MnLargeSymbols’164 MnLargeSymbols’171

# Small Dehn surgery and $SU(2)$

John A. Baldwin  Department of Mathematics
Boston College<EMAIL_ADDRESS>,  Zhenkun Li  Department of Mathematics
Stanford University<EMAIL_ADDRESS>,  Steven Sivek  Department of
Mathematics
Imperial College London<EMAIL_ADDRESS>and  Fan Ye  Department of
Pure Mathematics and Mathematical Statistics
University of Cambridge<EMAIL_ADDRESS>
###### Abstract.

We prove that the fundamental group of $3$-surgery on a nontrivial knot in
$S^{3}$ always admits an irreducible $SU(2)$-representation. This answers a
question of Kronheimer and Mrowka dating from their work on the Property P
conjecture. An important ingredient in our proof is a relationship between
instanton Floer homology and the symplectic Floer homology of genus-2 surface
diffeomorphisms, due to Ivan Smith. We use similar arguments at the end to
extend our main result to infinitely many surgery slopes in the interval
$[3,5)$.

JAB was supported by NSF FRG Grant DMS-1952707.

##  1\. Introduction

The $SU(2)$-representation variety

$R(Y)=\operatorname{Hom}(\pi_{1}(Y),SU(2))$

associated with a 3-manifold $Y$ is an object of central importance in
instanton gauge theory. A basic question about such varieties is whether they
contain irreducible representations. A homomorphism into $SU(2)$ is reducible
if and only if it has abelian image, inspiring the following definition:

######  Definition 1.1.

A 3-manifold $Y$ is _$SU(2)$ -abelian_ if every $\rho\in R(Y)$ has abelian
image.111If $b_{1}(Y)=0$ then $\rho$ has abelian image if and only if it has
cyclic image. In particular, $R(Y)$ contains an irreducible if and only if $Y$
is not $SU(2)$-abelian.

The classification of $SU(2)$-abelian 3-manifolds is a wide-open problem, even
among Dehn surgeries on knots.222Though see [LPCZ21, BS21] and [SZ19] for
progress on this problem in the cases of toroidal homology spheres and non-
hyperbolic geometric manifolds, respectively. In their proof of the Property P
conjecture [KM04b], Kronheimer and Mrowka proved that if $K$ is a nontrivial
knot in $S^{3}$ then $S^{3}_{1}(K)$ is not $SU(2)$-abelian (hence, not a
homotopy sphere). They then strengthened this result in [KM04a], proving under
the same hypothesis that $S^{3}_{r}(K)$ is not $SU(2)$-abelian for any
rational number $r\in[0,2]$.

It is natural to ask whether this also holds for larger values of $r$.
Kronheimer and Mrowka explicitly posed this question in [KM04a] for the next
two integers, $r=3$ and $4$, noting that it fails for $r=5$ because
$5$-surgery on the right-handed trefoil is a lens space. Baldwin and Sivek
answered half of this question in [BS19], proving that $S^{3}_{4}(K)$ is not
$SU(2)$-abelian for any nontrivial knot $K\subset S^{3}$. Our main theorem
answers the remaining half:

######  Theorem 1.2.

$S^{3}_{3}(K)$ is not $SU(2)$-abelian for any nontrivial knot $K\subset
S^{3}$.

######  Remark 1.3.

We ultimately expect that $r$-surgery on a nontrivial knot is not
$SU(2)$-abelian for any rational number $r\in[0,5)$. Our techniques can be
used to prove this for infinitely many additional slopes in the interval
$[3,5)$, as discussed in §1.2.

Our proof involves a novel blend of ingredients, including a relationship
between instanton Floer homology and the symplectic Floer homology of genus-2
surface diffeomorphisms, due to Smith [Smi12]; ideas of Baldwin–Hu–Sivek from
their recent work on cinquefoil detection [BHS21]; and new results of Li–Ye
relating the graded Euler characteristics of instanton and Heegaard knot Floer
homology [LY21b, LY21a]. Along the way, we partially characterize instanton
L-space knots of genus 2 (Theorem 2.1 and Remark 2.6), and prove several other
results (Proposition 3.2, and Theorems 4.1 and 5.1) which may be of
independent interest. We outline our proof of Theorem 1.2 in detail below.

###  1.1. Proof outline

Suppose $K\subset S^{3}$ is a nontrivial knot. Let us assume for a
contradiction that $S^{3}_{3}(K)$ is $SU(2)$-abelian. This surgered manifold
is thus an instanton L-space [BS18b]. It then follows from [BS19] that $K$ is
fibered and strongly quasipositive of genus 1 or 2. If the genus is 1, then
$K$ is the right-handed trefoil, but $3$-surgery on this trefoil is Seifert
fibered with base orbifold $S^{2}(2,3,3)$ and is therefore not
$SU(2)$-abelian, by [SZ19], a contradiction. We similarly rule out the
possibility that $K$ is the cinquefoil $T_{2,5}.$

Thus, $K\not\cong T_{2,5}$ is a genus-2 instanton L-space knot. Recent work of
Li and Ye [LY21c] then implies that $K$ has Alexander polynomial

$\Delta_{K}(t)=t^{2}-t+1-t^{-1}+t^{-2},$

from which we conclude as in [BHS21] that $K$ is a hyperbolic knot. Let
$(S,h)$ be an abstract open book corresponding to the fibration associated
with $K$. The hyperbolicity of $K$ implies that the monodromy $h$ is freely
isotopic to a pseudo-Anosov homeomorphism $\psi:S\to S.$

In [Smi12], Smith proved that (a version of) the instanton Floer homology of
the mapping torus of a genus-2 surface diffeomorphism encodes the symplectic
Floer cohomology of the diffeomorphism. We use Smith’s result, together with a
calculation of the framed instanton homology $I^{\\#}(S^{3}_{0}(K),\mu)$, to
prove that the homeomorphism $\psi$ has no fixed points, following ideas in
[BHS21]. We then apply results from [BHS21] to conclude that $K$ is doubly
periodic with unknotted quotient $A$ and axis $B$, where $A$ and $B$ have
linking number $5$.

We may therefore express $S^{3}_{3}(K)$ as the branched double cover of the
lens space

$L=S^{3}_{3/2}(A)\cong L(3,2),$

branched along the primitive knot $J\subset L$ induced by $B$ in this surgery
on $A$,

(1.1) $S^{3}_{3}(K)\cong\Sigma(L,J).$

Under the assumption that $S^{3}_{3}(K)$ is $SU(2)$-abelian, we prove that
every representation

$\rho:\pi_{1}(L\setminus N(J))\to SU(2)$

which sends a fixed meridian of $J$ to $i$ has finite cyclic image. There are
exactly $|H_{1}(L)|=3$ such representations. We prove that these
representations are nondegenerate as generators of the instanton knot homology
of $J$, and thereby conclude that $J$ is an _instanton Floer simple knot_ ,
meaning that

$\dim_{\mathbb{C}}\mathit{KHI}(L,J)=\dim_{\mathbb{C}}I^{\\#}(L)=3.$

We deduce that $J$ is rationally fibered, and that the graded Euler
characteristic of $\mathit{KHI}(L,J)$ is given by

$\chi_{\mathrm{gr}}(\mathit{KHI}(L,J))=t^{-n}+1+t^{n},$

for some $n=g(F)+1,$ where $F$ is a minimal-genus rational Seifert surface for
$J$.

In recent work [LY21b], Li and Ye proved an equality between the graded Euler
characteristics of instanton knot homology and (Heegaard) knot Floer homology,

$\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,J))=\pm\chi_{\mathrm{gr}}(\mathit{KHI}(L,J)).$

We combine this with constraints on the graded Euler characteristic of knot
Floer homology to conclude that $n=1$. Thus, $J$ has (rational) genus 0, which
implies that its fibered exterior is a solid torus. It follows that
$\Sigma(L,J)$, and hence $3$-surgery on $K$, is a lens space. Since $K$ is
doubly periodic and has a lens space surgery, a result of Wang and Zhou [WZ92]
says that $K$ must be a torus knot, a contradiction.

###  1.2. Other small surgeries

As mentioned above, we suspect that if $K\subset S^{3}$ is a nontrivial knot,
then $S^{3}_{r}(K)$ is not $SU(2)$-abelian for any rational $r\in[0,5)$. Long
after Kronheimer and Mrowka’s proof for $r\in[0,2]$, Baldwin and Sivek proved
this [BS19] for the dense subset of the interval $(2,3)$ consisting of
rationals with prime power numerators. Proving it for the corresponding subset
of $[3,5)$ is much harder, as explained in §6. Indeed, our main theorem
concerns the single case $r=3$. Nevertheless, our techniques can be used to
extend Theorem 1.2 to infinitely many surgery slopes in both $[3,4)$ and
$[4,5)$, and even to a special dense subset of slopes in $(5,7)$. Our main
result in this vein is:

######  Theorem 1.4.

Let $p$ and $q$ be coprime positive integers with $p/q\in[3,7)$, where $p$ is
a prime power. If either

  * •

$p$ is even,

  * •

$p$ is odd and $\gcd(p,5)=1$ and $p/q\in[4,5)$, or

  * •

$p$ is odd and $\gcd(p,5)=1$ and $p/q\in[3,4)$ with $23p-9\leq 80q\leq 25p+9$,

then $S^{3}_{p/q}(K)$ is not $SU(2)$-abelian for every nontrivial knot
$K\subset S^{3}$.

The inequality in Theorem 1.4 is satisfied for
$p/q\in[16/5,80/23)\subset[3,4)$. It is satisfied for $p/q=3$ as well, so this
theorem recovers Theorem 1.2. Note that Theorem 1.4 also applies to the slopes
$4,$ $7/2$, and $9/2$. Together with the results mentioned above, concerning
the intervals $[0,2]$ and $(2,3)$, this proves that:

######  Corollary 1.5.

If $K\subset S^{3}$ is a nontrivial knot then $S^{3}_{r}(K)$ is not
$SU(2)$-abelian for any integer or half-integer $r<5$.

###  1.3. Organization

In §2, we prove that if $K\not\cong T_{2,5}$ is an instanton L-space knot of
genus 2, then $K$ is doubly-periodic with unknotted quotient. We then use this
in §3 to show that if $3$-surgery on a nontrivial knot is $SU(2)$-abelian,
then this surgery is the branched double cover of a knot $J\subset L(3,2)$; in
fact, we prove something more general (Proposition 3.2). In §4, we prove that
such $J$ are instanton Floer simple (Theorem 4.1). In §5, we combine these
results with an understanding of instanton Floer simple knots in lens spaces
(Theorem 5.1) to prove Theorem 1.2. Finally, in §6, we extend our main theorem
to other surgery slopes in $[3,5)$, proving Theorem 1.4.

###  1.4. Acknowledgements

We thank Ying Hu for helping to pioneer the ideas in [BHS21] which inspired
many of the proofs in this paper. We also thank Ken Baker, Yi Ni, and Jake
Rasmussen for helpful discussions. Finally, thanks to Peter Kronheimer and Tom
Mrowka for posing the interesting question at the heart of this work, and for
their encouragement.

##  2\. Genus two L-space knots

Recall that a rational homology 3-sphere $Y$ is an _instanton L-space_ if its
framed instanton homology satisfies

$\dim_{\mathbb{C}}I^{\\#}(Y)=|H_{1}(Y)|.$

A knot $K\subset S^{3}$ is said to be an _instanton L-space knot_ if
$S^{3}_{r}(K)$ is an instanton L-space for some rational number $r>0$. Such
knots were studied extensively in [BS19]; in particular, it was shown there
[BS19, Theorem 1.15] that instanton L-space knots are fibered and strongly
quasipositive.

Our main goal in this section is to prove the following:

######  Theorem 2.1.

Let $K\subset S^{3}$ be an instanton L-space knot of genus 2. Then $K$ is
fibered and strongly quasipositive, with Alexander polynomial

$\Delta_{K}(t)=t^{2}-t+1-t^{-1}+t^{-2}.$

If $K\not\cong T_{2,5}$, then there exists a pseudo-Anosov 5-braid $\beta$
whose closure $B=\hat{\beta}$ is an unknot with braid axis $A$, such that $K$
is the lift of $A$ in the branched double cover

$\Sigma(S^{3},B)\cong S^{3}.$

In particular, $K$ is a doubly periodic knot with unknotted quotient $A$ and
axis $B$.

A Heegaard Floer analogue of this result was proved in [BHS21, §3]. Our proof
of Theorem 2.1 will largely follow the arguments there, and we will refer to
specific parts of [BHS21] for steps whose proofs are the same in either
setting. We begin with some preliminary results.

######  Lemma 2.2.

Let $K$ be an instanton L-space knot of genus 2. Then $K$ is fibered and
strongly quasipositive, with Alexander polynomial

$\Delta_{K}(t)=t^{2}-t+1-t^{-1}+t^{-2},$

and either $K\cong T_{2,5}$ or $K$ is hyperbolic.

###### Proof.

The claims that $K$ is fibered and strongly quasipositive are [BS19, Theorem
1.15], as discussed above. The Alexander polynomial $\Delta_{K}(t)$ is
determined by [LY21c, Theorem 1.9], which says that there are integers $k\geq
0$ and

$n_{k}>n_{k-1}>\dots>n_{0}=0>n_{-1}>\dots>n_{-k},$

with $n_{-j}=-n_{j}$ for all $j$, such that the instanton knot homology
$\mathit{KHI}(K)$ satisfies

$\dim_{\mathbb{C}}\mathit{KHI}(K,i)=\begin{cases}1,&i=n_{j}\text{ for some
j}\\\ 0,&\text{otherwise},\end{cases}$

with the nonzero summands $\mathit{KHI}(K,n_{j})$ supported in alternating
$\mathbb{Z}/2\mathbb{Z}$ gradings. We know that $\mathit{KHI}(K,2)\neq 0$
since $K$ has genus 2 [KM10b, Proposition 7.16], and then
$\mathit{KHI}(K,1)\neq 0$ since $K$ is fibered [BS18a, Theorem 1.7], so $k=2$
and $(n_{2},n_{1},n_{0})=(2,1,0)$. Then $\mathit{KHI}(K)$ has graded Euler
characteristic

$\sum_{j\in\mathbb{Z}}\chi(\mathit{KHI}(K,j))\cdot t^{j}=\pm\Delta_{K}(t)$

[KM10a, Lim10], so the normalization $\Delta_{K}(1)=1$ forces $\Delta_{K}(t)$
to be as claimed. Since $K$ is fibered with this particular Alexander
polynomial, it is either $T_{\pm 2,5}$ or hyperbolic by [BHS21, Lemma 3.2].
The case of $T_{-2,5}$ is ruled out as this knot is not strongly
quasipositive.∎

######  Lemma 2.3.

Let $K$ be an instanton L-space knot of genus 2, and let $w\to S^{3}_{0}(K)$
be a Hermitian line bundle such that $c_{1}(w)$ is Poincaré dual to the
meridian $\mu$ of $K$. Then

$\dim_{\mathbb{C}}I_{*}(S^{3}_{0}(K))_{w}=6,$

where $I_{*}$ refers here to the version of instanton homology for admissible
bundles described, for example, in [Don02, §5.6].

###### Proof.

We know that $3$-surgery on $K$ is an instanton L-space [BS19, Proposition
7.11], while neither $1$\- nor $2$-surgery on $K$ is an instanton L-space
[BS19, Proposition 7.12]. It follows in the language of [BS20] that
$\nu^{\sharp}(K)=r_{0}(K)=3$, meaning that

$\dim_{\mathbb{C}}I^{\\#}(S^{3}_{p/q}(K))=\begin{cases}p,&p/q\geq 3\\\
6q-p,&p/q<3\end{cases}$

for all coprime integers $p,q$ with $q\geq 1$. In particular we have

$\dim_{\mathbb{C}}I^{\\#}(S^{3}_{0}(K),\mu)=6$

by [BS20, Proposition 3.3].

To prove the lemma, we then relate $I_{*}(S^{3}_{0}(K))_{w}$ to
$I^{\\#}(S^{3}_{0}(K),\mu)$ using Fukaya’s connected sum theorem for instanton
homology [Sca15, Theorem 1.3]. To be precise, since $H_{2}(S^{3}_{0}(K))$ is
generated by a closed surface of genus 2 which is dual to $\mu$, the special
case of this theorem discussed in [Sca15, §9.8] says that

$I^{\\#}(S^{3}_{0}(K),\mu)\otimes H_{*}(S^{4})\cong
I_{*}(S^{3}_{0}(K))_{w}\otimes H_{*}(S^{3})$

as relatively $\mathbb{Z}/4\mathbb{Z}$-graded vector spaces over any field of
characteristic zero. It follows that
$\dim_{\mathbb{C}}I_{*}(S^{3}_{0}(K))_{w}=6$ as claimed. ∎

Recall for the proposition below that instanton L-space knots are fibered.

######  Proposition 2.4.

Let $K\not\cong T_{2,5}$ be an instanton L-space knot of genus 2. Let $(S,h)$
be an open book encoding the fibration of its complement. Then the monodromy
$h$ is freely isotopic to a pseudo-Anosov homeomorphism

$\psi:S\to S$

with no fixed points.

###### Proof.

The complement $S^{3}\setminus K$ is hyperbolic, by Lemma 2.2. It follows that
the monodromy $h$ is freely isotopic to a pseudo-Anosov homeomorphism

$\psi:S\to S,$

by work of Thurston [Thu98]. Since $K$ is strongly quasipositive, also by
Lemma 2.2, [BHS21, Lemma 3.3] says that the stable foliation of $\psi$ has
some number $n\geq 2$ of boundary prongs.

As discussed in [BHS21, §2], the fact that $n\geq 2$ means that the map $\psi$
extends naturally to a pseudo-Anosov homeomorphism

$\hat{\psi}:\hat{S}\to\hat{S}$

of the closed genus-2 surface obtained from $S$ by capping off its boundary
with a disk. The stable foliation of $\psi$ extends across the disk to a
stable invariant foliation for $\hat{\psi}$ in which the $n$ boundary prongs
extend to $n$ prongs meeting at a singularity $p$ in the capping disk (except
that $p$ is a smooth point if $n=2$). Moreover, $p$ is a fixed point of
$\hat{\psi}$.

Note that the mapping torus $Y_{\hat{\psi}}$ of $\hat{\psi}$ is homeomorphic
to $0$-surgery on $K$,

$Y_{\hat{\psi}}\cong S^{3}_{0}(K),$

by a map which identifies the suspension of $p$ with the meridian $\mu$ of $K$
in the $0$-surgery. Smith [Smi12, Corollary 1.8] proved in this case (or, more
generally, for any genus-2 mapping torus which is a homology $S^{1}\times
S^{2}$ with a distinguished section coming from a marked point on the surface)
that there is an isomorphism (with $\mathbb{C}$ coefficients) of the form

(2.1) $\mathit{HF}_{\mathrm{inst}}(E\to
Y_{\hat{\psi}})\cong\mathbb{C}\oplus\mathit{HF}_{\mathrm{symp}}^{\ast}(\hat{\psi})\oplus\mathbb{C}.$

Here, $E\to Y_{\hat{\psi}}$ is the nontrivial $SO(3)$ bundle with $w_{2}$ dual
to $\mu$; $\mathit{HF}^{\ast}_{\mathrm{symp}}(\hat{\psi})$ is the symplectic
Floer cohomology of $\hat{\psi}$, defined via any area-preserving
diffeomorphism (for any area form on $\hat{S}$) in the mapping class of
$\hat{\psi}$; and $\mathit{HF}_{\mathrm{inst}}$ is the version of instanton
homology used by Dostoglou and Salamon in [DS94]. The latter is defined using
a slightly larger gauge group than the determinant-1 gauge transformations
defining $I_{*}$, and is thus the quotient of the relatively
$\mathbb{Z}/8\mathbb{Z}$-graded $I_{*}$ by a degree-4 involution. It follows,
in particular, that

$\dim_{\mathbb{C}}\mathit{HF}_{\mathrm{inst}}(E\to
Y_{\hat{\psi}})=\frac{1}{2}\cdot\dim_{\mathbb{C}}I_{*}(S^{3}_{0}(K))_{w},$

where $w\to S^{3}_{0}(K)$ is a Hermitian line bundle with $c_{1}$ dual to
$\mu$, as in Lemma 2.3. The same lemma says that
$\dim_{\mathbb{C}}I_{*}(S^{3}_{0}(K))_{w}=6$, from which it follows that

$\dim_{\mathbb{C}}\mathit{HF}_{\mathrm{inst}}(E\to Y_{\hat{\psi}})=3,$

and therefore that

(2.2) $\dim_{\mathbb{C}}\mathit{HF}_{\mathrm{symp}}^{\ast}(\hat{\psi})=1,$

by the isomorphism (2.1).

The symplectic Floer cohomology of pseudo-Anosov homeomorphisms was computed
by Cotton-Clay in [CC09, §3]. In particular, given $\hat{\psi}$ as above,
Cotton-Clay defines in [CC09, §3.2] a canonical smooth representative
$\hat{\psi}_{{sm}}$ of $\hat{\psi}$ whose symplectic Floer chain complex

$\mathit{CF}_{\ast}^{\mathrm{symp}}(\hat{\psi}_{{sm}})$

is freely generated by the fixed points of $\hat{\psi}_{{sm}}$ and has trivial
differential. This implies that

(2.3)
$\mathit{HF}^{\ast}_{\mathrm{symp}}(\hat{\psi})\cong\mathit{CF}_{\ast}^{\mathrm{symp}}(\hat{\psi}_{{sm}}).$

(This is proved in [CC09] with $\mathbb{Z}/2\mathbb{Z}$ coefficients, but also
holds in characteristic zero since the differential vanishes for purely
topological reasons.) Furthermore, $\hat{\psi}_{{sm}}$ has the property that
each fixed point of $\hat{\psi}$ is also a fixed point of $\hat{\psi}_{{sm}}$.
The combination of (2.2) and (2.3) therefore implies that $\hat{\psi}$ has at
most one fixed point. On the other hand, we know that $p$ is a fixed point of
$\hat{\psi}$ from the discussion above. It follows that this is the only fixed
point of $\hat{\psi}$, which implies that the original map $\psi$ has no fixed
points. ∎

######  Remark 2.5.

The proof of Proposition 2.4 is notably easier than that of the analogous
[BHS21, Theorem 3.5], at least modulo deep theorems by others. The reason is
that the work of Lee and Taubes [LT12] used in [BHS21] requires a monotonicity
condition which only applies in genus 3 or higher, so a trick is required to
pass from the genus 2 case to the higher genus case. By contrast, the
references [DS94, Smi12] work perfectly well for genus-2 mapping tori, and in
fact [Smi12, Corollary 1.8] _only_ applies in genus 2.

######  Remark 2.6.

Suppose $K\not\cong T_{2,5}$ is a genus-2 instanton L-space knot, $(S,h)$ is
an open book associated to the fibration of its complement, and $\psi:S\to S$
a pseudo-Anosov map isotopic to $h$. Proposition 2.4 says that $\psi$ has no
fixed points. Thus we can apply [BHS21, Proposition 3.8], whose other
hypotheses follow from Lemma 2.2, to conclude the following:

  * •

the monodromy $h$ has fractional Dehn twist coefficient $c(h)=\frac{1}{4}$;

  * •

the stable foliation of $\psi$ has four prongs on $\partial S$, and two
interior 3-pronged singularities which are exchanged by $\psi$.

We will not need either of these properties in this article, but this may be
independently useful towards a characterization of genus-2 instanton L-space
knots.

At this point, the proof of Theorem 2.1 proceeds in exactly the same way as
that of its Heegaard Floer analogue [BHS21, Theorem 3.1].

###### Proof of Theorem 2.1.

The first few claims follow immediately from Lemma 2.2. To argue that if
$K\not\cong T_{2,5}$ then there is a pseudo-Anosov 5-braid $\beta$ with
unknotted closure $B=\hat{\beta}$ such that $K$ is the lift of the braid axis
$A$ to the branched double cover of $B$, we simply repeat the proof of [BHS21,
Theorem 3.1] verbatim. That proof relies only on the conclusions of
Proposition 2.4, so it applies equally well here. ∎

##  3\. $SU(2)$-abelian surgeries and branched double covers

The conditions of being an instanton L-space and being $SU(2)$-abelian are
closely related. For surgeries on knots, we have the following:

######  Lemma 3.1.

Let $K\subset S^{3}$ be a knot, and suppose $p$ and $q$ are positive coprime
integers such that $p$ is a prime power. If

$Y=S^{3}_{p/q}(K)$

is $SU(2)$-abelian, then $Y$ is an instanton L-space.

###### Proof.

An $SU(2)$-abelian rational homology sphere $Y$ is an instanton L-space
whenever $\pi_{1}(Y)$ is _cyclically finite_ , by [BS18b, Theorem 4.6]. This
cyclical finiteness is satisfied if, for example, $H_{1}(Y)$ is cyclic of
prime power order, by [BS18b, Proposition 4.9]. ∎

Our main goal is to prove that 3-surgery on a nontrivial knot in $S^{3}$ is
never $SU(2)$-abelian. In this section we use Lemma 3.1, together what we
proved about genus-2 instanton L-space knots in §2, to reduce the problem to
one about branched double covers of knots in the lens space $L(3,2)$, as
outlined in the introduction. In fact, we prove a substantially more general
result in Proposition 3.2 below. We will use this additional generality in §6
to address other surgeries.

######  Proposition 3.2.

Let $K\subset S^{3}$ be a nontrivial knot, and suppose $p,q$ are coprime
positive integers with $p/q<5$, where $p$ is an odd prime power. If

$Y=S^{3}_{p/q}(K)$

is $SU(2)$-abelian, then:

  * •

$Y$ is not Seifert fibered,

  * •

$K$ is hyperbolic, and doubly periodic with unknotted quotient $A$, and

  * •

$Y$ is homeomorphic to a branched double cover of the lens space

$L=S^{3}_{p/2q}(A)=L(p,2q)$

along some knot $J\subset L$ which is homologous to 5 times the core of the
solid torus $S^{3}\setminus N(A)\subset L$ in the genus-1 Heegaard splitting
of $L$ along the surface $\partial N(A)$.

###### Proof.

Lemma 3.1 implies that $Y$ is an instanton L-space; in particular, $K$ is an
instanton L-space knot. It then follows from [BS19, Theorem 1.15] that

$2g(K)-1\leq p/q<5,$

which implies that $K$ has genus 1 or 2.

We next observe that $K$ is neither $T_{2,3}$ nor $T_{2,5}$. Indeed, $p/q<5$
implies that

$|abq-p|>1,\textrm{ for }(a,b)=(2,3)\textrm{ and }(2,5).$

Then $(p/q)$-surgeries on $T_{2,3}$ and $T_{2,5}$ are Seifert manifolds which
are not lens spaces [Mos71]. Such surgeries cannot be $SU(2)$-abelian [SZ19,
Remark 1.3]. We conclude that $K$ is neither $T_{2,3}$ nor $T_{2,5}$. Since
$T_{2,3}$ is the only genus-1 instanton L-space knot [BS19, Proposition 7.12],
it follows that $K$ must be a genus-2 instanton L-space knot other than
$T_{2,5}$.

Theorem 2.1 then says that $K$ is hyperbolic, and doubly periodic with
unknotted quotient $A$ and axis $B$, where $A$ and $B$ have linking number
$5$. In particular, there is an involution $\tau:S^{3}\to S^{3}$ with
$\tau(K)=K$ whose fixed set is the axis $B$. This involution extends from the
exterior $S^{3}\setminus N(K)$ across the solid torus realizing the Dehn
surgery producing

$Y=S^{3}_{p/q}(K),$

and it acts freely on this solid torus since $p$ is odd. Thus, $\tau$ induces
an involution on $Y$ with fixed set $B$; the quotient of $Y$ by this action is
then the lens space

$L=S^{3}_{p/2q}(A)\cong L(p,2q)$

obtained as $(p/2q)$-surgery on the unknotted quotient $A$. In particular, $Y$
is homeomorphic to a branched double cover of $L$ along the image $J$ of $B$
in this lens space. The fact that $A$ and $B$ have linking number 5 implies
that $J$ is homologous in $H_{1}(L)$ to 5 times the core of the solid torus
$S^{3}\setminus N(A)$ in the genus-1 Heegaard splitting of $L$ along the
surface $\partial N(A)$.

Finally, we argue that

$Y=S^{3}_{p/q}(K)$

is not Seifert fibered. Suppose that it is. Then $Y$ is a lens space, by
[SZ19, Remark 1.3]. In this case, since $K$ is doubly periodic (and thus
admits a symmetry with 1-dimensional fixed point set which is not a strong
inversion) and has a nontrivial cyclic surgery, a theorem of Wang and Zhou
[WZ92, Proposition 3] says that $K$ must be a torus knot, a contradiction. ∎

##  4\. Branched double covers and Floer simple knots

Our goal in this section is to prove Theorem 4.1 below. This result, in
combination with Proposition 3.2, implies that if $3$-surgery on a nontrivial
knot is $SU(2)$-abelian, then the surgered manifold is the branched double
cover of an instanton Floer simple knot in $L(3,2)$. We will combine this with
an understanding of Floer simple knots in this lens space in §5 to complete
the proof of Theorem 1.2, as outlined in the introduction, and we will apply
Theorem 4.1 to infinitely many additional surgery slopes in §6. The various
results in this section may be of independent interest.

######  Theorem 4.1.

Suppose $L$ is a rational homology sphere in which $H_{1}(L)$ is cyclic of
order an odd prime power, and let $J\subset L$ be a primitive knot. If the
branched double cover $\Sigma(L,J)$ is $SU(2)$-abelian and satisfies

$|H_{1}(\Sigma(L,J))|=|H_{1}(L)|,$

then

$\dim_{\mathbb{C}}\mathit{KHI}(L,J)=|H_{1}(L)|.$

In other words, the knot $J\subset L$ is instanton Floer simple.

We start by clarifying some of the terminology and notation in the statement
of Theorem 4.1, explaining in particular why a primitive knot has a unique
branched double cover.

For the rest of this section, $L$ will refer to a rational homology sphere,
and $J$ to a knot in $L$. We say that $J\subset L$ is _primitive_ if its
homology class $[J]$ generates $H_{1}(L;\mathbb{Z})$. In this case, an
exercise in algebraic topology shows that

$H_{1}(L\setminus N(J))\cong\mathbb{Z}.$

This group is generated by the homology class of some peripheral curve
$\alpha\subset\partial N(J)$. Dehn filling along $\alpha$ produces an integral
homology sphere $Z$, and if $C$ is the core of this filling, then $\alpha$ is
a meridian $\mu_{C}$ of $C$, and we have a homeomorphism

$Z\setminus N(C)\cong L\setminus N(J).$

Since Dehn filling $Z\setminus N(C)$ along the meridian $\mu_{J}$ of $J$
recovers $L$, it follows that

$[\mu_{J}]=|H_{1}(L)|\cdot[\mu_{C}]$

in the first homology of this knot complement. We will use the notation
established in this paragraph throughout the rest of this section.

######  Lemma 4.2.

Suppose $H_{1}(L)$ has odd order and that $J\subset L$ is primitive, and let
$C\subset Z$ be as above. Then $J\subset L$ and $C\subset Z$ have unique
branched double covers, $\Sigma(L,J)$ and $\Sigma(Z,C)$, which satisfy

$H_{1}(\Sigma(L,J))\cong H_{1}(\Sigma(Z,C))\oplus H_{1}(L).$

###### Proof.

Since the knot $C\subset Z$ is nullhomologous, we can take a meridian
$\mu_{C}$ and longitude $\lambda_{C}$ in $\partial N(C)$ and write

$\mu_{J}=(\mu_{C})^{p}(\lambda_{C})^{q}$

in $\pi_{1}(\partial N(C))$ for some relatively prime integers $p$ and $q$,
where we reverse the orientation of $C$ if necessary to arrange $p\geq 0$.
Then $H_{1}(L)\cong\mathbb{Z}/p\mathbb{Z}$.

To define $\Sigma(L,J)$, we first consider the unique connected double cover
$M\to L\setminus N(J)$ of $L\setminus N(J)$, specified by the kernel of the
map

$\pi_{1}(L\setminus N(J))\to H_{1}(L\setminus
N(J))\cong\mathbb{Z}\xrightarrow{\text{mod }2}\mathbb{Z}/2\mathbb{Z}.$

Since $[\mu_{C}]$ generates $H_{1}(L\setminus N(J))$ and

$[\mu_{J}]=|H_{1}(L)|\cdot[\mu_{C}],$

with $|H_{1}(L)|$ odd, it follows that $\mu_{J}$ is sent to $1$ under this
map. Thus, $\mu_{J}^{2}$ lifts to a simple closed curve in $M$. The unique
branched double cover $\Sigma(L,J)$ is then formed by Dehn filling $M$ along
this lift of $\mu_{J}^{2}$. We similarly define $\Sigma(Z,C)$ by Dehn filling
$M$ along a lift of $\mu_{C}^{2}$.

Since $C$ is nullhomologous in $Z$, its lift $\tilde{C}\subset\Sigma(Z,C)$ is
also nullhomologous, so we define $\mu_{\tilde{C}}$ and $\lambda_{\tilde{C}}$
to be its meridian and longitude in the boundary of

$M=\Sigma(Z,C)\setminus N(\tilde{C}).$

These curves lift $\mu_{C}^{2}$ and $\lambda_{C}$. If
$\mu_{J}=(\mu_{C})^{p}(\lambda_{C})^{q}$ as above, then
$\mu_{J}^{2}=(\mu_{C}^{2})^{p}(\lambda_{C})^{2q}$ lifts to

$\mu_{\tilde{J}}=(\mu_{\tilde{C}})^{p}(\lambda_{\tilde{C}})^{2q}$

and so $\Sigma(L,J)$ is the result of $(p/2q)$-surgery on
$\tilde{C}\subset\Sigma(Z,C)$. Since $\tilde{C}$ is nullhomologous and
$H_{1}(L)\cong\mathbb{Z}/p\mathbb{Z}$, this completes the proof. ∎

######  Remark 4.3.

We will often use the shorthand $\Sigma(J)$ for $\Sigma(L,J)$ out of
convenience.

The lemma below will be useful in several places.

######  Lemma 4.4.

Suppose $H_{1}(L)$ has odd order and $J\subset L$ is primitive. Then

$|H_{1}(\Sigma(L,J))|=|H_{1}(L)|\cdot|\Delta_{J}(-1)|,$

where $\Delta_{J}(t)$ is the symmetrized Alexander polynomial of $J$.

###### Proof.

Let $C\subset Z$ be as above. By [BZ03, Theorem 8.21], we have that

$|H_{1}(\Sigma(Z,C))|=|\Delta_{C}(-1)|=|\Delta_{J}(-1)|,$

where the second equality follows from the fact that $C$ and $J$ have
homeomorphic complements and the Alexander polynomial depends only on the
complement. Then

$|H_{1}(\Sigma(L,J))|=|H_{1}(L)|\cdot|H_{1}(\Sigma(Z,C))|=|H_{1}(L)|\cdot|\Delta_{J}(-1)|,$

by Lemma 4.2. ∎

In the results below, we view $SU(2)$ as the group of unit quaternions.

######  Proposition 4.5.

Suppose $H_{1}(L)$ has odd order, that $J\subset L$ is primitive, and that

$|H_{1}(\Sigma(L,J))|=|H_{1}(L)|.$

Then every representation

$\rho:\pi_{1}(L\setminus N(J))\to SU(2)$

with image in the binary dihedral group
$\\{e^{i\theta}\\}\cup\\{e^{i\theta}j\\}$ actually has cyclic image. Moreover,
there are exactly $|H_{1}(L)|$ such representations satisfying
$\rho(\mu_{J})=i$.

###### Proof.

Let $C\subset Z$ be as above, so that $[\mu_{C}]$ generates $H_{1}(L\setminus
N(J))\cong\mathbb{Z}$ and

$[\mu_{J}]=|H_{1}(L)|\cdot[\mu_{C}].$

By Lemma 4.4, we have $|\Delta_{J}(-1)|=|\Delta_{C}(-1)|=1$. Therefore
$|H_{1}(\Sigma(C))|=1$. It follows that there are no non-abelian
representations

$\pi_{1}(Z\setminus N(C))\to SU(2)$

with binary dihedral image, since the number of such conjugacy classes is

$\frac{|H_{1}(\Sigma(C))|-1}{2}.$

This was originally proved for knots in $S^{3}$ by Klassen [Kla91, Theorem
10], whose proof makes essential use of the Wirtinger presentation; Boden and
Friedl [BF08, Corollary 1.3] give a generalization whose proof applies equally
well to knots in any integral homology sphere. In any case, since $L\setminus
N(J)\cong Z\setminus N(C)$, we conclude that there are no non-abelian binary
dihedral representations of $\pi_{1}(L\setminus N(J))$ either. In particular,
every binary dihedral representation of $\pi_{1}(L\setminus N(J))$ has image
in the cyclic group $\\{e^{i\theta}\\}$.

If $\rho$ is such a representation, then (since it has abelian image) it
factors through a map

$H_{1}(L\setminus N(J))\cong\mathbb{Z}\to SU(2),$

whose domain is generated by $[\mu_{C}]$. It follows that $\rho$ is uniquely
determined by its evaluation at $\mu_{C}$, and that

$\rho(\mu_{J})=\rho(\mu_{C})^{|H_{1}(L)|}.$

If we require that $\rho(\mu_{J})=i=e^{i\pi/2}$, then the possible $\rho$ are
specified by

$\rho(\mu_{C})=\exp\left(i\cdot\frac{\frac{\pi}{2}+2\pi
m}{|H_{1}(L)|}\right),\quad m=0,1,\dots,|H_{1}(L)|-1,$

and the proposition follows. ∎

The following is a mild adaptation of [Zen17, Proposition 3.1].

######  Lemma 4.6.

Suppose $H_{1}(L)$ has odd order, that $J\subset L$ is primitive, and that
$\Sigma(L,J)$ is an $SU(2)$-abelian rational homology sphere. Let

$\rho:\pi_{1}(L\setminus N(J))\to SU(2)$

be a representation satisfying $\rho(\mu_{J})=i$. Then the image of

$\operatorname{ad}\rho:\pi_{1}(L\setminus N(J))\to SO(3)$

is either cyclic or dihedral, and has order $2n$ for some odd integer $n$
dividing $|H_{1}(\Sigma(J))|$.

###### Proof.

We have a short exact sequence of groups

$1\to\pi_{1}(M)\xrightarrow{p_{*}}\pi_{1}(L\setminus
N(J))\xrightarrow{m}\mathbb{Z}/2\mathbb{Z}\to 1$

where

$M\xrightarrow{p}L\setminus N(J)$

is the unique connected double covering. The class $\mu_{J}^{2}$ belongs to
$\ker(m)=p_{*}\pi_{1}(M)$, since the homology class $[\mu_{J}^{2}]=2[\mu_{J}]$
is zero mod 2, so we can view the normal closure $\llangle\mu_{J}^{2}\rrangle$
as a subgroup of $\pi_{1}(M)$. Upon passing to quotients, we obtain a short
exact sequence

$1\to\pi_{1}(\Sigma(J))\to\frac{\pi_{1}(L\setminus
N(J))}{\llangle\mu_{J}^{2}\rrangle}\to\mathbb{Z}/2\mathbb{Z}\to 1.$

Since $\Sigma(J)$ is $SU(2)$-abelian, and $H_{1}(\Sigma(J))$ has odd order by
Lemma 4.2 and the oddness of $|H_{1}(L)|$, we may conclude [Zen17, Lemma 3.2]
that every representation

$\pi_{1}(\Sigma(J))\to SO(3)$

has cyclic image. (To see this, note that every representation
$\pi_{1}(\Sigma(J))\to SO(3)$ lifts to an $SU(2)$ representation, since the
obstruction to lifting belongs to

$H^{2}(\Sigma(J);\mathbb{Z}/2\mathbb{Z})=0.$

The image of this lift is abelian by assumption—and, hence, cyclic, since
$b_{1}(\Sigma(J))=0$—so the projection of this image back down to $SO(3)$ must
then be cyclic as well.) Any such representation therefore factors through
$H_{1}(\Sigma(J))$, which is finite of odd order, and so its image also has
odd order.

Now given

$\rho:\pi_{1}(L\setminus N(J))\to SU(2)$

with $\rho(\mu_{J})=i$, we observe that $\operatorname{ad}\rho$ sends
$\mu_{J}^{2}$ to $\operatorname{ad}(i^{2})=\operatorname{ad}(-1)=1$, hence
descends to a representation

$\tilde{\rho}:\pi_{1}(L\setminus N(J))/\llangle\mu_{J}^{2}\rrangle\to SO(3)$

with
$\operatorname{Im}(\tilde{\rho})=\operatorname{Im}(\operatorname{ad}\rho)$.
The short exact sequence above yields a commutative diagram

${1}$${\pi_{1}(\Sigma(J))}$${\frac{\pi_{1}(L\setminus
N(J))}{\llangle\mu_{J}^{2}\rrangle}}$${\mathbb{Z}/2\mathbb{Z}}$${1}$${SO(3)}$$\scriptstyle{r}$$\scriptstyle{\tilde{\rho}}$

in which the top row is exact, and the image of $r$ is an odd-order cyclic
subgroup of $SO(3)$. We will write $\operatorname{Im}(r)=\langle x\rangle$,
where $x\in SO(3)$ is an element of some odd order $n\geq 1$ and $n$ divides
$|H_{1}(\Sigma(J))|$.

Every element of $\pi_{1}(L\setminus N(J))/\llangle\mu_{J}^{2}\rrangle$ has
the form

$g\text{ or }\alpha\cdot g,\qquad g\in\pi_{1}(\Sigma(J))$

where $\alpha$ is some element of the nontrivial coset of the index-2 subgroup
$\pi_{1}(\Sigma(J))$. Thus, if we write $y=\tilde{\rho}(\alpha)$, then

$\operatorname{Im}(\tilde{\rho})=\langle x\rangle\cup\left(y\cdot\langle
x\rangle\right).$

If $y\in\langle x\rangle$ then $\operatorname{Im}(\tilde{\rho})$ is cyclic of
order $n$, but this is impossible since $\operatorname{Im}(\tilde{\rho})$
contains the element $\tilde{\rho}(\mu_{J})=\operatorname{ad}(i)$ of order 2.
Thus, $\langle x\rangle\cong\mathbb{Z}/n\mathbb{Z}$ is an index-2 subgroup of
$\operatorname{Im}(\tilde{\rho})\subset SO(3)$, so
$\operatorname{Im}(\tilde{\rho})$ has order $2n$, which is not a multiple of
4. Now, every finite subgroup of $SO(3)$ is either cyclic, dihedral,
tetrahedral, octahedral, or icosahedral, and the latter three have order 12,
24, or 60, so $\operatorname{Im}(\tilde{\rho})$ must be cyclic or dihedral, of
order $2n$. ∎

######  Proposition 4.7.

Suppose $H_{1}(L)$ has odd order, that $J\subset L$ is primitive, and that
$\Sigma(L,J)$ is an $SU(2)$-abelian rational homology sphere. Let

$\rho:\pi_{1}(L\setminus N(J))\to SU(2)$

be a representation satisfying $\rho(\mu_{J})=i$. Then
$\operatorname{Im}(\rho)$ is either cyclic or binary dihedral, and has order
$4n$, where $n$ divides $|H_{1}(\Sigma(L,J))|$. Moreover, if

$|H_{1}(\Sigma(L,J))|=|H_{1}(L)|,$

then $\rho$ has cyclic image.

###### Proof.

By Lemma 4.6, we know that the image of the composition

$\pi_{1}(L\setminus
N(J))\xrightarrow{\rho}SU(2)\xrightarrow{\operatorname{ad}}SO(3)$

is either cyclic or dihedral, and has order $2n$, where $n$ divides
$|H_{1}(\Sigma(J))|$ and is thus odd. Note that
$\operatorname{Im}(\rho)\subset SU(2)$ contains the kernel $\\{\pm 1\\}$ of
the map

$\operatorname{ad}:SU(2)\to SO(3),$

since $\rho(\mu_{J}^{2})=-1$. Therefore,

$\operatorname{Im}(\rho)=\operatorname{ad}^{-1}(\operatorname{Im}(\operatorname{ad}\rho))$

has order $4n$. If $\operatorname{Im}(\operatorname{ad}\rho)\subset SO(3)$ is
cyclic then so is $\operatorname{Im}(\rho)\subset SU(2)$, whereas if
$\operatorname{Im}(\operatorname{ad}\rho)\subset SO(3)$ is dihedral then
$\operatorname{Im}(\rho)\subset SU(2)$ is binary dihedral.

For the last assertion, if $|H_{1}(\Sigma(L,J))|=|H_{1}(L)|$ then Proposition
4.5 says that $\operatorname{Im}(\rho)$ cannot be non-abelian and binary
dihedral, so it must be cyclic. ∎

The instanton knot homology $\mathit{KHI}(L,J)$ is half of an instanton
homology group defined as the Morse homology of a Chern–Simons functional on
an associated space of connections. As discussed in [KM10b, §7], the space of
critical points of this functional can be identified with a double cover of
the representation variety

$R(J)=\\{\rho:\pi_{1}(L\setminus N(J))\to SU(2)\mid\rho(\mu_{J})=i\\}.$

If every element of $R(J)$ has cyclic image, then this is a trivial double
cover. If in addition each $\rho\in R(J)$ is nondegenerate (corresponds to
nondegenerate critical points of the Chern–Simons functional in a suitable
sense), then it follows that

(4.1) $\dim_{\mathbb{C}}\mathit{KHI}(L,J)\leq|R(J)|.$

We would thus like to know when each of the cyclic representations $\rho\in
R(J)$ is nondegenerate. This was addressed in the proof of [SZ17, Theorem
4.8]:

######  Lemma 4.8.

Let

$\rho:\pi_{1}(L\setminus N(J))\to SU(2)$

be a representation with cyclic image, satisfying $\rho(\mu_{J})=i$. If

$\dim_{\mathbb{R}}H^{1}(L\setminus N(J);\operatorname{ad}(\rho))=1,$

then $\rho$ is nondegenerate as a generator of $\mathit{KHI}(L,J)$.

###### Proof.

The proof of [SZ17, Theorem 4.8] establishes this for knots in $S^{3}$, in the
further situation where we have introduced some holonomy perturbation $\Phi$
and replaced the condition $\rho(\mu_{J})=i$ with a more general condition
described by some curve $C^{\prime}$ in the pillowcase orbifold (i.e., the
$SU(2)$ character variety of $T^{2}$). The current setting is simpler, though:
we take $\Phi=0$ and the original condition $\rho(\mu_{J})=i$, and then the
proof works verbatim. ∎

######  Proposition 4.9.

Suppose $H_{1}(L)$ has odd order and $J\subset L$ is primitive. Let

$\rho:\pi_{1}(L\setminus N(J))\to SU(2)$

be a representation with cyclic image, satisfying $\rho(\mu_{J})=i$. If the
$2|H_{1}(L)|$-fold cyclic cover of $L\setminus N(J)$ has first Betti number 1,
then $\rho$ is nondegenerate as a generator of $\mathit{KHI}(L,J)$.

###### Proof.

Let $C\subset Z$ be as above, so that $[\mu_{C}]$ generates $H_{1}(L\setminus
N(J))\cong\mathbb{Z}$ and

$[\mu_{J}]=|H_{1}(L)|\cdot[\mu_{C}].$

Since $\rho$ has cyclic image, it follows that $\operatorname{ad}\rho$ has
cyclic image as well and therefore factors through $H_{1}(L\setminus N(J))$.
Moreover, $\operatorname{Im}(\operatorname{ad}\rho)$ has order $n=2r$ for some
divisor $r$ of $|H_{1}(L)|$, because it sends
$|H_{1}(L)|\cdot[\mu_{C}]=[\mu_{J}]$ to the order-2 element
$\operatorname{ad}(i)$.

For each divisor $d$ of $n$, let $Y_{d}$ be the $d$-fold cyclic cover of
$L\setminus N(J)$ corresponding to the index-$d$ subgroup

$\ker\left(\pi_{1}(L\setminus
N(J))\xrightarrow{\operatorname{ad}\rho}SO(3)\xrightarrow{x\mapsto
x^{n/d}}SO(3)\right).$

Boyer and Nicas [BN90, Theorem 1.1 and Remark 1.2] proved that

$\dim H^{1}(L\setminus N(J);\operatorname{ad}\rho)=\dim H^{1}(L\setminus
N(J))+\frac{2}{\varphi(n)}\sum_{d\mid
n}\mu\left(\frac{n}{d}\right)b_{1}(Y_{d}),$

where $\mu$ denotes the Möbius function. If we can show that $b_{1}(Y_{d})=1$
for all $d\mid n$, then the sum on the right will be

$\sum_{d\mid n}\mu\left(\frac{n}{d}\right)=0$

since $n\geq 2$, and so we will have $\dim H^{1}(L\setminus
N(J);\operatorname{ad}\rho)=1$ as desired.

Note for each $d\mid n$ that $Y_{n}$ is a finite cover of $Y_{d}$, which in
turn is a finite cover of $L\setminus N(J)$. It follows by a transfer argument
that their first Betti numbers satisfy

$b_{1}(Y_{n})\geq b_{1}(Y_{d})\geq b_{1}(L\setminus N(J))=1,$

so if $b_{1}(Y_{n})=1$ then we will have $b_{1}(Y_{d})=1$ for all $d\mid n$
after all. Similarly, since $n$ divides $2|H_{1}(L)|$, we conclude that if the
$2|H_{1}(L)|$-fold cyclic cover of $L\setminus N(J)$ has $b_{1}=1$ then we
also have $b_{1}(Y_{n})=1$, and hence $\rho$ is nondegenerate. ∎

We now put together the various results of this section to prove Theorem 4.1.

###### Proof of Theorem 4.1.

Proposition 4.7 says that every representation

$\rho:\pi_{1}(L\setminus N(J))\to SU(2)$

with $\rho(\mu_{J})=i$ has cyclic image. By Proposition 4.5 there are
$|H_{1}(L)|$ such $\rho$. Thus, if they are all nondegenerate, then we will
have

$\dim_{\mathbb{C}}\mathit{KHI}(L,J)\leq|R(J)|=|H_{1}(L)|,$

as in (4.1). In the opposite direction, we know that

$|H_{1}(L)|\leq\dim_{\mathbb{C}}I^{\\#}(L)\leq\dim_{\mathbb{C}}\mathit{KHI}(L,J)$

by [LY20, Theorem 1.2], so equality will follow. It remains to establish the
nondegeneracy of these cyclic representations; by Proposition 4.9 it will
suffice to show that the $2|H_{1}(L)|$-fold cyclic cover of $L\setminus N(J)$
has first Betti number $1$.

Let $Y$ be the $n$-fold cyclic cover of $L\setminus N(J)\cong Z\setminus
N(C)$, where $n=2|H_{1}(L)|$. We can Dehn fill $Y$ along a lift $\tilde{\mu}$
of $\mu_{C}^{n}$ to get the $n$-fold branched cyclic cover of $C$, and this
Dehn filling decreases the first Betti number by $1$ since $[\tilde{\mu}]$ is
a non-torsion element of $H_{1}(Y)$. (Indeed, we can lift a Seifert surface
for $C$ to $Y$ to get a surface whose intersection pairing with $\tilde{\mu}$
is $1$.) It thus follows that

$b_{1}(Y)=1\text{ if and only if }b_{1}(\Sigma_{n}(Z,C))=0.$

But a theorem of Fox [BZ03, Theorem 8.21] says that

$|H_{1}(\Sigma_{n}(Z,C))|=\left|\prod_{k=1}^{n-1}\Delta_{C}(e^{2\pi
ik/n})\right|,$

where $\Sigma_{n}(Z,C)$ is the $n$-fold cyclic branched cover of $C\subset Z$.
It follows that $b_{1}(\Sigma_{n}(Z,C))=0$ if and only if no nontrivial $n$th
root of unity is a zero of the Alexander polynomial $\Delta_{C}(t)$.

We now wish to show that the cyclotomic polynomials $\Phi_{d}(t)$ do not
divide $\Delta_{C}(t)$ for all divisors $d\geq 2$ of $n=2|H_{1}(L)|$. If
$|H_{1}(L)|$ is an odd prime power, say $p^{e}$, then either

  * •

$d=2$, and $\Phi_{2}(t)=1+t$ implies that $\Phi_{d}(1)=2$;

  * •

$d=p^{k}$ with $1\leq k\leq e$, and then

$\Phi_{p^{k}}(t)=1+t^{p^{k}}+t^{2\cdot p^{k}}+\dots+t^{(p-1)\cdot p^{k}}$

implies that $\Phi_{d}(1)=p$; or

  * •

$d=2\cdot p^{k}$ with $1\leq k\leq e$, and

$\Phi_{2\cdot p^{k}}(t)=\Phi_{p^{k}}(-t)=1-t^{p^{k}}+t^{2\cdot
p^{k}}-\dots+t^{(p-1)\cdot p^{k}}$

implies that $\Phi_{d}(-1)=p$.

Thus, if some $\Phi_{d}(t)$ divides $\Delta_{C}(t)$ then at least one of
$|\Delta_{C}(1)|$ and $|\Delta_{C}(-1)|$ will be strictly greater than $1$.
But we know that $\Delta_{C}(1)=1$ for all knots in homology spheres, and we
know from Lemma 4.4 that

$|\Delta_{C}(-1)|=|H_{1}(\Sigma(Z,C))|=1,$

so we cannot have $\Phi_{d}(t)\mid\Delta_{C}(t)$ for any divisor $d\geq 2$ of
$2p^{e}$. We conclude that

$b_{1}(\Sigma_{n}(Z,C))=0,$

proving the required nondegeneracy. ∎

##  5\. Floer simple knots and Theorem 1.2

Recall that a knot in a lens space $L$ is _simple_ if it is isotopic to a
union of two arcs, each contained in a meridional disk for one of the two
solid tori in a genus-1 Heegaard splitting of $L$. For instance, the core of
each solid torus in such a splitting is a simple knot. Moreover, there is a
unique (oriented) simple knot in every homology class in $H_{1}(L)$. See
[Ras07, §2.1] and Section 6 for more on simple knots.

Our main result in this section is Theorem 5.1 below. This theorem is
substantially more general than is needed for the proof of Theorem 1.2, but we
will use this generality in the next section to extend the latter theorem to
additional slopes in the interval $[3,5)$.

######  Theorem 5.1.

Suppose $J$ is a primitive (oriented) knot in a lens space $L$ such that

$\dim_{\mathbb{C}}\mathit{KHI}(L,J)=|H_{1}(L)|.$

Then $J$ is rationally fibered of the same genus as the simple knot $S$ in its
homology class. Moreover, the symmetrized Alexander polynomials of $J$ and $S$
agree,

$\Delta_{J}(t)=\Delta_{S}(t).$

Finally, if

$g(J)\leq\frac{|H_{1}(L)|+1}{4}$

then $J$ is isotopic to $S$.

Before proving this theorem, we use it to prove our main result, Theorem 1.2.

###### Proof of Theorem 1.2.

Suppose, for a contradiction, that $S^{3}_{3}(K)$ is $SU(2)$-abelian for some
nontrivial knot $K$. Proposition 3.2 implies that this surgered manifold is
homeomorphic to a branched double cover of the lens space

$L=L(3,2)$

along a knot $J$ which is homologous to 5 times a core in a genus-1 Heegaard
splitting of $L$. In particular, the homology class of $J$ (with any
orientation) generates

$H_{1}(L)\cong\mathbb{Z}/3\mathbb{Z}$

since $\gcd(3,5)=1,$ meaning that $J\subset L$ is primitive. Its branched
double cover is therefore unique, by Lemma 4.2, so we may write

$S^{3}_{3}(K)\cong\Sigma(L,J).$

Since

$|H_{1}(\Sigma(L,J))|=|H_{1}(L)|=3,$

Theorem 4.1 implies that

$\dim_{\mathbb{C}}\mathit{KHI}(L,J)=|H_{1}(L)|=3.$

Then Theorem 5.1 says that $J$ is rationally fibered and has the same genus as
the simple knot in its homology class (for any orientation on $J$). But each
primitive simple knot in $L$ is homologous to, and hence isotopic to, a core
of a solid torus in a genus-1 Heegaard splitting of $L$, and therefore has
genus 0. It follows that $J$ has genus 0 as well. The fibered exterior of $J$
is then a solid torus, which implies that its branched double cover
$\Sigma(L,J)$ is a lens space. But Proposition 3.2 says that $S^{3}_{3}(K)$ is
not Seifert fibered, a contradiction.∎

It remains to prove Theorem 5.1. Before doing so, we review some facts about
instanton knot homology and Heegaard knot Floer homology. We will assume for
the discussion below that $J$ is a primitive knot in an irreducible rational
homology sphere $L$ with $|H_{1}(L)|=p>1$. In particular, the complement of
$J$ is irreducible.

As in §4, we can identify $L\setminus N(J)$ with $Z\setminus N(C)$ for a knot
$C$ in some homology sphere $Z$. Let

$(F,\partial F)\subset(L\setminus N(J),\partial N(J))$

be a genus-minimizing Seifert surface for $C$, so that $F$ generates
$H_{2}(L\setminus N(J),\partial N(J))$. Then $\partial F$ is a Seifert
longitude for $C$. Since

$[\mu_{J}]=|H_{1}(L)|\cdot[\mu_{C}]=p[\mu_{C}]\in H_{1}(L\setminus N(J))$

(see §4), $\partial F$ represents a primitive class on $\partial N(J)$, and
has algebraic intersection number $\pm p$ with the meridians of $J$.

Let $\gamma_{\mu}$ be the disjoint union of two oppositely-oriented meridians
of $J$ on $\partial N(J)$. Then $(L\setminus N(J),\gamma_{\mu})$ is a balanced
sutured manifold, and its sutured instanton homology defines the instanton
knot homology of $J\subset L$,

$\mathit{KHI}(L,J):=\mathit{SHI}(L\setminus N(J),\gamma_{\mu}),$

as in [KM10b, Definition 7.13]. Let us arrange that $\partial F$ intersects
$\gamma_{\mu}$ in $2p$ points. Then the construction in [LY21b, §2.3] defines
an integer-valued Alexander grading on $\mathit{KHI}(L,J)$,

$\mathit{KHI}(L,J)=\bigoplus_{i\in\mathbb{Z}}\mathit{KHI}(L,J,i).$

We will review below how this Alexander grading detects genus and fiberedness,
and yields a graded Euler characteristic which agrees with that of Heegaard
knot Floer homology.

Let

$n:=\frac{1}{2}(p-\chi(F))=g(F)+\frac{p-1}{2}.$

Then [LY21b, Theorem 2.30] says that

$\mathit{KHI}(L,J,i)=0\text{ for }|i|>n,$

and that if $(M_{\pm},\gamma_{\pm})$ are the sutured manifolds obtained by
decomposing $(L\setminus N(J),\gamma_{\mu})$ along $\pm F$, then

$\mathit{KHI}(L,J,\pm n)\cong\mathit{SHI}(M_{\pm},\gamma_{\pm}).$

Since $L\setminus N(J)$ is irreducible with toroidal boundary, and $F$ has
minimal genus and minimal intersection with $\gamma_{\mu}$, the sutured
manifolds $(M_{\pm},\gamma_{\pm})$ are both taut. Their sutured instanton
homologies are therefore nontrivial [KM10b, Theorem 7.12]; that is,

$\mathit{KHI}(L,J,\pm n)\neq 0.$

In this way, instanton knot homology detects the genus of $J$. Note, moreover,
that $M_{+}=M_{-}$ and $\gamma_{+}=-\gamma_{-}$. This implies by [LY21b,
Theorem 2.30] that the sutured instanton homologies of
$(M_{\pm},\gamma_{\pm})$ are isomorphic,

$\mathit{KHI}(L,J,n)\cong\mathit{KHI}(L,J,-n).$

The fact that

$H_{2}(L\setminus N(J))\cong H_{2}(Z\setminus N(C))=0$

implies that $H_{2}(M_{\pm})=0$ as well. We can thus apply [GL19, Theorem 1.2]
to conclude that

$\dim_{\mathbb{C}}\mathit{KHI}(L,J,\pm n)=1$

iff $(M_{\pm},\gamma_{\pm})$ are product sutured manifolds, or, equivalently,
iff $J$ is a rationally fibered knot with fiber surface $F$. In this way,
instanton knot homology detects whether $J$ is rationally fibered. Heegaard
knot Floer homology detects genus and fiberedness in the same way.

There is a $\mathbb{Z}/2\mathbb{Z}$-grading on instanton knot homology which
descends to a $\mathbb{Z}/2\mathbb{Z}$-grading on each Alexander-graded
summand $\mathit{KHI}(L,J,i)$, so we can define the graded Euler
characteristic of $\mathit{KHI}(L,J)$ by

$\chi_{\mathrm{gr}}(\mathit{KHI}(L,J))=\sum_{i\in\mathbb{Z}}\chi(\mathit{KHI}(L,J,i))\cdot
t^{i}.$

Recent work of Li and Ye [LY21b, Theorem 1.2] relates this with the graded
Euler characteristic of Heegaard knot Floer homology. Namely, they prove that

$\chi_{\mathrm{gr}}(\mathit{KHI}(L,J))=\pm
t^{s}\cdot\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,J)),$

for some $s\in\mathbb{Z}$, where the latter is defined with respect to the
corresponding Alexander and $\mathbb{Z}/2\mathbb{Z}$-gradings in the Heegaard
Floer setting.

Recall that

(5.1)
$\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,J))_{t=1}=\chi(\widehat{\mathit{HF}}(L))=|H_{1}(L)|=p.$

Recall further that the Alexander grading on the Heegaard knot Floer homology
of $J$ can be viewed as a grading by relative $\operatorname{Spin}^{c}$
structures in $\operatorname{Spin}^{c}(L,J)$. In the spectral sequence

$\widehat{\mathit{HFK}}(L,J)\implies\widehat{\mathit{HF}}(L),$

this grading projects to the $\operatorname{Spin}^{c}$ grading on the latter
via the natural map

$\operatorname{Spin}^{c}(L,J)\to\operatorname{Spin}^{c}(L).$

These sets of $\operatorname{Spin}^{c}$ structures are affine spaces over the
corresponding first homology groups, and the projection map above can be
identified with the quotient map

$H_{1}(L\setminus
N(J))\,\cong\,\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}\,\cong\,H_{1}(L).$

More concretely, there is an identification
$\operatorname{Spin}^{c}(L)\cong\mathbb{Z}/p\mathbb{Z}$ such that the spectral
sequence above restricts, for each $[i]\in\mathbb{Z}/p\mathbb{Z}$, to a
spectral sequence

(5.2) $\bigoplus_{m\equiv
i\\!\\!\\!\\!\\!\pmod{p}}\widehat{\mathit{HFK}}(L,J,m)\implies\widehat{\mathit{HF}}(L,[i]).$

In particular, for each $[i]\in\mathbb{Z}/p\mathbb{Z}$, we have

(5.3) $\sum_{m\equiv
i\\!\\!\\!\\!\\!\pmod{p}}\chi(\widehat{\mathit{HFK}}(L,J,m))=\chi(\widehat{\mathit{HF}}(L,[i]))=1.$

The following additional constraints on the graded Euler characteristic of
Heegaard knot Floer homology will be important in our proof of Theorem 5.1.

The first lemma below is proved in [Ras07, Corollary 5.3].

######  Lemma 5.2.

Let $J_{1}$ and $J_{2}$ be two homologous primitive knots in $L$. Then the
difference

$\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,J_{1}))-\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,J_{2}))$

is divisible by $(t^{p}-1)^{2}$.

The next lemma follows immediately from (5.3).

######  Lemma 5.3.

Let $J$ be a primitive knot in $L$. Then, for each
$[i]\in\mathbb{Z}/p\mathbb{Z}$, we have that

$\chi(\widehat{\mathit{HFK}}(L,J,m))\neq 0$

for some integer $m\equiv i\pmod{p}$.

The final lemma below is proved in [Ras07, Proof of Theorem 1]. We only need
it for the last claim in Theorem 5.1 in the case where $g(J)>0$. We do not
need it when $g(J)=0$ (in particular, we do not need it for the proof of
Theorem 1.2) because we know automatically in this case that $J$ is a core and
thus simple.

######  Lemma 5.4.

Let $J$ be a primitive knot in $L$. Recall that we have assumed that
$|H_{1}(L)|=p$. Suppose in addition that
$\dim_{\mathbb{Z}/2\mathbb{Z}}\widehat{\mathit{HF}}(L)=p$ and

$g(J)<\frac{p+1}{2}.$

Then $\dim_{\mathbb{Z}/2\mathbb{Z}}\widehat{\mathit{HFK}}(L,J)=p$. That is,
$J$ is Heegaard Floer simple.

We may now prove Theorem 5.1.

###### Proof of Theorem 5.1.

Let $J\subset L$ be as in the hypothesis of the theorem, and let $S$ denote
the simple knot in the same homology class as $J$. Simple knots admit Heegaard
diagrams for which the knot Floer complex has trivial differential; in
particular, they are Heegaard Floer simple.333Simple knots are also instanton
Floer simple, by [LY20]; see also [BLY20]. Thus,

$\dim_{\mathbb{Z}/2\mathbb{Z}}\widehat{\mathit{HFK}}(L,S)=p.$

It then follows from Lemma 5.3 and (5.1) that $S$ satisfies the property

($\ast$) $\begin{array}[]{l}\text{for each
}[i]\in\mathbb{Z}/p\mathbb{Z},\text{ there is a \emph{unique} integer }m\equiv
i\\!\\!\\!\pmod{p}\text{ such that:}\\\
\chi(\widehat{\mathit{HFK}}(L,S,m))=1,\textrm{ and
}\chi(\widehat{\mathit{HFK}}(L,S,k))=0\textrm{ for all other }k\equiv
i\\!\\!\\!\pmod{p}.\par\end{array}$

It also follows from Lemma 5.3 that

$\dim_{\mathbb{Z}/2\mathbb{Z}}\widehat{\mathit{HFK}}(L,S,m)=0\textrm{ or }1$

for each $m\in\mathbb{Z}$. Similarly, Lemma 5.3 applied to $J$, combined with
the facts that

$\dim_{\mathbb{C}}\mathit{KHI}(L,J)=p$

and that

(5.4) $\chi_{\mathrm{gr}}(\mathit{KHI}(L,J))=\pm
t^{s}\cdot\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,J))$

for some $s\in\mathbb{Z}$, implies that $J$ also satisfies ($\ast$ ‣ 5), and,
additionally, that

$\dim_{\mathbb{C}}\mathit{KHI}(L,J,m)=0\textrm{ or }1$

for each $m\in\mathbb{Z}$. In particular,

$\dim_{\mathbb{C}}\mathit{KHI}(L,J,\pm n)=1$

for

$n=g(J)+\frac{p-1}{2},$

which implies that $J$ is rationally fibered. Since

$\mathit{KHI}(L,J,i)=0$

for $|i|>n$, and likewise for the Heegaard knot Floer homology of $J$, and
since $\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,J))$ is symmetric under the
substitution $t\mapsto t^{-1}$, this then implies that $s=0$ in (5.4).

We next prove that $J$ and $S$ have the same genus. For this, we claim, as in
[Ras07, Proof of Theorem 2], that

(5.5)
$\overline{\Delta}_{J}(t):=\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,J))=\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,S))=:\overline{\Delta}_{S}(t).$

Indeed, the fact that $J$ and $S$ both satisfy property ($\ast$ ‣ 5) means
that we can write

$\displaystyle\overline{\Delta}_{S}(t)$
$\displaystyle=\sum_{[i]\in\mathbb{Z}/p\mathbb{Z}}t^{n_{[i]}}$
$\displaystyle\overline{\Delta}_{J}(t)$
$\displaystyle=\sum_{[i]\in\mathbb{Z}/p\mathbb{Z}}t^{n_{[i]}+p\cdot m_{[i]}}$

for some integers $n_{[i]}\equiv i\pmod{p}$ and $m_{[i]}$. Since $J$ and $S$
are in the same homology class, Lemma 5.2 says that $(t^{p}-1)^{2}$ divides
the difference

$\overline{\Delta}_{J}(t)-\overline{\Delta}_{S}(t)=\sum_{[i]\in\mathbb{Z}/p\mathbb{Z}}t^{n_{[i]}}(t^{p\cdot
m_{[i]}}-1).$

However, after dividing this difference by $t^{p}-1$, we are left with

$\sum_{\begin{subarray}{c}[i]\in\mathbb{Z}/p\mathbb{Z}\\\ m_{[i]}\neq
0\end{subarray}}t^{n_{[i]}}(1+t^{m_{[i]}}+\dots+t^{(p-1)\cdot m_{[i]}}).$

For this sum to be divisible by another factor of $t^{p}-1$, it must have $1$
as a root. But that is only possible if the sum is trivial—that is, if
$m_{[i]}=0$ for every $[i]\in\mathbb{Z}/p\mathbb{Z}$. This proves that
$\overline{\Delta}_{J}(t)=\overline{\Delta}_{S}(t)$ as desired.

Combining (5.4) (recall that $s=0$) with (5.5), we have therefore shown that

$\chi_{\mathrm{gr}}(\mathit{KHI}(L,J))=\pm\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,S)).$

Since the instanton Floer homology of $J$ has dimension at most 1 in each
Alexander grading, and likewise for the Heegaard knot Floer homology of $S$,
it follows that for each integer $i$,

$\mathit{KHI}(L,J,i)=0\,\textrm{ iff }\,\widehat{\mathit{HFK}}(L,S,i)=0.$

This implies that $J$ and $S$ have the same genus, since the Alexander
gradings in these two theories detect genus in the same way.

For the claim about Alexander polynomials, Rasmussen proves in [Ras07, §3.7]
that, for any primitive knot $J$ in a rational homology sphere $L$ of order
$p$, we have

$\overline{\Delta}_{J}(t)=\Delta_{J}(t)\cdot\frac{t^{p/2}-t^{-p/2}}{t^{1/2}-t^{-1/2}},$

where $\Delta_{J}(t)$ is the symmetrized Alexander polynomial of the
complement $L\setminus N(J)$. We have shown above that
$\overline{\Delta}_{J}(t)=\overline{\Delta}_{S}(t)$, from which it follows
that $\Delta_{J}(t)=\Delta_{S}(t)$ as well.

For the last claim of the theorem, suppose

$g(J)\leq\frac{|H_{1}(L)|+1}{4}=\frac{p+1}{4}.$

Then $J$ is Heegaard Floer simple, by Lemma 5.4. Furthermore, Baker proved in
[Bak06, Theorem 1.1] that any knot $J$ satisfying the genus bound above is a
1-bridge knot in $L$. Finally, Hedden proved in [Hed11, Proposition 3.3] that
any 1-bridge knot in a lens space which is also Heegaard Floer simple is
simple. Thus, $J\cong S$. ∎

##  6\. Other small surgeries and $SU(2)$

As mentioned in the introduction, we suspect that the following is true:

######  Conjecture 6.1.

If $K\subset S^{3}$ is a nontrivial knot, then $S^{3}_{r}(K)$ is not
$SU(2)$-abelian for any rational number $r\in[0,5)$.

Kronheimer–Mrowka proved this conjecture for all $r\in[0,2]$ in [KM04a].
Baldwin–Sivek proved it in [BS19, Theorem 1.8] for the dense subset of
rationals in $(2,3)$ given by

$r=p/q\in(2,3),$

where $p,q$ are coprime and $p$ is a prime power. Indeed, they proved that if
$S^{3}_{r}(K)$ is $SU(2)$-abelian in this case, then $K$ has genus 1, and is
fibered and strongly quasipositive. The only such knot, the right-handed
trefoil, does not admit $SU(2)$-abelian surgeries in this range.

As alluded to in §1.2, proving Conjecture 6.1 is much more difficult for the
corresponding dense subset of rationals in the interval $[3,5)$, given by

$r=p/q\in[3,5),$

where $p,q$ are coprime and $p$ is a prime power. If $S^{3}_{r}(K)$ is
$SU(2)$-abelian in this case, then it is an instanton L-space, by Lemma 3.1,
from which it follows that $K$ has genus at most 2, and is fibered and
strongly quasipositive [BS19]. One can rule out the trefoil, so $K$ has genus
2, in which case it also has the same Alexander polynomial as $T_{2,5}$, by
Lemma 2.2. The difficulty is that there are infinitely many such knots; see,
for example, [Mis17].

In this section, we illustrate how our techniques can be used to prove
Conjecture 6.1 for infinitely many slopes $r\in[3,5)$, culminating in the
proof of Theorem 1.4.

We start with a comparatively easy case, in which the numerator of $r$ is a
power of $2$, which is handled with a trick involving a result of Klassen. In
this case, we can actually deal with all slopes less than 7.

######  Theorem 6.2.

Let $p,q$ be coprime positive integers with $p/q<7$, where $p$ is a power of
$2$. Then $S^{3}_{p/q}(K)$ is not $SU(2)$-abelian for any nontrivial knot
$K\subset S^{3}$.

###### Proof.

Suppose $Y=S^{3}_{p/q}(K)$ is $SU(2)$-abelian. Then it is an instanton L-space
by Lemma 3.1. It then follows from [BS19, Theorem 1.15] that

$2g(K)-1\leq p/q<7,$

which implies that $g(K)\leq 3$. If $g(K)=1$ then $K$ is the right-handed
trefoil, and $Y$ is Seifert fibered with base orbifold $S^{2}(2,3,\Delta)$
where

$\Delta=|6q-p|.$

We cannot have $\Delta\leq 1$, because otherwise either
$\frac{p}{q}=\frac{6}{1}$ or

$\frac{p}{q}=6\pm\frac{1}{n}=\frac{6n\pm 1}{n}$

for some integer $n\geq 1$, and in these cases $p$ is not actually a power of
$2$. But then $\Delta>1$ and so $Y$ is not $SU(2)$-abelian [SZ19], a
contradiction. Therefore, $g(K)$ is either $2$ or $3$, which then implies that
$p/q\geq 3$. Since $p$ is a power of 2, it must be a multiple of 4.

Theorem 2.1 now tells us that $K$ has Alexander polynomial

$\Delta_{K}(t)=t^{2}-t+1-t^{-1}+t^{-2}$

if $g(K)=2$, whereas if $g(K)=3$, then the same argument as in the proof of
Lemma 2.2 (appealing to [LY21c, Theorem 1.9] as well as [KM10b, Proposition
7.16] and [BS18a, Theorem 1.7]) says that

$\displaystyle\Delta_{K}(t)$
$\displaystyle=t^{3}-t^{2}+t-1+t^{-1}-t^{-2}+t^{-3}$
$\displaystyle\phantom{=}\text{ or }t^{3}-t^{2}+1-t^{-2}+t^{-3}.$

In each of these cases we have

$\det(K)=|\Delta_{K}(-1)|=5\text{ or }7\text{ or }3,$

respectively. From here we argue exactly as in [BS19, Lemma 9.5]: Klassen
[Kla91, Theorem 10] proved that there are

$\frac{\det(K)-1}{2}=2\text{ or }3\text{ or }1,$

respectively, conjugacy classes of non-abelian representations

$\rho:\pi_{1}(S^{3}\setminus K)\to SU(2)$

with image in the binary dihedral group
$D_{\infty}=\\{e^{i\theta}\\}\cup\\{e^{i\theta}j\\}$. Let $\mu$ be a meridian
of $K$ and $\lambda$ the Seifert longitude. Following the proof of [Kla91,
Theorem 10], since $\rho$ has non-abelian image we must have that
$\rho(\mu)=j$ up to conjugacy, and hence $\rho(\mu^{4})=1$. Moreover, we know
that $\lambda$ lies in the second commutator subgroup of
$\pi_{1}(S^{3}\setminus K)$; thus $\rho(\lambda)$ lies in the second
commutator subgroup of $D_{\infty}$, and the latter is trivial, so
$\rho(\lambda)=1$. Since $p$ is a multiple of 4, it follows that
$\rho(\mu^{p}\lambda^{q})=1$, so $\rho$ descends to a representation

$\pi_{1}(S^{3}_{p/q}(K))\to SU(2)$

with non-abelian image, but we assumed $Y=S^{3}_{p/q}(K)$ is $SU(2)$-abelian,
a contradiction. ∎

We next turn to the much more difficult case in which the numerator of $r$ is
an odd prime power; this will require the full strength of the techniques in
this paper. To start, we review some facts about simple knots in lens spaces,
and then study these knots in greater depth.

If $A\subset S^{3}$ is an unknot, then the lens space $L(a,b)$ is given in our
notation by

$L(a,b)=S^{3}_{a/b}(A).$

The standard genus-1 Heegaard splitting of this lens space is given by the
union of the solid torus exterior $S^{3}\setminus N(A)$ with the surgery solid
torus. The core of each solid torus generates the first homology of the lens
space. Let $S(a,b,k)$ denote the unique (oriented) simple knot which is
homologous to $k$ times the core of the surgery solid torus, for some
orientation of this core.444This simple knot is denoted by $K(a,b,k)$ in
[Ras07]. Note that

$S(a,b,k)\cong S(a,b,k^{\prime})$

when $k\equiv k^{\prime}\pmod{a}.$ Moreover, $S(a,a-b,k)\subset-L(a,b)$ is the
orientation-reverse of the mirror of $S(a,b,k)$. Finally, $S(a,b,\pm 1)$ is
isotopic to a core and therefore has genus $0$.

Several of our results thus far, like Proposition 3.2, pertain to a knot
$J\subset L(p,2q)$ which is homologous to 5 times a core of the _other_ solid
torus $S^{3}\setminus N(A)$ in the genus-1 splitting of this lens space. It is
easy to check (for some orientation on $J$) that

$[J]=[S(p,2q,10q)]\in H_{1}(L(p,2q)).$

Note that when $p$ is odd, the knot $S(p,2q,10q)$ is primitive if and only if
$\gcd(p,5)=1$.

The main theorem of this section is as follows:

######  Theorem 6.3.

Let $p$ and $q$ be coprime positive integers with $p/q\in[3,5)$, where $p$ is
an odd prime power and $\gcd(p,5)=1$. If either $p/q\in(4,5)$ or

$g(S(p,2q,10q))\leq\frac{p+1}{4},$

then $S^{3}_{p/q}(K)$ is not $SU(2)$-abelian for any nontrivial knot $K\subset
S^{3}$.

Before proving this theorem, we establish the following proposition. In
addition to helping us prove Theorem 6.3, this proposition also provides the
genera of the relevant simple knots $S(p,2q,10q)$, making it easy to apply
Theorem 6.3 in practice. Indeed, we will apply this genus formula at the end
to prove Theorem 1.4.

######  Proposition 6.4.

Let $p$ and $q$ be coprime positive integers with $p/q\in[3,6]$, where $p$ is
odd and $\gcd(p,5)=1$. Then the branched double cover of

$S=S(p,2q,10q)\subset L(p,2q)=L$

is Seifert fibered with

$|H_{1}(\Sigma(L,S))|=\begin{cases}p,&p/q\in[3,4)\\\
5p,&p/q\in(4,6).\end{cases}$

Moreover, this primitive simple knot has genus

$g(S)=\big{|}\,|4p-10\min\\{2q,p-2q\\}|-2\,\big{|}.$

###### Proof.

Let $a,b$ be coprime positive integers with $b/a\in[1/3,2/3]$ and
$\gcd(a,5)=1$; eventually we will take $(a,b)=(p,2q)$. We will attempt to
understand the branched double cover and genus of the primitive simple knot

$S(a,b,5b)\subset L(a,b).$

Without loss of generality, we can assume that

$b/a\in[1/3,1/2]$

Indeed, if $b/a\in[1/2,2/3]$ then $(a-b)/a\in[1/3,1/2],$ and we can just
consider $S(a,a-b,5b)$ instead, as it is the orientation-reverse of the mirror
of $S(a,b,5b)$.

In [Ye20, §8], Ye determines when a simple knot in $L(a,b)$ is isotopic to a
torus knot on the Heegaard torus in the genus-1 splitting of the lens space.
More precisely, let us think of $L(a,b)$ as obtained via $(a/b)$-surgery on
the unknot $A\subset S^{3}$, with genus-1 Heegaard splitting given by the
union of

$H_{\alpha}=S^{3}\setminus N(A)$

with the surgery solid torus $H_{\beta}$, as above. In particular, the
meridian of $H_{\beta}$ is glued along the curve $a\mu+b\lambda$, with respect
to a meridian $\mu$ and Seifert longitude $\lambda$ of $A$ on $\partial
H_{\alpha}$, with $\mu\cdot\lambda=-1$. Let

$\Sigma=\partial H_{\alpha}=-\partial H_{\beta}$

be the Heegaard torus in this splitting. Let $T$ be the $(5,2)$-torus knot on
$\Sigma$—that is, the simple closed curve on $\Sigma$ which is homologous in
$\Sigma$ to $5\mu+2\lambda$. Since $b/a\in[1/2,1/3]$, [Ye20, Theorem 8.2] says
that

$S(a,b,5b)\cong T.$

Indeed, following the discussion above, it is easy to see that

$[S(a,b,5b)]=[T]\in H_{1}(L(a,b)),$

so it suffices to show that $T$ is simple. The basic idea behind Ye’s result,
inspired by [GLV18, §3.1], is that if we fix a Heegaard diagram
$(\Sigma,\alpha,\beta)$ for the Heegaard splitting

$L(a,b)=H_{\alpha}\cup_{\Sigma}-H_{\beta}$

where $\alpha$ is homologous to $\lambda$ and $\beta$ is homologous to
$a\mu+b\lambda$, then the slope $5/2$ is sufficiently close to $a/b$ in this
case that we can represent $T$ as the union of an arc in $\beta$ with an arc
in $\alpha$. Thus, $T$ is simple.

Our goals are then to show that the branched double cover of $T$ is Seifert
fibered, compute its genus, and compute the order of its branched double
cover.

For the former, note that one can Seifert fiber the solid tori $H_{\alpha}$
and $H_{\beta}$ so that $T$ is a regular fiber in both. These glue to give a
Seifert fibration of $L(a,b)$ with two singular fibers in which $T$ is a
regular fiber. The complement of a neighborhood of $T$ is thus Seifert
fibered, so the double cover of this complement is as well (each circle fiber
lifts to one or two circles). The Seifert fibration on this double cover then
extends across the solid torus we glue in to form the branched double cover of
$T$. (This is the same reasoning that shows that torus knots in $S^{3}$ have
Seifert fibered branched double covers.)

We turn next to the genus calculation. The knot Floer homology of $T$, and
therefore the genus of this knot, is determined by its graded Euler
characteristic since simple knots are Floer simple. Let $L=L(a,b).$ Following
Rasmussen in [Ras07, §3.7], we have that

$\chi_{\mathrm{gr}}(\widehat{\mathit{HFK}}(L,T))=\Delta_{T}(t)\cdot\frac{t^{a/2}-t^{-a/2}}{t^{1/2}-t^{-1/2}},$

where $\Delta_{T}(t)$ is the symmetrized Alexander polynomial of $T$. Letting
$\deg(\Delta_{T}(t))$ denote the top degree of this Laurent polynomial, it
follows that the top non-zero Alexander grading of
$\widehat{\mathit{HFK}}(L,T)$ is given by

$\deg(\Delta_{T}(t))+\frac{a-1}{2}.$

On the other hand, from the discussion in the previous section, this top
Alexander grading is also given by

$g(T)+\frac{a-1}{2}.$

Therefore,

$g(T)=\deg(\Delta_{T}(t)).$

It thus suffices to compute this Alexander polynomial.

The Alexander polynomial $\Delta_{T}(t)$ can be computed via Fox calculus,
from a presentation of the fundamental group of $L\setminus N(T)$, as
described in [Ras07, §3.7]. One finds a 2-generator, 1-relator presentation of
this group using Seifert–Van Kampen, exactly as for torus knots in $S^{3}$.
Namely, the complement of $T$ is obtained by gluing the solid tori
$H_{\alpha}$ and $H_{\beta}$ together along the annulus on $\Sigma$ given by

$C=\Sigma\setminus N(T).$

Let $x$ and $y$ be the homotopy classes of the cores of $H_{\alpha}$ and
$H_{\beta}$, respectively. The core of $C$ is homotopic to $x^{5}$ in
$\pi_{1}(H_{\alpha})$ and $y^{d}$ in $\pi_{1}(H_{\beta})$ (for some
orientation of these cores), where

$d=|\Delta(5/2,a/b)|=|2a-5b|.$

Therefore, we have a presentation

$\pi_{1}(L\setminus N(T))=\langle x,y\mid x^{5}=y^{d}\rangle.$

This is the same as the fundamental group of the complement of the torus knot

$T_{5,d}\subset S^{3},$

so Fox calculus tells us that $\Delta_{T}(t)$ agrees with the Alexander
polynomial of this torus knot. In particular,

$g(T)=\deg(\Delta_{T}(t))=\deg(\Delta_{T_{5,d}}(t))=|2(d-1)|=\big{|}\,|4a-10b|-2\,\big{|}.$

Recall that if $b/a\in[1/2,2/3]$ then we replace $b$ with $a-b$ in this
formula. In other words, we have that

$g(S(a,b,5b))=\big{|}\,|4a-10\min\\{b,a-b\\}|-2\,\big{|}.$

If $p,q$ are as in the hypotheses of this proposition, then letting $a=p$ and
$b=2q$ yields the desired genus formula. (Note that $p$ and $2q$ are coprime,
since $p$ and $q$ are and $p$ is odd.)

Finally, we compute the order of the branched double cover of

$S=S(p,2q,10q)\subset L(p,2q)=L,$

where $p$ and $q$ are as in the hypothesis of the proposition. By Lemma 4.4,
we have that

(6.1) $|H_{1}(\Sigma(L,S))|=p\cdot|\Delta_{S}(-1)|.$

Recall that

$\Delta_{S}(t)=\Delta_{T_{5,d}}(t),$

for some $d$, as explained above. If $p/q\in[3,4)$, then $2q/p\in(1/2,2/3]$,
and

$d=|2p-5(p-2q)|=|10q-3p|$

is odd since $p$ is odd. In this case, an easy calculation shows that
$|\Delta_{T_{5,d}}(-1)|=1$, which implies that $|H_{1}(\Sigma(L,S))|=p$ by
(6.1). If $p/q\in(4,6)$ instead, then $2q/p\in(1/3,1/2)$, and

$d=|2p-10q|$

is even. In this case, $|\Delta_{T_{5,d}}(-1)|=5$, and therefore
$|H_{1}(\Sigma(L,S))|=5p$. ∎

We remark that the formula for $g(S(p,2q,10q))$ from Proposition 6.4 can be
simplified for various values of $p/q$, as follows:

(6.2) $g(S(p,2q,10q))=\begin{cases}20q-6p-2,&3\leq p/q<10/3\\\
6p-20q-2,&10/3<p/q<4\\\ 20q-4p-2,&4<p/q<5\\\ 4p-20q-2,&5<p/q\leq 6\end{cases}$

We have omitted the cases $p/q=10/3,4,5$ above because of the requirement that
$p$ be odd and not a multiple of 5.

###### Proof of Theorem 6.3.

Suppose $Y=S^{3}_{p/q}(K)$ is $SU(2)$-abelian for $p$ and $q$ as in the
theorem. Proposition 3.2 implies that

$Y\cong\Sigma(L,J)$

for some primitive knot $J$ in the lens space $L=L(p,2q),$ where $J$ (for some
orientation) is homologous to the simple knot

$S=S(p,2q,10q),$

as described above. Proposition 3.2 also implies that $Y$ is not Seifert
fibered. From here, Theorem 4.1 tells us that

$\dim_{\mathbb{C}}\mathit{KHI}(L,J)=|H_{1}(L)|,$

and so Theorem 5.1 implies that $\Delta_{J}(t)=\Delta_{S}(t)\textrm{ and
}g(J)=g(S).$

If $p/q\in(4,5)$, then

$|H_{1}(\Sigma(L,J))|=p\cdot|\Delta_{J}(-1)|=p\cdot|\Delta_{S}(-1)|=|H_{1}(\Sigma(L,S))|,$

by Lemma 4.4. But the latter order is $5p$, by Proposition 6.4. This is a
contradiction, since $Y=\Sigma(L,J)$ has order $p$ first homology, as
$(p/q)$-surgery on a knot in $S^{3}$.

Finally, if

$g(S)\leq\frac{p+1}{4},$

then the same is true of the genus of $J$, in which case $J$ is isotopic to
$S,$ by Theorem 5.1. But then $Y\cong\Sigma(L,J)\cong\Sigma(L,S)$ is Seifert
fibered, by Proposition 6.4, a contradiction. ∎

Finally, we combine Theorems 6.2 and 6.3 and Proposition 6.4 to prove Theorem
1.4.

###### Proof of Theorem 1.4.

Let $p$ and $q$ be coprime positive integers with $p/q\in[3,7)$, where $p$ is
a prime power, as in the hypothesis of the theorem, and let $K\subset S^{3}$
be a nontrivial knot. We know that

$Y=S^{3}_{p/q}(K)$

is not $SU(2)$-abelian when $p$ is even, by Theorem 6.2.

Suppose now that $p$ is odd with $\gcd(p,5)=1$. Let us assume first that
$p/q\in[4,5)$. Then Theorem 6.3 says that $Y$ is not $SU(2)$-abelian. Next,
assume that $p/q\in[3,4)$ and

$23p-9\leq 80q\leq 25p+9.$

These inequalities are equivalent to the statement that both $20q-6p-2\textrm{
and }6p-20q-2$ are at most $(p+1)/4$. Thus,

$g(S(p,2q,10q))\leq\frac{p+1}{4},$

by (6.2). Therefore, Theorem 6.3 says that $Y$ is not $SU(2)$-abelian. ∎

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