# Examples of topologically unknotted tori

András Juhász  Mathematical Institute, University of Oxford, Andrew Wiles
Building, Radcliffe Observatory Quarter, Woodstock Road, Oxford, OX2 6GG, UK
<EMAIL_ADDRESS>and  Mark Powell  School of Mathematics and Statistics,
University of Glasgow, University Place, Glasgow, G12 8QQ, UK
<EMAIL_ADDRESS>
###### Abstract.

We show that certain smooth tori with group $\mathbb{Z}$ in $S^{4}$ have
exteriors with standard equivariant intersection forms, and so are
topologically unknotted. These include the turned 1-twist-spun tori in the
4-sphere constructed by Boyle, the union of the genus one Seifert surface of
Cochran and Davis, that has no slice derivative, with a ribbon disc, and any
torus with precisely four critical points. This gives evidence towards the
conjecture that all $\mathbb{Z}$-surfaces in $S^{4}$ are topologically
unknotted, which is open for genus one and two. It is unclear whether these
tori are smoothly unknotted. The double cover of $S^{4}$ branched along any of
these surfaces is a potentially exotic copy of $S^{2}\times S^{2}$ that cannot
be distinguished from $S^{2}\times S^{2}$ using Seiberg–Witten invariants.

###### Key words and phrases:

4-manifolds, embedded surfaces, isotopy

###### 1991 Mathematics Subject Classification:

57K40, 57K10, 57N35.

##  1\. Introduction

We study the topological isotopy problem for locally flat embeddings of
closed, orientable surfaces in $S^{4}$. Let $\Sigma\subseteq S^{4}$ be such a
surface of genus $g$. If $\pi_{1}(S^{4}\setminus\Sigma)\cong\mathbb{Z}$,
generated by a meridian of $\Sigma$, then we say that $\Sigma$ is a
_$\mathbb{Z}$ -surface_. The surface $\Sigma$ is _topologically unknotted_ if
it is topologically isotopic to the standard genus $g$ surface in $S^{3}$,
i.e. if it topologically bounds a genus $g$ handlebody. If $\Sigma$ has genus
one, then we say that $\Sigma$ is a $\mathbb{Z}$-torus. Conway and the second
author [CP23] have shown that any locally flat, embedded, closed, orientable
$\mathbb{Z}$-surface $\Sigma\subseteq S^{4}$ with $g(\Sigma)\not\in\\{1,2\\}$
is topologically unknotted. The topological unknotting conjecture states that
this also holds for $g\in\\{1,2\\}$.

Using the topological classification of $\mathbb{Z}$-surfaces in terms of the
equivariant intersection forms of their exteriors of Conway and the second
author [CP23], we show that certain $\mathbb{Z}$-tori $\Sigma$ are
topologically unknotted. Our three main results, for the three types of tori
we investigate, are described in Sections 1.1, 1.2, and 1.3 below.

We say that a pair of smoothly embedded surfaces in a 4-manifold are _exotic_
if they are topologically but not smoothly isotopic. This is equivalent to
saying that they are orientation-preserving ambiently homeomorphic but not
orientation-preserving ambiently diffeomorphic. Finashin, Kreck, and Viro
[FKV87] constructed infinitely many exotic copies of $\\#10\mathbb{R}P^{2}$ in
$S^{4}$. This was later improved by Finashin [Fin09] to $\\#6\mathbb{R}P^{2}$
in $S^{4}$. These are distinguished by their double branched covers. Further
improvements were recently announced in [Miy23, MOS23]. However, there is no
example known of an exotic pair of closed, orientable surfaces in $S^{4}$.

This work can be viewed as collecting evidence towards the genus one
topological unknotting conjecture. On the other hand, the examples we consider
are potentially exotic surfaces, i.e. so far as we know, they could be
counterexamples to the smooth unknotting conjecture for tori; cf. Section 8.
Furthermore, if we take the double cover $\Sigma(S^{4},T)$ of $S^{4}$ branched
along any of these tori $T$, we obtain a potentially exotic copy of
$S^{2}\times S^{2}$ that cannot be distinguished from $S^{2}\times S^{2}$
using Seiberg–Witten invariants. In two of the three constructions we consider
below, $\Sigma(S^{4},T)$ is diffeomorphic to $(S^{2}\times S^{2})\\#X$ for a
homotopy 4-sphere $X$.

###  1.1. Turned twisted tori

Given a knot $K$ in $S^{3}$, Artin [Art25] associated to it a 2-knot (i.e. an
embedded 2-sphere) in $S^{4}$ using a construction called _spinning_. This was
later extended by Zeeman [Zee65] to _twist-spinning_ , and then by Litherland
[Lit79] to _deform-spinning_. Building on this, in 1993, Boyle [Boy93]
associated to a knot $K$ in $S^{3}$ and integers $i$, $j$ the _$i$ -turned
$j$-twist-spun torus_ in $S^{4}$, which we will denote by $T_{K,i,j}$. See
Section 3 for the definition. The smooth isotopy class of $T_{K,i,j}$ only
depends on the parity of $i$. At the end of his paper, Boyle noted that the
torus $T_{K,1,1}$ is a $\mathbb{Z}$-torus and asked whether it is standard.

######  Theorem 1.1.

For every knot $K$ in $S^{3}$, the turned 1-twist spun torus $T_{K,1,1}$ is
topologically unknotted.

We also observe in Section 8 that the double cover of $S^{4}$ branched along
$T_{K,1,1}$ is $(S^{2}\times S^{2})\\#X$ for a homotopy 4-sphere $X$. So one
cannot use the Seiberg–Witten invariants of the double branched covers to
smoothly distinguish $T_{K,1,1}$ and the standard torus $T^{2}\subseteq
S^{3}\subseteq S^{4}$.

###  1.2. Seifert surface union a ribbon disc

Recall that, given a knot $K$ in $S^{3}$, a _homotopy ribbon disc_ $D$ for $K$
is a locally flat disc embedded in $D^{4}$ such that $\partial D=K$ and the
map

$\pi_{1}(S^{3}\setminus K)\to\pi_{1}(D^{4}\setminus D)$

is surjective. In particular, every ribbon disc is homotopy ribbon; see Gordon
[Gor81]. Let $K$ be a knot in $S^{3}$ with Seifert surface $F$ and homotopy
ribbon disc $D$. Then $\Sigma:=F\cup D\subseteq S^{4}$ is a
$\mathbb{Z}$-surface.

A _slice derivative_ for a genus $g$ Seifert surface $F$ for a knot $K$ in
$S^{3}$ is a collection of $g$ pairwise disjoint, homologically linearly
independent simple closed curves that form a smoothly slice link and have
self-linking zero [CHL10]. If $F$ has a slice derivative, we can perform
surgery on $F$ in $D^{4}$ along a slice derivative and obtain a smooth slice
disc $D$ for $K$. Then $F\cup D$ is smoothly unknotted; see Lemma 6.1. If $D$
is a ribbon disc that is not obtained from a Seifert surface $F$ by surgering
along a slice derivative, Gabai has asked whether $F\cup D$ is topologically
or smoothly unknotted.

Kauffman conjectured that every genus one Seifert surface for a slice knot
admits a slice derivative. Cochran and Davis [CD15] have shown that there
exists a ribbon knot admitting a genus one Seifert surface $F$ that has no
slice derivative. Moreover, there is an example where $F$ is the unique
minimal genus one Seifert surface, up to isotopy. The union of $F$ and a
ribbon disc is therefore a potentially smoothly knotted $\mathbb{Z}$-torus in
$S^{4}$. We show that it is topologically unknotted.

######  Theorem 1.2.

Let $K$ be the knot in $S^{3}$ with genus one Seifert surface $F$ and ribbon
disc $D$ constructed by Cochran and Davis. Then $\Sigma:=F\cup D$ is
topologically unknotted.

In 6.2, we note that a genus one Seifert surface union a slice disc for a knot
$K$ is a topologically unknotted $\mathbb{Z}$-torus also when
$\Delta_{K}\doteq 1$.

We will show in Section 8 that the double cover $M$ of $S^{4}$ branched along
$\Sigma$ can be obtained from $S^{2}\times S^{2}$ by knot surgery along a
0-homologous torus. By Theorem 1.2, the 4-manifolds $M$ and $S^{2}\times
S^{2}$ are homeomorphic. We do not know whether $M$ is diffeomorphic to
$S^{2}\times S^{2}$ or even whether it is irreducible. One cannot distinguish
$M$ from $S^{2}\times S^{2}$ using the knot surgery formula of Fintushel and
Stern [FS98], since it requires the torus to be homologically essential.

Theorem 1.2 follows from the following result on knot surgery along a torus in
the complement of an unknotted surface in $S^{4}$.

######  Theorem 1.3.

Let $\Sigma\subseteq S^{4}$ be a smoothly embedded, topologically unknotted,
oriented surface of genus $g$. Furthermore, let $T\subseteq S^{4}\setminus
N(\Sigma)$ be a smoothly embedded torus that is unknotted in $S^{4}$, and let
$J$ be a knot in $S^{3}$. Suppose that the push-off $s(m)$, for some
primitive, essential curve $m$ on $T$, is 0-homotopic in
$S^{4}\setminus(\Sigma\cup T)$. Then the $t$-twist, $r$-roll knot surgery
$\Sigma_{t,r}(T,J)$ is topologically unknotted for all $t$, $r\in\mathbb{Z}$.

For further details on the $t$-twist $r$-roll knot surgery construction
involved, see Section 5.

###  1.3. Tori with four critical points

Let $\Sigma\subseteq S^{4}$ be a smoothly embedded torus, and suppose that the
standard Morse function on $S^{4}$, with precisely two critical points,
restricts to $\Sigma$ to a Morse function $f\colon\Sigma\to\mathbb{R}$ with
precisely four critical points. Then we say that $\Sigma$ _has four critical
points_. In this case, the exterior of $\Sigma$ has a handle decomposition
with a unique 1-handle, so $\Sigma$ is a $\mathbb{Z}$-torus.

######  Theorem 1.4.

Let $\Sigma\subseteq S^{4}$ be a smoothly embedded torus with four critical
points. Then $\Sigma$ is topologically unknotted.

This answers Problem 4.30 of Kirby [Kir21] in the topological category.
Scharlemann [Sch85] has shown that every 2-sphere in $S^{4}$ with four
critical points is smoothly unknotted, and Bleiler–Scharlemann [BS88] have
shown that every embedded $\mathbb{RP}^{2}\subseteq S^{4}$ with three critical
points is smoothly unknotted. As far as we know, it is open whether every
torus with four critical points is smoothly unknotted. We are not aware of any
explicit potentially smoothly knotted examples associated with 1.4.

In Section 8, we will show that the double cover of $S^{4}$ branched along a
torus $\Sigma$ with four critical points is diffeomorphic to $(S^{2}\times
S^{2})\\#X$ for a homotopy 4-sphere $X$.

### Notation

Throughout, we write

$D^{3}_{r}:=\\{v\in\mathbb{R}^{3}:|v|\leq r\\}\text{ and
}S^{2}_{r}:=\\{v\in\mathbb{R}^{3}:|v|=r\\}.$

Let $B^{3}_{r}:=\operatorname{Int}(D^{3}_{r})$. As usual, $D^{3}:=D^{3}_{1}$,
$S^{2}:=S^{2}_{1}$, and $B^{3}:=B^{3}_{1}$. If $A$ is a submanifold of $B$,
then we write $N(A)$ for an open tubular neighbourhood of $A$ in $B$ and
$\nu_{A\subseteq B}$ for the normal bundle of $A$ in $B$. For smooth manifolds
$M$ and $N$, we write $M\cong N$ if they are diffeomorphic. Finally, let

$\mathbb{R}^{3}_{+}:=\\{(x,y,z)\in\mathbb{R}^{3}:z\geq 0\\}\text{ and
}\mathbb{R}^{3}_{>0}:=\\{(x,y,z)\in\mathbb{R}^{3}:z>0\\}.$

### Organisation

In Section 2, we set out the strategy that we will use for all our proofs,
which is to compute the equivariant intersection pairing of the surface knot
exterior, and apply [CP23]. In Section 3, we recall the construction of
turned, twisted tori. Then, in Section 4, we show that they are topologically
unknotted, proving 1.1. In Section 5, we prove 1.3 on knot surgeries. In
Section 6, we prove 1.2, and in Section 7, we prove 1.4. In Section 8, we
observe that using the double branched cover to detect that some turned
twisted torus is exotic, or to show that a torus with four critical points is
exotic, would entail finding a counterexample to the smooth 4-dimensional
Poincaré conjecture.

### Acknowledgements

We would like to thank Anthony Conway, Chris Davis, David Gabai, Jason Joseph,
Maggie Miller, András Stipsicz, Zoltán Szabó, and Ian Zemke for helpful
discussions and suggestions. MP was partially supported by EPSRC New
Investigator grant EP/T028335/2 and EPSRC New Horizons grant EP/V04821X/2.

##  2\. Standard intersection form implies unknotted

For a $\mathbb{Z}$-surface $\Sigma\subseteq S^{4}$, let

$E_{\Sigma}:=S^{4}\setminus N(\Sigma)$

be the exterior of $\Sigma$. Let $\Lambda$ be the group ring
$\mathbb{Z}[\mathbb{Z}]$, which is isomorphic to the ring of Laurent
polynomials $\mathbb{Z}[t,t^{-1}]$. This admits an involution determined by
$t\mapsto t^{-1}$. Consider the Hermitian form

$\mathcal{H}_{2}:=\left(\Lambda^{2},\begin{pmatrix}0&1-t\\\
1-t^{-1}&0\end{pmatrix}\right).$

The following is a consequence of [CP23, Theorem 1.4] and [CP23, Lemma 6.1].

######  Theorem 2.1.

Let $\Sigma\subseteq S^{4}$ be $\mathbb{Z}$-surface of genus $g$. Then the
$\Lambda$-intersection form of $E_{\Sigma}$ is isometric to

$\bigoplus_{g}\mathcal{H}_{2}$

if and only if $\Sigma$ is topologically unknotted.

So, to prove Theorems 1.1, 1.2, and 1.4, it suffices to compute the
equivariant intersection pairings of the surface exterior.

Next we record some homological facts about the exteriors of
$\mathbb{Z}$-surfaces in $S^{4}$. If $\Sigma\subseteq S^{4}$ is an embedded
surface and $\gamma\subseteq\Sigma$ is a simple closed curve, then we write
$T_{\gamma}$ for the total space of the restriction of the unit normal circle
bundle of $\Sigma$ to $\gamma$, which is known as the _rim torus_ over
$\gamma$.

######  Proposition 2.2.

Let $\Sigma\subseteq S^{4}$ be a closed, connected, and oriented
$\mathbb{Z}$-surface of genus $g$. Then

  1. (i)

$H_{0}(E_{\Sigma};\Lambda)\cong\mathbb{Z}$;

  2. (ii)

$H_{1}(E_{\Sigma};\Lambda)=0$;

  3. (iii)

$H_{1}(\partial E_{\Sigma};\Lambda)\cong H_{1}(\Sigma\times
S^{1};\Lambda)\cong(\Lambda/(1-t))^{\oplus 2g}$;

  4. (iv)

$H_{2}(E_{\Sigma};\mathbb{Z})\cong\mathbb{Z}^{2g}$, generated by rim tori;

  5. (v)

$H_{2}(E_{\Sigma};\Lambda)\cong\Lambda^{2g}$, generated by generically
immersed spheres that map to the classes of the rim tori under the surjective
map $H_{2}(E_{\Sigma};\Lambda)\to H_{2}(E_{\Sigma};\mathbb{Z})$ induced by
tensoring with $\mathbb{Z}$ over $\Lambda$;

  6. (vi)

$H_{3}(E_{\Sigma};\Lambda)=0$.

###### Proof.

Part (iii) is a special case of [CP23, Lemma 5.5]. Part (iv) follows from
Alexander duality. All the other parts follow from [CP23, Lemma 3.2], except
for (v), where some more argument is required, which we give now. In [CP23,
Lemma 3.2], it is shown that $H_{2}(E_{\Sigma};\Lambda)$ is free. The rank
follows from considering $\mathbb{Q}(t)$ coefficients: to see this, note that,
by (i), (ii), and (vi), the Euler characteristic over $\mathbb{Q}(t)$ equals
$\operatorname{rk}H_{2}(E_{\Sigma};\Lambda)$, whereas computing over
$\mathbb{Q}$ shows that $\chi(E_{\Sigma})=2g$. Hence
$\operatorname{rk}H_{2}(E_{\Sigma};\Lambda)=2g$.

The elements of the standard basis $\\{e_{i}\\}_{i=1}^{2g}$ of
$H_{2}(E_{\Sigma};\Lambda)\cong\Lambda^{2g}$ are represented by generically
immersed spheres, which we can see as follows. Apply successively the Hurewicz
theorem to the universal cover of $E_{\Sigma}$ to see that
$\pi_{2}(E_{\Sigma})\cong H_{2}(E_{\Sigma};\Lambda)$, the density of smooth
immersions in $C^{0}(S^{2},M)$ [Hir76, Theorems 2.2.6 and 2.2.12], and finally
the density of generic immersions in the space of smooth immersions [GG73,
Chapter III, Corollary 3.3]. Since the $\\{e_{i}\\}_{i=1}^{2g}$ are
represented by generically immersed spheres, so are $\\{e_{i}\otimes
1\\}_{i=1}^{2g}\subseteq
H_{2}(E_{\Sigma};\Lambda)\otimes_{\Lambda}\mathbb{Z}\cong\Lambda^{2g}\otimes_{\Lambda}\mathbb{Z}$.

It was shown in the proof of [CP23, Lemma 5.10] that the canonical map
$H_{2}(E_{\Sigma};\Lambda)\otimes_{\Lambda}\mathbb{Z}\xrightarrow{\cong}H_{2}(E_{\Sigma};\mathbb{Z})$
is an isomorphism. It follows that $H_{2}(E_{\Sigma};\Lambda)\to
H_{2}(E_{\Sigma};\mathbb{Z})$ is indeed surjective. Moreover, rim tori
generate the codomain $H_{2}(E_{\Sigma};\mathbb{Z})\cong\mathbb{Z}^{2g}$ by
(iv). The inverse images of the rim tori in
$H_{2}(E_{\Sigma};\Lambda)\otimes_{\Lambda}\mathbb{Z}$ are linear combinations
of the basis elements $\\{e_{i}\otimes 1\\}_{i=1}^{2g}$, which we can write as
$\sum_{j}a_{ij}(e_{j}\otimes 1)=\sum_{j}a_{ij}e_{j}\otimes 1$ for some
$a_{ij}\in\mathbb{Z}$ such that $(a_{ij})_{i,j=1}^{2g}$ is invertible over
$\mathbb{Z}$. Hence the elements
$\\{e_{i}^{\prime}:=\sum_{j}a_{ij}e_{j}\\}_{i=1}^{2g}$ generate
$H_{2}(E_{\Sigma};\Lambda)$, because $(a_{ij})_{i,j=1}^{2g}$ is also
invertible over $\Lambda$, and the $e_{i}^{\prime}$ map to the rim tori. By
taking parallel copies of the $e_{j}$ and tubing together, each of the
$e_{i}^{\prime}$ can also be represented by generically immersed spheres. ∎

######  Definition 2.3.

Let $E$ be a compact, connected, oriented, and based 4-manifold with
$\pi_{1}(E)\cong\mathbb{Z}$. We say that a closed, connected, and oriented
surface $S\looparrowright E$ immersed in $E$, and equipped with a basing arc
from $S$ to the basepoint of $E$, is _$\mathbb{Z}$ -trivial_ if the inclusion-
induced map $\pi_{1}(S)\to\pi_{1}(E)$ is the trivial homomorphism.

The next lemma encapsulates the strategy for the proofs of all of our
theorems.

######  Lemma 2.4.

Let $\Sigma\subseteq S^{4}$ be a closed, connected, and oriented
$\mathbb{Z}$-surface of genus $g$. Let $S_{1}$, $S_{2}\subseteq E_{\Sigma}$ be
immersed $\mathbb{Z}$-trivial surfaces, and let $\lambda$ be the
$\Lambda$-intersection form on $H_{2}(E_{\Sigma};\Lambda)$. Suppose that

$A:=[\lambda([S_{i}],[S_{j}])]_{i,j\in\\{1,\dots,2g\\}}=\bigoplus_{k=1}^{g}\begin{pmatrix}0&1-t\\\
1-t^{-1}&c_{k}(1-t)+\overline{c}_{k}(1-t^{-1})\end{pmatrix}$

for some $c_{k}\in\Lambda$. Then $\\{[S_{1}],\dots,[S_{2g}]\\}$ is a basis for
$H_{2}(E_{\Sigma};\Lambda)\cong\Lambda^{2g}$ and $\lambda$ is isometric to
$\bigoplus_{g}\mathcal{H}_{2}$. Hence $\Sigma$ is topologically unknotted by
2.1.

###### Proof.

Let $\langle e_{1},\dots,e_{2g}\rangle$ be a basis for
$H_{2}(E_{\Sigma};\Lambda)\cong\Lambda^{2g}$. Let $B$ be a matrix over
$\Lambda$ such that

(1) $([S_{1}],\dots,[S_{2g}])=B\cdot(e_{1},\dots,e_{2g}),$

i.e. $[S_{i}]=Be_{i}$ for $i\in\\{1,\dots,2g\\}$. By part (iii) of Proposition
2.2, the order of $H_{1}(\partial E_{\Sigma};\Lambda)$ is $(1-t)^{2g}$, which,
up to units in $\Lambda$, equals $(1-t)^{g}(1-t^{-1})^{g}$. Hence, up to
units,

(2)
$\det\left([\lambda(e_{i},e_{j})]_{i,j\in\\{1,2\\}}\right)\doteq(1-t)^{g}(1-t^{-1})^{g}.$

By a computation, the hypothesis, equation (1), another computation, and
finally using equation (2), we obtain that

$\begin{split}(-1)^{g}(1-t)^{g}(1-t^{-1})^{g}&=\det A\\\
&=\det\left([\lambda([S_{i}],[S_{j}])]_{i,j\in\\{1,\dots,2g\\}}\right)\\\
&=\det\left([\lambda(Be_{i},Be_{j})]_{i,j\in\\{1,\dots,2g\\}}\right)\\\
&=\det(B)\det\left([\lambda(e_{i},e_{j})]_{i,j\in\\{1,\dots,2g\\}}\right)\det\big{(}\overline{B^{T}}\big{)}\\\
&\doteq\det(B)\overline{\det(B)}(1-t)^{g}(1-t^{-1})^{g}.\end{split}$

The last equality is again up to units, so we can ignore the $(-1)^{g}$
factor, and deduce that

$(1-t)^{g}(1-t^{-1})^{g}\doteq\det(B)\overline{\det(B)}(1-t)^{g}(1-t^{-1})^{g}.$

Since $\Lambda$ is an integral domain, it follows that $\det(B)=\pm t^{k}$ for
some $k\in\mathbb{Z}$, so $B$ is invertible over $\Lambda$, and hence
$\\{Be_{1},\dots,Be_{2g}\\}$ is a basis, as claimed.

This means that the $\Lambda$-intersection form $\lambda$ of $E_{\Sigma}$ is
represented by $A$. By a change of basis, the form represented by $A$ is
isometric to $\bigoplus_{g}\mathcal{H}_{2}$:

$\begin{pmatrix}1&0\\\ -c_{k}&1\end{pmatrix}\begin{pmatrix}0&1-t\\\
1-t^{-1}&c_{k}(1-t)+\overline{c}_{k}(1-t^{-1})\end{pmatrix}\begin{pmatrix}1&-\overline{c}_{k}\\\
0&1\end{pmatrix}=\begin{pmatrix}0&1-t\\\ 1-t^{-1}&0\end{pmatrix}.$

Hence, the $\Lambda$-intersection form of $E_{\Sigma}$ is isometric to
$\bigoplus_{g}\mathcal{H}_{2}$, and so $\Sigma$ is topologically unknotted by
2.1. ∎

In each of Sections 4, 6, and 7, we will exhibit $\mathbb{Z}$-trivial surfaces
that generate a submodule with $\Lambda$-intersection form represented by $A$,
and then we will appeal to 2.4.

It will sometimes be helpful to compute the homological self-intersection
numbers $\lambda([S_{i}],[S_{i}])\in\Lambda$ in Lemma 2.4 using the Wall self-
intersection number $\mu(S_{i})$, so we recall this next; see [Wal99, Chapter
5]. Let $S\subseteq E_{\Sigma}$ be a $\mathbb{Z}$-trivial surface that is the
image of a generic immersion into $E_{\Sigma}$; i.e. all self-intersections
are transverse double points. For each double point $p\in S$, we obtain a sign
$\varepsilon_{p}$ and a double point loop $\gamma_{p}$ in $S$, with the latter
well-defined up to orientation and homotopy in $E_{\Sigma}$; see e.g. [BKK+21,
Section 11.3]. Thus each double point gives rise to a monomial $\pm
t^{k}\in\Lambda/\langle t^{r}\sim t^{-r}:r\in\mathbb{Z}\rangle$. To define the
target, consider $\Lambda$ as a free abelian group with basis
$\\{t^{r}:r\in\mathbb{Z}\\}$, and take the quotient abelian group where
$t^{r}$ and $t^{-r}$ are identified for all $r$. Taking the sum of
contributions from all the double points $p$ gives

$\mu(S):=\sum_{p}\varepsilon_{p}\gamma_{p}\in\Lambda/\langle t^{r}\sim
t^{-r}:r\in\mathbb{Z}\rangle.$

Let $e(S)\in\mathbb{Z}$ be the Euler number of the normal bundle of $S$, and
let $\iota\colon\mathbb{Z}\to\Lambda$ be the unique such ring homomorphism.
Then we have the useful formula [Wal99, Theorem 5.2]

(3) $\lambda([S],[S])=\mu(S)+\overline{\mu(S)}+\iota(e(S)).$

Here, we must fix a representative of $\mu(S)$ in $\Lambda$, and note that the
sum $\mu(S)+\overline{\mu(S)}\in\Lambda$ is independent of the choice of
representative. The next lemma will be useful for computing intersections
numbers.

######  Lemma 2.5.

Let $\Sigma\subseteq S^{4}$ be a $\mathbb{Z}$-surface and let $T\subseteq
E_{\Sigma}$ be a rim torus. Let $D\subseteq E_{\Sigma}$ be a
$\mathbb{Z}$-trivial immersed surface with one boundary component that is a
longitude of $T$. Suppose that the interior of $D$ intersects $T$
transversely, and suppose that a section of $\nu_{\partial D\subseteq T}$
extends to a nonvanishing section of $\nu_{D\subseteq E_{\Sigma}}$. Let $S$
denote the result of surgering $T$ along $D$. Let $G\subseteq E_{\Sigma}$ be
an immersed $\mathbb{Z}$-trivial surface disjoint from $T$ and intersecting
$D$ transversely in precisely one point, i.e. $|D\pitchfork G|=1$.

  1. (i)

Every intersection point of $\operatorname{Int}(D)$ with $T$ contributes $\pm
t^{k}(1-t)$ to $\mu(S)$, for some sign and some $k\in\mathbb{Z}$.

  2. (ii)

Possibly after changing the orientation and basing arc of $G$, we have that
$\lambda([S],[G])=1-t$.

  3. (iii)

Every self-intersection point of $D$ contributes $\pm t^{k}(1-t)(1-t^{-1})$ to
$\mu(S)$, for some sign and some $k\in\mathbb{Z}$.

  4. (iv)

The intersection number $\lambda([S],[S])$ has the form
$c(1-t)+\overline{c}(1-t^{-1})$, for some $c\in\Lambda$.

###### Proof.

The surgery cuts $T$ open along $\partial D$ and glues in two parallel copies
of $D$. Since $D$ is embedded, and we use the section of $\nu_{D\subseteq
E_{\Sigma}}$ from the statement, we obtain a surface with trivial normal
bundle.

First we prove part (i). Consider an intersection point
$p\in\operatorname{Int}(D)\pitchfork T$. This gives rise to two double points
$p_{+}$ and $p_{-}$ of $S$, which lie at either end of an arc $D^{1}\subseteq
T$ arising from the two parallel copies $D_{+}$ and $D_{-}$ of $D$. Since
$D_{+}$ and $D_{-}$ have opposite orientations induced on them from an
orientation of $S$, the signs of the double points $p_{+}$ and $p_{-}$ differ.
We can compare the corresponding group elements by taking a path $\gamma$ in
$D$, with $\gamma(0)=p$ and $\gamma(1)\in\partial D\subseteq T$. The path
$\gamma$ gives rise to $\gamma\times D^{1}$, with $\gamma\times S^{0}\subseteq
D_{+}\cup D_{-}$. Let $M$ be a meridional disc of $\Sigma$ with boundary on
$T$, such that $\gamma(1)\times D^{1}$ is an arc of $\partial M$. Then
$\delta:=(\gamma\times D^{1})\cup M\subseteq S^{4}$ is a disc with
$\partial\delta$ the union of:

  1. (a)

a meridian of $T$ minus $\gamma(1)\times D^{1}$;

  2. (b)

$\gamma\times S^{0}$;

  3. (c)

$\gamma(0)\times D^{1}$.

See Figure 1. The homotopy class of $\partial\delta$ in
$\pi_{1}(E_{\Sigma})\cong\mathbb{Z}$ measures the difference between the
double point loops of the two double points. Since $\delta$ intersects
$\Sigma$ exactly once, the difference in the contributions to $\mu(S)$ is
$t^{\pm 1}$. So, up to multiplication by $\pm t^{k}$, we have a contribution
$1-t$ or $1-t^{-1}$. But these are equal in $\Lambda/\langle t^{r}\sim
t^{-r}:r\in\mathbb{Z}\rangle$, so this completes the proof of part (i).

$\Sigma$
---
$D$
---
$T$
---
$T$
---
$T$
---
$+$
---
$-$
---
$S$
---
$\delta$
---
$\Sigma$
---
Figure 1. Left: a schematic showing a meridian of the rim torus $T$, a portion
of the surface $D$, and another sheet of $T$ intersecting $D$. Right: after
surgery, with the disc $\delta$ that is used to compute the difference in the
double point loops of the two intersection points in
$\operatorname{Int}(D)\pitchfork T$ shown.

The proof of part (ii) is analogous to the proof of part (i). The intersection
point in $D\pitchfork G$ gives rise to two intersection points between $S$ and
$G$. By the same argument as in the proof of part (i), they contribute $1-t$
or $1-t^{-1}$, up to multiplication by $\pm t^{k}$. But $-t(1-t^{-1})=1-t$,
so, up to units, there is again just one case. By changing the basing arc of
$G$ and its orientation, we can remove the up-to-units indeterminacy $\pm
t^{k}$.

To prove part (iii), note that every self-intersection point of $D$ gives rise
to four self-intersection points of $S$. Any intersections lying on parallel
local sheets differ by multiplication by $-t$ or $-t^{-1}$, by the same
argument as for the previous two parts. Their total contribution to $\mu(S)$
is therefore $\pm t^{k}(1-t)(1-t^{-1})$, for some sign and some
$k\in\mathbb{Z}$. See Figure 2. This proves part (iii).

$\Sigma$
---
$D$
---
$T$
---
$+1$
---
$-t$
---
$S$
---
$\Sigma$
---
+1
---
$-t^{-1}$
---
Figure 2. Each self-intersection point of $D$ gives rise to four self-
intersection points of $S$, which contribute $\pm t^{k}(1-t)(1-t^{-1})$ to
$\mu(S)$ for some $k\in\mathbb{Z}$.

To prove part (iv), first we note that $e(S)=0$, since $S$ arises by surgery
on a framed torus via a compatibly framed disc. By part (iii), the self-
intersection number $\mu(S)$ has a representative in $\Lambda$ that is
divisible by $1-t$, i.e. $\mu(S)=c(1-t)$ for some $c\in\Lambda$. Then, by
equation (3), we have that
$\lambda([S],[S])=\mu(S)+\overline{\mu(S)}=c(1-t)+\overline{c}(1-t^{-1})$, as
required. ∎

The topological unknotting conjecture for genus $g\in\\{1,2\\}$ is implied by
the following algebraic conjecture, which we advertise here. We say that a
hermitian, sesquilinear form on a free module over $\mathbb{Z}[\mathbb{Z}]$ is
_surface hyperbolic_ if it is isometric to $\mathcal{H}_{2}^{\oplus g}$ for
some $g$. Such a form $(P,\lambda)$ is _stably surface hyperbolic_ if there is
an isometry $(P,\lambda)\oplus\mathcal{H}_{2}^{\oplus
k}\cong\mathcal{H}_{2}^{\oplus g}$ for some $k$, $g$.

######  Conjecture 2.6.

Let $(P,\lambda)$ be a hermitian, sesquilinear form on a free
$\mathbb{Z}[\mathbb{Z}]$-module of rank $2g$, where $g\in\\{1,2\\}$, that is
stably surface hyperbolic. Then $(P,\lambda)$ is surface hyperbolic.

An important part of [CP23] was to prove the analogue of this conjecture for
$g\geq 3$.

##  3\. Turned twist-spun tori

In this section, we review Boyle’s construction [Boy93, Definition 5.1]. Let
$K$ be a knot in $D^{3}$ such that $K$ intersects $S^{2}_{1/2}$ in
$\\{(0,0,1/2),(0,0,-1/2)\\}$ transversely, and such that $K\setminus
B^{3}_{1/2}$ is a boundary-parallel arc in $D^{3}\setminus B^{3}_{1/2}$ with
endpoints on $S^{2}_{1/2}$. Furthermore, let $i$ and $j$ be integers.

Let $R_{\theta}$ be rotation of $\mathbb{R}^{3}$ about the $z$-axis by angle
$\theta$. Fix $\varepsilon\in(0,1/2)$. Choose an isotopy $\\{\Phi_{t}:t\in
I\\}$ of $D^{3}$ such that $\Phi_{t}|_{S^{2}_{r}}=\text{Id}_{S^{2}_{r}}$ for
all $r\in(1-\varepsilon,1]$, and

$\Phi_{t}|_{D^{3}_{1/2}}=R_{2\pi jt}|_{D^{3}_{1/2}}$

for every $t\in I$. Write $\phi_{j}:=\Phi_{1}$. In addition, we define the
diffeomorphism $\eta_{i}\colon S^{2}\times S^{1}\to S^{2}\times S^{1}$ via

$\eta_{i}|_{S^{2}\times\\{\theta\\}}:=R_{i\theta}|_{S^{2}}\times\text{Id}_{\\{\theta\\}}$

for each $\theta\in S^{1}$.

######  Definition 3.1.

The _$i$ -turned $j$-twist-spun torus of $K$_ is the pair $(S^{4},T_{K,i,j})$
given by

$\frac{(D^{3},K)\times I}{\\{(x,0)=(\phi_{j}(x),1):x\in
D^{3}\\}}\cup_{\eta_{i}}(S^{2}\times D^{2}),$

where $S^{4}=(D^{3}\times\partial D^{2})\cup_{\eta_{i}}(\partial D^{3}\times
D^{2})$ and $T_{K,i,j}:=K\times I$.

To see that the ambient manifold is indeed $S^{4}$, note that, if $i$ is odd,
then it is obtained from

$S^{4}\cong\partial D^{5}\cong\partial(D^{3}\times
D^{2})\cong(D^{3}\times\partial D^{2})\cup(\partial D^{3}\times D^{2})$

via a Gluck twist along the standard 2-sphere $S_{0}:=S^{2}\times\\{0\\}$. On
the other hand, if $i$ is even, then $\eta_{i}$ is isotopic to
$\text{Id}_{S^{2}\times S^{1}}$, and so we recover the standard $S^{4}$
directly.

Intuitively, we can describe $T_{K,i,j}$ as follows. Let
$K\subseteq\mathbb{R}^{3}_{>0}$ be a knot. Choose a 3-disc
$D\subseteq\mathbb{R}^{3}_{>0}$ such that $\partial D$ intersects $K$
transversely in a pair of antipodal points and $K\setminus D$ is an unknotted
arc. Then we obtain $T_{K,i,j}$ in $\mathbb{R}^{4}\subseteq S^{4}$ by rotating
$\mathbb{R}^{3}_{+}$ around $\mathbb{R}^{2}=\partial\mathbb{R}^{3}_{+}$, and
while doing so, rotating $K\cap D$ around the line through $K\cap\partial D$
and simultaneously rotating $K$ around the $z$-axis.

Boyle showed that the smooth isotopy class of $T_{K,i,j}$ only depends on the
parity of $i$. Indeed, the group of automorphisms of $S^{2}\times S^{1}$
modulo ones that extend to $S^{2}\times D^{2}$ is $\mathbb{Z}/2$, generated by
$\eta_{1}$. Boyle calls $T_{K,0,i}$ the _$j$ -twist-spun torus of $K$_ and
$T_{K,1,j}$ the _turned $j$-twist-spun torus of $K$._

Boyle also showed that $\pi_{1}(S^{4}\setminus T_{K,i,j})\cong G/[G,\mu^{j}]$,
where $G\cong\pi_{1}(S^{3}\setminus K)$ and $\mu\in G$ is the class of a
meridian of $K$. Since $\mu$ normally generates $G$, we have
$[G,\mu]\cong[G,G]$. This isomorphism can be seen using the commutator
identities $[g,\mu^{x}]=\bigl{[}g^{x^{-1}},\mu\bigr{]}^{x}$ and
$[g,\mu^{x}\mu^{y}]=[g,\mu^{x}][g,\mu^{y}]^{\mu^{x}}$, where superscripts
indicate conjugation. Hence

$\pi_{1}(S^{4}\setminus T_{K,i,1})\cong G/[G,G]\cong H_{1}(S^{3}\setminus
K)\cong\mathbb{Z}.$

In fact, $T_{K,0,1}$ is smoothly isotopic to $T^{2}$, since it is the 1-twist-
spun 2-knot obtained from $K$, which is the unknot by Zeeman [Zee65], with a
trivial tube attached. However, Boyle could not determine whether $T_{K,1,1}$
was smoothly equivalent to $T^{2}$, and this question is still open. Note that
one can obtain $T_{K,1,1}$ from $T_{K,0,1}$ via a Gluck twist along the
trivial 2-knot $S_{0}$ that links $T_{K,0,1}$ nontrivially.

##  4\. Every turned 1-twist spun torus is topologically unknotted

In this section, we study turned 1-twist-spun tori up to topological isotopy.
Our main result is the following, restated from the introduction.

See 1.1

###### Proof.

By 2.4, it suffices to find $\mathbb{Z}$-trivial surfaces in

$E:=E_{T_{K,1,1}}=S^{4}\setminus N(T_{K,1,1}),$

with respect to which the $\Lambda$-intersection form $\lambda$ of $E$ is
represented by a matrix of the form $A$ as in 2.4. Throughout, we use the
notation of Definition 3.1. Consider the sphere

$S_{0}:=\partial D^{3}\times\\{0\\}\subseteq S^{4}$

disjoint from $K$. This has self-intersection $0$.

Isotope the arc $K\setminus B^{3}_{1/2}$ relative to its boundary such that
its interior only intersects the $z$-axis in $\\{(0,0,3/4)\\}$, where it is
perpendicular to it. Let $\mu$ be a meridian of $K$ at $(0,0,3/4)$ of radius
$1/8$. Then

$T_{\ell}:=\frac{\mu\times I}{\\{(x,0)=(\phi_{1}(x),1):x\in\mu\\}}\subseteq E$

is a rim torus of $T_{K,1,1}$ over a longitude $\ell$ of $T_{K,1,1}$.

Since $T_{\ell}$ is not a $\mathbb{Z}$-trivial surface, we surger it to a
sphere. For this end, consider the arc

$J:=\\{0,0\\}\times[7/8,1]\subseteq D^{3}.$

Then

(4) $\begin{split}D&:=\frac{J\times I}{\\{(x,0)=(\phi_{1}(x),1):x\in
J\\}}\cup_{\eta_{1}}(\\{(0,0,1)\\}\times D^{2})\\\ &\subseteq\frac{D^{3}\times
I}{\\{(x,0)=(\phi_{1}(x),1):x\in D^{3}]\\}}\cup_{\eta_{1}}(S^{2}\times
D^{2})\end{split}$

is a disc in $S^{4}$ with boundary on $T_{\ell}$.

Since $\eta_{1}$ applies a $2\pi$ rotation about the $z$-axis to $\mu$, a
trivialisation of the normal bundle $\nu_{\partial D\subseteq T_{\ell}}$ does
not extend to a section of the normal bundle $\nu_{D\subseteq E}$. The
tangential framing of $\nu_{\partial D\subseteq T_{\ell}}$, transported to

$\frac{\partial D^{3}\times I}{\\{(x,0)=(\phi_{1}(x),1):x\in\partial
D^{3}\\}}=\frac{\partial D^{3}\times I}{\\{(x,0)=(x,1):x\in\partial
D^{3}\\}}\cong S^{2}\times S^{1},$

is a constant vector

$\underline{v}\times\\{\underline{0}\\}\in T_{(0,0,1)}S^{2}\times
T_{\theta}S^{1}\subseteq TS^{2}\times TS^{1}\cong T(S^{2}\times S^{1}).$

A nonvanishing section of the normal bundle of $\\{(0,0,1)\\}\times D^{2}$ in
$S^{2}\times D^{2}$ also gives rise to a constant vector, in the coordinates
of $S^{2}\times D^{2}$. However, the identification $\eta_{1}$ is a Gluck
twist, so introduces a full rotation of difference between the two. Hence $D$
is not compatibly framed with respect to the tangential framing of $\partial
D$.

However, if we perform a boundary twist on $D$ along the curve $\partial
D=\\{(0,0,7/8)\\}\times I$, then we can undo the twist. The resulting disc
$D_{1}$ intersects $T_{\ell}$ once, but now a trivialisation of $\nu_{\partial
D_{1}\subseteq T_{\ell}}$ does extend to $\nu_{D_{1}\subseteq E}$.

If we surger $T_{\ell}$ along $D_{1}$, we obtain an immersed sphere $S_{1}$
with two double points, and we claim that
$\mu([S_{1}])=\pm(1-t)\in\Lambda/\langle t^{r}\sim
t^{-r}:r\in\mathbb{Z}\rangle$. We have that $D_{1}$ is embedded and framed,
and intersects $T$ in one point, so, by part (i) of 2.5, we have that
$\mu(S_{1})=\pm t^{k}(1-t)$ for some $k$. Since the double points arose from a
boundary twist, it follows that one of the double points is local, i.e. the
entire double point loop is contained in a copy of $D^{4}\subseteq
E_{\Sigma}$. Hence, this double point loop is trivial. We deduce that
$\mu(S_{1})=\pm(1-t)$ or $\pm(1-t^{-1})$. Given the indeterminacy in
$\mu(S_{1})$, this completes the proof of the claim that
$\mu(S_{1})=\pm(1-t)$.

Furthermore, thanks to the boundary twist, $S_{1}$ has trivial normal bundle,
hence $e(S_{1})=0$ and so by equation (3) we have that

$\lambda([S_{1}],[S_{1}])=\mu([S_{1}])+\overline{\mu([S_{1}])}=\pm(1-t)\pm(1-t^{-1}).$

As $|S_{0}\pitchfork D_{1}|=1$, by part (ii) of 2.5, we have that
$\lambda([S_{0}],[S_{1}])=1-t$, after possibly changing the orientation of
$S_{0}$ and its basing arc. Since $\lambda$ is Hermitian,
$\lambda([S_{1}],[S_{0}])=1-t^{-1}$. Hence, the matrix of $\lambda$ on the
submodule generated by the classes $[S_{0}]$ and $[S_{1}]$ is

$\begin{pmatrix}0&1-t\\\ 1-t^{-1}&\pm(1-t)\pm(1-t^{-1})\end{pmatrix}.$

It follows from 2.4, with $g=1$ and $c_{1}=\pm 1$, that $([S_{0}],[S_{1}])$ is
indeed a basis of $H_{2}(E;\Lambda)\cong\Lambda^{2}$, that the equivariant
intersection form $\lambda$ is isometric to $\mathcal{H}_{2}$, and that
$\Sigma$ is topologically unknotted, as desired. ∎

######  Remark 4.1.

An alternative strategy to proving Theorem 1.1 is to observe that the 1-twist
spun torus $T_{K,1,1}$ can be obtained from the standard torus $T^{2}$ in
$S^{4}$ by performing 1-twist rim surgery along the curve with slope 1, and
then applying [CP23, Theorem 1.7].

##  5\. Knot surgery

In this section, we extend the results of Kim and Ruberman [KR08] to
$\mathbb{Z}$-surfaces in $S^{4}$. Let $\Sigma\subseteq S^{4}$ be a smoothly
embedded, topologically unknotted oriented surface of genus $g$. Furthermore,
let $T\subseteq S^{4}\setminus N(\Sigma)$ be a smoothly embedded torus that is
unknotted in $S^{4}$. If $J$ is a knot in $S^{3}$ with exterior $E_{J}$ and
$t$, $r\in\mathbb{Z}$, then we let

$S^{4}_{t,r}(T,J):=(S^{4}\setminus N(T))\cup_{f}(E_{J}\times S^{1})$

be the result of _$t$ -twist $r$-roll knot surgery along $T$ with pattern
$J$_, where we describe the gluing map $f$ next.

Fix a parametrisation $g\colon S^{1}\times S^{1}\to T$, and write
$\ell:=g(S^{1}\times\\{1\\})$ and $m:=g(\\{1\\}\times S^{1})$. Let $s\colon
T\to\partial\overline{N(T)}$ be a section such that $s(\ell)$ and $s(m)$ are
trivial in $H_{1}(S^{4}\setminus N(T))\cong\mathbb{Z}$. If $\mu_{T}$ is a
meridian of $T$, then

$\bigl{\\{}[s(\ell)],[s(m)],[\mu_{T}]\bigr{\\}}$

is a basis of $H_{1}\bigl{(}\partial\overline{N(T)}\bigr{)}\cong
H_{1}(T^{3})\cong\mathbb{Z}^{3}$.

Let $\mu_{J}\subseteq\partial E_{J}$ be a meridian and let
$\lambda_{J}\subseteq\partial E_{J}$ be a 0-framed longitude of $J$, and write
$\mu_{J}\cap\lambda_{J}=\\{b\\}$. Then

$\bigl{\\{}[\\{b\\}\times
S^{1}],[\mu_{J}\times\\{1\\}],[\lambda_{J}\times\\{1\\}]\bigr{\\}}$

is a basis of $H_{1}(\partial(E_{J}\times S^{1}))=H_{1}(\partial E_{J}\times
S^{1})\cong H_{1}(T^{3})\cong\mathbb{Z}^{3}$.

Note that
$\pi_{0}\bigl{(}\text{Diff}^{+}(T^{3})\bigr{)}\cong\text{SL}(3,\mathbb{Z})$,
with the isomorphism given by taking the induced map on $H_{1}(T^{3})$ by
Waldhausen [Wal68]. Hence, we can define the map
$f\colon\partial\overline{N(T)}\to\partial E_{J}\times S^{1}$ up to isotopy by
requiring that

$f(s(\ell))=\\{b\\}\times
S^{1}+t(\mu_{J}\times\\{1\\})+r(\lambda_{J}\times\\{1\\})\text{,
}f(s(m))=\mu_{J}\times\\{1\\}\text{, and
}f(\mu_{T})=\lambda_{J}\times\\{1\\}.$

Since $T$ is smoothly unknotted in $S^{4}$, there is a diffeomorphism
$\Theta\colon S^{4}_{t,r}(T,J)\to S^{4}$; see Kim–Ruberman [KR08]. We write

$\Sigma_{t,r}(T,J):=\Theta(\Sigma).$

When $T$ is a rim torus of $\Sigma$ and $m$ is a meridian of $\Sigma$, then
this is known as the $t$-twist $r$-roll rim surgery of $\Sigma$ with pattern
$J$, but in our application $T$ will not be a rim torus. We now restate our
main result on knot surgeries in surface complements from the introduction.

See 1.3

###### Proof.

Our discussion below works for arbitrary $g$, but it follows from the work of
Conway and the second author [CP23] that $\Sigma_{t,r}(T,J)$ is topologically
standard if it is a $\mathbb{Z}$-surface and $g\not\in\\{1,2\\}$. Let
$\boldsymbol{\alpha}=\\{\alpha_{1},\dots,\alpha_{g}\\}$ and
$\boldsymbol{\beta}=\\{\beta_{1},\dots,\beta_{g}\\}$ be collections of
pairwise disjoint simple closed curves on $\Sigma$ such that
$\boldsymbol{\alpha}\cup\boldsymbol{\beta}$ represents a basis of
$H_{1}(\Sigma)$. Furthermore, we can assume that
$|\alpha_{i}\pitchfork\beta_{j}|=\delta_{ij}$.

Let $R_{\gamma}\subseteq S^{4}\setminus\Sigma$ be a rim torus around $\gamma$
for $\gamma\in\boldsymbol{\alpha}\cup\boldsymbol{\beta}$. Let
$S_{\gamma}\subseteq S^{4}\setminus\Sigma$ be an embedded 2-sphere obtained by
surgering $R_{\gamma}$ along an embedded disc. This is possible since $\Sigma$
is a $\mathbb{Z}$-surface, hence any longitude of $R_{\gamma}$ becomes
0-homotopic in $S^{4}\setminus\Sigma$ after adding a suitable number of copies
of the meridian of $\Sigma$. Then
$\\{[S_{\alpha_{1}}],[S_{\beta_{1}}],\dots,[S_{\alpha_{g}}],[S_{\beta_{g}}]\\}$
is a basis of $H_{2}(S^{4}\setminus N(\Sigma);\Lambda)\cong\Lambda^{2g}$. In
this basis, the $\Lambda$-intersection form $\lambda_{S^{4}\setminus
N(\Sigma)}\cong\bigoplus_{g}\mathcal{H}_{2}$; see Section 2.

######  Lemma 5.1.

Suppose that $s(m)$ is 0-homotopic in $S^{4}\setminus(\Sigma\cup T)$. Then
$\Sigma_{t,r}(T,J)$ is a $\mathbb{Z}$-surface.

###### Proof.

Let $E:=S^{4}\setminus N(\Sigma_{t,r}(T,J))$ and $E_{\Sigma}:=S^{4}\setminus
N(\Sigma)$. Then

$E=(E_{\Sigma}\setminus N(T))\cup_{\partial\overline{N(T)}}(E_{J}\times
S^{1}).$

Hence, by the Seifert–van Kampen theorem,

$\pi_{1}(E)\cong\pi_{1}(E_{\Sigma}\setminus
N(T))\ast_{\mathbb{Z}^{3}}(\pi_{1}(E_{J})\times\mathbb{Z}),$

where $\pi_{1}(T^{3})\cong\mathbb{Z}^{3}$ is generated by the curves
$\mu_{J}\times\\{1\\}$, $\lambda_{J}\times\\{1\\}$, and $\\{b\\}\times S^{1}$
(recall that $\\{b\\}:=\mu_{J}\cap\lambda_{J}$). Since
$\mu_{J}\times\\{1\\}=f(s(m))$ and $s(m)$ is 0-homotopic in
$E_{\Sigma}\setminus N(T)$ by assumption, $\mu_{J}\times\\{1\\}$ is
0-homotopic in $E$. As $\mu_{J}$ normally generates $\pi_{1}(E_{J})$, the
curve $\lambda_{J}\times\\{1\\}$ is also 0-homotopic in $E$. However,
$f(\mu_{T})=\lambda_{J}\times\\{1\\}$, so $\mu_{T}$ is 0-homotopic in $E$.
Finally, as $f(s(\ell))=\\{b\\}\times
S^{1}+t(\mu_{J}\times\\{1\\})+r(\lambda_{J}\times\\{1\\})$ and
$\mu_{J}\times\\{1\\}$ and $\lambda_{J}\times\\{1\\}$ are 0-homotopic in $E$,
the curve $\\{b\\}\times S^{1}$ is homotopic to $s(\ell)$ in $E$. Hence

$\pi_{1}(E)\cong\pi_{1}(E_{\Sigma}\setminus
N(T))/\langle\mu_{T}\rangle\cong\pi_{1}(E_{\Sigma})\cong\mathbb{Z},$

where $\langle\mu_{T}\rangle$ is the normal subgroup generated by $\mu_{T}$,
and the second isomorphism follows from the decomposition
$E_{\Sigma}=(E_{\Sigma}\setminus N(T))\cup N(T)$ via the Seifert–van Kampen
theorem. ∎

For each $\gamma\in\boldsymbol{\alpha}\cup\boldsymbol{\beta}$, we construct
$S_{\gamma}^{\prime}\subseteq S^{4}_{t,r}(T,J)$ by first making $S_{\gamma}$
transverse to $T$, and replacing each component of $S_{\gamma}\cap N(T)$,
which is a disc, with a Seifert surface of $J$ in $E_{J}\times\\{p\\}$ for
some $p\in S^{1}$. We can assume that we have a different $p$ for every
component, so these Seifert surfaces are mutually disjoint.

Let us assume that $s(m)$ is 0-homotopic in $S^{4}\setminus(\Sigma\cup T)$. By
the proof of Lemma 5.1, the map
$\pi_{1}(E_{J})\to\pi_{1}(S^{4}_{t,r}(T,J)\setminus N(\Sigma))$ induced by the
embedding is trivial. Hence $S_{\gamma}^{\prime}$ is a $\mathbb{Z}$-surface in
$S^{4}_{t,r}(T,J)\setminus N(\Sigma)$.

We can arrange that $S_{\alpha_{i}}\cap S_{\beta_{j}}\subseteq S^{4}\setminus
N(\Sigma\cup T)$ for every $i$, $j\in\\{1,\dots,g\\}$. Then
$S_{\alpha_{i}}^{\prime}\cap S_{\beta_{j}}^{\prime}\subseteq S^{4}\setminus
N(\Sigma\cup T)$ for every $i$, $j\in\\{1,\dots,g\\}$ as well. Let
$U_{\gamma}:=\Theta(S_{\gamma}^{\prime})$ for
$\gamma\in\boldsymbol{\alpha}\cup\boldsymbol{\beta}$, oriented such that
$\Theta|_{S_{\gamma}^{\prime}}\colon S_{\gamma}^{\prime}\to U_{\gamma}$ is
orientation-preserving.

######  Proposition 5.2.

Let $q\in S_{\alpha_{i}}\cap S_{\beta_{i}}$ be an intersection point with sign
$\varepsilon\in\\{+,-\\}$ and double point group element $t^{k}\in\langle
t\rangle\cong C_{\infty}\cong\mathbb{Z}$. Since $q\in S^{4}\setminus
N(\Sigma\cup T)$, we can consider $\Theta(q)\in U_{\alpha_{i}}\cap
U_{\beta_{i}}\subseteq S^{4}\setminus N(\Sigma_{t,r}(T,J)$. Then $\Theta(q)$
also has sign $\varepsilon$ and double point group element $t^{k}$.

###### Proof.

As $\Theta$ is orientation-preserving and the orientations of $U_{\alpha_{i}}$
and $U_{\beta_{i}}$ are induced by $\Theta$, the intersection point
$\Theta(q)$ has the same sign as $q$.

Let $V\subseteq S^{4}$ be a disc such that $\partial V$ is a double point loop
for $q$. Then $|V\pitchfork\Sigma|=k$. We can arrange that $V$ is transverse
to $T$. Replace $V\cap N(T)$ with Seifert surfaces of $J$ to obtain
$V^{\prime}\subseteq S^{4}_{t,r}(T,J)$, and write
$W:=\Theta(V^{\prime})\subseteq S^{4}$. Then $\partial W$ is a double point
loop for $\Theta(q)$, and $|W\pitchfork\Sigma_{t,r}(T,J))|=k$. As $W$ is a
$\mathbb{Z}$-surface, it follows that the double point group element of
$\Theta(q)$ is $t^{k}$. ∎

Let $E:=S^{4}\setminus N(\Sigma_{t,r}(T,J))$. As noted above,
$\lambda_{E_{\Sigma}}\cong\bigoplus_{g}\mathcal{H}_{2}$. Hence, by Proposition
5.2, the matrix of $\lambda_{E}$ on
$\\{U_{\alpha_{1}},U_{\beta_{1}},\dots,U_{\alpha_{g}},U_{\beta_{g}}\\}$ is
$\bigoplus_{g}\mathcal{H}_{2}$. It follows from Lemma 2.4 that
$\\{U_{\alpha_{1}},U_{\beta_{1}},\dots,U_{\alpha_{g}},U_{\beta_{g}}\\}$ is a
basis of $H_{2}(E;\Lambda)$, and that

$\lambda_{E}\cong\bigoplus_{g}\mathcal{H}_{2},$

so $\Sigma_{T,J}$ is topologically unknotted by Theorem 2.1. This concludes
the proof of Theorem 1.3. ∎

##  6\. The Cochran–Davis Seifert surface union a ribbon disc

A natural idea for constructing potentially nontrivial, closed, orientable
$\mathbb{Z}$-surfaces, smoothly embedded in $S^{4}$, is to take the union of a
Seifert surface $F$ for a slice knot $K$ in $S^{3}$ and a homotopy ribbon disc
$D$ for $K$. However, by the following lemma, this is smoothly unknotted if
$D$ is obtained from $F$ by compressing along a slice derivative, which is
often the case.

######  Lemma 6.1.

Let $F\subseteq S^{3}$ be a genus $g$ Seifert surface, and suppose that
$c\subseteq F$ is a slice derivative. Write $D$ for the slice disc obtained by
surgering $F$ along $c$. Then $\Sigma:=F\cup D$ is smoothly unknotted.

###### Proof.

Since $D$ is obtained by surgering $F$ along $c$, it follows that $\Sigma$
bounds an embedded copy of $F\times I$ with a collection of 3-dimensional
2-handles attached along $c\times\\{0\\}$, which we denote by $H$. Since the
components of $c$ are $g$ homologically linearly independent simple closed
curves in $F$, the 3-manifold $H$ is a genus $g$ handlebody. As $\partial
H=\Sigma$, it follows that $\Sigma$ is smoothly unknotted. ∎

Kauffman conjectured that every genus one Seifert surface for a slice knot
admits a slice derivative, but Cochran and Davis [CD15] constructed a
counterexample. This motivates 1.2, which we restate for the reader’s
convenience.

See 1.2

Before beginning the proof, we recall the Cochran–Davis construction [CD15].
It starts with the ribbon knot $R$ on the left of Figure 3, and a ribbon disc
$\Delta\subseteq D^{4}$ for it. The ribbon disc $\Delta\subseteq D^{4}$ for
$R$ is obtained by cutting the left band with a saddle move, as shown in
Figure 4, and then capping off the resulting 2-component unlink with discs.
This is equivalent to surgering $F^{\prime}$ along the slice derivative
$\alpha^{\prime}$, since cutting the left band of $F^{\prime}$ results in an
annulus with core $\alpha^{\prime}$.

$R$
---
$F^{\prime}$
---
$+3$
---
$\alpha^{\prime}$
---
$R$
---
$+3$
---
$\eta_{1}$
---
$\eta_{2}$
---
Figure 3. Left: a ribbon knot $R$, together with a Seifert surface
$F^{\prime}$ and a slice derivative $\alpha^{\prime}$ on $F^{\prime}$. Right:
the same knot $R$, with two embedded curves $\eta_{1}$ and $\eta_{2}$ in
$S^{3}\setminus R$ such that $\eta_{1}\sqcup\eta_{2}$ is an unlink. See [CD15,
Figure 6].

The curves $\eta_{1}\sqcup\eta_{2}$ shown on the right of Figure 3 form an
unlink, and cobound an annulus $A$ smoothly embedded in $D^{4}\setminus\Delta$
by [CD15, Proposition 5.1]. To construct this annulus, perform a saddle move
on $\eta_{1}\sqcup\eta_{2}$ and then on $R$, as shown in Figure 4. Figure 4
now shows a 3-component unlink, so the result of the saddle moves can be
capped off with disjoint discs to produce the rest of $\Delta$ and the rest of
$A$.

$R$
---
$+3$
---
Figure 4. The saddle moves for constructing the ribbon disc $\Delta$ and the
annulus $A$.

Cochran and Davis perform a satellite construction on $D^{4}$ along the
annulus $A\subseteq D^{4}$ with $\partial A=\eta_{1}\sqcup\eta_{2}\subseteq
S^{3}$. We write

$S^{4}\cong S^{4}_{+}\cup_{S^{3}}S^{4}_{-},$

where $S^{4}_{\pm}\cong D^{4}$. Using the identification $S^{4}_{+}\cong
D^{4}$, we have $\Delta\subseteq S^{4}_{+}$. We push the Seifert surface
$F^{\prime}$ for $R$ into $S^{4}_{-}$. Consider the surface
$\Sigma^{\prime}:=\Delta\cup F^{\prime}\subseteq S^{4}_{+}\cup S^{4}_{-}$.
Relabel $A$ as $A_{+}\subseteq S^{4}_{+}$, and let $A_{-}$ denote another copy
of $A$ in $S^{4}_{-}$. By only pushing $F^{\prime}$ into $S^{4}_{-}$ a small
amount, so it lives in a collar of the boundary $S^{3}$, we may assume that
$A_{-}\cap F^{\prime}=\emptyset$.

Our goal is to perform the satellite construction mentioned above (and
explained in detail below) simultaneously on $A_{+}$ and $A_{-}$, which will
produce the knot $K$, with a ribbon disc $D$ and a Seifert surface $F$. Then
$\Sigma:=D\cup F\subseteq S^{4}$ is the $\mathbb{Z}$-torus that is the subject
of 1.2.

Let $J$ be one of the knots in $S^{3}$ considered by Cochran and Davis, for
example the left-handed trefoil, and let $E_{J}$ be its exterior. Cochran and
Davis parametrise $A$ as $S^{1}\times[0,1]$, and then perform a 1-parameter
satellite construction on $D^{4}$, removing a neighbourhood
$S^{1}\times[0,1]\times B^{2}\cong S^{1}\times B^{2}\times[0,1]$, and glueing
in $E_{J}\times[0,1]$ instead, in such a way that each meridian $\\{p\\}\times
S^{1}\times\\{t\\}$ is identified with a longitude of $J$ in
$E_{J}\times\\{t\\}$ for $p\in S^{1}$ and $t\in I$. They show that the result
of this construction is a manifold $W$ that comes with a diffeomorphism
$\Phi\colon W\to D^{4}$; see [CD15, Theorem 3.1]. They define the knot $K$ to
be $\Phi(R)$, the disc $D$ to be $\Phi(\Delta)$, and the Seifert surface $F$
to be $\Phi(F^{\prime})$.

The knot $K$ can be obtained with a purely 3-dimensional construction, as
follows. Perform two simultaneous infections (satellite constructions) on the
curves $\eta_{1}$ and $\eta_{2}$, with companion knots $J$ and $-J$
respectively. The resulting knot $K:=R_{\eta_{1},\eta_{2}}(J,-J)$ is the
_Cochran–Davis knot_.

###### Proof of 1.2.

We consider the decomposition $S^{4}=S^{4}_{+}\cup_{S^{3}}S^{4}_{-}$. We
perform the satellite construction on both $S^{4}_{+}$ and $S^{4}_{-}$ along
the annuli $A_{+}$ and $A_{-}$, respectively. This produces 4-manifolds
$W_{\pm}$ with $\partial W_{\pm}\cong S^{3}$ and diffeomorphisms
$\Phi_{\pm}\colon W_{\pm}\to S^{4}_{\pm}$. The knot $R$, the pushed-in Seifert
surface $F^{\prime}$, and the disc $\Delta$ are all disjoint from $A_{\pm}$,
and so represent analogous objects in $W_{\pm}$. We then have the pushed-in
Seifert surface $F:=\Phi_{-}(F^{\prime})\subseteq S^{4}_{-}$ for the knot
$K:=\Phi_{\pm}(R)\subseteq S^{3}$, and ribbon disc
$D:=\Phi_{+}(\Delta)\subseteq S^{4}_{+}$. So

$\Sigma:=F\cup D=\Phi_{-}(F^{\prime})\cup\Phi_{+}(\Delta).$

Hence $\Sigma$ is obtained from $\Sigma^{\prime}=F^{\prime}\cup\Delta$ via
knot surgery along the torus $T_{A}:=A_{+}\cup A_{-}$ with pattern the knot
$J$.

Note that $\pi_{1}(D^{4}\setminus F^{\prime})\cong\mathbb{Z}$, generated by a
meridian by Chen [Che21, Lemma 3.1]. Since $\Delta$ is homotopy ribbon, it
follows from the Seifert–van Kampen theorem that $\Sigma^{\prime}$ is a
$\mathbb{Z}$-torus.

As $\eta_{1}\sqcup\eta_{2}$ is an unlink, and we obtain $A$ from it by
attaching a trivial band and capping off the resulting unknot with a disc, it
follows that $A$ and hence $T_{A}$ are smoothly unknotted. The surface
$\Sigma^{\prime}$ is also unknotted by Lemma 6.1, since one can obtain
$\Delta$ by compressing $F^{\prime}$ along the slice derivative
$\alpha^{\prime}$ shown on the left of Figure 3.

The annulus $A_{-}$ is obtained by pushing $A$ into $S^{4}_{-}$ deeper than
$F_{-}$, hence the meridian $m$ of $A_{-}$ bounds a disc whose interior is
disjoint from $A_{-}\cup F_{-}$. So the meridian of $T_{A}$ also bounds a disc
in $S^{4}\setminus\Sigma^{\prime}$ whose interior lies in
$S^{4}\setminus(\Sigma^{\prime}\cup T_{A})$.

So $\Sigma^{\prime}$ and $T_{A}$ satisfy all the assumptions of Theorem 1.3,
and hence the result $\Sigma$ of knot surgery on $\Sigma^{\prime}$ along
$T_{A}$ with pattern $J$ is topologically unknotted, as claimed. ∎

We close this section by noting another general case when a genus one Seifert
surface union a slice disc is a topologically unknotted $\mathbb{Z}$-torus,
suggested by Anthony Conway. Let $K\subseteq S^{3}\subseteq
S^{4}=S^{4}_{+}\cup S^{4}_{-}$ be a genus one knot with Alexander polynomial
$\Delta_{K}\doteq 1$. Let $D\subseteq S^{4}_{+}$ be a $\mathbb{Z}$-slice disc
for $K$, and let $F$ be a genus one Seifert surface for $K$ pushed into
$S^{4}_{-}$. Let $\Sigma:=F\cup_{K}D$.

######  Proposition 6.2.

The torus $\Sigma\subseteq S^{4}$ is topologically unknotted.

###### Proof.

Let $E_{D}:=S^{4}_{+}\setminus N(D)$, $E_{K}:=S^{3}\setminus N(K)$, and
$E_{F}:=S^{4}_{-}\setminus N(F)$. Then $E_{\Sigma}:=S^{4}\setminus
N(\Sigma)=E_{D}\cup_{E_{K}}E_{F}$ and $\pi_{1}(E_{\Sigma})\cong\mathbb{Z}$.
With $\Lambda$-coefficients, $H_{i}(E_{K};\Lambda)=H_{i}(E_{D};\Lambda)=0$ for
$i\in\\{1,2\\}$. The Mayer–Vietoris sequence implies that the inclusion-
induced map $H_{2}(E_{F};\Lambda)\xrightarrow{\cong}H_{2}(E_{\Sigma};\Lambda)$
is an isomorphism. By [CP23, Proposition 7.10], the intersection form of
$E_{F}$ is isometric to $\mathcal{H}_{2}$. It follows that intersection form
of $E_{\Sigma}$ is also isometric to $\mathcal{H}_{2}$, and hence $\Sigma$ is
topologically unknotted by 2.1. ∎

##  7\. Tori with four critical points

We prove our final theorem, which we recall here.

See 1.4

###### Proof.

Denote the critical points of $\Sigma$ by $c_{0}$, $c_{1}^{a}$, $c_{1}^{b}$,
and $c_{2}$, where the subscript denotes the index. As before, write
$S^{4}=S^{4}_{+}\cup_{S^{3}}S^{4}_{-}$, a union of two copies of $D^{4}$.
Arrange the critical points of $\Sigma$ such that $c_{0}$ and $c_{1}^{a}$ lie
in $S^{4}_{-}\cong D^{4}$, and $c_{1}^{b}$ and $c_{2}$ lie in $S^{4}_{+}\cong
D^{4}$. The intersection of $\Sigma$ with $S^{4}_{+}\cap S^{4}_{-}=S^{3}$ is a
two-component link $L=L_{1}\sqcup L_{2}$, with the property that a single
fusion band move yields the unknot, in two different ways.

A theorem of Scharlemann [Sch85, Main Theorem] implies that $L$ is the unlink
and the bands are both trivial. We do not know whether the bands can be
simultaneously trivialised (if we knew that, we would know that $\Sigma$ is
smoothly unknotted). Let $E_{L}:=S^{3}\setminus N(L)$. By an isotopy, we can
and shall assume that $L$ is standardly embedded, as is the band corresponding
to $c_{1}^{a}$.

Let $S_{0}\subseteq S^{3}$ be an embedded 2-sphere that represents a generator
of $H_{2}(E_{L})\cong\mathbb{Z}$, i.e. $S_{0}$ separates $L_{1}$ and $L_{2}$.
Let $B\cong D^{1}\times D^{1}\subseteq S^{3}$ denote the band corresponding to
$c_{1}^{a}$. Its boundary consists of four arcs. Two of these are $S^{0}\times
D^{1}\subseteq L$. The other two, $D^{1}\times S^{0}$, are parallel paths
connecting $L_{1}$ and $L_{2}$. Let $\gamma\subseteq E_{L}$ denote a simple
closed curve that is obtained by rotating $B$ by $\pi/2$ around
$D^{1}\times\\{0\\}$, and attaching meridians of $L_{1}$ and $L_{2}$ to either
end. The fundamental group of $E_{L}$ is free on two generators: meridians
$\mu_{1}$ and $\mu_{2}$ to $L_{1}$ and $L_{2}$, respectively. The curve
$\gamma$ represents $\mu_{1}\mu_{2}^{-1}$. See Figure 5.

$L_{1}$
---
$L_{2}$
---
$S_{0}$
---
$\gamma$
---
Figure 5. An unlink $L=L_{1}\cup L_{2}$, the curve $\gamma$, and the 2-sphere
$S_{0}$, all in the equatorial $S^{3}$.

Sending $\mu_{i}\mapsto t$ for $i\in\\{1,2\\}$ gives rise to a
$\Lambda$-coefficient system that extends over both $S^{4}_{+}\setminus
N(\Sigma)$ and $S^{4}_{-}\setminus N(\Sigma)$. We consider the Mayer–Vietoris
sequence

$\begin{split}&H_{2}(S^{4}_{+}\setminus N(\Sigma);\Lambda)\oplus
H_{2}(S^{4}_{-}\setminus N(\Sigma);\Lambda)\to H_{2}(E_{\Sigma};\Lambda)\to
H_{1}(E_{L};\Lambda)\to\\\ &H_{1}(S^{4}_{+}\setminus N(\Sigma);\Lambda)\oplus
H_{1}(S^{4}_{-}\setminus N(\Sigma);\Lambda).\end{split}$

By 2.2, we have $H_{2}(E_{\Sigma};\Lambda)\cong\Lambda^{2}$. Since
$E_{L}\simeq S^{1}\vee S^{1}\vee S^{2}$, it is easy to compute that
$H_{1}(E_{L};\Lambda)\cong\Lambda$, generated by the curve $\gamma$. By the
rising water principle [GS99], the surface exterior $S^{4}_{\pm}\setminus
N(\Sigma)$ has a handle decomposition consisting of a single handle in each of
the indices 0, 1, and 2. We deduce that $\pi_{1}(S^{4}_{\pm}\setminus
N(\Sigma))\cong\mathbb{Z}$ and that the handle chain complex of
$S^{4}_{\pm}\setminus N(\Sigma)$ with $\Lambda$ coefficients has the form

$\Lambda\xrightarrow{(0)}\Lambda\xrightarrow{(1-t)}\Lambda,$

and hence $H_{1}(S^{4}_{\pm}\setminus N(\Sigma);\Lambda)=0$ and
$H_{2}(S^{4}_{\pm}\setminus N(\Sigma);\Lambda)\cong\Lambda$ (we will not use
the latter fact).

We therefore have the exact sequence

$H_{2}(S^{4}_{+}\setminus N(\Sigma);\Lambda)\oplus H_{2}(S^{4}_{-}\setminus
N(\Sigma);\Lambda)\to\Lambda^{2}\to\Lambda\to 0.$

The sequence splits, so $H_{1}(E_{L};\Lambda)=\langle\gamma\rangle$ is a
summand of $H_{2}(E_{\Sigma};\Lambda)\cong\Lambda^{2}$.

We realise this summand as a surgered rim torus. Let
$\delta_{\pm}\subseteq\Sigma\cap S^{4}_{\pm}$ be embedded paths connecting the
two endpoints of the core of the band $B\subseteq S^{3}$ corresponding to
$c_{1}^{a}$; see Figure 6. Let $T_{\delta}$ be the rim torus over
$\delta:=\delta_{+}\cup\delta_{-}$. Let $\sigma_{\pm}\subseteq
S^{4}_{\pm}\setminus N(\Sigma)$ denote simple closed curves, each of which is
obtained by taking an embedded arc (half of a longitude) on
$T_{\delta_{\pm}}:=T_{\delta}\cap S^{4}_{\pm}$ that is a push-off of
$\delta_{\pm}$ along a section of the normal bundle of $\Sigma\cap
S^{4}_{\pm}$, where the section points along $B$ at the endpoints, and joining
the endpoints with the (slightly shortened) core of the band $B$. We may
assume, by winding around the meridian, that $\sigma_{\pm}$ is $0$-homotopic
in $S^{4}_{\pm}\setminus N(\Sigma)$. Let $\tau_{\pm}$ denote a choice of
$0$-homotopy, perturbed to be a generic immersion. Then the union
$\tau:=\tau_{+}\cup\tau_{-}$ is a generic immersion of a copy of $D^{2}$ with
boundary a 0-linking longitude on the rim torus $T_{\delta}$. If necessary,
perform boundary twists on $\tau$ to arrange that $\tau$ is framed with
respect to $T_{\delta}$. Surger $T_{\delta}$ using $\tau$ to obtain a framed,
immersed 2-sphere $S_{1}$ in $E_{\Sigma}$, such that

$S_{1}\cap S^{3}=\gamma.$

$S^{4}_{+}$
---
$S^{4}_{-}$
---
$\sigma_{+}$
---
$\sigma_{-}$
---
$\delta_{-}$
---
$\delta_{+}$
---
$\Sigma$
---
$\tau_{+}$
---
$B$
---
$\tau_{-}$
---
Figure 6. A schematic diagram showing the paths $\delta_{\pm}$, the paths
$\sigma_{\pm}\subseteq T_{\delta}$, the band $B$, and the $0$-homotopies
$\tau_{\pm}$. The union $\tau:=\tau_{+}\cup\tau_{-}$ is a disc that we use to
surger the rim torus $T_{\delta}$ into the sphere $S_{1}$.

######  Lemma 7.1.

After possibly changing basing arcs and orientations, we have the
$\Lambda$-intersection numbers

  1. (a)

$\lambda([S_{0}],[S_{0}])=0$,

  2. (b)

$\lambda([S_{1}],[S_{1}])=c(1-t)+\overline{c}(1-t^{-1})$ for some
$c\in\Lambda$, and

  3. (c)

$\lambda([S_{0}],[S_{1}])=1-t=\overline{\lambda([S_{1}],[S_{0}])}$.

###### Proof.

The sphere $S_{0}$ is embedded and framed, so $\mu(S_{0})=0$ and $e(S_{0})=0$.
Therefore $\lambda([S_{0}],[S_{0}])=0$ by equation (3), proving (a).

The sphere $S_{1}$ is obtained from performing surgery on the rim torus
$T_{\beta}$ along a compatibly framed disc. Therefore (b) follows directly
from 2.5 (iv).

Since $S_{0}\subseteq S^{3}$ and $S_{1}\cap S^{3}=\gamma$, we can compute
$\lambda([S_{0}],[S_{1}])$ by computing the intersections between $S_{0}$ and
$\gamma$. They intersect in two points, which come from the unique
intersection point $\tau\pitchfork S_{0}$ before surgery; see Figure 5. As
usual, by 2.5 (ii), the intersection points have opposite signs and the double
point loops differ by multiplication by $t^{\pm 1}$. Up to possibly changing
the basing arc and the orientation of $S_{0}$, we indeed have that
$\lambda([S_{0}],[S_{1}])=1-t$. The final statement, that
$\overline{\lambda([S_{1}],[S_{0}])}=1-t$, and so
$\lambda([S_{1}],[S_{0}])=1-t^{-1}$, follows because $\lambda$ is Hermitian. ∎

It follows that restricting the intersection pairing $\lambda$ to the
submodule generated by $[S_{0}]$ and $[S_{1}]$ gives rise to a pairing
represented by a matrix

$\begin{pmatrix}0&1-t\\\ 1-t^{-1}&c(1-t)+\overline{c}(1-t^{-1})\end{pmatrix},$

for some $c\in\Lambda$. By 2.4, with $g=1$ and $c_{1}=c$, we deduce that
$\\{[S_{0}],[S_{1}]\\}$ is a basis for
$H_{2}(E_{\Sigma};\Lambda)\cong\Lambda^{2}$, the $\Lambda$-intersection form
is congruent to $\mathcal{H}_{2}$, and $\Sigma$ is topologically unknotted.
This conclude the proof of 1.4. ∎

##  8\. Double branched covers

Given a surface $S$ in a 4-manifold $X$, we denote the double cover of $X$
branched along $S$ by $\Sigma(X,S)$. As in the case of the exotic copies of
$5\mathbb{R}P^{2}$, $\\#6\mathbb{R}P^{2}$, and $\\#10\mathbb{R}P^{2}$
constructed by Matić–Öztürk–Stipsicz [MOS23], Finashin [Fin09], and
Finashin–Kreck–Viro [FKV87], respectively, one might try to distinguish the
turned 1-twist-spun torus $T_{K,1,1}$ and the standard torus $T^{2}$ using
their double branched covers. Note that $\Sigma(S^{4},T^{2})$ is diffeomorphic
to $S^{2}\times S^{2}$, where the covering involution is given by reflection
in $S^{1}$ in both $S^{2}$ factors.

######  Proposition 8.1.

Let $K$ be a knot in $D^{3}$. Then $\Sigma(S^{4},T_{K,1,1})$ is diffeomorphic
to $(S^{2}\times S^{2})\\#X$ for a homotopy 4-sphere $X$.

###### Proof.

Using the notation of Definition 3.1, note that both $\phi_{j}$ and
$\eta_{i}|_{S^{2}\times\\{\theta\\}}$ for $\theta\in S^{1}$ fix the $z$-axis
pointwise. Let

$J:=\\{0,0\\}\times[1/2,1]\subseteq D^{3}.$

Isotope $K$ such that $K\cap J=\\{(0,0,1/2)\\}$. Then

$D:=\frac{J\times I}{\\{(x,0)=(\phi_{1}(x),1):x\in
J\\}}\cup_{\eta_{1}}(\\{(0,0,1)\\}\times D^{2})$

is a disc in $S^{4}$ with boundary on $T_{K,1,1}$. Furthermore, consider the
sphere

$S_{0}:=\partial D^{3}\times\\{0\\}\subseteq S^{4}$

disjoint from $K$. Then $S_{0}$ lifts to a pair of spheres in
$\Sigma(S^{4},K)$, one of which we denote by $S$. Since $S_{0}$ has self-
intersection $0$, so does $S$. The disc $D$ lifts to a sphere $S^{\prime}$ in
$\Sigma(S^{4},K)$, which has even self-intersection (in fact, it is $4$ by the
proof of Theorem 1.1). Furthermore, $|S\pitchfork S^{\prime}|=1$, so $N(S\cup
S^{\prime})$ is a plumbing of two spheres of self-intersection 0 and 4,
respectively, and is hence diffeomorphic to $(S^{2}\times S^{2})\setminus
D^{4}$; see Neumann–Weintraub [NW78].

As $T_{K,1,1}$ and $T^{2}$ are topologically isotopic by Theorem 1.1 and
$\Sigma(S^{4},T^{2})\cong S^{2}\times S^{2}$, it follows that
$\Sigma(S^{4},T_{K,1,1})$ is homeomorphic to $S^{2}\times S^{2}$.
Consequently,

$\Sigma(S^{4},T_{K,1,1})\setminus N(S\cup S^{\prime})$

is a homotopy 4-disc, and the result follows. ∎

######  Proposition 8.2.

Let $T$ be a smoothly embedded torus in $S^{4}$ with four critical points.
Then $\Sigma(S^{4},T)$ is diffeomorphic to $(S^{2}\times S^{2})\\#X$ for a
homotopy 4-sphere $X$.

###### Proof.

We use the notation of Section 7. The sphere $S_{0}$ has two lifts to
$\Sigma(S^{4},T)$, one of which we denote by $S$. Since $S_{0}$ is 0-framed,
so is $S$.

Consider the immersed disc $\tau\subseteq S^{4}\setminus N(T)$, extended such
that $\partial\tau=\delta$ (we do not perform boundary twists). Then $\tau$
lifts to an immersed sphere $S_{\tau}\subseteq\Sigma(S^{4},T)$. Since
$\tau\cap S^{3}$ is the core of the band $B$, which intersects $S_{0}\subseteq
S^{3}$ transversely in a single point, $|S_{\tau}\pitchfork S|=1$. Hence, we
can connect the double points of $S_{\tau}$ to parallel copies of $S$ with
tubes, to obtain an embedded sphere $S^{\prime}$ such that $|S\pitchfork
S^{\prime}|=1$.

By Theorem 1.4, the double branched cover $\Sigma(S^{4},T)$ is homeomorphic to
$S^{2}\times S^{2}$, and hence has even intersection form. So $S^{\prime}$ has
even self-intersection, and so $\Sigma(S^{4},T)\cong(S^{2}\times S^{2})\\#X$
for a homotopy 4-sphere $X$, analogously to the proof of Proposition 8.1. ∎

We have the following consequence of Propositions 8.1 and 8.2.

######  Corollary 8.3.

Let $K$ be a knot in $S^{3}$, and let $T$ be a smoothly embedded torus in
$S^{4}$ with four critical points. Then the Seiberg–Witten invariants of
$\Sigma(S^{4},T_{K,1,1})$, $\Sigma(S^{4},T)$, and $S^{2}\times S^{2}$ agree.

###### Proof.

For a 4-manifold $M$ with $b_{2}^{+}(M)=1$ and $b_{2}^{-}(M)<10$, the
Seiberg–Witten invariant is well-defined in the metric chamber; see Szabó
[Sza96, Lemma 3.2]. In particular, if $M$ is $(S^{2}\times S^{2})\\#X$ for a
homotopy 4-sphere $X$, then, for every characteristic element $K\in H^{2}(X)$,
Riemannian metric $g$, and small enough perturbing 2-form $h$, the
Seiberg–Witten invariant $\text{SW}_{M}(K,g,h)$ is well-defined, and it
vanishes by the adjunction inequality applied to the self-intersection zero
2-sphere $S^{2}\times\\{p\\}$ for $p\in S^{2}$ in $M$. Hence, the
Seiberg–Witten invariants of $\Sigma(S^{4},T_{K,1,1})$, $\Sigma(S^{4},T)$, and
$S^{2}\times S^{2}$ agree in all chambers by the wall-crossing formula. ∎

In particular, we cannot use the Seiberg–Witten invariants of the double
branched covers to distinguish $T_{K,1,1}$ or $T$ from $T^{2}$ up to
diffeomorphism of pairs.

We finally consider the double cover $M$ of $S^{4}$ branched along the
Cochran–Davis Seifert surface union a ribbon disc. By Theorem 1.2, the
4-manifold $M$ is homeomorphic to $S^{2}\times S^{2}$. On the other hand, we
have the following.

######  Proposition 8.4.

Let $M$ be the double cover of $S^{4}$ branched along the surface $\Sigma$
constructed in Theorem 1.2. Then $M$ can be obtained from $S^{2}\times S^{2}$
using knot surgery along a 0-homologous torus.

###### Proof.

We use the notation of Section 6. Let us write

$S^{4}_{\pm}:=\frac{S^{3}\times I_{\pm}}{S^{3}\times\\{\pm 1\\}},$

where $I_{+}=[0,1]$ and $I_{-}=[-1,0]$. Let the Seifert surface pushed into
$S^{4}_{-}$ be

$\digamma^{\prime}:=(R\times[-1/8,0])\cup(F^{\prime}\times\\{-1/8\\}).$

  1. (i)

We can arrange that $A_{-}\cap(S^{3}\times[-1/2,0])$ takes the form
$(\eta_{1}\sqcup\eta_{2})\times[-1/2,0]$, i.e. $A_{-}$ is a product in this
collar.

  2. (ii)

We assume that the saddle moves used to construct $A_{-}$ are performed in
$S^{3}\times[-3/4,-1/2]$, and the disc capping it off (i.e. the minimum of the
projection to $[-1,0]$ restricted to $A_{-}$) lies in $S^{3}\times[-1,-3/4]$.

  3. (iii)

We suppose that $A_{+}\cap(S^{3}\times[0,1/2])$ takes the form
$(\eta_{1}\sqcup\eta_{2})\times[0,1/2]$.

  4. (iv)

We assume that the saddle moves used to construct $A_{+}$ and $\Delta$ are
performed in $S^{3}\times[1/2,3/4]$, and the discs capping them off (i.e. the
maxima of the projection to $[0,1]$ restricted to $A_{+}\cup\Delta$) lie in
$S^{3}\times[3/4,1]$.

Consider the torus $T_{A}:=A_{+}\cup A_{-}$ in
$S^{4}\setminus\Sigma^{\prime}$, where

$\Sigma^{\prime}=\Delta\cup\digamma^{\prime}\subseteq S^{4}_{+}\cup
S^{4}_{-}.$

Let $p\colon M\to S^{2}\times S^{2}$ be the covering map. We now show that
$T:=p^{-1}(T_{A})$ is connected, and hence a torus. The curve $\eta_{1}$ is a
meridian of $T_{A}$. Since $\text{lk}(\eta_{1},R)=0$ and $\Sigma^{\prime}\cap
S^{3}=R$, the meridian of $T_{A}$ is 0-homologous in
$S^{4}\setminus\Sigma^{\prime}$.

We claim that a longitude of $T_{A}$ represents a generator of
$H_{1}(S^{4}\setminus\Sigma^{\prime})\cong\mathbb{Z}$. Recall that the annulus
$A_{\pm}$ is obtained by performing a saddle move on $\eta_{1}\cup\eta_{2}$
and capping off the resulting unknot with a disc. Suppose that the saddle is
attached at points $x_{1}\in\eta_{1}$ and $x_{2}\in\eta_{2}$; see Figure 7. We
can assume that $x_{1}$ and $x_{2}$ lie on opposite sides of the Seifert
surface $F^{\prime}$. Suppose that the saddle point of $A_{\pm}$ lies in
$S^{3}\times\\{t_{\pm}\\}$, and let $c_{\pm}$ be the core of the band that is
attached. Then

$\ell:=\bigl{(}\\{x_{1},x_{2}\\}\times[t_{-},t_{+}]\bigr{)}\cup c_{+}\cup
c_{-}$

is a longitude of $T_{A}$.

$R$
---
$+3$
---
$\eta_{1}$
---
$\eta_{2}$
---
$x_{1}$
---
$x_{2}$
---
$a$
---
$F^{\prime}$
---
Figure 7. The points $x_{1}\in\eta_{1}$ and $x_{2}\in\eta_{2}$, together with
the arc $a$ connecting them that intersects $F^{\prime}$ in a single point.

Let $a$ be a straight arc in $S^{3}$ connecting $x_{1}$ and $x_{2}$ such that
$|a\pitchfork F^{\prime}|=1$; see Figure 7. Then $D_{A}:=a\times[t_{-},t_{+}]$
gives a disc in $S^{4}$ with boundary $\ell$. We have
$D_{A}\pitchfork\digamma^{\prime}=\\{(a\cap F^{\prime},-1/8)\\}$. Furthermore,
$D_{A}\cap\Delta=\emptyset$, because we attach the saddle to
$\eta_{1}\cup\eta_{2}$ to obtain $A_{+}$ before we attach the saddle to $R$
when forming $\Delta$. It follows that $|D_{A}\pitchfork\Sigma^{\prime}|=1$.
Hence $\ell$ represents a generator of $H_{1}(S^{4}\setminus\Sigma^{\prime})$,
as claimed. Consequently, the pre-image $p^{-1}(T_{A})$ is connected.

By construction, we obtain $\Sigma$ from $\Sigma^{\prime}$ by performing knot
surgery along $T_{A}$ with pattern some knot $J$ in $S^{3}$ (in the main
Cochran–Davis example, $J$ is the left-handed trefoil). So we can obtain $M$
from $S^{2}\times S^{2}$ by knot surgery along the torus $T$. Since
$H_{2}(S^{4})=0$, the torus $T_{A}$ bounds a 3-manifold $N$ in $S^{4}$. Make
$N$ transverse to $\Sigma^{\prime}$. Then $p^{-1}(N)$ is a 3-manifold in $M$
with boundary $T$. Hence $T$ is 0-homologous, as claimed. ∎

The knot surgery formula of Fintushel and Stern [FS98] only applies when the
torus along which we surger is homologically essential. Furthermore,
$S^{2}\times S^{2}$ has vanishing Seiberg–Witten invariants in the relevant
chamber. Hence, we cannot use the knot surgery formula to distinguish $M$ from
$S^{2}\times S^{2}$, and consequently $\Sigma$ from $T^{2}$.

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