Equivariantly slicing strongly negative amphichiral knots

We prove obstructions to a strongly negative amphichiral knot bounding an equivariant slice disk in the 4-ball using the determinant, Spinc-structures and Donaldson's theorem. Of the 16 slice strongly negative amphichiral knots with 12 or fewer crossings, our obstructions show that 8 are not equivariantly slice, we exhibit equivariant ribbon diagrams for 5 others, and the remaining 3 are unknown. Finally, we give an obstruction to a knot being strongly negative amphichiral in terms of Heegaard Floer correction terms.


Introduction
A strongly negative amphichiral knot .K; / is a smooth knot K S 3 along with a smooth (orientationreversing) involution W S 3 !S 3 such that .K/ D K and has exactly two fixed points, both of which lie on K; see Figure 1.A knot K S 3 is slice if it bounds a smooth disk (the slice disk) properly embedded in B 4 .Our main goal is to study when there exists an equivariant slice disk for a strongly negative amphichiral knot .K; /.Specifically, we are interested in the following property: Definition 1.1 A strongly negative amphichiral knot .K; / is equivariantly slice if there is a smooth slice disk D and a smooth involution 0 W B 4 !B 4 with 0 .D/ D D which restricts to on @B 4 D S 3 .Figure 1 gives an example of a strongly negative amphichiral diagram, that is, a knot diagram with the strongly negative amphichiral symmetry given by -rotation around an axis perpendicular to the page followed by reflection across the plane of the diagram.Furthermore, the knot in Figure 1 is equivariantly slice.The slice disk is given by performing the pair of equivariant band moves shown in red, then equivariantly capping off the resulting 3-component unlink in B 4 .Among nontrivial prime knots with 12 or fewer crossings, there are 16 slice strongly negative amphichiral knots.For five of them, namely 8 9 , 10 99 , 12a 819 , 12a 1269 and 12n 462 , we found similar equivariant ribbon diagrams; see the table in Section 7.
Strongly negative amphichiral knots, and in particular the equivariant surfaces they bound in the 4-ball, have been studied less than their more popular orientation-preserving cousins: strongly invertible knots (see for example Boyle and Issa [2] and Sakuma [23]) and periodic knots (see for example [2], Cha and Ko [5], Davis and Naik [6], and Grove and Jabuka [14] among others).Many of the obstructions used Figure 1: A strongly negative amphichiral diagram for 8 9 .The symmetry is given by -rotation around an axis perpendicular to the page, followed by a reflection across the plane of the diagram.An equivariant slice disk can be seen by performing the band moves shown in red.
in the strongly invertible and periodic settings do not adapt to the strongly negative amphichiral case.In fact, even showing that the (nonequivariant) 4-genus for strongly negative amphichiral knots can be arbitrarily large was only recently accomplished by Miller [20].
Our first equivariant slice obstruction comes from studying the knot determinant.It was shown by Goeritz [10] that the determinant of an amphichiral knot is the sum of two squares (see also Friedl,Miller and Powell [9] for a partial generalization and Stoimenow [24] for the converse).We prove the following strengthening of this determinant condition in the case that K bounds an equivariant slice disk: Theorem 1.2 If K is an equivariantly slice strongly negative amphichiral knot, then det.K/ is the square of a sum of two squares.
Our second obstruction, which applies to knots with an alternating strongly negative amphichiral diagram, comes from applying Donaldson's theorem [8].Donaldson's theorem can often be used to obstruct the existence of slice disks (see for example Lisca [18]).More recently, it has also been used to obstruct equivariant slice disks for strongly invertible and periodic knots [2].A key ingredient in that obstruction is the existence of an invariant definite spanning surface for the knot.In contrast, strongly negative amphichiral knots do not bound invariant spanning surfaces in S 3 .Instead, we use the fact that, if K bounds an equivariant slice disk D, then the subset S of Spin c -structures on the double branched cover Y D †.S 3 ; K/ that extend over †.B 4 ; D/ is Q -invariant, where Q is a lift of the symmetry to Y ; see Proposition 4.1 and the discussion following its proof.Donaldson's theorem can be used to obtain restrictions on S .Using the interplay between the pair of checkerboard surfaces exchanged by the symmetry, we carefully keep track of Spin c -structures, allowing us to compute the Q -action on Spin c .Y /.This results in a nice combinatorial description of the Q -action on Spin c .Y / in terms of the oriented incidence matrices of the checkerboard graphs for an alternating symmetric diagram.Specifically, we prove the following theorem: Algebraic & Geometric Topology, Volume 24 (2024) Equivariantly slicing strongly negative amphichiral knots 899 Theorem 1.3 Let .K; / be a knot with an alternating strongly negative amphichiral diagram and let Y D †.S 3 ; K/.Let F ˙be the positive and negative definite checkerboard surfaces, let J ˙be compatible oriented incidence matrices with a row removed1 for the checkerboard graphs of F ˙, and let A ˙D J ˙.J ˙/T 2 M n .Z/ be the Goeritz matrices for F ˙. Then there is a lift Our analysis of the Q -action on Spin c .†.S 3 ; K// also leads us to the following obstruction to strongly negative amphichirality in terms of Heegaard Floer correction terms.
Theorem 1.4 Let .K; / be a strongly negative amphichiral knot and let Q be a lift of to Y WD †.S 3 ; K/ (see Proposition 2.1).Then the orbits of Spin c .Y / under the action of Q take the following form: (1) There is exactly one orbit fs 0 g of order 1 with d.Y; s 0 / D 0.

«
. We checked that, for all 2-bridge knots with 12 or fewer crossings, the d -invariants have this structure precisely when the knot is strongly negative amphichiral, leading us to the following conjecture: Conjecture 1.5 Let p; q 2 N with p odd and .p;q/ D 1.The following are equivalent: (1) The Heegaard Floer correction terms of the lens space L.p; q/ can be partitioned into multisets, each of the form fr; r; r; r g for some r 2 Q, and a single set f0g.
We note that (2), ( 3) and ( 4) are known to be equivalent (see for example Bonahon

Open questions
We conclude the introduction with a list of interesting open questions for further exploration.
Question 1.6 Is there a nonslice strongly negative amphichiral knot with equivariant 4-genus larger than its 4-genus?
Question 1.7 Is there a strongly negative amphichiral knot which is topologically equivariantly slice but not smoothly equivariantly slice?
Question 1.8 Is every strongly negative amphichiral knot with Alexander polynomial 1 topologically equivariantly slice?Question 1.9 If a strongly negative amphichiral knot is smoothly equivariantly slice, then must the knot admit an equivariant ribbon diagram, as in Figure 1?
2 Lifting the action to the double branched cover In this section we show that the strongly negative amphichiral involution on S 3 lifts to the double branched cover †.S 3 ; K/.Since we are interested in equivariant slice disks for K, we also show that this lift Q can be extended to †.B 4 ; S/ for any equivariant surface S B 4 with @S D K. Specifically, we have the following proposition, which is similar to [2, Proposition 12].However, in our situation there are no fixed points disjoint from the branch set; the amphichiral involution lifts to an order-4 symmetry on the double branched cover., @E can be identified with the unit normal bundle of S, where preserves S 1 fibers.Lifting this bundle structure to @ z E, p gives a bijection between the set of fibers of @E and the set of fibers of @ z E (p restricts to a two-to-one covering on each fiber).In particular, Q preserves the set of S 1 fibers on the S 1 -bundle boundary of z E. By extending this action over each D 2 fiber, we can (smoothly) extend Q to the tubular neighborhood p 1 .N .S// †.S 4 ; S/ such that p ı Q D ı p.
Finally, p ı Q D ı p implies that p ı Q 2 D 2 ı p D p, so that Q 2 is either the identity map, or else the nontrivial deck transformation on †.B 4 ; S/.Note that, in either case, Q 4 is the identity map.However, acts by -rotation on an equivariant meridian ˛of a fixed point of .Indeed, if acted by reflection or identity on ˛, then there would be fixed points disjoint from S. In the branched cover we then have that Q acts by 2 -rotation on p 1 .˛/.Thus Q has order 4 and Q 2 D , as desired.Finally, we note that there are exactly two lifts, Q and ı Q D Q 3 , one for each choice of Q t.
Corollary 2.2 Let .K; / be a strongly negative amphichiral knot with double branched cover †.S 3 ; K/.Let S B 4 be a smooth properly embedded surface with boundary K which is invariant under an extension of to B 4 (which we again call ).Then there is a lift Q W †.B 4 ; S/ !†.B 4 ; S/ such that Q 2 D (and hence Q 4 D Id) and p ı Q D ı p.In fact, there are exactly two such lifts, namely Q and Q 3 .
Proof Take the double of †.B 4 ; S / to obtain a closed connected surface in S 4 , then apply Proposition 2.1 and restrict to †.B 4 ; S/.Proposition 2.3 Every strongly negative amphichiral knot .K; / bounds a smooth properly embedded surface S B 4 which is invariant under the cone of .
Proof First we fix a symmetric diagram for .K; /, from which we will produce an equivariant unknotting sequence.Since each equivariant pair of crossing changes produces an equivariant genus-2 cobordism, this will imply that .K; / is equivariantly cobordant to the unknot.Then we note that the unknot bounds a smooth disk in B 4 (given by the cone of the unknot), which is invariant under the cone of .
For the equivariant unknotting sequence, separate K at the two fixed points of into two arcs, ˛and ˇ.
Now, for each equivariant pair of crossings between ˛and ˇ, either ˛is the overstrand in both crossings, or ˇis.Hence we can perform equivariant crossing changes so that ˛is always the overstrand in crossings between ˛and ˇ.Then we can pull ˛and ˇapart to get a knot of the form J # J , where the symmetry exchanges J and J .Finally, any unknotting sequence for J produces an equivariant unknotting sequence for J # J , as desired.
We conclude by lifting to the double branched cover of K.
Proposition 2.4 Let .K; / be a strongly negative amphichiral knot.Then there exist exactly two lifts of to †.S 3 ; K/.Moreover, each such lift Q has Q 2 D , where W †.S 3 ; K/ !†.S 3 ; K/ is the nontrivial deck transformation action, and hence Q has order 4.
Proof The proof is essentially the same as that of Proposition 2.1.It can also be obtained by restricting the lifts in Corollary 2.2 to the boundary †.S 3 ; K/, using the surface guaranteed by Proposition 2.3.

A condition on the determinant
It is implicit in the work of Goeritz [10] that the determinant of an amphichiral knot can be written as the sum of two squares (see also [24] for the converse and [9] for a partial generalization).In this section we reprove this theorem for strongly negative amphichiral knots, and show that the same condition must hold on the square root of the determinant if K is equivariantly slice.
Theorem 1.2 Let .K; / be a strongly negative amphichiral knot.Then det.K/ is a sum of two squares.Furthermore, if .K; / is equivariantly slice, then det.K/ is the square of a sum of two squares.
Before we give a proof of the theorem, we need a few lemmas.
Lemma 3.1 Let A be an abelian group, and let †.X; Y / be the double cover of a manifold X (possibly with boundary), branched over a properly embedded submanifold Y X with nontrivial deck transformation involution W †.X; Y / !†.X; Y /.Suppose that H n .X I A/ D 0. Then .x/D x for all x 2 H n .†.X; Y /I A/.
Proof Since H n .X I A/ D 0, the image of the transfer homomorphism T W H n .X I A/ !H n .†.X; Y /I A/ is 0. For any x 2 H n .†.X; Y /I A/, we have that x C .x/ is in the image of T and hence is 0. Thus .x/D x.
Letting .X; Y / D .S 3 ; K/ in Lemma 3.1, we observe that fixes only the identity element since H 1 .†.S 3 ; K/I A/ has no elements of order 2. Proof The proof is as in [4, Lemma 3], noting that since A is torsion free the universal coefficient theorem does not introduce any unwanted Tor terms.Lemma 3.3 Suppose .K; / has an equivariant slice disk D. Then the kernel of the map induced by inclusion, is invariant under the induced action of any lift Q W †.S 3 ; K/ !†.S 3 ; K/ of on homology.
Proof Let x 2 ker.i / so that x is a boundary in †.B 4 ; D/.By Corollary 2.2, there is an extension of the lift Q to †.B 4 ; D/.Hence Q .x/ is also a boundary, and hence contained in ker.i /.
Proof of Theorem 1.2 By Proposition 2.4, lifts to an order-4 action Q on †.S 3 ; K/ with Q 2 D .In particular, Lemma 3.1 implies that all orbits of Q W H 1 .†.S 3 ; K/I A/ !H 1 .†.S 3 ; K/I A/ have order 4, except the orbit consisting of the identity element.Taking coefficients A as the p-adic integers Z p for some prime p, we have jH 1 .†.S 3 ; K/I Z p /j Á 1 .mod4/: For p Á 3 .mod4/, this implies that jH 1 .†.S 3 ; K/I Z p /j is an even power of p.However, by the universal coefficient theorem, H 1 .†.S 3 ; K/I Z p / Š H 1 .†.S 3 ; K/I Z/ ˝Zp and hence the prime decomposition of jH 1 .†.S 3 ; K/I Z/j D det.K/ contains an even power of p.By the sum of two squares theorem, we then have that det.K/ is the sum of two squares.Now suppose that .K; / has an equivariant slice disk D B 4 .By Lemma 3.2 with p-adic coefficients, the kernel of H 1 .†.S 3 ; K/I Z p / !H 1 .†.B 4 ; D/I Z p / is a square-root order subgroup of H 1 .†.S 3 ; K/I Z p /, and by Lemma 3.3, this subgroup is invariant under the action of Q .In particular this subgroup must consist of the identity plus a (finite) collection of order-4 orbits, so that As above, we then have that p det.K/ can be written as the sum of two squares.

An obstruction on Spin c -structures
In this section we prove Theorem 1.3, giving an obstruction to an alternating strongly negative amphichiral knot bounding an equivariant slice disk D in B 4 .We do so by considering Spin c -structures on the double branched cover and applying Donaldson's theorem.This obstruction is based on the following observation: where W Spin c .Y / !Spin c .Y / is the induced map on the Spin c -structures on the boundary.
In order to use this proposition, take Y D †.We recall the following characterization of Spin c -structures in terms of characteristic covectors which we will use throughout this section.Let X be a smooth 4-manifold which is either closed with no 2-torsion in H 1 .X /, or constructed by attaching 2-handles to the 4-ball with @X a rational homology sphere.Let Q be the intersection form on X and Spin c .X / be the set of Spin c -structures of X .Then the first Chern class gives a bijection between the Spin c -structures on X and the characteristic covectors of H 2 .X /; see [11,Proposition 2.4.16].More precisely, Spin c .X / Š Char.H 2 .X // WD fu 2 H 2 .X / j u.x/ Á Q.x; x/ .mod2/ for all x 2 H 2 .X /g: In the case that @X ¤ ∅ this identification induces a bijection Spin c .@X / Š Char.H 2 .X //=2i .H 2 .X //; where i W H 2 .X / !H 2 .X / is given by x 7 !Q.x; / (see for example [21,Section 2.3]).
The following proposition gives restrictions on the set of Spin c -structures on a 3-manifold which extend over a Z=2Z-homology 4-ball which it bounds.Analogous statements are discussed in [13, Section 2] and [7, Theorem 5.1].Proposition 4.2 Let X be a positive-definite smooth 4-manifold obtained by attaching 2-handles to the 4-ball and with @X a rational homology sphere Y .Suppose that Y also bounds a Z=2Z-homology 4-ball W .The inclusion map X !X [ Y W induces an embedding Ã W .H 2 .X /; Q/ !.Z n ; Id/, where Q is the intersection form of X .Choosing a basis for H 2 .X /, Ã is given by an n n matrix A, and the Spin c -structures on Y which extend over W are those of the form where v 2 Z n is any vector with all odd entries, and where elements of Char.H 2 .X // Hom.H 2 .X /; Z/ are written in the dual basis.
Proof Let Z D X [ Y W , and note that Z is positive definite (see eg [16,Proposition 7]).Hence, by Donaldson's theorem, there is an isomorphism of intersection forms .H 2 .Z/= Tor; Q Z / Š .Z n ; Id/, where n D b 2 .X /.We then have a map Ã W .H 2 .X /; Q/ !.Z n ; Id/ induced by the inclusion ÃW X ,! Z.Note that we may identify Char.H 2 .Z// with Spin c .Z/ (since H 1 .Z/ has no 2-torsion), and similarly Char.H 2 .X // with Spin c .X /; see the discussion preceding Proposition 4.2.Applying Hom. ; Z/ gives the map Ã W H2 .Z/= Tor !H 2 .X /, which induces a map Ã W Char.H 2 .Z// !Char.H 2 .X // on Spin cstructures.Recall as well that the restriction r W Spin c .X / !Spin c .Y / is given by the quotient map where i W H 2 .X / !Hom.H 2 .X /; Z/ is given by x 7 !Q.x; /.Hence the restriction map from Spin c .Z/ to Spin c .Y / is given by r ı Ã .We then claim that the image of r ı Ã is precisely the Spin c -structures on Y which extend over W . Indeed r is surjective, so all Spin c -structures on Y extend over X , and hence a Spin c -structure on Y extends over W if and only if it extends over all of Z.
Combinatorially, we can compute this restriction as follows.Choose a basis for H 2 .X /, and the dual basis for Hom.H 2 .X /; Z/.Then Ã is given by a matrix A, and Ã is given by A T .The characteristic covectors of H 2 .Z/ are given by vectors v in Z n with all odd entries.Then the image of Ã consists of elements of all vectors of the form written in the dual basis for Hom.H 2 .X /; Z/ Char.H 2 .X //.The image of r ı Ã then consists of these vectors modulo the column space of 2Q.
We now turn to computing Q W Spin c .†.S3 ; K// !Spin c .†.S 3 ; K//.To do so, begin with a strongly negative amphichiral alternating diagram for K, and let F C and F be the pair of checkerboard surfaces with F C and F positive and negative definite, respectively.Note that F C and F are exchanged by the strongly negative amphichiral symmetry.Definition 4.3 Take S 4 as the unit sphere in R 5 .Define swap W S 4 !S 4 as the involution .x 1 ; x 2 ; x 3 ; x 4 ; x 5 / 7 !.x 1 ; x 2 ; x 3 ; x 4 ; x 5 /: On the equatorial S 3 D f.x 1 ; x 2 ; x 3 ; x 4 ; 0/ j x 2 1 C x 2 2 C x 2 3 C x 2 4 D 1g, swap restricts to the (unique 2 ) amphichiral symmetry with two fixed points .˙1;0; 0; 0; 0/.Finally, note that swap is orientationpreserving and exchanges the two hemispheres of S 4 .
With respect to this involution swap , we can push F C and F equivariantly into distinct hemispheres of S 4 .By Proposition 2.1 there are two lifts, Q swap and Q 0 swap , of swap to an order-4 symmetry of †.S 4 ; F C [ F /. We have that Proof By construction, .Q swap j †.S 3 ;K / / D Q .Hence the map   graphs, embedded as dual planar graphs.The graphs G c .F ˙/ are compatibly oriented if their edges are oriented so that intersecting dual edges satisfy the right-hand rule, as in the left of Figure 2. Suppose G c .F ˙/ are compatibly oriented, order the vertices of each of G c .F ˙/ so that the strongly negative amphichiral symmetry respects the orderings and enumerate the edges of each graph so that intersecting edges have the same index; see Figure 6 for an example.We call the oriented incidence matrices J ˙for G c .F ˙/ compatible.We use the notation J C (resp.J ) to denote the matrix J C (resp.J ) with the last row removed.Recall that, in an oriented incidence matrix A, A i;j D 8 < : 1 if the j th edge begins at the i th vertex; 1 if the j th edge terminates at the i th vertex; 0 otherwise: The following proposition can be used to combinatorially compute the maps r C and r from Proposition 4.5 in terms of oriented incidence matrices; see Remark 4.9.
Proposition 4.7 Let D be an alternating knot diagram with positive and negative definite checkerboard surfaces F C and F , respectively, and let G c .F ˙/ be compatibly oriented complementary checkerboard graphs (see Definition 4.6).Then there is an orthonormal basis fe i g of H 2 .†.S 4 ; F C [ F // in bijection with the crossings of D for which the maps H 2 .˙†.B 4 ; F ˙// !H 2 .†.S 4 ; F C [ F //, induced by inclusion, are given by the transposes .J ˙/T of the oriented incidence matrices of G c .F ˙/ with respect to the vertex generating sets for H 2 .˙†.B 4 ; F ˙//.There is an isomorphism Note that e 1 ; : : : ; e k are in correspondence with the edges of G c .F C / (and G c .F /).Furthermore, the orientation on an edge E i in G c .F C / induces an orientation on the corresponding e i as follows.First, orient the arc ˛i going into the page of the knot diagram (away from the reader).Next, push the interior of ˛i into the region corresponding to the terminal vertex of E i and then out of the page of the diagram (toward the reader) so that it is disjoint from F C [ F .Call the resulting arc ˛0 i ; see Observe that e i and e j are disjoint for i ¤ j , so it is enough to show that e i e i D 1. Consider the arcs .˛i/˙shown in Figure 3, where is the starting endpoint of an edge corresponding to e j , then e j v i D 1.The magenta loop is the boundary of the gray disk e j \ N C , and is oriented so that the arc coming out of the page is isotopic (keeping the endpoints on K) to ˛0 j (see Figure 2) in the complement of F C [F .at a single point.Observe that the preimages .
There is an isotopy in S 3 between .˛i/C and .˛i/intersecting ˛i in a single point, which induces an isotopy between .Q ˛i/ C and .Q ˛i/ .Gluing D 2 C .˛i/ 0 to D 2 .˛i/ 0along the (image of the) isotopy in †.S 3 ; K/ defines a push-off of e i which has a single positive transverse intersection with e i .
Recall that an element v i 2 H 2 .†.B 4 ; F C // of the vertex generating set is represented by a sphere which intersects Hence v i e j can be computed locally in N C .Diagrammatically (see Figure 4), we draw @D 1 D S 3 and think of N C as a neighborhood of F C S 3 .Specifically, v i e j D 0 if the edge corresponding to e j and v i are not incident, v i e j D 1 if the edge corresponding to e j begins at v i , and v i e j D 1 if the edge corresponding to e j terminates at v i .A similar argument applies to the vertex generating set of H 2 .†.B 4 ; F //. Remark 4.9 Proposition 4.7 combinatorially determines the maps r ˙W Spin c .†.S 4 ; F C [ F // !Spin c .˙ †.B 4 ; F ˙// from Proposition 4.5.Specifically, the maps r ˙are given by taking the duals of then restricting to characteristic vectors.
We conclude the section with a proof of Theorem 1.3 from the introduction: Proof of Theorem 1.3 Let Y D †.S 3 ; K/ and X ˙D †.B 4 ; F ˙/.We identify each of H 2 .X ˙/ with the Z-span of Vert.G c .F ˙//nfv ˙g, where fv C ; v g is the pair of -invariant vertices removed when defining J ˙.Note that X ˙can be constructed by attaching 2-handles to the 4-ball (see for example the proof of Lemma 3.6 in [22]).Hence, using the dual basis for H 2 .X ˙/ , we may identify Spin

An alternating slice strongly negative amphichiral example
In this section we give an example of a strongly negative amphichiral knot which Theorem 1.3 shows is not equivariantly slice.

:
We will show that this collection S of Spin c -structures on †.S 3 ; K/ is not Q -invariant.Specifically, we will show that the Spin c -structure represented by the second vector s D .3;3; 2; 0; 1; 2/ T is mapped by Q to a Spin c -structure not contained in S . Consider

Heegaard Floer correction terms
In this section we give a necessary condition on the Heegaard Floer correction terms d. †.S 3 ; K/; s/, also known as d-invariants, for a knot to be strongly negative amphichiral.In the case of periodic knots, a similar type of condition was proved by Jabuka and Naik in [17].As in the case of periodic knots, we first need invariance of the d-invariants.Proof This follows directly from the diffeomorphism invariance of Heegaard Floer homology.
Along with the following lemma, this implies our final theorem below.Lemma 6.2For any knot K S 3 , the deck transformation involution of the double branched cover †.S 3 ; K/ acts on the set of Spin c -structures by conjugation.
Proof The first Chern class c 1 W Spin c .†.S 3 ; K// !H 2 .†.S 3 ; K/I Z/ is an isomorphism, since †.S 3 ; K/ is a Z=2Z homology sphere, and by Poincaré duality we also have an isomorphism H 2 .†.S 3 ; K/I Z/ Š H 1 .†.S 3 ; K/I Z/: By Lemma 3.1, acts as the negative of the identity on H 1 .†.S 3 ; K/I Z/, which then induces conjugation on the set of Spin c -structures under these natural isomorphisms.
Theorem 1.4 Let .K; / be a strongly negative amphichiral knot and let Q be a lift of to †.S 3 ; K/ (see Proposition 2.1).Then the orbits of the d -invariants of †.S 3 ; K/ under the action of Q satisfy: There is exactly one orbit fs 0 g of order 1.Moreover, d. †.S 3 ; K/; s 0 / D 0.
Proof Since Q has order 4, the Q -orbits of the Spin c -structures will have order 1, 2 or 4. Let D Q 2 be the deck transformation action on †.S 3 ; K/, and note that acts on the set of Spin c -structures by conjugation by Lemma 6.2.Hence, if a Spin c -structure is not fixed by conjugation, then it will have a Q -orbit of length 4. On the other hand, since †.S 3 ; K/ is a Z=2Z-homology sphere, there is a unique Spin c -structure s 0 fixed by conjugation.Furthermore, since jH 1 .†.S 3 ; K/j is odd there are an odd number of Spin c -structures, and hence s 0 has a Q -orbit of length 1.

S 3 ;
K/, X D †.B 4 ; D/ and D Q W †.B 4 ; D/ !†.B 4 ; D/ a lift of the strongly negative amphichiral symmetry from Corollary 2.2.In order to rule out that Q .Spin c .X /j Y / D Spin c .X /j Y , we will need to compute Q W Spin c .Y / !Spin c .Y / and also restrict the possible subsets Spin c .X /j Y Spin c .Y / using Donaldson's theorem.Propositions 4.5 and 4.7 combined allow us to compute Q W Spin c .Y / !Spin c .Y /, and Proposition 4.2 gives restrictions on Spin c .X /j Y Spin c .Y /.See Section 5 for an example.

Definition 4 . 6
Fix a strongly negative amphichiral alternating knot diagram, let F ˙be the positive and negative definite checkerboard surfaces and let G c .F ˙/ be the corresponding complementary checkerboard

Figure 2 :
Figure 2: An oriented edge of G c .F C / in black intersecting an edge of G c .F / in red (left).The orientation on the red edge is induced by the right-hand rule.On the right is the oriented arc ˛0 i induced from the oriented edge of G c .F C / in black.

Remark 4 . 8 Figure 3 :
Figure3: The arcs .˛i/C and .˛i/are contained in the horizontal and vertical checkerboard surfaces, respectively.The green arrow indicates an isotopy between them in S 3 .Lifting this to †.S 3 ; K/, we see that the self pairing of the sphere e i is 1.
and Q C is the intersection form, which is given as follows.Letting a be a 1-cycle in F C , .OEa/ D OE.cone on a in D 1 / .cone on Ã.a/ in D 2 /: In their interiors, the surfaces F C and F in S 3 intersect in a collection of k arcs ˛1; : : : ; ˛k, one for each crossing of D. The I -subbundle of N C over ˛i is a disk D 2 C .˛i/D 1 with boundary Q ˛i, the preimage of ˛i in †.S 3 ; K/.(The disk D 2 C .˛i/ is also the trace of ˛i under the isotopy pushing int.F C / into int.B 4 /.)Note that D 2 C .˛i/ is properly embedded in †.B 4 ; F C /. Similarly, there is a disk D 2 .˛i/properly embedded in †.B 4 ; F /, and gluing these disks along Q ˛i gives a sphere e i in †.S 4 ; F C [ F /.

Figure 2 .
Recall that †.B 4 ; F C / D D 1 [ D 2 as an oriented manifold.Then the orientation of ˛0 i D 1 determines an orientation on the union of ˛0 i D 1 with ˛0 i D 2 , which is locally isotopic within †.S 3 ; K/ to Q ˛i.This orientation on Q ˛i then determines an orientation on D 2 C .˛i/ as its oriented boundary, and this orientation on D 2 C .˛i/ extends to an orientation on e i D D 2 C .˛i/ [ D 2 .˛i/.We now show that fe 1 ; : : : ; e k g is an orthonormal basis for H 2 .†.S 4 ; F C [ F //.Note that b 2 .†.S 4 ; F C [ F // D b 2 .†.B 4 ; F C // C b 2 .†.B 4 ; F //; since †.S 3 ; K/ is a rational homology sphere.However, b 2 .†.B 4 ; F ˙// D n ˙ 1, where n ˙is the number of vertices of G c .F ˙/.From the Euler characteristic of the sphere of the knot diagram, we get 2 D n C k Cn since G c .F C / and G c .F / are dual graphs.Hence b 2 .†.S 4 ; F C [F // D k.Thus it suffices to show that e 1 ; : : : ; e k are orthonormal.

4 . 9 )
, the maps r ˙in Proposition 4.5 are given by J ˙. Proposition 4.5 then shows that the map Q W Spin c .Y / !Spin c .Y / is determined by Q OEJ C v D OEJ v for all v 2 Z 2n with v Á .1;1; : : : ; 1/ T .mod2/: Finally, let D be an equivariant slice disk for K.By Proposition 4.2, the set of Spin c -structures of Y which extend over †.B 4 ; D/ is given by S D fOEu 2 Spin c .Y / j u D A T v for some v 2 Z n with v Á .1;1; : : : ; 1/ T .mod2/g; and by Corollary 2.2 there is a lift †.B 4 ; D/ !†.B 4 ; D/ which restricts to the lift Q on Y .Hence, by Proposition 4.1, S is Q -invariant.

Figure 5 :
Figure 5: A strongly negative amphichiral symmetry on 12a 1105 .The symmetry is -rotation within the plane of the diagram followed by a reflection across the plane of the diagram.

12 Figure 6 :
Figure 6: The pair of complementary checkerboard graphs of the alternating diagram for 12a 1105 in Figure 5.They are exchanged by the strongly negative amphichiral symmetry.G c .F C / is black and G c .F / is red.The fe i g correspond to crossings in the knot diagram.
Proposition 2.1 Let S S 4 be a closed connected smoothly embedded surface and let W .S Furthermore, Q 2 D , and there are exactly two such lifts, namely Q and Q 3 .Proof Let N .S/ be an equivariant tubular neighborhood of S and E D S 4 nN .S/ be the surface exterior.Denote by z E the double cover of E corresponding to the kernel G of 1 .E/ !H 1 .EI Z=2Z/.
r .N s/; where the final equality holds since Q swap exchanges †.B 4 ; F C / and †.B 4 ; F / in †.S 4 ; F C [ F /.We now consider the complementary checkerboard graph G c .F C /, which has a vertex v i corresponding to each planar region of the knot diagram complementary to F C and an edge corresponding to each crossing in the knot diagram.Let i be the simple loop in F C running once counterclockwise around the region corresponding to v i .Applying the isomorphism H 1 .F C / Š H 2 .†.B 4 ; F C // from [12, Theorem 3], we get an element v i 2 H 2 .†.B 4 ; F C //.We call fv i g the vertex generating set of H 2 .†.B 4 ; F C //, and we declare the vertex generating set of H 2 .†.B 4 ; F C // to be f v i g.

:
Neither matrix satisfies the Q -invariance condition in Theorem 1.3.We will show this for the matrix A 1 ; the computation for A 2 is similar.For A 1 , we compute that the set T .mod2/g consists of the 17 classes represented by the following vectors: S D fOEu 2 Spin c .Y / j u D A T 1 v for some v 2 Z n with v Á .1;1; : : : ; 1/ Multiplying, we see that J C .Q s/ D s and J .Q s/ D .1;3;16;8; 3; 2/ T .A straightforward linear algebra computation shows that .1;3;16;8; 3; 2/ T is not equivalent modulo 2A C to any of the 17 vectors in S.Hence Q OEJ C .Q s/ D OEJ .Q s/ is not in S.Along with a similar computation for A 2 , this implies that K is not equivariantly slice, by Theorem 1.3.