# Brieskorn spheres, cyclic group actions and the Milnor conjecture

David Baraglia  School of Mathematical Sciences, The University of Adelaide,
Adelaide SA 5005, Australia<EMAIL_ADDRESS>and  Pedram
Hekmati  Department of Mathematics, The University of Auckland, Auckland,
1010, New Zealand<EMAIL_ADDRESS>
###### Abstract.

In this paper we further develop the theory of equivariant
Seiberg–Witten–Floer cohomology of the two authors, with an emphasis on
Brieskorn homology spheres. We obtain a number of applications. First, we show
that the knot concordance invariants $\theta^{(c)}$ defined by the first
author satisfy $\theta^{(c)}(T_{a,b})=(a-1)(b-1)/2$ for torus knots, whenever
$c$ is a prime not dividing $ab$. Since $\theta^{(c)}$ is a lower bound for
the slice genus, this gives a new proof of the Milnor conjecture. Second, we
prove that a free cyclic group action on a Brieskorn homology $3$-sphere
$Y=\Sigma(a_{1},\dots,a_{r})$ does not extend smoothly to any homology
$4$-ball bounding $Y$. In the case of a non-free cyclic group action of prime
order, we prove that if the rank of $HF_{red}^{+}(Y)$ is greater than $p$
times the rank of $HF_{red}^{+}(Y/\mathbb{Z}_{p})$, then the
$\mathbb{Z}_{p}$-action on $Y$ does not extend smoothly to any homology
$4$-ball bounding $Y$. Third, we prove that for all but finitely many primes a
similar non-extension result holds in the case that the bounding $4$-manifold
has positive definite intersection form. Finally, we also prove non-extension
results for equivariant connected sums of Brieskorn homology spheres.

##  1\. Introduction

In [5], we introduced the theory of equivariant Seiberg–Witten–Floer
cohomology and established its basic properties. In this paper we further
develop this theory, with a particular emphasis on Brieskorn homology spheres.
Applications include a new proof of the Milnor conjecture and obstructions to
extending group actions over a bounding $4$-manifold.

For pairwise coprime positive integers $a_{1},\dots,a_{r}>1$, the Brieskorn
manifold $Y=\Sigma(a_{1},\dots,a_{r})$ is an integral homology Seifert
$3$-manifold. The Seifert structure defines a circle action on $Y$.
Restricting the circle action to finite subgroups, we obtain an action of the
cyclic group $\mathbb{Z}_{p}$ on $Y$ for each integer $p>1$. We obtain our
main results by considering the equivariant Seiberg–Witten–Floer cohomology of
$Y$ with respect to such $\mathbb{Z}_{p}$-actions.

###  1.1. Knot concordance invariants and the Milnor conjecture

Let $p$ be a prime number. One way of producing $\mathbb{Z}_{p}$-actions on
rational homology $3$-spheres is to take $Y=\Sigma_{p}(K)$, the cyclic
$p$-fold cover of $S^{3}$ branched over a knot $K$. From the equivariant
Seiberg–Witten–Floer cohomology of $Y$ one may extract invariants of the knot
$K$. In [4], this construction was used to obtain a series of knot concordance
invariants $\theta^{(p)}(K)$. These invariants are lower bounds for the slice
genus $g_{4}(K)$, that is, $g_{4}(K)\geq\theta^{(p)}(K)$ for all primes $p$.
More generally, the invariants $\theta^{(p)}$ can be used to bound the genus
of surfaces bounding $K$ in negative definite $4$-manifolds with $S^{3}$
boundary.

We are interested in the case that $K$ is a torus knot. If $K=T_{a,b}$ is an
$(a,b)$ torus knot and $c$ is a prime not dividing $ab$, then
$\Sigma_{c}(T_{a,b})$ is the Brieskorn homology sphere $\Sigma(a,b,c)$ and the
$\mathbb{Z}_{c}$-action arising from the branched covering construction
coincides with the restriction to $\mathbb{Z}_{c}$ of the Seifert circle
action. By studying the equivariant Seiberg–Witten–Floer homology of
$\Sigma(a,b,c)$, we deduce the following:

######  Theorem 1.1.

Let $a,b>1$ be coprime integers and let $c$ be a prime not dividing $ab$. Then
$\theta^{(c)}(T_{a,b})=\dfrac{1}{2}(a-1)(b-1)$.

Since $\theta^{(c)}$ is a lower bound for the slice genus, we obtain the
Milnor conjecture as an immediate corollary:

######  Corollary 1.2.

Let $a,b>1$ be coprime. Then $g_{4}(T_{a,b})=\dfrac{1}{2}(a-1)(b-1)$.

###### Proof.

The Milnor fibre of the singularity $x^{a}=y^{b}$ has genus $(a-1)(b-1)/2$
[29], hence $g_{4}(T_{a,b})\leq(a-1)(b-1)/2$. On the other hand, if we let $c$
be any prime not dividing $ab$, then Theorem 1.1 gives
$g_{4}(T_{a,b})\geq(a-1)(b-1)/2$. ∎

The original proof of the Milnor conjecture due to Kronheimer and Mrowka uses
gauge theory and adjunction inequalities [18]. The result was proven again by
Ozsváth and Szabó using the $\tau$-invariant of Knot Floer homology [33] and
by Rasmussen using the $s$-invariant of Khovanov homology [36]. Although our
proof uses gauge theory, it does not use adjunction inequalities but rather is
based on Floer theoretic methods. Thus our proof has more in common with
Ozsváth–Szabó and Rasmussen than with Kronheimer–Mrowka. It is interesting to
note that our proof, like those of Ozsváth–Szabó and Rasmussen, is based on
finding a knot concordance invariant which bounds the slice genus and equals
$(a-1)(b-1)/2$ for the torus knot $T_{a,b}$.

###  1.2. Equivariant delta invariants of Brieskorn homology spheres

Our next result concerns the equivariant delta invariants of Brieskorn
homology spheres. The equivariant delta invariants, introduced in [5], are a
certain equivariant generalisation of the Ozsváth–Szabó $d$-invariant and are
equivariant homology cobordism invariants. Given a rational homology
$3$-sphere $Y$, an action of $\mathbb{Z}_{p}$ on $Y$ by orientation preserving
diffeomorphisms and a $\mathbb{Z}_{p}$-invariant spinc-structure
$\mathfrak{s}$, we obtain a sequence of invariants
$\delta_{j}^{(p)}(Y,\mathfrak{s})\in\mathbb{Q}$ indexed by a non-negative
integer $j$. We call $\delta_{j}^{(p)}(Y,\mathfrak{s})$ the equivariant delta
invariants of $(Y,\mathfrak{s})$. When $Y$ is an integral homology $3$-sphere,
it has a unique spinc-structure which is automatically
$\mathbb{Z}_{p}$-invariant. In this case we may write the invariants as
$\delta^{(p)}_{j}(Y)$. The most important property of these invariants is that
they satisfy an equivariant version of the Froyshov inequality [5]. In
particular, this implies that they are invariant under equivariant homology
cobordism. Consequently, the $\delta_{j}^{(p)}$ define obstructions to
extending the $\mathbb{Z}_{p}$-action over an integral or rational homology
$4$-ball bounding $Y$:

######  Proposition 1.3 (Proposition 7.6, [5]).

Let $Y$ be an integral homology $3$-sphere on which $\mathbb{Z}_{p}$ acts by
orientation preserving diffeomorphisms. Suppose that $Y$ is bounded by a
smooth integer homology $4$-ball $W$. If the $\mathbb{Z}_{p}$-action extends
smoothly over $W$ then $\delta_{j}^{(p)}(Y)=\delta_{j}^{(p)}(-Y)=0$ for all
$j\geq 0$.

In fact, we can relax the assumption that $W$ is an integer homology $4$-ball
to simply being a rational homology $4$-ball provided that $W$ admits a
$\mathbb{Z}_{p}$-invariant spinc-structure. This is automatically true if $p$
does not divide the order of $H^{2}(W;\mathbb{Z})$, for then $\mathbb{Z}_{p}$
can’t act freely on the set of spinc-structures.

The sequence of invariants $\\{\delta_{j}^{(p)}(Y)\\}_{j\geq 0}$ is decreasing
and eventually constant. We set
$\delta_{\infty}^{(p)}(Y)=\lim_{j\to\infty}\delta_{j}^{(p)}(Y)$.

Let $Y=\Sigma(a_{1},\dots,a_{r})$ be a Brieskorn homology sphere and let $p$
be any prime. We assume $a_{1},\dots,a_{r}>1$ and $r\geq 3$ so that $Y\neq
S^{3}$. In §3.3 we prove the following results (see Proposition 3.6):

  * (1)

$\delta_{j}^{(p)}(Y)=\delta_{\infty}^{(p)}(Y)$ for all $j\geq 0$.

  * (2)

$\delta(Y)\leq\delta_{\infty}^{(p)}(Y)\leq-\lambda(Y)$.

  * (3)

$\lambda(Y)\leq\delta_{j}^{(p)}(-Y)\leq-\delta(Y)$ for all $j\geq 0$.

Here $\lambda(Y)$ is the Casson invariant of $Y$ and $\delta(Y)=d(Y)/2$ is
half the Ozsváth–Szabó $d$-invariant. In particular, if $Y$ is a Brieskorn
homology sphere which bounds a contractible $4$-manifold, then $\delta(Y)=0$,
$\delta_{j}^{(p)}(Y)=\delta_{\infty}^{(p)}(Y)$ and

$\delta_{\infty}^{(p)}(-Y)\leq\delta_{j}^{(p)}(-Y)\leq 0$

for all $j\geq 0$. Thus $\delta_{j}^{(p)}(\pm Y)=0$ for all $j$ if and only if
$\delta_{\infty}^{(p)}(\pm Y)=0$. This justifies restricting attention to the
invariants $\delta_{\infty}^{(p)}(\pm Y)$.

We first consider the case of free $\mathbb{Z}_{p}$-actions. Given a prime
$p$, the restriction of the circle action on $Y=\Sigma(a_{1},\dots,a_{r})$ to
$\mathbb{Z}_{p}$ acts freely if and only if $p$ does not divide $a_{1}\cdots
a_{r}$. Hence the action is free for all but finitely many primes. In fact any
free action of a finite group on $Y$ is conjugate to a finite subgroup of the
Seifert circle action [27, Proposition 4.3]. In the free case we have:

######  Theorem 1.4.

Let $Y=\Sigma(a_{1},\dots,a_{r})$ be a Brieskorn homology sphere and let $p$
be a prime not dividing $a_{1}\cdots a_{r}$. Set $Y_{0}=Y/\mathbb{Z}_{p}$.
Then for any spinc-structure $\mathfrak{s}_{0}$ on $Y_{0}$, we have

$\delta^{(p)}_{\infty}(Y)-\delta(Y)={\rm rk}(HF_{red}^{+}(Y))-{\rm
rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0})).$

Furthermore, we have:

######  Theorem 1.5.

We have that ${\rm rk}(HF_{red}^{+}(Y))>{\rm
rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0}))$ except in the following cases:

  * (1)

$Y=\Sigma(2,3,5)$ and $p$ is any prime.

  * (2)

$Y=\Sigma(2,3,11)$ and $p=5$.

In case (1) we have ${\rm rk}(HF_{red}^{+}(Y))={\rm
rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0}))=0$ and in case (2) we have ${\rm
rk}(HF_{red}^{+}(Y))={\rm rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0}))=1$.

Combining these two results gives:

######  Corollary 1.6.

Let $Y=\Sigma(a_{1},a_{2},\dots,a_{r})$ be a Brieskorn homology sphere and let
$p$ be a prime not dividing $a_{1}\cdots a_{r}$. Then
$\delta_{\infty}^{(p)}(Y)>\delta(Y)$ except in the following cases:

  * (1)

$Y=\Sigma(2,3,5)$ and $p$ is any prime.

  * (2)

$Y=\Sigma(2,3,11)$ and $p=5$.

In both cases we have $\delta_{\infty}^{(p)}(Y)=\delta(Y)=1$.

######  Corollary 1.7.

Let $Y=\Sigma(a_{1},a_{2},\dots,a_{r})$ be a Brieskorn homology sphere and let
$m>1$ be an integer not dividing $a_{1}\cdots a_{r}$. Suppose that $W$ is
smooth rational homology $4$-ball bounding $Y$ and that $m$ does not divide
$|H^{2}(X;\mathbb{Z})|$. Then the $\mathbb{Z}_{m}$-action on $Y$ does not
extend smoothly to $W$.

###### Proof.

Let $p$ be a prime divisor of $m$ which does not divide
$|H^{2}(X;\mathbb{Z})|$. It suffices to show that the subgroup
$\mathbb{Z}_{p}\subseteq\mathbb{Z}_{m}$ does not extend over $W$. Since $Y$ is
bounded by a rational homology $4$-ball we have $\delta(Y)=0$. Then Corollary
1.6 implies that $\delta_{\infty}^{(p)}(Y)>0$, unless $Y=\Sigma(2,3,5)$ or
$Y=\Sigma(2,3,11)$ and $p=5$. However these cases do not bound rational
homology $4$-balls as they have $\delta(Y)=1$. So $\delta_{\infty}^{(p)}(Y)>0$
which implies that the $\mathbb{Z}_{p}$-action does not extend smoothly to
$W$. ∎

The $r=3$ case of the above result was proven by Anvari–Hambleton [2], under
the assumption that the bounding manifold is contractible. We note that there
is a conjecture that Brieskorn spheres with $r>3$ can not bound contractible
manifolds (see, for example [39, Problem I]). On the other hand, there are
many examples of Brieskorn spheres which bound rational homology balls, but
not integer homology balls [14, 1, 38]. Thus Corollary 1.7 is a non-trivial
result.

We can also show that for all sufficiently large primes,
$\delta_{\infty}^{(p)}(Y)$ equals minus the Casson invariant $-\lambda(Y)$.

######  Theorem 1.8.

Let $Y=\Sigma(a_{1},a_{2},\dots,a_{r})$ be a Brieskorn homology sphere and let
$p$ be a prime not dividing $a_{1}\cdots a_{r}$. Suppose that $p>N$ where

$N=a_{1}\cdots a_{r}\left((r-2)-\sum_{i=1}^{r}\frac{1}{a_{i}}\right).$

Then $\delta_{\infty}^{(p)}(Y)=-\lambda(Y)$.

###### Proof.

This follows from Theorem 1.4 and Proposition 6.2. ∎

We now consider the case of branched coverings. Let
$Y=\Sigma(a_{1},\dots,a_{r})$ be a Brieskorn homology sphere and let $p$ be a
prime dividing $a_{1}\cdots a_{r}$. Without loss of generality we may assume
that $p$ divides $a_{1}$. Then the quotient space $Y_{0}=Y/\mathbb{Z}_{p}$ is
the Brieskorn homology sphere $\Sigma(a_{1}/p,a_{2},\dots,a_{r})$ and $Y\to
Y_{0}$ is a cyclic branched covering. Our main result is the following:

######  Theorem 1.9.

We have that

$\delta(-Y)-\delta^{(p)}_{\infty}(-Y)\geq{\rm rk}(HF_{red}^{+}(Y))-p\,{\rm
rk}(HF_{red}^{+}(Y_{0})).$

######  Remark 1.10.

Karakurt–Lidman have shown that ${\rm rk}(HF_{red}^{+}(Y))\geq p\,{\rm
rk}(HF_{red}^{+}(Y_{0}))$. Thus the right hand side of the inequality in
Theorem 1.9 is non-negative. Moreover, if
$\delta_{\infty}^{(p)}(-Y)=\delta(-Y)$ then we must have an equality: ${\rm
rk}(HF_{red}^{+}(Y))=p\,{\rm rk}(HF_{red}^{+}(Y_{0}))$.

######  Theorem 1.11.

Let $Y=\Sigma(a_{1},a_{2},\dots,a_{r})$ be a Brieskorn homology sphere and let
$p$ be a prime dividing $a_{1}\cdots a_{r}$. Suppose that $W$ is a rational
homology $4$-ball bounding $Y$ and that $p$ does not divide the order of
$H^{2}(W;\mathbb{Z})$. If ${\rm rk}(HF_{red}^{+}(Y))>p\,{\rm
rk}(HF_{red}^{+}(Y_{0}))$, then the $\mathbb{Z}_{p}$-action on $Y$ does not
extend smoothly to $W$.

The $r=3$ case of this result was proved by Anvari–Hambleton [3] (in the
integral homology case) without requiring the assumption that ${\rm
rk}(HF_{red}^{+}(Y))>p\,{\rm rk}(HF_{red}^{+}(Y_{0}))$.

We expect that the condition ${\rm rk}(HF_{red}^{+}(Y))=p\,{\rm
rk}(HF_{red}^{+}(Y_{0}))$ is rarely satisfied, however there are some cases
where it does hold. One family of examples is given by $Y=\Sigma(2,3,30n+5)$
and $p=5$, in which case $Y_{0}=\Sigma(2,3,6n+1)$ and ${\rm
rk}(HF_{red}^{+}(Y))=5\,{\rm rk}(HF_{red}^{+}(Y_{0}))=5n$. All of these
examples have $\delta(Y)=1$, so such a $Y$ can not bound a contractible
$4$-manifold. We suspect that there are no examples where ${\rm
rk}(HF_{red}^{+}(Y))=p\,{\rm rk}(HF_{red}^{+}(Y_{0}))$ and $Y$ bounds an
integral homology $4$-ball.

###  1.3. Non-extension results for positive definite $4$-manifolds

Our equivariant $\delta$-invariants can also be used to obstruct the extension
of the $\mathbb{Z}_{p}$-action over a positive definite $4$-manifold bounding
$Y$. First, we have the following result, which is a consequence of [5,
Theorem 5.3]:

######  Proposition 1.12.

Let $Y$ be an integral homology $3$-sphere on which $\mathbb{Z}_{p}$ acts by
orientation preserving diffeomorphisms. Suppose that $Y$ is bounded by a
smooth, compact, oriented, $4$-manifold $W$ with positive definite
intersection form and with $b_{1}(W)=0$. Suppose that the $\mathbb{Z}_{p}$
extends to a smooth, homologically trivial action on $W$. Then

$\min_{c}\left\\{\frac{c^{2}-rk(H^{2}(W;\mathbb{Z}))}{8}\right\\}\geq\delta_{0}^{(p)}(Y)$

where the minimum is taken over all characteristic elements of
$H^{2}(W;\mathbb{Z})$.

###### Proof.

Suppose $\mathbb{Z}_{p}$ extends smoothly and homologically trivially to $W$.
Take any characteristic $c\in H^{2}(X;\mathbb{Z})$. Then there is a unique
spinc-structure $\mathfrak{s}$ with $\mathfrak{s}=c$. Since the action is
homologically trivial it follows that $\mathfrak{s}$ is
$\mathbb{Z}_{p}$-invariant. Now we apply [5, Theorem 5.3] to $-W$ with $Y$
regarded as an ingoing boundary to obtain:
$(c^{2}-rk(H^{2}(W;\mathbb{Z})))/8\geq\delta_{0}^{(p)}(Y)$. Taking the minimum
over all characteristics gives the result. ∎

######  Corollary 1.13.

Let $W$ and $Y$ be as in Proposition 1.12. If $\delta_{0}^{(p)}(Y)>0$ or
$\delta^{(p)}_{\infty}(-Y)<0$. Then the $\mathbb{Z}_{p}$-action on $Y$ does
not extend smoothly and homologically trivially to $W$.

###### Proof.

First note that
$\delta_{0}^{(p)}(Y)+\delta_{\infty}^{(p)}(-Y)\geq\delta_{\infty}^{(p)}(Y)+\delta_{\infty}^{(p)}(-Y)\geq
0$ [5]. So if $\delta_{\infty}^{(p)}(-Y)<0$, then $\delta_{0}^{(p)}(Y)>0$. So
we can assume that $\delta_{0}^{(p)}(Y)>0$. If the $\mathbb{Z}_{p}$-action
extends smoothly and homologically trivially to $W$, then

$\min_{c}\left\\{\frac{c^{2}-rk(H^{2}(W;\mathbb{Z}))}{8}\right\\}\geq\delta_{0}^{(p)}(Y)>0.$

But $Y$ is an integral homology $3$-sphere, so $H^{2}(W;\mathbb{Z})$ is a
unimodular integral lattice. A result of Elkies [13] implies that
$\min_{c}\\{(c^{2}-rk(H^{2}(W;\mathbb{Z})))/8\\}\leq 0$, which is a
contradiction. ∎

Combined with our calculation of $\delta_{0}(Y),\delta_{\infty}(-Y)$ for
Brieskorn spheres, we obtain the following non-extension result:

######  Corollary 1.14.

Let $Y=\Sigma(a_{1},\dots,a_{r})$ be a Brieskorn homology sphere and $p$ any
prime. If $\delta_{\infty}(Y)>0$ or $\delta_{\infty}(-Y)>0$, then the
$\mathbb{Z}_{p}$-action on $Y$ does not extend smoothly and homologically
trivial to any smooth, compact, oriented, $4$-manifold $W$ with positive
definite intersection form and with $b_{1}(W)=0$ bounding $Y$.

In particular, if $p$ does not divide $a_{1}\cdots a_{r}$ and $p>N=a_{1}\cdots
a_{r}\left((r-2)-\sum_{i=1}^{r}\frac{1}{a_{i}}\right)$, then
$\delta_{\infty}(Y)=-\lambda(Y)>0$ by Theorem 6.4. Hence for a given $Y$, the
above non-extension result applies to all but finitely many primes.

Note that such a non-extension result does not exist if $W$ has a negative
definite intersection form. Indeed for any $p$, the $\mathbb{Z}_{p}$-action on
$Y$ extends smoothly and homologically trivially to any negative definite
plumbing bounding $Y$.

###  1.4. Non-extension results for connected sums

Suppose that $Y_{1},\dots,Y_{m}$ are Brieskorn homology spheres and $p$ is a
prime such that for each $i$, the $\mathbb{Z}_{p}$-action on $Y_{i}$ is not
free. Then we can form an equivariant connected sum $Y=Y_{1}\\#\cdots\\#Y_{m}$
by attaching the summands to each other along fixed points. From [4,
Proposition 3.1] we have that
$\delta_{j_{1}+\cdots+j_{m}}^{(p)}(-Y)\leq\sum_{k=1}^{m}\delta_{j_{k}}^{(p)}(-Y_{k})$.
Taking $j_{1},\dots,j_{m}$ sufficiently large, we obtain

$\delta_{\infty}^{(p)}(-Y)\leq\sum_{k=1}^{m}\delta_{\infty}^{(p)}(-Y_{k}).$

Then Theorem 1.9 implies that

(1.1) $\delta(-Y)-\delta_{\infty}^{(p)}(-Y)\geq\sum_{k=1}^{m}\left({\rm
rk}(HF_{red}^{+}(Y_{k}))-p\,{\rm
rk}(HF_{red}^{+}(Y_{k}/\mathbb{Z}_{p}))\right),$

which gives us the following result:

######  Corollary 1.15.

Let $Y_{1},\dots,Y_{m}$ be Brieskorn homology spheres and let $p$ be a prime
such that for each $i$, the $\mathbb{Z}_{p}$-action on $Y_{i}$ is not free.
Let $Y=Y_{1}\\#\cdots\\#Y_{m}$ be the equivariant connected sum. Suppose that
$W$ is a rational homology $4$-ball bounding $Y$ and that $p$ does not divide
the order of $H^{2}(W;\mathbb{Z})$. If ${\rm rk}(HF_{red}^{+}(Y_{i}))>p\,{\rm
rk}(HF_{red}^{+}(Y_{i}/\mathbb{Z}_{p}))$ for some $i$, then the
$\mathbb{Z}_{p}$-action on $Y$ does not extend smoothly to $W$.

Corollary 1.13 and (1.1) also give us a non-extension result over positive-
definite $4$-manifolds:

######  Corollary 1.16.

Let $Y_{1},\dots,Y_{m}$ be Brieskorn homology spheres and let $p$ be a prime
such that for each $i$, the $\mathbb{Z}_{p}$-action on $Y_{i}$ is not free.
Let $Y=Y_{1}\\#\cdots\\#Y_{m}$ be the equivariant connected sum. Suppose that
$W$ is a smooth, compact, oriented, $4$-manifold bounding $Y$, with positive
definite intersection form and with $b_{1}(W)=0$. If

$\delta(Y)+\sum_{k=1}^{m}\left({\rm rk}(HF_{red}^{+}(Y_{k}))-p\,{\rm
rk}(HF_{red}^{+}(Y_{k}/\mathbb{Z}_{p}))\right)>0$

then the $\mathbb{Z}_{p}$-action on $Y$ does not extend smoothly and
homologically trivially to $W$.

###  1.5. Structure of the paper

The paper is structured as follows. In §2 we recall the basic results on
equivariant Seiberg–Witten–Floer cohomology from [5] and the associated knot
concordance invariants. In §3 we examine in great detail the spectral sequence
relating equivariant and non-equivariant Floer cohomology and use this to
deduce more refined information about the equivariant delta invariants. In §4,
we study the Floer homology of Brieskorn spheres $\Sigma(a,b,c)$ and use this
to compute the invariants $\theta^{(c)}(T_{a,b})$ leading to the proof of
Theorem 1.1. In §5, we consider the case where $\mathbb{Z}_{p}$ acts non-
freely on $\Sigma(a_{1},\dots,a_{r})$ and prove Theorem 1.9. Finally in §6, we
consider the case where $\mathbb{Z}_{p}$ acts freely on
$\Sigma(a_{1},\dots,a_{r})$ and prove Theorems 1.4, 1.5 and 6.4.

##  2\. Equivariant Seiberg–Witten–Floer cohomology and knot concordance
invariants

###  2.1. Seiberg–Witten–Floer cohomology

Let $Y$ be a rational homology $3$-sphere and $\mathfrak{s}$ a spinc-
structure. For such a pair $(Y,\mathfrak{s})$, Manolescu constructed an
$S^{1}$-equivariant stable homotopy type whose equivariant (co)homology groups
are isomorphic to the Heegaard Floer or Monopole Floer (co)homology groups of
$(Y,\mathfrak{s})$ [28]. We denote the $S^{1}$-equivariant reduced cohomology
groups with coefficients in $\mathbb{F}$ by
$HSW^{*}(Y,\mathfrak{s};\mathbb{F})$ and refer to them as the
Seiberg–Witten–Floer cohomology of $(Y,\mathfrak{s})$. If the coefficient
group is understood then we will write $HSW^{*}(Y,\mathfrak{s})$. If $Y$ is an
integral homology $3$-sphere, then it has a unique spinc-structure and in this
case we simply write $HSW^{*}(Y)$. We have that $HSW^{*}(Y,\mathfrak{s})$ is a
graded module over the ring $H^{*}_{S^{1}}=H^{*}_{S^{1}}(pt;\mathbb{F})$. Note
that $H^{*}_{S^{1}}\cong\mathbb{F}[U]$, where $deg(U)=2$.

There exists a chain of isomorphisms relating Seiberg–Witten–Floer homology to
monopole Floer homology [25] and to Heegaard Floer homology [19, 20, 21, 22,
23, 7, 8, 9, 40]. In particular we have isomorphisms

$HSW^{*}(Y,\mathfrak{s})\cong HF_{+}^{*}(Y,\mathfrak{s})$

where $HF_{+}^{*}(Y,\mathfrak{s})$ denotes the plus version of Heegaard Floer
cohomology with coefficients in $\mathbb{F}$. Unless stated otherwise, we will
take our coefficient group $\mathbb{F}$ to be a field. Then the universal
coefficient theorem implies that the Heegaard Floer cohomology
$HF_{+}^{*}(Y,\mathfrak{s})$ is isomorphic to the Heegaard Floer homology
$HF^{+}_{*}(Y,\mathfrak{s})$, except that the action of $\mathbb{F}[U]$ on
$HF^{+}_{*}(Y,\mathfrak{s})$ gets replaced by its dual, so $deg(U)=2$ whereas
in Heegaard Floer homology one has $deg(U)=-2$. We will frequently identify
$HSW^{*}(Y,\mathfrak{s})$ with $HF^{+}_{*}(Y,\mathfrak{s})$, equipped with the
dual $\mathbb{F}[U]$-module structure.

Let $d(Y,\mathfrak{s})$ denote the Ozsváth–Szabó $d$-invariant. Due to the
isomorphism $HSW^{*}(Y,\mathfrak{s})\cong HF^{+}_{*}(Y,\mathfrak{s})$, we have
that $d(Y,\mathfrak{s})$ is the minimal degree $i$ for which there exists an
$x\in HSW^{i}(Y,\mathfrak{s})$ with $U^{k}x\neq 0$ for all $k\geq 0$. For
notational convenience we define $\delta(Y,\mathfrak{s})=d(Y,\mathfrak{s})/2$.

We define $HSW_{red}^{*}(Y,\mathfrak{s})=\\{x\in
HSW^{*}(Y,\mathfrak{s})\;|\;U^{k}x=0\text{ for some }k\geq 0\\}$. More
generally, given an $\mathbb{F}[U]$-module $M$, we write $M_{red}$ for the
submodule of elements $x\in M$ such that $U^{k}x=0$ for some $k\geq 0$.

Recall that for any $(Y,\mathfrak{s})$, there is an isomorphism of
$\mathbb{F}[U]$-modules

$HF^{+}(Y,\mathfrak{s})\cong\mathbb{F}[U]_{d(Y,\mathfrak{s})}\oplus
HF^{+}_{red}(Y,\mathfrak{s}),$

where for any $\mathbb{F}[U]$-module $M^{*}$ and any $d\in\mathbb{Q}$, we
define $M^{*}_{d}$ by $(M^{i}_{d})=M^{i-d}$. It follows that we similarly have
an isomorphism

$HSW^{*}(Y,\mathfrak{s})\cong\mathbb{F}[U]_{d(Y,\mathfrak{s})}\oplus
HSW_{red}(Y,\mathfrak{s}).$

###  2.2. Equivariant Seiberg–Witten–Floer cohomology

Let $Y$ be a rational homology $3$-sphere and suppose that $\tau\colon Y\to Y$
an orientation preserving diffeomorphism of order $p$, where $p$ is prime.
This gives an action of the finite group $G=\mathbb{Z}_{p}$ on $Y$ generated
by $\tau$. Let $\mathfrak{s}$ be a spinc-structure preserved by $\tau$. In
[5], the authors constructed the equivariant Seiberg–Witten–Floer cohomology
groups $HSW^{*}_{G}(Y,\mathfrak{s})$. Except where stated otherwise, we take
Floer cohomology with respect to the coefficient field
$\mathbb{F}=\mathbb{Z}_{p}$. Then $HSW^{*}_{G}(Y,\mathfrak{s})$ is a module
over the ring $H^{*}_{S^{1}\times G}=H^{*}_{S^{1}\times G}(pt;\mathbb{F})$. If
$p=2$, then $H^{*}_{S^{1}\times G}\cong\mathbb{F}[U,Q]$, where $deg(U)=2$,
$deg(Q)=1$. If $p$ is odd, then $H^{*}_{S^{1}\times
G}\cong\mathbb{F}[U,R,S]/(R^{2})$, where $deg(U)=2$, $deg(R)=1$, $deg(S)=2$.
As in the non-equivariant case, the grading on $HSW^{*}_{G}(Y,\mathfrak{s})$
can in general take rational values. However, if $Y$ is an integral homology
sphere then the grading is integer-valued.

The localisation theorem in equivariant cohomology implies that the
localisation $U^{-1}HSW^{*}_{G}(Y,\mathfrak{s})$ is a free
$U^{-1}H^{*}_{S^{1}\times G}$-module of rank $1$. Letting $\mu$ denote a
generator of $U^{-1}HSW^{*}_{G}(Y,\mathfrak{s})$, we have an isomorphism of
the form

$\iota\colon U^{-1}HSW^{*}_{G}(Y,\mathfrak{s})\to\mathbb{F}[Q,U,U^{-1}]\mu$

for $p=2$ and

$\iota\colon
U^{-1}HSW^{*}_{G}(Y,\mathfrak{s})\to\frac{\mathbb{F}[R,S,U,U^{-1}]}{(R^{2})}\mu$

for $p$ odd. Following [5, §3] we define a sequence of equivariant
$\delta$-invariants as follows. The cases $p=2$ and $p\neq 2$ need to be
treated separately. First suppose $p=2$. For each $j\geq 0$, we define
$\delta^{(p)}_{j}(Y,\mathfrak{s},\tau)$ to be $i/2-j/2$, where $i$ is the
least degree for which there exists an element $x\in
HSW^{i}_{G}(Y,\mathfrak{s})$ and a $k\in\mathbb{Z}$ such that

$\iota x=Q^{j}U^{k}\tau\;({\rm mod}\;Q^{j+1}).$

If $p\neq 2$, then for each $j\geq 0$, we define
$\delta_{j}^{(p)}(Y,\mathfrak{s},\tau)$ to be $i/2-j$, where $i$ is the least
degree for which there exists an element $x\in HSW^{i}_{G}(Y,\mathfrak{s})$
and a $k\in\mathbb{Z}$ such that

$\iota x=S^{j}U^{k}\tau\;({\rm mod}\;S^{j+1},RS^{j}).$

When the diffeomorphism $\tau$ is understood we will omit it from the notation
and simply write the delta invariants as $\delta_{j}^{(p)}(Y,\mathfrak{s})$.

Various properties of the $\delta$-invariants are shown in [5]. In particular,
we have:

  * (1)

$\delta^{(p)}_{0}(Y,\mathfrak{s})\geq\delta(Y,\mathfrak{s})$, where
$\delta(Y,\mathfrak{s})=d(Y,\mathfrak{s})/2$ and $d(Y,\mathfrak{s})$ is the
Ozsváth–Szabó $d$-invariant.

  * (2)

$\delta_{j+1}^{(p)}(Y,\mathfrak{s})\leq\delta_{j}^{(p)}(Y,\mathfrak{s})$ for
all $j\geq 0$.

  * (3)

The sequence $\\{\delta_{j}^{(p)}(Y,\mathfrak{s})\\}_{j\geq 0}$ is eventually
constant.

Using property (3), we may define two additional invariants of
$(Y,\mathfrak{s},\tau)$ as follows. We define
$\delta_{\infty}^{(p)}(Y,\mathfrak{s},\tau)=\lim_{j\to\infty}\delta_{j}^{(p)}(Y,\mathfrak{s},\tau)$
and we define $j^{(p)}(Y,\mathfrak{s},\tau)$ to be the smallest $j$ such that
$\delta_{j}^{(p)}(Y,\mathfrak{s},\tau)=\delta_{\infty}^{(p)}(Y,\mathfrak{s},\tau)$.
If $\tau$ is understood we will simply write
$\delta_{\infty}^{(p)}(Y,\mathfrak{s})$ and $j^{(p)}(Y,\mathfrak{s})$.

###  2.3. Knot concordance invariants

Given a knot $K\subset S^{3}$ and a prime number $p$, we let $Y=\Sigma_{p}(K)$
denote the degree $p$ cyclic cover of $S^{3}$ branched over $K$. Then $Y$ is a
rational homology $3$-sphere [26, Corollary 3.2] and it comes equipped with a
natural $\mathbb{Z}_{p}$-action. Let $\pi\colon Y\to S^{3}$ denote the
covering map. From [16, Corollary 2.2], any spinc-structure on
$Y\setminus\pi^{-1}(K)$ uniquely extends to $Y$. Then since
$H^{2}(S^{3}\setminus K;\mathbb{Z})=0$, there is a unique spinc-structure on
$S^{3}\setminus K$. The pullback of this spinc-structure under $\pi$ extends
uniquely to a spinc-structure on $Y$. Following [16], we denote this spinc-
structure by $\mathfrak{s}_{0}=\mathfrak{s}_{0}(K,p)$. Uniqueness of the
extension implies that $\mathfrak{s}_{0}$ is $\mathbb{Z}_{p}$-invariant. Thus
for any prime $p$ and any $j\geq 0$, we obtain a knot invariant
$\delta^{(p)}_{j}(K)$ by setting

$\delta_{j}^{(p)}(K)=4\delta_{j}^{(p)}(\Sigma_{p}(K),\mathfrak{s}_{0}).$

The invariants $\delta^{(p)}_{j}(K)$ are knot concordance invariants and they
satisfy a number of properties [5], [4]. In particular, for $p=2$ we have:

  * (1)

$\delta^{(2)}_{0}(K)\geq\delta^{(2)}(K)$, where $\delta^{(2)}(K)$ is the
Manolescu–Owens invariant [30].

  * (2)

$\delta^{(2)}_{j+1}(K)\leq\delta^{(2)}_{j}(K)$ for all $j\geq 0$.

  * (3)

$\delta^{(2)}_{j}(K)\geq-\sigma(K)/2$ for all $j\geq 0$ and
$\delta^{(2)}_{j}(K)=-\sigma(K)/2$ for $j\geq g_{4}(K)-\sigma(K)/2$.

Here $g_{4}(K)$ is the slice genus of $K$ and $\sigma(K)$ is the signature.
Similarly for $p\neq 2$, we have:

  * (1)

$\delta^{(p)}_{0}(K)\geq\delta^{(p)}(K)$, where $\delta^{(p)}(K)$ is the
Jabuka invariant [16].

  * (2)

$\delta^{(p)}_{j+1}(K)\leq\delta^{(p)}_{j}(K)$ for all $j\geq 0$.

  * (3)

$\delta^{(p)}_{j}(K)\geq-\sigma^{(p)}(K)/2$ for all $j\geq 0$ and
$\delta^{(p)}_{j}(K)=-\sigma^{(p)}(K)/2$ for
$2j\geq(p-1)g_{4}(K)-\sigma^{(p)}(K)/2$.

Here $\sigma^{(p)}(K)$ is defined as

$\sigma^{(p)}(K)=\sum_{j=1}^{p-1}\sigma_{K}(e^{2\pi ij/p}),$

where $\sigma_{K}(\omega)$ is the Levine–Tristram signature of $K$. Define
$j^{(p)}(K)$ to be the smallest $j$ such that $\delta^{(p)}_{j}(K)$ attains
its minimum. Thus

$j^{(p)}(K)=j^{(p)}(\Sigma_{p}(K),\mathfrak{s}_{0}).$

From property (3), we see that the minimum value of $\delta^{(2)}_{j}(K)$ is
precisely $-\sigma(K)/2$ and that

$j^{(2)}(K)\leq g_{4}(K)-\sigma(K)/2.$

This can be re-arranged into a lower bound for the slice genus:

$g_{4}(K)\geq j^{(2)}(K)+\frac{\sigma(K)}{2}.$

Similarly for $p\neq 2$, the minimum value of $\delta^{(p)}_{j}(K)$ is
$-\sigma^{(p)}(K)/2$ and that

$2j^{(p)}(K)\leq(p-1)g_{4}(K)-\frac{\sigma^{(p)}(K)}{2}.$

This can be re-arranged to

$g_{4}(K)\geq\frac{2j^{(p)}(K)}{(p-1)}+\frac{\sigma^{(p)}(K)}{2(p-1)}.$

Replacing $K$ by its mirror $-K$, we obtain the following slice genus bounds:

$g_{4}(K)\geq j^{(2)}(-K)-\frac{\sigma(K)}{2}$

for $p=2$, and

$g_{4}(K)\geq\frac{2j^{(p)}(-K)}{(p-1)}-\frac{\sigma^{(p)}(K)}{2(p-1)}$

for $p\neq 2$. Following [4], we define knot concordance invariants
$\theta^{(p)}(K)$ for all primes $p$ by setting

$\theta^{(2)}(K)=\max\left\\{0,j^{(2)}(-K)-\frac{\sigma(K)}{2}\right\\}$

for $p=2$, and

$\theta^{(p)}(K)=\max\left\\{0,\frac{2j^{(p)}(-K)}{(p-1)}-\frac{\sigma^{(p)}(K)}{2(p-1)}\right\\}$

for $p\neq 2$. Then we have the slice genus bounds
$g_{4}(K)\geq\theta^{(p)}(K)$ for all $p$.

##  3\. Further properties of the delta invariants

In this section we will use the spectral sequence of [5] relating equivariant
and non-equivariant Seiberg–Witten–Floer cohomologies in order to deduce more
precise information on the $\delta$-invariants. In particular, we will be able
to determine the value of $j^{(p)}(Y)$ under certain assumptions on the Floer
homology of $Y$.

###  3.1. Spectral sequence

Let $Y$ be a rational homology $3$-sphere and $\tau\colon Y\to Y$ an
orientation preserving diffeomorphism of prime order $p$. Let $\mathfrak{s}$
be a spinc-structure preserved by $\tau$. Let the coefficient group be
$\mathbb{F}=\mathbb{Z}_{p}$ and let $G=\langle\tau\rangle=\mathbb{Z}_{p}$.

Recall that $H^{*}_{G}$ is isomorphic to $\mathbb{F}[Q]$, where $deg(Q)=1$ if
$p=2$ and is isomorphic to $\mathbb{F}[R,S]/(R^{2})$ where $deg(R)=1$,
$deg(S)=2$ if $p$ is odd. Thus in either case we have
$H^{ev}_{G}\cong\mathbb{F}[S]$, where in the $p=2$ case we set $S=Q^{2}$.

Let $\\{E_{r}^{p,q},d_{r}\\}$ denote the spectral sequence for the
$G$-equivariant Seiberg–Witten–Floer cohomology $HSW^{*}_{G}(Y,\mathfrak{s})$
[5, Theorem 3.2]. In detail, this means there is a filtration
$\\{\mathcal{F}_{j}^{*}\\}_{j\geq 0}$ on $HSW^{*}_{G}(Y,\mathfrak{s})$ such
that $E_{\infty}$ is isomorphic to the associated graded group

$E_{\infty}\cong Gr(HSW^{*}_{G}(Y,\mathfrak{s}))=\bigoplus_{j\geq
0}\mathcal{F}^{*}_{j}/\mathcal{F}^{*}_{j+1}$

as $H^{*}_{S^{1}\times G}$-modules. In terms of bigrading this means that
$E^{p,q}_{r}\cong\mathcal{F}_{p}^{p+q}/\mathcal{F}_{p+1}^{p+q}$. Furthermore,
we have that

$E_{2}^{p,q}=H^{p}(\mathbb{Z}_{p};HSW^{q}(Y,\mathfrak{s})).$

###  3.2. Behaviour of the delta invariants

Let $(Y,\mathfrak{s},\tau)$ be as in Section 3.1. We will examine the spectral
sequence $\\{E_{r}^{p,q}\\}$ to deduce some properties of the delta invariants
$\\{\delta_{j}^{(p)}(Y,\mathfrak{s})\\}$.

To simplify notation, we will set $H^{i}=HSW^{i}(Y,\mathfrak{s})$ and
$\widehat{H}^{i}=HSW^{i}_{G}(Y,\mathfrak{s})$. Let $d=d(Y,\mathfrak{s})$ and
$\delta=d/2$. We will make the assumption that $H^{i}=0$ unless $i=d\;({\rm
mod}\;2)$. This assumption is satisfied if $Y$ is a Seifert homology sphere
oriented so that $-Y$ is the boundary of a negative definite plumbing.

######  Lemma 3.1.

If $p=2$ and if $HSW^{*}(Y,\mathfrak{s})$ is concentrated in degrees equal to
$d(Y,\mathfrak{s})$ mod $2$, then for all $j\geq 0$ we have
$\delta_{2j+1}^{(2)}(Y,\mathfrak{s})=\delta_{2j}^{(2)}(Y,\mathfrak{s})$.

###### Proof.

Adapting the proof of [5, Lemma 5.7] to the $p=2$ case, one can show that
$Q\colon E_{\infty}^{p,q}\to E_{\infty}^{p+1,q}$ is surjective for all $p,q$.
This implies that $Q\colon\mathcal{F}_{j}\to\mathcal{F}_{j+1}$ is surjective.
Now let $x\in HSW_{\mathbb{Z}_{2}}^{i}(Y,\mathfrak{s})$ satisfy $\iota
x=Q^{2j+1}U^{k}\mu\;({\rm mod}\;Q^{2j+2})$ for some $k$ and assume that $x$
has the minimal possible degree, so

$\delta_{2j+1}^{(2)}(Y,\mathfrak{s})=deg(x)/2-(2j+1)/2=i/2-j-1/2.$

Note that since $deg(Q)=1$, $deg(U)=2$, we have that
$deg(x)=1+d(Y,\mathfrak{s})\;({\rm mod}\;2)$. This means that the image of $x$
in $HSW_{\mathbb{Z}_{2}}^{i}(Y,\mathfrak{s})/\mathcal{F}_{1}=E_{\infty}^{0,q}$
is zero, because $E_{\infty}^{0,q}\subseteq
E_{2}^{0,q}=H^{0}(\mathbb{Z}_{2};HSW^{q}(Y,\mathfrak{s}))$ is concentrated in
degrees equal to $d(Y,\mathfrak{s})$ mod $2$. Hence $x\in\mathcal{F}_{1}$.
This implies that $x=Qy$ for some $y\in
HSW^{i-1}_{\mathbb{Z}_{2}}(Y,\mathfrak{s})$. Then since $x=Qy$, we have

$\iota x=Q\iota y=Q^{2j+1}U^{k}\mu\;({\rm mod}\;Q^{2j+2}).$

Moreover, $Q$ is injective on
$U^{-1}HSW_{\mathbb{Z}_{2}}^{*}(Y,\mathfrak{s})\cong\mathbb{F}[Q,U,U^{-1}]$,
so we deduce that

$\iota y=Q^{2j}U^{k}\mu\;({\rm mod}\;Q^{2j+1}).$

Therefore, from the definition of $\delta_{2j}^{(2)}(Y,\mathfrak{s})$, it
follows that

$\delta_{2j}^{(2)}(Y,\mathfrak{s})\leq
deg(y)/2-j=deg(x)/2-1/2-j=\delta_{2j+1}^{(2)}(Y,\mathfrak{s}).$

But we always have
$\delta_{2j+1}^{(2)}(Y,\mathfrak{s})\leq\delta_{2j}^{(2)}(Y,\mathfrak{s})$, so
we must have an equality
$\delta_{2j}^{(2)}(Y,\mathfrak{s})=\delta_{2j+1}^{(2)}(Y,\mathfrak{s})$. ∎

######  Remark 3.2.

By Lemma 3.1, if $p=2$ and $HSW^{*}(Y,\mathfrak{s})$ is concentrated in
degrees equal to $d(Y,\mathfrak{s})$ mod $2$, then $j^{(2)}(Y,\mathfrak{s})$
is even.

In what follows, we find it more convenient to consider not the whole
equivariant Seiberg–Witten–Floer cohomology but only the part concentrated in
degrees equal to $d(Y,\mathfrak{s})$ mod $2$. For this reason we consider even
counterparts of the relevant ingredients, such as the spectral sequence and
filtration. One advantage is that it allows us to treat the $p=2$ and $p\neq
2$ cases simultaneously.

Let $H^{ev}_{S^{1}\times G}$ denote the subring of $H^{*}_{S^{1}\times G}$
given by elements of even degree. Then $H^{ev}_{S^{1}\times
G}\cong\mathbb{F}[U,S]$. Let $\widehat{H}^{ev}$ denote the
$H^{ev}_{S^{1}\times G}$-submodule of $\widehat{H}^{*}$ given by elements of
degree equal to $d$ mod $2$. The filtration $\\{\mathcal{F}_{j}^{*}\\}$
defines a filtration $\\{\mathcal{F}_{j}^{ev}\\}$ on $\widehat{H}^{ev}$, where
$\mathcal{F}_{j}^{ev}=\mathcal{F}_{2j}\cap\widehat{H}^{ev}$. The associated
graded $H^{ev}_{S^{1}\times G}$-module of this filtration is isomorphic to

$E_{\infty}^{ev}=\bigoplus_{p,q}E_{\infty}^{2p,q}.$

By [5, Lemma 5.7], the map $S\colon E_{\infty}^{2p,q}\to E_{\infty}^{2p+2,q}$
is surjective for all $p,q$. It follows that
$S\colon\mathcal{F}^{ev}_{j}\to\mathcal{F}^{ev}_{j+1}$ is surjective for each
$j\geq 0$. In particular, since $\mathcal{F}^{ev}_{0}=\widehat{H}^{ev}$, we
see that $\mathcal{F}^{ev}_{j}=S^{j}\widehat{H}^{ev}$.

Recall the localisation isomorphism

$\iota\colon U^{-1}\widehat{H}^{*}\cong H^{*}_{G}[U,U^{-1}]\mu$

for some $\mu$. Restricting to $\widehat{H}^{ev}$, we get a localisation
isomorphism

$\iota\colon U^{-1}\widehat{H}^{ev}\cong
H^{ev}_{G}[U,U^{-1}]\mu\cong\mathbb{F}[S,U,U^{-1}]\mu.$

Let $\delta_{j}$ denote $\delta_{j}^{(p)}(Y,\mathfrak{s})$ if $p$ is odd and
$\delta_{2j}^{(2)}(Y,\mathfrak{s})$ if $p=2$. Since $S=Q^{2}$ in the $p=2$
case, it follows that regardless of whether $p$ is even or odd, we have
$\delta_{j}=a/2-j$, where $a$ is the least degree such that there exists an
element $x\in\widehat{H}^{a}$ with

$\iota x=S^{j}U^{k}\mu\;({\rm mod}\;S^{j+1})$

for some $k\in\mathbb{Z}$.

Observe that $E_{2}^{0,q}=H^{0}(\mathbb{Z}_{p};H^{q})$ is a submodule of
$H^{q}$. Then since $E_{r+1}^{0,q}$ is a submodule of $E_{r}^{0,q}$ for each
$r\geq 2$, we see that
$E_{\infty}^{0,*}=\mathcal{F}^{ev}_{0}/\mathcal{F}^{ev}_{1}$ can be identified
with an $\mathbb{F}[U]$-submodule of $H^{*}$. We will denote this submodule by
$J^{*}\subset H^{*}$. Now since $H^{*}\cong\mathbb{F}[U]\lambda\oplus
H^{*}_{red}$, where $\lambda$ has degree $d$ and $H^{*}_{red}=\\{x\in
H^{*}\;|\;U^{m}x=0\text{ for some }m\geq 0\\}$. It follows that
$J^{*}\cong\mathbb{F}[U]\theta\oplus J^{*}_{red}$, where $\theta$ has degree
at least $d$ and $J^{*}_{red}\subseteq H^{*}_{red}$. Since $U^{m}\theta\neq 0$
for all $m\geq 0$, we see that the image of $\theta$ under the localisation
map $\iota$ is non-zero. More precisely,

$\iota\theta=U^{\alpha}\mu\;({\rm mod}\;S)$

for some $\alpha\in\mathbb{Z}$. On the other hand, any $x\in J^{*}$ has
$U^{m}x=0\;({\rm mod}\;\mathcal{F}_{1}^{ev})$ for some $m\geq 0$. This means
that $U^{m}x=Sy$ for some $y$ and thus $\iota x$ is a multiple of $S$. It
follows that $deg(\theta)=\delta_{0}$.

By [5, Proposition 3.14], we then have that $\delta_{j}=a/2-j$, where $a$ is
the least degree such that there exists an element $x\in\widehat{H}^{a}$ such
that

$U^{m}x=S^{j}U^{k}\theta\;({\rm mod}\;\mathcal{F}_{j+1}^{ev})$

for some $m,k\geq 0$.

Consider again the sequence
$\delta_{0}\geq\delta_{1}\geq\delta_{2}\geq\cdots$. This sequence is
eventually constant, hence there exists a finite set of indices
$0<j_{1}<j_{2}<\cdots<j_{r}$ such that for each $j>0$,
$\delta_{j}<\delta_{j-1}$ if and only if $j=j_{i}$ for some $i$. We let
$n_{i}=\delta_{j_{i-1}}-\delta_{j_{i}}>0$ for $1<i\leq r$ and
$n_{1}=\delta_{j_{1}}-\delta_{0}$ for $i=1$. Thus
$\delta_{j_{i}}=\delta_{0}-(n_{1}+\cdots+n_{i})$.

######  Lemma 3.3.

There exists $x_{1},\dots,x_{r}\in J^{*}_{red}$ such that:

  * (1)

The set $\\{U^{a}x_{i}\;|\;1\leq i\leq r,0\leq a\leq n_{i}-1\\}$ is linearly
independent over $\mathbb{F}$. In particular, $U^{a}x_{i}\neq 0$ for
$a<n_{i}$.

  * (2)

We have $\delta_{j_{i}}=deg(x_{i})/2-j_{i}$.

  * (3)

Each $x_{i}$ has a lift to an element $y_{i}\in\widehat{H}^{i}$ satisfying

$U^{m_{i}}y_{i}=S^{j_{i}}U^{k_{i}}\theta\;({\rm
mod}\;\mathcal{F}_{j_{i}+1}^{ev})$

for some $m_{i},k_{i}\geq 0$.

###### Proof.

By the definition of $\delta_{j_{i}}$, there exists $y_{i}\in\widehat{H}^{*}$
such that $\delta_{j_{i}}=deg(y_{i})/2-j_{i}$ and
$U^{m_{i}}y_{i}=S^{j_{i}}U^{k_{i}}\theta\;({\rm
mod}\;\mathcal{F}_{j_{i}+1}^{ev})$ for some $m_{i},k_{i}$. Let $x_{i}\in
H^{*}$ be the image of $y_{i}$ under the natural map
$\widehat{H}^{ev}\to\widehat{H}^{ev}/\mathcal{F}_{1}^{ev}\cong J^{*}$. We
claim that $x_{i}\in J^{*}_{red}$. If not, then $x_{i}=cU^{k}\theta+w$ for
some $c\in\mathbb{F}^{*}$, $k\geq 0$ and $w\in J^{*}_{red}$. Then
$U^{m}x_{i}=cU^{k+m}\theta$ for some $m,k\geq 0$ and hence
$U^{m}y_{i}=cU^{k+m}\theta\;({\rm mod}\;\mathcal{F}^{ev}_{1})$. This
contradicts $U^{m_{i}}y_{i}=S^{j_{i}}U^{k_{i}}\theta\;({\rm
mod}\;\mathcal{F}_{j_{i}+1}^{ev})$ as $S^{j_{i}}U^{k_{i}}\theta=0\;({\rm
mod}\;\mathcal{F}^{ev}_{1})$ (since $j_{i}>0$). So $x_{i}\in J^{*}_{red}$. It
is now evident that (2) and (3) are satisfied. We claim that (1) also holds,
that is, the set $\\{U^{a}x_{i}\;|\;1\leq i\leq r,0\leq a\leq n_{i}-1\\}$ is
linearly independent over $\mathbb{F}$.

Suppose we have a non-trivial linear relation amongst elements of
$\mathcal{B}=\\{U^{a}x_{i}\;|\;1\leq i\leq r,0\leq a\leq n_{i}-1\\}$ of degree
$s$. So

$\sum_{i}c_{i}U^{\beta_{i}}x_{i}=0$

for some $c_{i}\in\mathbb{F}$, not all zero and $\beta_{i}=(s-deg(x_{i}))/2$.
Furthermore, the sum is restricted to those $i$ such that $0\leq\beta_{i}\leq
n_{i}-1$ (so that $U^{\beta_{i}}x_{i}$ belongs to $\mathcal{B}$). Lifting this
to $\widehat{H}^{ev}$, we have
$\sum_{i}c_{i}U^{\beta_{i}}y_{i}\in\mathcal{F}^{ev}_{1}$ and hence

(3.1) $\sum_{i}c_{i}U^{\beta_{i}}y_{i}=Sw$

for some $w\in\widehat{H}^{s-2}$. Let $u$ be the smallest value of $i$ such
that $c_{u}\neq 0$. Multiplying by a sufficiently large power of $U$ and
taking the result modulo $\mathcal{F}^{ev}_{j_{u}+1}$, we see that

$SU^{m}w=c_{i}S^{j_{u}}U^{b}\theta\;({\rm mod}\;\mathcal{F}^{ev}_{j_{u}+1})$

for some $m,b\geq 0$. Hence
$S\iota(U^{m}w)=S\iota(c_{i}S^{j_{u}-1}U^{b}\theta)\;({\rm
mod}\;S^{j_{u}+1})$, where $\iota$ is the localisation map. Since $S$ is
injective on $\mathbb{F}[S,U,U^{-1}]\mu$, we also have $\iota
w=c_{i}S^{j_{u}-1}U^{b-m}\iota\theta\;({\rm mod}\;S^{j_{u}})$. Now we recall
that $\iota\theta=U^{\alpha}\mu\;({\rm mod}\;S)$ for some $\alpha$, hence
$\iota(c_{i}^{-1}w)=S^{j_{u}-1}U^{b-m+\alpha}\mu\;({\rm mod}\;S^{j_{u}})$. By
the definition of $\delta_{j_{u}-1}$, we must therefore have
$deg(w)/2\geq\delta_{j_{u}-1}+(j_{u}-1)$. But $deg(w)=s-2$, so
$s\geq\delta_{j_{u}-1}+j_{u}$. On the other hand, equating degrees in Equation
(3.1) gives $s=\beta_{u}+deg(x_{u})/2=\beta_{u}+\delta_{j_{u}}+j_{u}$. This
gives

$\beta_{u}=s-\delta_{j_{u}}-j_{u}\geq\delta_{j_{u}-1}-\delta_{j_{u}}=n_{u}.$

But this contradicts $\beta_{u}\leq n_{u}-1$. So no such linear relation can
exist. ∎

######  Corollary 3.4.

We have that $\delta_{0}-\delta_{j}\leq dim_{\mathbb{F}}(J_{red}^{*})\leq
dim_{\mathbb{F}}(HF_{red}^{*}(Y,\mathfrak{s}))$ for all $j\geq 0$.

###### Proof.

Since $\delta_{j}=\delta_{j_{r}}$ for all $j\geq j_{r}$, we have that
$\delta_{0}-\delta_{j}\leq\delta_{0}-\delta_{j_{r}}$ for all $j$. But the
subspace $span_{\mathbb{F}}\\{U^{a}x_{i}\;|\;1\leq i\leq r,0\leq a\leq
n_{i}-1\\}\subseteq J^{*}$ of Lemma 3.3 has dimension
$n_{1}+\cdots+n_{r}=\delta_{0}-\delta_{j_{r}}$. ∎

Let $j^{\prime}(Y,\mathfrak{s})$ denote the smallest $j$ such that
$\delta_{j}$ attains its minimum. Thus
$j^{\prime}(Y,\mathfrak{s})=j^{(p)}(Y,\mathfrak{s})$ if $p$ is odd and
$j^{\prime}(Y,\mathfrak{s})=j^{(2)}(Y,\mathfrak{s})/2$ if $p=2$, by Remark
3.2.

######  Proposition 3.5.

Let $Y$ be a rational homology $3$-sphere, $\tau\colon Y\to Y$ an orientation
preserving diffeomorphism of order $p$, where $p$ is prime. Let $\mathfrak{s}$
be a spinc-structure preserved by $\tau$. Suppose that the following
conditions hold:

  * (1)

$HF^{+}(Y,\mathfrak{s})$ is non-zero only in degrees
$i=d(Y,\mathfrak{s})\;({\rm mod}\;2)$.

  * (2)

$\delta_{0}-\delta_{j}=dim_{\mathbb{F}}(HF_{red}^{*}(Y,\mathfrak{s}))$ for
some $j\geq 0$.

  * (3)

$HF_{red}^{*}(Y,\mathfrak{s})\neq 0$.

  * (4)

Let $\ell^{+}(Y,\mathfrak{s})$ denote the highest non-zero degree in
$HF^{+}_{red}(Y,\mathfrak{s})$. Any non-zero element in the image of $U\colon
HF_{red}^{+}(Y,\mathfrak{s})\to HF_{red}^{+}(Y,\mathfrak{s})$ has degree
strictly less than $\ell^{+}(Y,\mathfrak{s})$.

Then

$j^{\prime}(Y,\mathfrak{s})=\ell^{+}(Y,\mathfrak{s})/2-\delta_{0}(Y,\mathfrak{s})+dim_{\mathbb{F}}(HF_{red}^{*}(Y,\mathfrak{s})).$

Moreover, if $Y$ is an integral homology sphere and $\delta(Y)\in
2\mathbb{Z}$, then

$j^{\prime}(Y)=\ell^{+}(Y)/2-\delta_{0}(Y)+\delta(Y)+\lambda(Y),$

where $\lambda(Y)$ is the Casson invariant of $Y$.

###### Proof.

Let $x_{1},\dots,x_{r}\in J_{red}^{*}$ be as in Lemma 3.3. It follows that
$j^{\prime}(Y,\mathfrak{s})=j_{r}$. Choose $j$ such that
$\delta_{0}-\delta_{j}=dim_{\mathbb{F}}(HF_{red}^{*}(Y,\mathfrak{s}))$. From
Corollary 3.4, we have

$dim_{\mathbb{F}}(HF_{red}^{*}(Y,\mathfrak{s}))=\delta_{0}-\delta_{j}\leq\delta_{0}-\delta_{j_{r}}=n_{1}+\cdots+n_{r}\leq
dim_{\mathbb{F}}(J_{red}^{*})\leq
dim_{\mathbb{F}}(HF_{red}^{*}(Y,\mathfrak{s})).$

It follows that we must have equalities throughout. So
$\delta_{0}-\delta_{j_{r}}=n_{1}+\cdots+n_{r}=dim_{\mathbb{F}}(HF_{red}^{*}(Y,\mathfrak{s}))$.
So $\\{U^{a}x_{i}\;|\;1\leq i\leq r,0\leq a\leq n_{i}-1\\}$ is a basis for
$HF_{red}^{*}(Y,\mathfrak{s})$. Set $\alpha_{i}=deg(x_{i})/2$. Lemma 3.3 (2)
implies that $\delta_{j_{i}}=\alpha_{i}-j_{i}$ for $1\leq i\leq r$. Setting
$\alpha_{0}=\delta_{0}$ and $j_{0}=0$, we also have
$\delta_{j_{i}}=\alpha_{i}-j_{i}$ for $i=0$. Further, we have that
$\delta_{j_{i}}-\delta_{j_{i-1}}=n_{i}$ for $1\leq i\leq r$. Hence we obtain
$(\alpha_{i}+n_{i}-1)-\alpha_{i-1}=j_{i}-j_{i-1}-1$. But $j_{i}>j_{i-1}$, so
$j_{i}-j_{i-1}-1\geq 0$, giving

(3.2) $(\alpha_{i}+n_{i}-1)\geq\alpha_{i-1}.$

Now we observe that $2(\alpha_{i}+n_{i}-1)$ is the degree of
$U^{n_{i}-1}x_{i}$ and $2\alpha_{i-1}$ is the degree of $\alpha_{i-1}$.
Consider the space $A=HF^{\ell^{+}(Y,\mathfrak{s})}_{red}(Y,\mathfrak{s})$ of
top degree elements in $HF^{+}_{red}(Y,\mathfrak{s})$. Let
$a=dim_{\mathbb{F}}(A)>0$. By assumption (4), no non-zero element of $A$ is in
the image of $U$. Hence there must exist indices $k_{1}<k_{2}<\dots<k_{a}$
such that $x_{k_{1}},\dots,x_{k_{a}}$ is a basis for $A$. In fact,
$K=\\{k_{1},\dots,k_{a}\\}$ is precisely the set of indices $i$ such that
$2\alpha_{i}=\ell^{+}(Y,\mathfrak{s})$. Suppose $i\in K$ and $i<r$. Then (3.2)
gives $2(\alpha_{i+1}+n_{i+1}-1)\geq 2\alpha_{i}=\ell^{+}(Y,\mathfrak{s})$.
But $2(\alpha_{i+1}+n_{i+1}-1)$ is the degree of $U^{n_{i+1}-1}x_{i+1}$ and
$2\ell^{+}(Y,\mathfrak{s})$ is the highest degree in
$HF^{+}_{red}(Y,\mathfrak{s})$. So we must have
$2(\alpha_{i+1}+n_{i+1}-1)=\ell^{+}(Y,\mathfrak{s})$. But from assumption (4),
this can only happen if $n_{i+1}=1$, hence
$2\alpha_{i+1}=\ell^{+}(Y,\mathfrak{s})$ and so $i+1\in K$. So if $i\in K$ and
$i<r$, then $i+1\in K$. It follows that $K=\\{r-a+1,r-a+2,\dots,r\\}$. In
particular, $r\in K$ and $\alpha_{r}=\ell^{+}(Y,\mathfrak{s})/2$. Therefore

$j^{\prime}(Y,\mathfrak{s})=j_{r}=\alpha_{r}-\delta_{j_{r}}=\ell^{+}(Y,\mathfrak{s})/2-\delta_{0}(Y,\mathfrak{s})+dim_{\mathbb{F}}(HF_{red}^{*}(Y,\mathfrak{s})),$

where the last equality holds since
$\delta_{0}-\delta_{j_{r}}=dim_{\mathbb{F}}(HF_{red}^{*}(Y,\mathfrak{s}))$.

Lastly, suppose that $Y$ is an integral homology sphere and $\delta(Y)\in
2\mathbb{Z}$. From assumption (1), it follows that $HF_{red}^{+}(Y)$ is non-
zero only in even degrees. Then from [35, Theorem 1.3], it follows that
$dim_{\mathbb{F}}(HF_{red}^{*}(Y))=\delta(Y)+\lambda(Y)$, where $\lambda(Y)$
is the Casson invariant. Hence we have that
$j^{\prime}(Y)=\ell^{+}(Y)/2-\delta_{0}(Y)+\delta(Y)+\lambda(Y)$. ∎

###  3.3. Delta invariants of Brieskorn spheres

Given pairwise coprime integers $a_{1},\dots,a_{r}$ with $r\geq 3$ define the
Brieskorn homology sphere $\Sigma(a_{1},\dots,a_{r})$ to be the link of the
singularity at the origin of the variey
$\\{(z_{1},\dots,z_{r})\in\mathbb{C}^{n}\;|\;b_{i1}z_{1}^{a_{1}}+b_{i2}z_{2}^{a_{2}}+\cdots+b_{ir}z_{r}^{a_{r}}=0,\;1\leq
i\leq r-2\\}$, where $(b_{ij})$ is a sufficiently generic $(r-2)\times r$
matrix. Then $\Sigma(a_{1},a_{2},\dots,a_{r})$ is the Seifert homology sphere
$M(e_{0},(a_{1},b_{1}),\dots,(a_{r},b_{r}))$, where $e_{0},b_{1},\dots,b_{r}$
are uniquely determined by the conditions that $0<b_{j}<a_{j}$ and

(3.3) $e_{0}+\sum_{j=1}^{r}\frac{b_{j}}{a_{j}}=-\frac{1}{a_{1}a_{2}\cdots
a_{r}}.$

Note that $e_{0}<0$ because $b_{j}>0$ for all $j$.

Let $Y=\Sigma(a_{1},\dots,a_{r})$ be a Brieskorn homology sphere. The Seifert
structure of $Y$ gives a circle action. For any prime $p$, the restriction of
the circle action to $\mathbb{Z}_{p}\subset S^{1}$ defines an action of the
finite cyclic group $G=\mathbb{Z}_{p}$. We are interested in studying the
equivariant Seiberg–Witten–Floer cohomology groups of $Y$ and $-Y$ with
respect to this action. Since $Y$ has a unique spinc-structure we will omit it
from the notation and write $HSW^{*}_{G}(Y)$ and $HSW^{*}_{G}(-Y)$. Similarly
the $\delta$-invariants will be denoted $\delta_{j}^{(p)}(Y)$ and
$\delta_{j}^{(p)}(-Y)$.

To understand the equivariant Seiberg–Witten–Floer cohomology groups of $Y$
and $-Y$, we will use the spectral sequence relating it to the corresponding
non-equivariant groups. We have

$\displaystyle HSW^{*}(Y)\cong HF^{+}(Y)$
$\displaystyle\cong\mathbb{F}[U]_{d(Y)}\oplus HF_{red}^{+}(Y)$ $\displaystyle
HSW^{*}(-Y)\cong HF^{+}(-Y)$ $\displaystyle\cong\mathbb{F}[U]_{-d(Y)}\oplus
HF_{red}^{+}(-Y)$

From [34], we have that $HF^{+}(-Y)$ is concentrated in even degrees. It
follows that $d(Y)$ is even and $HF_{red}^{+}(Y)$ is concentrated in odd
degrees. Furthermore $HF^{+}(-Y)$ can be computed from the graded roots
algorithm [31]. This algorithm implies that $HF_{red}^{+}(-Y)$ is concentrated
in degrees at least $-d(Y)$. Dually this implies that $HF_{red}^{+}(Y)$ is
concentrated in degrees at most $d(Y)-1$.

######  Proposition 3.6.

Let $Y=\Sigma(a_{1},\dots,a_{r})$ be a Brieskorn homology sphere. We have:

  * (1)

${\rm rk}(HF_{red}^{+}(Y))=-\delta(Y)-\lambda(Y)$ where $\lambda(Y)$ is the
Casson invariant of $Y$.

  * (2)

$\delta_{j}^{(p)}(Y)=\delta_{\infty}^{(p)}(Y)$ for all $j\geq 0$.

  * (3)

$\delta(Y)\leq\delta_{\infty}^{(p)}(Y)\leq-\lambda(Y)$ and
$\lambda(Y)\leq\delta_{j}^{(p)}(-Y)\leq-\delta(Y)$ for all $j\geq 0$.

###### Proof.

From [35, Theorem 1.3] we have that
$\chi(HF_{red}^{+}(Y))=\delta(Y)+\lambda(Y)$. But $HF_{red}^{+}(Y)$ is
concentrated in odd degrees, which gives (1).

We will prove (2) and (3) in the case that $p$ is odd. The case $p=2$ is
similar and omitted for brevity. Recall that for odd primes we have
$H^{*}_{\mathbb{Z}_{p}}\cong\mathbb{F}[R,S]/(R^{2})$ with $deg(R)=1$,
$deg(S)=2$.

As in Section 3.1, let $\\{E_{r}^{p,q}(Y),d_{r}\\}$ denote the spectral
sequence for the equivariant Seiberg–Witten–Floer cohomology
$HSW^{*}_{\mathbb{Z}_{p}}(Y)$. We have a filtration
$\\{\mathcal{F}_{j}^{*}(Y)\\}_{j\geq 0}$ on $HSW^{*}_{\mathbb{Z}_{p}}(Y)$ such
that $E_{\infty}(Y)$ is isomorphic to the associated graded group

$E_{\infty}(Y)\cong Gr(HSW^{*}_{\mathbb{Z}_{p}}(Y))=\bigoplus_{j\geq
0}\mathcal{F}_{j}(Y)/\mathcal{F}_{j+1}(Y)$

as $H^{*}_{S^{1}\times\mathbb{Z}_{p}}$-modules. Furthermore, we have that

$E_{2}^{p,q}(Y)=H^{p}(\mathbb{Z}_{p};HF^{+}_{q}(Y)).$

Now since the $\mathbb{Z}_{p}$-action on $Y$ is contained in a circle action,
the generator of the $\mathbb{Z}_{p}$-action is smoothly isotopic to the
identity and acts trivially on $HF^{+}(Y)$. Hence we further have

$E_{2}^{p,q}(Y)\cong HF^{+}_{q}(Y)\otimes_{\mathbb{F}}H^{p}_{\mathbb{Z}_{p}}.$

Choose an element $\theta\in HF^{+}_{d(Y)}(Y)$ such that $U^{k}\theta\neq 0$
for all $k\geq 0$. Then $HF^{+}(Y)\cong\mathbb{F}[U]\theta\oplus
HF^{+}_{red}(Y)$ and hence

$\displaystyle E_{2}(Y)$ $\displaystyle\cong
H^{*}_{\mathbb{Z}_{p}}[U]\theta\oplus\left(HF_{red}^{+}(Y)\otimes_{\mathbb{F}}H^{*}_{\mathbb{Z}_{p}}\right)$
$\displaystyle\cong\frac{\mathbb{F}[U,R,S]}{(R^{2})}\theta\oplus\frac{HF_{red}^{+}(Y)[R,S]}{(R^{2})}.$

We have a similar spectral sequence $E_{r}^{p,q}(-Y)$ and filtration
$\\{\mathcal{F}_{j}(-Y)\\}$ on $HSW^{*}_{\mathbb{Z}_{p}}(-Y)$ such that
$E_{\infty}^{*,*}(-Y)$ is isomorphic to the associated graded
$H^{*}_{S^{1}\times\mathbb{Z}_{p}}$-module of the filtration. We have

$E_{2}(-Y)\cong\frac{\mathbb{F}[U,R,S]}{(R^{2})}\omega\oplus\frac{HF_{red}^{+}(-Y)[R,S]}{(R^{2})}$

for some $\omega$ of bi-degree $(0,-d(Y))$. Now recall that $HF_{red}^{+}(-Y)$
is concentrated in degrees at least $-d(Y)$. Hence $d_{r}(\omega)=0$ for all
$r\geq 2$. It immediately follows that $\delta_{0}^{(p)}(-Y)=-\delta(Y)$ and
hence $\delta_{j}^{(p)}(-Y)\leq-\delta(Y)$ for all $j\geq 0$.

Define $E_{r}(Y)_{red}$ to be the subgroup of $E_{r}(Y)$ consisting of
elements $x$ such that $U^{k}x=0$ for some $k\geq 0$. So for $r=2$ we have
$E_{2}(Y)_{red}\cong HF_{red}^{+}(Y)[R,S]/(R^{2})$ and

$E_{2}(Y)\cong\frac{\mathbb{F}[U,R,S]}{(R^{2})}\theta\oplus E_{2}(Y)_{red}.$

As explained in [5, §5], the image of the differential $d_{r}$ is contained in
$E_{r}(Y)_{red}$. It follows that for each $r\geq 2$, $E_{r}(Y)$ is of the
form

$E_{r}(Y)\cong\frac{\mathbb{F}[U,R,S]}{(R^{2})}U^{m_{r}}\theta\oplus
E_{r}(Y)_{red}$

for some increasing sequence $0=m_{0}\leq m_{1}\leq\dots$. It is further shown
in [5, §5] that the sequence $m_{r}$ must be eventually constant. Hence we
have

$E_{\infty}(Y)\cong\frac{\mathbb{F}[U,R,S]}{(R^{2})}U^{m}\theta\oplus
E_{\infty}(Y)_{red}$

where $m=\lim_{r\to\infty}m_{r}$.

Since we work over a field $\mathbb{F}$, it is possible to (non-canonically)
split the filtration $\\{\mathcal{F}_{j}(Y)\\}$ giving an isomorphism of
$\mathbb{F}$-vector spaces

$HSW^{*}_{\mathbb{Z}_{p}}(Y)\cong E_{\infty}^{*,*}(Y).$

Under this isomorphism $\mathcal{F}_{j}(Y)$ corresponds to the subspace of
$E_{\infty}(Y)$ spanned by homogeneous elements of bi-degree $(p,q)$ where
$p\geq j$. We fix a choice of such an isomorphism $\varphi\colon
HSW^{*}_{\mathbb{Z}_{p}}(Y)\to E_{\infty}^{*,*}(Y)$ and henceforth identify
$HSW^{*}_{\mathbb{Z}_{p}}(Y)$ with $E_{\infty}^{*,*}(Y)$. This isomorphism
will typically not be an isomorphism of
$H^{*}_{S^{1}\times\mathbb{Z}_{p}}$-modules. Nevertheless we can use the
isomorphim $\varphi$ to induce a new
$H^{*}_{S^{1}\times\mathbb{Z}_{p}}$-module action on $E_{\infty}^{*,*}(Y)$
corresponding to the one on $HSW^{*}_{\mathbb{Z}_{p}}(Y)$. To be precise, if
$c\in H^{*}_{S^{1}\times\mathbb{Z}_{p}}$, then we define $\widehat{c}\colon
E_{\infty}\to E_{\infty}$ by $\widehat{c}\,x=\varphi(c\varphi^{-1}(x))$.
Consider in particular $\widehat{U}$. The action of $\widehat{U}$ will
typically not respect the bi-grading, but we can decompose it into homogeneous
components as

$\widehat{U}=U_{(0,2)}+U_{(1,1)}+U_{(2,0)}+U_{(3,-1)}+\cdots$

where $U_{j,2-j}\colon E_{\infty}^{p,q}(Y)\to E_{\infty}^{p+j,2-j}(Y)$. Note
that there are only terms of bi-degree $(j,2-j)$ for $j\geq 0$ as
$\widehat{U}$ respects the filtration $\\{\mathcal{F}_{j}(Y)\\}$. Note also
that $U_{(0,2)}=U$ because taking the associated graded module of the
filtration recovers the original $H^{*}_{S^{1}\times\mathbb{Z}_{p}}$-module
structure on $E_{\infty}(Y)$.

Recall that $\delta_{j}^{(p)}(Y)$ is given by $i/2-j$ where $i$ is the least
degree such that there exists an $x\in HSW^{i}_{\mathbb{Z}_{p}}(Y)$ with
$U^{k}x=S^{j}U^{l}\theta\;({\rm mod}\;\mathcal{F}_{2j+1}(Y))$ for some
$k,l\geq 0$.

Recall that
$E_{\infty}(Y)\cong\mathbb{F}[U,R,S]/(R^{2})U^{m}\theta\,\oplus\,(E_{\infty}(Y))_{red}$.
Since $\theta$ has bi-degree $(0,d(Y))$, each homogeneous element in
$\mathbb{F}[U,R,S]/(R^{2})U^{m}\theta$ has bi-degree $(p,q)$ where $q\geq
d(Y)+2m$. On the other hand, since $E_{\infty}(Y)_{red}$ is a subquotient of
$HF_{red}^{+}(Y)[R,S]/(R^{2})$ and $HF_{red}^{+}(Y)$ is concentrated in
degrees at most $d(Y)-1$, we see that each homogeneous element $x$ in
$E_{\infty}(Y)_{red}$ has bi-degree $(p,q)$, where $q\leq d(Y)-1$. Then
$Ux=U_{(0,2)}x+U_{(1,1)}x+\cdots$ is a sum of homogeneous terms $U_{(j,2-j)}x$
of bidegree $(p+j,q+2-j)$. Consider first $U_{(0,2)}x$. Since $U_{(0,2)}=U$,
we see that $U_{(0,2)}x\in E_{\infty}(Y)_{red}$. Each of the remaining terms
$U_{(j,2-j)}x$ for $j\geq 1$ has bi-degree of the form
$(p^{\prime},q^{\prime})=(p+j,q+2-j)$, hence $q^{\prime}=q+2-j\leq q+1\leq
d(Y)$. Thus if $m>0$, then $U_{(j,2-j)}x$ must belong again to
$E_{\infty}(Y)_{red}$. This implies that $E_{\infty}(Y)_{red}$ is an
$\mathbb{F}[\widehat{U}]$-submodule of $E_{\infty}(Y)$. It follows easily from
this that for any fixed $j\geq 0$, there exists a $k\geq 0$ such that
$U^{k}E_{\infty}(Y)_{red}\subseteq\mathcal{F}_{2j+1}(Y)$. This further implies
that $x=S^{j}U^{m}\theta$ is a minimal degree element such that
$U^{k}x=S^{j}U^{l}\theta\;({\rm mod}\;\mathcal{F}_{2j+1}(Y))$ for some
$k,l\geq 0$, hence $\delta_{j}^{(p)}(Y)=m+d(Y)/2$ for all $j\geq 0$. So
$\delta_{j}^{(p)}(Y)=\delta_{\infty}^{(p)}(Y)=m+\delta(Y)>\delta(Y)$. From [5,
Proposition 5.10], we also have that $m\leq{\rm
rk}(HF_{red}^{+}(Y))=-\delta(Y)-\lambda(Y)$, so
$\delta_{\infty}^{(p)}(Y)=m+\delta(Y)\leq-\lambda(Y)$.

If $m=0$, then we still have that $\widehat{U}E_{\infty}(Y)_{red}\subseteq
E_{\infty}(Y)_{red}$, for if this were not the case then we would have some
$x\in E_{\infty}(Y)_{red}^{a,d(Y)-1}$ such that $U_{(1,1)}x\notin
E_{\infty}(Y)_{red}$. Further, since $d(Y)-1$ is the highest degree in
$HF_{red}^{+}(Y)$, it follows that $U_{(0,2)}x=0$ and that
$U_{(1,1)}x=c\theta$ for some non-zero element $c\in
H_{\mathbb{Z}_{p}}^{a+1}$. Then since $U_{(j,2-j)}x\in\mathcal{F}_{a+2}(Y)$
for $j\geq 2$, it follows that

$\widehat{U}x=c\theta\;({\rm mod}\;\mathcal{F}_{a+2}(Y)).$

Hence $\delta_{\mathbb{Z}_{p},c}(Y)<\delta(Y)$. On the other hand, we have
already shown that
$\delta_{\mathbb{Z}_{p},1}=\delta_{0}^{(p)}(-Y)=-\delta(Y)$. Using this and
[5, Theorem 4.4], we have

$0\leq\delta_{\mathbb{Z}_{p},c}(Y)+\delta_{\mathbb{Z}_{p},1}(-Y)<\delta(Y)-\delta(Y)=0,$

a contradiction. So even in the $m=0$ case we still have that
$\widehat{U}E_{\infty}(Y)_{red}\subseteq E_{\infty}(Y)_{red}$ and so we again
conclude that $\delta_{j}^{(p)}(Y)=\delta_{\infty}^{(p)}(Y)$ for all $j\geq 0$
and that $\delta_{\infty}^{(p)}(Y)\leq-\lambda(Y)$. This proves (2).

We have proven everything except the inequality
$\delta_{j}^{(p)}(-Y)\geq\lambda(Y)$. But from
$\delta_{0}^{(p)}(Y)\leq-\lambda(Y)$ and [5, Theorem 4.4], it follows that

$0\leq\delta_{j}^{(p)}(-Y)+\delta_{0}^{(p)}(Y)\leq\delta_{j}^{(p)}(-Y)-\lambda(Y),$

for all $j\geq 0$. Hence $\delta_{j}^{(p)}(-Y)\geq\lambda(Y)$. ∎

##  4\. Computation of $\theta^{(c)}(T_{a,b})$

Let $a,b,c>1$ be coprime integers and suppose also that $c$ is prime. In this
section we will prove that $\theta^{(c)}(T_{a,b})=(a-1)(b-1)/2$, where
$T_{a,b}$ is the $(a,b)$-torus knot. Since
$\Sigma_{c}(T_{a,b})=\Sigma(a,b,c)$, we will be interested in computing the
invariant $j^{(c)}(\Sigma(a,b,c))$ of the Brieskorn sphere $\Sigma(a,b,c)$.
Consequently we are interested in studying the structure of the Floer homology
of $\Sigma(a,b,c)$. Combined with the results of Section 3, we will be able to
carry out the computation of $\theta^{(c)}(T_{a,b})$.

###  4.1. Floer homology of $-\Sigma(a,b,c)$

Let $1<a<b<c$ be pairwise coprime integers and let $\Sigma(a,b,c)$ denote the
Brieskorn homology $3$-sphere oriented so that it is the link of the
singularity $x^{a}+y^{b}+z^{c}=0$ in $\mathbb{C}^{3}$. The graded roots
algorithm of Némethi can be used to compute the Floer homology
$HF^{+}(-\Sigma(a,b,c))$ [31]. In this case the algorithm produces a $\tau$
function $\tau\colon\mathbb{Z}_{\geq 0}\to\mathbb{Z}$ from which a graded root
$(R,\chi)$ may be constructed [31, Example 3.4 (3)]. The graded root
$(R,\chi)$ gives rise to an associated $\mathbb{F}[U]$-module
$\mathbb{H}(R,\chi)$, which up to a grading shift coincides with
$HF^{+}(-\Sigma(a,b,c))$.

For a Brieskorn sphere $\Sigma(a,b,c)$, the $\tau$ function $\tau(i)$ is given
as follows [31, §11], [6]. Let $e_{0},p_{1},p_{2},p_{3}$ be the unique
integers with $0<p_{1}<a$, $0<p_{2}<b$, $0<p_{3}<c$ and

$abce_{0}+p_{1}bc+ap_{2}c+abp_{3}=-1.$

Then

(4.1) $\tau(i)=\sum_{n=0}^{i-1}\Delta(n),$

where

(4.2)
$\Delta(n)=1-e_{0}n-\left\lceil\frac{p_{1}n}{a}\right\rceil-\left\lceil\frac{p_{2}n}{b}\right\rceil-\left\lceil\frac{p_{3}n}{c}\right\rceil.$

By [6, Theorem 1.3], we have $\tau(n+1)\geq\tau(n)$ for $n>N$, where

$N=abc-bc-ac-ab.$

Note that $N>0$ except when $(a,b,c)=(2,3,5)$. We will exclude this case from
the discussion unless stated otherwise. For the purpose of computing the Floer
homology, it suffices to consider the $\tau$ function up until the point where
it is increasing. So we only need the restricted $\tau$ function
$\tau\colon[0,N+1]\to\mathbb{Z}$, or equivalently, it suffices to determine
the $\Delta$-function $\Delta\colon[0,N]\to\mathbb{Z}$. Furthermore, the
function $\Delta$ is completely determined on $[0,N]$ as follows [6, Theorem
1.3]. Let

$G=\\{g\in\mathbb{Z}_{\geq 0}\;|\;g=bci+acj+abk\text{ for some
}i,j,k\in\mathbb{Z}_{\geq 0}\\}$

be the additive semigroup generated by $bc,ac,ab$. We have that
$\Delta(n)\in\\{-1,0,1\\}$ for all $n\in[0,N]$. Moreover, for any $n\in[0,N]$,
we have that

$\Delta(n)=\begin{cases}\;\;\,1&n\in G,\\\ -1&N-n\in G,\\\
\;\;\,0&\text{otherwise}.\end{cases}$

It follow that for $n\in[0,N]$, we have

(4.3) $\tau(n+1)=|G\cap[0,n]|-|G\cap[N-n,N]|$

where $|S|$ denotes the cardinality of a finite set $S$. From this expression
it is clear that $\tau$ satisfies $\tau(N+1-n)=\tau(n)$ for all $n\in[0,N]$.

######  Remark 4.1.

Note that since $0<N<abc$, it follows from the Chinese remainder theorem that
any $n\in G\cap[0,N]$ can be written as $abc(i/a+j/b+k/c)$ for uniquely
determined integers $i,j,k$ satisfying $0<i<a$, $0<j<b$, $0<k<c$.

We briefly explain how to obtain $HF^{+}(-\Sigma(a,b,c))$ from the $\tau$
function. First we recall the definition of a graded root:

######  Definition 4.2 ([31]).

Let $R$ be an infinite tree with vertices $\mathcal{V}$ and edges
$\mathcal{E}$. Denote by $[u,v]$ the edge joining $u$ and $v$. Edges are
unordered, so $[u,v]=[v,u]$. We say that $R$ is a graded root with grading
$\chi\colon\mathcal{V}\to\mathbb{Z}$ if

  * (a)

$\chi(u)-\chi(v)=\pm 1$ for all $[u,v]\in\mathcal{E}$,

  * (b)

$\chi(u)>\min\\{\chi(v),\chi(w)\\}$ if $[u,v],[u,w]\in\mathcal{E}$ and $v\neq
w$,

  * (c)

$\chi$ is bounded from below, $\chi^{-1}(k)$ is finite for any
$k\in\mathbb{Z}$ and $|\chi^{-1}(k)|=1$ for all sufficiently large $k$.

Define a partial relation $\succeq$ on $(R,\chi)$ by declaring $v\succeq w$ if
there exists a sequence of vertices $w=w_{0},w_{1},\dots,w_{k}=v$ such that
$[w_{i-1},w_{i}]\in\mathcal{E}$ and $\chi(w_{i})-\chi(w_{i-1})=1$ for
$i=1,\dots,k$. From the axioms (a)-(c), it is easily seen that for each
$u\in\mathcal{V}$ there is a unique vertex $v\in\mathcal{V}$ with
$[u,v]\in\mathcal{E}$ and $\chi(v)=\chi(u)+1$. From this and axiom (c), it is
seen that for any two vertices $u,v\in\mathcal{V}$ there is a unique
$\succeq$-minimal element $w\in\mathcal{V}$ with $w\succeq v$ and $w\succeq
u$. We denote this element by $\sup(u,v)$.

For $v\in\mathcal{V}$, let $\delta_{v}$ be the number of edges which have $v$
as an endpoint. Clearly $\delta_{v}\geq 1$ and $\delta_{v}=1$ if and only if
$v$ is $\succeq$-minimal.

Given a graded root $(R,\chi)$ define $\mathbb{H}(R,\chi)$ to be the graded
$\mathbb{F}[U]$-module constructed as follows. A basis for
$\mathbb{H}(R,\chi)$ is given by $\\{e_{v}\;|\;v\in\mathcal{V}\\}$ where
$deg(e_{v})=2\chi(v)$ and $Ue_{v}=e_{u}$, where $u$ is the unique
$u\in\mathcal{V}$ such that $[u,v]\in\mathcal{E}$ and $\chi(u)=\chi(v)+1$.
Note that our definition of $\mathbb{H}(R,\chi)$ differs slightly from the one
in [31]. This is because we are using Floer cohomology rather than homology.
The underlying graded abelian group of $\mathbb{H}(R,\chi)$ is the same as
that in [31], but our $\mathbb{F}[U]$-module action is the transpose of the
one in [31].

For a positive integer $n$ let $\mathcal{T}^{+}(n)$ denote the
$\mathbb{F}[U]$-module $\mathbb{F}[U]/(U^{n})$. For any integer $d$ and any
$\mathbb{F}[U]$-module $M$, let $M_{d}$ be the degree-shift of $M$ defined by
$(M_{d})^{j}=M^{j-d}$. In particular, for any positive integer $n$ and any
integer $d$, we have the $\mathbb{F}[U]$-module
$\mathcal{T}^{+}_{d}(n)=(\mathcal{T}^{+}(n))_{d}$. We refer to
$\mathcal{T}^{+}_{d}(n)$ as a tower of length $n$. The lowest degree in
$\mathcal{T}^{+}_{d}(n)$ is $d$ and the highest degree is $d+2n-2$.

######  Proposition 4.3 ([31], Proposition 3.5.2).

Let $(R,\chi)$ be a graded root. Set
$I=\\{v\in\mathcal{V}\;|\;\delta_{v}=1\\}$. We choose an ordering on the set
$I$ as follows. The first element $v_{1}\in I$ is chosen so that $\chi(v_{1})$
is the minimal value of $\chi$. If $v_{1},\dots,v_{k}\in I$ are already
determined and $J=\\{v_{1},\dots,v_{k}\\}\neq I$, then $v_{k+1}$ is chosen
from $I\setminus J$ such that $\chi(v_{k+1})=\min_{v\in I\setminus J}\chi(v)$.
Let $w_{k+1}\in\mathcal{V}$ be the unique $\succeq$-minimal vertex of $R$
which dominates both $v_{k+1}$ and some element of $J$. Then we have an
isomorphism of $\mathbb{F}[U]$-modules:

$\mathbb{H}(R,\chi)\cong\mathbb{F}[U]_{2\chi(v_{1})}\oplus\bigoplus_{k\geq
2}\mathcal{T}^{+}_{2\chi(v_{k})}(\chi(w_{k})-\chi(v_{k})).$

In particular we have an isomorphism

$\mathbb{H}_{red}(R,\chi)\cong\bigoplus_{k\geq
2}\mathcal{T}^{+}_{2\chi(v_{k})}(\chi(w_{k})-\chi(v_{k})).$

Let $\tau\colon\\{0,1,\dots,l\\}\to\mathbb{Z}$ be any function and suppose
there is an $l$ such that $\tau(i+1)\geq\tau(i)$ for all $i\geq l$. Following
[31, Example 3.4], we construct a graded root $(R,\chi)$ from $\tau$ as
follows. Start with the vertex set $\widetilde{\mathcal{V}}=\\{v_{i}^{k}\\}$
where $0\leq i\leq l$ and $k\geq\tau(i)$, edge set
$\widetilde{\mathcal{E}}=[v_{i}^{k},v_{i}^{k+1}]$ and define
$\widetilde{\chi}$ by $\widetilde{\chi}(v_{i}^{k})=k$. Now define
$\mathcal{V}=\widetilde{\mathcal{V}}/\\!\\!\sim$ and
$\mathcal{E}=\widetilde{\mathcal{E}}/\\!\\!\sim$ where for $i\leq j$ we set
$v_{i}^{k}\sim v_{j}^{k}$ and
$[v_{i}^{k},v_{i}^{k+1}]\sim[v_{j}^{k},v_{j}^{k+1}]$ if $k\geq\max_{i\leq
u\leq j}\tau(u)$. We define $\chi$ to be the function on $\mathcal{V}$ induced
by $\widetilde{\chi}$. It is easily checked that
$(\mathcal{V},\mathcal{E},\chi)$ is a graded root. We call it the graded root
associated to $\tau$.

Let $1<a<b<c$ be pairwise coprime integers and let $\tau$ be the
$\tau$-function defined as in Equations (4.1)-(4.2). Let
$(R_{\tau},\chi_{\tau})$ be the associated graded root. Then from [31] it
follows that up to a grading shift $HF^{+}(-\Sigma(a,b,c))$ is isomorphic to
$\mathbb{H}(R_{\tau},\chi_{\tau})$:

$HF^{+}(-\Sigma(a,b,c))\cong\mathbb{H}(R_{\tau},\chi_{\tau})[u].$

The precise value of the grading shift $u$ can be computed [31, Proposition
4.7], but we will not need this.

Observe that $\tau(0)=\tau(N+1)=0$ and that $\tau(1)=\tau(N)=1$. We will see
in Proposition 4.5 that $\tau(n)\leq 1$ for all $n\in[0,N+1]$, hence $1$ and
$N$ are global maxima of $\tau$. Let $n\in[0,N+1]$ be a global maximum of
$\tau$, that is, $\tau(n)=1$. We will say $n$ is a trivial maximum if we
either have that $\tau(i)=1$ for all $1\leq i\leq n$ or $\tau(i)=1$ for all
$n\leq i\leq N$.

######  Proposition 4.4.

Suppose that all global maxima of $\tau$ are trivial. Let
$\ell^{+}(-\Sigma(a,b,c))$ denote the highest non-zero degree in
$HF^{+}_{red}(-\Sigma(a,b,c))$. Any non-zero element in the image of $U\colon
HF^{+}_{red}(-\Sigma(a,b,c))\to HF_{red}^{+}(-\Sigma(a,b,c))$ has degree
strictly less than $\ell^{+}(-\Sigma(a,b,c))$.

###### Proof.

Let $(R_{\tau},\chi_{\tau})$ be the graded root associated to $\tau$. From the
definition of $(R_{\tau},\chi_{\tau})$, it is clear that
$(R_{\tau},\chi_{\tau})$ depends only on the values of the local minima and
local maxima of $\tau$. More precisely, let $S^{min}$ be the set of
$m\in[0,N+1]$ such that $m=0$, $m=N+1$, or $\tau(m-1)>\tau(m)$ and there
exists a $k\geq 0$ such that $\tau(i)=\tau(m)$ for $m\leq i\leq m+k$ and
$\tau(m+k+1)>\tau(m)$. Similarly, let $S^{max}$ be the set of $M\in[1,N]$ such
that $\tau(M)>\tau(M-1)$ and there exists a $k\geq 0$ such that
$\tau(i)=\tau(M)$ for $M\leq i\leq M+k$ and $\tau(M+k+1)<\tau(M)$. Write the
elements of $S^{min}$ and $S^{max}$ in increasing order as
$S^{min}=\\{m_{0},m_{1},\dots,m_{r}\\}$,
$S^{max}=\\{M_{1},M_{2},\dots,M_{r}\\}$ where $0=m_{0}<m_{1}<\cdots<m_{r}=N+1$
and $1=M_{1}<M_{2}<\cdots<M_{r}$. The local minima and maxima must occur in
alternating order, so we have

$0=m_{0}<M_{1}<m_{1}<M_{2}<\cdots<m_{r-1}<M_{r}<m_{r}=N+1.$

Furthermore, $M_{1}$ and $M_{r}$ correspond to the trivial global maxima. By
assumption these are the only global maxima. Hence $\tau(M_{j})\leq 0$ for
$1<j<r$.

Recall that $(R_{\tau},\chi_{\tau})$ is constructed as follows. Start with
$\widetilde{\mathcal{V}}=\\{v_{i}^{k}\\}$ where $0\leq i\leq N+1$ and
$k\geq\tau(i)$, edge set $\widetilde{\mathcal{E}}=[v_{i}^{k},v_{i}^{k+1}]$ and
define $\widetilde{\chi}$ by $\widetilde{\chi}(v_{i}^{k})=k$. Then
$(R_{\tau},\chi_{\tau})$ is obtained by taking the quotient of this by the
equivalence relation $\sim$, where for $i\leq j$ we set $v_{i}^{k}\sim
v_{j}^{k}$ and $[v_{i}^{k},v_{i}^{k+1}]\sim[v_{j}^{k},v_{j}^{k+1}]$ if
$k\geq\max_{i\leq u\leq j}\tau(u)$.

It is easily seen that the minima of $\succeq$ on $(R_{\tau},\chi_{\tau})$ are
precisely the elements $I=\\{v_{m_{i}}^{\tau(m_{i})}\\}_{0\leq i\leq r}$. From
Proposition 4.5 we have that $\tau(n)\leq 1$ for all $n\in[0,N+1]$ and hence
$\tau(m_{i})\leq 0$ for any local minimum. Hence we may choose a permutation
$\sigma\colon\\{0,1,\dots,r\\}\to\\{0,1,\dots,r\\}$ such that
$\tau(m_{\sigma(i)})\leq\tau(m_{\sigma(i+1)})$ for $0\leq i\leq r-1$ and
$\sigma(r-1)=0$, $\sigma(r)=r$. This gives an ordering of the set $I$ as in
the statement of Proposition 4.3, namely
$I=\\{v^{\prime}_{0},v^{\prime}_{1},v^{\prime}_{2},\dots,v^{\prime}_{r}\\}$
where $v^{\prime}_{i}=v_{m_{\sigma(i)}}^{\tau(m_{\sigma(i)})}$. For each
$i\in\\{1,\dots,r\\}$ and each $j\in\\{0,1,\dots,i-1\\}$ one finds that
$sup(v^{\prime}_{i},v^{\prime}_{j})=v_{m_{\sigma}(i)}^{K_{ij}}$, where
$K_{ij}$ is the maximum of $\tau(M_{a})$ for all $a$ such that $M_{a}$ lies in
the interval joining $m_{\sigma(i)}$ and $m_{\sigma(j)}$. Now let
$w_{1},w_{2},\dots,w_{r}$ be defined as in the statement of Proposition 4.3.
Then it follows that $w_{i}=v_{m_{\sigma(i)}}^{K_{i}}$, where
$K_{i}=\min_{0\leq j\leq i-i}K_{ij}$. Now suppose that $i\neq r-1,r$. Then for
all $j\in\\{0,\dots,i-1\\}$ we have that $\sigma(i)\neq 0,r$ and
$\sigma(j)\neq 0,r$. It follows that each maximum $M_{a}$ in the interval
joining $m_{\sigma(i)}$ and $m_{\sigma(j)}$ is not trivial and hence
$\tau(M_{a})\leq 0$. Hence $K_{ij}\leq 0$ for all $i\leq r-2$ and $j<i$. This
also implies that $\chi(w_{i})=K_{i}\leq 0$ for all $i\leq r-2$. On the other
hand since $\tau(M_{1})=\tau(M_{r})=1$, it follows that $K_{r-1}=K_{r}=1$ and
thus $\chi(w_{r-1})=\chi(w_{r})=1$.

From Proposition 4.3, we have an isomorphism

$\mathbb{H}_{red}(R_{\tau},\chi_{\tau})\cong\bigoplus_{k=1}^{r}\mathcal{T}^{+}_{2\chi(v^{\prime}_{k})}(\chi(w_{k})-\chi(v^{\prime}_{k})).$

The highest degree in the tower
$\mathcal{T}^{+}_{2\chi(v^{\prime}_{k})}(\chi(w_{k})-\chi(v^{\prime}_{k}))$ is
$2\chi(w_{k})-2$. For $k\neq r-1,r$ we have $2\chi(w_{k})-2\leq-2$ whereas for
$k=r-1,r$ we have $2\chi(w_{k})-2=0$. Thus the highest non-zero degree in
$\mathbb{H}_{red}(R,\chi)$ is $0$ and only the towers for $k=r-1,r$ attain
this degree. Moreover the length of the tower
$\mathcal{T}^{+}_{2\chi(v^{\prime}_{k})}(\chi(w_{k})-\chi(v^{\prime}_{k}))$ is
$\chi(w_{k})-\chi(v^{\prime}_{k})$ which for $k=r-1,r$ equals $1$, since
$v^{\prime}_{r-1}=v_{\sigma(r-1)}^{\tau(m_{\sigma(r-1)})}=v_{0}^{\tau(0)}=v_{0}^{0}$
and
$v^{\prime}_{r}=v_{\sigma(r)}^{\tau(m_{\sigma(r)})}=v_{r}^{\tau(m_{r})}=v_{r}^{0}$,
so $\chi(v^{\prime}_{r-1})=\chi(v^{\prime}_{r})=0$. It follows that any non-
zero element in the image of
$U\colon\mathbb{H}_{red}(R_{\tau},\chi_{\tau})\to\mathbb{H}_{red}(R_{\tau},\chi_{\tau})$
has degree strictly less than $0$. Now since $HF_{red}^{+}(-\Sigma(a,b,c))$ is
isomorphic to a grading shift of $\mathbb{H}_{red}(R_{\tau},\chi_{\tau})$, it
also follows that any non-zero element in the image of $U\colon
HF^{+}_{red}(-\Sigma(a,b,c))\to HF_{red}^{+}(-\Sigma(a,b,c))$ has degree
strictly less than $\ell^{+}(-\Sigma(a,b,c))$. ∎

######  Proposition 4.5.

Let $1<a<b<c$ be pairwise coprime integers and assume that
$(a,b,c)\neq(2,3,5)$. Then $\tau(n+1)\leq 1$ for all $n\in[0,N]$. Furthermore,
all maxima of $\tau$ are trivial, except in the following cases:
$(a,b,c)=(2,3,6n-1)$, $n\geq 2$ or $(a,b,c)=(2,3,6n+1)$, $n\geq 1$.

###### Proof.

For $n\in[0,N]$, let $\alpha_{n}$ be the largest element of $G$ less than or
equal to $N-n$. Suppose that $j\in G\cap[0,n]\setminus\\{0\\}$. We claim that
$j+\alpha_{n}\in G\cap[N-n,N]$. Clearly $j+\alpha_{n}\in G$ because
$j,\alpha_{n}\in G$. Since $j\leq n$ and $\alpha_{n}\leq N-n$, we also have
$j+\alpha_{n}\leq N$. Lastly, since $j>0$, we have $j+\alpha_{n}\geq N-n$ for
if not then $\alpha_{n}$ is not the largest element of $G$ less than or equal
to $N-n$. Therefore $j+\alpha_{n}\in G\cap[N-n,N]$. So we have constructed an
injective map $\phi_{n}:G\cap[0,n]\setminus\\{0\\}\to G\cap[N-n,N]$, given by
$\phi(j)=j+\alpha_{n}$. Therefore $|G\cap[0,n]|\leq 1+|G\cap[N-n,N]$.
Comparing with Equation (4.3), we have shown that $\tau(n)\leq 1$.

Now suppose that $n_{0}\in[1,N]$ is a maximum of $\tau$, so $\tau(n_{0})=1$.
By the symmetry $\tau(N+1+i)=\tau(i)$ of the $\tau$ function, it suffices to
consider only maxima $n_{0}$ such that $n_{0}\geq(N+1)/2$. Since $\tau$ can
only change by $\pm 1$, there exists an $n\geq n_{0}$ such that $\tau(i)=1$
for $n_{0}\leq i\leq n$ and $\tau(n+1)=0$. Since $\tau(N+1)=0$, we see that
$n\leq N$. Also we have $n\geq n_{0}\geq(N+1)/2$.

Since $\tau(n)=1$ and $\tau(n+1)=0$, we have that
$\Delta(n)=\tau(n+1)-\tau(n)=-1$. Hence $N-n\in G$ and $n\notin G$. Since
$N-n\in G$, it follows that $\alpha_{n}=N-n$. Consider the map $\phi\colon
G\cap[0,n]\to G\cap[N-n,N]$ given by $\phi(j)=j+N-n$ (we already saw that
$\phi(j)\in G\cap[N-n,N]$ for $j>0$ and we also have $\phi(0)=N-n\in
G\cap[N-n,N]$). Since $\tau(n+1)=0$, it follows that $\phi$ is a bijection.
Thus every $g\in G\cap[N-n,N]$ can be written uniquely as $g=(N-n)+g^{\prime}$
for some $g\in G\cap[0,n]$. Since $N-n\in G$, we may write it in the form

$N-n=abc\left(\frac{i_{0}}{a}+\frac{j_{0}}{b}+\frac{k_{0}}{c}\right)$

for some $i_{0},j_{0},k_{0}\geq 0$. It follows that any $g\in G\cap[N-n,N]$
has the form

(4.4) $g=abc\left(\frac{i}{a}+\frac{j}{b}+\frac{k}{c}\right)$

where $i\geq i_{0}$, $j\geq j_{0}$, $k\geq k_{0}$. Furthermore, by Remark 4.1,
this is a necessary condition. That is, if $g\in G\cap[N-n,N]$ is written as
in (4.4) then we must have $i\geq i_{0}$, $j\geq j_{0}$, $k\geq k_{0}$.

Suppose that $i_{0}>0$. There exists an integer $k$ such that $1/a$ lies in
the interval $[(k-1)/c,k/c]$. Thus

$\frac{1}{a}\leq\frac{k}{c}\leq\frac{1}{a}+\frac{1}{c}.$

Consider

$g=(N-n)+abc\left(-\frac{1}{a}+\frac{k}{c}\right)=abc\left(\frac{i_{0}-1}{a}+\frac{j_{0}}{b}+\frac{k_{0}+k}{c}\right).$

Then $g\in G$ and $g\geq N-n$ since $k/c-1/a\geq 0$. But $g$ is not of the
form $abc(i^{\prime}/a+j^{\prime}/b+k^{\prime}/c)$ with $i^{\prime}\geq
i_{0}$, $j^{\prime}\geq j_{0}$, $k^{\prime}\geq k_{0}$, hence $g\notin
G\cap[N-n,N]$. The only way this can happen is that $g>N$. Thus

$g=(N-n)+abc\left(-\frac{1}{a}+\frac{k}{c}\right)>N,$

which implies that $n<abc(-1/a+k/c)$. But $k/c\leq 1/a+1/c$, so $n<abc/c=ab$.
Note that since $a<b<c$, $ab$ is the smallest positive element of $G$. Thus
$i\notin G$ for all $1\leq i\leq n$. So $\Delta(i)\leq 0$ for $1\leq i\leq n$.
But we also have that

$\tau(n)=1=\sum_{i=0}^{n-1}\Delta(i)=1+\sum_{i=1}^{n-1}\Delta(i).$

Since $\Delta(i)\leq 0$ for $1\leq i\leq n$, the only way we can have equality
is that $\Delta(i)=0$ for $1\leq i\leq n$ and hence $\tau(i)=1$ for $1\leq
i\leq n$. This means that $n$ is a trivial maximum.

Next, suppose that $i_{0}=0$ and $j_{0}>0$. Then by a similar argument to the
$i_{0}>0$ case, there exists a $k$ such that $1/b\leq k/c\leq 1/b+1/c$.
Consider

$g=(N-n)+abc\left(-\frac{1}{b}+\frac{k}{c}\right).$

Arguing as in the $i_{0}>0$ case, we see that $n$ is again a trivial maximum.

Now consider the case that $i_{0}=j_{0}=0$. Hence $N-n=abc(k_{0}/c)$ for some
$k_{0}\geq 0$. We will assume that $k_{0}>0$ for if $k_{0}=0$, then $n=N$ is a
trivial maximum. Furthermore, recall that we are assuming $n\geq(N+1)/2>N/2$.
Therefore

(4.5)
$\frac{k_{0}}{c}<\frac{N}{2abc}=\frac{1}{2}-\frac{1}{2a}-\frac{1}{2b}-\frac{1}{2c}.$

Recall that each element of $G\cap[N-n,N]$ has the form given by (4.4) with
$i\geq 0$, $j\geq 0$, $k\geq k_{0}$. It follows that there does not exist a
solution to

(4.6) $\frac{k_{0}}{c}\leq\frac{j}{b}+\frac{k_{0}-1}{c}\leq
1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$

with $j\geq 0$. For if such a $j$ exists, we would have

$N-n=abc\left(\frac{k_{0}}{c}\right)<abc\left(\frac{j}{b}+\frac{k_{0}-1}{c}\right)\leq
N$

and thus $g=abc(j/b+(k_{0}-1)/c)$ would be an element of $G\cap[N-n,N]$ not of
the form $abc(i^{\prime}/a+j^{\prime}/b+k^{\prime}/c)$ with $i^{\prime}\geq
i_{0}$, $j^{\prime}\geq j_{0}$, $k^{\prime}\geq k_{0}$. In particular, $j=1$
is not a solution. But since $b<c$, we have $1/b+(k_{0}-1)/c>k_{0}/c$. So it
must be the second inequality in (4.6) that is violated. That is, we must have

$\frac{j}{b}+\frac{k_{0}-1}{c}>1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}.$

Rearranging and using (4.5), we find that

(4.7) $\frac{1}{a}+\frac{3}{b}-\frac{1}{c}>1.$

If this condition is not satisfied, then all maxima of $\tau$ are trivial.

If $a\geq 4$, then $b\geq 5$ and

$\frac{1}{a}+\frac{3}{b}-\frac{1}{c}<\frac{1}{4}+\frac{3}{5}<1.$

So (4.7) implies $a<4$, hence $a=2$ or $3$.

If $a=3$, then since $1/3+3/5<1$, (4.7) implies that $b\leq 5$. So $b=4$ and
then (4.7) implies that $c>12$.

If $a=2$, then since $1/2+3/6-1/c<1$, we must have $b<6$. Hence $b=3$ or $5$.

If $a=2,b=5$, then (4.7) is satisfied for any $c>10$.

If $a=2,b=3$, then (4.7) is satisfied for any $c>3$. But we are excluding
$(2,3,5)$ so $c>5$.

To summarise, (4.7) is satisfied only in the following cases:

  * (1)

$(a,b,c)=(3,4,c)$, $c>12$.

  * (2)

$(a,b,c)=(2,5,c)$, $c>10$.

  * (3)

$(a,b,c)=(2,3,c)$, $c>5$.

It remains to show in cases (1) and (2) we still have that all maxima of
$\tau$ are trivial. We do this by a direct computation of the $\tau$ function.

In case (1) there are four subcases:

  * (1a)

$(a,b,c)=(3,4,12k+1)$, $k\geq 1$.

  * (1b)

$(a,b,c)=(3,4,12k+5)$, $k\geq 1$.

  * (1c)

$(a,b,c)=(3,4,12k+7)$, $k\geq 1$.

  * (1d)

$(a,b,c)=(3,4,12k+11)$, $k\geq 1$.

Similarly in case (2) there are four subcases:

  * (2a)

$(a,b,c)=(2,5,10k+1)$, $k\geq 1$.

  * (2b)

$(a,b,c)=(2,5,10k+3)$, $k\geq 1$.

  * (2c)

$(a,b,c)=(2,5,10k+7)$, $k\geq 1$.

  * (2d)

$(a,b,c)=(2,5,10k+9)$, $k\geq 1$.

Case (1a): $N=60k-7$, $G$ is generated by $\\{12,36k+3,48k+4\\}$. By the
symmetry of the $\tau$ function, it suffices to only look for maxima of
$\tau(n+1)$ with $n\in[0,N/2]$. We consider the intersection of $G$ with
$[0,N/2]$. Since $36k+3>N/2$, all elements of $G\cap[0,N]$ have the form $12j$
for some $j$. Thus

$G\cap[0,N/2]=\\{12j\\}_{0\leq j\leq\lfloor N/24\rfloor}.$

Now we consider the intersection of $N-G$ with $[0,N]$. Elements of $N-G$ in
the range $[0,N]$ have the form $N-12u$, $N-(36k+3)-12u$ or $N-(48k+4)-12u$
for some $u\geq 0$. In the first case, we have

$N-12u=12(5k-1-u)+5.$

In the second case, we have

$N-(36k+3)-12u=12(2k-1-u)+2$

and in the third case, we have

$N-(48+4)-12u=12(k-1-u)+1.$

Therefore

$(N-G)\cap[0,N]=\\{12j+5\\}_{0\leq j\leq 5k-1}\cup\\{12j+2\\}_{0\leq j\leq
2k-1}\cup\\{12j+1\\}_{0\leq j\leq k-1}.$

Partition $[0,N]$ into subintervals $I_{j}=[12j,12j+11]$, $0\leq j\leq 5k-2$
and $I_{5k-1}=[60k-12,60k-7]$. We are only interested in the subintervals
which intersect with $[0,N/2]$, so we can assume $j\leq\lfloor N/24\rfloor$.
Under this condition, we have $G\cap I_{j}\cap G=\\{12j\\}$ and

$(N-G)\cap I_{j}=\begin{cases}12j+1,12j+2,12j+5&0\leq j\leq k-1,\\\
12j+2,12j+5&k\leq j\leq 2k-1,\\\ 12j+5&2k\leq j.\end{cases}$

It follows easily that $\tau$ has only trivial maxima.

Case (1b): $N=60k+13$, $G$ is generated by $12,36k+15,48k+20$. Arguing
similarly to case (1a), we find

$G\cap[0,N/2]=\\{12j\\}_{0\leq j\leq\lfloor N/24\rfloor}$

and

$(N-G)\cap[0,N]=\\{12j+1\\}_{0\leq j\leq 5k+1}\cup\\{12j+10\\}_{0\leq j\leq
2k-1}\cup\\{12j+5\\}_{0\leq j\leq k-1}.$

By a similar argument, we see that $\tau$ has only trivial maxima.

Case (1c): $N=60k+23$, $G$ is generated by $12,36k+21,48k+28$. We find

$G\cap[0,N/2]=\\{12j\\}_{0\leq j\leq\lfloor N/24\rfloor}$

and

$(N-G)\cap[0,N]=\\{12j+11\\}_{0\leq j\leq 5k+1}\cup\\{12j+2\\}_{0\leq j\leq
2k}\cup\\{12j+7\\}_{0\leq j\leq k-1}.$

We then see that $\tau$ has only trivial maxima.

Case (1d): $N=60k+43$, $G$ is generated by $12,36k+33,48k+44$. We find

$G\cap[0,N/2]=\\{12j\\}_{0\leq j\leq\lfloor N/24\rfloor}$

and

$(N-G)\cap[0,N]=\\{12j+7\\}_{0\leq j\leq 5k+3}\cup\\{12j+10\\}_{0\leq j\leq
2k}\cup\\{12j+11\\}_{0\leq j\leq k-1}.$

We then see that $\tau$ has only trivial maxima.

Case (2a): $N=30k-7$, $G$ is generated by $10,20k+2,50k+5$. We find

$G\cap[0,N/2]=\\{10j\\}_{0\leq j\leq\lfloor N/20\rfloor}$

and

$(N-G)\cap[0,N]=\\{10j+3\\}_{0\leq j\leq 3k-1}\cup\\{10j+1\\}_{0\leq j\leq
k-1}$

We then see that $\tau$ has only trivial maxima.

Case (2b): $N=30k-1$, $G$ is generated by $10,20k+6,50k+15$. We find

$G\cap[0,N/2]=\\{10j\\}_{0\leq j\leq\lfloor N/20\rfloor}$

and

$(N-G)\cap[0,N]=\\{10j+9\\}_{0\leq j\leq 3k-1}\cup\\{10j+3\\}_{0\leq j\leq
k-1}$

We then see that $\tau$ has only trivial maxima.

Case (2c): $N=30k+11$, $G$ is generated by $10,20k+14,50k+35$. We find

$G\cap[0,N/2]=\\{10j\\}_{0\leq j\leq\lfloor N/20\rfloor}$

and

$(N-G)\cap[0,N]=\\{10j+1\\}_{0\leq j\leq 3k+1}\cup\\{10j+7\\}_{0\leq j\leq
k-1}$

We then see that $\tau$ has only trivial maxima.

Case (2d): $N=30k+17$, $G$ is generated by $10,20k+18,50k+45$. We find

$G\cap[0,N/2]=\\{10j\\}_{0\leq j\leq\lfloor N/20\rfloor}$

and

$(N-G)\cap[0,N]=\\{10j+7\\}_{0\leq j\leq 3k+1}\cup\\{10j+9\\}_{0\leq j\leq
k-1}$

We then see that $\tau$ has only trivial maxima.

∎

######  Proposition 4.6.

Let $1<a<b<c$ be pairwise coprime integers and assume $(a,b,c)\neq(2,3,5)$.
Let $Y=-\Sigma(a,b,c)$ and let $\ell^{+}(Y)$ denote the highest non-zero
degree in $HF^{+}_{red}(Y)$. Then $\ell^{+}(Y)=2\delta(Y)-2\min\\{\tau\\}$.
Moreover, any non-zero element in the image of $U\colon HF_{red}^{+}(Y)\to
HF_{red}^{+}(Y)$ has degree strictly less than $\ell^{+}(Y)$.

###### Proof.

Recall that $HF^{+}(Y)$ is isomorphic to $\mathbb{H}(R_{\tau},\chi_{\tau})$ up
to an overall grading shift. The lowest degree in $HF^{+}(-\Sigma(a,b,c))$ is
$2\delta(Y)$ and the lowest degree in $\mathbb{H}(R_{\tau},\chi_{\tau})$ is
$2\min\\{\tau\\}$, hence the grading shift is $2\delta(Y)-2\min\\{\tau\\}$.
The highest non-zero degree in $\mathbb{H}(R_{\tau},\chi_{\tau})$ is easily
seen to be $0$, hence $\ell^{+}(Y)=2\delta(Y)-2\min\\{\tau\\}$.

If $(a,b,c)\neq(2,3,6n\pm 1)$ for any $n$, then all global maxima of $\tau$
are trivial by Proposition 4.5. Then by Proposition 4.4 we have that any non-
zero element in the image of $U\colon HF_{red}^{+}(Y)\to HF_{red}^{+}(Y)$ has
degree strictly less than $\ell^{+}(Y)$.

If $(a,b,c)=(2,3,6n\pm 1)$ for some $n$, then it is easily seen that
$HF_{red}^{+}(Y)$ is concentrated in a single degree and hence $U$ acts
trivially on $HF_{red}^{+}(Y)$.

∎

###  4.2. Computation of $j^{(c)}(T_{a,b})$ and $\theta^{(c)}(T_{a,b})$

Following [6], define $\kappa(a,b,c)$ to be the cardinality of $G\cap[0,N]$.

######  Lemma 4.7.

Let $\tau_{1}(a,b,c)$ denote the number of integers $x,y,z$ with $0<x<a$,
$0<y<b$, $0<z<c$ and $x/a+y/b+z/c<1$. Then $\tau_{1}(a,b,c)=\kappa(a,b,c)$.

###### Proof.

By Remark 4.1, any $x\in G$ with $x\leq N$ has a unique representation
$xbc+yac+zab$ with $0\leq x<a$, $0\leq y<b$, $0\leq z<c$. Thus $\kappa(a,b,c)$
is the number of points $(x,y,z)\in\mathbb{Z}^{3}_{\geq 0}$ such that

$xbc+yac+zab\leq abc-bc-ac-ab.$

From [6, Theorem 1.3] it follows that $N\notin G$, so $\kappa(a,b,c)$ is also
the number of $x,y,z\geq 0$ such that

$xbc+yac+zab<abc-bc-ac-ab.$

Dividing through by $abc$, this is equivalent to

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}<1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$

which can be rewritten as

$\frac{x+1}{a}+\frac{y+1}{b}+\frac{z+1}{q}<1.$

Setting $x^{\prime}=x+1,y^{\prime}=y+1,z^{\prime}=z+1$, we see that
$\kappa(a,b,c)$ is equal to $\tau_{1}(a,b,c)$. ∎

Recall that $\lambda$ denotes the Casson invariant. From [37, §19], we have
$8\lambda(\Sigma(a,b,c))=-(a-1)(b-1)(c-1)+4\tau_{1}(a,b,c)$. Thus Lemma 4.7
gives:

(4.8) $8\lambda(\Sigma(a,b,c))=-(a-1)(b-1)(c-1)+4\kappa(a,b,c).$

From [15], [10], we also have that

$8\lambda(\Sigma(a,b,c))=\sum_{j=1}^{c-1}\sigma_{T_{b,c}}(j/c)=\sigma^{(c)}(T_{a,b}).$

######  Theorem 4.8.

Let $a,b,c>1$ be coprime integers and suppose that $c$ is a prime number. Then

$j^{(c)}(-T_{a,b})=\begin{cases}\kappa(a,b,c)&\text{if }c\text{ is odd},\\\
2\kappa(a,b,c)&\text{if }c=2\end{cases}$

and

$\theta^{(c)}(T_{a,b})=(a-1)(b-1)/2.$

###### Proof.

Let $Y=-\Sigma(a,b,c)$. If $(a,b,c)$ is a permutation of $(2,3,5)$, then
$HF^{+}_{red}(Y)=0$ and $\kappa(2,3,5)=0$. Then from [5, Proposition 3.16], it
follows that $j^{(c)}(-T_{a,b})=0$ and
$\theta^{(c)}(T_{a,b})=-\sigma^{(c)}(T_{a,b})/(c-1)$. But since
$\kappa(2,3,5)=0$, Equation (4.8) gives
$\sigma^{(c)}(T_{a,b})=8\lambda(\Sigma(a,b,c))=-(a-1)(b-1)(c-1)$. Hence
$\theta^{(c)}(T_{a,b})=(a-1)(b-1)/2$.

Henceforth we assume that $(a,b,c)$ is not a permutation of $(2,3,5)$ and
hence $HF^{+}_{red}(Y)\neq 0$. Recall that $\Sigma(a,b,c)$ is the boundary of
a negative definite plumbing [32] whose plumbing graph has only one bad vertex
in the sense of [34]. Then from [34, Corollary 1.4], we have that $HF^{+}(Y)$
is concentrated in even degrees. As explained in [5, §7] we have
$\Sigma(a,b,c)=\Sigma_{c}(T_{a,b})$ and the generator of the
$\mathbb{Z}_{c}$-action on $\Sigma(a,b,c)$ is isotopic to the identity.
Therefore in the spectral sequence $\\{E_{r}^{p,q},d_{r}\\}$ for the
equivariant Seiberg–Witten–Floer cohomology of $Y$, we have
$E_{2}^{0,q}=H^{0}(\mathbb{Z}_{c};HSW^{q}(Y))\cong HSW^{q}(Y)$. Furthermore,
the graded roots algorithm implies that $HF_{red}^{+}(Y)$ is concentrated in
degrees $d(Y)$ and above. It follows that there can be no differentials in the
spectral sequence and hence $\delta_{\mathbb{Z}_{c},S^{0}}(Y)=\delta(Y)$.
Recall that $\delta^{(c)}_{j}(K)=-\sigma^{(c)}(K)/2$ for all sufficiently
large $j$. Then since $Y=-\Sigma(a,b,c)=\Sigma_{c}(-T_{a,b})$, it follows that

$\delta_{\mathbb{Z}_{c},S^{j}}(Y)=-\sigma^{(c)}(-T_{a,b})/8=\lambda(\Sigma(a,b,c))=-\lambda(Y)$

for sufficiently large $j$. But since $HF^{*}_{red}(Y)$ is concentrated in
even degrees, we also have
$dim_{\mathbb{F}}(HF^{*}_{red}(Y))=\delta(Y)+\lambda(Y)$. Hence for large
enough $j$, we have
$\delta_{\mathbb{Z}_{c},S^{0}}(Y)-\delta_{\mathbb{Z}_{c},S^{j}}(Y)=\delta(Y)+\lambda(Y)=dim_{\mathbb{F}}(HF^{*}_{red}(Y))$.
From Proposition 4.6, it follows that any element in the image of $U\colon
HF_{red}^{+}(Y,\mathfrak{s})\to HF_{red}^{+}(Y,\mathfrak{s})$ has degree
strictly less than $\ell^{+}(Y,\mathfrak{s})$. Thus conditions (1)-(4) of
Proposition 3.5 are met. Therefore (since $c$ is an odd prime and $Y$ is an
integral homology sphere) we have

$j^{\prime}(Y)=\ell^{+}(Y)/2-\delta_{\mathbb{Z}_{c},S^{0}}(Y)+\delta(Y)+\lambda(Y)=\ell^{+}(Y)/2+\lambda(Y),$

where $j^{\prime}(Y)=j^{(c)}(Y)$ if $c$ is odd and
$j^{\prime}(Y)=j^{(2)}(Y)/2$ if $c=2$. Moreover, Proposition 4.6 also gives
$\ell^{+}(Y)=2(\delta(Y)-\min(\tau))$. From [6, §5], we also have that
$-\min(\tau)+\delta(Y)=\kappa(a,b,c)-\lambda(Y)$. Thus
$\ell^{+}(Y)=2(\kappa-\lambda(Y))$. Hence

$j^{\prime}(-T_{a,b})=j^{(c)}(Y)=\ell^{+}(Y)/2+\lambda(Y)=\kappa(a,b,c),$

which gives $j^{(c)}(-T_{a,b})=\kappa(a,b,c)$ if $c$ is odd and
$j^{(2)}(-T_{a,b})=2\kappa(a,b,c)$ if $c=2$. We also have
$\sigma^{(c)}(T_{a,b})=-8\lambda(Y)=8\lambda(\Sigma(a,b,c))$ and thus we have

$\displaystyle\theta^{(c)}(T_{a,b})$
$\displaystyle=\max\left\\{0,\frac{2j^{\prime}(Y)}{(c-1)}-\frac{\sigma^{(c)}(T_{a,b})}{2(c-1)}\right\\}$
$\displaystyle=\max\left\\{0,\frac{2\kappa(a,b,c)-4\lambda(\Sigma(a,b,c))}{(c-1)}\right\\}$
$\displaystyle=\frac{1}{2}(a-1)(b-1)$

where the last line follows from Equation (4.8). ∎

##  5\. Branched covers

Let $Y=\Sigma(a^{\prime}_{1},a^{\prime}_{2},\dots,a^{\prime}_{r})$ be a
Brieskorn homology sphere and $p$ a prime such that $p$ divides
$a^{\prime}_{1}a^{\prime}_{2}\cdots a^{\prime}_{r}$. Without loss of
generality we may assume $p$ divides $a^{\prime}_{1}$. We set
$a_{1}=a^{\prime}_{1}/p$ and $a_{j}=a^{\prime}_{j}$ for $j>1$. So
$Y=\Sigma(pa_{1},a_{2},\dots,a_{r})$. Then $\mathbb{Z}_{p}$ acts on $Y$ with
quotient space $Y_{0}=Y/\mathbb{Z}_{p}=\Sigma(a_{1},\dots,a_{r})$ and the
quotient map $Y\to Y_{0}$ is a $p$-fold cyclic branched cover.

######  Theorem 5.1.

We have that

$\delta(-Y)-\delta^{(p)}_{\infty}(-Y)\geq{\rm rk}(HF_{red}^{+}(Y))-p\,{\rm
rk}(HF_{red}^{+}(Y_{0})).$

###### Proof.

Let $W_{0}$ be the negative definite plumbing bounded by $Y_{0}$. Since
$Y_{0}$ is an integral homology sphere it follows that
$H_{1}(W_{0};\mathbb{Z})=0$. Let $k\subset Y_{0}$ denote the branch locus of
$Y\to Y_{0}$. Then $k$ is a knot in $Y_{0}$. Let $\Sigma\subset W_{0}$ be the
pushoff of a Seifert surface for $k$, so $\Sigma$ is a properly embedded
surface in $W_{0}$ which meets $\partial W_{0}=Y_{0}$ in $k$. Let $W\to W_{0}$
be the $p$-fold cyclic cover of $W_{0}$ branched over $\Sigma$. Then $W$ has
boundary $Y$ and the $\mathbb{Z}_{p}$-action on $Y$ extends to $W$. By [4,
Proposition 2.5], for any characteristic $c\in H^{2}(W_{0};\mathbb{Z})$ there
exists a $\mathbb{Z}_{p}$-invariant spinc-structure $\mathfrak{s}$ on $W$ such
that $c_{1}(\mathfrak{s})=\pi^{*}(c)$ in $H^{2}(W;\mathbb{Q})$. Now we apply
the equivariant Froyshov inequality [5, Theorem 5.3] to $W$ giving
$\delta_{\infty}(-Y)+\delta(W,\mathfrak{s})\leq 0$, where

$\displaystyle\delta(W,\mathfrak{s})$
$\displaystyle=\frac{c_{1}(\mathfrak{s})^{2}-\sigma(W)}{8}$
$\displaystyle=\frac{pc^{2}-\sigma(W)}{8}$
$\displaystyle=p\left(\frac{c^{2}-\sigma(W_{0})}{8}\right)+\frac{p\sigma(W_{0})-\sigma(W)}{8}.$

The maximum of $(c^{2}-\sigma(W_{0}))/8$ over all characteristics of
$H^{2}(W_{0};\mathbb{Z})$ equals $\delta(Y_{0})$ [31, Theorem 8.3]. Hence we
obtain

$\delta^{(p)}_{\infty}(-Y)+p\delta(Y_{0})+\frac{p\sigma(W_{0})-\sigma(W)}{8}\leq
0,$

which we may rewrite as

$\delta(-Y)-\delta^{(p)}_{\infty}(-Y)\geq-(\delta(Y)-p\delta(Y_{0}))+\frac{p\sigma(W_{0})-\sigma(W)}{8}.$

Next, we claim that $(p\sigma(W_{0})-\sigma(W))/8=p\lambda(Y_{0})-\lambda(Y)$,
where $\lambda(Y_{0}),\lambda(Y)$ are the Casson invariants of $Y_{0}$ and
$Y$. Assuming this claim for the moment, our inquality becomes

$\delta(-Y)-\delta^{(p)}_{\infty}(-Y)\geq-(\delta(Y)+\lambda(Y))+p(\delta(Y_{0})+\lambda(Y_{0})).$

But since $HF_{red}^{+}(Y)$ and $HF_{red}^{+}(Y_{0})$ are concentrated in odd
degrees, [35, Theorem 1.3] gives

${\rm rk}(HF_{red}^{+}(Y))=-\delta(Y)-\lambda(Y),\quad{\rm
rk}(HF_{red}^{+}(Y_{0}))=-\delta(Y_{0})-\lambda(Y_{0}),$

and hence we obtain

$\delta(-Y)-\delta^{(p)}_{\infty}(-Y)\geq{\rm rk}(HF_{red}(-Y))-p\,{\rm
rk}(HF_{red}(-Y_{0})).$

It remains to prove the claim that
$(p\sigma(W_{0})-\sigma(W))/8=p\lambda(Y_{0})-\lambda(Y)$. From [11, Theorem
2], we have that

$\lambda^{\mathbb{Z}/p}(Y)-p\lambda(Y_{0})=\frac{\sigma(W)-p\sigma(W_{0})}{8}$

where $\lambda^{\mathbb{Z}/p}(Y)$ is the equivariant Casson invariant of $Y$
with respect to the $\mathbb{Z}_{p}$-action [11]. Furthermore [11, Theorem 3]
implies that $\lambda^{\mathbb{Z}/p}(Y)=\lambda(Y)$, because $k\subset
Y_{0}=\Sigma(a_{1},\dots,a_{r})$ is a fibre of the Seifert fibration on
$Y_{0}$, so it is a graph knot in the terminology of [11, §5]. This proves the
claim that $\lambda(Y)-p\lambda(Y_{0})=(\sigma(W)-p\sigma(W_{0}))/8$. ∎

Let $Y,Y_{0}$ be as in Theorem 5.1. Then by [17, Theorem 1.1], we have an
inequality ${\rm rk}(HF_{red}^{+}(Y))\geq p\,{\rm rk}(HF_{red}^{+}(Y_{0}))$.
Combined with Theorem 5.1, this gives:

$p\,{\rm rk}(HF_{red}^{+}(Y_{0}))\leq{\rm rk}(HF_{red}^{+}(Y))\leq p\,{\rm
rk}(HF_{red}^{+}(Y_{0}))+(\delta(-Y)-\delta^{(p)}_{\infty}(-Y)).$

In particular, the equality $\delta^{(p)}_{\infty}(-Y)=\delta(-Y)$ can only
happen if ${\rm rk}(HF_{red}^{+}(Y))=p\,{\rm rk}(HF_{red}^{+}(Y_{0}))$. Hence
we obtain:

######  Corollary 5.2.

Let $Y,Y_{0}$ be as in Theorem 5.1. If ${\rm rk}(HF_{red}^{+}(Y))>p\,{\rm
rk}(HF_{red}^{+}(Y_{0}))$ then $\delta^{(p)}_{\infty}(-Y)<\delta(-Y)$.

##  6\. Free actions

Suppose $p$ does not divide $a_{1}a_{2}\cdots a_{r}$. Then the restriction of
the $S^{1}$-action on $Y=\Sigma(a_{1},\dots,a_{r})$ acts freely on $Y$. Let
$Y_{0}=Y/\mathbb{Z}_{p}$ be the quotient. Then $Y$ is a rational homology
sphere. In fact if $Y=M(e_{0},(a_{1},b_{1}),\dots,(a_{r},b_{r}))$, then it is
easily seen that $Y_{0}=M(pe_{0},(a_{1},pb_{1}),\dots,(a_{r},pb_{r}))$. It is
easy to see that $H_{1}(Y;\mathbb{Z})\cong\mathbb{Z}_{p}$ and thus $Y_{0}$ has
exactly $p$ spinc-structures. Further, the pullback to $Y$ of any spinc-
structure on $Y_{0}$ must coincide with the unique spinc-structure on $Y$.

######  Theorem 6.1.

For any spinc-structure $\mathfrak{s}_{0}$ on $Y_{0}$, we have

$\delta^{(p)}_{\infty}(Y)-\delta(Y)={\rm rk}(HF_{red}^{+}(Y))-{\rm
rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0})).$

###### Proof.

Set $G=\mathbb{Z}_{p}$ and $H^{*}_{G}=H^{*}_{G}(pt;\mathbb{F})$. By choosing a
$G$-invariant metric on $Y$, we may construct a $S^{1}\times G$-equivariant
Conley index for $(Y,\mathfrak{s})$, which we denote by $I(Y,\mathfrak{s})$.
As shown in [24, §3], the Conley index $I(Y_{0},\mathfrak{s}_{0})$ for
$(Y_{0},\mathfrak{s}_{0})$ can be identified with the $\mathbb{Z}_{p}$-fixed
point set $I(Y,\mathfrak{s})^{\mathbb{Z}_{p}}$ of $I(Y,\mathfrak{s})$. Thus we
have isomorphisms

$\displaystyle\widetilde{H}^{*}_{S^{1}\times
G}(I(Y,\mathfrak{s})^{\mathbb{Z}_{p}};\mathbb{F})$
$\displaystyle\cong\widetilde{H}^{*}_{S^{1}\times
G}(I(Y_{0},\mathfrak{s}_{0});\mathbb{F})$
$\displaystyle\cong\widetilde{H}^{*}_{S^{1}}(I(Y_{0},\mathfrak{s}_{0}),\mathbb{F})\otimes_{\mathbb{F}}H^{*}_{G}$
$\displaystyle\cong
HSW^{*}(Y_{0},\mathfrak{s}_{0})\otimes_{\mathbb{F}}H^{*}_{G}.$

It should be noted that the above isomorphisms only preserve relative
gradings. Recall that $H^{*}_{G}\cong\mathbb{F}[Q]$, $deg(Q)=1$ if $p=2$ and
$H^{*}_{G}\cong\mathbb{F}[R,S]/(R^{2})$, $deg(R)=1$, $deg(S)=2$ if $p$ is odd.
In the case $p=2$, define $S=Q^{2}$. Let $\mathcal{S}=\\{1,S,S^{2},\dots\\}$.
Then $\mathcal{S}$ is a multiplicative subset of $H^{*}_{S^{1}\times
G}=H^{*}_{S^{1}\times G}(pt;\mathbb{F})\cong H^{*}_{G}[U]$. The localisation
theorem in equivariant cohomology [12, III, Theorem 3.8] applied to the pair
$(I(Y,\mathfrak{s}),I(Y,\mathfrak{s})^{\mathbb{Z}_{p}})$ and multiplicative
set $\mathcal{S}$ implies that the inclusion
$I(Y,\mathfrak{s})^{\mathbb{Z}_{p}}\to I(Y,\mathfrak{s})$ induces an
isomorphism

(6.1) $S^{-1}\widetilde{H}^{*}_{S^{1}\times
G}(I(Y,\mathfrak{s});\mathbb{F})\to S^{-1}\widetilde{H}^{*}_{S^{1}\times
G}(I(Y,\mathfrak{s})^{\mathbb{Z}_{p}};\mathbb{F}).$

Since $\widetilde{H}^{*}_{S^{1}\times G}(I(Y,\mathfrak{s});\mathbb{F})\cong
HSW^{*}_{\mathbb{Z}_{p}}$ and as we have shown above,
$\widetilde{H}^{*}_{S^{1}\times
G}(I(Y,\mathfrak{s})^{\mathbb{Z}_{p}};\mathbb{F})\cong
HSW^{*}(Y_{0},\mathfrak{s}_{0})\otimes_{\mathbb{F}}H^{*}_{G}$, we get an
isomorphism

$S^{-1}HSW^{*}_{\mathbb{Z}_{p}}(Y,\mathfrak{s})\cong
HSW^{*}(Y_{0},\mathfrak{s}_{0})\otimes_{\mathbb{F}}S^{-1}H^{*}_{G}.$

This is an isomorphism of relatively graded $H^{*}_{S^{1}\times G}$-modules.

For the rest of the proof we restrict to the case that $p$ is odd. The proof
in the case that $p=2$ is similar. We have

$HSW^{*}(Y_{0},\mathfrak{s}_{0})\cong
HF^{+}_{*}(Y_{0},\mathfrak{s}_{0})\cong\mathbb{F}[U]\theta_{0}\oplus
HF_{red}^{+}(Y_{0},\mathfrak{s}_{0})$

for some $\theta_{0}$. Combined with (6.1), we have an isomorphism

(6.2)
$S^{-1}HSW^{*}_{\mathbb{Z}_{p}}(Y,\mathfrak{s})\cong\frac{\mathbb{F}[U,R,S,S^{-1}]}{(R^{2})}\theta_{0}\oplus\frac{HF_{red}^{+}(Y_{0},\mathfrak{s}_{0})[R,S,S^{-1}]}{(R^{2})}.$

On the other hand, the proof of Proposition 3.6 gives an isomorphism

$HSW^{*}_{\mathbb{Z}_{p}}(Y,\mathfrak{s})\cong
E_{\infty}(Y)\cong\frac{\mathbb{F}[U,R,S]}{(R^{2})}\theta\oplus
E_{\infty}(Y)_{red}$

under which the $\mathbb{F}[U]$-module structure is given by an endomorphism
of the form $\widehat{U}=U_{(0,2)}+U_{(1,1)}+\cdots$ with $U_{(0,2)}=U$.
Localising with respect to $S$ gives an isomorphism

$S^{-1}HSW^{*}_{\mathbb{Z}_{p}}(Y,\mathfrak{s})\cong
E_{\infty}(Y)\cong\frac{\mathbb{F}[U,R,S,S^{-1}]}{(R^{2})}\theta\oplus
S^{-1}E_{\infty}(Y)_{red}.$

Define $W\subseteq S^{-1}HSW^{*}_{\mathbb{Z}_{p}}(Y,\mathfrak{s}))$ to be the
set of $x\in S^{-1}HSW^{*}_{\mathbb{Z}_{p}}(Y,\mathfrak{s}))$ such that for
each $j\geq 0$, there exists a $k\geq 0$ for which
$U^{k}x\in\mathcal{F}_{j}(Y)$. The proof of Proposition 3.6 demonstrates that
$W\cong S^{-1}E_{\infty}(Y)_{red}$. On the other hand, the isomorphism (6.2)
clearly shows that $W\cong
HF_{red}^{+}(Y_{0},\mathfrak{s}_{0})[R,S,S^{-1}]/(R^{2})$. Combining these, we
have an isomorphism

$S^{-1}E_{\infty}(Y)_{red}\cong
HF_{red}^{+}(Y_{0},\mathfrak{s}_{0})[R,S,S^{-1}]/(R^{2}).$

In any fixed degree $j$, the rank of
$(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0})[R,S,S^{-1}]/(R^{2}))^{j}$ is equal to
${\rm rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0}))$, hence the same is true of
$S^{-1}E_{\infty}(Y)_{red}$. From [5, Lemma 5.7], we have that $S\colon
E_{r}^{p,q}(Y)\to E_{r}^{p+2,q}(Y)$ is an isomorphism for all large enough
$p$. Hence for large enough $p$, $E_{r}^{2p,*}(Y)$ is independent of $p$ and
we denote the resulting group by $M_{r}^{*}$. We similarly define
$M_{\infty}^{*}$. Clearly the rank of $(S^{-1}E_{\infty}(Y)_{red})^{j}$ for
any $j$ equals ${\rm rk}(M_{\infty})_{red}$. So we have proven that

${\rm rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0}))={\rm rk}(M_{\infty})_{red}.$

Set $s_{r}={\rm rk}(M_{r})_{red}$. It follows from [5, Lemma 5.8] that
$(M_{r+1})_{red}$ is a subquotient of $(M_{r})_{red}$. Hence the sequence
$s_{2},s_{3},\dots$ is decreasing and equals ${\rm rk}(M_{\infty})_{red}$ for
sufficiently large $r$. Furthermore, $M_{2}\cong HF^{+}(Y)$, so $s_{2}={\rm
rk}(HF_{red}^{+}(Y))$.

Recall from the proof of Proposition 3.6 that

$E_{r}(Y)\cong\frac{\mathbb{F}[U,R,S]}{(R^{2})}U^{m_{r}}\theta\oplus
E_{r}(Y)_{red}$

for some increasing sequence $0=m_{0}\leq m_{1}\leq\dots$. Suppose
$m_{r+1}>m_{r}$. Then $d_{r}(U^{m_{r}+j}\theta)\neq 0$ for $0\leq j\leq
m_{r+1}-m_{r}$. Notice that $U^{m_{r}+j}\theta$ has bi-degree $(0,a)$ where
$a=2m_{r}+2j+d(Y)$ is even, hence $d_{r}(U^{m_{r}+j}\theta)$ has bi-degree
$(r,a+(1-r))$. But all elements of $E_{r}(Y)_{red}$ have bi-degree $(u,v)$
with $v$ odd (because $E_{r}(Y)_{red}$ is a subquotient of
$HF_{red}^{+}(Y)[R,S]/(R^{2})$ and $HF_{red}^{+}(Y)$ is concentrated in odd
degrees). Hence $a+(1-r)$ must be odd, which means that $r$ is even. So we can
regard $d_{r}$ as mapping into $(M_{r})_{red}$. Hence the rank of
$(M_{r+1})_{red}$ is at least $m_{r+1}-m_{r}$ less than the rank of
$(M_{r})_{red}$, that is,

(6.3) $s_{r}-s_{r+1}\geq m_{r+1}-m_{r}.$

We claim that this inequality is actually an equality. Equivalently $M_{r+1}$
is the quotient of $M_{r}$ by the span of $\\{d_{r}(U^{m_{r}+j})\\}$, $0\leq
j\leq m_{r+1}-m_{r}$. This is also equivalent to saying that $d_{r}(x)=0$ for
all $r\geq 2$ and all $x\in E_{r}(Y)_{red}$. Consider a non-zero homogeneous
element $x\in E_{r}(Y)_{red}^{a,b}$. Since $E_{r}(Y)_{red}$ is a subquotient
of $HF_{red}^{+}(Y)[R,S]/(R^{2})$ and $HF_{red}^{+}(Y)$ is concentrated in odd
degrees, we have that $b$ is odd. Then $d_{r}(x)$ has bi-degree $(a+r,b+1-r)$.
If $d_{r}(x)\neq 0$, then $b+1-r$ must be off and so $r$ must be even. Thus in
order to prove the claim, it is sufficient to show that $d_{r}=0$ for all even
$r$.

Consider the equivariant Seiberg–Witten–Floer cohomology of $Y$ with integral
coefficients $HSW^{*}_{\mathbb{Z}_{p}}(Y;\mathbb{Z})$. This is a module over
$H^{*}_{\mathbb{Z}_{p}}(pt;\mathbb{Z})\cong\mathbb{Z}[S]/(pS)$ where
$deg(S)=2$. The key point to observe here is that
$H^{*}_{\mathbb{Z}_{p}}(pt;\mathbb{Z})$ is concentrated in even degrees. There
is a spectral sequence $(E_{r}^{p,q}(Y;\mathbb{Z}),d_{r})$ and a filtration
$\\{\mathcal{F}_{j}(Y;\mathbb{Z})\\}$ such that $E_{\infty}(Y;\mathbb{Z})$ is
the associated graded module of the filtration. The mod $p$ reduction map
$\mathbb{Z}\to\mathbb{Z}_{p}=\mathbb{F}$ induces a morphism of
$HSW^{*}(Y;\mathbb{Z})\to HSW^{*}(Y;\mathbb{F})$. There is also a reduction
map of the equivariant Floer cohomologies, the filtrations and the spectral
sequences. The results of [34] also hold with integer coefficients so

$HF^{+}(Y;\mathbb{Z})\cong\mathbb{Z}[U]_{d(Y)}\oplus
HF_{red}^{+}(Y;\mathbb{Z})$

and $HF_{red}^{+}(Y;\mathbb{Z})$ is concentrated in odd degrees. By the
universal coefficient theorem $HF_{red}^{+}(Y;\mathbb{F})\cong
HF_{red}^{+}(Y;\mathbb{Z})\otimes_{\mathbb{Z}}\mathbb{F}$. Furthermore, we
have that

$\displaystyle E_{2}(Y;\mathbb{Z})$ $\displaystyle\cong
H^{*}(\mathbb{Z}_{p};HSW^{*}(Y;\mathbb{Z}))$
$\displaystyle\cong\frac{\mathbb{Z}[U,S]}{(pS)}\theta\oplus\frac{HF_{red}^{+}(Y;\mathbb{Z})[S]}{(pS)}.$

Clearly every homogeneous element in $\mathbb{Z}[U,S]/(pS)\theta$ has bi-
degree $(a,b)$ where $a$ and $b$ are even and every homogeneous element in
$HF_{red}^{+}(Y;\mathbb{Z})[S]/(pS)$ has bi-degree $(a,b)$ where $a$ is even
and $b$ is odd. From this and the fact that the image of $d_{r}\colon
E_{r}(Y;\mathbb{Z})\to E_{r}(Y;\mathbb{Z})$ is contained in
$E_{r}(Y;\mathbb{Z})_{red}$ it follows easily that $d_{r}\colon
E_{r}(Y;\mathbb{Z})\to E_{r}(Y;\mathbb{Z})$ is zero for even $r$.

Now using induction on $r$ one shows the following properties hold: (1) any
$x\in E_{r}^{a,b}(Y)$ with $a$ even is in the image of the reduction map
$E_{r}^{a,b}(Y;\mathbb{Z})\to E_{r}^{a,b}(Y)$, (2) every $x\in E_{r}^{a,b}(Y)$
with $a$ odd is of the form $x=Ry$ where $y$ is in the image of the reduction
map $E_{r}^{a,b}(Y;\mathbb{Z})\to E_{r}^{a,b}(Y)$ and (3) $d_{r}\colon
E_{r}(Y)\to E_{r}(Y)$ is zero for even $r$. In particular, this proves the
claim that $d_{r}\colon E_{r}(Y)\to E_{r}(Y)$ is zero for all even $r$ and
hence the inequality (6.3) is actually an equality:

(6.4) $s_{r}-s_{r+1}=m_{r+1}-m_{r}.$

Let $s={\rm rk}((M_{\infty})_{red})={\rm
rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0}))$ and let $m=\lim_{r\to\infty}m_{r}$.
Summing (6.4) from $r=2$ to infinity gives $s_{2}-s=m-m_{2}$. But $m_{2}=0$
and $s_{2}={\rm rk}(HF_{red}^{+}(Y))$, so we get $m={\rm
rk}(HF_{red}^{+}(Y))-{\rm rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0}))$. But
recall from the proof of Proposition 3.6 that
$\delta_{\infty}^{(p)}(Y)=\delta(Y)+m$, hence we get

$\delta_{\infty}^{(p)}(Y)=\delta(Y)+{\rm rk}(HF_{red}^{+}(Y))-{\rm
rk}(HF_{red}^{+}(Y_{0},\mathfrak{s}_{0})).$

∎

We will take $\mathfrak{s}_{0}$ to be the restriction to $Y_{0}$ of the
canonical spinc-structure on the negative definite plumbing which $Y_{0}$
bounds. With this choice of spinc-structure, the computation of
$HF^{+}(-Y_{0},\mathfrak{s}_{0})$ is easily obtained through the graded roots
algorithm [31]. Let $\Delta_{p}(n)$ denote the delta function for
$(Y_{0},\mathfrak{s}_{0})$. Then (since $e_{0}<0$) we have

(6.5)
$\Delta_{p}(n)=1+npe_{0}-\sum_{j=1}^{r}\left\lceil\frac{npb_{j}}{a_{j}}\right\rceil$

######  Proposition 6.2.

We have that $\Delta_{p}(n)\geq 0$ for $n>N/p$. In particular, if $p>N$, then
$HF_{red}^{+}(-Y_{0},\mathfrak{s}_{0})=0$.

###### Proof.

Following [6, §3], we let $f(x)=\lceil x\rceil-x$. Then from Equations (3.3)
and (6.5), we see that

$\Delta_{p}(n)=1+\frac{np}{a_{1}\cdots
a_{r}}-\sum_{j=1}^{r}f\left(\frac{npb_{j}}{a_{j}}\right).$

For any integer $m$ we have $f(m/a_{j})\leq 1-1/a_{j}$, hence

(6.6) $\Delta_{p}(n)\geq 1+\frac{np}{a_{1}\cdots
a_{r}}-r+\sum_{j=1}^{r}\frac{1}{a_{j}}.$

Now suppose that $pn>N$. Thus

$pn>a_{1}\cdots a_{n}\left((r-2)-\sum_{j=1}^{r}\frac{1}{a_{j}}\right).$

Re-arranging, we get

$(1-r)+\frac{np}{a_{1}\cdots a_{r}}+\sum_{j=1}^{r}\frac{1}{a_{j}}>-1.$

Combined with (6.6), we get $\Delta_{p}(n)>-1$. But $\Delta_{p}$ is integer-
valued, so $\Delta_{p}(n)\geq 0$. ∎

######  Lemma 6.3.

We have that ${\rm rk}(HF_{red}^{+}(-Y))>{\rm
rk}(HF_{red}^{+}(-Y_{0},\mathfrak{s}_{0}))$ unless $r=3$ and up to reordering
$(a_{1},a_{2},a_{3})$ is one of the following:

$(2,3,5),(2,3,7),(2,3,11),(2,3,13),(2,3,17),(2,5,7),(2,5,9),(3,4,5),(3,4,7).$

###### Proof.

We assume that $Y$ is not $\Sigma(2,3,5)$ and hence $N>0$. Let $N_{p}=\lfloor
N/p\rfloor$. By Proposition 6.2 we only need to consider $\Delta_{p}(x)$ for
$x\in[0,N_{p}]$.

In the proof we make use of abstract delta sequences [17, §3]. Let
$P_{p}=G\cap[0,N_{p}]$. Consider the map $\phi\colon P_{p}\to[0,N]$ given by
$\phi(n)=np$. From Equation (6.5) it is immediately clear that
$\Delta(np)=\Delta_{p}(n)$, hence $\phi$ identifies the delta sequence
$(P_{p},\Delta_{p})$ with a subsequence of the delta sequence $(P,\Delta)$.
This gives the inequality ${\rm rk}(HF_{red}^{+}(-Y))\geq{\rm
rk}(HF_{red}^{+}(-Y_{0},\mathfrak{s}_{0})$ by [17, Proposition 3.5]. To refine
this to a strict inequality we will show that (except for the cases listed in
the statement of the lemma) there exists an $x\in P\setminus\phi(P_{p})$ such
that $2x\leq N$. In this case $\\{x,N-x\\}$ defines a delta subsequence of
$(P,\Delta)$ disjoint from $\phi(P_{p})$ and then by [17, Corollary 3.12] we
obtain a strict inequality ${\rm rk}(HF_{red}^{+}(-Y))>{\rm
rk}(HF_{red}^{+}(-Y_{0},\mathfrak{s}_{0})$, as the module $\mathbb{H}_{red}$
corresponding to $\\{x,N-x\\}$ has rank $1$.

We seek an $x\in P\setminus\phi(P_{p})$ such that $2x\leq N$. Equivalently
$x\in P$, $p$ does not divide $x$ and $2x\leq N$. Reorder $a_{1},\dots,a_{r}$
such that $a_{1}>a_{2}>\cdots>a_{r}$. Consider $x=a_{2}a_{3}\cdots a_{r}$. It
follows that $p$ does not divide $x$ as $p$ is coprime to $a_{2},\dots,a_{r}$
and we also have that $x\in P$. Hence we obtain a strict rank inequality
provided that $2x\leq N$. Suppose on the contrary that $2x>N$, that is,

$2a_{2}\cdots a_{r}>a_{1}\cdots
a_{r}\left((r-2)-\sum_{j=1}^{r}\frac{1}{a_{j}}\right).$

Re-arranging, this is equivalent to

(6.7)
$\frac{3}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{r}}\geq(r-2).$

Since $a_{j}\geq 2$ for all $j$ and at most one can be equal to $2$, we get
that

$\frac{3}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{r}}<\frac{3}{a_{1}}+\frac{(r-1)}{2},$

hence

$\frac{3}{a_{1}}+\frac{(r-1)}{2}>(r-2).$

From $a_{1}>a_{2}>\cdots>a_{r}\geq 2$, we see that $a_{r}\geq r+1$, hence

$\frac{3}{r+1}+\frac{(r-1)}{2}>(r-2)$

which simplifies to $(r+1)(r-3)<6$. If $r\geq 5$, then $(r+1)(r-3)\geq 12$, so
this leaves only the cases $r=3,4$.

If $r=4$, then since $a_{4}\geq 2$, $a_{3}\geq 3$, $a_{2}\geq 4$, $a_{1}\geq
5$, we have

$\frac{3}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}\leq\frac{3}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}=\frac{101}{60}<2.$

Hence (6.7) can not be satisfied with $r=4$.

Now suppose that $r=3$. If $a_{3}\geq 4$, then $a_{2}\geq 5$, $a_{1}\geq 6$
and so

$\frac{3}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}\leq\frac{3}{6}+\frac{1}{5}+\frac{1}{4}=\frac{19}{20}<1$

and so (6.7) can only be satisfied if $a_{3}=2$ or $3$.

If $a_{3}=3$ and $a_{4}\geq 5$, then $a_{1}\geq 7$ (since $a_{1}$ must be
coprime to $a_{3}=3$) and so

$\frac{3}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}\leq\frac{3}{7}+\frac{1}{5}+\frac{1}{3}=\frac{101}{105}<1.$

So (6.7) implies that $a_{4}=4$. Then since $3/8+1/4+1/3=23/24<1$, it follows
that $a_{1}<8$. Hence if $a_{3}=3$, then $(a_{1},a_{2},a_{3})=(5,4,3)$ or
$(7,4,3)$.

Lastly, suppose that $a_{3}=2$. If $a_{2}>5$, then $a_{2}\geq 7$ and
$a_{1}\geq 9$ (since $a_{1},a_{2}$ must be coprime to $a_{3}$). So

$\frac{3}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}\leq\frac{3}{9}+\frac{1}{7}+\frac{1}{2}=\frac{41}{42}<1.$

So $a_{2}\leq 5$, hence $a_{2}=3$ or $5$.

If $a_{3}=2$ and $a_{2}=5$, then (6.7) implies $a_{1}\leq 10$, hence
$(a_{1},a_{2},a_{3})=(7,5,2)$ or $(9,5,2)$.

If $a_{3}=2$ and $a_{2}=3$, then (6.7) implies $a_{1}\leq 18$, hence
$(a_{1},a_{2},a_{3})$ is one of $(5,3,2),(7,3,2),(11,3,2),(13,3,2),(17,3,2)$.
∎

$(a_{1},a_{2},a_{3})$ | $N$ | ${\rm rk}(HF_{red}^{+}(-Y))$ | $p$ | ${\rm rk}(HF_{red}^{+}(-Y_{0}))$
---|---|---|---|---
$(2,3,7)$ | $1$ | $1$ | none |
$(2,3,11)$ | $5$ | $1$ | $5$ | $1$
$(2,3,13)$ | $7$ | $2$ | $5$ | $0$
|  |  | $7$ | $1$
$(2,3,17)$ | $11$ | $2$ | $5$ | $1$
|  |  | $7$ | $0$
|  |  | $11$ | $1$
$(2,5,7)$ | $11$ | $2$ | $3$ | $0$
|  |  | $11$ | $1$
$(2,5,9)$ | $17$ | $2$ | $7$ | $1$
|  |  | $11$ | $0$
|  |  | $13$ | $0$
|  |  | $17$ | $1$
$(3,4,5)$ | $13$ | $2$ | $2$ | $0$
|  |  | $7$ | $0$
|  |  | $11$ | $0$
|  |  | $13$ | $1$
$(3,4,7)$ | $23$ | $2$ | $2$ | $1$
|  |  | $5$ | $0$
|  |  | $11$ | $1$
|  |  | $13$ | $0$
|  |  | $17$ | $0$
|  |  | $19$ | $0$
|  |  | $23$ | $1$
Table 1.

######  Theorem 6.4.

We have that ${\rm rk}(HF_{red}^{+}(-Y))>{\rm
rk}(HF_{red}^{+}(-Y_{0},\mathfrak{s}_{0}))$ except in the following cases:

  * (1)

$Y=\Sigma(2,3,5)$ and $p$ is any prime.

  * (2)

$Y=\Sigma(2,3,11)$ and $p=5$.

In case (1) we have ${\rm rk}(HF_{red}^{+}(-Y))={\rm
rk}(HF_{red}^{+}(-Y_{0},\mathfrak{s}_{0}))=0$ and in case (2) we have ${\rm
rk}(HF_{red}^{+}(-Y))={\rm rk}(HF_{red}^{+}(-Y_{0},\mathfrak{s}_{0}))=1$.

###### Proof.

By Lemma 6.3, we only need to consider the case that $r=3$ and up to
reordering $(a_{1},a_{2},a_{3})$ is one of:

$(2,3,5),(2,3,7),(2,3,11),(2,3,13),(2,3,17),(2,5,7),(2,5,9),(3,4,5),(3,4,7).$

In the case $Y=\Sigma(2,3,5)$ we have that ${\rm rk}(HF_{red}^{+}(-Y))=0$ and
$N<0$. Hence we also have ${\rm rk}(HF_{red}^{+}(-Y_{0}))=0$ by Proposition
6.2.

In the remaining cases we have ${\rm rk}(HF_{red}^{+}(-Y))>0$ and $N>0$. Hence
if $p>N$ then ${\rm rk}(HF_{red}^{+}(-Y))>{\rm rk}(HF_{red}^{+}(-Y_{0}))=0$,
by Proposition 6.2. So we only need to consider primes such that $p\leq N$ and
coprime to $a_{1},a_{2},a_{3}$. This leaves only finitely many cases of
$4$-tuples $\\{(a_{1},a_{2},a_{3},p)\\}$ to consider. We check each of these
cases by directly computing the ranks of $HF_{red}^{+}(-Y)$ and
$HF_{red}^{+}(-Y_{0})$. The results are shown in Table 1. By inspection we see
that ${\rm rk}(HF_{red}^{+}(-Y))>{\rm
rk}(HF_{red}^{+}(-Y_{0},\mathfrak{s}_{0}))$ in all cases except for
$Y=\Sigma(2,3,11)$ and $p=5$.

∎

######  Theorem 6.5.

Let $Y=\Sigma(a_{1},a_{2},\dots,a_{r})$ be a Brieskorn homology sphere and let
$p$ be a prime not dividing $a_{1}\cdots a_{r}$. Then
$\delta_{\infty}^{(p)}(Y)>\delta(Y)$ except in the following cases:

  * (1)

$Y=\Sigma(2,3,5)$ and $p$ is any prime.

  * (2)

$Y=\Sigma(2,3,11)$ and $p=5$.

In both cases we have $\delta_{\infty}^{(p)}(Y)=\delta(Y)=1$.

###### Proof.

Theorems 6.1 and 6.4 imply that $\delta_{\infty}^{(p)}(Y)\geq\delta(Y)$ with
equality only in the cases listed. In these cases, we have
$\delta_{\infty}^{(p)}(Y)=\delta(Y)$ and $\delta(Y)=1$ [35, §8]. ∎

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