# Obstructing Two-torsion in the Rational Knot Concordance Group

Jaewon Lee  Department of Mathematical Sciences, Korea Advanced Institute for
Science and Technology<EMAIL_ADDRESS>
###### Abstract.

It is well known that there are many 2-torsion elements in the classical knot
concordance group. On the other hand, it is not known if there is any torsion
element in the rational knot concordance group $\mathcal{C}_{\mathbb{Q}}$. Cha
defined the algebraic rational concordance group $\mathcal{AC}_{\mathbb{Q}}$,
an analogue of the classical algebraic concordance group, and showed that
$\mathcal{AC}_{\mathbb{Q}}\cong\mathbb{Z}^{\infty}\oplus\mathbb{Z}_{2}^{\infty}\oplus\mathbb{Z}_{4}^{\infty}$.
The knots that represent 2-torsions in $\mathcal{AC}_{\mathbb{Q}}$ potentially
have order $2$ in $\mathcal{C}_{\mathbb{Q}}$. In this paper, we provide an
obstruction for knots of order $2$ in $\mathcal{AC}_{\mathbb{Q}}$ from being
of finite order in $\mathcal{C}_{\mathbb{Q}}$. Moreover, we give a family
consisting of such knots that generates an infinite rank subgroup of
$\mathcal{C}_{\mathbb{Q}}$. We also note that Cha proved that in higher
dimensions, the algebraic rational concordance order is the same as the
rational knot concordance order. Our obstruction is based on the localized von
Neumann $\rho$-invariant.

###### 1991 Mathematics Subject Classification:

57K10

##  1\. Introduction

Two knots in $S^{3}$ are said to be concordant if they cobound a properly
embedded locally flat annulus in $S^{3}\times I$. The concordance classes form
an abelian group $\mathcal{C}$ under connected sum, called the knot
concordance group. The identity element of $\mathcal{C}$ is the concordance
class of knots bounding an embedded locally flat disk in $B^{4}$, called slice
knots. Since the inverse for a knot $K$ in $\mathcal{C}$ is represented by its
reversed mirror image $r\overline{K}$, any non-slice knot $K$ which is
negative amphichiral, i.e., $K$ is isotopic to $r\overline{K}$, represents a
2-torsion element in $\mathcal{C}$. There are many such knots like the figure-
eight knot, and in fact, we know that $\mathcal{C}$ contains
$\mathbb{Z}_{2}^{\infty}$ as a subgroup. We make a remark that Gordon [Kir97,
Problem 1.94] conjectured that any 2-torsion element in $\mathcal{C}$ would be
concordant to a negative amphichiral knot. Naturally, we can ask if there are
any other torsions in $\mathcal{C}$.

######  Question 1.

[Kir97, Problem 1.32] Does $\mathcal{C}$ contain a torsion element of order
other than $2$? In particular, does $\mathcal{C}$ contain an order $4$
element?

A good candidate comes from the algebraic concordance group $\mathcal{AC}$.
Recall that Levine [Lev69] defined the following surjective homomorphism:

$\phi:\mathcal{C}\twoheadrightarrow\mathcal{AC}$

and showed that $\mathcal{AC}$ is isomorphic to
$\mathbb{Z}^{\infty}\oplus\mathbb{Z}_{2}^{\infty}\oplus\mathbb{Z}_{4}^{\infty}$.
Moreover, he proved that for odd higher-dimension, $\phi$ is an isomorphism.

A knot is said to be algebraically slice if its image in $\mathcal{AC}$ is
trivial. In contrast to higher dimensions, $\ker\phi$ turned out to be
nontrivial in the classical dimension, as proved by Casson and Gordon [CG78,
CG86]. The provided examples were shown by Jiang [Jia81] to generate an
infinite rank subgroup in $\ker\phi\leq\mathcal{C}$. An element in
$\mathcal{C}$ whose image under $\phi$ has order $4$ is a potential candidate
for a 4-torsion in $\mathcal{C}$. Livingston and Naik [LN99, LN01] (see also
[JN07, GL21]) proved that certain knots having order $4$ in $\mathcal{AC}$ are
of infinite order in $\mathcal{C}$.

In this article, we consider an analogous question for the rational knot
concordance group. A knot in $S^{3}$ is called rationally slice if it bounds a
locally flat embedded disk in some rational homology $4$-ball. The rational
knot concordance group, or briefly the rational concordance group,
$\mathcal{C}_{\mathbb{Q}}$ is formed by the quotient of $\mathcal{C}$ modulo
rationally slice knots. A negative amphichiral knot is called strongly
negative amphichiral if the isotopy from $K$ to $r\overline{K}$ is an
involution on $S^{3}$. The figure-eight knot is the simplest nontrivial
example and was proved to be rationally slice by Cochran, based on [FS84].
Kawauchi [Kaw09] proved that any strongly negative amphichiral knot is
rationally slice.111Such a knot is smoothly rationally slice, i.e., it bounds
a smoothly embedded disk in a rational ball. Kim and Wu [KW18] showed that
negative amphichiral knots satisfying certain properties are (smoothly)
rationally slice. It is not known if there exists a negative amphichiral knot
which is not rationally slice. This motivates the following question:

######  Question 2.

Does $\mathcal{C}_{\mathbb{Q}}$ contain any torsion element? In particular,
does $\mathcal{C}_{\mathbb{Q}}$ contain an order $2$ element?

Again, a good candidate for 2-torsion comes from the algebraic rational
concordance group $\mathcal{AC}_{\mathbb{Q}}$ defined by Cha [Cha07]. He also
defined the surjective homomorphism
$\phi_{\mathbb{Q}}:\mathcal{C}_{\mathbb{Q}}\twoheadrightarrow\mathcal{AC}_{\mathbb{Q}}$
and proved that

$\mathcal{AC}_{\mathbb{Q}}\cong\mathbb{Z}^{\infty}\oplus\mathbb{Z}_{2}^{\infty}\oplus\mathbb{Z}_{4}^{\infty}.$

We remark that Cha also proved that
$\phi_{\mathbb{Q}}:\mathcal{C}_{\mathbb{Q}}\twoheadrightarrow\mathcal{AC}_{\mathbb{Q}}$
is an isomorphism in higher dimensions. As one can observe in the following
commutative diagram:

${\mathcal{C}}$${\mathcal{C}_{\mathbb{Q}}}$${\mathcal{AC}}$${\mathcal{AC}_{\mathbb{Q}}}$$\scriptstyle{\psi}$$\scriptstyle{\phi}$$\scriptstyle{\phi_{\mathbb{Q}}}$$\scriptstyle{\overline{\psi}}$

the knots that represent 2-torsions in $\mathcal{AC}_{\mathbb{Q}}$ potentially
have order $2$ in $\mathcal{C}_{\mathbb{Q}}$. We obstruct some of them from
being actually torsion elements.

######  Theorem 1.1.

There are knots of order $2$ in $\mathcal{AC}_{\mathbb{Q}}$ which generate
$\mathbb{Z}^{\infty}$ in $\mathcal{C}_{\mathbb{Q}}$.

Theorem 1.1 implies a difference between higher dimensional rational
concordance groups and the rational concordance group in classical dimension.

To prove Theorem 1.1, we construct a rational slice obstruction by employing
the von Neumann $\rho$-invariant. It was developed as a representation
invariant of 3-manifolds [CG85]. Cochran, Orr, and Teichner [COT03] introduced
the $\rho$-invariant of the $0$-surgery of a knot as a knot concordance
invariant to find a filtration structure, called solvable filtration, on
$\mathcal{C}$. Cochran, Harvey, and Leidy [CHL09] defined the first order
signature $\rho^{(1)}(K)$ for a knot $K$, based on the von Neumann
$\rho$-invariant, and used it to prove that each layer of the filtration
contains an infinite rank subgroup (see also [COT04, Cha07, Har08, CHL11a,
CHL11b, Dav14, Kim20, DPR21]). Using the first order signature, our
obstruction can be summarized as follows:

######  Theorem 1.2.

Let $\\{K_{i}\\}$ be a set of knots whose order are finite in
$\mathcal{AC}_{\mathbb{Q}}$ and $\Delta_{i}$ be the Alexander polynomial of
$K_{i}$. If $\Delta_{i}$’s are strongly irreducible and pairwisely strongly
coprime, and $\rho^{(1)}(K_{i})\neq 0$, then $\\{K_{i}\\}$ is linearly
independent in $\mathcal{C}_{\mathbb{Q}}$.

Notice that the finite order condition in Theorem 1.2 does not include the
algebraically rationally slice cases because of strong irreducibility of
$\Delta_{i}$, by generalized Fox-Milnor condition [Cha07, CFHH13].

Figure 1. $n$-twist knot with a positive clasp.

We obtain an infinite family of knots satisfying the condition in Theorem 1.2,
thus Theorem 1.1 follows. Such knots of finite order in
$\mathcal{AC}_{\mathbb{Q}}$ arise from the $n$-twist knots $K_{n}$ with
nonnegative $n$. The $n$-twist knot is the positive Whitehead double of the
unknot with $n$ twists as illustrated in the Figure 1. Levine [Lev69]
classified their algebraic concordance order. Bullock and Davis [BD12] proved
that whenever $n$ is not a perfect power and $K_{n}$ is not algebraically
slice, the Alexander polynomials $\Delta_{n}$ of $K_{n}$ are strongly
irreducible. They also showed that the $\Delta_{n}$’s are pairwisely strongly
coprime. Moreover, Davis [Dav12a, Dav12b] proved that the first order
signature $\rho^{(1)}(K_{n})$ does not vanish for infinitely many $K_{n}$ of
order $2$ in $\mathcal{AC}$.

######  Theorem 1.3.

$n$-twist knots with order 2 in $\mathcal{AC}_{\mathbb{Q}}$ generate an
infinite rank subgroup in $\mathcal{C}_{\mathbb{Q}}$.

It was previously shown by Davis [Dav12a, Dav12b] that $K_{n}$ which are of
order 2 in $\mathcal{AC}$ generate an infinite rank subgroup in $\mathcal{C}$.
(See also [Kim05, Ilt22].) As one can see, the Theorem 1.2 covers the case for
order $4$ in $\mathcal{AC}_{\mathbb{Q}}$, but our examples only involve order
2 cases due to the general difficulty in computing $\rho^{(1)}$. We do not
know if there is a knot of order $4$ in $\mathcal{AC}_{\mathbb{Q}}$ that
satisfies the conditions of Theorem 1.2.

We make a remark on our twist knot $K_{n}$. Recall that there are several
tools to obstruct rational sliceness. For example, the knot signature and the
Levine-Tristram signature function vanish for the rationally slice knots.
[CO93, CK02]. Moreover, many Heegaard Floer theoretic invariants such as
$\tau$ [OS03, Ras03], $\epsilon$ [Hom14], $\nu^{+}$ [HW16], and $\Upsilon$
[OSS17] also vanish for the rationally slice knots, as the $1$-surgery of a
rationally slice knot bounds a rational ball.

For our $n$-twist knots $K_{n}$ in target, however, all the aforementioned
invariants vanish. 222One can take our $K_{n}$ with infection by a knot whose
$\tau$ is non-zero and use $\Upsilon$ to prove Theorem 1.1 in smooth category.
This is because the $n$-twist knots $K_{n}$ with $n\geq 0$ are $0$-bipolar
[CHH13], and in particular, they are BPH-slice [MP23]. In other words,

$\sigma(K_{n})=\sigma_{K_{n}}(\omega)=\tau(K_{n})=\epsilon(K_{n})=\nu^{+}(K_{n})=\Upsilon_{K_{n}}(t)=0.$

On the other hand, we introduce the notion of complexity [COT03, Cha07] on the
von Neumann $\rho$-invariant to prove Theorem 1.3. The complexity of a
rationally slice knot is, roughly speaking, the positive integer $c$ counting
with sign how many times meridian meets the slice disk. We can introduce
complexity $c$ on Alexander module, denoted as $\mathcal{A}_{c}$, and define
corresponding von Neumann $\rho$-invariant $\rho_{c}$ to obstruct a knot from
being rationally slice with complexity $c$. Since this invariant is defined
for each complexity $c$, as a rational slice obstruction, it is necessary to
verify that all values $\rho_{c}$ do not vanish. However, localization on the
rational Alexander module at a polynomial $p$, denoted as $\mathcal{A}_{c,p}$,
can provide a single value that does not depend on $c$. The main ingredient of
our construction involves the localization on Alexander module with complexity
and its corresponding $\rho_{c,p}$-invariant.

Note that a knot $K$ is rationally slice with complexity $c$ if and only if
the $(c,1)$-cable $K_{c,1}$ is strongly rationally slice, i.e., rationally
slice with complexity $1$ [CFHH13]. Thus, our $\rho_{c,p}$ for $K$ gives an
obstruction for $K_{c,1}$ from being strongly rationally slice. Moreover,

######  Corollary 1.4.

Let $K_{n}$ be an $n$-twist knot with order $2$ in
$\mathcal{AC}_{\mathbb{Q}}$. If $\rho^{(1)}(K_{n})\neq 0$, then any cable
$(K_{n})_{c,1}$ has infinite order in $\mathcal{C}$.

On the other hand, $K_{n}$ when $n$ is a square is algebraically rationally
slice, and moreover, $K_{1}$, which is the figure-eight knot, is rationally
slice. It is not known if $(K_{1})_{c,1}$ has infinite order in $\mathcal{C}$.
In contrast, in the smooth knot concordance group, it is known that
$(K_{1})_{c,1}$ has infinite order for odd $c$ [HKPS22], for $c=2$ [DKM+22]
(see also [ACM+23]), and for $c=2n$ when $n$ is odd [KPT24]. In particular,
such knots are examples of infinite order rationally slice knots in the smooth
category, whereas it is not known if such a knot exists in the topological
category. We close the introduction with the following analogous question in
the topological category:

######  Question 3.

Does there exist a rationally slice knot of infinite order in $\mathcal{C}$?

Organization In Section 2, we recall definition of von Neumann
$\rho$-invariant and its basic properties. After that, we introduce
localization and complexity separately on Alexander module and define the
corresponding $\rho$-invariants. In Section 3, we define the localized
$\rho$-invariant with complexity and check its properties including
additivity. In Section 4, we give a constraint for a knot and show that the
invariant vanishes for rationally slice knots satisfying the constraint. Then
we prove Theorem 1.2. In Section 5, we prove Theorem 1.3 by applying Theorem
1.2. We also prove 1.4 and ask some questions about rational concordance order
of certain knots and concordance order of their cable. In Section 6, we
briefly recall the notions of 0-bipolar and BPH-slice, and discuss some
relations of 0-bipolar, BPH-slice, and rational slice.

Acknowledgement The author is grateful to his advisor JungHwan Park for
generous advice. The author would also like to thank Jae Choon Cha and Taehee
Kim for helpful comments. This work is partially supported by Samsung Science
and Technology Foundation (SSTF-BA2102-02) and the POSCO TJ Park Science
Fellowship.

##  2\. Background

In this section, we first review the definition of von Neumann
$\rho$-invariant by using $L^{2}$-signature defect and see its basic
properties. Then we briefly recall some ordinary concordance invariants based
on the $\rho$-invariant such as the zeroth and first order signatures. After
that, we explain the localization of the Alexander module, the extended
Blanchfield form, and the corresponding $\rho$-invariant. At last, we recall
the notion of complexity, introduce it on Alexander module and Blanchfield
form, and define a rational version of the first signatures. Notice that the
last two subsections will be the main ingredients of our construction. We
claim no own result in this section. Throughout this section, every manifold
is assumed to be compact and oriented, unless otherwise stated.

###  2.1. von Neumann $\rho$-invariant

A von Neumann $\rho$-invariant is an invariant of a 3-manifold $M$ and its
representation $\phi:\pi_{1}(M)\rightarrow\Gamma$ for a discrete group
$\Gamma$. It can be obtained by the signature of the bordism class of
$(M,\phi)$. Hence, the $L^{2}$-signature defect (the difference between the
$L^{2}$-signature and the ordinary signature) of a 4-manifold $W$ bounded by
$M$ with extension $\psi:\pi_{1}(W)\rightarrow\Gamma$ of $\phi$, can be used
to define von Neumann $\rho$-invariant. For the definition of
$L^{2}$-signature $\sigma^{(2)}$, see [Lü02], [COT03, Section 5], and [Cha08].

######  Lemma 2.1.

[CW03, Appendix], [Cha08, below of Lemma 2.7] For a closed, connected
3-manifold $M$ with a representation $\phi:\pi_{1}(M)\rightarrow\Gamma$, there
exists a 4-manifold $W$ bounded by $k$ copies $M_{i}$ of $M$ for some $k$ and
$\psi:\pi_{1}(W)\rightarrow\Lambda$ for a group $\Lambda$ of which $\Gamma$ is
a subgroup such that below diagram commutes for each $i=1,\cdots,k$.

${\pi_{1}(M_{i})}$${\Gamma}$${\Lambda}$${\pi_{1}(W)}$$\scriptstyle{\phi}$$\scriptstyle{\psi}$

######  Definition 2.2.

The von Neumann $\rho$-invariant for above $(M,\phi)$ is defined by

$\rho(M,\phi)=\frac{1}{k}\left(\sigma^{(2)}(W,\psi)-\sigma(W)\right),$

where $k$ and $(W,\psi)$ are given in Lemma 2.1.

This value is independent of the choice of $(W,\psi)$ by [Har08, Lemma 3.6]
and the following lemma, called subgroup property:

######  Proposition 2.3.

[COT03, Proposition 5.13] Given $\phi:\pi_{1}(M)\rightarrow\Gamma$, if
$j:\Gamma\rightarrow\Lambda$ is injective, then

$\rho(M,j\circ\phi)=\rho(M,\phi).$

The value defined above coincides with the originally defined von Neumann
$\rho$-invariant by Cheeger and Gromov [CG85], and the following lemma
follows.

######  Lemma 2.4.

[COT03, Remark 5.10] Suppose that closed connected 3-manifolds
$M_{1},\cdots,M_{n}$ cobound a 4-manifold $W$ and the following diagram
commutes for each $i$:

${\pi_{1}(M_{i})}$${\Gamma_{i}}$${\pi_{1}(W)}$${\Lambda}$$\scriptstyle{\phi_{i}}$$\scriptstyle{j_{i}}$$\scriptstyle{\psi}$

If each $j_{i}$ is injective, then we obtain

$\sum\limits_{i=1}^{n}\rho(M_{i},\phi_{i})=\sigma^{(2)}(W,\psi)-\sigma(W).$

In this paper, we mainly have an interest in a cobordism between 3-manifolds
which may not be homeomorphic. Thus, Lemma 2.4, combined with the following
lemma, will be frequently used to compute a $\rho$-invariant of 3-manifold.
Below proposition immediately comes from the trivial cobordism $M\times I$ by
Lemma 2.4.

######  Proposition 2.5.

Let $-M$ be the orientation-reversed one of $M$. By abusing of notation, we
still use $\phi$ for the induced representation of $-M$. Then,

$\rho(-M,\phi)=-\rho(M,\phi).$

We close this subsection by introducing a useful lemma to compute
$\sigma^{(2)}(W,\phi)$ when $\Gamma$ is a poly-torsion-free-abelian (PTFA)
group. Note that all concerned group for $\rho$-invariant in this paper are
PTFA. For the definition of PTFA group, see [COT03, Definition 2.1].

######  Lemma 2.6.

Let $\Gamma$ be a PTFA group, and $\mathcal{K}$ be the fraction field of the
group ring $\mathbb{Q}\Gamma$. Then

$|\sigma^{(2)}(W,\psi:\pi_{1}(W)\rightarrow\Gamma)|\leq\operatorname{rk}_{\mathcal{K}}H_{2}(W;\mathcal{K}).$

In particular, if $H_{2}(W;\mathcal{K})=0$, then the $L^{2}$-signature
$\sigma^{(2)}(W,\psi)$ vanishes.

###### Proof.

By definition of $L^{2}$-signature [Lü02, COT03],
$|\sigma^{(2)}(W,\psi)|\leq\operatorname{rk}_{\mathbb{Q}\Gamma}H_{2}(W;\mathbb{Q}\Gamma)$.
Since
$\operatorname{rk}_{\mathbb{Q}\Gamma}H_{2}(W;\mathbb{Q}\Gamma)=\operatorname{rk}_{\mathcal{K}}H_{2}(W;\mathcal{K})$
(for example, see [Cha08, Lemma 2.4]), the result holds. ∎

###  2.2. Knot invariants based on $\rho$

The von Neumann $\rho$-invariant of the 0-surgery $M_{K}$ of a knot $K$ with a
specified representation can be adopted as a knot concordance invariant. In
this subsection, we recall several invariants defined in such a way.

######  Definition 2.7.

[CHL10, Definition 3.1] The zeroth order signature $\rho^{(0)}(K)$ of a knot
$K$ is defined by $\rho(M_{K},\phi_{0})$ where
$\phi_{0}:\pi_{1}(M_{K})\rightarrow\mathbb{Z}$ sends a meridian to $1$.

Note that $\phi$ is just the abelianization, so above $\mathbb{Z}$ is indeed
the first homology $H_{1}(M_{K};\mathbb{Z})$ of $M_{K}$. Although we will not
compute this invariant directly in this paper, we present how this abelian
invariant can be related with the Levine-Tristram signature function. For
simplicity, we normalize $\int_{S^{1}}d\omega=1$.

######  Theorem 2.8.

[COT04] Let $\sigma_{K}$ be the Levine-Tristram signature function for a knot
$K$. Then,

$\rho^{(0)}(K)=\int_{S^{1}}\sigma_{K}(\omega)d\omega.$

In particular, if $K$ is algebraically slice, then $\rho^{(0)}(K)=0$.

To distinguish non-slice knots among the algebraically slice knots, we need a
finer obstruction. Notice that the PTFA group $\mathbb{Z}$, say $\Gamma_{0}$,
used to define $\rho^{(0)}(K)$ is

$\Gamma_{0}=\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(1)}}.$

Thus, $\rho^{(0)}(K)$ captures homological information of the infinite cyclic
cover of $M_{K}$, which can be interpreted as

$H_{1}(M_{K};\mathbb{Z}\Gamma_{0})\cong\frac{\pi_{1}(M_{K})^{(1)}}{\pi_{1}(M_{K})^{(2)}}.$

To catch information lying deeper inside such as $\pi_{1}(M_{K})^{(2)}$, one
may construct a larger PTFA group $\Gamma_{1}$ and see the corresponding
higher cover. Before going on, we recall the definition of Blanchfield form.
From now on, we denote the Alexander module $H_{1}(M_{K};{\mathbb{Q}[t^{\pm
1}]})$ by $\mathcal{A}(K)$. To avoid confusion, note that $\mathcal{A}(K)$ is
induced by the trivial inclusion ${\mathbb{Z}[t^{\pm
1}]}\hookrightarrow{\mathbb{Q}[t^{\pm 1}]}$. Since $\mathcal{A}(K)$ is a
torsion ${\mathbb{Q}[t^{\pm 1}]}$-module, the linking form is well-defined.

######  Definition 2.9.

[COT03] The Blanchfield form
$\mathcal{B}\ell:\mathcal{A}(K)\times\mathcal{A}(K)\rightarrow\mathbb{Q}(t)/{\mathbb{Q}[t^{\pm
1}]}$ is defined to make the following diagram commutes:

${\mathcal{A}(K)}$${\operatorname{Hom}_{{\mathbb{Q}[t^{\pm
1}]}}\left(\mathcal{A}(K),\frac{\mathbb{Q}(t)}{{\mathbb{Q}[t^{\pm
1}]}}\right)}$${H^{2}(M_{K};{\mathbb{Q}[t^{\pm
1}]})}$${H^{1}\left(M_{K};\frac{\mathbb{Q}(t)}{{\mathbb{Q}[t^{\pm
1}]}}\right)}$$\scriptstyle{\cong}$$\scriptstyle{PD}$$\scriptstyle{x\mapsto\mathcal{B}\ell(x,\cdot)}$$\scriptstyle{\cong}$$\scriptstyle{B^{-1}}$$\scriptstyle{\cong}$$\scriptstyle{\kappa}$

$PD$ is the Poincaré duality, $B^{-1}$ is the inverse of Bockstein
homomorphism, and the $\kappa$ is the Kronecker evaluation map in the
universal coefficient spectral sequence. $B$ is isomorphism since
$H^{2}(M_{K};{\mathbb{Q}[t^{\pm 1}]})$ is torsion, and the isomorphism of
$\kappa$ comes from the universal coefficient spectral sequence theorem
[Lev77].

We say a submodule $P\subset\mathcal{A}(K)$ is isotropic if $P\subset
P^{\perp}$ with respect to $\mathcal{B}\ell$. If $P=P^{\perp}$, it is called
Lagrangian.

######  Definition 2.10.

[CHL10, Definition 4.1] Let $P$ be an isotropic submodule of $\mathcal{A}(K)$.
A first order $L^{2}$-signature $\rho^{(1)}(K,P)$ is defined by
$\rho(M_{K},\phi_{P})$ where $\phi_{P}$ is the quotient map by the kernel of:

$\pi_{1}(M_{K})^{(1)}\rightarrow\frac{\pi_{1}(M_{K})^{(1)}}{\pi_{1}(M_{K})^{(2)}}\hookrightarrow\mathcal{A}(K)\rightarrow\mathcal{A}(K)/P.$

In particular, when $P=0$, by subgroup property,

$\rho(M_{K},\phi_{P})=\rho\left(M_{K},\phi_{1}:\pi_{1}(M_{K})\rightarrow\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(2)}}\right).$

For this case, we call it by the first order signature and write it as
$\rho^{(1)}(K)$.

Observe that the following sequence is exact.

$0\rightarrow\frac{\pi_{1}(M_{K})^{(1)}}{\pi_{1}(M_{K})^{(2)}}\rightarrow\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(2)}}\rightarrow\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(1)}}\rightarrow
0$

Since $\frac{\pi_{1}(M_{K})^{(1)}}{\pi_{1}(M_{K})^{(2)}}$ can be identified
with the integral Alexander module
$\mathcal{A}^{\mathbb{Z}}=H_{1}(M_{K};{\mathbb{Z}[t^{\pm 1}]})$ and
$\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(1)}}\cong\mathbb{Z}$, our
$\Gamma_{1}=\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(2)}}$ can be regarded as
$\mathcal{A}^{\mathbb{Z}}\rtimes\mathbb{Z}$.

A first order $L^{2}$-signature is not a concordance invariant by itself, but
the set of such values can work as a slice obstruction.

######  Theorem 2.11.

[COT03, Theorem 4.2, 4.4] [CHL10, Theorem 4.2] If $K$ is slice, then there
exists a Lagrangian submodule $P$ of $\mathcal{A}(K)$ such that
$\rho^{(1)}(K,P)=0$. For a slice disk $\Delta$, above $P$ is given by

$P=\ker(H_{1}(M_{K};{\mathbb{Q}[t^{\pm 1}]})\rightarrow
H_{1}(B^{4}-\Delta;{\mathbb{Q}[t^{\pm 1}]})).$

######  Theorem 2.12.

[CHL09, Proposition 5.8] If $K$ is rationally slice, then one of the first
order $L^{2}$-signatures of $K$ vanishes. In other words, there exists an
isotropic submodule $P$ of $\mathcal{A}(K)$ such that $\rho^{(1)}(K,P)=0$.

###  2.3. Localized $\rho$-invariant

From the perspective of obstructing a knot $K$ from being slice, it is
required to check that for all isotropic submodule $P\subset\mathcal{A}$,
$\rho^{(1)}(K,P)\neq 0$. Instead of looking through all such a submodule, one
strategy is to impose a strong condition for Alexander module $\mathcal{A}(K)$
and check a single $\rho$-invariant. A localization technique fits well for
this strategy. This subsection is based on the work by Davis [Dav12b]. We
assume that every polynomial $p$ be a symmetric Laurent polynomial with
rational coefficients.

######  Definition 2.13.

Let
$R_{p}=\left\\{\frac{f}{g}\in\mathbb{Q}(t):(p,g)=1,f,g\in{\mathbb{Q}[t^{\pm
1}]}\right\\}$. The localized Alexander module at $p$ is

$\mathcal{A}_{p}(K)=\mathcal{A}(K)\otimes R_{p},$

where the ${\mathbb{Q}[t^{\pm 1}]}$-tensor product is the usual
multiplication.

######  Definition 2.14.

Let $\pi_{1}(M_{K})^{(2)}_{p}$ be the kernel of the following map:

$\pi_{1}(M_{K})^{(1)}\rightarrow\frac{\pi_{1}(M_{K})^{(1)}}{\pi_{1}(M_{K})^{(2)}}=H_{1}(M_{K};\mathbb{Z}[t^{\pm
1}])\hookrightarrow\mathcal{A}(K)\rightarrow\mathcal{A}(K)\otimes
R_{p}=\mathcal{A}_{p}(K).$

The localized $\rho$-invariant of $K$ at $p$ is defined as:

$\rho_{p}^{(1)}(K)=\rho\left(M_{K},\phi_{p}:\pi_{1}(M_{K})\rightarrow\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(2)}_{p}}\right).$

After localization at $p$, any submodule of $\mathcal{A}(K)$ whose order is
relatively prime with $p$ is annihilated in $\mathcal{A}_{p}(K)$. Then we may
expect some isotropic submodules to be killed. Since $p$ is assumed to be
symmetric, the Blanchfield form $\mathcal{B}\ell$ naturally extends to
$\mathcal{B}\ell_{p}$ on $\mathcal{A}_{p}(K)$ as

$\mathcal{B}\ell_{p}(x\otimes f(t),y\otimes
g(t))=f(t)\cdot\mathcal{B}\ell(x,y)\cdot g(t^{-1}).$

######  Definition 2.15.

A knot $K$ is $p$-anisotropic if $\mathcal{A}_{p}(K)$ has no nontrivial
isotropic submodule with respect to $\mathcal{B}\ell_{p}$.

Now $p$-anisotropy condition for $K$ makes $\rho_{p}^{(1)}(K)$ be enough to
obstruct $K$ from being slice. Moreover,

######  Theorem 2.16.

[Dav12b, Theorem 4.1] Let $K_{1},\cdots,K_{n}$ be $p$-anisotropic knots. If
$K=K_{1}\\#\cdots\\#K_{n}$ is slice, then $\rho_{p}^{(1)}(K)=0$.

Note that even if each $K_{i}$ is $p$-anisotropic, the connected sum
$K_{1}\\#\cdots\\#K_{n}$ may not be $p$-anisotropic.

The vanishing of the signature defect of the slice disk complement bounded by
$M_{K}$ is anticipated. The essence of localization and $p$-anisotropy is that
we expect all possible Lagrangian (or generally isotropic) submodule of
$\mathcal{A}(K)$ to be killed in $\mathcal{A}_{p}(K)$. The key role of
$p$-anisotropy is to make $\mathcal{A}_{p}(K)$ inject to
$\mathcal{A}_{p}(W)=H_{1}(W;{\mathbb{Q}[t^{\pm 1}]})\otimes R_{p}$ so that
$\rho^{(1)}_{p}(K)$ can be well-defined as the signature defect of $W$. Then
it works well as a single-valued concordance invariant, without computing all
$\rho^{(1)}(K,P)$ for all isotropic submodule $P$. One trade-off is that
$\rho_{p}^{(1)}(K)$ can only be applied to the subgroup generated by the
$p$-anisotropic knots. Before moving on to the next subsection, we give a
sufficient condition for $p$-anisotropy.

######  Proposition 2.17.

If each factor of $p$ is symmetric and divides the Alexander polynomial of $K$
with multiplicity at most 1, then $K$ is $p$-anisotropic.

###  2.4. $\rho$-invariant with complexity

Now suppose that $K$ is slice in a rational homology ball $V$ via a disk
$\Delta^{2}$. We can consider the complexity of $c$ from the injection:

$H_{1}(M_{K};\mathbb{Z})\xrightarrow{\times
c}H_{1}(V-\Delta;\mathbb{Z})/torsion.$

Sometimes we just say a boundary component $M$ of a 4-manifold $W$ has
complexity $c$ if

$H_{1}(M;\mathbb{Q})\cong H_{1}(W;\mathbb{Q})\cong\mathbb{Q}$

and the inclusion induces multiplication by $c$ on the free part of first
integral homology. Note that $c$ can be always adjusted to be positive, so we
assume that $c>0$. Complexity can be introduced on Alexander module and the
Blanchfield form, and then the corresponding $\rho$-invariant can be defined.
Readers can find this notion in [Cha07], and also in [COT03] with different
terminology “multiplicity”.

######  Definition 2.18.

The Alexander module with complexity $c$ is defined as:

$\mathcal{A}_{c}(K)=\mathcal{A}(K)\otimes_{c}{\mathbb{Q}[t^{\pm 1}]},$

where the ${\mathbb{Q}[t^{\pm 1}]}$-tensor product $\otimes_{c}$ is given by
the right action of $t\in{\mathbb{Q}[t^{\pm 1}]}$ as multiplication by
$t^{c}$.

In the sense that $\mathcal{A}(K)$ is the first homology with rational
coefficient of the infinite cyclic cover corresponding
$\pi_{1}(M_{K})\xrightarrow{\phi_{0}}\mathbb{Z}=\langle t\rangle$, we can
regard $\mathcal{A}_{c}(K)$ as the one similarly obtained from the top row
composition:

${\pi_{1}(M_{K})}$${\mathbb{Z}}$${\mathbb{Z}}$${\pi_{1}(W)}$$\scriptstyle{\phi_{0}}$$\scriptstyle{t\mapsto
t^{c}}$

Thus, if $M_{K}$ bounds $W$ with complexity $c$, then the coefficient system
extends to $\pi_{1}(W)\rightarrow\mathbb{Z}$ as described in above diagram so
that the inclusion induces the natural ${\mathbb{Q}[t^{\pm 1}]}$-module
homomorphism:

$j_{c}:\mathcal{A}_{c}(K)\rightarrow\mathcal{A}(W)=H_{1}(W;{\mathbb{Q}[t^{\pm
1}]}).$

Notice that if $\Delta_{K}(t)$ is the Alexander polynomial, then the order of
$\mathcal{A}_{c}(K)$ is $\Delta_{K}(t^{c})$ and the map
$\mathcal{A}(K)\rightarrow\mathcal{A}_{c}(K)$ sends $t$ to $t^{c}$.

The Blanchfield form $\mathcal{B}\ell_{c}$ on $\mathcal{A}_{c}(K)$ is defined
similarly as Definition 2.9, but equivalently, can be also defined as a
natural extension of the ordinary $\mathcal{B}\ell$.

######  Proposition 2.19.

[Cha07] Let $x,y\in\mathcal{A}(K)$ and $f(t),g(t)\in{\mathbb{Q}[t^{\pm 1}]}$.
Then the rational Blanchfield form $\mathcal{B}\ell_{c}$ on
$\mathcal{A}_{c}(K)$ extends $\mathcal{B}\ell$ in the sense that:

$\mathcal{B}\ell_{c}(x\otimes f(t),y\otimes g(t))=f(t)\cdot
h(\mathcal{B}\ell(x,y))\cdot g(t^{-1})$

where $h:\frac{\mathbb{Q}(t)}{{\mathbb{Q}[t^{\pm 1}]}}\xrightarrow{t\mapsto
t^{c}}\frac{\mathbb{Q}(t)}{{\mathbb{Q}[t^{\pm 1}]}}$.

A previously used $\rho$-invariant rational obstruction [Cha07, Kim23] is
defined by introducing complexity on the $\rho$-invariant determined by $x\in
P\subset\mathcal{A}_{c}(K)$ in [COT03]. Our obstruction is slightly different
than those. We introduce complexity on the first order $L^{2}$-signatures.
However, by the solvability of the quotient group, below vanishing theorem
directly comes from [COT03, Theorem 4.2]. Although these invariants would not
be used explicitly by themselves in this paper, except for the case when
$P=0$, we contain these for compatibility with above obstructions and better
understanding of the usage of localization in later.

######  Definition 2.20.

Let $P$ be an isotropic submodule of $\mathcal{A}_{c}(K)$ with respect to
$\mathcal{B}\ell_{c}$. A first order $L^{2}$-signature $\rho^{(1)}_{c}(K,P)$
with complexity $c$ is defined by $\rho(M_{K},\phi)$ where $\phi$ is the
quotient map by the kernel of:

$\pi_{1}(M_{K})^{(1)}\rightarrow\frac{\pi_{1}(M_{K})^{(1)}}{\pi_{1}(M_{K})^{(2)}}\rightarrow\mathcal{A}(K)\rightarrow\mathcal{A}_{c}(K)\rightarrow\mathcal{A}_{c}(K)/P.$

######  Theorem 2.21.

If $K$ is rationally slice in a rational homology ball $V^{4}$ with complexity
$c$, then there exists a Lagrangian submodule $P$ of $\mathcal{A}_{c}(K)$ such
that $\rho_{c}(K,P)=0.$ For a slice disk $\Delta$, above $P$ is given by

$P=\ker(\mathcal{A}_{c}(K)\rightarrow H_{1}(V-\Delta;{\mathbb{Q}[t^{\pm
1}]})).$

###### Proof.

Note that the complement $W=V-\nu\Sigma$ is a rational $n$-solution for any
positive (half) integer $n$ of the zero surgery $M_{K}$ (For example, see
[CHL09, Remark 5.5 (6)]). Then $P$ is Lagrangian with respect to
$\mathcal{B}\ell_{c}$ [Cha07, Theorem 5.13]. Consider the induced map
$j_{c}:\mathcal{A}_{c}(K)\rightarrow\mathcal{A}(W)$. Let $P=\ker{j_{c}}$. Then
the following diagram commutes:

${\frac{\pi_{1}(M)^{(1)}}{\pi_{1}(M)^{(2)}}}$${\mathcal{A}_{c}(K)/P}$${\frac{\pi_{1}(M)^{(1)}}{\ker\phi}}$${\frac{\pi_{1}(W)^{(1)}}{\pi_{1}(W)^{(2)}}}$${\mathcal{A}(W)}$$\scriptstyle{j_{*}}$$\scriptstyle{j_{c}}$

The dashed arrow is naturally induced since any $\alpha\in\ker\phi$ is sent to
$0\in\mathcal{A}(W)$ so that it goes to $\pi_{1}(W)^{(2)}$. It follows from
injectivity of the dashed map that $\phi$ extends to
$\pi_{1}(W)\rightarrow\frac{\pi_{1}(W)}{\pi_{1}(W)^{(2)}}$. This extension
argument is proved in Lemma 3.6 in detail. Now by [COT03, Theorem 4.2],
$\rho_{c}(K,P)$ vanishes. ∎

######  Remark 2.22.

Readers may compare Theorem 2.21 with Theorem 2.11 and Theorem 2.12. Theorem
2.21 is a rational version of Theorem 2.11, and Theorem 2.12 also obstructs
rational sliceness.

If the set of all Lagrangian submodules can be controlled as complexity $c$
varies, it might be possible to obstruct with $\rho_{c}$ for only finitely
many $c$. This phenomenon is based on the subgroup property and is remarked in
[Cha07, Section 5.2] for the originally defined von Neumann $\rho$-invariants
determined by $x\in\mathcal{A}_{c}(K)$.

##  3\. Localized $\rho$-invariant with complexity

To control the set of all Lagrangian submodules of $\mathcal{A}_{c}$, we
introduce localization on the rational Alexander module $\mathcal{A}_{c}$.
Then we define a $\rho$-invariant $\rho^{(1)}_{c,p}$ and prove basic
properties. The proof for additivity is rather long, so a reader may skip it.
Throughout this section, $K$ denotes a knot in $S^{3}$.

######  Definition 3.1.

Let $R_{p}$ be the ${\mathbb{Q}[t^{\pm 1}]}$-module defined in Section 2.3.
The localized Alexander module with complexity is

$\mathcal{A}_{c,p}(K)=\mathcal{A}_{c}(K)\otimes R_{p},$

where the ${\mathbb{Q}[t^{\pm 1}]}$-tensor product is the usual
multiplication.

As in Section 2.3, here we similarly expect that some isotropic submodule of
$\mathcal{A}_{c}(K)$ would be killed since the torsion submodules of order
relatively prime with $p$ are annihilated in $\mathcal{A}_{c,p}(K)$.

######  Definition 3.2.

Let $\pi_{1}(M_{K})^{(2)}_{c,p}$ be the kernel of the following map:

$\pi_{1}(M_{K})^{(1)}\rightarrow\frac{\pi_{1}(M_{K})^{(1)}}{\pi_{1}(M_{K})^{(2)}}\hookrightarrow\mathcal{A}(K)\hookrightarrow\mathcal{A}_{c}(K)\rightarrow\mathcal{A}_{c,p}(K).$

A localized $\rho$-invariant of $K$ at $p$ with complexity $c$ is defined as:

$\rho^{(1)}_{c,p}(K)=\rho\left(M_{K},\phi:\pi_{1}(M_{K})\rightarrow\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(2)}_{c,p}}\right).$

######  Remark 3.3.

Since $R_{p}$ is a free ${\mathbb{Q}[t^{\pm 1}]}$-module, it is flat so that
$\otimes_{{\mathbb{Q}[t^{\pm 1}]}}R_{p}$ is exact.

######  Proposition 3.4.

Let $p\in{\mathbb{Q}[t^{\pm 1}]}$, $c$ be a positive integer, and $\Delta_{K}$
be the Alexander polynomial of $K$.

  * (a)

If $\Delta_{K}(t^{c})$ and $p(t)$ are coprime, then
$\rho^{(1)}_{c,p}(K)=\rho^{(0)}(K)$.

  * (b)

If $\Delta_{K}(t^{c})=p(t)$, then $\rho^{(1)}_{c,p}(K)=\rho^{(1)}(K)$.

###### Proof.

  * (a)

Let $f(t)$ be an element of $\mathcal{A}_{c,p}(K)$. Since $\mathcal{A}_{c}(K)$
is annihilated by $\Delta_{K}(t^{c})$ and $\Delta_{K}(t^{c})$ is invertible in
$\mathcal{A}_{c,p}(K)$,

$f(t)=f(t)\otimes
1=f(t)\Delta_{K}(t^{c})\otimes\frac{1}{\Delta_{K}(t^{c})}=0.$

Thus, $\mathcal{A}_{c,p}(K)=0$, which implies that the kernel
$\pi_{1}(M_{K})^{(2)}_{c,p}$ is $\pi_{1}(M_{K})^{(1)}$. Therefore,

$\rho_{c,p}^{(1)}(K)=\rho\left(M_{K},\phi_{0}:\pi_{1}(M_{K})\rightarrow\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(1)}}\right)=\rho^{(0)}(K).$

  * (b)

Let $g(t)$ be an element of the kernel of
$\mathcal{A}_{c}(K)\rightarrow\mathcal{A}_{c,p}(K)$. Then there exists
$h(t)\in{\mathbb{Q}[t^{\pm 1}]}$ such that

$g(t)\otimes 1=g(t)h(t)\otimes\frac{1}{h(t)}=0\otimes\frac{1}{h(t)}=0,$

where $(h(t),\Delta_{K}(t^{c}))=1$. Since $g(t)h(t)=0$ in
$\mathcal{A}_{c}(K)$, $g(t)$ has a factor $\Delta_{K}(t^{c})$. Thus, $g(t)=0$
in $\mathcal{A}_{c}(K)$, which implies that $\mathcal{A}_{c}(K)$ injects to
$\mathcal{A}_{c,p}(K)$. Then $\pi_{1}(M_{K})^{(2)}_{c,p}$ is the same as the
kernel $\pi_{1}(M_{K})^{(2)}$ of the quotient
$\pi_{1}(M_{K})^{(1)}\rightarrow\frac{\pi_{1}(M_{K})^{(1)}}{\pi_{1}(M_{K})^{(2)}}$.
Thus,

$\rho_{c,p}^{(1)}(K)=\rho(M_{K},\phi_{1}:\pi_{1}(M_{K})\rightarrow\frac{\pi_{1}(M_{K})}{\pi_{1}(M_{K})^{(2)}})=\rho^{(1)}(K).$

∎

Proposition 3.4 is where the localization plays a role. Each condition on $p$
with respect to the Alexander polynomial in (a) and (b) makes $\phi$ be
independent of the complexity, so we can obtain a fixed value as $c$ varies.

We prove several propositions which will be used later. Instead of the
$0$-surgery $M_{K}$, we consider a boundary component $M^{3}$ of $W^{4}$ with
complexity $c$, and use notations such as $\mathcal{A}(M)$,
$\mathcal{A}_{c}(M)$, and $\mathcal{A}_{c,p}(M)$ to denote various versions of
Alexander module of $M$.

######  Proposition 3.5.

Suppose that $M$ is a boundary component of $W$ with complexity $c$. Then the
inclusion induces a well-defined homomorphism

$i_{c}:\frac{\pi_{1}(M)}{\pi_{1}(M)^{(2)}_{c,p}}\rightarrow\frac{\pi_{1}(W)}{\pi_{1}(W)^{(2)}_{p}}.$

###### Proof.

As mentioned in Section 2.4, there is a well-defined ${\mathbb{Q}[t^{\pm
1}]}$-homomorphism $j_{c}:\mathcal{A}_{c}(M)\rightarrow\mathcal{A}(W)$, and
this induces $j_{c}\otimes
id:\mathcal{A}_{c,p}(M)\rightarrow\mathcal{A}_{p}(W)$. Then the diagram below
commutes.

${\frac{\pi_{1}(M)^{(1)}}{\pi_{1}(M)^{(2)}}}$${\mathcal{A}(M)}$${\mathcal{A}_{c}(M)}$${\mathcal{A}_{c,p}(M)}$${\frac{\pi_{1}(W)^{(1)}}{\pi_{1}(W)^{(2)}}}$${\mathcal{A}(W)}$${\mathcal{A}_{p}(W)}$$\scriptstyle{j_{c}}$$\scriptstyle{j_{c}\otimes
id}$

Since any $x\in\pi_{1}(M)^{(2)}_{c,p}$ is $0$ in $\mathcal{A}_{c,p}(M)$, it
goes to $0$ in $\mathcal{A}_{p}(W)$ via $j_{c}\otimes id$. Thus, the image of
$x$ under $\pi_{1}(M)\rightarrow\pi_{1}(W)$ is contained in
$\pi_{1}(W)_{p}^{(2)}$. ∎

For the value $\rho(M_{K},\phi)$ to be obtained by the signature defect of $W$
as Lemma 2.4, it is necessary for the above induced map to be injective. We
provide a sufficient condition for such a map to be injective.

######  Lemma 3.6.

For a boundary component $M$ of a 4-manifold $W$ with complexity $c$, if the
inclusion induced map $j_{c}\otimes
id:\mathcal{A}_{c,p}(M)\rightarrow\mathcal{A}_{p}(W)$ is injective, then so is
$i_{c}$:

$\frac{\pi_{1}(M)}{\pi_{1}(M)^{(2)}_{c,p}}\xhookrightarrow{i_{c}}\frac{\pi_{1}(W)}{\pi_{1}(W)^{(2)}_{p}}.$

###### Proof.

Consider the following commutative diagram with the exact rows and all
vertical maps induced by inclusions.

${0}$${\frac{\pi_{1}(M)^{(1)}}{\pi_{1}(M)^{(2)}_{c,p}}}$${\frac{\pi_{1}(M)}{\pi_{1}(M)^{(2)}_{c,p}}}$${\frac{\pi_{1}(M)}{\pi_{1}(M)^{(1)}}}$${0}$${0}$${\frac{\pi_{1}(W)^{(1)}}{\pi_{1}(W)^{(2)}_{p}}}$${\frac{\pi_{1}(W)}{\pi_{1}(W)^{(2)}_{p}}}$${\frac{\pi_{1}(W)}{\pi_{1}(W)^{(1)}}}$${0}$$\scriptstyle{j_{*}}$$\scriptstyle{i_{c}}$$\scriptstyle{k_{*}}$

Note that $k_{*}$ is the map between the first homology with coefficient
$\mathbb{Z}$ so that it is injective. Now observe that $j_{*}$ is injective
because the below diagram commutes.

${\frac{\pi_{1}(M)^{(1)}}{\pi_{1}(M)^{(2)}_{c,p}}}$${\mathcal{A}_{c,p}(M)}$${\frac{\pi_{1}(W)^{(1)}}{\pi_{1}(W)^{(2)}_{p}}}$${\mathcal{A}_{p}(W)}$$\scriptstyle{j_{*}}$$\scriptstyle{j_{c}\otimes
id}$

Since $j_{*}$ and $k_{*}$ are injective, so is $i_{c}$. ∎

To obstruct a knot from being of finite order, one particularly important
aspect of localization is that this value is additive under connected sum. It
is not expected in Theorem 2.11 and Theorem 2.21.

######  Proposition 3.7.

$\rho^{(1)}_{c,p}(J\\#K)=\rho^{(1)}_{c,p}(J)+\rho^{(1)}_{c,p}(K)$.

###### Proof.

We first construct a cobordism $W$ between $M_{J}\amalg M_{K}$ and
$-M_{J\\#K}$. Then we show that for $M=M_{J}$, $M_{K}$, and $M_{J\\#K}$, the
induced map $i_{c}$ given in Proposition 3.5 is injective so that we obtain
$\rho_{c,p}^{(1)}(J)+\rho_{c,p}^{(1)}(K)-\rho_{c,p}^{(1)}(J\\#K)=\sigma^{(2)}\left(W,\psi:\pi_{1}(W)\rightarrow\frac{\pi_{1}(W)}{\pi_{1}(W)^{(2)}_{p}}\right)-\sigma(W)$
by the Lemma 2.4. See the commutative diagram below.

${\pi_{1}(M)}$${\frac{\pi_{1}(M)^{(1)}}{\pi_{1}(M)^{(2)}_{c,p}}}$${\pi_{1}(W)}$${\frac{\pi_{1}(W)^{(1)}}{\pi_{1}(W)^{(2)}_{c,p}}}$$\scriptstyle{\phi}$$\scriptstyle{i_{c}}$$\scriptstyle{\psi}$

To complete the proof, we show that $\sigma^{(2)}(W,\psi)$ and $\sigma(W)$
vanish. After construction, most of the remains are just routines, but the
proof is long because our cobordism is obtained through several times of
glueing. Readers who are familiar with this kind of homology computation may
skip those parts. The schematic picture for $W$ is given in Figure 5. For any
knot $K$, the $0$-surgery for $K$ is denoted by $M_{K}$, and its meridian is
denoted by $\mu_{K}$. We abuse notation by removing the minus sign for
orientation-reversed manifolds on the top boundary of cobordisms until Step 3.

Step 1. Construct a cobordism $W$ between $M_{J}\amalg M_{K}$ and $M_{J\\#K}$.

Let $V$ be the 4-manifold obtained by attaching a 1-handle to $M_{J}\times
1\subset M_{J}\times I$ with its dotted circle as an unknot unlinked with $J$
in the Kirby diagram. Isotope the dotted circle to form the $(c,1)$-cable
pattern $P$. Now let $W_{1}$ be the 4-manifold obtained by attaching a
$0$-framed 2-handle to $V$ along the circle $\alpha$ in the leftmost of Figure
2. By Kirby calculus illustrated as Figure 2, the top boundary is
$M_{J_{c,1}}$. This cobordism $W_{1}$ between $M_{J}$ and $M_{J_{c,1}}$ is a
special case of rational homology cobordisms between the 0-surgery for a knot
and the 0-surgery for its satellite with a $\mathbb{Z}[\frac{1}{c}]$-slice
pattern constructed in [CFHH13]. As proved there, we have isomorphisms

$\displaystyle H_{1}(M_{J};\mathbb{Q})$ $\displaystyle\cong
H_{1}(W_{1};\mathbb{Q})\cong H_{1}(M_{J_{c,1}};\mathbb{Q}),$ $\displaystyle
H_{2}(W_{1};\mathbb{Q})$ $\displaystyle\cong H_{2}(M_{P};\mathbb{Q})\oplus
H_{2}(M_{J};\mathbb{Q}),$

where the first homology is generated by $c[\mu_{J_{c,1}}]=[\mu_{J}]$.

Figure 2. Slide $c$ strands highlighted in the top right over the 0-framed
2-handle knotted by $J$ and cancel the 1-handle. The top boundary is
$\partial_{+}W_{1}\cong M_{J_{c,1}}$. Figure 3. $\partial_{+}W_{2}\cong
M_{(J\\#K)_{c,1}}$.

Take a meridian $\mu_{J_{c,1}}$ in $M_{J_{c,1}}=\partial_{+}W_{1}$ and a
meridian $\mu_{K}$ in $M_{K}\times 1\subset M_{K}\times I$, and let $W_{2}$ be
the 4-manifold obtained by glueing $W_{1}$ and $M_{K}\times I$ along tubular
neighborhoods of a simple closed curve representing $c[\mu_{J_{c,1}}]$ and
$\mu_{K}$. This connected 4-manifold can be also obtained by attaching a
1-handle to join $W_{1}$ and $M_{K}\times I$ and a 0-framed 2-handle along the
curve $\beta$ in Figure 3 to identify $c[\mu_{J_{c,1}}]$ and $\mu_{K}$. By a
similar Kirby calculus with Figure 2, we obtain the top boundary
$\partial_{+}W_{2}$ as $M_{(J\\#K)_{c,1}}$, and hence, $W_{2}$ is a cobordism
between $M_{J_{c,1}}\amalg M_{K}$ and $M_{(J\\#K)_{c,1}}$. On the long exact
sequence for the pair $(W_{2},M_{J_{c,1}}\amalg M_{K})$,

$\cdots\xrightarrow{j_{n}}H_{n}(W_{2},M_{J_{c,1}}\amalg
M_{K};\mathbb{Q})\xrightarrow{\partial_{n}}H_{n-1}(M_{J_{c,1}}\amalg
M_{K};\mathbb{Q})\xrightarrow{i_{n-1}}H_{n-1}(W_{2};\mathbb{Q})\xrightarrow{j_{n-1}}\cdots$

the boundary map $\partial_{2}$ sends the attached 2-handle, which is the
generator, to $c[\mu_{J_{c,1}}]-[\mu_{K}]$. Thus, we have

$H_{1}(W_{2};\mathbb{Q})\cong\frac{H_{1}(M_{J_{c,1}};\mathbb{Q})\oplus
H_{1}(M_{K};\mathbb{Q})}{(c[\mu_{J_{c,1}}]-[\mu_{K}])}\cong
H_{1}(M_{J_{c,1}};\mathbb{Q})\cong H_{1}(M_{J};\mathbb{Q})\cong
H_{1}(M_{K};\mathbb{Q}),$

which are generated by $c[\mu_{J_{c,1}}]=[\mu_{J}]=[\mu_{K}]$. Moreover, since
$\partial_{2}$ is injective, $j_{2}$ is the zero map, and then the inclusion
induced map $i_{2}$ is surjective:

$i_{2}:H_{2}(M_{J_{c,1}}\amalg M_{K};\mathbb{Q})\twoheadrightarrow
H_{2}(W_{2};\mathbb{Q}).$

Figure 4. After attaching $0$-framed 2-handle, slide $c$ strands of
$(J\\#K)_{c,1}$ over the 2-handle to make it unknotted. Canceling the
unknotted 2-handle by attaching 3-handle, $\partial_{+}W_{3}\cong M_{J\\#K}$.

Let $W_{3}$ be the rational homology cobordism from $M_{(J\\#K)_{c,1}}$ to
$M_{J\\#K}$ constructed in a similar but upside down way with $W_{1}$. In
other words, the attached handles are now a 2-handle and a 3-handle as in
Figure 4. The Kirby calculus in Figure 2 and Figure 4 are explicitly
illustrated in [DPR21].

Glueing $W_{3}$ to $W_{2}$ along $M_{(J\\#K)_{c,1}}$, the resulting manifold
$W$ is a cobordism between $M_{J}\amalg M_{K}$ and $M_{J\\#K}$. By the Meyer-
Vietoris sequence below,

$\cdots\xrightarrow{j_{n}}H_{n}(W;\mathbb{Q})\xrightarrow{\partial_{n}}H_{n-1}(M_{(J\\#K)_{c,1}};\mathbb{Q})\xrightarrow{i_{n-1}}H_{n-1}(W_{2};\mathbb{Q})\oplus
H_{n-1}(W_{3};\mathbb{Q})\xrightarrow{j_{n-1}}\cdots.$

We have an isomorphism

$H_{1}(W;\mathbb{Q})\cong\frac{H_{1}(W_{2};\mathbb{Q})\oplus
H_{1}(W_{3};\mathbb{Q})}{([\mu_{(J\\#K)_{c,1}}]-c[\mu_{J\\#K})]}\cong
H_{1}(W_{2};\mathbb{Q})\cong H_{1}(W_{3};\mathbb{Q})\cong
H_{1}(M_{J\\#K};\mathbb{Q}).$

The generator is $[\mu_{J\\#K}]=c[\mu_{(J\\#K)_{c,1}}]=[\mu_{J}]=[\mu_{K}]$.
Thus, the inclusion of each boundary component $M_{J}$, $M_{K}$, and
$M_{J\\#K}$ induces isomorphisms on the first rational homology. Since $i_{1}$
is injective, the boundary map $\partial_{2}$ is the zero map, so $j_{2}$ is
surjective:

$j_{2}:H_{2}(W_{2};\mathbb{Q})\oplus H_{2}(W_{3};\mathbb{Q})\twoheadrightarrow
H_{2}(W;\mathbb{Q}).$

Since we have seen that $H_{2}(M_{J};\mathbb{Q})\oplus
H_{2}(M_{K};\mathbb{Q})$ surjects onto $H_{2}(W_{2};\mathbb{Q})$ and
$H_{2}(M_{J\\#K};\mathbb{Q})$ surjects onto $H_{2}(W_{3};\mathbb{Q})$, we
conclude that every second homology element of $W$ comes from its boundary,
namely, the inclusion induced map of boundary is surjective:

$H_{2}(\partial W;\mathbb{Q})\twoheadrightarrow H_{2}(W;\mathbb{Q}).$

Figure 5. A schematic picture of the cobordism $W$ between $M_{J}\amalg M_{K}$
and $M_{J\\#K}$.

Step 2. Check that
$\frac{\pi_{1}(M)}{\pi_{1}(M)_{c,p}}\rightarrow\frac{\pi_{1}(W)}{\pi_{1}(W)_{p}}$
is injective for each boundary component $M$ of $W$.

Notice that $H_{1}(M_{K};{\mathbb{Q}[t^{\pm 1}]})=\mathcal{A}_{c}(K)$ with
complexity $c$, but $H_{1}(M_{(J\\#K)_{c,1}};{\mathbb{Q}[t^{\pm
1}]})=\mathcal{A}((J\\#K)_{c,1})$ without complexity, i.e., complexity $c=1$.
In Step 1, we have shown that each inclusion for $M=M_{J},M_{K}$, and
$M_{J\\#K}$ into $W$ induces an isomorphism $H_{1}(M;\mathbb{Q})\rightarrow
H_{1}(W;\mathbb{Q})$. For $\frac{\pi_{1}(M)}{\pi_{1}(M)^{(2)}_{c,p}}$ to
inject into $\frac{\pi_{1}(W)}{\pi_{1}(W)^{(2)}_{p}}$, by Lemma 3.6, it is
enough to check that each Alexander module $\mathcal{A}_{c,p}(M)$ injects to
$\mathcal{A}_{p}(W)$.

We give an another description for $W_{1}$ in a similar way as we have done
for $W_{2}$. Let $W_{0}$ be the complement of an unknotted slice disk
$\Delta_{P}$ of the unknotted $(c,1)$-cable pattern $P$ in $B^{4}$, which is
bounded by the 0-surgery $M_{P}$. Just as we have identified the simple closed
curve representing $c[\mu_{J_{c,1}}]\subset\partial_{+}W_{1}$ and
$\mu_{K}\subset M_{K}\times 1\subset M_{K}\times I$, here we take a simple
closed curve representing $c[\mu_{P}]$ in $M_{P}$ and the meridian $\mu_{J}$
in $M_{J}\times 1\subset M_{J}\times I$. $W_{1}$ is the glueing result of
$W_{0}$ and $M_{J}\times I$ along the tubular neighborhood of such curves
$\nu(c\mu_{P})=\nu(\mu_{J})\cong S^{1}\times D^{2}$. Then we have the
following Mayer-Vietoris sequences with coefficient system $\mathbb{Q}[t^{\pm
1}]$:

$\cdots\rightarrow H_{n}(S^{1}\times D^{2})\rightarrow H_{n}(W_{0})\oplus
H_{n}(M_{J}\times I)\rightarrow H_{n}(W_{1})\rightarrow\cdots$

Note that $W_{0}$ is the 4-ball with a 1-handle attached, so $W_{0}$ is
diffeomorphic to $S^{1}\times D^{3}$. Since $S^{1}\times D^{2}$ and $W_{0}$
have contractible infinite cyclic covers as $\mathbb{R}\times D^{2}$ and
$\mathbb{R}\times D^{3}$ respectively, their homologies with
${\mathbb{Q}[t^{\pm 1}]}$-coefficient are trivial. Thus,

$H_{n}(W_{1};{\mathbb{Q}[t^{\pm 1}]})\cong H_{n}(M_{J};{\mathbb{Q}[t^{\pm
1}]})=\mathcal{A}_{c}(J).$

Similarly, $W_{2}$ is $W_{1}\cup(M_{K}\times I)$ with intersection
$\nu(c\mu_{J_{c,1}})=\nu\mu_{K}\cong S^{1}\times D^{2}$. By the exact sequence
below,

$\cdots\rightarrow H_{n}(S^{1}\times D^{2})\rightarrow H_{n}(W_{1})\oplus
H_{n}(M_{K}\times I)\rightarrow H_{n}(W_{2})\rightarrow\cdots$

we obtain isomorphisms:

$\displaystyle H_{n}(W_{2};{\mathbb{Q}[t^{\pm 1}]})$ $\displaystyle\cong
H_{n}(W_{1};{\mathbb{Q}[t^{\pm 1}]})\oplus H_{n}(M_{K};{\mathbb{Q}[t^{\pm
1}]})$ $\displaystyle\cong H_{n}(M_{J};{\mathbb{Q}[t^{\pm 1}]})\oplus
H_{n}(M_{K};{\mathbb{Q}[t^{\pm 1}]}).$

Furthermore, by Poincare duality and the universal coefficient theorem,

$\displaystyle 0=H_{n}(W_{2},M_{J}\amalg M_{K};{\mathbb{Q}[t^{\pm 1}]})$
$\displaystyle\cong H^{4-n}(W_{2},M_{(J\\#K)_{c,1}};{\mathbb{Q}[t^{\pm 1}]})$
$\displaystyle\cong H_{4-n}(W_{2},M_{(J\\#K)_{c,1}};{\mathbb{Q}[t^{\pm 1}]}).$

Then, by the long exact sequence on the pair $(W_{2},M_{(J\\#K)_{c,1}})$, we
have

$H_{n}(M_{(J\\#K)_{c,1}};{\mathbb{Q}[t^{\pm 1}]})\cong
H_{n}(W_{2};{\mathbb{Q}[t^{\pm 1}]}).$

In particular, the first homology $\mathcal{A}_{c}(J),\mathcal{A}_{c}(K)$, and
$\mathcal{A}((J\\#K)_{c,1})$ inject into $\mathcal{A}(W_{2})$. Since $\otimes
R_{p}$ is exact, the localized Alexander module
$\mathcal{A}_{c,p}(J),\mathcal{A}_{c,p}(K)$, and
$\mathcal{A}_{p}((J\\#K)_{c,1})$ inject into $\mathcal{A}_{p}(W_{2})$.

It remains to show that $\mathcal{A}_{p}(W_{2})$ and
$\mathcal{A}_{c,p}(M_{J\\#K})$ inject into $\mathcal{A}_{p}(W)$. For the pair
$(W,W_{2})$, by excision, Poincare duality, and universal coefficient theorem
as previous,

$\displaystyle H_{n}(W,W_{2};{\mathbb{Q}[t^{\pm 1}]})$ $\displaystyle\cong
H_{n}(W_{3},M_{(J\\#K)_{c,1}};{\mathbb{Q}[t^{\pm 1}]})$ $\displaystyle\cong
H_{4-n}(W_{3},M_{J\\#K};{\mathbb{Q}[t^{\pm 1}]}),$

but we can find $H_{4-n}(M_{J\\#K};{\mathbb{Q}[t^{\pm 1}]})\cong
H_{4-n}(W_{3};{\mathbb{Q}[t^{\pm 1}]})$ in a similar Meyer-Vietoris sequence
with the case for $W_{1}$ so that $H_{n}(W,W_{2};{\mathbb{Q}[t^{\pm 1}]})=0$.
Thus, $H_{n}(W_{2};{\mathbb{Q}[t^{\pm 1}]})\cong H_{n}(W;{\mathbb{Q}[t^{\pm
1}]})$. In particular, $\mathcal{A}_{p}(W_{2})$ injects into
$\mathcal{A}_{p}(W)$. Moreover, by an excision on the pair $(W,W_{3})$, we
have

$H_{n}(W,W_{3};{\mathbb{Q}[t^{\pm 1}]})\cong
H_{n}(W_{2},M_{(J\\#K)_{c,1}};{\mathbb{Q}[t^{\pm 1}]})=0$

so that

$H_{n}(W;{\mathbb{Q}[t^{\pm 1}]})\cong H_{n}(W_{3};{\mathbb{Q}[t^{\pm
1}]})\cong H_{n}(M_{J\\#K};{\mathbb{Q}[t^{\pm 1}]}).$

Therefore, by flatness of $R_{p}$, the desired condition for injectivity
$\mathcal{A}_{c,p}(M)\hookrightarrow\mathcal{A}_{p}(W)$ for each
$M=M_{J},M_{K}$, and $M_{J\\#K}$ holds.

Step 3. Compute the signature defect of $W$.

Note that $\partial W=M_{J}\amalg M_{K}\amalg-M_{J\\#K}$. Since
$\frac{\pi_{1}(M)}{\pi_{1}(M)^{(2)}_{c,p}}$ injects into
$\frac{\pi_{1}(W)}{\pi_{1}(W)^{(2)}_{p}}$ for each $M=M_{J},M_{K}$, and
$-M_{J\\#K}$, by Lemma 2.4 and Proposition 2.5, we have

$\rho_{c,p}^{(1)}(J)+\rho_{c,p}^{(1)}(K)-\rho_{c,p}^{(1)}(J\\#K)=\sigma^{(2)}(W,\psi:\pi_{1}(W)\rightarrow\frac{\pi_{1}(W)}{\pi_{1}(W)_{p}})-\sigma(W).$

Let $\mathcal{K}$ be the fraction field of the Ore domain
$\mathbb{Q}\left[\frac{\pi_{1}(W)}{\pi_{1}(W)_{p}^{2}}\right]$. Consider the
first Meyer-Vietoris sequence in Step 2 with coefficient $\mathcal{K}$. As in
[Coc04],

$H_{*}(S^{1}\times D^{2};\mathcal{K})\cong H_{*}(M_{J};\mathcal{K})\cong
H_{*}(W_{0};\mathcal{K})\cong 0.$

Thus, $H_{*}(W_{1};\mathcal{K})=0$, and since $H_{*}(M_{K};\mathcal{K})=0$ as
well, it follows that $H_{*}(W_{2};\mathcal{K})=0$ from the second sequence in
Step 2. Since $W_{3}$ is constructed in the same way with $W_{1}$, it is also
true that $H_{*}(W_{3};\mathcal{K})=0$ and hence,

$H_{2}(W;\mathcal{K})=0.$

Thus, by Lemma 2.6, the $L^{2}$-signature $\sigma^{(2)}(W,\psi)$ vanishes.
Moreover, while we compute the rational homology of $W$ in Step 1, we have
already seen that $H_{2}(W;\mathbb{Q})$ are supported by the boundary
components. Therefore, $\sigma(W)$ vanishes as well, so

$\rho_{c,p}^{(1)}(J)+\rho_{c,p}^{(1)}(K)-\rho_{c,p}^{(1)}(J\\#K)=0.$

∎

######  Proposition 3.8.

Let $-K$ be the mirror image of $K$. Then,

$\rho_{c,p}^{(1)}(-K)=-\rho_{c,p}^{(1)}(K).$

###### Proof.

Since $S^{3}_{0}(-K)$ and $-S^{3}_{0}(K)$ are orientation-preserving
diffeomorphic, it holds by Proposition 2.5. ∎

######  Remark 3.9.

Consider the $(c,1)$-cable $K_{c,1}$ of $K$. Recall the localized
$\rho$-invariant $\rho_{p}$ defined in Section 2.3. One can easily see that

$\rho^{(1)}_{p}(K_{c,1})=\rho^{(1)}_{1,p}(K_{c,1})=\rho^{(1)}_{c,p}(K).$

The first equality is by definition. The second one can be derived from the
cobordism $W_{1}$ constructed in the proof of Proposition 3.7, between $M_{J}$
and $M_{J_{c,1}}$. See also [CFHH13].

##  4\. Rational Slice Obstruction

In this section, we verify that our invariant defined in Section 3 gives a
rational slice obstruction. To use this, we impose a rather strong condition,
called $(c,p)$-anisotropy, for the localized $\mathcal{A}_{c,p}$ to have no
nontrivial isotropic submodule with respect to $\mathcal{B}\ell_{c,p}$. For
such a knot $K$, we show that if $K$ is rationally slice with complexity $c$,
then $\rho_{c,p}(K)$ vanishes. Moreover, as Proposition 2.16, it can obstruct
any connected sum of those knots. After providing several sufficient
conditions for $(c,p)$-anisotropy, we prove Theorem 1.2. Throughout this
section, $p$ is assume to be a Laurent polynomial with rational coefficients.
We also assume that $p$ be symmetric so that the rational Blanchfield form
$\mathcal{B}\ell_{c}$ naturally extends to $\mathcal{B}\ell_{c,p}$ on the
localized Alexander module.

######  Definition 4.1.

A knot $K$ is $(c,p)$-anisotropic if $\mathcal{A}_{c,p}(K)$ has no nontrivial
isotropic submodule with respect to the Blanchfield form
$\mathcal{B}\ell_{c,p}$.

######  Theorem 4.2.

Let $K_{1},\cdots,K_{n}$ be $(c,p)$-anisotropic knots. If
$K=K_{1}\\#\cdots\\#K_{n}$ is rationally slice with complexity $c$, then
$\rho_{c,p}(K)=0$.

###### Proof.

Suppose that $K=K_{1}\\#\cdots\\#K_{n}$ with $(c,p)$-anisopropic $K_{i}$,
possibly $K_{i}=\pm K_{j}$. Let $M_{i}=M_{K_{i}}$. Here we similarly go on
with the proof of Proposition 3.7 by constructing a 4-manifold $X$ bounded by
$M_{1},\cdots M_{n}$ satisfying certain properties. More precisely, the
resulting $X$ is required to satisfy the condition in the Lemma 3.6 so that
$\rho_{c,p}^{(1)}(K)=\sigma^{(2)}(X,\psi)-\sigma(X)$. After that, it is easily
verified that the signature defect for $X$ vanishes.

Step 1. Construct a 4-manifold $X$ bounded by $M_{1},\cdots,M_{n}$.

Take $M_{i}\times I$ and let $\mu_{i}$ be a meridian of $M_{i}\times 1\subset
M_{i}\times I$ for each $i$. By identifying all tubular neighborhoods of
$\mu_{i}$, we obtain a cobordism $W$ between $M_{1}\amalg\cdots\amalg M_{n}$
and $M_{K}$. This can be also described as the result attaching $n-1$ copies
of a tube $S^{1}\times B^{2}\times I$, say $V_{i}\times I$ along $V_{i}\times
0=\nu\mu_{i}$ and $V_{i}\times 1=\nu\mu_{i+1}$. Using a simple Mayer-Vietoris
argument, one can check that the inclusion of each boundary component induces
an isomorphism. See [Dav12b, Theorem 4.1] for the detail.

$H_{1}(M_{i};\mathbb{Q})\cong H_{1}(W;\mathbb{Q})\cong
H_{1}(M_{K};\mathbb{Q}).$

which is generated by $[\mu_{1}]=\cdots=[\mu_{n}]=[\mu_{K}]$. We also obtain
the following surjection:

$H_{2}(\amalg_{i=1}^{n}M_{i};\mathbb{Q})\oplus
H_{2}(M_{K};\mathbb{Q})\twoheadrightarrow H_{2}(W;\mathbb{Q}).$

Let $E$ be a slice disk complement for $K$ in a rational 4-ball. Then
$H_{*}(E;\mathbb{Q})\cong H_{*}(S^{1};\mathbb{Q})$ and
$H_{1}(E;\mathbb{Q})\cong H_{1}(M_{K};\mathbb{Q})$ generated by $c[\mu_{K}]$.
Glueing $W$ and $E$ along $M_{K}$, let $X$ be the resulting 4-manifold with
$\partial X=M_{1}\amalg\cdots\amalg M_{n}$. Then $X$ satisfies the required
conditions for injectivity on the first and surjectivity on the second
homology with $\mathbb{Q}$ coefficient by the Meyer-Vietoris sequence for
$X=W\cup E$.

Step 2. Show that
$\frac{\pi_{1}(M_{i})}{\pi_{1}(M_{i})_{c,p}}\rightarrow\frac{\pi_{1}(W)}{\pi_{1}(W)_{p}}$
is injective.

It is enough to show that the localized Alexander module
$\mathcal{A}_{c,p}(K_{i})$ injects into $\mathcal{A}_{p}(X)$ by inclusion
induced map. It comes from the following lemma.

######  Lemma 4.3.

The kernel of the inclusion induced map
$\mathcal{A}_{c,p}(K_{i})\rightarrow\mathcal{A}_{p}(X)$ is isotropic with
respect to the localized Blanchfield form $\mathcal{B}\ell_{c,p}$ on $\partial
X$.

###### Proof.

Note that every isomorphism before localization is preserved since $\otimes
R_{p}$ is exact. Consider the following commutative diagram. Each map is the
inclusion induced map.

${\mathcal{A}_{c,p}(K_{i})}$${\mathcal{A}_{p}(X)}$${\mathcal{A}_{p}(W)}$${\mathcal{A}_{c,p}(K)}$${\mathcal{A}_{p}(E)}$$\scriptstyle{\cong}$$\scriptstyle{i_{*}\otimes
id}$$\scriptstyle{\cong}$$\scriptstyle{j_{*}\otimes id}$$\scriptstyle{\cong}$

Via the vertical isomorphisms, the statement is equivalent with that the
kernel of $j_{*}\otimes id$ is isotropic with respect to
$\mathcal{B}\ell_{c,p}$ on $M_{K}$. Let $f(t),g(t)$ be elements in that
kernel. Then $f(t)=\frac{f_{1}(t)}{f_{2}(t)}$ and
$g(t)=\frac{g_{1}(t)}{g_{2}(t)}$ where
$f_{1}(t),g_{1}(t)\in\mathcal{A}_{c}(K)$ and each of
$f_{2}(t),g_{2}(t)\in{\mathbb{Q}[t^{\pm 1}]}$ is coprime with $p$. Since
$\ker(j_{*}\otimes id)=(\ker j_{*})\otimes{R_{p}}$, $f_{1}(t)$ and $g_{1}(t)$
should lie in the kernel $P=\ker j_{*}$ so that
$\mathcal{B}\ell_{c}(f_{1}(t),g_{1}(t))=0$.

By [COT03], since $E$ is a rational $(1.5)$-solution, $P$ is isotropic with
respect to the Blanchfield form $\mathcal{B}\ell_{c}$ on $\mathcal{A}_{c}(K)$.
Choose $k(t)\in{\mathbb{Q}[t^{\pm 1}]}$ such that it represents

$\mathcal{B}\ell_{c}(f_{1}(t),g_{1}(t))=[k(t)]=0\in\frac{\mathbb{Q}(t)}{{\mathbb{Q}[t^{\pm
1}]}}.$

Now we compute the localized Blanchfield form
$\mathcal{B}\ell_{c,p}(f(t),g(t))$ on $\mathcal{A}_{c,p}(K)$:

$\displaystyle\mathcal{B}\ell_{c,p}(f(t),g(t))$
$\displaystyle=\frac{1}{f_{2}(t)}\mathcal{B}\ell_{c}(f_{1}(t),g_{1}(t))\frac{1}{\overline{g_{2}}(t)}$
$\displaystyle=\left[\frac{k(t)}{f_{2}(t)g_{2}(t^{-1})}\right]=0.$

The last equality comes from the fact that
$\frac{k(t)}{f_{2}(t)g_{2}(t^{-1})}\in R_{p}$. ∎

If the kernel of $\mathcal{A}_{c,p}(K_{i})\rightarrow\mathcal{A}_{p}(X)$ were
nontrivial, then, since it is isotropic by Lemma 4.3, it contradicts with the
$(c,p)$-anisotropy assumption for each $K_{i}$. Thus, each
$\mathcal{A}_{c,p}(K_{i})$ successfully injects into $\mathcal{A}_{p}(X)$. Now
by Lemma 3.6, $\frac{\pi_{1}(M_{i})}{\pi_{1}(M_{i})^{(2)}_{c,p}}$ injects into
$\frac{\pi_{1}(X)}{\pi_{1}(X)^{(2)}_{p}}$. By Proposition 3.7 (additivity) and
Lemma 2.4, the invariant is obtained by the signature defect of $X$ as:

$\displaystyle\rho^{(2)}_{c,p}(K)$
$\displaystyle=\rho^{(2)}_{c,p}(K_{1})+\cdots+\rho^{(2)}(K_{n})$
$\displaystyle=\sigma^{(2)}\left(X,\psi:\pi_{1}(X)\rightarrow\frac{\pi_{1}(X)}{\pi_{1}(X)_{p}^{(1)}}\right)-\sigma(X).$

Step 3. Compute the signature defect of $X$.

As [Dav12b, Lemma 4.9], $H_{*}(X;\mathcal{K})=0$, where $\mathcal{K}$ is the
fraction field of the Ore domain
$\mathbb{Q}\left[\frac{\pi_{1}(X)}{\pi_{1}(X)_{p}^{2}}\right]$. Thus, by Lemma
2.6, the $L^{2}$-signature $\sigma^{(2)}(X,\psi)$ vanishes. Since
$H_{2}(\partial X;\mathbb{Q})\rightarrow H_{2}(X;\mathbb{Q})$ is surjective,
the ordinary signature $\sigma(X)$ vanishes as well. ∎

######  Remark 4.4.

Recall $\rho^{(1)}_{1,p}(K_{c,1})=\rho^{(1)}_{c,p}(K)$ as in Remark 3.9. Thus,
if a $(c,p)$-anisotropic knot $K$ has $\rho^{(1)}_{c,p}(K)\neq 0$, then it
also obstructs $K_{c,1}$ from being strongly rationally slice, namely
rationally slice with complexity $1$. This reminds us of the fact that $K$ is
rationally slice with complexity $c$ if and only if $K_{c,1}$ is strongly
rationally slice [CFHH13].

Now we provide a sufficient condition for $K$ to be $(c,p)$-anisotropic.

######  Proposition 4.5.

Let $K$ be a knot and $\Delta_{K}$ be its Alexander polynomial. If each factor
of $p$ is symmetric and divides $\Delta_{K}(t^{c})$ with multiplicity at most
$1$, then $K$ is $(c,p)$-anisotropic.

###### Proof.

Since $(\mathcal{A}_{c}(K),\mathcal{B}\ell_{c})$ is equivalent with
$(\mathcal{A}(K_{c,1}),\mathcal{B}\ell)$, the $(c,p)$-anisotropy condition of
$K$ is equivalent with the $p$-anisotropy condition of $K_{c,1}$. Also note
that $\Delta_{K}(t^{c})=\Delta_{K_{c,1}}(t)$. Thus, the result follows from
the fact that $p$-anisotropy of $K_{c,1}$ holds by the Lemma 2.17. ∎

Then the followings are clear.

######  Proposition 4.6.

Suppose that each factor of $p$ is symmetric. $K$ is $(c,p)$-anisotropic if
one of the following holds:

  * (a)

$\Delta_{K}(t^{c})$ is squarefree.

  * (b)

$\Delta_{K}(t^{c})$ and $p$ are coprime.

A stronger but simpler condition is irreducibility for $\Delta_{K}(t^{c})$,
which follows from Proposition 4.6(a). We say a polynomial $p(t)$ is strongly
irreducible if $p(t^{c})$ is irreducible for all positive integers $c$. $p(t)$
and $q(t)$ are said to be strongly coprime if $p(t^{c})$ and $q(t^{d})$ are
coprime for all positive integers $c$, $d$.

######  Proposition 4.7.

If $\Delta_{K}$ is strongly irreducible, then $K$ is
$(c,\Delta_{K}(t^{c}))$-anisotropic for all $c$.

Now we are ready to prove the obstructive Theorem 1.2 which is stated without
$(c,p)$-anisotropy condition.

######  Theorem 1.2.

Let $\\{K_{i}\\}$ be a set of knots whose order is finite in
$\mathcal{AC}_{\mathbb{Q}}$ and $\Delta_{i}$ be the Alexander polynomial of
$K_{i}$. If $\Delta_{i}$’s are strongly irreducible and pairwisely strongly
coprime, and $\rho^{(1)}(K_{i})\neq 0$, then $\\{K_{i}\\}$ is linearly
independent in $\mathcal{C}_{\mathbb{Q}}$.

###### Proof.

Since $K_{i}$ has finite order in $\mathcal{AC}_{\mathbb{Q}}$, the Levine-
Tristram signature function $\sigma_{K_{i}}(\omega)=0$ by [CO93], [CK02].
Thus, by Theorem 2.8, the zeroth order signature vanishes:

$\rho^{(0)}(K_{i})=0.$

Fix any index $j$, and let $p(t)=\Delta_{j}(t^{c})$. Note that $K_{i}$ is
$(c,p)$-anisotropic for either $i=j$ or $i\neq j$ by Proposition 4.6(b) and
Proposition 4.7. By Proposition 3.4,

$\displaystyle\rho_{c,p}(K_{i})$ $\displaystyle=\rho^{(0)}(K_{i})=0\text{ if
}i\neq j.$ $\displaystyle\rho_{c,p}(K_{i})$
$\displaystyle=\rho^{(1)}(K_{i})\neq 0\text{ if }i=j.$

Given $J_{1},\cdots,J_{n}$ arbitrarily chosen in $\\{K_{i}\\}$, suppose that
$a_{1}J_{1}\\#\cdots\\#a_{n}J_{n}$ is rationally slice with some complexity
$c$. Let $p(t)=\Delta_{k}(t^{c})$ for a fixed $k\in\\{1,\cdots,n\\}$. Then by
Theorem 4.2, Proposition 3.7 and the above result for $\rho_{c,p}$,

$\displaystyle\rho_{c,p}^{(1)}(a_{1}J_{1}\\#\cdots a_{n}J_{n})$
$\displaystyle=a_{1}\rho_{c,p}^{(1)}(J_{1})+\cdots
a_{n}\rho_{c,p}^{(1)}(J_{n})$ $\displaystyle=\sum\limits_{i\neq
k}a_{i}\rho^{(0)}(J_{i})+a_{k}\rho^{(1)}(J_{k})=0.$

Since $\rho^{(0)}(K_{i})=0$ and $\rho^{(1)}(K_{i})\neq 0$ for all $i$, we have
$a_{k}=0$. In this way, $a_{1}=\cdots=a_{n}=0$. ∎

On the other hand, below corollary immediately follows from Remark 4.4 in the
exactly same way.

######  Corollary 4.8.

Suppose $\\{K_{i}\\}_{i\in I}$ satisfies all conditions in above Theorem 1.2.
Then $\\{(K_{i})_{c,1}:i\in I,c\in\mathbb{Z}^{+}\\}$ is linearly independent
in the classical knot concordance group $\mathcal{C}$.

We remark that Corollary 4.8 can be restated in slightly stronger way by
replacing $\mathcal{C}$ with the strong rational concordance group. The latter
means the quotient of $\mathcal{C}$ by strongly rationally slice knots.

##  5\. Examples and Remark on $(c,1)$-cable

Recall that the $n$-twist knot is the positive Whitehead double of the unknot
with $n$ twists, illustrated in Figure 1. In this section, we prove Theorem
1.3 by applying our rational slice obstruction Theorem 1.2 to $n$-twist knots
which are of order 2 in the algebraic rational concordance group
$\mathcal{AC}_{\mathbb{Q}}$. Let $K_{n}$ be such an $n$-twist knot and
$\Delta_{n}$ be its Alexander polynomial. Since they are algebraically
torsions, their zeroth order signatures $\rho^{(0)}(K)$ vanish by Theorem 2.8.
Davis [Dav12a, Dav12b] computed their first order signature
$\rho^{(1)}(K_{n})$ except for finitely many (thirty nine) $n$. The remaining
thing is just to show that there exist infinitely many such $K_{n}$’s which
satisfy the required condition on $\Delta_{n}$ in Theorem 1.2.

######  Theorem 5.1.

[Dav12a, Lemma 7.1] (cf. [Lev69]) $K_{n}$ is order $2$ in $\mathcal{AC}$ if
and only if there exist positive integers $a$ and $b$ such that
$n=a^{2}-a+b^{2}$ but there does not exist an integer $m$ such that
$n=m^{2}-m$.

######  Proposition 5.2.

[BD12]

  * (a)

$\Delta_{n}$ and $\Delta_{m}$ are strongly coprime for distinct integers $n$
and $m$.

  * (b)

$\Delta_{n}$ is strongly irreducible if $n$ is neither $m(m-1)$ nor $m^{k}$
for $k>1$ for some $m$.

Since the order $2$ cases in Theorem 5.1 may include some perfect power cases,
we should exclude those to exploit strong irreducibility.

######  Proposition 5.3.

There are infinitely many positive integer $n$ such that $K_{n}$ is of order 2
in $\mathcal{AC}$ and $\Delta_{n}$ is strongly irreducible. In particular,
such $K_{n}$ has order $2$ in $\mathcal{AC}_{\mathbb{Q}}$.

###### Proof.

Let $k$ be any odd integer and $n=36k^{2}-6k+4$. Then $n=6k(6k-1)+4\equiv 1$
mod $3$, but any number of the form $m(m-1)$ is congruent to either $0$ or $2$
mod $3$. Thus, $K_{n}$ has order 2 in $\mathcal{AC}$. Moreover,
$n=2(18k^{2}-3k+2)$ has a factor $2$ with multiplicity $1$ so that it cannot
be any perfect power. Thus, $\Delta_{n}$ is strongly irreducible so that, by
generalized Fox-Milnor condition, $K_{n}$ is not algebraically rationally
slice. ∎

Now we are ready to prove Theorem 1.3.

###### Proof of Theorem 1.3.

Choose infinitely many $K_{n}$ obtained from Proposition 5.3. They are of
order $2$ in $\mathcal{AC}_{\mathbb{Q}}$. Their Alexander polynomials are
strongly irreducible, and pairwisely strongly coprime by Proposition 5.2. For
such $n$’s with finitely many exceptions, Davis showed that
$\rho^{(1)}(K_{n})\neq 0$ in [Dav12a, Dav12b]. Thus, the conditions in Theorem
1.2 for $\\{K_{i}\\}$ to generate $\mathbb{Z}^{\infty}$ in
$\mathcal{C}_{\mathbb{Q}}$ are satisfied. ∎

On the other hand, when $n=m^{2}$, $K_{n}$ is of order $2$ in $\mathcal{AC}$
while it is of order $1$ in $\mathcal{AC}_{\mathbb{Q}}$ [BD12]. Indeed,
$\Delta_{n}(t^{2})$ splits into symmetric pairs. When $m>8$,
$\rho^{(1)}(K_{n})\neq 0$ as in [Dav12a, Dav12b]. By Theorem 2.12, such
$K_{n}$ is not rationally slice. Nevertheless, we cannot conclude that it is
of infinite order in $\mathcal{C}_{\mathbb{Q}}$ since $K_{n}$ for such
$n=m^{2}$ is not $(2,\Delta_{n}(t^{2}))$-anisotropic.

######  Question 4.

Let $n=m^{2}$. Is $K_{n}$ of infinite order in $\mathcal{C}_{\mathbb{Q}}$?
Even when $m\leq 8$, is $K_{n}$ not rationally slice?

Both answers for $n=m=1$ are “no” because $K_{1}$ is the figure-eight knot
$4_{1}$, which is rationally slice.

On the other hand, we also obtain linear independence of their cable in the
classical knot concordance group $\mathcal{C}$. In particular, we restate
Corollary 1.4 in a more general way. Its proof is done by Corollary 4.8.

######  Corollary 1.4.

$\\{(K_{n})_{c,1}:K_{n}\text{ is of order }2\text{ in
}\mathcal{AC}_{\mathbb{Q}}\text{ with }\rho^{(1)}(K_{n})\neq
0,c\in\mathbb{Z}_{+}\\}$ generates an infinite rank subgroup in $\mathcal{C}$.

Corollary 1.4 reminds us of the following question, a stronger version of
[Miy94, Question 3], asked by Kang-Park [KP22].

######  Question 5.

If $K$ is not slice, then does $K_{c,1}$ have infinite order in $\mathcal{C}$?

The twist knots $K_{n}$ in Corollary 1.4, which are not algebraically
rationally slice, give a partially positive answer for above question. In
particular, such cable $(K_{n})_{c,1}$ is of infinite order in $\mathcal{C}$.
For other cases, in particular, the case when $n=1$, it is hard to tell that
its cable has infinite order in $\mathcal{C}$ since $K_{1}=4_{1}$ itself is
not of infinite order in $\mathcal{C}$. It turned out that $(4_{1})_{c,1}$ for
odd $c>1$ are linearly independent in smooth category [HKPS22] (see also
[KP22]). This case is especially important because those cables are members of
infinite order in the kernel of
$\psi:\mathcal{C}\rightarrow\mathcal{C}_{\mathbb{Q}}$. However, there is no
known analogue in the topological category. Question 3 is asked here again
with more specific candidates:

######  Question 6.

Is $(4_{1})_{\text{odd},1}$ of infinite order in $\mathcal{C}$?

On the other hand, when $c$ is even, $(4_{1})_{c,1}$ is strongly rationally
slice [Kaw80]. When $c$ is $2n$ for any odd $n$, it is known that they have
infinite order in smooth category [DKM+22, KPT24] (see also [ACM+23]). In
particular, they are examples of strongly rationally slice but not smoothly
slice. It still remains unknown whether they are topologically slice or not.
Even if $\rho$ can be used to obstruct a knot from being topologically slice,
our obstruction is far from being able to obstruct topological sliceness of
$(4_{1})_{c,1}$.

######  Question 7.

Does there exist a strongly rationally slice but not topologically slice knot?

##  6\. Remark on BPH-slice knots

In this section, we consider smooth category. A knot $K$ is called H-slice in
a closed, smooth, orientable 4-manifold $W$ if $K\subset
S^{3}=\partial(W-B^{4})$ bounds a smoothly embedded nullhomologous disk in
$W-B^{4}$. $K$ is said to be $0$-positive (negative, resp) if it is H-slice in
a simply-connected positive (negative, resp) definite $W^{+}$ ($W^{-}$, resp).
The following facts are well known:

######  Theorem 6.1.

[CHH13] If $K$ is 0-positive, then

  * (a)

the signature $\sigma(K)\leq 0$,

  * (b)

the Levine-Tristram signature function $\sigma_{K}(\omega)\leq 0$,

  * (c)

Ozsváth-Szabó’s $\tau(K)\geq 0$.

  * (d)

Ozsváth-Szabó’s correction term $d(S^{3}_{p/q}(K),i)\leq d(L(p,q),i)$.

If $K$ is 0-negative, then the inequalities above are given in the opposite
way.

A knot $K$ is called $0$-bipolar if it is both $0$-positive and $0$-negative.
In such a particular case, Hom’s $\epsilon(K)=0$ [CHH13]. Since
$d(S^{3}_{p/q}(K),i)=d(L(p,q),i)$ for all coprime integers $p$ and $q$,
$\nu^{+}(K)=0$ by [NW15, Theorem 2.5] and [HW16, Proposition 2.3]. It follows
from $\nu^{+}(K)=0$ that $\Upsilon_{K}(t)=0$ by [OSS17, Proof of Theorem 1.1].
In summary, if $K$ is $0$-bipolar, then we have (see also [CK21]):

$\sigma(K)=\sigma_{K}(\omega)=\tau(K)=\epsilon(K)=\Upsilon_{K}(t)=\nu^{+}(K)=0.$

On the other hand, a $0$-bipolar knot is called biprojective H-slice (or, BPH-
slice) if definite 4-manifolds $W^{\pm}$ in which $K$ is H-slice are specified
as $n\mathbb{CP}^{2}$ and $n\overline{\mathbb{CP}^{2}}$. Recall the
Rasmussen’s $s$-invariant [Ras10]. Recently it has turned out that:

######  Theorem 6.2.

[MMSW23] If $K$ is H-slice in $n\mathbb{CP}^{2}$
($n\overline{\mathbb{CP}^{2}}$, resp) for some $n$, then $s(K)\geq 0$ ($\leq
0$, resp). In particular, if $K$ is BPH-slice, then

$s(K)=0.$

By works of Donaldson [Don83] and Freedman [Fre82], every simply-connected
positive definite smooth 4-manifold is homeomorphic to $n\mathbb{CP}^{2}$
while it is not known if there exists an exotic $n\mathbb{CP}^{2}$ or
$n\overline{\mathbb{CP}^{2}}$. Therefore, it is natural to ask the following
questions:

######  Question 8.

Does there exist a 0-bipolar but not BPH-slice knot?

If such a knot exists, then it implies that there exists a simply-connected
closed exotic definite 4-manifold. For existence of non-simply-connected case,
see [LLP23, SS23]. The notion of H-slice and BPH-slice has been introduced
recently. H-slice knots can be used to detect an exotic indefinite 4-manifolds
in gauge theoretic way [MMP24], and BPH-slice knots are used to construct a
potential exotic $n\mathbb{CP}^{2}$, including $S^{4}$ for the case when $n=0$
[MP23]. Such homotopy $n\mathbb{CP}^{2}$’s are shown to be standard [Nak23].

It is interesting to compare the set of rationally slice knots and the set of
BPH slice knots. We make a remark that every known examples of rationally
slice knots are BPH-slice. On the other hand, not all BPH-slice knots are
rationally slice knots (see [CHH13]). As a corollary of Theorem 1.3, we can
even take twist knots as such examples:

######  Corollary 6.3.

There are infinitely many BPH-slice twist knots that generate
$\mathbb{Z}^{\infty}$ subgroup of $\mathcal{C}_{\mathbb{Q}}$.

Note that if $K$ is rationally slice, then:

$\sigma(K)=\sigma_{K}(\omega)=\tau(K)=\epsilon(K)=\Upsilon_{K}(t)=\nu^{+}(K)=0.$

On the other hand, it is not known if the Rasmussen’s $s$-invariant vanishes
for rationally slice knots. We end this article with the following question.

######  Question 9.

Does there exist a rationally slice but not BPH-slice knot?

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