Branched covers and rational homology balls

The concordance group of knots in the three-sphere contains an infinite subgroup generated by elements of order two, each one of which is represented by a knot K with the property that for every n>0, the n-fold cyclic cover of S^3 branched over K bounds a rational homology ball. This implies that the kernel of the canonical homomorphism from the knot concordance group to the infinite direct sum of rational homology cobordism groups (defined via prime-power branched covers) contains an infinitely generated two-torsion subgroup.


Introduction
There is a homomorphism where C is the smooth concordance group of knots in S 3 , Q is the set of prime power integers, and Θ 3 Q is the rational homology cobordism group.For a knot K and q ∈ Q, the q-component of ϕ(K) is the class of M q (K), the q-fold cyclic cover of S 3 branched over K.
In the paper [1], Aceto, Meier, A. Miller, M. Miller, Park, and Stipsicz proved that ker ϕ contains a subgroup isomorphic to (Z 2 ) 5 .Here we will prove that ker ϕ contains a subgroup isomorphic to (Z 2 ) ∞ .Our examples are of the form K # −K r , where −K denotes the concordance inverse of K (the mirror image of K with string orientation reversed), and K r is formed from K by reversing its string orientation.Such knots are easily seen to be in the kernel of ϕ; the more difficult work is to find nontrivial examples of order two.
The first known example of a nontrivial element in ker ϕ was represented by the knot K 1 = 8 17 # −8 r 17 , which is of order two in C. That K 1 is of order at most two is elementary; that K 1 is nontrivial in C is one of the main results of [9], proved using twisted Alexander polynomials.
The results of [7] provide an infinitely generated free subgroup of ker ϕ.
1.1.Main result.Figure 1 illustrates a knot P n in a solid torus, where J n represents the braid illustrated on the right in the case of n = 5; n will always be odd.We let K n denote the satellite of 8 17 built from P n .In standard notation, K n = P n (8 17 ).For future reference, we illustrate the braid J * n formed by rotating J n around the vertical axis.
Theorem 1.Let K n = P n (8 17 ).For all odd n, the knot There is an infinite set of prime integers P for which L α = L β ∈ C for all α = β in P. In particular, the set of knots {L n } n∈P generates a subgroup of ker ϕ that is isomorphic to The rest of the paper presents a proof of this theorem.The first two claims are easily dealt with in Sections 2 and 3.The more difficult step of the proof calls on an analysis of twisted Alexander polynomials and their relevance to knot slicing; a review of twisted polynomials is included in Section 4. The proof of Theorem 1 is completed in Section 5, with the exception of a number theoretic result that is described Appendix A. Acknowledgements.Thanks to Darrell Haile for assisting me in proof of the number theoretic result in Appendix A. Allison Miller provided valuable feedback about an early draft of this paper.

Proof:
denote the knot formed using the braid J * n in Figure 2.For any knot K, let P * n (K) denote the satellite of K built using P * n .It should be clear that P n and P * n are orientation preserving isotopic, and thus for all knots K, P n (K) = P * n (K). Figure 2 illustrates for an arbitrary knot K, the connected sum P n (K) # P * n (K) = P n (K) # P n (K) in the case of n = 5.Performing n − 1 band moves in the evident way yields the (0, n)-cable of K # K. Thus, if K # K = 0 ∈ C, then the n components of this link can be capped off with parallel copies of the slice disk for K # K, implying that P n (K) # P n (K) = 0 ∈ C. In particular, 2K n = 0 ∈ C and 2K r n = 0 ∈ C.

Proof L n ∈ ker ϕ
We prove a stronger statement: For all odd n, and for all positive integers q, M q (L n ) is a rational homology sphere that represents 0 ∈ Θ 3 Q .The q-fold cyclic cover of S 3 branched over K n # −K r n is the same space as the q-fold cyclic cover of S 3 branched over K n # −K n .A slice disk for K n # −K n is built from (S 3 × I, K n × I) by removing a copy of B 3 × I. Taking the q-fold branched cover shows that the q-fold cyclic cover of B 4 branched over that slice disk is diffeomorphic to M q (K n ) * × I, where M q (K n ) * denotes a punctured copy of M q (K n ).It remains to show that M q (K n ) is a rational homology 3-sphere.
A formula of Fox [5] and Goeritz [6] states that the order of the first homology of M q (K n ) is given by the product of values ∆ Kn (ω i q ), where ∆ Kn (t) denotes the Alexander polynomial, ω q is a primitive q-root of unity, and i runs from 1 to q − 1.
A result of Seifert [11] shows that ∆ , where U is the unknot.We have that A numeric computation confirms that this polynomial does not have roots on the unit complex circle, and hence ∆ 817 (t n ) has no roots on the unit complex circle.From this is follows that ∆ Kn (ω i q ) = 0 for all i; thus the order of the homology of M q (K n ) is finite.

Review of twisted polynomials and 8 17
In this section we review twisted Alexander polynomials and their application in [8,9] showing that 8 17 # −8 r 17 = 0 ∈ C. Let (X, B) → (S 3 , K) be the q-fold cyclic branched cover of a knot K with q a prime power.In particular, X is rational homology sphere.There is a canonical surjection : H 1 (X−B) → Z. Suppose that ρ : H 1 (X) → Z p is homomorphism for some prime p. Then there is an associated twisted polynomial ∆ It is well-defined, up to factors of the form at k , where a = 0 ∈ Q(ω p ).These polynomials are discriminants of Casson-Gordon invariants, first defined in [3].
In the case of K = 8 17 and q = 3, we have H 1 (X) ∼ = Z 13 ⊕ Z 13 , and as a Z 13 -vector space this splits as a direct sum of a 3-eigenspace and a 9-eigenspace under the order three action of the deck transformation.Both eigenspaces are 1-dimensional.We denote this splitting as E 3 ⊕E 9 .There are corresponding characters ρ 3 and ρ 9 of H 1 (X) onto Z 13 ; these are defined as the quotient maps onto H 1 (X)/E 3 and onto H 1 (X)/E 9 .We let ρ 0 denote the trivial Z 13 -valued character.
The values of ∆ 817, ,ρi (t) are given in [9], duplicated here in Appendix B. For i = 0 it is polynomial in . An essential observation is that for 8 r 17 , the roles of ρ 3 and ρ 9 are reversed.All three of the polynomials are irreducible in their respective polynomial rings, once any factors of (1 − t) and t are removed.
In [9] the proof that 8 17 # −8 r 17 is not slice comes down to the observation that no product of the form (That is, these products are not norms in the polynomial ring Q(ω 13 )[t, t −1 ], modulo powers of (1 − t) and t.)Here i = 0 or i = 9 and j = 0 or j = 3.The number a is in Q(ω) and the σ ν are Galois automorphisms of Q(ω p ) (which acts by sending ω p to ω ν p ). Showing that the product of the polynomials does not factor in this way is elementary once it is established that ∆ 817, ,ρ3 (t) and ∆ 817, ,ρ9 (t) are irreducible and not Galois conjugate.

Main Proof
Using the fact that −P n (8 17 ) r = P n (8 17 ) r , the knot L α # L β can be expanded as We begin by analyzing the 3-fold cover of S 3 branched over P n (8 17 ) and assume that 3 does not divide n.This cover is M 3 (P n (8 17 )) and we denote the branch set in the cover by B.
There is the obvious separating torus T in S 3 \ P n (8 17 ).Since 3 does not divide n, T has a connected separating lift T ⊂ M 3 (P n (8 17 )).One sees that T splits M 3 (P n (8 17 )) into two components: X, the 3-fold cyclic cover of S 3 \ 8 17 and Y , the 3-fold cyclic branched cover of S 1 × B 2 , branched over P n .A simple exercise shows that since P n (U ) is unknotted, Y is the complement of some knot J n ⊂ S 3 .
A Mayer-Vietoris argument shows that H 1 (M 3 (P n (8 17 ))) ∼ = Z 13 ⊕ Z 13 and the two canonical representations ρ 3 and ρ 9 that are defined on X extend trivially on Y and so to M 3 (P n (8 17 )).We denote these extension ρ 3 and ρ 9 .Let be the canonical surjective homomorphism : H 1 (M 3 (P n (8 17 ))) \ B) → Z. Restricted to X we have (x) = (nx), where was the canonical representation to Z defined for the cover of S 3 \ 8 17 .
In [8,Theorem 3.7] there is a discussion of twisted Alexander polynomials of satellite knots in S 3 , working in the greater generality of homomorphisms to the unitary group U (m). (A map to Z p can be viewed as a representation to U (1)).The proof of that theorem, which relies on the multiplicativity of Reidemeister torsion, applies in the current setting, yielding the following lemma.
As described in [8,9] Casson-Gordon theory implies that if L α # L β is slice, then for some 3-eigenvector or for some 9-eigenvector the corresponding twisted Alexander polynomial is a norm; that is, it factors as at k f (t)f (t −1 ), modulo multiples of (1 − t).If it is a 3-eigenvector, the relevant polynomial is of the form where one of x, y, z, or w is equal to 1, and each of the others are either 1 or 0. The four Q(ω 13 )[t]-polynomials that appear here, ∆ 817, ,ρ3 (t α ), ∆ 817, ,ρ9 (t α ), ∆ 817, ,ρ3 (t β ), and ∆ 817, ,ρ9 (t β ), and all their Galois conjugates are easily seen to be distinct for any pair α = β.The following number theoretic result implies that there is an infinite set of primes such that if α ∈ P and β ∈ P, then no product as given in Equation 1 can be a norm in Q(ω 13 )[t], proving that the connected sum L α # L β is not slice.We will present a proof in Appendix A, Lemma 3. Let f (t) ∈ Z(ω p )[t] be an irreducible monic polynomial.If there exists ζ ∈ C such that f (ζ) = 0 and ζ n = 1 for all n > 0, then the set of primes p for which f (t p ) is reducible is finite.
Proof of Theorem 1.The last factor in Equation 1 involving the J n is a norm, so it can be ignored in determining if the product is a norm.A numeric computation shows that the twisted polynomials ∆ 817, ,ρi (t) for i = 3 and i = 9 do not have roots on the unit circle, so Lemma 3 can be applied with F = Q(ω 13 ).Let P be the infinite set of primes with the property that if p ∈ P then ∆ 817, ,ρ3 (t p ) and ∆ 817, ,ρ9 (t p ) are irreducible.Consider the case of x = 1 in Equation 1.Then, assuming that α ∈ P and β ∈ P, the term σ a (∆ 817, ,ρ3 )(t α ) that appears in Equation 1 is relatively prime to the remaining factors and all the factors are irreducible, modulo powers of t and (1 − t).Hence, the product cannot be of the form t k (1 − t) j f (t)f (t −1 ) for any f (t) ∈ Q(ω 13 )[t].The cases of y, z, or w = 1 are the same.
In this appendix we prove Lemma 3, stated in somewhat more generality as Lemma 4 below.We first summarize some background material.Further details can be found in any graduate textbook on algebraic number theory.
• A ⊂ C denotes the ring of algebraic integers.This is the ring consisting of all roots of monic polynomials in Z[t].• For an extension field F/Q, the ring of algebraic integers in F is defined by F is defined to be the set of units in O F .• The norm of an element x ∈ O F is defined as N (x) = x i ∈ Z where the x i are the complex Galois conjugates of x.This map satisfies • The Dirichlet Unit Theorem states that for a finite extension F/Q, the abelian group O × F is finitely generated, isomorphic to G ⊕ Z r+s−1 , where G is finite cyclic, r is the number of embeddings of F in R, and 2s is the number of non-real embeddings of F in C.
This follows immediately from the assumption that f (t) is monic. Step By the degree calculation just given, t p − ζ has an irreducible factor g(t) ∈ F(ζ)[t] of degree l < p.We can write g(t) = (t − ω i p ξ) where the product is over some proper subset of {0, . . ., p − 1}.Multiplying this out, one finds that the constant term is of the form ω j p ξ l ∈ F(ζ) for some j and l < p.Since l and p are relatively prime, there are integers u and v such that ul + vp = 1.Thus, (ω j p ξ l ) u (ξ p ) v = ω s p ξ for some s.In particular, for some s, we have ω s p ξ ∈ F(ζ).We let η = ω s p ξ and find that η p = (ω s p ) p ξ p = ζ.Finally, η satisfies the monic polynomial f (t p ) and thus is in O F(ζ) .Comments.The argument just given is based on a summary of the proof for the case F = Q presented in MathOverflow by Dimitrove [4].Step (2) is a special case of the Vahlen-Capelli Theorem, proved in the case of F = Q by Vahlen and for fields of characteristic 0 by Capelli [2].A proof for fields of finite characteristic is given in the book by Rédei [10].

Lemma 4 .
Let F be a finite extension of Q and let f (t) ∈ O F [t] be an irreducible monic polynomial.If there exists ζ ∈ C such that f (ζ) = 0 and ζ n = 1 for all n > 0, then the set of primes p for which

Step 3 .
The set of primes p such that ζ = η p for someη ∈ O F(ζ) is finite.If ζ = η p then N (ζ) = N (η) p .If N (ζ) = ±1, then the set of p for which N (ζ) = a p for some integer a is finite.If N (ζ) = ±1, then ζ ∈ O × F(ζ).Hence ζ represents a non-torsion element in a finitely generated abelian group, and thus it has a finite number of roots.
Since f (t) is irreducible of degree n and f (t p ) is reducible, we have the degrees of extensions satisfying [F(ζ) : F] = n and [F(ξ) : F] < np.It follows from the multiplicity of degrees of extensions that [F(ξ) : F(ζ)] < p.The polynomial t p − ζ ∈ F(ζ)[t] has ξ as a root.For all i, ω i p ξ is also a root, so t p − ζ factors completely in C[t] as t 2. Suppose that f (t) ∈ F[t] is irreducible and f (ζ) = 0.If for some prime p, f (t p ) is reducible over F, then ζ = η p for some η ∈ O F(ζ) .Let ξ ∈ C satisfy ξ p = ζ.