Topologically slice (1,1) -knots which are not smoothly slice
TTopologically slice (1 , -knots which are not smoothly slice Zipei NieJanuary 24, 2019
Abstract
We prove that there are infinitely many (1 , -knots which are topologically slice, but notsmoothly slice, which was a conjecture proposed by Béla András Rácz. As defined in [1], a knot K is called a ( g, b ) -knot in S , if there is a Heegaard splitting S = U ∪ V ofgenus g , such that each of K ∩ U and K ∩ V consists of b trivial arcs. By definition, the (0 , b ) -knotsare the b -bridge knots. So the (1 , -knots can be seen as -bridge knots on the standard torus.All -bridge knots and torus knots are (1 , -knots.The smooth (resp., topological) slice genus of a knot K is the minimal genus of a connected,orientable -manifold smoothly (resp., locally flatly) embedded in the -ball D whose boundary isthe knot K . A knot K is called smoothly (resp.,topologically) slice, if its smooth (resp., topological)slice genus is zero. While it is known that [5] there exist infinitely many topologically slice knotswhich are not smoothly slice, we prove that it is still true if we restrict the knots to be (1 , -knots,which was a conjecture proposed [7] by Rácz. In other words, our main theorem is the following. Theorem.
There are infinitely many (1 , -knots which are topologically slice, but not smoothlyslice. To prove the main theorem, we will construct a one-parameter family of (1 , -knots K n ( n =0 , , . . . ) as a valid example.For a (1 , -knot K , the intersection of K and the standard torus consists two basepoints w and z . The information we need to determines the knot is the trivial arcs connecting w and z inside and outside the standard torus. However, knowing how to embed the two trivial arcs on thestandard torus is more than enough. The meridian disks of the -handles inside and outside thestandard torus which does not intersect K determines the knot K . The boundary of the meridiandisks are called the α curve and the β curve. A torus with a pair of basepoints w and z and a pairof curves α and β on it, is called a (1 , -diagram, if α and β are embedded closed curves on thecomplement of { w, z } and have the algebraic intersection number ± . We can use four parametersto specify a (1 , -diagram or a (1 , -knot, which is called the Rasmussen’s notation K ( p, q, r, s ) ,as in [8].In Section 2, we construct the family of knots and prove they are indeed (1 , -knots by derivingthe (1 , -diagrams for them.Given a (1 , -knot K = K ( p, q, r, s ) , we can find [4] the boundary operator of the chain complex CF K ∞ ( S , K ) . Moreover, the invariant τ ( K ) defined in [6] can be found because it only relies on CF K ∞ ( S , K ) . This invariant is closely related to the smooth slice genus g ( K ) by the inequality | τ ( K ) | ≤ g ( K ) , as demonstrated in [6].In Section 3, we compute the knot Floer homology of each K n in the above way. From that weprove these knots have trivial Conway polynomials, and therefore are topologically slice.In Section 4, we first compute the τ invariant of K in the demonstrated way. By certaininequalities for τ invariants in [6], we prove the τ invariant and the smooth slice genus of each K n are , and therefore these knots are not smoothly slice.The pictures of knots are generated by the software KnotPlot [9].The author would like to thank Zoltán Szabó for his help and support.1 a r X i v : . [ m a t h . G T ] J a n The (1 , -knots K n For each non-negative integer n , let K n be the knot with the following planar projection.Figure 1: Knot K as shown. The knot K n has n full left hand twists inside the white circle. Theorem 1. K n is the (1 , -knot K (64 n +31 , n +12 , n +6 , n +18) in Rasmussen’s notation.Proof. A (1 , -diagram of K n is essentially a doubly-pointed Heegaard diagram (Σ , α, β, w, z ) ,where (Σ , α, β ) is a genus one Heegaard splitting of S and w, z are base points on the torus Σ .The following figure gives the torus Σ as the blue torus and the base points w, z as labeled. The α circle is immediately known. The only thing left to compute is the β circle. This can be foundvia a sequence of isotopies of the space.Figure 2: First Step.Then, we move z along a longitude of the blue torus clockwise and move z around w clockwise.Now we get the following figure. 2igure 3: Second Step.Then, we move z along a longitude of the blue torus counterclockwise, and then move z alonga meridian of the blue torus into the paper. Now we get the following figure.Figure 4: Third Step.Now we move w around z clockwise n times, and we get the following figure.Figure 5: Fourth Step.Then, we move w along a longitude of the blue torus clockwise and and then move w along ameridian of the blue torus out of the paper. Then we get the following figure. Here an arc with ared number s on it represents a family of s parallel curves.3igure 6: Fifth Step.Before we proceed to find the β curve, we simplify the α curve without moving the basepoints w and z and get the following diagram.Figure 7: Sixth Step.Finally, it is straightforward to get the triviality of the outside arc, and a β curve can beconstructed as follows (the blue curve).Figure 8: Seventh Step.4y straightening the β curve, we obtain the following (1 , -diagram.Figure 9: Eighth Step.Equivalently, we have the following (1 , -diagram. In Rasmussen’s notation, K n is the (1 , -knot K (64 n + 31 , n + 12 , n + 6 , n + 18) .Figure 10: Ninth Step. K n Theorem 2.
The Poincare polynomial of the knot Floer homology of K n isHF K n ( q, t ) = (cid:88) m,a ∈ Z q m t a rank Z ( (cid:100) HFK m ( S , K n , a ))= − q − + (2 n + 1) q − t − (1 + q )(1 + qt ) . Proof.
It is clear that the total rank of the knot Floer homology of K n is n + 31 , where thecrossing points x i ( i = 1 , . . . , n + 31 ) between the α curve and the β curve represent a basis ofthe knot Floer homology. To compute the knot Floer homology, we need to find the Alexandergrading a i and the Maslov grading m i of each crossing point x i ( i = 1 , . . . , n + 31 ).From each Whitney disk of unit Maslov index, we derive a equation for the Alexander andMaslov grading of the vertices of the disk. With sufficient number of such disks, we can obtain therelative Alexander and Maslov gradings. 5irst, there is a disk from x i to x n +25 − i for each i = 1 , . . . , n + 12 , with a single basepoint z inside. And there is a disk from x i to x n +39 − i for each i = 16 n + 8 , . . . , n + 19 , with a singlebasepoint w inside.Then, we concentrate on the disk from x n +12 to x n +13 , and try to extend it into largerdisks while keeping the basepoint on it unchanged. Sequentially, we obtain disks from x n − i )+2 to x n − i )+1 and disks from x n + i )+9 to x n + i )+10 for i = 0 , . . . , n − . Then we obtain a diskfrom x to x . After that, we sequentially obtain disks from x n − i )+5 to x n − i )+4 and disksfrom x i +6 to x i +7 for i = 0 , . . . , n . All the disks obtained here are Whitney disk of unit Maslovindex with a single basepoint z inside.Finally, we keep extending to obtain the disk from x n +31 to x n +25 . To argue that the lastone we got is an embedded disk in the universal covering space, we notice that, in every previousstep, the segment where the endpoints lie is moving diagonally in the universal covering space, butin the last step, it is moving vertically, so the disk has no self intersections. Therefore the last diskwe obtained is a Whitney disk of unit Maslov index with two z ’s inside.Similarly, we extend the disk from x n +20 to x n +19 in the same way. Then we sequentiallyobtain disks from x n + i )+5 to x n + i )+6 and disks from x n − i ) − to x n − i ) − for i = 0 , . . . , n − .And then we obtain a disk from x n +5 to x n +6 . Then we sequentially obtain disks from x i +2 to x i +3 and disks from x n − i )+1 to x n − i ) for i = 0 , . . . , n − . And then we obtain a diskfrom x n +2 to x n +3 . All the disks obtained here are Whitney disk of unit Maslov index with asingle basepoint w inside. And similarly, we obtain a Whitney disk of unit Maslov index with two w ’s inside from x to x n +7 in the final step.Now we extend the disk from x n +10 to x n +15 in the same way. Then we sequentially obtaindisks from x n − i )+4 to x n − i ) − and disks from x n + i )+7 to x n + i )+12 for i = 0 , . . . , n − . Allthe disks obtained here are Whitney disk of unit Maslov index with a single basepoint z inside.Similarly, we extend the disk from x n +22 to x n +17 in the same way. Then we sequentiallyobtain disks from x n + i )+3 to x n + i )+8 and disks from x n − i ) to x n − i ) − for i = 0 , . . . , n − .Then we obtain a disk from x n +3 to x n +8 . All the disks obtained here are Whitney disk ofunit Maslov index with a single basepoint w inside.Now we have found sufficient Whitney disks to get the relative Alexander and Maslov gradingsfor the first n + 8 crossing points. By symmetry, we also get the relative Alexander and Maslovgradings for the last n + 8 crossings. Via the Whitney disks from x i to x n +39 − i ( i = 16 n +8 , . . . n + 19 ), because we have a common element x n +8 in the sets of crossing points, we getthe relative Alexander and Maslov gradings for the first n + 15 crossing points. By using thesymmetry and the Whitney disks for a second time, we get the relative Alexander and Maslovgradings for all the crossing points.The following is a solution to the relative Alexander gradings. ( a i +1 , a i +2 , a i +3 , a i +4 , a i +5 , a i +6 , a i +7 , a i +8 ) = ( − , , , , , , − , for i = 0 , . . . , n − . a n +1 = − . ( a n +8 i +2 , a n +8 i +3 , a n +8 i +4 , a n +8 i +5 , a n +8 i +6 , a n +8 i +7 , a n +8 i +8 , a n +8 i +9 )=(0 , , , , , , , for i = 0 , . . . , n . ( a n +8 i +10 , a n +8 i +11 , a n +8 i +12 , a n +8 i +13 , a n +8 i +14 , a n +8 i +15 , a n +8 i +16 , a n +8 i +17 )=(0 , , , − , , − , , for i = 0 , . . . , n − . ( a n +10 , a n +11 , a n +12 , a n +13 , a n +14 , a n +15 , a n +16 ,a n +17 , a n +18 , a n +19 , a n +20 , a n +21 , a n +22 )=(0 , , , − , , − , , , , , , − , . ( a n +8 i +23 , a n +8 i +24 , a n +8 i +25 , a n +8 i +26 , a n +8 i +27 , a n +8 i +28 , a n +8 i +29 , a n +8 i +30 )=( − , , , , , , − , i = 0 , . . . , n − . ( a n +8 i +23 , a n +8 i +24 , a n +8 i +25 , a n +8 i +26 , a n +8 i +27 , a n +8 i +28 , a n +8 i +29 , a n +8 i +30 )=( − , − , − , − , − , , − , for i = 0 , . . . , n . a n +31 = 1 . ( a n +8 i +32 , a n +8 i +33 , a n +8 i +34 , a n +8 i +35 , a n +8 i +36 , a n +8 i +37 , a n +8 i +38 , a n +8 i +39 )=(0 , , , − , , − , , for i = 0 , . . . , n − .By symmetry, the above is also a solution to the absolute Alexander gradings.The following is a solution to the relative Maslov gradings. ( m i +1 , m i +2 , m i +3 , m i +4 , m i +5 , m i +6 , m i +7 , m i +8 ) = ( − , − , , , , , − , − for i = 0 , . . . , n − . m n +1 = − . ( m n +8 i +2 , m n +8 i +3 , m n +8 i +4 , m n +8 i +5 , m n +8 i +6 , m n +8 i +7 , m n +8 i +8 , m n +8 i +9 )=( − , , , , , , , for i = 0 , . . . , n . ( m n +8 i +10 , m n +8 i +11 , m n +8 i +12 , m n +8 i +13 , m n +8 i +14 , m n +8 i +15 , m n +8 i +16 , m n +8 i +17 )=( − , , , , , , , for i = 0 , . . . , n − . ( m n +8 i +10 , m n +8 i +11 , m n +8 i +12 , m n +8 i +13 , m n +8 i +14 , m n +8 i +15 , m n +8 i +16 , m n +8 i +17 )=( − , , , − , , − , − , for i = 0 , . . . , n − . ( m n +10 , m n +11 , m n +12 , m n +13 , m n +14 , m n +15 , m n +16 ,m n +17 , m n +18 , m n +19 , m n +20 , m n +21 , m n +22 )=( − , , , − , , − , − , , , , , − , − . ( m n +8 i +23 , m n +8 i +24 , m n +8 i +25 , m n +8 i +26 , m n +8 i +27 , m n +8 i +28 , m n +8 i +29 , m n +8 i +30 )=( − , − , , , , , − , − for i = 0 , . . . , n − . ( m n +8 i +23 , m n +8 i +24 , m n +8 i +25 , m n +8 i +26 , m n +8 i +27 , m n +8 i +28 , m n +8 i +29 , m n +8 i +30 )=( − , − , − , − , − , , − , − for i = 0 , . . . , n − . ( m n +8 i +23 , m n +8 i +24 , m n +8 i +25 , m n +8 i +26 , m n +8 i +27 , m n +8 i +28 , m n +8 i +29 , m n +8 i +30 )=( − , − , − , − , − , , − , − for i = 0 , . . . , n . a n +31 = 0 . ( m n +8 i +32 , m n +8 i +33 , m n +8 i +34 , m n +8 i +35 , m n +8 i +36 , m n +8 i +37 , m n +8 i +38 , m n +8 i +39 )=( − , , , − , , − , − , i = 0 , . . . , n − .To find the absolute Maslov gradings, we take a step back to look at Figure 9. If we removebasepoint z , most of the crossing points can be easily reduced. In fact, only the last three crossingpoints might survive. By analyzing the twisting part of the α curve, we find that the crossingpoints x n +29 and x n +30 can be reduced from above. Therefore, the only survived crossingpoint x n +31 has absolute Maslov grading zero. So the relative Maslov gradings we got is also theabsolute Maslov gradings.The Poincare polynomial is derived from counting the number of crossing points with givenAlexander and Maslov grading. Corollary 1. K n and K m are non-isomorphic if m (cid:54) = n .Proof. This is because they have different knot Floer homology.
Corollary 2.
The Conway polynomial of K n is .Proof. The Conway polynomial is the graded Euler characteristic of the knot Floer homology, sowe have ∆ K n ( t ) = HF K n ( − , t ) = 1 . Corollary 3. K n is topologically slice.Proof. Any knot with trivial Conway polynomial is topologically slice, as proved in [2, 3]. τ invariant and the smooth slice genus of K n Lemma 1.
The invariant τ ( K ) is .Proof. To calculate the invariant τ ( K ) , we first compute the chain complex CFK ∞ ( S , K ) . Bydefinition, the generators of CFK ∞ ( S , K ) is given by [ x i , j, j + a i ] ( i = 1 , . . . , n + 31 , j ∈ Z ),and the boundary operator on CFK ∞ ( S , K ) is given by ∂ [ x , i + 1 , i ] = [ x , i − , i ] − [ x , i, i ] + [ x , i, i − − [ x , i + 1 , i − ,∂ [ x , i + 1 , i + 1] = [ x , i + 1 , i ] − [ x , i, i + 1] − [ x , i, i + 1] − [ x , i + 1 , i ] ,∂ [ x , i, i + 1] = [ x , i − , i ] − [ x , i − , i + 1] + [ x , i, i − − [ x , i, i ] ,∂ [ x , i, i ] = − [ x , i − , i ] − [ x , i, i − ,∂ [ x , i, i + 1] = [ x , i, i ] − [ x , i − , i + 1] − [ x , i, i ] ,∂ [ x , i − , i + 1] = − [ x , i − , i ] − [ x , i − , i ] ,∂ [ x , i − , i ] = − [ x , i − , i − ,∂ [ x , i − , i + 1] = − [ x , i − , i ] ,∂ [ x , i, i + 1] = [ x , i − , i + 1] − [ x , i, i ] − [ x , i, i ] ,∂ [ x , i, i ] = − [ x , i, i − ,∂ [ x , i − , i ] = − [ x , i − , i − ,∂ [ x , i, i ] = [ x , i − , i ] − [ x , i, i − ,∂ [ x , i, i −
1] = − [ x , i − , i − ,∂ [ x , i − , i −
1] = 0 ,∂ [ x , i, i −
1] = 0 ,∂ [ x , i, i ] = [ x , i, i − − [ x , i − , i ] ,∂ [ x , i − , i ] = 0 ,∂ [ x , i − , i −
1] = 0 ,∂ [ x , i − , i ] = [ x , i − , i − ,∂ [ x , i, i ] = [ x , i − , i ] − [ x , i, i − ,∂ [ x , i, i −
1] = [ x , i − , i − ,∂ [ x , i, i ] = [ x , i − , i ] , [ x , i + 1 , i ] = [ x , i, i ] + [ x , i, i ] − [ x , i + 1 , i − ,∂ [ x , i + 1 , i −
1] = [ x , i, i − ,∂ [ x , i, i −
1] = [ x , i − , i − ,∂ [ x , i + 1 , i −
1] = [ x , i, i −
1] + [ x , i, i − ,∂ [ x , i + 1 , i ] = [ x , i, i ] + [ x , i + 1 , i − − [ x , i, i ] ,∂ [ x , i, i ] = [ x , i − , i ] + [ x , i, i − ,∂ [ x , i + 1 , i ] = [ x , i, i ] − [ x , i − , i ] + [ x , i + 1 , i − − [ x , i, i − ,∂ [ x , i + 1 , i + 1] = [ x , i, i + 1] + [ x , i + 1 , i ] + [ x , i + 1 , i ] − [ x , i, i + 1] ,∂ [ x , i, i + 1] = [ x , i − , i + 1] − [ x , i − , i ] + [ x , i, i ] − [ x , i, i − , by counting all Whitney disks of unit Maslov index, where the signs are by an orientation of the β curve.By definition, the chain complex (cid:99) CF ( S ) is generated by [ x i , , a i ] ( i = 1 , . . . , n + 31 ), andthe boundary operator on (cid:99) CF ( S ) is given by ∂ [ x , , −
1] = − [ x , , − ,∂ [ x , ,
0] = [ x , , − − [ x , , − ,∂ [ x , ,
1] = [ x , , − − [ x , , ,∂ [ x , ,
0] = − [ x , , − ,∂ [ x , ,
1] = [ x , , − [ x , , ,∂ [ x , ,
2] = − [ x , , − [ x , , ,∂ [ x , ,
1] = − [ x , , ,∂ [ x , ,
2] = − [ x , , ,∂ [ x , ,
1] = − [ x , , − [ x , , ,∂ [ x , ,
0] = − [ x , , − ,∂ [ x , ,
1] = − [ x , , ,∂ [ x , ,
0] = − [ x , , − ,∂ [ x , , −
1] = 0 ,∂ [ x , ,
0] = 0 ,∂ [ x , , −
1] = 0 ,∂ [ x , ,
0] = [ x , , − ,∂ [ x , ,
1] = 0 ,∂ [ x , ,
0] = 0 ,∂ [ x , ,
1] = [ x , , ,∂ [ x , ,
0] = − [ x , , − ,∂ [ x , , −
1] = 0 ,∂ [ x , ,
0] = 0 ,∂ [ x , , −
1] = − [ x , , − ,∂ [ x , , −
2] = 0 ,∂ [ x , , −
1] = 0 ,∂ [ x , , −
2] = 0 ,∂ [ x , , −
1] = [ x , , − ,∂ [ x , ,
0] = [ x , , − ,∂ [ x , , −
1] = [ x , , − ,∂ [ x , ,
0] = [ x , , −
1] + [ x , , − ,∂ [ x , ,
1] = [ x , , − [ x , , − . The homology of is generated by the cycle [ x , ,
1] + [ x , ,
0] + [ x , ,
0] + [ x , , . Since [ x , ,
1] + [ x , ,
0] + [ x , ,
0] + [ x , , ∈ F ( K , , we have τ ( K ) ≤ .9onsider the projection p : (cid:99) CF ( S ) → Z that sends all generators to zero except for p ([ x , , . Then p is a chain map which induces isomorphism on homology. Since p maps all elements in F ( K , to zero, we have τ ( K ) ≥ .Therefore the invariant of τ ( K ) is . Theorem 3.
The τ invariant and the smooth slice genus of K n are .Proof. After changing n of the n positive crossings in white circle in Figure 1 to negative crossings,the new knot becomes K , therefore by [6] we have g ( K n ) ≥ τ ( K n ) ≥ τ ( K ) = 1 .By resolving one of the n positive crossings in white circle and the crossing a in Figure 1, weget a new knot as follows.Figure 11: The knot obtained by resolving two crossings in K n .By straightening the blue part and the green part in Figure 11, we get the following isotopicknot. Figure 12: The knot obtained by resolving two crossings in K n .By straightening the green part in Figure 12, it is now clear that our new knot is the unknot.Hence, by resolving two crossings, we can construct a torus cobordism (a split cobordismfollowed by a merge cobordism) between K n and the unknot. By definition, we have g ( K n ) ≤ .Therefore, we have τ ( K n ) = g ( K n ) = 1 . 10 heorem 4. There are infinitely many (1 , -knots which are topologically slice, but not smoothlyslice.Proof. The family of knots K n serves as an example. References [1] Helmut Doll. "A generalized bridge number for links in 3-manifolds." Mathematische Annalen294.1 (1992): 701-717.[2] Michael H. Freedman, and Frank Quinn. Topology of 4-manifolds. Princeton Univ. Press, 1990.[3] Stavros Garoufalidis, and Peter Teichner. "On knots with trivial Alexander polynomial." Jour-nal of Differential Geometry 67.1 (2004): 167-193.[4] Hiroshi Goda, Hiroshi Matsuda, and Takayuki Morifuji. "Knot Floer homology of (1, 1)-knots."Geometriae Dedicata 112.1 (2005): 197-214.[5] Robert E Gompf. "Smooth concordance of topologically slice knots." Topology 25.3 (1986):353-373.[6] Peter Ozsváth, and Zoltán Szabó. "Knot Floer homology and the four-ball genus." Geometry& Topology 7.2 (2003): 615-639.[7] Béla András Rácz. "Geometry of (1, 1)-Knots and Knot Floer Homology." (2015).[8] Jacob Rasmussen. "Knot polynomials and knot homologies." Geometry and topology of mani-folds 47 (2005): 261-280.[9] Robert Scharein, Knotplot, https://knotplot.com/https://knotplot.com/