Proof Let C be a homotopy ribbon concordance from J to K and let $X_C$ be its exterior, i.e., the complement of an open tubular neighborhood of C in $S^3\times [0,1]$ . See [Reference Friedl, Nagel, Orson and PowellFNOP19] for a thorough treatment of tubular neighborhoods of submanifolds of a topological $4$ -manifold. The boundary $\partial X_C$ is homeomorphic to the gluing of $X_{-J}$ and $X_K$ along their boundary tori, gluing meridian to meridian and longitude to longitude.

The composition $j\colon X_{J} \to X_C$ of orientation reversal and inclusion induces a surjection $j_*\colon H_1(X_J;\mathbb {Z}[t^{\pm 1}]) \to H_1(X_C;\mathbb {Z}[t^{\pm 1}])$ , while the inclusion $k\colon X_K \to X_C$ induces an injection $k_*\colon H_1(X_K;\mathbb {Z}[t^{\pm 1}]) \to H_1(X_C;\mathbb {Z}[t^{\pm 1}])$ . This was shown in [Reference Friedl and PowellFP20]; the proof (also for twisted homology) can be found in Section 4.

The surjectivity of $j_*$ implies that the kernel of the homomorphism induced by the inclusion $\partial X_C\to X_C$ is a metabolizer of $-\operatorname {\mathrm {Bl}}_J \oplus \operatorname {\mathrm {Bl}}_K$ (see [Reference Friedl, Leidy, Nagel and PowellFLNP17] for additivity of the Blanchfield form and [Reference FriedlF04, Proposition 8.2] for the proof that the kernel is a metabolizer). The remainder of the proof is an algebraic argument that we outsource to the following lemma, applied with $e = j_*$ , $m = k_*$ , $\lambda _E = \operatorname {\mathrm {Bl}}_J$ , $\lambda _M = \operatorname {\mathrm {Bl}}_K$ , and $R = \mathbb {Z}[t^{\pm 1}]$ . Such a submodule $G \subset H_1(X_J; \mathbb {Z}[t^{\pm 1}])$ as in the theorem can be chosen to be $\operatorname {Ker} (j_*)$ .▪

Proof Let $S = e^{-1}(\operatorname {\mathrm {Im}}(m)) \subset E$ and define $g\colon S\to M$ as $m^{-1}\circ e$ . Let us check that g respects the pairings. Let $a_1, a_2 \in S$ be given. Then for $i\in \{1,2\}$ , we have $e(a_i) = m(g(a_i))$ , so $a_i - g(a_i) \in \operatorname {Ker}(e + m)$ . Since $\operatorname {Ker}(e + m)$ is a metabolizer for $-\lambda _E \oplus \lambda _M$ , we have

$$ \begin{align*} -\lambda_E(a_1, a_2) + \lambda_M(-g(a_1), -g(a_2)) = 0 ~\Rightarrow~ \lambda_E(a_1, a_2) = \lambda_M(g(a_1), g(a_2)) \end{align*} $$

as claimed. Next, g is clearly surjective and has kernel $G = \operatorname {Ker} e$ , so it induces an isometry $S/G \to M$ . Let us now prove that $S = G^{\bot }$ .

To show $S\subset G^{\bot }$ , let $a\in S$ . If $b\in G$ , then

$$ \begin{align*} \lambda_E(a,b) = - (-\lambda_E(a,b) + \lambda_M(-g(a),0)), \end{align*} $$

which equals zero since $a - g(a)$ and b are both contained in the metabolizer $\operatorname {Ker}(e + m)$ .

To show $G^{\bot } \subset S$ , let $c\in G^{\bot }$ be given. Consider the homomorphism $S/G \to Q(R)/R$ given by $[a] \mapsto \lambda _E(a,c)$ . The pairing induced on $S/G$ by $\lambda _E$ is nonsingular since it is isometric to $\lambda _M$ . So its adjoint map is an isomorphism, whence there exists $[c'] \in S/G$ such that for all $[a] \in S/G$ , we have $\lambda _E(a,c) = \lambda _E(a,c')$ . Thus for all $a\in S$ , we have $\lambda _E(a, c-c') = 0$ . If $a\in E$ and $b\in M$ with $e(a) + m(b) = 0$ are given, then $a\in S$ , and so $-\lambda _E(a, c-c') + \lambda _M(b,0) = 0$ . Thus $c-c' \in \operatorname {Ker}(e + m)^{\bot } = \operatorname {Ker}(e + m)$ , and thus $c - c' \in \operatorname {Ker}(e + m) \cap E = G \subset S$ . Since $c'$ is also in S, it follows that $c\in S$ .▪

Question 3.7 Does $[10_{99}] \geq _S [12n_{582}]$ hold? If so, is $10_{99}$ homotopy ribbon concordant to $12n_{582}$ ?

Proof of Proposition 3.3 First, we appeal to two technical results dealing with change of base ring for the Blanchfield form, whose proofs we have relegated to Appendix A. The $\mathbb {Z}[t^{\pm 1}]$ -algebra $R:= \mathbb {Z}[t^{\pm 1}]/(t + 1)$ has only $(t+1)$ -torsion, and is thus admissible in the terminology of Appendix A. So, by Proposition A.1, for all knots L, there is a Blanchfield pairing $\operatorname {\mathrm {Bl}}^R_L$ defined on $H_1(X_L; R)$ , which is determined by $\operatorname {\mathrm {Bl}}_L$ . Moreover, by Proposition A.2, $[J] \geq _S [K]$ implies that there exists an R-submodule $G \subset H_1(X_J; R)$ such that $G\subset G^\bot $ , and $G^{\bot } / G$ is isomorphic to $H_1(X_K;R)$ .

Note that $R \to \mathbb {Z}, t \mapsto -1$ is an isomorphism of $\mathbb {Z}[t^{\pm 1}]$ -algebras (with trivial involution), where t acts on $\mathbb {Z}$ by $-1$ . In particular, as a ring $\mathbb {Z}[t^{\pm 1}]/(t + 1)$ is a PID. By Lemma 3.10 (iii) below, there are two short exact sequences

Note that for all knots L, $H_1(\Sigma _{2,L}; \mathbb {Z})$ and $H_1(X_L; R)$ are isomorphic abelian groups. Indeed, since $H_1(X_L;R) \cong H_1(X_L;\mathbb {Z}[t^{\pm 1}]) \otimes _{\mathbb {Z}[t^{\pm 1}]} R \cong H_1(X_L;\mathbb {Z}[t^{\pm 1}])/(t+1)$ by a straightforward application of the universal coefficient theorem (noting $\displaystyle {\operatorname {\mathrm {Tor}}_1^{\mathbb {Z}[t^{\pm 1}]}(\mathbb {Z},R) =0}$ ), both modules are presented by the matrix $(tV-V^T)|_{t=-1} = -V-V^T$ , where V is a Seifert matrix for L. Thus, we have indeed constructed the two short exact sequences (3.2).▪

Proof (i) Cyclic torsion modules are (noncanonically) isomorphic to $R / (a)$ for some $a\in R\setminus \{0\}$ , and an isomorphism $R/(a) \to \operatorname {Hom}(R/(a), Q(R)/R)$ is given by $1\mapsto (1\mapsto 1/a)$ . This suffices, since M decomposes as a sum of cyclic torsion modules, and $\operatorname {Hom}(N\oplus N', Q(R)/R) \cong \operatorname {Hom}(N, Q(R)/R)\oplus \operatorname {Hom}(N', Q(R)/R)$ .

(ii) There is a short exact sequence

(3.3)

where the second map is the inclusion, and the third map is the composition of the adjoint map $A\to \operatorname {\overline {Hom}}(A,Q(R)/R)$ of $\lambda $ (which is an isomorphism) and the map $\operatorname {\overline {Hom}}(A,Q(R)/R) \to \operatorname {\overline {Hom}}(M,Q(R)/R)$ , which is surjective since $Q(R)/R$ is an injective R-module, and so the contravariant functor $\operatorname {\overline {Hom}}(-,Q(R)/R)$ is exact. The orders of modules in a short exact sequence are multiplicative, so $\operatorname {\mathrm {ord}} M^{\bot } \cdot \operatorname {\mathrm {ord}} \operatorname {\overline {Hom}}(M,Q(R)/R) \doteq \operatorname {\mathrm {ord}} A$ , which implies $\operatorname {\mathrm {ord}} M^{\bot } \cdot \operatorname {\mathrm {ord}} \overline {M} \doteq \operatorname {\mathrm {ord}} A$ using (i). In the same way, we obtain $\operatorname {\mathrm {ord}} M^{\bot \bot } \cdot \operatorname {\mathrm {ord}} \overline {M^{\bot }} \doteq \operatorname {\mathrm {ord}} A$ . Thus we get $\operatorname {\mathrm {ord}} M \doteq \operatorname {\mathrm {ord}} M^{\bot \bot }$ . The inclusion $M\subset M^{\bot \bot }$ follows directly from the definition of $^\bot $ , and since M and $M^{\bot \bot }$ have the same order, they are equal.

(iii) Let us write $\iota $ for inclusions of submodules of A into A, and $\psi \colon A\to \operatorname {\overline {Hom}}(A,Q(R)/R)$ for the adjoint of $\lambda $ . Let $\beta $ be the composition of the maps

The kernel of $\beta $ is $G^{\bot \bot } = G$ . So $\beta $ induces an injection

$$ \begin{align*} \tilde\beta\colon {G^{\bot} / G} \to \operatorname{\overline{Hom}}(G^{\bot},Q(R)/R). \end{align*} $$

Fix an isomorphism $\alpha \colon {\operatorname {\overline {Hom}}(G^{\bot },Q(R)/R)}\to \overline {G^{\bot }}$ . We get the following commutative diagram, in which the two rows and the first column are exact; this induces the dashed maps, such that the second column is also exact.

The second row and second column of this diagram form the desired diagram, using that $\overline {A}\cong A$ and setting $W = \overline {A}/\operatorname {\mathrm {Im}} \iota \alpha \tilde \beta $ . It just remains to check that $\overline {A/G^{\bot }}\cong G$ and $\overline {G^{\bot }} / \operatorname {\mathrm {Im}} \alpha \tilde \beta \cong \overline {G}$ .

An isomorphism $\overline {A/G^{\bot }} \cong \operatorname {Hom}(G,Q(R)/R)$ is given by considering the short exact sequence (3.3), and $\operatorname {Hom}(G,Q(R)/R)$ is isomorphic to ${G}$ by (i).

In the following commutative diagram, the rows are exact (the surjectivity of the second map in the top row follows from the injectivity of the module $Q(R)/R$ as in the proof of (ii) above), and the first two vertical maps are isomorphisms.

This induces the desired isomorphism $G \cong \operatorname {\overline {Hom}}(G,Q(R)/R) \to \overline {G^{\bot }} / \operatorname {\mathrm {Im}} \alpha \tilde \beta $ , drawn dashed.▪

Proof By definition of $\geq _S$ , there exists a submodule $G\subset H_1(X_J; \mathbb {Z}[t^{\pm 1}])$ such that $G\subset G^{\bot }$ and the pairing induced by $\operatorname {\mathrm {Bl}}_J$ on $G^{\bot } / G$ is isometric to $\operatorname {\mathrm {Bl}}_K$ . Let $\widetilde G = G \otimes \mathbb {Q}$ . By Proposition A.2, $\widetilde G^{\bot } / \widetilde G$ is isomorphic to $H_1(X_K; \mathbb {Q}[t^{\pm 1}])$ , so

$$ \begin{align*} \operatorname{\mathrm{ord}} \widetilde G \cdot \Delta_K \doteq \operatorname{\mathrm{ord}}\widetilde G^{\bot}. \end{align*} $$

Moreover, by Lemma 3.10 (ii), we have

$$ \begin{align*} \operatorname{\mathrm{ord}} \widetilde G \cdot \operatorname{\mathrm{ord}} \widetilde G^{\bot} \doteq \Delta_J. \end{align*} $$

Since $\Delta _K \doteq \Delta _J$ by assumption, those two equations imply that $\operatorname {\mathrm {ord}} \widetilde G \doteq 1$ , which means that $\widetilde G$ is trivial. This is equivalent to G being $\mathbb {Z}$ -torsion. However, $H_1(X_J; \mathbb {Z}[t^{\pm 1}])$ is $\mathbb {Z}$ -torsion free (see Lemma A.7), and thus its submodule G is as well. Hence G is trivial, and the pairing induced by $\operatorname {\mathrm {Bl}}_J$ on $G^{\bot } / G \cong H_1(X_J; \mathbb {Z}[t^{\pm 1}])$ , to which $\operatorname {\mathrm {Bl}}_K$ is isometric, is just $\operatorname {\mathrm {Bl}}_J$ itself.▪

Proof Since $\zeta $ is irreducible, the ideal $(\zeta )$ is prime. Now, R is the localization of a PID at a prime ideal, and as such R is a DVR.▪

Proof (i) The sum of the inclusion maps gives a map $B \oplus B^{\bot } \to A$ that clearly preserves pairings. It just remains to see that this map is bijective. Injectivity follows from $B \cap B^{\bot } = \{0\}$ . Surjectivity follows since $\operatorname {\mathrm {ord}} B \oplus B^{\bot } \doteq \operatorname {\mathrm {ord}} A$ by Lemma 3.10 (ii), using that R is a PID.

(ii) By (i), we just have to check that $\langle a\rangle \cap \langle a\rangle ^{\bot } = \{0\}$ . Take an element $ra \in \langle a\rangle $ with $r\in R$ . If $ra\in \langle a\rangle ^{\bot }$ , then $\lambda (ra, a) = 0$ , and so $r\lambda (a,a) = 0$ . Hence $r \in (\tau ^{\nu \lambda (a,a)}) = (\tau ^{\nu a})$ , and thus $ra = 0$ .

(iii) If $a = 0$ , take $b = 0$ . If $a \neq 0$ , let $\tilde a = \tau ^{-1 + \nu a} a$ . Since $\lambda $ is nonsingular, there exists $b\in A$ such that $\lambda (\tilde a, b) \neq 0$ . It follows that $\nu \lambda (\tilde a, b) = 1$ and $\nu \lambda (a,b) = \nu a$ as desired.

(iv) Pick any $b\in A$ of maximal $\nu b = k$ . If $\nu \lambda (b,b) = k$ , just take $a = b$ . Else, by (iii), we may pick $c\in A$ such that $\nu \lambda (b,c) = k$ , i.e., $\lambda (b,c) = r\tau ^{-k}$ for $r\in R$ a unit. Note this implies $\nu c = k$ . Again, if $\nu \lambda (c,c) = k$ , just set $a = c$ . Else we have $\nu \lambda (b,b) < k$ and $\nu \lambda (c,c) < k$ . Pick $s\in R$ such that $s + \overline {s} = 1$ and set $a = b + s\overline {r}^{-1}c$ . This element a satisfies $\nu \lambda (a,a) = k$ , as one computes:

$$ \begin{align*} \begin{split} \tau^{k-1}\lambda(b + s\overline{r}^{-1}c,b + s\overline{r}^{-1}c) &= \tau^{k-1}(\lambda(b, s\overline{r}^{-1}c) + \lambda(s\overline{r}^{-1}c, b)) \\ &= \tau^{k-1}(\overline{s}\tau^{-k} + s\overline{\tau}^{-k}) = \tau^{-1}. \end{split} \end{align*} $$

(v) Note that $\operatorname {\mathrm {ord}} A \doteq \tau ^j$ for some $j\geq 0$ . The proof goes by induction over j. For $j = 0$ , A is trivial and the statement holds. If A is nontrivial, by (iv), there exist a nontrivial $a\in A$ with maximal $\nu a$ and $\nu \lambda (a,a) = \nu a$ . By (ii), A is isometric to $\langle a\rangle \oplus \langle a\rangle ^{\bot }$ , and by induction $\langle a\rangle ^{\bot }$ is isometric to an orthogonal sum of cyclic modules.

For the second statement, note that $\tau ^{\nu e_i}\lambda (e_i, e_i) = 0$ , so $\nu \lambda (e_i, e_i) \leq \nu e_i$ . Moreover, $\lambda $ is nonsingular, so there exists some $a\in A$ such that $\lambda (\tau ^{\nu e_i - 1} e_i, a) \neq 0$ . Because of the orthogonality of the sum, we have $a = a_i e_i + b$ with $a_i\in R\setminus \{0\}$ and $b\in \langle e_1, \ldots , e_{i-1}, e_{i+1}, \ldots , e_n\rangle $ . Then $\lambda (\tau ^{\nu e_i - 1} e_i, a) = a_i \tau ^{\nu e_i - 1} \lambda (e_i, e_i) \neq 0$ , and so $\nu \lambda (e_i, e_i) \geq \nu e_i$ .

(vi) Using (v), we may identify $(A, \lambda )$ with the orthogonal sum over cyclic modules with generators $e_1, \ldots , e_n$ , so we write elements of A as vectors with respect to that generator set. In this notation, the ith coordinate of a is $r_i \tau ^{-1+\nu e_i}$ for $r_i \in R$ a unit or 0. Since $\nu a=1$ , $a \neq 0$ , and by maximality of $\nu \tilde {a}$ , there exists i such that $r_i \neq 0$ and $\nu e_i = \nu \tilde a$ . This implies that the ith coordinate of $\tilde a$ is $r_i$ . Set b to be $e_i$ . Then $\nu b = \nu \tilde a$ and by the second part of (v), $\nu \lambda (e_i,e_i) = \nu e_i$ , so $\lambda (\tilde a,b) = \lambda (r_i e_i, e_i)$ , where $\nu \lambda (r_i e_i, e_i) = \nu e_i$ . Likewise $\nu \lambda (b,b) = \nu \lambda (e_i, e_i) = \nu e_i$ .

(vii) Choose $e_1, \ldots , e_n$ as in (vi). Let the ith coordinate of a be $r_i \tau ^{k_i}$ with $r_i\in R$ a unit or $0$ , and $k_i \geq 0$ . Reorder the $e_i$ such that there exists $j\in \{1, \ldots , n\}$ with $\nu (r_i \tau ^{k_i} e_i) = \nu (a)$ for $i \leq j$ and $\nu (r_i \tau ^{k_i} e_i) < \nu (a)$ for $i> j$ . If $k_i = 0$ for some $i\in \{1,\ldots , j\}$ , then $\nu (a) = \nu (e_i) = \nu \lambda (e_i, a)$ , contradicting the assumption that no such $b = e_i$ exists. So $k_i \geq 1$ for all $i\in \{1,\ldots , j\}$ . Set

$$ \begin{align*} c = -\sum_{i=1}^j r_i \tau^{k_i - 1} e_i. \end{align*} $$

Then, clearly, $\nu c = 1 + \nu a$ and $\nu (a + \tau c) < \nu a$ .▪

Proof Throughout this proof, let us write $\lambda \geq _{\mathfrak M} \lambda '$ if $T_n$ and $h_n$ as above exist. It is straightforward to check that the relation $\geq _{\mathfrak M}$ on $\mathfrak {M}(\mathbb {F})$ is transitive.

Now, $\lambda \geq _{\mathfrak M} \lambda '$ holds for trivial G. The main part of the proof is to show the statement for $\operatorname {\mathrm {ord}} G \doteq \tau $ . Having established that, let $\operatorname {\mathrm {ord}} G \doteq \tau ^j$ with $j\geq 2$ , and proceed by induction over j as follows. Pick $g\in G$ with $\nu g = 1$ . Let $\mu $ be the pairing induced on $\langle g\rangle ^{\bot } / \langle g\rangle $ by $\lambda $ . Note that the pairing induced on $(G / \langle g\rangle )^{\bot } / (G / \langle g\rangle )$ by $\mu $ is isometric to $\lambda '$ . Since $\operatorname {\mathrm {ord}}\langle g\rangle = \tau $ and $\operatorname {\mathrm {ord}} G / \langle g\rangle = \tau ^{j-1}$ , we may apply the induction hypothesis twice, and get $\lambda \geq _{\mathfrak M} \mu \geq _{\mathfrak M} \lambda '$ .

So let us now deal with the case that $\operatorname {\mathrm {ord}} G \doteq \tau $ , i.e., $G = \langle g\rangle $ with $g\in \Phi _1(A)\setminus \{0\}$ . Let $\tilde g \in A$ have maximal $\nu \tilde g =: k$ such that $g = \tau ^{k-1}\tilde g$ . We will distinguish two cases, depending on whether

(3.6) $$ \begin{align} \nu \lambda(\tilde g, \tilde g) = k \end{align} $$

holds. If it does, set $B = \langle \tilde g\rangle $ . Otherwise, we have $\nu \lambda (\tilde g, \tilde g) < k$ and pick $b\in A$ as in Lemma 3.13(vi), i.e.,

(3.7) $$ \begin{align} \nu b = \nu \lambda(\tilde g, b) = \nu \lambda(b,b) = k \end{align} $$

and set $B = \langle \tilde g, b\rangle \subseteq A$ . Let us check that $B \cap B^{\bot } = \{0\}$ . In case (3.6) holds, this is true by Lemma 3.13 (i). If (3.6) does not hold, let $a \in B \cap B^{\bot }$ be given. Since $a\in B$ , we have $a = r \tau ^{\ell } \tilde g + s \tau ^m b$ for some units $r,s \in R$ and $\ell , m\geq 0$ . We want to show $a = 0$ . For that, it is sufficient that $\ell \geq k$ and $m\geq k$ , since both $\tilde g$ and b are annihilated by $\tau ^k$ . We have $\lambda (a, b) = 0$ , since $a \in B^{\bot }$ and $b \in B$ , and thus

$$ \begin{align*} r \tau^{\ell} \lambda(\tilde g, b) + s \tau^m \lambda(b, b) = 0. \end{align*} $$

If both summands are zero, then $\ell \geq k$ and $m \geq k$ using (3.7), and so we are done. Otherwise, the two summands must have the same $\nu $ , and so $k - \ell = k - m$ , implying that $\ell = m$ . Now $\lambda (a, \tilde g) = 0$ implies that

$$ \begin{align*} r \tau^{\ell} \lambda(\tilde g, \tilde g) + s \tau^m \lambda(b, \tilde g) = 0. \end{align*} $$

If the second summand is zero, then $\ell \geq k$ , and we are done. Otherwise, the two summands must have the same $\nu $ , i.e., $\nu \lambda (\tilde g, \tilde g) - \ell = k - \ell $ , implying $\nu \lambda (\tilde g, \tilde g) = k$ contradicting the assumption that (3.6) does not hold. We have thus shown $B \cap B^{\bot } = \{0\}$ .

By Lemma 3.13 (i), it follows that A is isometric to $B \oplus B^{\bot }$ . Since $\langle g\rangle \subset B$ , we have $B^{\bot } \subset \langle g\rangle ^{\bot }$ . Thus, $\langle g\rangle ^{\bot } / \langle g\rangle $ is isometric to $C \oplus B^{\bot }$ where $C = (B \cap \langle g\rangle ^{\bot }) / \langle g\rangle $ with the pairing induced by $\lambda $ . Hence it suffices to show $\lambda |_{B\times B} \geq _{\mathfrak {M}} \lambda |_{C\times C}$ .

If (3.6) holds, then B is cyclically generated by $\tilde g$ with $\nu \tilde g = k$ , where $k\geq 2$ follows. So, $\Phi _n(B)$ is trivial for $n \neq k$ , and $\Phi _k(\lambda _{B\times B})$ is isometric to the pairing on $\mathbb {F}$ sending $(e_1,e_1)\mapsto \tau ^k\lambda (\tilde g, \tilde g) \in R / (\tau ) = \mathbb {F}$ . Observe that C is also cyclic, generated by $a = [\tau \tilde g]$ , with $\nu a = k - 2$ and $\lambda (a,a) = \tau ^2 \lambda (\tilde g, \tilde g)$ . So $\Phi _n(C)$ is trivial for all $n\geq k-2$ ; and $\Phi _{k-2}(\lambda _{C\times C}) = \Phi _k(\lambda _{B\times B})$ if $k> 2$ (if $k = 2$ , then C is trivial). To demonstrate $\lambda |_{B\times B} \geq _{\mathfrak {M}} \lambda |_{C\times C}$ , one may take $h_n = 0$ for all n, $T_n = 0$ for all $n\neq k$ , and $T_k = \Phi _k(\lambda _{B\times B})$ .

If (3.6) does not hold, recall that $B = \langle \tilde g, b\rangle $ . So $\Phi _n(B)$ is trivial for $n\neq k$ and $\Phi _k(B)$ is two-dimensional, with basis $[\tilde g], [b]$ . The pairing $\Phi _k(\lambda _{B\times B})$ is metabolic, since $([\tilde g], [\tilde g])$ is sent to $0$ . So $\Phi _k(\lambda _{B\times B}) = H$ . One calculates that $\langle g\rangle ^{\bot } \cap B = \langle \tilde g, \tau b\rangle $ , so $\Phi _n(C)$ is trivial for $n \geq k - 1$ ; and $\Phi _{k-1}(\lambda _{C\times C}) = H$ if $k> 1$ (if $k = 1$ , then C is trivial). To demonstrate $\lambda |_{B\times B} \geq _{\mathfrak {M}} \lambda |_{C\times C}$ , one may take $T_n = 0$ for all n, $h_n = 1$ if $n \leq k$ and $n \equiv k \pmod {2}$ , and $h_n = 0$ otherwise.▪

Proof of Proposition 3.4 If $e^{\pi i x}$ is not the root of the Alexander polynomial of any knot, then $\sigma _x$ is a topological concordance invariant. For such x, the stated inequalities thus simply read $0\geq 0\geq 0$ . So from now on, we may assume that $e^{\pi ix}$ is the root of some Alexander polynomial. In fact, it will be sufficient to assume $e^{\pi i x}\notin \mathbb {R}$ (equivalently, that $x\notin \mathbb {Z}$ ). Let $\zeta (t) = t^{-1} - 2\cos x + t$ . The localization $\mathbb {R}[t^{\pm 1}]_{(\zeta )}$ of $\mathbb {R}[t^{\pm 1}]$ at $(\zeta )$ is a DVR satisfying conditions (3.4) and (3.5) by Lemma 3.12.

Note that $\mathbb {R}[t^{\pm 1}] / (\zeta )$ is isomorphic to $\mathbb {C}$ via the isomorphism sending t to $e^{\pi ix}$ . So $\mathfrak {M}(\mathbb {R}[t^{\pm 1}]/(\zeta ))$ is isomorphic to the monoid $\mathbb {N}^2$ via the signature. Indeed, send $[\mu ]\in \mathfrak {M}(\mathbb {R}[t^{\pm 1}]/(\zeta ))$ to $(\mu _1, \mu _{-1}) \in \mathbb {N}^2$ , where $\mu _1$ and $\mu _{-1}$ are the dimensions of maximal subspaces on which $\mu $ is positive definite and negative definite, respectively. Let us write $\sigma (\mu ) = \mu _1 - \mu _{-1}$ .

For a knot L, let us abbreviate $\Psi _i(L) := \Phi _i(\operatorname {\mathrm {Bl}}^{\mathbb {R}[t^{\pm 1}]_{(\zeta )}}_L)$ . Denote the difference between the signature and the averaged signature at $e^{\pi ix}$ by $d_x(L)$ :

$$ \begin{align*} d_x(L) = \sigma_x(L) - \tfrac12\lim_{\varepsilon\to 0} \sigma_{x+\varepsilon}(L) + \sigma_{x-\varepsilon}(L). \end{align*} $$

Note that $d_x(J) - d_x(K) = \sigma _x(J) - \sigma _x(K)$ , since $\sigma _{x\pm \varepsilon }(J) = \sigma _{x\pm \varepsilon }(K)$ for almost all $\varepsilon $ . One may read $\deg _x(L), \eta _x(L)$ , and $d_x(L)$ from $\Psi _i(L)$ (see [Reference LevineLev89], where the first and second equations are implicit, and the third equation is given in Theorem 2.3):

$$ \begin{align*} \deg_x(L) = \sum_{i=1}^\infty i\cdot \dim \Psi_i(L),\quad \eta_x(L) = \sum_{i=1}^\infty \dim \Psi_i(L),\quad d_x(L) = -\sum_{i=1}^\infty \sigma(\Psi_{2i}(L)). \end{align*} $$

For the two knots $J, K$ with $[J]\geq _S [K]$ , Proposition A.1 yields a submodule $\tilde G \subset H_1(X_J; \mathbb {R}[t^{\pm 1}]_{(\zeta )})$ such that the pairing induced on $\tilde G^{\bot } / \tilde G$ by $\operatorname {\mathrm {Bl}}^{\mathbb {R}[t^{\pm 1}]_{(\zeta )}}_J$ is isometric to $\operatorname {\mathrm {Bl}}^{\mathbb {R}[t^{\pm 1}]_{(\zeta )}}_K$ . By Lemma 3.14, this implies that for all $n\geq 1$ , there are $T_n\in \mathfrak {M}(\mathbb {F})$ and $h_n\in \mathbb {N}$ such that $T_1$ is trivial and $h_{n} \geq h_{n+2}$ , and

$$ \begin{align*} T_n \oplus h_n \cdot H \oplus \bigoplus_{i=0}^{\infty} \Psi_{n+2i}(K) = h_{n+1}\cdot H \oplus \bigoplus_{i=0}^{\infty} \Psi_{n+2i}(J). \end{align*} $$

For $n = 2$ , this implies

(3.8) $$ \begin{align} \sigma(T_2) + \sum_{i=1}^{\infty} \sigma(\Psi_{2i}(K)) = \sum_{i=1}^{\infty} \sigma(\Psi_{2i}(J)). \end{align} $$

It also implies

(3.9) $$ \begin{align} T_n \oplus T_{n+1} \oplus (h_n - h_{n+2})\cdot H \oplus \bigoplus_{i=0}^{\infty} \Psi_{n+i}(K) = \bigoplus_{i=0}^{\infty} \Psi_{n + i}(J) \end{align} $$

for all $n\geq 1$ . It follows that

$$ \begin{align*} \begin{split} &|\sigma_x(J) - \sigma_x(K)| = |d_x(J) - d_x(K)| = \Bigl|\sum_{i=1}^{\infty} \sigma(\Psi_{2i}(J)) - \sigma(\Psi_{2i}(K))\Bigr| \stackrel{(3.8)}{=} |\sigma(T_2)| \\ &\leq\kern-1pt \dim(T_2)\kern-1pt \leq\kern-1pt \dim(T_2\kern-1pt +\kern-1pt (h_1\kern-1pt -\kern-1pt h_3)\kern-1pt\cdot\kern-1pt H)\kern-1pt \stackrel{(3.9)}{=}\kern-1pt \sum_{i=1}^{\infty} \dim \Psi_i(J) - \dim \Psi_i(K) = \eta_x(J) - \eta_x(K), \end{split} \end{align*} $$

proving the second desired inequality. To prove the first inequality:

(■) $$ \begin{align*} \bigoplus_{n=2}^{\infty}\bigg(T_n \oplus T_{n+1} \oplus (h_n - h_{n+2})\cdot H \oplus \bigoplus_{i=0}^{\infty} \Psi_{n+i}(K)\bigg) & \stackrel{(3.9)}{=} \bigoplus_{n=2}^{\infty}\bigoplus_{i=0}^{\infty} \Psi_{n + i}(J) \\ \Rightarrow\quad (h_2 + h_3)\cdot H \oplus T_2 \oplus \bigoplus_{i=3}^{\infty}2T_i \oplus \bigoplus_{i=1}^{\infty} (i - 1)\Psi_{i}(K) & = \bigoplus_{i=1}^{\infty} (i - 1)\Psi_{i}(J) \\ \Rightarrow\quad \sum_{i=1}^{\infty} (i - 1)\dim \Psi_{i}(K) & \leq \sum_{i=1}^{\infty} (i - 1)\dim\Psi_{i}(J) \\[1.5ex] \Rightarrow\quad \eta_x(J) - \eta_x(K) & \leq \deg_x(J) - \deg_x(K). \end{align*} $$

▪

Proof We consider the homology long exact sequence for $(X_C, X_J)$ :

$$ \begin{align*} \begin{split} H_1(X_C; \mathbb{Z} \pi_C) & \to H_1(X_C, X_J; \mathbb{Z} \pi_C) \to \\ H_0(X_J; \mathbb{Z} \pi_C) \to H_0(X_C; \mathbb{Z} \pi_C) &\to H_0(X_C, X_J; \mathbb{Z} \pi_C) \to 0. \end{split} \end{align*} $$

Since $\iota _J$ is surjective, the induced homomorphism $H_0(X_J; \mathbb {Z} \pi _C) \to H_0(X_C; \mathbb {Z} \pi _C)$ is an isomorphism. Also, we have $H_1(X_C; \mathbb {Z} \pi _C) = H_1(\widetilde {X}_C; \mathbb {Z}) = 0$ , which proves the lemma.▪

Proof We consider the universal coefficient spectral sequence for homology groups [Reference RotmanRo09, Theorem 10.90]:

$$ \begin{align*} E_{p, q}^2 = \operatorname{\mathrm{Tor}}_p^{\mathbb{Z} \pi_C}(H_q(X_C, X_J; \mathbb{Z} \pi_C), R[t^{\pm 1}]^n) \Rightarrow H_{p+q}^{\alpha \otimes (\pm \phi_C)}(X_C, X_J; R[t^{\pm 1}]^n), \end{align*} $$

where the $\operatorname {\mathrm {Tor}}$ group on the left hand side uses $\alpha \otimes (\pm \phi _C)$ . It follows from Lemma 4.2 that

$$ \begin{align*} \operatorname{\mathrm{Tor}}_1^{\mathbb{Z} \pi_C}(H_0(X_C, X_J; \mathbb{Z} \pi_C), R[t^{\pm 1}]^n) = H_1(X_C, X_J; \mathbb{Z} \pi_C) \otimes_{\mathbb{Z} \pi_C} R[t^{\pm 1}]^n = 0. \end{align*} $$

Thus, all the terms on the line $p+q=1$ of $E^2$ page of the spectral sequence vanish, and the desired equation holds.▪

Proof of Proposition 4.1 Consider the long exact sequence of pairs, and apply Lemma 4.3.▪

Proof Since the proof of (i) is similar to that of (ii), we only prove (ii). By Poincaré–Lefschetz duality we have an isomorphism

$$ \begin{align*} H_{4-i}^{\alpha \otimes \phi_C}(X_C, X_K; R[t^{\pm 1}]^n) \cong H_{\alpha \otimes \phi_C}^i(X_C, X_J; R[t^{\pm 1}]^n) \end{align*} $$

(See for instance [Reference Friedl, Nagel, Orson and PowellFNOP19, Theorem A.15] for the case of twisted coefficients.) Also, we have an isomorphism

$$ \begin{align*} \operatorname{Hom}_{\mathbb{Z} \pi_C}(C_*(\widetilde{X}_C, \widetilde{X}_J), R[t^{\pm 1}]^n) \cong \operatorname{Hom}_{R[t^{\pm 1}]}(C_*(\widetilde{X}_C, \widetilde{X}_J) \otimes_{\mathbb{Z} \pi_C} R[t^{\pm 1}]^n, R[t^{\pm 1}]), \end{align*} $$

where the tensor product on the right hand side uses $\alpha ^\dagger \otimes (-\phi _C)$ , given by

$$ \begin{align*} f \mapsto ( (c \otimes v) \mapsto f(c)^T \bar{v}) \end{align*} $$

for $f \in \operatorname {Hom}_{\mathbb {Z} \pi _C}(C_*(\widetilde {X}_C, \widetilde {X}_J), R[t^{\pm 1}]^n)$ , $c \in C_*(\widetilde {X}_C, \widetilde {X}_J)$ and $v \in R[t^{\pm 1}]^n$ . Now the lemma follows from the universal coefficient spectral sequence for cohomology groups [Reference LevineLev77, Theorem 2.3].▪

Proof We first prove (i). We consider the homology long exact sequence for $(X_C, X_K)$ :

$$ \begin{align*} H_1^{\alpha \otimes \phi_C}(X_C; R[t^{\pm 1}]^n) \to H_1^{\alpha \otimes \phi_C}(X_C, X_K; R[t^{\pm 1}]^n) \to H_0^{\alpha \circ \iota_K \otimes \phi_K}(X_K; R[t^{\pm 1}]^n). \end{align*} $$

It is easy to check that $H_0^{\alpha \circ \iota _K \otimes \phi _K}(X_K; R[t^{\pm 1}]^n)$ is a torsion module. Since $\Delta _J^{\alpha \circ \iota _J} \neq 0$ , the module $H_1^{\alpha \circ \iota _J \otimes \phi _J}(X_J; R[t^{\pm 1}]^n)$ is torsion, and it follows from Proposition 4.1 that $H_1^{\alpha \otimes \phi _C}(X_C; R[t^{\pm 1}]^n)$ is also a torsion module. Therefore (i) follows.

Next, we prove (iii). It follows from Lemma 4.5 and (i) that

$$ \begin{align*} \operatorname{\mathrm{Ext}}_1^{R[t^{\pm 1}]}(H_0^{\alpha \otimes \phi_C}(X_C, X_K; R[t^{\pm 1}]^n), R[t^{\pm 1}]) &= 0, \\ \operatorname{Hom}_{R[t^{\pm 1}]}(H_1^{\alpha \otimes \phi_C}(X_C, X_K; R[t^{\pm 1}]^n), R[t^{\pm 1}]) &=0. \end{align*} $$

Thus all the terms on the line $p+q=1$ of $E^2$ page of the spectral sequence of Lemma 4.6 (i) vanish. Therefore, (iii) follows.

Finally, we prove (ii). Let $(X, Y)$ be a finite CW-pair homotopy equivalent to $(X_C, X_J)$ , and we identify $\pi _1(X)$ with $\pi _C$ . By (iii) and Lemmas 4.3 and 4.5, we have

$$ \begin{align*} &\operatorname{\mathrm{rank}}_{R[t^{\pm 1}]} H_2^{\alpha^{\dagger} \otimes (-\phi_C)}(X_C, X_J; R[t^{\pm 1}]^n)\\ &\quad = \sum_{i=0}^3 (-1)^i \operatorname{\mathrm{rank}}_{R[t^{\pm 1}]} H_i^{\alpha^{\dagger} \otimes (-\phi_C)}(X_C, X_J; R[t^{\pm 1}]^n) \\ &\quad = \sum_{i=0}^3 (-1)^i \operatorname{\mathrm{rank}}_{R[t^{\pm 1}]} H_i^{\alpha^{\dagger} \otimes (-\phi_C)}(X, Y; R[t^{\pm 1}]^n) \\ &\quad = \sum_{i=0}^3 (-1)^i \operatorname{\mathrm{rank}}_{R[t^{\pm 1}]} \overline{C}_i(\widetilde{X}, \widetilde{Y}) \otimes_{\mathbb{Z} \pi_C} R[t^{\pm 1}]^n \\ &\quad = n \sum_{i=0}^3 (-1)^i \operatorname{\mathrm{rank}}_{\mathbb{Z}} C_i(X, Y) \\ &\quad = n \chi(X, Y) = n \chi(X_C, X_J) = n ( \chi(X_C) - \chi(X_J)) = 0, \end{align*} $$

where we consider the cellular chain complexes $C_*(X, Y)$ and $C_*(\widetilde {X}, \widetilde {Y})$ . Therefore, (ii) follows, which completes the proof.▪

Proof of Proposition 4.4 It follows from Lemmas 4.3, 4.5 and 4.7(ii) that

$$ \begin{align*} \operatorname{\mathrm{Ext}}_2^{R[t^{\pm 1}]}(H_0^{\alpha^\dagger \otimes (-\phi_C)}(X_C, X_J; R[t^{\pm 1}]), R[t^{\pm 1}]^n) &=0, \\ \operatorname{\mathrm{Ext}}_1^{R[t^{\pm 1}]}(H_1^{\alpha^\dagger \otimes (-\phi_C)}(X_C, X_J; R[t^{\pm 1}]), R[t^{\pm 1}]^n) &=0, \\ \operatorname{Hom}_{R[t^{\pm 1}]}(H_2^{\alpha^\dagger \otimes (-\phi_C)}(X_C, X_J; R[t^{\pm 1}]), R[t^{\pm 1}]^n) &=0. \end{align*} $$

Thus, all the terms on the line $p+q=2$ of $E^2$ page of the spectral sequence of Lemma 4.6(ii) vanish. Therefore, $H_2^{\alpha \otimes \phi _C}(X_C, X_K; R[t^{\pm 1}]^n) = 0$ , and the proposition follows.▪

Proof We may suppose that $\Delta _J^{\alpha \circ \iota _J} \neq 0$ , since $g~|~0$ for every $g \in R[t^{\pm 1}]$ , so if $\Delta _J^{\alpha \circ \iota _J} = 0$ then the theorem trivially holds. Let $\Delta _C^{\alpha }(t)$ be the order of $H_1^{\alpha \otimes \phi _C}(X_C; R[t^{\pm 1}]^n)$ . Since orders are multiplicative for a short exact sequence of finitely generated $R[t^{\pm 1}]$ -modules, it follows from Propositions 4.1 and 4.4 that

$$ \begin{align*} \Delta_C^{\alpha} ~|~ \Delta_J^{\alpha \circ \iota_J} ~\text{and}~ \Delta_K^{\alpha \circ \iota_K} ~|~ \Delta_C^{\alpha}, \end{align*} $$

which proves the theorem.▪

Proof Since A is disjoint from F, we have the image $F'$ of F by the inclusion map $X_A \to X_S$ . We see at once that $F'$ is a Seifert surface of S, and its Seifert form is isomorphic to V, which proves the lemma.▪

Proof First, we prove (i). A ribbon concordance from J to the unknot can be encoded as a sequence of n saddle moves from J to the unlink with $n+1$ components. If A bounds a disc in $S^3$ intersecting K in w points, then one may find a sequence of $wn$ saddle moves from S to the disjoint union of K with a $wn$ -component unlink. To find them, for each saddle move on J, perform w parallel copies of it on S as in Figure 2. This converts S to a disjoint union of K and a $wn$ -component unlink. Cap off the unlink with $wn$ discs to complete the construction of a ribbon concordance from S to K. This completes the proof of (i).

Figure 2:

Parallel copies of a saddle move in the case $w=4$ .

Now we prove (ii). Since J is a homotopy ribbon knot, J is homotopy ribbon concordant to the unknot U, via a homotopy ribbon concordance C. So the exterior $X_C$ has boundary $X_J \cup S^1 \times S^1 \times I \cup -X_U$ . Recall that $X_{S(K,J,A)} = X_{K \cup A} \cup _{\partial \overline {\nu A}} X_J$ and that, trivially, $X_K = X_{K \cup A} \cup _{\partial \overline {\nu A}} X_U$ . Glue to form:

$$ \begin{align*} Y:= X_{K \cup A} \times I \cup_{\partial \overline{\nu A} \times I} X_C \end{align*} $$

along the $S^1 \times S^1 \times I$ part of the boundary of $X_C$ . Note that $\partial Y = X_S \cup -X_K$ and Y is a $\mathbb {Z}$ –homology cobordism with $\pi _1(Y)$ normally generated by a meridian of K or of S. Therefore, $Y \cup S^1 \times D^2 \times I$ is an 4-dimensional h-cobordism from $S^3$ to itself. The five-dimensional h-cobordism theorem [Reference Freedman and QuinnFQ90, Theorem 7.1A] may be applied to deduce that $Y \cup S^1 \times D^2 \times I \cong S^3 \times I$ , and therefore Y is a concordance exterior. The image of $S^1 \times {0} \times I$ under this homeomorphism gives rise to a concordance from $S(K,J,A)$ to K. The fact that $X_C$ is a homotopy ribbon concordance exterior easily implies that Y is too. This completes the proof of (ii) and of the lemma.▪

Proof The groups $\Gamma _J^{r, p}$ and $\Gamma _K^{r, p}$ are quotients of $\pi _J / \pi _J^{(2)}$ and $\pi _K / \pi _K^{(2)}$ , respectively, and $\psi $ induces an isomorphism $\Gamma _J^{r, p} \to \Gamma _K^{r, p}$ . Since the actions of $\pi _J / \pi _J^{(2)}$ on $\Gamma _J^{r, p}$ and $\pi _K / \pi _K^{(2)}$ on $\Gamma _K^{r, p}$ are compatible with $\psi $ and the induced isomorphism, $\overline {\alpha }_K^{r, p} \circ \psi $ and $\overline {\alpha }_J^{r, p}$ are conjugate, and so are $\psi ^* \alpha _K^{r, p}$ and $\alpha _J^{r, p}$ . Therefore, the lemma follows from Remark 2.1 (2).▪

Proof of Theorem 6.4 The proof is similar to that of [Reference Friedl and PowellFP12, Proposition 4.1].

Let $\mu $ be a meridian in $X_K$ . Since the induced homomorphism $H_*(\mu ; \mathbb {F}_p) \to H_*(X_K; \mathbb {F}_p)$ is an isomorphism, we have $H_*(X_K, \mu ; \mathbb {F}_p) = 0$ . We first prove that $H_*^{\alpha _K^{r, p} \otimes \phi _K}(X_K, \mu ; \mathbb {Q}(t)^n) = 0$ , following a standard argument of chain homotopy lifting (see for instance [Reference Cochran, Orr and TeichnerCOT03, Proposition 2.10]).

We pick a CW complex structure of $X_K$ containing $\mu $ as a subcomplex, and set $C_*$ and $\widetilde {C}_*$ to be the cellular chain complexes $C_*(X_K, \mu ) \otimes _{\mathbb {Z}} \mathbb {F}_p$ and $C_*(\widetilde {X}_K, \widetilde {\mu })$ , respectively, where $\widetilde {\mu }$ is the pullback of $\mu $ in the universal cover $\widetilde {X}_K$ of $X_K$ . Let $h_i \colon C_i \to C_{i+1}$ be a chain contraction for $C_*$ , so that $\partial h_i + h_{i-1} \partial $ is the identity map on $C_i$ . Since the natural map $\widetilde {C}_* \to C_*$ is surjective and all modules in $C_*$ and $\widetilde {C}_*$ are free, there exists a lift $\tilde {h}_i \colon \widetilde {C}_i \to \widetilde {C}_{i+1}$ of $h_i$ for each i. We write $f_i \colon \widetilde {C}_i \to \widetilde {C}_i$ for the chain map $\partial \tilde {h}_i + \tilde {h}_{i-1} \partial $ .

Since the chain map $C_i \to C_i$ induced by $f_i$ is the identity map and $\alpha _K^{r, p}|_{\operatorname {Ker} \phi _K}$ factors through the p-group

$$ \begin{align*} H_K \otimes_{\mathbb{Z}[t^{\pm 1}]} \mathbb{F}_p[t^{\pm 1}] / (t^r-1), \end{align*} $$

it follows from Theorem 6.5 that the chain map

$$ \begin{align*} \widetilde{C}_i \otimes_{\mathbb{Z} \pi_K} \mathbb{Q}(t)^n \to \widetilde{C}_i \otimes_{\mathbb{Z} \pi_K} \mathbb{Q}(t)^n \end{align*} $$

induced by $f_i$ is injective, and is therefore an isomorphism. Thus, $\tilde {h}_i$ induces a chain homotopy between the zero map and the chain isomorphism. Therefore, $H_*^{\alpha _K^{r, p} \otimes \phi _K}(X_K, \mu ; \mathbb {Q}(t)^n) = 0$ .

By a direct computation we have $H_*^{(\alpha _K^{r, p} \otimes \phi _K) \circ \iota }(\mu ; \mathbb {Q}(t)^n) = 0$ , where $\iota \colon \pi _1(\mu ) \to \pi _K$ is the induced homomorphism. Now it follows from the homology long exact sequence for $(X_K, \mu )$ that $H_*^{\alpha _{r, p} \otimes \phi _K}(X_K; \mathbb {Q}(t)^n) = 0$ . Hence $H_1^{\alpha _{r, p} \otimes \phi _K}(X_K; \mathbb {Z}[t^{\pm 1}]^n)$ is a torsion module, which completes the proof.▪