An explicit description of (1,1) L-space knots, and non-left-orderable surgeries
AAn explicit description of (1 ,
1) L-space knots, andnon-left-orderable surgeries
Zipei NieFebruary 23, 2021
Abstract
Greene, Lewallen and Vafaee characterized (1 ,
1) L-space knots in S and lens space inthe notation of coherent reduced (1 , , S has non-left-orderable fundamental group. An L-space is a rational homology 3-sphere with minimal Heegaard Floer homology,that is, dim (cid:100) HF = | H ( Y ) | . A nice topological property of L-spaces is that [18] they donot admit co-orientable taut foliations, and its converse statement is only partially verified.Another conjectural property of L-spaces is the non-left-orderability of fundamental groups[2], that is, there does not exist a total order ≤ on the fundamental group such that g ≤ h implies f g ≤ f h . Although we have multiple computational tools, the Heegaard Floer datais not easy to utilize. Therefore, a better characterization of L-spaces would be helpful.One way to construct L-spaces is via Dehn surgeries. A knot K is called an L-spaceknot, if it admits an L-space surgery. It is a positive (resp. negative) L-space knot if itadmits a positive (resp. negative) L-space surgery. The Dehn surgery along a nontrivialpositive L-space knot K in S with slope pq yields an L-space if and only if pq ≥ g ( K ) − Y , we say that a knot K in Y is a ( g, b )-knot, if thereexists a Heegaard splitting Y = U ∪ U of genus g , such that each of K ∩ U and K ∩ U consists of b trivial arcs. The family of (1 , , , , K in the three-sphere or lens space Y a doubly-pointedHeegaard diagram (Σ , α, β, w, z ), which consists of two simple closed curves α and β onthe torus Σ and two basepoints w and z in Σ − α − β . The diagram (Σ , α, β, w, z ) is calledreduced if each bigon contains a basepoint. In this case, the diagram can be specified[20] by four parameters p, q, r, s . Via successive isotopies to removing empty bigons, every(1 , , , a r X i v : . [ m a t h . G T ] F e b revious Work. [9, Theorem 1.2] A reduced (1 , -diagram represents an L-space knot ifand only if it is coherent, that is, there exist orientations on α and β that induce coherentorientations on the boundary of every embedded bigon ( D, ∂D ) ⊆ (Σ , α ∪ β ) . It representsa positive or negative L-space knot according to the sign of α · β with coherent orientation. Building on their work, we describe the family of (1 ,
1) L-space knots explicitly asfollows.
Theorem 1.
Let Y = U ∪ Σ U be a genus one Heegaard splitting of a three-sphere orlens space, with standard geometry. A knot in Y is a (1 , L-space knot if and only if it isisotopic to a union of three arcs ρ ∪ τ ∪ τ , such that(a) ρ is a geodesic of Σ ;(b) τ is properly embedded in some meridional disk of U ;(c) τ is properly embedded in some meridional disk of U . Note that, if τ or τ is of length zero, then by definition, the knot is a 1-bridge braidin Y . The study of 1-bridge braids originates from the classification of knots in a solidtorus with nontrivial solid torus surgeries [1, 6, 7], where it is shown that every such knotis a torus knot or a 1-bridge braid. If we put solid torus in the standard position in S ,these knots has nontrivial lens space surgeries. And as its name suggests, any lens space isan L-space. It is also proved that any 1-bridge braid in the three-sphere or lens space isan L-space knot [9], and the L-spaces obtained by surgeries along 1-bridge braids in S has non-left-orderable fundamental groups [17]. In line with these researches, we deducesimilar properties of (1 ,
1) L-space knots in S . Theorem 2.
A nontrivial positive (1 , L-space knot in S can be represented as theclosure of the braid ( σ ω σ ω − · · · σ ω − b +1 ) ( σ ω σ ω − · · · σ ) b ( σ ω − σ ω − · · · σ ) t − b in the braid group B ω +1 on ω + 1 strands, where ≤ b ≤ ω and ≤ b ≤ t . An example is shown in Figure 1 below.
PQ Q (cid:48)(cid:48) R Figure 1: The braid when ( ω, t, b , b ) = (6 , , , In [17], the author introduced the property (D) as follows.
Definition 3.
For a nontrivial knot K in S with µ and λ representing a meridian and alongitude in the knot group, we say K has property (D) if1. for any homomorphism ρ from π ( S − K ) to Homeo + ( R ), if s ∈ R is a commonfixed point of ρ ( µ ) and ρ ( λ ), then s is a fixed point of every element in π ( S − K );2. µ is in the root-closed, conjugacy-closed submonoid generated by µ g ( K ) − λ and µ − . he author proved that [17, Theorem 1.3] nontrivial knots which are closures of positive1-bridge braids have property (D). And by [17, Theorem 4.1], it implies the non-left-orderability of the fundamental groups of the L-spaces obtained by Dehn surgeries onclosures of 1-bridge braids. In this paper, we prove the following result in a similar way.Thanks to the additional symmetry, our proof is simplified compared to the proof of [17,Theorem 1.3]. Theorem 4.
Nontrivial positive (1 , L-space knots in S have property (D). Therefore, by [17, Theorem 4.1], we have the following conclusion.
Theorem 5.
The fundamental group of an L-space obtained by Dehn surgery on a (1 , -knot in S is not left orderable. Because a (1 , , Acknowledgement
The author thanks Fan Ye for helpful discussions. (1 , L-space knots
This section is dedicated to prove Theorem 1.Let (Σ , α, β, w, z ) denote a reduced (1 , Y . Since the α curve is simple, it represents a primitive elementin H (Σ). We can straighten out the α curve via a self-homeomorphism of Σ. Assumethat α curve is horizontal, with an orientation from left to right. The β curve is cut intostrands of two bands and two rainbows by the α curve. By [9, Theorem 1.2], we can choosean orientation of β which induces an orientation from left to right on every rainbow strand,which is the opposite of the coherent orientation. Assume that w is on the left side of α ,and z is on the right side of α . We define the positive curves on the torus Σ as follows.
Definition 6.
An oriented curve γ on Σ is called positive, if at each inner intersectionpoint of γ and α ∪ β , the γ curve goes from the right side of the α or β curve to the leftside transversally.Our first step is to construct a positive curve γ connecting w to z .Let S be the set of endpoints of all positive curves originating from w . If z ∈ S , our firststep is completed. Otherwise, we assume z (cid:54)∈ S . Let S w be the connected component of S − α containing w . Since w ∈ S w , each point on a rainbow strand around w is an interiorpoint of S w . Since z (cid:54)∈ S , each point on a rainbow strand around z is in the exterior of S w . Therefore, the shape of S w is a rectangle. Let the vertices of S w be P , P , P , P counterclockwise, with P P , P P being parts of the α curve, P P , P P being parts of he β curve, and the basepoint w being close to the edge P P . From the definition of S w ,we can derive the orientation of P P and P P as shown in Figure 2. wP P P P Figure 2: The rectangle S w . If there exists another embedded open rectangle P P P P on the left of P P whichdoes not contain any basepoints, then we replace P P P P by the immersed rectangle P P P P and try the same extension again. Because the β curve is connected, the edge P P extends to the right strand of the basepoint z in finite steps. Hence, we assume thatthe sequence of extensions ends at an immersed rectangle P P Q Q , as shown in Figure 3. wP P Q Q P P Figure 3: The immersed rectangle P P Q Q . There are two possibilities for not able to extend the immersed rectangle: one of theedge P Q and P Q extends to a rainbow strand, or the embedded rectangle on the leftof Q Q contains at least one basepoint. If the embedded rectangle on the left of Q Q contains the basepoint z . Then by the definition of S , we have z ∈ S . Otherwise, the edge Q Q intersects with the edge P P on the torus Σ.By the definition of S w , the strands P P and P P are on the boundary of S , so wehave Q Q ⊆ P P . If P = Q or P = Q , then the edge Q P or the edge P Q coversthe β curve. In that case, Q P or P Q contains the left and right strand of the basepoint z , which is a contradiction. Therefore, the edge Q Q lies in the interior of the edge P P .Suppose that there are q rainbow strands in the middle, r ≥ Q P ) and r ≥ P Q ) in the immersedrectangle P P Q Q . Suppose that the i -th intersection point on Q Q is the ( i + k )-thintersection point on P P for 1 ≤ i ≤ r + r . Then we have 1 ≤ k ≤ q − ≤ i ≤ q + r + r , let ε i = 1 if the β curve goes from the right side of the α curve to the left side at the i -th intersection point on P P . Otherwise, let ε i = −
1. Then e have ε i = ε i + k , if 1 ≤ i ≤ r ;1 , if r + 1 ≤ i ≤ r + q ; − r + q + 1 ≤ i ≤ r + 2 q ; ε i − q + k , if 2 q + r + 1 ≤ i ≤ q + r + r . If 1 ≤ k ≤ q , then we have ε = 1 by induction. If q + 1 ≤ k ≤ q −
1, then we have ε q + r + r = − We have constructed a positive curve γ connecting w to z . By eliminating self-loops,we assume that γ is simple and intersects each connected component of Σ − α − β at mostonce. Our second step is to construct a positive simple closed curve γ passing through w and z .Let T , T , . . . , T p be all intersection points between the α curve and the β curve,ordered along the orientation of α . Via a self-homeomorphism of Σ, we assume thefollowing condition: for 1 ≤ i ≤ p , if the α -segment T i T i +1 does not intersect with the γ curve, then it has unit length; otherwise, it has length 2 q + 1, where q is the number ofstrands in each rainbow.For a downward-oriented band strand e and an upward-oriented band strand e onthe β curve, there exists an embedded open rectangle R with two edges being e and e and the other two edges e , e on the α curve. We can further assume that the rectangle ison the left of e , e , e and on the right of e .Let l i denote the length of e i for i = 3 ,
4, then l i = | e i ∩ β | + 2 q | e i ∩ γ | − . The difference | e ∩ β | − | e ∩ β | depends on whether each basepoint lies in R , that is, | e ∩ β | − | e ∩ β | = 2 q |{ z } ∩ R | − q |{ w } ∩ R | . The difference | e ∩ γ | − | e ∩ γ | depends on how γ intersects R . Since γ is positive,if it intersects e , e or e at a point, then it enters R there; if it intersects e at a point,then it exits R there. Hence we have | e ∩ γ | − | e ∩ γ | = | e ∩ γ | + | e ∩ γ | + |{ w } ∩ R | − |{ z } ∩ R | . By combining these equations, we get l ≤ l .Therefore, there exists a linear foliation F of the torus Σ, such that, up to isotopy, eachstrand of the β curve is contained in a leaf of F or transverse to F in a fixed direction.Via another isotopy, we can assume that the entire β curve is either contained in a leaf of F or transverse to F . In either case, we can assume that the slope of F is irrational undera perturbation of the foliation, so the leaves of F are dense. We extend the curve in bothdirections from the basepoint w along a leaf of F until the endpoints reach the connectedcomponent of Σ − α − β containing the basepoint z . After closing the curve by connectingtwo endpoints within the connected component, we get the positive simple closed curve γ passing through w and z . .3 The third and the last steps Our third step is to complete the proof of the “only if” part of Theorem 1. Viaa self-homeomorphism of Σ, we abandon the horizontality of the α curve, and assumethat the γ curve is horizontal instead. Since γ is positive, we can either assume α is a geodesic or assume β is a geodesic, but not simultaneously. In fact, there existsan isotopy f : ( α ∪ β ∪ γ ) × [0 , → Σ, such that f ( z, t ) , f ( γ, t ) are independent of t ,and f ( α, , f ( β, , f ( γ,
0) are geodesics. Let ρ be the curve in Σ × [0 ,
1] defined by ρ ( t ) = ( f ( w, t ) , t ). Let τ (resp. τ ) be a geodesic in (Σ ,
0) (resp. (Σ , f ( α, ,
0) (resp. ( f ( β, , U and U to the boundary components of Σ × [0 , f ( α, ,
0) (resp. ( f ( β, , U (resp. U ), we recover the knot ρ ∪ τ ∪ τ from the (1 , , S and lens space to utilize [9, Theorem 1.2], as shown in Figure 4. We makea tiny change in the construction here: the basepoint z is no longer restricted to be thestarting point of the γ (cid:48) , but can be any point in Σ − α − β . The topological meaning ofthe diagram is as explained in the previous paragraph: if we move the basepoint w along ageodesic γ (cid:48) , we can untwist the β curve at the expense of twisting the α curve. With thischange, a (1 , , α and β curves induce coherent orientations on the boundary of every embedded bigon( D, ∂D ) ⊆ (Σ , α ∪ β ). Each isotopy to remove an empty bigon preserves the coherence, sowe get a reduced (1 , z wα γ (cid:48) β β β Figure 4: The construction of a coherent diagram of the 1-bridge braid K ( − , ,
7) in S ,modified from [9, Figure 3]. In this subsection, we prove Theorem 2 and derive the genus formula.Let K be a nontrivial positive (1 ,
1) L-space knot in S . Let S = U ∪ Σ U be a genusone Heegaard splitting with standard geometry. By Theorem 1, K is isotopic to ρ ∪ τ ∪ τ , here ρ is a geodesic of Σ, and τ (resp. τ ) is properly embedded in some meridional diskof U (resp. U ).An orientation on the geodesic ρ induces orientations on the cores of U and U . Ifthe cores of U and U are negatively linked, then the construction in Subsection 2.3yields a negative coherent reduced (1 , K is a negative(1 ,
1) L-space knot, which contradicts the assumption that K is a nontrivial positive (1 , S . Thus, the cores of U and U with induced orientations are positivelylinked. So ρ can be realized as a part of a positive braid. After appending the arcs τ and τ , we get a positive braid as shown in Figure 1.Let K be the closure of the positive braid represented by( σ ω σ ω − · · · σ ω − b +1 ) ( σ ω σ ω − · · · σ ) b ( σ ω − σ ω − · · · σ ) t − b . If b = b = 0, then K has an unknot component, which is not allowed. If b = 0, we candecrease t and b by one and set b to ω . If b = 0, we can decrease ω and b by one andset b to t . For the representation with minimal t + ω , we have 1 ≤ b ≤ ω and 1 ≤ b ≤ t .Therefore, Theorem 2 holds.A minimal genus Seifert surface is obtained [5] by applying Seifert’s algorithm to apositive diagram, so the genus of K is g ( K ) = 12 ( − b + b ω + ( t − b )( ω − − ( ω + 1) + 1)= 12 ( tω − t − ω + b + b ) . In this subsection, we investigate the knot group π ( S − K ). As a (1 , x , y , x , y in the knot group instead.Let D (resp. D ) be the meridional disk of U (resp. U ) containing τ (resp. τ ).Then D (resp. D ) is divided by τ (resp. τ ) into two disks D x, and D y, (resp. D x, and D y, ). Let the points P, Q, R on Σ be ρ ∩ τ , τ ∩ τ , τ ∩ ρ , respectively. Orientthe knot K so that P, Q, R appears in order. Orient the cores of U , U and the disks D , D , D x, , D y, , D x, , D y, accordingly. Let Q (cid:48) be a point near Q in Σ − ρ − D − D ,so that Q (cid:48) is on the negative side of D and on the positive side of D . Let Q (cid:48)(cid:48) be a pointin ρ on the boundary of the connected component of Σ − ρ − D − D containing Q (cid:48) , asshown in Figure 1.The fundamental group of U − τ (resp. U − τ ) is freely generated by two elements x , y (resp. x , y ), where x (resp. y , x , y ) is represented by a loop based at Q (cid:48) intersecting D x, (resp. D y, , D x, , D y, ) once positively and not intersecting other disks.Then π ( S − K ) based at Q (cid:48) is generated by x , y , x , y .Without loss of generality, we assume x (resp. x ) has larger norm than y (resp. y )in H ( S − K ). Then µ = x y − = y − x represents a meridian of K around Q . he boundary of D (resp. D ) intersects ρ in t (resp. ω ) points, not counting P , Q and R . Starting from Q along positive direction, let the points be R t , R t − , . . . , R (resp. P ω , P ω − , . . . , P ) in order. For each i with 1 ≤ i ≤ t (resp. 1 ≤ i ≤ ω ), let g i (resp. h − i )represent the loop based at Q (cid:48) in U − τ (resp. U − τ ) which first travels to R i (resp. P i ) without intersecting D (resp. D ), then follows ρ in positive (resp. negative) directionto P (resp. R ) but not past it, lastly travels back to Q (cid:48) without intersecting D (resp. D ).Then each g i (resp. h i ) represented by a word in x and y (resp. x and y ), and we have y = (cid:0) g µg − (cid:1) (cid:0) g µg − (cid:1) · · · (cid:0) g t µg − t (cid:1) ,y = (cid:0) h − µh (cid:1) (cid:0) h − µh (cid:1) · · · (cid:0) h − ω µh ω (cid:1) . Since b , b (cid:54) = 0, the point Q (cid:48)(cid:48) is on the arc R P ω ⊂ ρ which is a part of boundary of theconnected component of Σ − ρ − D − D containing Q (cid:48) . The longitude λ of K startingfrom Q is determined by µ k λ = h ω g . The integer k can be found by counting the crossings between K and a loop representedby h ω g on a planar diagram. The loop represented by h ω g differs from the blackboardframing of K as shown in Figure 1 by 2 t additional positive crossings, so we have k = t = b + b ω + ( t − b )( ω −
1) + t = tω + b + b . Therefore we have µ tω + b + b λ = h ω g , and µ g ( K ) − λ = h ω g µ − t − ω − . Furthermore, the word representing g starts with an x , and the word representing h ω ends with an x . In this subsection, we prove that K has property (D).The first part of the property (D) is the following. Lemma 7.
For any homomorphism ρ from π ( S − K ) to Homeo + ( R ) , if s ∈ R is acommon fixed point of ρ ( µ ) and ρ ( λ ) , then s is a fixed point of every element in π ( S − K ) .Proof. Since s is a common fixed point of ρ ( µ ) and ρ ( λ ), it is a common fixed point of ρ ( x y − ), ρ ( y − x ) and ρ ( h ω g ). Without loss of generality, we assume ρ ( x ) s ≥ s , thenwe have ρ ( y ) s ≥ s . We also have ρ ( x ) s ≥ s (resp. ρ ( x ) s ≤ s ) if and only if ρ ( y ) s ≥ s (resp. ρ ( y ) s ≤ s ).Starting from the base point Q (cid:48) , we construct a geodesic γ on Σ − ρ parallel to ρ .Because K is nontrivial, the arc ρ is not parallel to ∂D or ∂D . Extend γ until it crosseseach disk D , D at least once and reaches the connected component of Σ − ρ − D − D containing Q (cid:48) again. Then we close up the curve to obtain the knot group element g ,which can be represented by a nontrivial word in x and y , and also by a nontrivial word n x and y . By the first condition, we have ρ ( g ) s ≥ s . By the second condition, we have ρ ( x ) s ≥ s and ρ ( y ) s ≥ s . Because h ω g is represented by a word in x , y , x , y with atleast one x and one x , we have ρ ( x ) s = ρ ( y ) s = ρ ( x ) s = ρ ( y ) s = s . Therefore s is afixed point of every element in π ( S − K ). Remark.
This lemma is the starting point of our research. Suppose we are given a (1 , x , y and one relation. To prove this lemma, we have to construct another element suchas y in terms of x and y . The algebraic construction is highly nontrivial, and better tobe done topologically as Theorem 1.Then we prove the second part of the property (D). Lemma 8.
The element µ is in the root-closed, conjugacy-closed submonoid generated by µ g ( K ) − λ and µ − .Proof. As in [17, Section 3], we define the preorder ≤ k generated by µ and ( µ g ( K ) − λ ) − on π ( S − K ). Since µ = x y − = y − x , we have x ≥ k y and x ≥ k y . Since µ g ( K ) − λ = h ω g µ − t − ω − , we have h ω g ≤ k µ t + ω +1 .Let ˜ g = 1 , ˜ g , . . . , ˜ g t (cid:48) = g be all suffixes of g , and ˜ h , ˜ h , . . . , ˜ h ω (cid:48) = h ω be all prefixesof h ω , ordered by length. Suppose that ˜ g i appears m i times in g , g , . . . , g t for each0 ≤ i < t (cid:48) , and ˜ h i appears n i times in h , h , . . . , h ω for each 0 ≤ i < ω (cid:48) . Then we have y = (cid:0) g µg − (cid:1) (cid:0) g µg − (cid:1) · · · (cid:0) g t µg − t (cid:1) ≥ k (cid:0) g µg − (cid:1) (cid:0) ˜ g i µ ˜ g − i (cid:1) m i , and y = (cid:0) h − µh (cid:1) (cid:0) h − µh (cid:1) · · · (cid:0) h − ω µh ω (cid:1) ≥ k (cid:16) ˜ h − i µ ˜ h i (cid:17) n i (cid:0) h − ω µh ω (cid:1) . For each 0 ≤ i < t (cid:48) , we have either ˜ g i +1 = y ˜ g i or ˜ g i +1 = x ˜ g i = µy ˜ g i . And for i = t (cid:48) −
1, it is necessarily the latter case. So we have˜ g i +1 ≥ k y ˜ g i ≥ k (cid:0) g µg − (cid:1) ˜ g i µ m i . By induction, we have g = ˜ g t (cid:48) = µy ˜ g t (cid:48) − ≥ k µg µ t (cid:48) g − µ (cid:80) t (cid:48)− i =0 m i . By symmetry, we have h ω ≥ k µ (cid:80) ω (cid:48)− i =0 n i h − ω µ ω (cid:48) h ω µ. Here t (cid:48) (resp. ω (cid:48) ) is the number of intersection points between Q (cid:48)(cid:48) P ⊂ ρ and D (resp. RQ (cid:48)(cid:48) ⊂ ρ and D ), not counting P and R . And (cid:80) t (cid:48) − i =0 m i (resp. (cid:80) ω (cid:48) − i =0 n i ) is the numberof intersection points between Q (cid:48)(cid:48) P ⊂ ρ and D (resp. RQ (cid:48)(cid:48) ⊂ ρ and D ). So we have t (cid:48) − (cid:88) i =0 m i = ω − ω (cid:48) nd ω (cid:48) − (cid:88) i =0 n i = t − t (cid:48) . Then we have h ω ≥ k µ t − t (cid:48) h − ω µ ω (cid:48) h ω µ ≥ k µ t − t (cid:48) h − ω µ ω (cid:48) h ω , So h ω ≥ k µ t − t (cid:48) + ω (cid:48) . Because h ω g ≤ k µ t + ω +1 , we have g ≤ k µ ω + t (cid:48) − ω (cid:48) +1 . Then we have g ≥ k µg µ t (cid:48) g − µ ω − ω (cid:48) ≥ k µg µ − Since g ≥ k µg µ − , we get g µ t (cid:48) g − ≥ k µ t (cid:48) . So we have g ≥ k µg µ t (cid:48) g − µ ω − ω (cid:48) ≥ k µ ω + t (cid:48) − ω (cid:48) +1 . By symmetry, we have h ω ≥ k µ t − t (cid:48) + ω (cid:48) +1 . By h ω g ≤ k µ t + ω +1 , we have µ ≤ k
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