along D n , ..., D j − .There is a symmetry between the two pairs ( x, u ) and ( y, v ) with respectto D so we have C B,S ′′ ( x, u ) = C B,S ′′ ( v, y ) , where C B,S ′′ is defined as in definition 3.1. Since S ′′ represents a generatorof H ( V, ∂V ), we have C B,S ′′ ( u, v ) = − . Using equality (1), we have C B,S ′′ ( x, y ) = C B,S ′′ ( x, u ) · C B,S ′′ ( u, v ) · C B,S ′′ ( v, y ) = − . Thus the two components of l are in different components of B \ S ′′ . Pick aconnected component S ′′ which separates the two components of l , we canapply lemma 3.3 to get a contradiction.9 EFERENCES REFERENCES
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