Integer surgeries rational homology cobordant to lens spaces

Antony T.H. Fung

Abstract.

The Cyclic Surgery Theorem and Moser’s work on surgeries on torus knots imply that for any non-trivial knot in $S^{3}$ , there are at most two integer surgeries that produce a lens space. This paper investigates how many positive integer surgeries on a given knot in $S^{3}$ can produce a manifold rational homology cobordant to a lens space. Tools include Greene and McCoy’s work on changemaker lattices which come from Heegaard Floer $d$ -invariants, and Aceto-Celoria-Park’s work on rational cobordisms and integral homology which is based on Lisca’s work on lens spaces.

1. Introduction

Lens spaces are one of the simplest classes of 3-manifolds, and Dehn surgery is one of the simplest ways of constructing 3-manifolds. There are a lot of famous theorems and conjectures that concern when surgery on a knot in $S^{3}$ produces a lens space. For example, the Berge conjecture concerns “which knots”, the Cyclic Surgery Theorem [7] concerns “which slopes”, and Greene’s work on the lens space realization problem [9] answers the question of “which lens spaces”.

This paper addresses a generalization of these questions, namely when surgery on a knot in $S^{3}$ is smoothly rational homology cobordant to a lens space. The following definition is central to this paper:

Definition 1.1.

A lensbordant surgery on a knot in an integer homology 3-sphere is a positive integer surgery which is smoothly rational homology cobordant to a lens space.

We make the following conjecture, which is an analog of the Cyclic Surgery Theorem.

Conjecture 1.2.

There exists a positive integer $N$ such that for all knots $K$ in $S^{3}$ satisfying $\nu^{+}(K)\neq 0$ , there are less than $N$ lensbordant surgeries on $K$ .

The non-negative integer $\nu^{+}(K)$ is a lower bound on the 4-ball genus $g_{4}(K)$ coming from work by Rasmussen [15] and Ni and Wu [14]. When $K$ is an $L$ -space knot, $\nu^{+}(K)$ coincides with the genus $g(K)$ .

We are not aware of any knot $K$ in $S^{3}$ satisfying $\nu^{+}(K)\neq 0$ having more than 5 lensbordant surgeries, though we also do not have a good reason to believe that such a knot cannot exist. There are examples of knots with 5 known lensbordant surgeries: When $K$ is the torus knot $T_{p,p+1}$ , the $(p^{2}+p+1)$ -surgery and the $(p^{2}+p-1)$ -surgery are both lens spaces, while the $p^{2}$ -surgery and the $(p+1)^{2}$ -surgery both bound a smooth rational ball [2, Th. 1.4], and hence being smoothly rational homology cobordant to $L(4,3)$ . This gives 4 lensbordant surgeries. In particular, when $p=2$ , the 8-surgery is smoothly rational homology cobordant to $L(2,1)$ [3, Th. 1.3]. When $p=3$ , the 12-surgery produces $L(4,1)\#L(3,2)$ , which is smoothly rational homology coboardant to $L(3,2)$ . When $p=4$ , the 20-surgery produces $L(5,1)\#L(4,3)$ , which is smoothly rational homology coboardant to $L(5,1)$ . Hence, when $p=2,3,4$ , there are 5 known lensbordant surgeries. Note that in this paper, we take the convention where $L(p,q)$ is the $-\tfrac{p}{q}$ -surgery on the unknot in $S^{3}$ .

The first main result of this paper is

Theorem 1.3.

Let $K$ be a knot in $S^{3}$ . If $\nu^{+}(K)\neq 0$ , there is a finite number of lensbordant surgeries on $K$ .

This is implied by Theorem 1.7 and Corollary 1.9, which are stated later in this section and proved in Section 4 and Section 5 respectively.

We do not know whether the condition $\nu^{+}(K)\neq 0$ can be weakened. However, we know that it is necessary to have some condition because the conclusion of the theorem is false whenever $K$ is a slice knot. When $K$ is a slice knot, all positive integer surgeries are integer homology cobordant to a lens space. We do not know whether there is any non-slice knot with an infinite number of lensbordant surgeries.

As a complement to Theorem 1.3, we have the following result about knots with vanishing $\nu^{+}$ :

Theorem 1.4.

Let $K$ be a knot in $S^{3}$ with $\nu^{+}(K)=0$ and let $m$ be a positive integer. If the $m$ -surgery on $K$ is lensbordant, then it is smoothly rational homology cobordant to $L(m,m-1)$ .

Remark: When $m=1$ , “ $L(m,m-1)$ ” is referring to $S^{3}$ .

Theorem 1.4 is proved in Section 4.

Aceto, Celoria, and Park defined the notion of a reduced lens space [1, Def. 2.2]. Any lens space is smoothly rational homology cobordant to a reduced lens space. For simplicity, in this paper we say that $S^{3}$ is reduced with $p=1$ , so when we say “reduced lens space” it also includes $S^{3}$ .

Theorem 1.5.

If a positive integer surgery on a knot in $S^{3}$ is smoothly rational homology cobordant to a reduced lens space, then that reduced lens space must also be a positive integer surgery on a knot in $S^{3}$ .

Section 3 of this paper is dedicated to proving Theorem 1.5.

Questions: Is there any interesting relationship between the two knots mentioned in Theorem 1.5? Is there any nice construction for knots in $S^{3}$ with lensbordant surgeries similar to what Berge [4] did for knots in $S^{3}$ with lens space surgeries?

If a positive integer surgery on a knot $K$ in an integer homology 3-sphere is smoothly rational cobordant to a reduced $L(p,q)$ , the surgery slope must be $r^{2}p$ for some positive integer $r$ [1, Th 1.1]. This motivates the following definition:

Definition 1.6.

Let $r,p$ be positive integers. A surgery on a knot in an integer homology 3-sphere is $(r,p)$ -lensbordant if the surgery slope is $r^{2}p$ and the resulting manifold is smoothly rational homology cobordant to some reduced $L(p,q)$ .

Our next result is an analog of the Cyclic Surgery Theorem in the direction of Conjecture 1.2. The result states that for most knots $K$ , when $r$ is fixed, the number of possible surgery slopes is very limited. More precisely:

Theorem 1.7.

Let $K$ be a knot in $S^{3}$ with $\nu^{+}(K)\neq 0$ . Then,
(1) When $r=1$ , there are at most 3 $(r,p)$ -lensbordant surgeries on $K$ .
(2) When $r$ is even, there are at most 2 $(r,p)$ -lensbordant surgeries on $K$ .
(3) When $r\geq 3$ is odd, there is at most 1 $(r,p)$ -lensbordant surgery on $K$ .

Section 4 of this paper is dedicated to proving Theorem 1.7.

The proofs of Theorem 1.5 and Theorem 1.7 involve the changemaker vector $\sigma$ described in [9]. Every lens space $L(p,q)$ is the boundary of a canonical negative definite plumbing 4-manifold $X(p,q)$ that arises from the Hirzebruch-Jung continued fraction of $\tfrac{p}{q}$ . Let $n$ be the rank of $H_{2}(X(p,q);\mathbb{Z})$ . When $L(p,q)$ is a positive integer surgery on a knot in $S^{3}$ , the intersection form of $X(p,q)$ embeds in $\mathbb{Z}^{n+1}$ as the orthogonal complement of some vector $\sigma$ with entries that satisfy certain combinatorial properties [10]. We say that $\sigma$ is a changemaker associated with $L(p,q)$ . Note that $L(p,q)$ can have multiple associated changemakers. In this paper, whenever we say that a lensbordant surgery is associated with a changemaker $\sigma$ , we mean that the surgery is rational homology cobordant to a reduced $L(p,q)$ associated with $\sigma$ . Because of Theorem 1.5, every lensbordant surgery is associated with some changemaker $\sigma$ .

In Section 5, we establish some bounds for the surgery slope $r^{2}p$ . At the start of Section 5, we show that

Theorem 1.8.

Suppose there is an $(r,p)$ -lensbordant surgery on a knot $K$ in $S^{3}$ with associated changemaker $\sigma$ , then

2\nu^{+}(K)+2(r-1)\geq r^{2}p-r|\sigma|_{1}\geq 2\nu^{+}(K).

Using the fact that $|\sigma|_{1}\leq p$ , this implies

Corollary 1.9.

Suppose there is an $(r,p)$ -lensbordant surgery on a knot $K$ in $S^{3}$ with associated changemaker $\sigma$ , then
(1) If $r\geq 3$ , then $\dfrac{2r^{2}}{(r-1)(r-2)}\nu^{+}(K)\geq r^{2}p\geq 2\nu^{+}(K)+r$
(2) If $r\geq 2$ , then $4\nu^{+}(K)+5\geq r^{2}p\geq 2\nu^{+}(K)+r$

Theorem 1.8 and Corollary 1.9 are proved in Section 5.

For every knot $K$ in $S^{3}$ , based on work by Rasmussen [15] and Ni and Wu [14], there is an associated sequence of non-negative integers $V_{i}(K)$ . The non-negative integer $\nu^{+}(K)$ is the smallest index $i$ such that $V_{i}(K)=0$ . Also, when $K$ is an $L$ -space knot, the coefficients $V_{i}(K)$ coincide with the torsion coefficients $t_{i}(K)$ which can be computed from the Alexander polynomial.

In Section 6, we will look in depth into the combinatorial conditions established in Section 4. We will show that

Theorem 1.10.

Suppose there is an $(r,p)$ -lensbordant surgery on a knot $K$ in $S^{3}$ . If $r,p$ are both even, the surgery slope $r^{2}p$ must equal $8V_{0}(K)$ .

Note that when $r,p$ are not both even, the surgery slope $r^{2}p$ is still very close to $8V_{0}(K)$ . See Theorem 6.1 in Section 6 for a precise statement.

Because of Theorem 1.5, we can focus on lens spaces that are positive integer surgeries on knots in $S^{3}$ . These lens spaces are classified. Berge [4] found examples of integer surgeries on knots in $S^{3}$ producing lens spaces, and Rasmussen [16] organized them nicely into a list of Berge types. Greene [9] showed that this is the complete list of lens spaces that are integer surgeries on knots in $S^{3}$ . He showed this by analyzing the possible structures of changemakers that arise from positive integer surgeries on knots in $S^{3}$ producing lens spaces. Greene concluded that such a changemaker must be in one of the 33 forms that he wrote down in [9], and then noted that for all 33 such structures, the resulting lens spaces are within this list of Berge types. Among these 33 structures in [9], 26 are classified as “small families” and 7 are classified as “large families”.

One way to tackle Conjecture 1.2 is to show that for each of these 33 changemaker structures described by Greene, there exists some $N$ such that for all knots $K$ in $S^{3}$ , there are at most $N$ lensbordant surgeries associated with a changemaker having that structure. In Section 7, we do this for one of the small families. This small family structure has changemakers in the form

\sigma=\begin{pmatrix}4s+3,&2s+1,&s+1,&s,&\smash{\underbrace{\begin{matrix}1,&\dots&,&1\end{matrix}}_{s\text{ copies of }1\text{'s}}}\end{pmatrix}

where $s\geq 2$ .

In this paper, we show that

Theorem 1.11.

Any knot in $S^{3}$ has at most 7 lensbordant surgeries with an associated changemaker coming from family mentioned above.

Theorem 1.11 is proved in Section 7.

We believe that similar techniques can be use on all 26 small families. However, we do not see how this approach extends to the 7 large families. Some large families are expected to be difficult to tackle because they contain infinite families of non-reduced lens spaces, which are being studied by the author in a separate project.

So far, everything we have discussed addresses surgeries on knots in $S^{3}$ . One may ask whether this generalizes to knots in other integer homology 3-spheres. Caudell addressed the lens space realization problem for knots in the Poincaré homology sphere [6] $\mathcal{P}$ . In $\mathcal{P}$ , we need to make more assumptions. We will show that

Theorem 1.12.

Suppose $r^{2}p$ -surgery on a knot $K\subset\mathcal{P}$ produces an L-space that is smoothly $\mathbb{Z}_{2}$ -homology cobordant to a reduced $L(p,q)$ . Let

i(K,r,p)=\begin{cases}2g(K)+r\text{ if }p\text{ is odd}\\ 2g(K)\quad\ \ \text{ if }p\text{ is even}\end{cases}

If $r^{2}p\geq i(K,r,p)$ , then $L(p,q)$ must be a positive integer surgery on a knot in $\mathcal{P}$ .
If $r^{2}p<i(K,r,p)$ , then $L(p,q)$ must be a positive integer surgery on a knot in $S^{3}$ .

Theorem 1.12 is proved in Section 8. Note that we do not know whether the conditions in Theorem 1.12 can be weakened.

Question: Does Theorem 1.12 generalize to knots in other integer homology spheres? One potentially interesting candidate to consider would be $\Sigma(2,3,7)$ . In [19], Tange gave some examples of positive integer surgeries on knots in $\Sigma(2,3,7)$ yielding lens spaces.

Convention: In this paper, unless specified, homology and cohomology always take integer coefficients.

Acknowledgements: The author would like to thank his adviser Joshua Greene for inspiring this project and also for many helpful conversations throughout the project. The author would also like to thank Duncan McCoy for pointing out that using $V$ coefficients instead of torsion coefficients allows results to be stated for general knots in $S^{3}$ instead of just L-space knots in $S^{3}$ , and thank Paolo Aceto for mentioning examples of knots in $S^{3}$ with 5 lensbordant surgeries.

2. Analyzing the Aceto-Celoria-Park map

In [1], Aceto, Celoria, and Park showed that when a closed 3-manifold $Y$ is smoothly rational homology cobordant to a reduced connected sum of lens spaces $L_{Y}$ (see [1, Def. 2.2]), there is an injective map $H_{1}(L_{Y})\rightarrow H_{1}(Y)$ . They proved this by showing that $H_{1}(L_{Y})$ injects into a group that $H_{1}(Y)$ surjects onto, and note that if $G\rightarrow G^{\prime}$ is a surjective map between finite abelian groups then there exists an injection $G^{\prime}\rightarrow G$ .

The goal of this section is to show that this interacts well with the $\text{spin}^{\text{c}}$ structures (and hence the $d$ -invariants).

We use the notations in [1]. Let $W$ be the smooth rational cobordism between $L_{Y}$ and $Y$ . Let $P$ be the canonical negative definite plumbed 4-manifold bounded by $L_{Y}$ . Let $X:=P\cup_{L}W$ .

Aceto, Celoria, and Park showed that the intersection forms of $P$ and $X$ are isomorphic [1, Prop. 4.1], and its proof implies that the isomorphism is induced by the inclusion $P\rightarrow X$ . Since $H_{2}(P)$ is free, we have the decomposition

H_{2}(X)=Free(H_{2}(X))\oplus Torsion(H_{2}(X))

with $Free(H_{2}(X))$ being the image of $H_{2}(P)\xrightarrow{i_{*}}H_{2}(X)$ .

Lemma 2.1.

The map $H_{1}(L_{Y})\xrightarrow{i_{*}}H_{1}(W)$ induced by inclusion is injective.

Proof.

By excision and the neighborhood collar theorem, the inclusion $(P,L_{Y})\rightarrow(X,W)$ induces isomorphisms $H_{*}(P,L_{Y})\cong H_{*}(X,W)$ .

Consider the long exact seqence of the pair $(X,W)$ :

\rightarrow H_{2}(W)\xrightarrow{i_{*}}H_{2}(X)\rightarrow H_{2}(X,W)\rightarrow

Since $H_{1}(P)=0$ , $H_{2}(X,W)\cong H_{2}(P,L_{Y})\cong H^{2}(P)$ is free, and hence the image of the map $H_{2}(W)\xrightarrow{i_{*}}H_{2}(X)$ contains all torsion elements of $H_{2}(X)$ . Now we consider the Mayer-Vietoris sequence

0\rightarrow H_{2}(P)\oplus H_{2}(W)\xrightarrow{i_{*}-i_{*}}H_{2}(X)\rightarrow H_{1}(L_{Y})\xrightarrow{i_{*}}H_{1}(W)\rightarrow

Since the image of the map $H_{2}(P)\xrightarrow{i_{*}}H_{2}(X)$ is $Free(H_{2}(X))$ and the image of the map $H_{2}(W)\xrightarrow{i_{*}}H_{2}(X)$ contains all torsion elements, we know that the map $H_{2}(P)\oplus H_{2}(W)\xrightarrow{i_{*}-i_{*}}H_{2}(X)$ is surjective, and hence the map $H_{1}(L_{Y})\xrightarrow{i_{*}}H_{1}(W)$ is injective. ∎

Consider the diagram

The two rows are the long exact sequences of the pairs $(X,Y)$ and $(P,L_{Y})$ . $i_{*}$ is induced by the inclusion $P\rightarrow X$ . The map $H_{2}(X,Y)\rightarrow H_{2}(X,W)$ comes from the long exact sequence of the triple $(Y,W,X)$ . The map $H_{2}(P,L_{Y})\rightarrow H_{2}(X,W)$ is excision.

For notational convenience in later parts of this section, we label the map $H_{2}(X)\rightarrow H_{2}(X,Y)$ as $\phi$ , the map $H_{2}(P)\rightarrow H_{2}(P,L_{Y})$ as $h$ , and the map $H_{2}(X,Y)\rightarrow H_{1}(Y)$ as $b$ .

Lemma 2.2.

This diagram commutes.

Proof.

Given a cycle $\alpha\in C_{2}(P)$ , both directions map $\alpha$ to $[\alpha]\in\dfrac{C_{2}(X)}{C_{2}(W)}$ , where we view $\alpha\in C_{2}(P)\subset C_{2}(X)$ .
∎

As mentioned in the proof of Lemma 2.1, $H_{2}(X,W)$ is free. Hence, in the map $H_{2}(X,Y)\rightarrow H_{2}(X,W)$ , all torsion elements of $H_{2}(X,Y)$ are mapped to $0$ . Hence, it induces a map

f:\dfrac{H_{2}(X,Y)}{Torsion(H_{2}(X,Y))}\rightarrow H_{2}(X,W)

Treating $Free(H_{2}(X))\cong Free(H_{2}(X))\oplus\{0\}$ as a subgroup of $H_{2}(X)$ , we obtain the following new commutative diagram where everything except $H_{1}(L_{Y})$ is free.

The map on top, which we labeled as $g$ , is obtained by composing $\phi|_{Free(H_{2}(X))}$ with the projection

H_{2}(X,Y)\rightarrow\dfrac{H_{2}(X,Y)}{Torsion(H_{2}(X,Y))}

Using notations in [1, §4], let $Q_{X}$ be the matrix representation of the intersection form of $X$ , which also represents the map $g$ under some suitable basis. Similarly, let $Q_{P}$ be the matrix representation of the intersection form of $P$ , which also represents the map $H_{2}(P)\rightarrow H_{2}(P,L_{Y})$ under some suitable basis.

Since $(H_{2}(P),Q_{P})$ and $(H_{2}(X),Q_{X})$ are isomorphic [1, Prop. 4.1], we know that $det(Q_{P})=det(Q_{X})$ . Since $|H_{1}(L_{Y})|$ is finite, we know that $det(Q_{P})\neq 0$ . Therefore, $f$ has has determinant 1, and hence we know that $f$ is an isomorphism.

By the commutativity of the diagram, we know that the image of the map $f\circ g$ coincides with the image of the map $H_{2}(P)\rightarrow H_{2}(P,L_{Y})\cong H_{2}(X,W)$ . Hence, $f$ induces an isomorphism

\bar{f}:\dfrac{H_{2}(X,Y)}{G}\rightarrow\dfrac{H_{2}(P,L_{Y})}{h(H_{2}(P))}

where $G$ is the subgroup of $H_{2}(X,Y)$ generated by elements in $\phi(Free(H_{2}(X)))$ and $Torsion(H_{2}(X,Y))$ .

A homomorphism cannot map a torsion element to a non-torsion element. Therefore, the image of $Torsion(H_{2}(X))$ under $\phi$ lies within $Torsion(H_{2}(X,Y))$ . Hence, $G$ is exactly the subgroup of $H_{2}(X,Y)$ generated by elements in $\phi(H_{2}(X))$ and $Torsion(H_{2}(X,Y))$ .

We go back to our first diagram

Let $\overline{H_{1}(Y)}$ be the image of $b$ , which is a subgroup of $H_{1}(Y)$ .

We see that $H_{1}(L_{Y})\cong\dfrac{H_{2}(P,L_{Y})}{h(H_{2}(P))}$ and $\dfrac{H_{2}(X,Y)}{\phi(H_{2}(X))}\cong\overline{H_{1}(Y)}$ . Hence, we can view $\bar{f}$ as an isomorphism

\dfrac{\overline{H_{1}(Y)}}{b(Torsion(H_{2}(X,Y)))}\rightarrow H_{1}(L_{Y})

Now we compose this with the projection

\overline{H_{1}(Y)}\rightarrow\dfrac{\overline{H_{1}(Y)}}{b(Torsion(H_{2}(X,Y)))}

We obtain a surjective map $\overline{H_{1}(Y)}\rightarrow H_{1}(L_{Y})$ which we denote as $s$ .

Now, we consider the maps $H_{1}(L_{Y})\rightarrow H_{1}(W)$ and $H_{1}(Y)\rightarrow H_{1}(W)$ induced by inclusion.

Lemma 2.3.

The diagram commutes

Proof.

Let $y\in\overline{H_{1}(Y)}$ .

Let $\alpha\in C_{2}(X)$ be such that $\alpha$ represents a cycle in $C_{2}(X,Y)$ that represents an element in $H_{2}(X,Y)$ that gets mapped to $y$ under $b$ . By considering how the map $b$ works in the long exact sequence of a pair, we see that $y$ is represented by $\partial\alpha\in C_{1}(Y)\subset C_{1}(X)$ .

By considering the chain maps being used in defining the Mayer-Vietoris sequence for $P\cup W=X$ , we can write $\alpha=\beta-\gamma$ , where $\beta\in C_{2}(P)$ and $\gamma\in C_{2}(W)$ . Under

H_{2}(X,Y)\rightarrow H_{2}(X,W)\cong H_{2}(P,L_{Y})\rightarrow H_{1}(L_{Y})

The element $[\alpha]\in H_{2}(X,Y)$ is being mapped to $[\beta]\in H_{2}(P,L_{Y})$ , which is then being mapped to $[\partial\beta]\in H_{1}(L_{Y})$ , where $\partial\beta\in C_{1}(L_{Y})\subset C_{1}(P)$ .

Now we have $s(y)=[\partial\beta]$ . From $\alpha=\beta-\gamma$ , we know that $\partial\gamma=\partial\beta-\partial\alpha$ . Hence, inside $C_{1}(W)$ , $\partial\beta$ and $\partial\alpha$ differ by a boundary element, and therefore represent the same element in $H_{1}(W)$ . ∎

Lemma 2.4.

Let $A_{1},\dots,A_{k}$ be distinct $\text{spin}^{\text{c}}$ structures on $L_{Y}$ . Then there exists distinct $\text{spin}^{\text{c}}$ structures $B_{1},\dots,B_{k}$ on $Y$ such that for each $i$ , $A_{i}$ and $B_{i}$ are restrictions of the same $\text{spin}^{\text{c}}$ structure on $W$ .

Proof.

Consider the long exact sequence

Let $\mathfrak{s}$ be a $\text{spin}^{\text{c}}$ structure on $W$ . For any $l$ , $A_{l}=a_{l}\cdot\mathfrak{s}|_{L_{Y}}$ for some $a_{l}\in H^{2}(L_{Y})\cong H_{1}(L_{Y})$ . By Lemma 2.3 and the surjectivity of $s$ , we know that there exists some $b_{l}\in\overline{H_{1}(Y)}\subset H_{1}(Y)\cong H^{2}(Y)$ such that $(a_{l},b_{l})$ is in the kernal of $i_{*}$ , and hence it is in the image of $i^{*}$ . Let $c_{l}\in H^{2}(W)$ be such that $i^{*}(c_{l})=(a_{l},b_{l})$ . We set $B_{l}:=b_{l}\cdot\mathfrak{s}|_{Y}$ . Then $c_{l}\cdot\mathfrak{s}$ restricts to $A_{l}$ and $B_{l}$ respectively. The $B_{l}$ ’s are distinct because given any $b_{l}\in\overline{H_{1}(Y)}$ , we can use $s$ to find the corresponding $a_{l}$ in $H_{1}(L_{Y})$ . ∎

Corollary 2.5.

The list of $d$ -invariants of $Y$ is extended from the list of $d$ -invariants of $L_{Y}$ .

Proof.

That is because $d$ -invariants are invariant under smooth rational homology cobordism. ∎

3. Proof of Theorem 1.5

This section is dedicated to proving Theorem 1.5.

Theorem 1.5.

We continue the work in Section 2, using the same notations. But now $L_{Y}$ is a reduced $L(p,q)$ , and $Y$ is the $r^{2}p$ -surgery on $K$ . We denote the $r^{2}p$ -surgery on $K$ as $K_{r^{2}p}$ .

The goal of this section is to show that $L(p,q)$ is a positive integer surgery on a knot in $S^{3}$ . To show that, we will use Greene’s work on the lens space realization problem [9, Th. 1.7]. To state the statement of [9, Th. 1.7], we need the following definition:

Definition 3.1.

[9, Def. 1.5] A changemaker vector is a vector $\sigma=(\sigma_{0},\dots,\sigma_{n})\in\mathbb{Z}^{n+1}_{>0}$ such that for all $k\in\mathbb{Z}$ satisfying $0\leq k\leq\sigma_{0}+\dots+\sigma_{n}$ , there exists some $A\subseteq\{\sigma_{0},\dots,\sigma_{n}\}$ such that the sum of the numbers in $A$ equals to $k$ .

Note that we requires the $\sigma$ to be in $\mathbb{Z}^{n+1}_{>0}$ , which implies the entries of $\sigma$ to be all non-zero. This is consistent with [9], but different from some other papers (for example [10]).

In this paper, we will choose a basis in a way such that $\sigma_{0}\geq\dots\geq\sigma_{n}$ , which is the opposite of the ordering used in [9, Def. 1.5].

Let $q^{\prime}$ be such that $qq^{{}^{\prime}}\equiv 1$ (mod $p$ ). Let $\Lambda(p,q)$ be the lattice coming from the intersection pairing on $P$ . Recall from Section 2 that $P$ is the canonical negative definite plumbing bounded by $L_{Y}$ , and in this section we are setting $L_{Y}=L(p,q)$ .

Greene showed that $\Lambda(p,q)$ embeds as the orthogonal complement to a changemaker vector in $\mathbb{Z}^{n+1}$ if and only if at least one of $L(p,q)$ , $L(p,q^{{}^{\prime}})$ appears on Berge’s list of lens spaces obtained from a positive integer surgery on a knot in $S^{3}$ [9, Th. 1.7].

Since $L(p,q^{{}^{\prime}})$ is homeomorphic to $L(p,q)$ in an orientation preserving manner, to prove Theorem 1.5, it suffices to show that $\Lambda(p,q)$ embeds as the orthogonal complement to a changemaker vector in $\mathbb{Z}^{n+1}$ . This is what we will do in this section.

Similar to the set up in [10, §2], we let $W_{r^{2}p}$ be the 4-manifold obtained by attaching a $r^{2}p$ -framed 2-handle along $K\subset S^{3}=\partial D^{4}$ . By construction, we have $H_{1}(W_{r^{2}p})=0$ . The manifold $W_{r^{2}p}$ has boundary $K_{r^{2}p}$ . Let $Z$ be the closed 4-manifold $X\cup_{K_{r^{2}p}}W_{-r^{2}p}$ . $H_{2}(-W_{r^{2}p})$ is generated by the class of a surface $\Sigma$ obtained by gluing the core of the 2-handle to a Seifert surface of $K$ . $\Sigma$ has the property that the intersection pairing of $[\Sigma]$ with itself gives $-r^{2}p$ . Each $\text{spin}^{\text{c}}$ structure on $K_{r^{2}p}$ comes with a label $i\in\{0,\dots,r^{2}p-1\}$ . This label has the property that for all $\mathfrak{s}\in\text{Spin}^{\text{c}}(-W_{r^{2}p})$ that extends the $\text{spin}^{\text{c}}$ structure on $K_{r^{2}p}$ with label $i$ , we have $\langle c_{1}(\mathfrak{s}),[\Sigma]\rangle+r^{2}p\equiv 2i$ (mod $2r^{2}p$ ).

Note that we do not completely follow the notations in [10, §2]. We defined $W$ to be the cobordism in order to be consistent with [1], and later in this chapter we will define $\sigma$ slightly differently from [10, §2].

Figure 1. The set up of Section 3.

Lemma 3.2.

$H_{2}(P)$ embeds primitively in $H_{2}(Z)$ , in the sense that for any $k\in\mathbb{Z}$ and $z\in H_{2}(Z)$ , if $kz\in H_{2}(P)$ , then $z$ must equals to $p+\tau$ for some $p\in H_{2}(P)$ and $\tau$ being a torsion element in $H_{2}(Z)$ .

Proof.

Suppose $kz$ is in $H_{2}(P)$ .

Consider the Mayer-Vietoris sequence

0\rightarrow H_{2}(P)\oplus H_{2}(W)\rightarrow H_{2}(P\cup W)\rightarrow H_{1}(L(p,q))\rightarrow H_{1}(P)\oplus H_{1}(W)\rightarrow\dots

By Lemma 2.1, we know that the map $H_{2}(P)\oplus H_{2}(W)\rightarrow H_{2}(P\cup W)$ is an isomorphism.

Since $H_{2}(P)$ embeds in $H_{2}(P\cup W)$ , we can also treat $kz$ as an element of $H_{2}(P\cup W)$ . Now, consider the exact sequence of a pair

\dots\rightarrow H_{3}(Z,P\cup W)\rightarrow H_{2}(P\cup W)\rightarrow H_{2}(Z)\rightarrow H_{2}(Z,P\cup W)\rightarrow\dots

By excision, we have

H_{3}(Z,P\cup W)\cong H_{3}(-W_{r^{2}p},K_{r^{2}p})\cong H^{1}(-W_{r^{2}p})\cong 0

Also by excsion, we have

H_{2}(Z,P\cup W)\cong H_{2}(-W_{r^{2}p},K_{r^{2}p})\cong H^{2}(-W_{r^{2}p})\cong\mathbb{Z}

So, the exact sequence simlpifies to

0\rightarrow H_{2}(P\cup W)\rightarrow H_{2}(Z)\rightarrow\mathbb{Z}\rightarrow\dots

Since $kz$ is in $H_{2}(P\cup W)$ , it gets mapped to $0$ under the map $H_{2}(Z)\rightarrow\mathbb{Z}$ . Since $\mathbb{Z}$ is free, $z$ also gets mapped to $0$ under the map $H_{2}(Z)\rightarrow\mathbb{Z}$ . Hence, $z\in H_{2}(P\cup W)$ .

Since $H_{2}(P)\oplus H_{2}(W)\rightarrow H_{2}(P\cup W)$ is an isomorphism, we must have $z=p+\tau$ for some $p\in H_{2}(P)$ and $\tau\in H_{2}(W)$ . ∎

Lemma 3.3.

In the intersection form of $Z$ , $H_{2}(P)$ is the orthogonal copmlement of $H_{2}(W\cup-W_{r^{2}p})$ .

Proof.

From the Mayer-Vietoris sequence

0\rightarrow H_{2}(P)\oplus H_{2}(W\cup-W_{r^{2}p})\rightarrow H_{2}(Z)\rightarrow H_{1}(L(p,q))\rightarrow\dots

We know that $H_{2}(P)\oplus H_{2}(W\cup-W_{r^{2}p})\rightarrow H_{2}(Z)$ is a full rank embedding. Since $H_{2}(P)$ and $H_{2}(W\cup-W_{r^{2}p})$ are orthogonal to each other, Lemma 3.2 implies that $H_{2}(P)$ is the orthogonal complement of $H_{2}(W\cup-W_{r^{2}p})$ . ∎

Lemma 3.4.

The image of the map $H_{2}(W)\rightarrow H_{2}(W\cup-W_{r^{2}p})$ induced by inclusion is exactly the torsion subgroup of $H_{2}(W\cup-W_{r^{2}p})$ .

Proof.

By excision and the neighborhood collar theorem, we have $H_{2}(W\cup-W_{r^{2}p},W)\cong H_{2}(-W_{r^{2}p},K_{r^{2}p})\cong H^{2}(-W_{r^{2}p})$ , which has no torsion because $H_{1}(-W_{r^{2}p})=0$ .

Consider the exact sequence of a pair

\rightarrow H_{2}(W)\rightarrow H_{2}(W\cup-W_{r^{2}p})\rightarrow H_{2}(W\cup-W_{r^{2}p},W)\rightarrow

Since $H_{2}(W\cup-W_{r^{2}p},W)$ is free, the image of the map $H_{2}(W)\rightarrow H_{2}(W\cup-W_{r^{2}p})$ must contains all torsion elements. Since $H_{2}(W)$ is finite, the image is exactly the torsion subgroup. ∎

Lemma 3.5.

The map $H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ induced by the inclusion $L(p,q)\cup K_{r^{2}p}\rightarrow W$ has kernal of size $rp$ and image of size $rp$ .

Proof.

Consider the exact sequence of a pair

		$\displaystyle 0\rightarrow H_{2}(W)\rightarrow H_{2}(W,L(p,q)\cup K_{r^{2}p})\rightarrow H_{1}(L(p,q)\cup K_{r^{2}p})\rightarrow H_{1}(W)$
		$\displaystyle\rightarrow H_{1}(W,L(p,q)\cup K_{r^{2}p})\rightarrow\mathbb{Z}^{2}\rightarrow\dots$

Using Poincaré-Lefschetz duality and universal coefficient theorem, we get

		$\displaystyle 0\rightarrow H_{2}(W)\rightarrow H_{1}(W)\rightarrow H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$
		$\displaystyle\rightarrow H_{2}(W)\oplus\mathbb{Z}\rightarrow\mathbb{Z}^{2}\rightarrow\dots$

which simplifies to

0\rightarrow H_{2}(W)\rightarrow H_{1}(W)\rightarrow H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)\rightarrow H_{2}(W)\rightarrow 0

Hence, the map $H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ has kernal of size $\dfrac{|H_{1}(W)|}{|H_{2}(W)|}$ and image of size $\dfrac{|H_{1}(W)|}{|H_{2}(W)|}$ . Since the kernal and the image has the same size, their sizes must both be $\sqrt{|H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})|}=rp$ . ∎

Lemma 3.6.

The kernel of the map $H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ induced by inclusion is isomorphic to $\mathbb{Z}_{r}$ (cyclic group of order $r$ ).

Proof.

Due to the surjectivitiy of the map labeled as “ $s$ ” in Lemma 2.3, we know that the image of the map $H_{1}(L(p,q))\rightarrow H_{1}(W)$ is contained in the image of the map $H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ . Hence, the image of the map $H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ is the same as the image of the map $H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ .

Therefore, by Lemma 3.5, the map $H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ has an image of size $rp$ and a kernal of size $\dfrac{r^{2}p}{rp}=r$ . Since $H_{1}(K_{r^{2}p})$ is cyclic, the kernal is a cyclic group of order $r$ . ∎

Recall that $H_{2}(-W_{r^{2}p})$ is generated by $[\Sigma]$ . The follow Lemma is about the generator of the free part of $H_{2}(W\cup-W_{r^{2}p})$ :

Lemma 3.7.

The map $H_{2}(-W_{r^{2}p})\rightarrow H_{2}(W\cup-W_{r^{2}p})$ induced by inclusion maps $[\Sigma]$ to $r$ times a generator of the free part of $H_{2}(W\cup-W_{r^{2}p})$ plus a torsion element (can be $0$ ).

Proof.

Consider the Mayer-Vietoris sequence

0\rightarrow H_{2}(W)\oplus H_{2}(-W_{r^{2}p})\rightarrow H_{2}(W\cup-W_{r^{2}p})\rightarrow H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)\rightarrow\dots

By Lemma 3.6, we can simplify this into

0\rightarrow H_{2}(W)\oplus H_{2}(-W_{r^{2}p})\rightarrow H_{2}(W\cup-W_{r^{2}p})\rightarrow\mathbb{Z}_{r}\rightarrow 0

Combining this with Lemma 3.4, we conclude that $H_{2}(-W_{r^{2}p})\rightarrow H_{2}(W\cup-W_{r^{2}p})$ has to map $[\Sigma]$ to $r$ times a generator of the free part of $H_{2}(W\cup-W_{r^{2}p})$ plus a torsion element (can be $0$ ). ∎

Let $\sigma\in H_{2}(W\cup-W_{r^{2}p})$ be a generator of the free part such that $[\Sigma]=r\sigma+torsion$ .

Recall that the intersection pairing of $[\Sigma]$ with itself gives $-r^{2}p$ . So we have

\langle\sigma,\sigma\rangle=\dfrac{1}{r^{2}}\langle[\Sigma],[\Sigma]\rangle=\dfrac{-r^{2}p}{r^{2}}=-p

Since $P$ is negative definite, by Lemma 3.3, we know that $Z$ is negative definite. By Donaldson’s Theorem [8], we know that the intersection pairing on $Z$ is $-\mathbb{Z}^{n+1}$ , where $n$ is the rank of $H_{2}(P)$ . Moreover, under the identification $H_{2}(Z)\cong-\mathbb{Z}^{n+1}$ , the image of the first Chern class map $c_{1}:\text{Spin}^{\text{c}}(Z)\rightarrow H^{2}(Z)\cong H_{2}(Z)\cong-\mathbb{Z}^{n+1}$ is exactly the set of characteristic vectors

\text{Char}(-\mathbb{Z}^{n+1}):=\{v\in-\mathbb{Z}^{n+1}|\forall w\in-\mathbb{Z}^{n+1}\ \langle v,w\rangle\equiv\langle w,w\rangle\text{ (mod 2)}\}

where $H^{2}(Z)\cong H_{2}(Z)$ is Poincaré duality.

In any orthonormal basis, the set of characteristic vectors of $-\mathbb{Z}^{n+1}$ is exactly the set of vectors with all entries being odd.

Now we can prove the following:

Lemma 3.8.

A $\text{spin}^{\text{c}}$ structure on $K_{r^{2}p}$ extends to $W$ if and only if the label $i$ on the $\text{spin}^{\text{c}}$ structure satisfies $i\equiv\begin{cases}\dfrac{r}{2}\quad(mod\ r)\text{ if }r\text{ is even and }p\text{ is odd}\\ 0\quad(mod\ r)\text{ otherwise}\end{cases}$ .

Note that since $P$ and $-W_{r^{2}p}$ have trivial $H_{1}$ , a $\text{spin}^{\text{c}}$ structure on on $K_{r^{2}p}$ extends to $W$ if and only if it extends to $Z$ . So, in the statement of Lemma 3.8, we can always replace the $W$ with $Z$ .

Proof of Lemma 3.8.

Consider the map $H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ mentioned in Lemma 3.5. By Lemma 3.5, it has a kernal of size $rp$ . Since the map $H_{1}(L(p,q))\rightarrow H_{1}(W)$ is injective, the kernal of $H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ cannot contain two different elements with the same $H_{1}(K_{r^{2}p})$ component. Hence, the $rp$ elements in the kernal of $H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)$ all have different $H_{1}(K_{r^{2}p})$ components.

Therefore, by applying Poincaré-Lefschetz duality on the exact sequence of a pair

\dots\rightarrow H_{2}(W,L(p,q)\cup K_{r^{2}p})\rightarrow H_{1}(L(p,q))\oplus H_{1}(K_{r^{2}p})\rightarrow H_{1}(W)\rightarrow\dots

we can conclude that the map $H^{2}(W)\rightarrow H^{2}(K_{r^{2}p})$ has image of size $rp$ . Therefore there are exactly $rp$ $\text{spin}^{\text{c}}$ structures on $K_{r^{2}p}$ that extend to $W$ .

Notice that there are exactly $rp$ possible values of $i\ (mod\ r^{2}p)$ satisfying

(1)

i\equiv\begin{cases}\dfrac{r}{2}\quad(mod\ r)\text{ if }r\text{ is even and }p\text{ is odd}\\ 0\quad(mod\ r)\text{ otherwise}\end{cases}

Therefore, to prove Lemma 3.8, it suffices to show that a $\text{spin}^{\text{c}}$ structure on $K_{r^{2}p}$ extends to $W$ implies that its label $i$ satisfies (1). By the definition of the labeling, this is equivalent to showing that for any $\text{spin}^{\text{c}}$ structure $\mathfrak{s}$ on $Z$ ,

\langle c_{1}(\mathfrak{s}),[\Sigma]\rangle+r^{2}p\equiv\begin{cases}r\quad(mod\ 2r)\text{ if }r\text{ is even and }p\text{ is odd}\\ 0\quad(mod\ 2r)\text{ otherwise}\end{cases}

Since $\langle c_{1}(\mathfrak{s}),[\Sigma]\rangle+r^{2}p=r(\langle c_{1}(\mathfrak{s}),\sigma\rangle+rp)$ , we only need to figure out whether $\langle c_{1}(\mathfrak{s}),\sigma\rangle+rp$ is odd or even.

Since $c_{1}(\mathfrak{s})$ is a characteristic vector, we have

\langle c_{1}(\mathfrak{s}),\sigma\rangle\equiv\langle\sigma,\sigma\rangle=-p\quad(mod\ 2)

Hence, $\langle c_{1}(\mathfrak{s}),\sigma\rangle+rp$ is odd when $r$ is even and $p$ is odd, and it is even otherwise. ∎

By [9, Th. 1.7] and Lemma 3.3, to prove Theorem 1.5, we only need to show that $\sigma$ embeds as a changemaker vector in the lattice $-\mathbb{Z}^{n+1}$ .

By choosing suitable orthonormal basis, we let $\sigma=(\sigma_{0},\dots,\sigma_{n})$ with

0\leq\sigma_{n}\leq\dots\leq\sigma_{0}

For every knot $K$ in $S^{3}$ , based on work by Rasmussen [15] and also Ni and Wu [14], there is a sequence of non-negative integers $V_{i}(K)$ satisfying the following properties:

•

$\forall i\geq 0$ , $V_{i}(K)-V_{i+1}(K)\in\{0,1\}$
•

$V_{g(K)}(K)=0$ , where $g(K)$ is the genus of the knot
•

For all positive integer $k$ , for all $i$ with $0\leq i\leq k-1$ , we have

$-2V_{\text{min}\{i,k-i\}}=d(K_{k},i)-d(U_{k},i)$

where $K_{k}$ is the $k$ -surgery on $K$ , and $U_{k}$ is the $k$ -surgery on the unknot (which is $L(k,k-1)$ ). The $i$ behind denotes the $\text{spin}^{\text{c}}$ structure with label $i$ , and the $d$ is the $d$ -invariant.

The last property implies that for $0\leq i\leq\dfrac{r^{2}p}{2}$ , we have

-2V_{i}(K)=d(K_{r^{2}p},i)-d(U_{r^{2}p},i)

The smooth concordance invariant $\nu^{+}(K)$ is the smallest $i$ such that $V_{i}(K)=0$ . It is a lower bound of the 4-ball genus of $K$ [11, Prop. 2.4]. When $K$ is an L-space knot, $\nu^{+}$ coincides with the genus of $K$ and these $V$ coefficients coincide with the torsion coefficients that can be computed from the Alexander polynomial.

For notational simplicity, we often drop the $(K)$ and only write $V_{i}$ and $\nu^{+}$ .

We now proceed to prove that $\sigma$ is a changemaker vector. Our proof will follow the argument in [10, §2-3] that shows that $\sigma$ is a changemaker. The only difference is that we use the $V$ coefficients instead of torsion coefficients because we are not assuming $K$ to be an L-space knot, and there are some other minor changes because of the $r$ factor coming from our cobordism which did not exist in [10, §2-3].

Based on Lemma 3.8, we make the following definition:

Definition 3.9.

We say that the coefficient $V_{i}$ is a relevant coefficient if the index satisfies $0\leq i\leq\dfrac{r^{2}p}{2}$ and

i\equiv\begin{cases}\dfrac{r}{2}\quad(mod\ r)\text{ if }r\text{ is even and }p\text{ is odd}\\ 0\quad(mod\ r)\text{ otherwise}\end{cases}

In that case, we say that the index $i$ is relevant.

Lemma 3.10.

Let $i$ be relevant. Then

-8V_{i}=\max_{\begin{subarray}{c}\mathfrak{c}\in Char(-\mathbb{Z}^{n+1})\\ \langle\mathfrak{c},[\Sigma]\rangle+r^{2}p=2i\end{subarray}}(\mathfrak{c}^{2}+(n+1))

Proof of Lemma 3.10.

Let $i$ be relevant. There is a $\text{spin}^{\text{c}}$ structure on $K_{r^{2}p}$ with label $i$ . By Lemma 3.8, this $\text{spin}^{\text{c}}$ structure extends to some $\text{spin}^{\text{c}}$ structure $\mathfrak{s}^{W}_{i}$ on $W$ . Let $\mathfrak{s}^{L(p,q)}_{i}$ be the restriction of $\mathfrak{s}^{W}_{i}$ on $L(p,q)$ . (Warning: In the literature, there is a convention of labeling $\text{spin}^{\text{c}}$ structures on lens spaces with numbers $0,\dots,p-1$ . The label of $\mathfrak{s}^{L(p,q)}_{i}$ under such a convention may not necessarily be $i$ .)

Now we recreate [10, Lemma 2.3], but under the context of this paper. Since the canonical negative plumbing of a lens space is always sharp (see [10, Lemma 2.1]), the argument in the proof of [10, Lemma 2.3] implies that for any $\text{spin}^{\text{c}}$ structure $\mathfrak{s}^{P}_{i}$ on $P$ that extends $\mathfrak{s}^{L(p,q)}_{i}$ , we have

c_{1}(\mathfrak{s}^{P}_{i})^{2}+n\leq 4d(\mathfrak{s}^{L(p,q)}_{i})

And for any $\text{spin}^{\text{c}}$ structure $\mathfrak{s}^{-W_{r^{2}p}}_{i}$ on $-W_{r^{2}p}$ that extends $(K_{r^{2}p},i)$ , we have

c_{1}(\mathfrak{s}^{-W_{r^{2}p}}_{i})^{2}+1\leq-4d(U_{r^{2}p},i)

Furthermore, $\mathfrak{s}^{P}_{i}$ and $\mathfrak{s}^{-W_{r^{2}p}}_{i}$ can be chosen such that equality holds. Adding these two inequalities together, we have

c_{1}(\mathfrak{s}^{P}_{i})^{2}+c_{1}(\mathfrak{s}^{-W_{r^{2}p}}_{i})^{2}+(n+1)\leq 4d(\mathfrak{s}^{L(p,q)}_{i})-4d(U_{r^{2}p},i)

Now we look at the cobordism $W$ that exists in this paper but not in [10]. Since the $d$ -invariant is invariant under smooth rational homology cobordism, we know that

d(\mathfrak{s}^{L(p,q)}_{i})=d(K_{r^{2}p},i)

Also, since $L(p,q)$ and $K_{r^{2}p}$ are rational homology spheres and $H_{2}(W)$ is finite, we have

c_{1}(\mathfrak{s})^{2}=c_{1}(\mathfrak{s}^{P}_{i})^{2}+c_{1}(\mathfrak{s}^{-W_{r^{2}p}}_{i})^{2}

Hence, for any $\text{spin}^{\text{c}}$ structure $\mathfrak{s}$ on $Z$ that extends $(K_{r^{2}p},i)$ , we have

c_{1}(\mathfrak{s})^{2}+(n+1)\leq 4d(K_{r^{2}p},i)-4d(U_{r^{2}p},i)=-8V_{i}

and for any relevant $i$ , $\mathfrak{s}$ can be chosen such that equality holds.

Due to the way the labeling of $\text{spin}^{\text{c}}$ structures on $K_{r^{2}p}$ works, we know that $\text{spin}^{\text{c}}$ structure $\mathfrak{s}$ on $Z$ that extends $(K_{r^{2}p},i)$ are exactly the ones that satisfies

(2)

\langle c_{1}(\mathfrak{s}),[\Sigma]\rangle+r^{2}p\equiv 2i\text{ (mod }2r^{2}p\text{)}

Hence, we conclude that

(3)

-8V_{i}=\max_{\begin{subarray}{c}\mathfrak{c}\in Char(-\mathbb{Z}^{n+1})\\ \langle\mathfrak{c},[\Sigma]\rangle+r^{2}p\equiv 2i\text{ (mod }2r^{2}p\text{)}\end{subarray}}(\mathfrak{c}^{2}+(n+1))

(3) is very close to the statement of Lemma 3.10. The only difference is that in Lemma 3.10, it says “ $=2i$ ” instead of “ $\equiv 2i$ (mod $2r^{2}p$ )” under the $\max$ .

To finish the proof of Lemma 3.10, it suffices to show that for any $\mathfrak{c}=c_{1}(\mathfrak{s})\in Char(-\mathbb{Z}^{n+1})$ satisfying (2), there exists some $\mathfrak{c}^{\prime}$ with ${\mathfrak{c}^{\prime}}^{2}\geq\mathfrak{c}^{2}$ satisfying

\langle\mathfrak{c}^{\prime},[\Sigma]\rangle+r^{2}p=2i

To do this, we apply the argument in the proof of [12, Lemma 2.9]. Let $\mathfrak{c}=c_{1}(\mathfrak{s})\in Char(-\mathbb{Z}^{n+1})$ satisfying (2). Then

\langle\mathfrak{c},[\Sigma]\rangle+r^{2}p=2i+2mr^{2}p

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

Let $\mathfrak{c}^{\prime}:=\mathfrak{c}+2m[\Sigma]$ . Then we have

\langle\mathfrak{c}^{\prime},[\Sigma]\rangle+r^{2}p=2i+2mr^{2}p+2m\langle[\Sigma],[\Sigma]\rangle=2i

And also

	$\displaystyle{\mathfrak{c}^{\prime}}^{2}$	$\displaystyle=\mathfrak{c}^{2}+4m\langle\mathfrak{c},[\Sigma]\rangle+4m^{2}\langle[\Sigma],[\Sigma]\rangle$
		$\displaystyle=\mathfrak{c}^{2}+4m(2i+2mr^{2}p-r^{2}p)+4m^{2}(-r^{2}p)$
		$\displaystyle=\mathfrak{c}^{2}+4r^{2}pm^{2}+(8i-4r^{2}p)m$

Viewing $4r^{2}pm^{2}+(8i-4r^{2}p)m$ as a quadratic in $m$ , the roots are 0 and $(1-\dfrac{2i}{r^{2}p})$ . Since $i$ is relevant, we have $0\leq 2i\leq r^{2}p$ . Hence, there is no integers strictly between $0$ and $(1-\dfrac{2i}{r^{2}p})$ . Therefore, since $m$ is an integer, we have $4r^{2}pm^{2}+(8i-4r^{2}p)m\geq 0$ , and hence ${\mathfrak{c}^{\prime}}^{2}\geq\mathfrak{c}^{2}$ . ∎

Lemma 3.11.

$\sigma$ is a changemaker vector.

Proof.

First of all, similar to the argument in [9, §3.4], by Lemma 3.3 and the indecomposability of linear lattices [9, Cor. 3.5], we know that all entries of $\sigma$ are non-zero.

When $\mathfrak{c}=(1,\dots,1)$ , we have

	$\displaystyle\langle\mathfrak{c},[\Sigma]\rangle+r^{2}p$	$\displaystyle=r\langle\mathfrak{c},\sigma\rangle+r^{2}p$
		$\displaystyle=-r\|\sigma\|_{1}+r^{2}p$
		$\displaystyle=2\left(\dfrac{r^{2}p-r\|\sigma\|_{1}}{2}\right)$

Also, $\mathfrak{c}^{2}+(n+1)=0$ .

By Lemma 3.10, since those $V_{i}$ coefficients are all non-negative, we can conclude that

V_{\tfrac{1}{2}r(rp-|\sigma|_{1})}=0

Therefore, $\nu^{+}\leq\tfrac{1}{2}r(rp-|\sigma|_{1})$ . Hence, for all $i\geq\tfrac{1}{2}r(rp-|\sigma|_{1})$ , we have $V_{i}=0$ .

By Lemma 3.10, we know that for all relevant $i\geq\tfrac{1}{2}r(rp-|\sigma|_{1})$ , there exists some $\mathfrak{c}\in Char(-\mathbb{Z}^{n+1})$ such that $\mathfrak{c}^{2}=-(n+1)$ and $\langle\mathfrak{c},[\Sigma]\rangle+r^{2}p=2i$ . This is equivalent to saying that there exists some $\mathfrak{c}\in\{\pm 1\}^{n+1}$ such that

r\langle\mathfrak{c},\sigma\rangle+r^{2}p=2i

For the rest of the proof, we will separate the cases according to Lemma 3.8. The proofs for the two cases are almost identical.

Case 1: $r$ is odd or $p$ is even
Let $i$ be relevant and $i\geq\tfrac{1}{2}r(rp-|\sigma|_{1})$ . When $r$ is odd or $p$ is even, that is equivalent to saying that $i$ is a multiple of $r$ that satisfies

\tfrac{1}{2}r(rp-|\sigma|_{1})\leq i\leq\tfrac{r^{2}p}{2}

Let $i^{\prime}:=\dfrac{i}{r}$ .

Then we have $\tfrac{rp-|\sigma^{\prime}|_{1}}{2}\leq i^{\prime}\leq\tfrac{rp}{2}$ , and there exists $\mathfrak{c}\in\{\pm 1\}^{n+1}$ such that

r\langle\mathfrak{c},\sigma\rangle+r^{2}p=2i^{\prime}r

which simplifies to

(4)

\langle\mathfrak{c},\sigma\rangle+rp=2i^{\prime}

Let $i^{\prime\prime}:=rp-i^{\prime}$ . Then we have $\tfrac{rp}{2}\leq i^{\prime\prime}\leq\tfrac{rp+|\sigma|_{1}}{2}$ . Substituting $i^{\prime}=rp-i^{\prime\prime}$ into (4) and rearranging, we get

\langle-\mathfrak{c},\sigma\rangle+rp=2i^{\prime\prime}

Notice that $\mathfrak{c}\in\{\pm 1\}^{n+1}$ implies $-\mathfrak{c}\in\{\pm 1\}^{n+1}$ . Hence, we can conclude that for any integer $i^{\prime}$ satisfying $\tfrac{rp-|\sigma^{\prime}|_{1}}{2}\leq i^{\prime}\leq\tfrac{rp+|\sigma^{\prime}|_{1}}{2}$ , there exists $\mathfrak{c}\in\{\pm 1\}^{n+1}$ such that (4) holds.

Note that since $|\sigma|_{1}\equiv\langle\sigma,\sigma\rangle=-p$ (mod $2$ ), $\tfrac{rp\pm|\sigma|_{1}}{2}$ must be integers under Case 1.

Setting $j:=i^{\prime}-\tfrac{rp+|\sigma|_{1}}{2}$ and substituting it into (4), we arrive to the conclusion that for any integer $j$ satisfying $-|\sigma|_{1}\leq j\leq 0$ , there exists $\mathfrak{c}\in\{\pm 1\}^{n+1}$ such that

\langle\mathfrak{c},\sigma\rangle=2j+|\sigma|_{1}

Setting $\mathfrak{c}=(-1,\dots,-1)+2\chi$ for some $\chi\in\{0,1\}^{n+1}$ , we arrive to the conclusion that for any integer $j$ satisfying $-|\sigma|_{1}\leq j\leq 0$ , there exists $\chi\in\{0,1\}^{n+1}$ such that

\langle\chi,\sigma\rangle=j

Together with the fact that all entries of $\sigma$ are non-zero, this implies $\sigma$ being a changemaker vector.

Case 2: $r$ is even and $p$ is odd
Let $i$ be relevant and $i\geq\tfrac{1}{2}r(rp-|\sigma|_{1})$ . When $r$ is even and $p$ is odd, that is equivalent to saying that $i$ satisfies $i\equiv\tfrac{r}{2}$ (mod $r$ ) and

\tfrac{1}{2}r(rp-|\sigma|_{1})\leq i\leq\tfrac{r^{2}p}{2}

Let $i^{\prime}:=\tfrac{i}{r}-\tfrac{1}{2}$ .

Then we have $\tfrac{rp-|\sigma^{\prime}|_{1}}{2}\leq i^{\prime}+\tfrac{1}{2}\leq\tfrac{rp}{2}$ , and there exists $\mathfrak{c}\in\{\pm 1\}^{n+1}$ such that

r\langle\mathfrak{c},\sigma\rangle+r^{2}p=2\left(i^{\prime}+\tfrac{1}{2}\right)r

which simplifies to

(5)

\langle\mathfrak{c},\sigma\rangle+rp=2i^{\prime}+1

Let $i^{\prime\prime}:=rp-i^{\prime}-1$ . Then $\tfrac{rp}{2}\leq i^{\prime\prime}+\tfrac{1}{2}\leq\tfrac{rp+|\sigma|_{1}}{2}$ and

\langle-\mathfrak{c},\sigma\rangle+rp=2i^{\prime\prime}+1

Hence (5) holds for $\tfrac{rp-|\sigma|_{1}}{2}\leq i^{\prime\prime}+\tfrac{1}{2}\leq\tfrac{rp+|\sigma|_{1}}{2}$ .

Note that in Case 2, $rp+|\sigma|_{1}$ is odd. Therefore, $\tfrac{rp+|\sigma|_{1}-1}{2}$ is an integer.

Setting $j:=i^{\prime}-\tfrac{rp+|\sigma^{\prime}|_{1}-1}{2}$ and substituting it into (5), we see that for any integer $j$ satisfying $-|\sigma^{\prime}|_{1}\leq j\leq 0$ , there exists $\mathfrak{c}\in\{\pm 1\}^{n+1}$ such that

\langle\mathfrak{c},\sigma\rangle=2j+|\sigma|_{1}

The rest of the proof is the same as Case 1. ∎

Proof of Theorem 1.5.

Theorem 1.5 follows from [9, Th. 1.7], Lemma 3.3, and Lemma 3.11. ∎

4. Number of possible surgeries when fixing $r$

In this section we will show that when we fix $K\subset S^{3}$ and a value of $r$ , the number of $(r,p)$ -lensbordant surgeries is very limited. We will follow an argument similar to [12, §2], with some changes to take $r$ into account.

Recall that we defined the coefficient $V_{i}$ to relevant if $0\leq i\leq\dfrac{r^{2}p}{2}$ and

i\equiv\begin{cases}\dfrac{r}{2}\quad(mod\ r)\text{ if }r\text{ is even and }p\text{ is odd}\\ 0\quad(mod\ r)\text{ otherwise}\end{cases}

Sometimes it is convenient to consider the relevant coefficients as a sequence itself:

Definition 4.1.

For $i$ satisfying

0\leq i\leq\begin{cases}\frac{rp}{2}-1\text{ if }r\text{ is even and }p\text{ is odd}\\ \frac{rp-1}{2}\text{ if }r,p\text{ are both odd}\\ \frac{rp}{2}\text{ if }p\text{ is even}\end{cases}

we define

V^{rel}_{i}:=\begin{cases}V_{\frac{r}{2}+ir}\text{ if }r\text{ is even and }p\text{ is odd}\\ V_{ir}\text{ otherwise}\end{cases}

This range of $i$ for $V^{rel}_{i}$ corresponds to the range of $i$ for $V_{i}$ to be relevant.

Since the $V$ coefficients of a knot are always monotonic decreasing, the relevant coefficients are also monotonic decreasing. However, consecutive relevant coefficients can differ by more than 1.

We define $\nu^{+rel}$ to be the number of non-zero relevant coefficients. Equivalently,

\nu^{+rel}:=min\{i|V^{rel}_{i}=0\}

Similar to how McCoy [12] defined $T_{m}$ , we define $T^{rel}_{m}$ be the number of relevant coefficients less than or equal to $m$ . Equivalently,

T^{rel}_{m}:=|\{0\leq i<\nu^{+rel}\mid 0<V^{rel}_{i}\leq m\}|

Note that due to monotonicity, we have $\nu^{+rel}=T^{rel}_{V^{rel}_{0}}$ .

We define $S_{m}=\{\alpha\in\mathbb{Z}^{dim(\sigma)}_{\geq 0}\mid\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)=2m\}$ . This is exactly the same as how McCoy defined $S_{m}$ in [12].

We also define $S_{\leq m}=\{\alpha\in\mathbb{Z}^{dim(\sigma)}_{\geq 0}\mid\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)\leq 2m\}$ . Due to the fact that consecutive relevant coefficients can decrease by more than 1, some arguments later in this section will not work if we use $S_{m}$ like how it is used in [12]. Instead, we will need to consider $S_{\leq m}$ .

First we prove the following properties, which is an analog of [12, Lemma 2.10]:

Lemma 4.2.

The following three statements are true:
(1) For all $m$ satisfying $0\leq m<V^{rel}_{0}$ , we have $T^{rel}_{m}=\displaystyle\max_{\alpha\in S_{\leq m}}\sigma\cdot\alpha$

(2) $\nu^{+rel}=\begin{cases}-\dfrac{1}{2}+\dfrac{1}{2}(rp-|\sigma|_{1})\text{ if }r\text{ is even and }p\text{ is odd}\\ \dfrac{1}{2}(rp-|\sigma|_{1})\text{ otherwise}\end{cases}$

(3) $\nu^{+rel}=T^{rel}_{V^{rel}_{0}}\leq\displaystyle\max_{\alpha\in S_{V^{rel}_{0}}}\sigma\cdot\alpha$

Note that when $r\geq 2$ , statement (1) holds for $m=V^{rel}_{0}$ as well. We will prove this in Section 6 (Corollary 6.13) after we study more about the behavior of relevant coefficients.

Proof of Lemma 4.2.

Lemma 3.10 states when for any relevant $i$ ,

-8V_{i}=\max_{\begin{subarray}{c}\mathfrak{c}\in Char(-\mathbb{Z}^{n+1})\\ \langle\mathfrak{c},[\Sigma]\rangle+r^{2}p=2i\end{subarray}}(\mathfrak{c}^{2}+(n+1))

which can be rewritten as

	$\displaystyle-8V^{rel}_{i}$	$\displaystyle=\max_{\begin{subarray}{c}\mathfrak{c}\in Char(-\mathbb{Z}^{n+1})\\ r\langle\mathfrak{c},\sigma\rangle+r^{2}p=2ir+r\end{subarray}}(\mathfrak{c}^{2}+(n+1))\text{ if }r\text{ is even and }p\text{ is odd}$
	$\displaystyle-8V^{rel}_{i}$	$\displaystyle=\max_{\begin{subarray}{c}\mathfrak{c}\in Char(-\mathbb{Z}^{n+1})\\ r\langle\mathfrak{c},\sigma\rangle+r^{2}p=2ir\end{subarray}}(\mathfrak{c}^{2}+(n+1))\text{ otherwise}$

The condition of $\mathfrak{c}$ being a characteristic vector can be stated as $\mathfrak{c}=2\alpha+(1,\dots,1)$ for some $\alpha\in\mathbb{Z}^{n+1}$ . Now we can rewrite the above as

	$\displaystyle-8V^{rel}_{i}$	$\displaystyle=\max_{rp-\|\sigma\|_{1}-2\sum\limits_{j}\sigma_{j}\alpha_{j}=2i+1}((n+1)-\sum\limits_{j}((2\alpha_{j}+1)^{2}))\text{ if }r\text{ is even and }p\text{ is odd}$
	$\displaystyle-8V^{rel}_{i}$	$\displaystyle=\max_{rp-\|\sigma\|_{1}-2\sum\limits_{j}\sigma_{j}\alpha_{j}=2i}((n+1)-\sum\limits_{j}((2\alpha_{j}+1)^{2}))\text{ otherwise}$

This can further simplify to

(6)		$\displaystyle 2V^{rel}_{i}$	$\displaystyle=\min_{rp-\|\sigma\|_{1}-2\sum\limits_{j}\sigma_{j}\alpha_{j}=2i+1}\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)\text{ if }r\text{ is even and }p\text{ is odd}$
	$\displaystyle 2V^{rel}_{i}$	$\displaystyle=\min_{rp-\|\sigma\|_{1}-2\sum\limits_{j}\sigma_{j}\alpha_{j}=2i}\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)\text{ otherwise}$

For any $m$ with $0\leq m<V^{rel}_{0}$ , consider the minimal integer $i$ such that there exists $\alpha$ that satisfies

(7)

rp-|\sigma|_{1}-2\sum\limits_{j}\sigma_{j}\alpha_{j}=\begin{cases}2i+1\text{ if }r\text{ is even and }p\text{ is odd}\\ 2i\text{ otherwise}\end{cases}

and

\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)\leq 2m

Notice that all entries of such an $\alpha$ must be non-negative because if there is any negative entry $\alpha_{j}$ , we can replace it with $-1-\alpha_{j}$ which lowers $i$ but doesn’t change $\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)$ . Hence, such an $\alpha$ must be in $S_{\leq m}$ . (Note that the entries being non-negative is in the definition of $\alpha\in S_{\leq m}$ )

From the argument above, we see that if such an $i$ is non-negative, we have $i=min\{t\mid V^{rel}_{t}\leq m\}$ .

Now we argue that $m<V^{rel}_{0}$ implies such an $i$ being non-negative: Assume the contrary and suppose such a minimal $i$ is negative. Let $k$ be the smallest positive integer such that there exists $\alpha\in S_{\leq m}$ with

rp-|\sigma|_{1}-2\sum\limits_{j}\sigma_{j}\alpha_{j}=\begin{cases}2(-k)+1\text{ if }r\text{ is even and }p\text{ is odd}\\ 2(-k)\text{ otherwise}\end{cases}

Let $l$ be maximal such that $\alpha_{l}\neq 0$ . If $k>\sigma_{l}$ , we can decrease $\alpha_{l}$ by 1 to produce a new $\alpha$ that is still in $S_{\leq m}$ but $k$ decreases by $\sigma_{l}$ , contradicting with the minimality of $k$ . Hence, we know that $k\leq\sigma_{l}$ . Since $\sigma$ is a changemaker, $\exists A\subseteq\{l,\dots,n\}$ such that $k=\sum\limits_{j\in A}\sigma_{j}$ . Consider $\alpha^{\prime}$ where

\alpha^{\prime}_{j}:=\begin{cases}\alpha_{j}-1\text{ if }j\in A\\ \alpha_{j}\text{ otherwise}\end{cases}

Then $\alpha^{\prime}\in\mathbb{Z}_{\geq-1}^{n+1}$ satisfies $\sum\limits_{j}\alpha^{\prime}_{j}(\alpha^{\prime}_{j}+1)\leq m$ and

rp-|\sigma|_{1}-2\sum\limits_{j}\sigma_{j}\alpha^{\prime}_{j}=\begin{cases}1\text{ if }r\text{ is even and }p\text{ is odd}\\ 0\text{ otherwise}\end{cases}

(6) implies that $V_{0}^{rel}\leq m$ , which contradicts with $m<V^{rel}_{0}$ . Hence, we can conclude that $m<V^{rel}_{0}$ implies such an $i$ being non-negative, and therefore we have $i=min\{t\mid V^{rel}_{t}\leq m\}$ . So, we can conclude that

min\{t\mid V^{rel}_{t}\leq m\}=\begin{cases}\displaystyle\min_{\alpha\in S_{\leq m}}\frac{1}{2}\left(rp-|\sigma|_{1}-1-2\sum\limits_{j}\sigma_{j}\alpha_{j}\right)\text{ if }r\text{ is even and }p\text{ is odd}\\ \displaystyle\min_{\alpha\in S_{\leq m}}\frac{1}{2}\left(rp-|\sigma|_{1}-2\sum\limits_{j}\sigma_{j}\alpha_{j}\right)\text{ otherwise}\end{cases}

Setting $m=0$ gives statement (2) of Lemma 4.2. By monotonicity of relevant coefficients, we know that $T^{rel}_{m}=\nu^{+rel}-min\{t\mid V^{rel}_{t}\leq m\}$ . That gives statement (1) of Lemma 4.2.

This argument does not fully apply to the $m=V^{rel}_{0}$ because of the issue of $i$ potentially being negative, causing it to not represent a relevant coefficient. However, we can still apply (6) to see that there exists some $\alpha\in\mathbb{Z}^{n+1}$ such that $2V^{rel}_{0}=\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)$ and

rp-|\sigma|_{1}-2\sum\limits_{j}\sigma_{j}\alpha_{j}=\begin{cases}1\text{ if }r\text{ is even and }p\text{ is odd}\\ 0\text{ otherwise}\end{cases}

We see that $\nu^{+rel}=T^{rel}_{V^{rel}_{0}}=\sigma\cdot\alpha$ for this particular $\alpha$ . Hence,

\nu^{+rel}=T^{rel}_{V^{rel}_{0}}\leq\displaystyle\max_{\begin{subarray}{c}\alpha\in\mathbb{Z}^{n+1}\\ 2V^{rel}_{0}=\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)\end{subarray}}\sigma\cdot\alpha

If $\alpha$ has any negative entry $\alpha_{j}$ , we can replace it with $-1-\alpha_{j}$ which increases $\sigma\cdot\alpha$ but keeping $\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)$ unchanged. Hence, we can take the max to be over $\alpha\in S_{V^{rel}_{0}}$ , which gives statement (3) of Lemma 4.2. ∎

Remark 1: Since relevant coefficients can skip numbers, $2m$ may not equal to $V^{rel}_{i}$ for any $i$ , and hence we need to introduce $S_{\leq m}$ in this paper instead of just $S_{m}$ as in [12, §2].

Remark 2: The reason why the $m=V^{rel}_{0}$ case was dealt with separately in the proof is because of the issue of $i$ potentially being negative. If there is some condition that implies $\max\limits_{\alpha\in S_{V^{rel}_{0}}}\sigma\cdot\alpha-\max\limits_{\alpha\in S_{V^{rel}_{0}-1}}\sigma\cdot\alpha$ to be at most 1, since the minimal $i$ for $m=V^{rel}_{0}-1$ in (7) is positive, such an $i$ for $m=V^{rel}_{0}$ must be non-negative. Hence, under such condition, statement (1) of Lemma 4.2 would also apply to $m=V^{rel}_{0}$ . In Section 6, we will show that $r\geq 2$ is such a condition.

Most of the remaining content of this section is for proving:

Theorem 1.7.

Many ideas that go into the proof come from [12, §2].

First, we prove

Lemma 4.3.

Statement (1) of Theorem 1.7 is true.

Proof.

When $r=1$ , the conditions in Lemma 4.2 become the conditions in [12, Lemma 2.10]. Therefore, arguments in [12, §2-4] apply directly. ∎

For statement (2) and (3), notice that specifying $K$ and with the value of $r$ , and also specifying the parity of $p$ when $r$ is even, together determines all the relevant coefficients. So it suffices to show that the relevant coefficients and the value of $r$ , and the parity of $p$ when $r$ is even, together determines $\sigma$ . That is because $\sigma$ determines $p$ , which determines the surgery slope $r^{2}p$ when we know the value of $r$ .

We deal with the trivial cases first.

Lemma 4.4.

If $V^{rel}_{0}=0$ and $r\geq 2$ , then $r=2$ and $\sigma=(1)$ .

If $V^{rel}_{0}=1$ and $r\geq 2$ , then it must be one of the following cases:
$r=2$ and $\sigma=(1,1)\ or\ (1,1,1)$
$r=3$ and $\sigma=(1)$
$r=4$ and $\sigma=(1)$

If $V^{rel}_{0}=2$ and $r\geq 2$ and $\sigma_{0}=2$ and $\sigma_{1}=1$ , then $r=2$ and $\sigma=(2,1)$ .

Proof.

If $V^{rel}_{0}=0$ , that means there are no non-zero relevant coefficients. Therefore

0=\nu^{+rel}=\begin{cases}-\dfrac{1}{2}+\dfrac{1}{2}(rp-|\sigma|_{1})\text{ if }r\text{ is even and }p\text{ is odd}\\ \dfrac{1}{2}(rp-|\sigma|_{1})\text{ otherwise}\end{cases}

Since $p\geq|\sigma|_{1}$ , given that $r\geq 2$ , the only way this expression can be 0 is when $r=2$ and $p=|\sigma|_{1}=1$ .

If $V^{rel}_{0}=1$ , that means all non-zero relevant coefficients are $1$ . Hence we have $\nu^{+rel}=T^{rel}_{1}$ . Now we have

0\neq\nu^{+rel}=T^{rel}_{1}\leq\max_{\alpha\in S_{1}}\sigma\cdot\alpha=\sigma_{0}

Hence we have

\begin{cases}-\dfrac{1}{2}+\dfrac{1}{2}(r\sigma_{0}^{2}-\sigma_{0})\leq-\dfrac{1}{2}+\dfrac{1}{2}(rp-|\sigma|_{1})=g^{rel}\leq\sigma_{0}\text{ if }r\text{ is even and }p\text{ is odd}\\ \dfrac{1}{2}(r\sigma_{0}^{2}-\sigma_{0})\leq\dfrac{1}{2}(rp-|\sigma|_{1})=g^{rel}\leq\sigma_{0}\text{ otherwise}\end{cases}

Rearranging, we have

r\sigma_{0}\leq\begin{cases}3+\frac{1}{\sigma_{0}}\text{ if }r\text{ is even and }p\text{ is odd}\\ 3\text{ otherwise}\end{cases}

So, we have $r\leq 4$ . Given that $r\geq 2$ , we also have $\sigma_{0}=1$ . So we can conclude that $\nu^{+rel}\leq 1$ and all entries of $\sigma$ are $1$ . Since $\nu^{+rel}\neq 0$ , we have $\nu^{+rel}=1$ . By considering the expression for $\nu^{+rel}$ in terms of $r,p,|\sigma|_{1}$ , we can see that the only possibilities are as stated in the lemma.

If $V^{rel}_{0}=2$ and $r\geq 2$ and $\sigma_{0}=2$ and $\sigma_{1}=1$ , then

\nu^{+rel}=T^{rel}_{2}\leq\max_{\alpha\in S_{2}}\sigma\cdot\alpha=\sigma_{0}+\sigma_{1}=3

Let $x$ be the number of 1’s in $\sigma$ . Then

	$\displaystyle\nu^{+rel}$	$\displaystyle\geq-\dfrac{1}{2}+\dfrac{1}{2}(rp-\|\sigma\|_{1})$
		$\displaystyle=-\dfrac{1}{2}+\dfrac{1}{2}\left(r(4+x)-(2+x)\right)$
		$\displaystyle=2r-1+\dfrac{(r-1)x-1}{2}$

Hence, we must have $r=2$ and $x=1$ . ∎

Theorem 1.4 comes as a corollary.

Theorem 1.4.

Let $K$ be a knot in $S^{3}$ with $\nu^{+}(K)=0$ and let $m$ be a positive integer. If the $m$ -surgery on $K$ is lensbordant, then it is smoothly rational homology cobordant to $L(m,m-1)$ .

Proof of Theorem 1.4.

When $\nu^{+}=0$ , Lemma 4.4 implies that either $r=1$ , or $r=2$ with $\sigma=(1)$ .

The changemaker $\sigma=(1)$ represents $S^{3}$ . Hence, in the $r=2$ with $\sigma=(1)$ case, $m=4$ and $Y$ is smoothly rational homology cobordant to $S^{3}$ . Since $S^{3}$ is smoothly rational homology cobordant to $L(4,3)$ , $Y$ is also smoothly rational homology cobordant to $L(4,3)$ .

When $r=1$ , the conditions in Lemma 4.2 become the conditions in [12, Lemma 2.10]. Therefore, when $\nu^{+}(K)=0$ , all entries of $\sigma$ are 1’s [12, Lemma 2.12]. These changemakers correspond to $L(m,m-1)$ . ∎

Given Lemma 4.3 and 4.4, from this point onwards, until the end of this section, we assume that $r\geq 2$ and $V^{rel}_{0}\geq 2$ .

Now we can define

\mu:=\min_{1\leq i<V^{rel}_{0}}(T^{rel}_{i}-T^{rel}_{i-1})

Note that with $r\geq 2$ , $\mu$ can be 0 here. This is very different from what McCoy did in [12] where $\mu$ must be at least 1.

Lemma 4.5.

V^{rel}_{0}\geq\begin{cases}\frac{1}{8}((r^{2}p-2r+1)-(n+1))\text{ if }r\text{ is even and }p\text{ is odd}\\ \frac{1}{8}(r^{2}p-(n+1))\text{ otherwise}\end{cases}

Proof.

At the start of the proof of Lemma 4.2, we see that

-8V^{rel}_{0}=\mathfrak{c}^{2}+(n+1)

for some characteristic vector $\mathfrak{c}$ satisfying

\langle\mathfrak{c},\sigma\rangle+rp=\begin{cases}1\text{ if }r\text{ is even and }p\text{ is odd}\\ 0\text{ otherwise}\end{cases}

By Cauchy-Schwarz inequality and the fact that $\sigma^{2}=-p$ , we have

|\mathfrak{c}^{2}|=\frac{|\mathfrak{c}^{2}||\sigma^{2}|}{p}\geq\frac{|\langle\mathfrak{c},\sigma\rangle|^{2}}{p}\geq\begin{cases}\frac{(rp-1)^{2}}{p}\text{ if }r\text{ is even and }p\text{ is odd}\\ \frac{(rp)^{2}}{p}\text{ otherwise}\end{cases}

Simplifying the expressions and considering the fact that $|\mathfrak{c}^{2}|\in\mathbb{Z}$ , we have

|\mathfrak{c}^{2}|\geq\begin{cases}r^{2}p-2r+1\text{ if }r\text{ is even and }p\text{ is odd}\\ r^{2}p\text{ otherwise}\end{cases}

Result follows by substituting this back to $-8V^{rel}_{0}=\mathfrak{c}^{2}+(n+1)$ . ∎

Lemma 4.6.

If $\mu\geq 2$ , then $V_{0}^{rel}=2$ and $\sigma_{0}=2$ and $\sigma_{1}=1$ .

Proof.

Suppose $\mu\geq 2$ . We define

M:=\begin{cases}\lceil\frac{1}{8}((r^{2}p-2r+1)-(n+1))\rceil-1\text{ if }r\text{ is even and }p\text{ is odd}\\ \lceil\frac{1}{8}(r^{2}p-(n+1))\rceil-1\text{ otherwise}\end{cases}

which is an analogue of $N-1$ in [12, §2].

By Lemma 4.5, we have $V_{0}^{rel}>M$ . Hence, by Lemma 4.2, we know that

T^{rel}_{M}=\displaystyle\max_{\alpha\in S_{\leq M}}\sigma\cdot\alpha

Take such an $\alpha$ .

First of all, we argue that $\alpha\in S_{M}$ (without the $\leq$ ). If $\alpha\not\in S_{M}$ , then we must have $M\geq 1$ and $\max\limits_{\alpha\in S_{\leq M}}\sigma\cdot\alpha=\max\limits_{\alpha\in S_{\leq M-1}}\sigma\cdot\alpha$ . This implies $T^{rel}_{M}=T^{rel}_{M-1}$ , which contradicts with $\mu\geq 2$ . Hence, $\alpha\in S_{M}$ .

For any index $l$ with $\alpha_{l}>0$ , if we consider $\alpha^{{}^{\prime}}$ defined with $\alpha^{{}^{\prime}}_{i}=\begin{cases}\alpha_{i}\text{ if }i\neq l\\ \alpha_{i}-1\text{ if }i=l\end{cases}$ , we have $\alpha^{{}^{\prime}}\in S_{\leq M-\alpha_{l}}$ . So we have

T^{rel}_{M-\alpha_{l}}\geq\alpha^{{}^{\prime}}\cdot\sigma=\alpha\cdot\sigma-\sigma_{l}=T^{rel}_{M}-\sigma_{l}

Hence,

\sigma_{l}\geq T^{rel}_{M}-T^{rel}_{M-\alpha_{l}}\geq\alpha_{l}\mu

Since $\mu\geq 2$ , from our discussion above, for all index $l$ we have $\sigma_{l}\geq 2\alpha_{l}$ (this also holds when $\alpha_{l}=0$ because the right hand side becomes $0$ ). Hence,

	$\displaystyle M$	$\displaystyle=\dfrac{1}{2}\displaystyle\sum_{l}\alpha_{l}(\alpha_{l}+1)$
		$\displaystyle\leq\dfrac{1}{2}\displaystyle\sum_{l}\dfrac{\sigma_{l}}{2}\left(\dfrac{\sigma_{l}}{2}+1\right)$
		$\displaystyle=\dfrac{1}{8}\displaystyle\sum_{l}\sigma_{l}(\sigma_{l}+2)$
		$\displaystyle=\dfrac{1}{8}(p+2\|\sigma\|_{1})$

Combining this with the definition of $M$ , we have

\dfrac{1}{8}(p+2|\sigma|_{1})\geq\begin{cases}\left\lceil\frac{1}{8}((r^{2}p-2r+1)-(n+1))\right\rceil-1\text{ if }r\text{ is even and }p\text{ is odd}\\ \left\lceil\frac{1}{8}(r^{2}p-(n+1))\right\rceil-1\text{ otherwise}\end{cases}

Since $r\geq 2$ and $|\sigma|_{1}\geq(n+1)$ (because $\sigma$ has no $0$ entries), we have

\dfrac{1}{8}(p+2|\sigma|_{1})\geq\begin{cases}\frac{1}{8}((4p-4+1)-|\sigma|_{1})-1\text{ if }r\text{ is even and }p\text{ is odd}\\ \frac{1}{8}(4p-|\sigma|_{1})-1\text{ otherwise}\end{cases}

Rearranging this, we get

\begin{cases}\dfrac{11}{3}\geq p-|\sigma|_{1}\text{ if }r\text{ is even and }p\text{ is odd}\\[12.91663pt] \dfrac{8}{3}\geq p-|\sigma|_{1}\text{ otherwise}\end{cases}

Since $p-|\sigma|_{1}$ is an integer, this implies

3\geq\displaystyle\sum_{l}(\sigma_{l}^{2}-\sigma_{l})

Therefore, the entries of $\sigma$ is either a 2 followed by a bunch of 1’s or simply all 1’s.

Since $V^{rel}_{0}\geq 2$ , we have $T^{rel}_{0}=0$ and $T^{rel}_{1}=\sigma_{0}$ . Also, if $V^{rel}_{0}\geq 3$ , we have $T^{rel}_{2}=\sigma_{0}+\sigma_{1}$ . Therefore, the only way for $\mu$ to be at least 2 is when $V^{rel}_{0}=2$ and $\sigma$ is a 2 followed by a bunch of 1’s. ∎

Now we can prove Theorem 1.7. Statement (1) of Theorem 1.7 has already been proven in Lemma 4.3. So we only need to prove statements (2) and (3) of Theorem 1.7.

Proof of statement (2) and (3) of Theorem 1.7.

Fix the knot $K$ . Fix $r\geq 2$ . Also, if $r$ is even, we fix the parity of $p$ (odd or even).

We want to show that these together determines $p$ , which determines the surgery slope $r^{2}p$ since $r$ is fixed.

These data determines the relevant coefficients. So, it suffices to show that the relevant coefficients together with the value of $r$ determines $\sigma$ (which determines $p$ ).

The algorithm McCoy described in the proof of [12, Lemma 2.16] shows that statement (1) of our Lemma 4.2 determines all entries of $\sigma$ strictly greater than $\mu$ . Together with Lemma 4.6, this implies that all entries of $\sigma$ except for the 1’s are determined, except for a special case which was addressed in Lemma 4.4. Given that $r\geq 2$ , statement (2) of Lemma 4.2 determines the number of 1’s. ∎

Figure 2. Flowchart showing McCoy’s algorithm and argument, but in notations used in this paper.

5. Some bounds on the surgery slope $r^{2}p$

In this section, we establish some inequalities involving the surgery slope $r^{2}p$ .

Theorem 1.8.

Suppose there is an $(r,p)$ -lensbordant surgery on a knot $K$ in $S^{3}$ with associated changemaker $\sigma$ , then

2\nu^{+}(K)+2(r-1)\geq r^{2}p-r|\sigma|_{1}\geq 2\nu^{+}(K)

Proof.

When $\nu^{+rel}=0$ , Lemma 4.4 implies that either $r=2$ with $\sigma=(1)$ , or $r=1$ . When $r=2$ and $\sigma=(1)$ , the inequality becomes $2\geq 2\geq 0$ . When $r=1$ , $V_{0}=V^{rel}_{0}=0$ , and hence $\nu^{+}=0$ . As discussed in the proof of Theorem 1.4, $\sigma$ must be $(1,\dots,1)$ in this case, and the inequality becomes $0\geq 0\geq 0$ .

Now we assume that $\nu^{+rel}\neq 0$

Case 1: $r$ is odd or $p$ is even
We have $\nu^{+rel}=\dfrac{1}{2}(rp-|\sigma|_{1})$ . Hence,

V^{rel}_{\tfrac{1}{2}(rp-|\sigma|_{1})}=0\quad\text{and}\quad V^{rel}_{\tfrac{1}{2}(rp-|\sigma|_{1})-1}\neq 0

So we have

V_{\tfrac{1}{2}(r^{2}p-r|\sigma|_{1})}=0\quad\text{and}\quad V_{\tfrac{1}{2}(r^{2}p-r|\sigma|_{1})-r}\neq 0

This implies

\dfrac{1}{2}(r^{2}p-r|\sigma|_{1})\geq\nu^{+}>\dfrac{1}{2}(r^{2}p-r|\sigma|_{1})-r

Rearranging, we have

2\nu^{+}+2r>r^{2}p-r|\sigma|_{1}\geq 2\nu^{+}

Since $p\equiv|\sigma|_{1}$ (mod 2), the value $r^{2}p-r|\sigma|_{1}$ must be an even integer. Hence

2\nu^{+}+2(r-1)\geq r^{2}p-r|\sigma|_{1}\geq 2\nu^{+}

Case 2: $r$ is even and $p$ is odd
We have $\nu^{+rel}=-\dfrac{1}{2}+\dfrac{1}{2}(rp-|\sigma|_{1})$ . Hence,

V^{rel}_{-\tfrac{1}{2}+\tfrac{1}{2}(rp-|\sigma|_{1})}=0\quad\text{and}\quad V^{rel}_{-\tfrac{3}{2}+\tfrac{1}{2}(rp-|\sigma|_{1})}\neq 0

So we have

V_{\tfrac{1}{2}(r^{2}p-r|\sigma|_{1})}=0\quad\text{and}\quad V_{\tfrac{1}{2}(r^{2}p-r|\sigma|_{1})-r}\neq 0

The rest of the proof is the same as Case 1. ∎

Now we prove Corollary 1.9.

Corollary 1.9.

Proof.

Since $|\sigma|_{1}\geq 1$ , Theorem 1.8 implies that $r^{2}p\geq 2\nu^{+}+r$ .

Since $p\geq|\sigma|_{1}$ , Theorem 1.8 implies that $2\nu^{+}+2(r-1)\geq r(r-1)p$ . So we have

(8)

\dfrac{2r}{r-1}\nu^{+}+2r\geq r^{2}p

Then we have

\dfrac{r-2}{r}r^{2}p=r^{2}p-\dfrac{2}{r}r^{2}p\leq\left(\dfrac{2r}{r-1}\nu^{+}+2r\right)-2r=\dfrac{2r}{r-1}\nu^{+}

Statement (1) follows.

When $r\geq 6$ , $\dfrac{2r^{2}}{(r-1)(r-2)}<4$ . So, we only need to check statement (2) for $r=2,3,4,5$ .

When $r=2$ , (8) implies $r^{2}p\leq 4\nu^{+}+4$

By Lemma 4.4, when $r\geq 3$ , we have $\nu^{+}\geq 1$ .
When $r=3$ , (8) implies $r^{2}p\leq 3\nu^{+}+6\leq 4\nu^{+}+5$ .

When $r=4$ and $\nu^{+}\geq 3$ , (8) implies $r^{2}p\leq\tfrac{8}{3}\nu^{+}+8=4\nu^{+}+4-\tfrac{4}{3}(\nu^{+}-3)\leq 4\nu^{+}+4$ .
When $r=4$ and $\nu^{+}\leq 2$ , statement (1) implies $r^{2}p\leq\tfrac{16}{3}\nu^{+}=4\nu^{+}+\tfrac{8}{3}-\tfrac{4}{3}(2-\nu^{+})\leq 4\nu^{+}+\tfrac{8}{3}$ .

When $r=5$ and $\nu^{+}\geq 4$ , (8) implies $r^{2}p\leq\tfrac{5}{2}\nu^{+}+10=4\nu^{+}+4-\tfrac{3}{2}(\nu^{+}-4)\leq 4\nu^{+}+4$ .
When $r=5$ and $\nu^{+}\leq 3$ , statement (1) implies $r^{2}p\leq\tfrac{25}{6}\nu^{+}=4\nu^{+}+\tfrac{1}{2}-\tfrac{1}{6}(3-\nu^{+})\leq 4\nu^{+}+\tfrac{1}{2}$ . ∎

6. Structure of relevant coefficients

In the introduction, the remark after the statement of Theorem 1.10 states that the surgery slope $r^{2}p$ is very close to $8V_{0}(K)$ . The precise statement is

Theorem 6.1.

(1) If $r,p$ are both even, then $r^{2}p=8V_{0}(K)$ .
(2) If $r$ is even and $p$ is odd, then $8V_{0}(K)-2r\leq r^{2}p\leq 8V_{0}(K)+2r$ .
(3) If $r$ is odd, then $8V_{0}(K)-3odd(\sigma)\leq r^{2}p\leq 8V_{0}(K)+odd(\sigma)$
where $odd(\sigma)$ is the number of odd entries in $\sigma$ .

Furthermore, in the case when $r$ is odd, $r^{2}p=8V_{0}(K)+odd(\sigma)$ if and only if the even entries of $\sigma$ can be partitioned into two sets with equal sum.

Note that Theorem 1.10 is statement (1) of Theorem 6.1. The aim of this section is to gain a deeper understanding on relevant coefficients, and then prove Theorem 6.1.

	$r$ even	$r$ odd
$p$ even	$r^{2}p=8V_{0}$	$8V_{0}-3odd(\sigma)\leq r^{2}p\leq 8V_{0}+odd(\sigma)$
		upper bound is equality if and only if
$p$ odd	$8V_{0}-2r\leq r^{2}p\leq 8V_{0}+2r$	the even entries of $\sigma$ can be partitioned
		into two sets with equal sum

Table 1. Table describing Theorem 6.1.

The $T^{rel}_{m}$ in statement (1) of Lemma 4.2 is defined with the relevant coefficients. Besides $m$ , it also depends on the knot $K$ , the value of $r$ , and the parity of $p$ if $r$ is even. However, the right hand side $\displaystyle\max_{\alpha\in S_{\leq m}}\sigma\cdot\alpha$ only depends on $m$ and $\sigma$ . Hence, it makes sense to think of that as some coefficients coming from $\sigma$ itself.

Definition 6.2.

For $m\geq 0$ , we define

T^{\sigma}_{m}:=\displaystyle\max_{\alpha\in S_{\leq m}}\sigma\cdot\alpha

and we define the relevant coefficients $V^{\sigma}_{i}$ (for $i>0$ ) of the changemaker vector $\sigma$ to be the monotonic increasing sequence of positive integers such that $T^{\sigma}_{m}$ is the number of relevant coefficients less than or equal to $m$ . Equivalently,

V^{\sigma}_{i}:=\min\{m\mid i\leq T^{\sigma}_{m}\},\quad i\in\mathbb{Z}_{>0}

We also define $V^{\sigma}_{0}$ to be $0$ , but we do not call it a relevant coefficient to avoid making the above definition more complicated.

Now statement (1) and (3) of Lemma 4.2 can be rewritten as:

T^{rel}_{m}=T^{\sigma}_{m}\text{ for all }m<V^{rel}_{0}

and

T^{rel}_{V^{rel}_{0}}\leq T^{\sigma}_{V^{rel}_{0}}

Together they imply

V^{rel}_{i}=V^{\sigma}_{\nu^{+rel}-i}\text{ for all }0\leq i<\nu^{+rel}

Recall that the definition of $S_{\leq m}$ is

S_{\leq m}:=\{\alpha\in\mathbb{Z}^{dim(\sigma)}_{\geq 0}\mid\Sigma\alpha_{j}(\alpha_{j}+1)\leq 2m\}

Notice that $\tfrac{1}{2}\alpha_{j}(\alpha_{j}+1)$ is exactly $1+2+\dots+\alpha_{j}$ . Hence, we can interpret $T^{\sigma}_{m}$ as the answer to the following combinatorial problem:

Consider the following game. There is a shop where you can spend coins to buy items. The available items are labeled $I_{0},\dots,I_{n}$ . You can buy any non-negative integer amount of each of the items. For each individual item, it costs 1 coin to buy the first one, 2 coins to buy the second one, 3 coins to buy the third one, etc. After finish doing all the purchases, each $I_{i}$ you own gives you $\sigma_{i}$ points. If you start with $m$ coins, what is the highest amount of points that you can get?

A way to approach this game is to consider the coin cost per point in each purchase option. Purchasing the item $I_{i}$ for the $k^{th}$ time costs $k$ coins to obtain $\sigma_{i}$ points, hence it is a $\dfrac{k}{\sigma_{i}}$ coin per point purchase. One intuitive thing to do is the use a greedy algorithm by always making the purchase with the lowest coin cost per point among all available purchase options until running out of coins.

This greedy algorithm does not always give the optimal solution. At the end, you might have some coins left that is not enough to purchase the next lowest coin per point purchase option and are forced to purchase a cheaper option with a higher cost per point, or simply having some leftover coins with no available purchase options.

However, consider a modified version of the game where you are allowed to make fractional purchases under the same coin per point. For example, being able to buy $(4+\tfrac{2}{3})$ copies of $I_{i}$ by spending $(1+2+3+4+\tfrac{2}{3}\times 5)$ coins, gaining $(4+\tfrac{2}{3})\sigma_{i}$ points.

In this modified version of the game, a greedy algorithm will not result in any leftover coins, and all the purchased options have a lower or equal coin per point than any unpurchased option. Hence, a greedy algorithm must give the optimal solution.

Definition 6.3.

We call a greedy algorithm on the modified version of the game a rational greedy algorithm. We write $T^{\sigma,\mathbb{Q}}_{m}$ to denote the amount of points obtained with $m$ coins under the rational greedy algorithm.

Given $m,\sigma$ , we say that the resulting rational greedy algorithm is up to $x\in\mathbb{Q}$ coin per point if the resulting rational greedy algorithm makes a non-zero a amount of purchase options with $x$ coin per point and no purchase options with $>x$ coin per point.

Because of the way rational greedy algorithm works, a rational greedy algorithm up to $x$ coin per point must make all available purchases that are $<x$ coin per point.

Also, given $\sigma$ and $m$ , if a rational greedy algorithm is up to $x$ coin per point, then all other optimal solutions to the modified version of the game must also take the same form: Purchasing all available purchases that are $<x$ coin per point, and not purchasing any options that are $>x$ coin per point. That is because any other way of spending all $m$ coins will result in having a higher average coin per point, and thus a lower total point at the end.

Hence, in this modified version of the game, all optimal solutions must be rational greedy algorithms up to $x$ coin per cost. The only difference allowed in different optimal solutions is picking different $x$ coin per point purchase options.

Definition 6.4.

If an optimal solution in the modified version of the game does not include fractional purchases, we say that the solution is integral. Equivalently, an integral optimal solution in the modified version of the game is a solution that is valid in the original version of the game.

If an optimal solution in the modified version of the game is a rational greedy algorithm up to $x$ coin per point that purchases all options that are $x$ coin per point, we say that the solution is full.

Clearly, a full rational greedy algorithm must also be integral.

All purchase options in the original version of the game are valid purchase options in the modified version of the game. Hence, $T^{\sigma,\mathbb{Q}}_{m}\geq T^{\sigma}_{m}$ . In particular, $T^{\sigma,\mathbb{Q}}_{m}=T^{\sigma}_{m}$ if and only if there exist an integral implementation of the rational greedy algorithm to the modified version of the game.

In this section we will use these ideas to help investigating the structure of relevant coefficients.

Lemma 6.5.

For any positive integer $x$ ,

T^{\sigma}_{\tfrac{x^{2}p+x|\sigma|_{1}}{2}}=xp

Proof.

Consider the full rational greedy algorithm up to $x$ coin per point. So, for each $i$ , it purchases $x\sigma_{i}$ copies of the item $I_{i}$ . The total amount of coins spent is

	$\displaystyle\dfrac{1}{2}\sum\limits_{i}(x\sigma_{i})(x\sigma_{i}+1)$	$\displaystyle=\dfrac{1}{2}x^{2}\sum\limits_{i}\sigma_{i}+\dfrac{1}{2}x\sum\limits_{i}\sigma_{i}$
		$\displaystyle=\dfrac{x^{2}p+x\|\sigma\|_{1}}{2}$

And the number of points gained is

\sum\limits_{i}(x\sigma_{i})\sigma_{i}=xp

Hence, $T^{\sigma,\mathbb{Q}}_{\tfrac{x^{2}p+x|\sigma|_{1}}{2}}=xp$ . Since the greedy algorithm is full, it must be integral. Therefore,

T^{\sigma,\mathbb{Q}}_{\tfrac{x^{2}p+x|\sigma|_{1}}{2}}=T^{\sigma}_{\tfrac{x^{2}p+x|\sigma|_{1}}{2}}

∎

Corollary 6.6.

For any positive integer $x$ ,

V^{\sigma}_{xp}=\dfrac{x^{2}p+x|\sigma|_{1}}{2}

Proof.

From Lemma 6.5, we know there is exactly $xp$ relevant coefficients less than or equal to $\dfrac{x^{2}p+x|\sigma|_{1}}{2}$ . Since the relevant coefficients of a changemaker vector are monotonic increasing, we know that

V^{\sigma}_{xp}\leq\dfrac{x^{2}p+x|\sigma|_{1}}{2}

To prove $V^{\sigma}_{xp}=\dfrac{x^{2}p+x|\sigma|_{1}}{2}$ , it suffices to show that

T^{\sigma}_{\tfrac{x^{2}p+x|\sigma|_{1}}{2}-1}<xp

i.e. there are strictly less than $xp$ relevant coefficients that are less than or equal to $\dfrac{x^{2}p+x|\sigma|_{1}}{2}-1$ (which implies that $V^{\sigma}_{xp}$ cannot be less than or equal to $\tfrac{x^{2}p+x|\sigma|_{1}}{2}-1$ )

This follows from the way Lemma 6.5 is proved. $xp$ points is obtained from a rational greedy algorithm with $\tfrac{x^{2}p+x|\sigma|_{1}}{2}$ coins. Since every coin contributes to some points under a rational greedy algorithm, strictly less than $xp$ points is obtained from a rational greedy algorithm with strictly less than $\tfrac{x^{2}p+x|\sigma|_{1}}{2}$ coins. Hence

T^{\sigma}_{\tfrac{x^{2}p+x|\sigma|_{1}}{2}-1}\leq T^{\sigma,\mathbb{Q}}_{\tfrac{x^{2}p+x|\sigma|_{1}}{2}-1}<xp

∎

Roughly speaking, Corollary 6.6 reveals that the relevant coefficents increase in a quadratic rate. The next thing we aim to show is that the first $p$ relevant coefficients of $\sigma$ determines the rest of them in a simple manner. Precisely, we want to show that for all positive integer $x$ and $0\leq k<p$ ,

(9)

V^{\sigma}_{xp+k}=\dfrac{x^{2}p+x|\sigma|_{1}}{2}+xk+V^{\sigma}_{k}

This is particularly useful for those who wish to do some programming on this topic because the size of $S_{\leq m}$ increases quickly with $m$ , which can cause run-time and memory issues. Based on data from some programming efforts related to this paper, when $m=100$ , there are 157,452 $\alpha\in\mathbb{Z}^{something}_{>0}$ such that $\alpha_{0}\geq\alpha_{1}\geq\dots$ and $\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)\leq 2m$ . When $m=200$ , the number becomes 13,552,451.

To help proving (9), we prove some more lemmas first.

Lemma 6.7.

For any positive integer $x$ ,

T^{\sigma}_{\tfrac{x^{2}p-x|\sigma|_{1}}{2}}=xp-|\sigma|_{1}

Proof.

Consider the full rational greedy algorithm that purchases all options strictly less than $x$ coin per point. So, for each $i$ , it purchases $x\sigma_{i}-1$ copies of the item $I_{i}$ . The total amount of coins spent is

\dfrac{1}{2}\sum\limits_{i}(x\sigma_{i}-1)(x\sigma_{i})=\dfrac{x^{2}p-x|\sigma|_{1}}{2}

And the number of points gained is

\sum\limits_{i}(x\sigma_{i}-1)\sigma_{i}=xp-|\sigma|_{1}

Result follows from the fact that the rational greedy algorithm is full, and therefore integral. ∎

Corollary 6.8.

V^{\sigma}_{xp-|\sigma|_{1}}=\dfrac{x^{2}p-x|\sigma|_{1}}{2}

Proof.

Since Lemma 6.7 came from an integral rational greedy algorithm, the argument used in the proof of Corollary 6.6 can be applied. ∎

Lemma 6.9.

For any positive integer $x$ , if $a\leq\dfrac{x^{2}p+x|\sigma|_{1}}{2}$ and $T^{\sigma}_{a-1}<T^{\sigma}_{a}$ and $T^{\sigma}_{b}=T^{\sigma}_{a-1}$ , then $a-b\leq x$ .

Proof.

Let $\alpha\in S_{\leq b}$ be such that $\alpha\cdot\sigma=T^{\sigma}_{b}$ .

To show that $a-b\leq x$ , it suffices to show that $\exists\alpha^{\prime}\in S_{\leq b+x}$ such that $\alpha^{\prime}\cdot\sigma\geq T^{\sigma}_{b}+1$ because the existence of such an $\alpha^{\prime}$ implies $T^{\sigma}_{\leq b+x}\geq T^{\sigma}_{b}+1>T^{\sigma}_{b}=T^{\sigma}_{a-1}$ , which implies $b+x>a-1$ , which implies $a-b\leq x$ .

Let $i$ be minimal such that $\alpha_{i}<x\sigma_{i}$ . Such an $i$ much exists, otherwise

\dfrac{1}{2}\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)\geq\dfrac{x^{2}p+x|\sigma|_{1}}{2}\geq a>b

which contradicts $\alpha\in S_{\leq b}$ .

Since $\sigma$ is a changemaker, $\exists\sigma_{i_{1}},\dots,\sigma_{i_{k}}$ such that $\sigma_{i_{1}}+\dots+\sigma_{i_{k}}=\sigma_{i}-1$ . (when $\sigma_{i}=1$ , we set $k=0$ and this is an empty sum)

Define $\alpha^{\prime}$ be such that

\alpha^{\prime}_{j}=\begin{cases}\alpha_{j}+1\text{ if }j=i\\ \alpha_{j}-1\text{ if }j=i_{1},\dots,i_{k}\\ \alpha_{j}\text{ otherwise}\end{cases}

Then

	$\displaystyle\alpha^{\prime}\cdot\sigma$	$\displaystyle=\alpha\cdot\sigma+\sigma_{i}-\sigma_{i_{1}}-\dots-\sigma_{i_{k}}$
		$\displaystyle=\alpha\cdot\sigma+1$
		$\displaystyle=T^{\sigma}_{b}+1$

and also

\alpha^{\prime}\in S_{\leq b+(\alpha_{i}+1)-\alpha_{i_{1}}-\dots-\alpha_{i_{k}}}

We complete the proof by noting

	$\displaystyle b+(\alpha_{i}+1)-\alpha_{i_{1}}-\dots-\alpha_{i_{k}}$	$\displaystyle<b+(n\sigma_{i}+1)-n\sigma_{i_{1}}-\dots-n\sigma_{i_{k}}$
		$\displaystyle=b+x+1$

Hence

b+(\alpha_{i}+1)-\alpha_{i_{1}}-\dots-\alpha_{i_{k}}\leq b+x

∎

Corollary 6.10.

For any positive integer $x$ , if $0<a\leq xp$ , then $V^{\sigma}_{a}-V^{\sigma}_{a-1}\leq x$

Proof.

If $V^{\sigma}_{a}=V^{\sigma}_{a-1}$ , then the result is trivial. So, we assume $V^{\sigma}_{a}>V^{\sigma}_{a-1}$ .

By Corollary 6.6, we know that $V_{a}\leq\tfrac{x^{2}p+x|\sigma|_{1}}{2}$ . Since there is at least one relevant coefficient with value $V^{\sigma}_{a}$ (namely $V^{\sigma}_{a}$ itself), we know that $T^{\sigma}_{V_{a}-1}<T^{\sigma}_{V_{a}}$ . Also, since $V^{\sigma}_{a}>V^{\sigma}_{a-1}$ , we have $T^{\sigma}_{V_{a-1}}=T^{\sigma}_{V_{a}-1}=a-1$ . Result follows from Lemma 6.9. ∎

Lemma 6.11.

For any positive integer $x$ , if $a>\dfrac{x^{2}p-x|\sigma|_{1}}{2}$ and $T^{\sigma}_{a}>T^{\sigma}_{a-1}$ and $T^{\sigma}_{a}-T^{\sigma}_{b}\geq 2$ , then $a-b\geq x+1$ .

Proof.

Let $\alpha\in S_{\leq a}$ be such that $\alpha\cdot\sigma=T^{\sigma}_{a}$ .

To show that $a-b\geq x+1$ , it suffices to show that $\exists\alpha^{\prime}\in S_{\leq a-x}$ such that $\alpha^{\prime}\cdot\sigma\geq T^{\sigma}_{a}-1$ because the existence of such an $\alpha^{\prime}$ implies $T^{\sigma}_{\leq a-x}\geq T^{\sigma}_{a}-1\geq T^{\sigma}_{b}+1>T^{\sigma}_{b}$ , which implies $a-x>b$ , which implies $a-b\geq x+1$ .

Let $i$ be minimal such that $\alpha_{i}\geq x\sigma_{i}$ . Such an $i$ much exists, otherwise

	$\displaystyle\dfrac{1}{2}\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)$	$\displaystyle\leq\dfrac{1}{2}\sum\limits_{j}(x\sigma_{j}-1)(x\sigma_{j})$
		$\displaystyle=\dfrac{x^{2}p-x\|\sigma\|_{1}}{2}$
		$\displaystyle<a$

which implies $\alpha\in S_{\leq a-1}$ , which contradicts with $T^{\sigma}_{a}>T^{\sigma}_{a-1}$ .

Since $\sigma$ is a changemaker, $\exists\sigma_{i_{1}},\dots,\sigma_{i_{k}}$ such that $\sigma_{i_{1}}+\dots+\sigma_{i_{k}}=\sigma_{i}-1$ . (when $\sigma_{i}=1$ , we set $k=0$ and this is an empty sum)

Define $\alpha^{\prime}$ be such that

\alpha^{\prime}_{j}=\begin{cases}\alpha_{j}-1\text{ if }j=i\\ \alpha_{j}+1\text{ if }j=i_{1},\dots,i_{k}\\ \alpha_{j}\text{ otherwise}\end{cases}

Then

	$\displaystyle\alpha^{\prime}\cdot\sigma$	$\displaystyle=\alpha\cdot\sigma-\sigma_{i}+\sigma_{i_{1}}+\dots+\sigma_{i_{k}}$
		$\displaystyle=\alpha\cdot\sigma-1$
		$\displaystyle=T^{\sigma}_{a}-1$

and also

\alpha^{\prime}\in S_{\leq a-\alpha_{i}+(\alpha_{i_{1}}+1)+\dots+(\alpha_{i_{k}}+1)}

We complete the proof by noting

	$\displaystyle a-\alpha_{i}+(\alpha_{i_{1}}+1)+\dots+(\alpha_{i_{k}}+1)$	$\displaystyle\leq a-\alpha_{i}+n\sigma_{i_{1}}+\dots+n\sigma_{i_{k}}$
		$\displaystyle=a-n-(\alpha_{i}-n\sigma_{i})$
		$\displaystyle\leq a-n$

∎

Corollary 6.12.

For any positive integer $x$ , if $a\geq xp-|\sigma|_{1}$ , then $V^{\sigma}_{a+1}-V^{\sigma}_{a}\geq x$

Proof.

By Lemma 6.7, there are only $xp-|\sigma|_{1}$ relevant coefficients less than or equal to $\tfrac{x^{2}p-x|\sigma|_{1}}{2}$ . So we know that

V^{\sigma}_{a+1}>\dfrac{x^{2}p-x|\sigma|_{1}}{2}

Also, $V^{\sigma}_{a}$ and $V^{\sigma}_{a+1}$ are relevant coefficients less than or equal to $V^{\sigma}_{a+1}$ but not less than or equal to $V^{\sigma}_{a}-1$ . Hence

T^{\sigma}_{V^{\sigma}_{a+1}}-T^{\sigma}_{V^{\sigma}_{a}-1}\geq 2

Since $V^{\sigma}_{a+1}$ is a relevant coefficient, we have $T^{\sigma}_{V^{\sigma}_{a+1}}>T^{\sigma}_{V^{\sigma}_{a+1}-1}$ . By Lemma 6.11, we know that

V^{\sigma}_{a+1}-(V^{\sigma}_{a}-1)\geq x+1

∎

Another corollary of Lemma 6.11 lemma is that

Corollary 6.13.

When $r\geq 2$ , statement (3) of Lemma 4.2 is an equality. In other words

T^{rel}_{V_{0}^{rel}}=T^{\sigma}_{V_{0}^{rel}}

Proof.

When $r=2$ and $\sigma=(1)$ , statement (2) of Lemma 4.2 says that $\nu^{+rel}=0$ . Hence $V_{0}^{rel}=0$ and the result is trivial. Now we assume that it is not this special case.

By Remark 2 after the proof of Lemma 4.2, it suffices to show that

(10)

T^{\sigma}_{V_{0}^{rel}}-T^{\sigma}_{V_{0}^{rel}-1}\leq 1

We assume that $T^{\sigma}_{V_{0}^{rel}}>T^{\sigma}_{V_{0}^{rel}-1}$ , otherwise (10) is trivially true.

Whenever $r\geq 2$ , apart from the special case we dealt with at the start of this proof, statement (2) of Lemma 4.2 implies that

\nu^{+rel}>p-|\sigma|_{1}

Lemma 6.7 implies that

V^{rel}_{0}=V^{\sigma}_{\nu^{+rel}}>\dfrac{p-|\sigma|_{1}}{2}

Now we look at the statement of Lemma 6.11. When $a=V^{rel}_{0}$ , $b=V^{rel}_{0}-1$ , $x=1$ , the first two conditions of Lemma 6.11 are true but the conclusion is false. Therefore, the third condition

T^{\sigma}_{V_{0}^{rel}}-T^{\sigma}_{V_{0}^{rel}-1}\geq 2

must be false. ∎

Note that Corollary 6.10 implies that when $0\leq k<p$ , $V^{\sigma}_{k+1}=V^{\sigma}_{k}\text{ or }V^{\sigma}_{k}+1$ .

Also, note that Corollary 6.10 and Corollary 6.12 together imply that for any positive integer $x$ , when $0\leq k<p$ , $V^{\sigma}_{xp+k+1}-V^{\sigma}_{xp+k}$ must be either $x$ or $x+1$ .

To prove equation (9), by induction on $k$ , it suffices to show that

V^{\sigma}_{xp+k+1}-V^{\sigma}_{xp+k}=\begin{cases}x+1\text{ if }V^{\sigma}_{k+1}=V^{\sigma}_{k}+1\\ x\phantom{+11}\text{ if }V^{\sigma}_{k+1}=V^{\sigma}_{k}\end{cases}

Lemma 6.14.

Let $x$ be a positive integer, $0\leq k<p$ . If $V^{\sigma}_{k}=V^{\sigma}_{k+1}$ , then

V^{\sigma}_{xp+k+1}-V^{\sigma}_{xp+k}=x

Proof.

By Corollary 6.12, we know that $V^{\sigma}_{xp+k+1}>V^{\sigma}_{xp+k}$ , and therefore $T^{\sigma}_{V^{\sigma}_{xp+k}}=xp+k$ .

Let $\alpha\in S_{\leq V^{\sigma}_{xp+k}}$ be such that $\alpha\cdot\sigma=T^{\sigma}_{V^{\sigma}_{xp+k}}=xp+k$ . Since $V^{\sigma}_{xp+k}$ is a relevant coefficient, we know that $T^{\sigma}_{V^{\sigma}_{xp+k}}>T^{\sigma}_{V^{\sigma}_{xp+k}-1}$ , and hence $\alpha\not\in S_{\leq V^{\sigma}_{xp+k}-1}$ . So, we know that

\dfrac{1}{2}\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)=V^{\sigma}_{xp+k}

Since $V^{\sigma}_{xp+k+1}$ is a relevant coefficient, we know that $T^{\sigma}_{V^{\sigma}_{xp+k+1}}>T^{\sigma}_{V^{\sigma}_{xp+k+1}-1}$ . Since there are no relevant coefficients between $V^{\sigma}_{xp+k+1}$ and $V^{\sigma}_{xp+k}$ , we also know that $T^{\sigma}_{V^{\sigma}_{xp+k}}=T^{\sigma}_{V^{\sigma}_{xp+k+1}-1}$ .

Now we try to apply Lemma 6.9 with $a=V^{\sigma}_{xp+k+1}$ , $b=V^{\sigma}_{xp+k}$ . We do not have the first condition $a\leq\tfrac{x^{2}p+x|\sigma|_{1}}{2}$ . However, in the proof of Lemma 6.9, the only place where the condition is used is to show that there exists some $i$ such that $\alpha_{i}<x\sigma_{i}$ . If such an $i$ exists, even if the condition $a\leq\tfrac{x^{2}p+x|\sigma|_{1}}{2}$ is not satisfied, the proof of Lemma 6.9 still goes through and we can conclude that

V^{\sigma}_{xp+k+1}-V^{\sigma}_{xp+k}\leq x

Hence, we can assume that such an $i$ does not exist. In that case, $\alpha=\beta+x\sigma$ for some $\beta$ with all entries being non-negative. $\beta$ satisfies

	$\displaystyle\beta\cdot\sigma$	$\displaystyle=(\alpha-x\sigma)\cdot\sigma$
		$\displaystyle=\alpha\cdot\sigma-xp$
		$\displaystyle=k$

$V^{\sigma}_{k}-1$ is less than the $k^{th}$ relevant coefficient $V_{k}$ . Therefore, $T^{\sigma}_{V^{\sigma}_{k}-1}<k$ and we can conclude that

\dfrac{1}{2}\sum\limits_{j}\beta_{j}(\beta_{j}+1)\geq V^{\sigma}_{k}

Now we have

	$\displaystyle V^{\sigma}_{xp+k}$	$\displaystyle=\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)$
		$\displaystyle=\sum\limits_{j}(\beta_{j}+x\sigma_{j})(\beta_{j}+x\sigma_{j}+1)$
		$\displaystyle=\dfrac{x^{2}p+x\|\sigma\|_{1}}{2}+x(\beta\cdot\sigma)+\dfrac{1}{2}\sum\limits_{j}\beta_{j}(\beta_{j}+1)$
		$\displaystyle\geq\dfrac{x^{2}p+x\|\sigma\|_{1}}{2}+xk+V^{\sigma}_{k}$

Let $c$ be maximal such that $V^{\sigma}_{k}=V^{\sigma}_{k+c}$ .

By the maximality of $c$ , $T^{\sigma}_{V^{\sigma}_{k}}=k+c$ . Let $\beta^{\prime}\in S_{\leq V_{k}}$ be such that $\beta^{\prime}\cdot\sigma=k+c$ .

Let $\alpha^{\prime}:=\beta^{\prime}+x\sigma$ . Then we have $\alpha^{\prime}\cdot\sigma=xp+k+c$ . Also

	$\displaystyle\sum\limits_{j}\alpha^{\prime}_{j}(\alpha^{\prime}_{j}+1)$	$\displaystyle=\sum\limits_{j}(\beta^{\prime}_{j}+x\sigma_{j})(\beta^{\prime}_{j}+x\sigma_{j}+1)$
		$\displaystyle=\dfrac{x^{2}p+x\|\sigma\|_{1}}{2}+x(\beta^{\prime}\cdot\sigma)+\dfrac{1}{2}\sum\limits_{j}\beta^{\prime}_{j}(\beta^{\prime}_{j}+1)$
		$\displaystyle\leq\dfrac{x^{2}p+x\|\sigma\|_{1}}{2}+xk+xc+V^{\sigma}_{k}$
		$\displaystyle\leq V^{\sigma}_{xp+k}+xc$

Hence, $\alpha^{\prime}\in S_{V^{\sigma}_{xp+k}+xc}$ and we can conclude that

T^{\sigma}_{V^{\sigma}_{xp+k}+xc}\geq xp+k+c

So, there are at least $xp+k+c$ relevant coefficients less than or equal to $V^{\sigma}_{xp+k}+xc$ . Hence, we can conclude that

V^{\sigma}_{xp+k+c}\leq V^{\sigma}_{xp+k}+xc

which can be arranged into

V^{\sigma}_{xp+k+c}-V^{\sigma}_{xp+k}\leq xc

By Corollary 6.12, we know that

	$\displaystyle V^{\sigma}_{xp+k+1}-V^{\sigma}_{xp+k}$	$\displaystyle\geq x$
	$\displaystyle V^{\sigma}_{xp+k+2}-V^{\sigma}_{xp+k+1}$	$\displaystyle\geq x$
	$\displaystyle\dots$
	$\displaystyle V^{\sigma}_{xp+k+c}-V^{\sigma}_{xp+k+c-1}$	$\displaystyle\geq x$

The only way that $V^{\sigma}_{xp+k+c}-V^{\sigma}_{xp+k}\leq xc$ can hold is that all these inequalities are all equalities. ∎

Now we prove equation (9):

Lemma 6.15.

Let $x$ be a positive integer and $0\leq k<p$ . Then

V^{\sigma}_{xp+k}=\dfrac{x^{2}p+x|\sigma|_{1}}{2}+xk+V^{\sigma}_{k}

Proof.

First of all, by Corollary 6.6, we know this is true when $k=0$ . Now we consider

	$\displaystyle V^{\sigma}_{xp+p}-V^{\sigma}_{xp}$	$\displaystyle=\dfrac{(x+1)^{2}p+(x+1)\|\sigma\|_{1}}{2}-\dfrac{x^{2}p+x\|\sigma\|_{1}}{2}$
		$\displaystyle=xp+\dfrac{p+\|\sigma\|_{1}}{2}$

Therefore, by considering Corollary 6.10 and Corollary 6.12, we know that exactly $\tfrac{p+|\sigma|_{1}}{2}$ values of $k\in\{0,\dots,p-1\}$ satisfy $V^{\sigma}_{xp+k+1}-V^{\sigma}_{xp+k}=x+1$ , and the rest of the $k$ values satisfies $V^{\sigma}_{xp+k+1}-V^{\sigma}_{xp+k}=x$ .

Since $V^{\sigma}_{p}=\tfrac{p+|\sigma|_{1}}{2}$ , by Corollary 6.10 we know that exactly $\tfrac{p+|\sigma|_{1}}{2}$ values of $k\in\{0,\dots,p-1\}$ satisfy $V^{\sigma}_{k+1}=V^{\sigma}_{k}+1$ . Therefore, by considering Lemma 6.14, we conclude that for $k\in\{0,\dots,p-1\}$ ,

V^{\sigma}_{xp+k+1}-V^{\sigma}_{xp+k}=x+1\text{ if and only if }V^{\sigma}_{k+1}=V^{\sigma}_{k}+1

and

V^{\sigma}_{xp+k+1}-V^{\sigma}_{xp+k}=x\text{ if and only if }V^{\sigma}_{k+1}=V^{\sigma}_{k}

Result follows from induction on $k$ . ∎

Before we prove Theorem 6.1, we make a few more evaluations of relevant coefficients.

Let $odd(\sigma)$ be the number of odd entries in $\sigma$ .

Lemma 6.16.

The following five statements are true:
(1) If $p$ is even, $V^{\sigma}_{\tfrac{2p-|\sigma|_{1}}{2}}=\tfrac{p}{2}$ .
(2) If $p$ is odd, $V^{\sigma}_{\tfrac{2p-|\sigma|_{1}-1}{2}}=\tfrac{p-1}{2}$ .
(3) $V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}}\geq\tfrac{p-odd(\sigma)}{8}$
(4) In (3), equality holds if and only if the even entries of $\sigma$ can be partitioned into two sets with equal sum.
(5) $V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}}\leq\tfrac{p+3odd(\sigma)}{8}$

Proof.

Corollary 6.6 and Corollary 6.10 and Corollary 6.12 together implies that for any $0\leq k\leq|\sigma|_{1}$ ,

V^{\sigma}_{p-k}=\dfrac{p+|\sigma|_{1}}{2}-k

This implies statements (1) and (2) of Lemma 6.16.

Now we investigate $V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}}$ .

Consider the full rational greedy algorithm that purchases all options strictly less than $\tfrac{1}{2}$ coin per point. The amount of coins spent is

	$\displaystyle\sum\limits_{\sigma_{i}\ even}\dfrac{\left(\tfrac{\sigma_{i}}{2}-1\right)\left(\tfrac{\sigma_{i}}{2}\right)}{2}-\sum\limits_{\sigma_{i}\ odd}\dfrac{\left(\tfrac{\sigma_{i}-1}{2}\right)\left(\tfrac{\sigma_{i}-1}{2}+1\right)}{2}$	$\displaystyle=\sum\limits_{\sigma_{i}\ even}\left(\dfrac{\sigma_{i}^{2}}{8}-\dfrac{\sigma_{i}}{4}\right)+\sum\limits_{\sigma_{i}\ odd}\left(\dfrac{\sigma_{i}^{2}}{8}-\dfrac{1}{8}\right)$
		$\displaystyle=\dfrac{p-odd(\sigma)}{8}-\dfrac{1}{4}\sum\limits_{\sigma_{i}\ even}\sigma_{i}$

The amount of points obtained is

	$\displaystyle\sum\limits_{\sigma_{i}\ even}\left(\dfrac{\sigma_{i}}{2}-1\right)\sigma_{i}+\sum\limits_{\sigma_{i}\ odd}\left(\dfrac{\sigma_{i}-1}{2}\right)\sigma_{i}$	$\displaystyle=\sum\limits_{\sigma_{i}\ even}\left(\dfrac{\sigma^{2}_{i}}{2}-\sigma_{i}\right)+\sum\limits_{\sigma_{i}\ odd}\left(\dfrac{\sigma^{2}_{i}}{2}-\dfrac{\sigma_{i}}{2}\right)$
		$\displaystyle=\dfrac{p-\|\sigma\|_{1}}{2}-\dfrac{1}{2}\sum\limits_{\sigma_{i}\ even}\sigma_{i}$

Hence, $T^{\sigma}_{\tfrac{p-odd(\sigma)}{8}-\tfrac{1}{4}\sum\limits_{\sigma_{i}\ even}\sigma_{i}}=\dfrac{p-|\sigma|_{1}}{2}-\dfrac{1}{2}\sum\limits_{\sigma_{i}\ even}\sigma_{i}$ .

By also considering the purchase options that are exactly $\tfrac{1}{2}$ coin per point, we conclude that for any $0\leq k\leq\tfrac{1}{2}\sum\limits_{\sigma_{i}\ even}\sigma_{i}$ ,

T^{\sigma,\mathbb{Q}}_{\tfrac{p-odd(\sigma)}{8}-\tfrac{1}{4}\sum\limits_{\sigma_{i}\ even}\sigma_{i}+k}=\dfrac{p-|\sigma|_{1}}{2}-\dfrac{1}{2}\sum\limits_{\sigma_{i}\ even}\sigma_{i}+\dfrac{k}{2}

where $k$ represents the amount of coins spent on purchasing options that exactly $\tfrac{1}{2}$ coin per point under a rational greedy algorithm.

Setting $k=\tfrac{1}{4}\sum\limits_{\sigma_{i}\ even}\sigma_{i}$ , we conclude that

T^{\sigma,\mathbb{Q}}_{\tfrac{p-odd(\sigma)}{8}}=\dfrac{p-|\sigma|_{1}}{2}

In particular $T^{\sigma,\mathbb{Q}}_{\tfrac{p-odd(\sigma)}{8}}=T^{\sigma}_{\tfrac{p-odd(\sigma)}{8}}$ if and only if there is an integral implementation of a greedy algorithm that spends exactly $\tfrac{1}{4}\sum\limits_{\sigma_{i}\ even}\sigma_{i}$ coins on purchasing options that are exactly $\tfrac{1}{2}$ coin per point. Such an integral implementation is possible if and only if there is a subset of even $\sigma_{i}$ ’s that sum to $\tfrac{1}{2}\sum\limits_{\sigma_{i}\ even}$ , which is equivalent to saying that the even entries of $\sigma$ can be partitioned into two sets with equal sum. Therefore,

T^{\sigma}_{\tfrac{p-odd(\sigma)}{8}}\leq\dfrac{p-|\sigma|_{1}}{2}

with equality if and only if the even entries of $\sigma$ can be partitioned into two sets with equal sum.

Note that $T^{\sigma}_{\tfrac{p-odd(\sigma)}{8}}=\tfrac{p-|\sigma|_{1}}{2}$ implies $V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}}\leq\tfrac{p-odd(\sigma)}{8}$ and $V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}+1}>\tfrac{p-odd(\sigma)}{8}$ , which implies $V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}}=\tfrac{p-odd(\sigma)}{8}$ because of Corollary 6.10. Also, $T^{\sigma}_{\tfrac{p-odd(\sigma)}{8}}<\tfrac{p-|\sigma|_{1}}{2}$ implies $V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}}>T^{\sigma}_{\tfrac{p-odd(\sigma)}{8}}$ . Therefore, this implies statements (3) and (4) of Lemma 6.16.

Since $\sigma$ is a changemaker vector, there is some $A\subseteq\{0,\dots,n\}$ such that $\sum\limits_{i\in A}\sigma_{i}=\tfrac{1}{2}\sum\limits_{\sigma_{i}\ even}\sigma_{i}$ . Consider $\alpha$ defined with

\alpha_{i}=\begin{cases}\tfrac{\sigma_{i}}{2}-1\text{ if }\sigma_{i}\text{ is even and }i\not\in A\\ \tfrac{\sigma_{i}}{2}\text{ if }\sigma_{i}\text{ is even and }i\in A\\ \tfrac{\sigma_{i}-1}{2}\text{ if }\sigma_{i}\text{ is odd and }i\not\in A\\ \tfrac{\sigma_{i}-1}{2}+1\text{ if }\sigma_{i}\text{ is odd and }i\in A\end{cases}

Then

	$\displaystyle\alpha\cdot\sigma$	$\displaystyle=\dfrac{p-\|\sigma\|_{1}}{2}-\dfrac{1}{2}\sum\limits_{\sigma_{i}\ even}\sigma_{i}+\sum\limits_{i\in A}\sigma_{i}$
		$\displaystyle=\dfrac{p-\|\sigma\|_{1}}{2}$

and

	$\displaystyle\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)$	$\displaystyle=\dfrac{p-odd(\sigma)}{8}-\dfrac{1}{4}\sum\limits_{\sigma_{i}\ even}\sigma_{i}+\sum_{\begin{subarray}{c}\sigma_{i}\ even\\ i\in A\end{subarray}}\dfrac{\sigma_{i}}{2}+\sum_{\begin{subarray}{c}\sigma_{i}\ odd\\ i\in A\end{subarray}}\dfrac{\sigma_{i}+1}{2}$
		$\displaystyle=\dfrac{p-odd(\sigma)}{8}-\dfrac{1}{4}\sum\limits_{\sigma_{i}\ even}\sigma_{i}+\dfrac{1}{2}\sum_{i\in A}\sigma_{i}+\sum_{\begin{subarray}{c}\sigma_{i}\ odd\\ i\in A\end{subarray}}\dfrac{1}{2}$
		$\displaystyle=\dfrac{p-odd(\sigma)}{8}+\dfrac{\text{number of }i\in A\text{ such that }\sigma_{i}\text{ is odd}}{2}$

Let $odd_{A}$ denote the number of $i\in A$ such that $\sigma_{i}$ is odd. Now we have

T^{\sigma}_{\tfrac{p-odd(\sigma)}{8}+\tfrac{odd_{A}}{2}}\geq\dfrac{p-|\sigma|_{1}}{2}

Therefore,

V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}}\leq\tfrac{p+4odd_{A}-odd(\sigma)}{8}

Since $odd_{A}\leq odd(\sigma)$ , we conclude that

V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}}\leq\tfrac{p+3odd(\sigma)}{8}

∎

Now we can prove Theorem 6.1

Proof or Theorem 6.1.

When $r,p$ are both even, let $r=2x$ .

	$\displaystyle\nu^{+rel}$	$\displaystyle=\dfrac{2xp-\|\sigma\|_{1}}{2}$
		$\displaystyle=(x-1)p+\dfrac{2p-\|\sigma\|_{1}}{2}$

	$\displaystyle V_{0}$	$\displaystyle=V^{rel}_{0}$
		$\displaystyle=V^{\sigma}_{(x-1)p+\tfrac{2p-\|\sigma\|_{1}}{2}}$
		$\displaystyle=\dfrac{(x-1)^{2}p+(x-1)\|\sigma\|_{1}}{2}+(x-1)\dfrac{2p-\|\sigma\|_{1}}{2}+V^{\sigma}_{\tfrac{2p-\|\sigma\|_{1}}{2}}$
		$\displaystyle=\dfrac{\left(\tfrac{r}{2}-1\right)^{2}p+\left(\tfrac{r}{2}-1\right)\|\sigma\|_{1}}{2}+\left(\dfrac{r}{2}-1\right)\dfrac{2p-\|\sigma\|_{1}}{2}+\dfrac{p}{2}$

which simplifies to $V_{0}=\dfrac{r^{2}p}{8}$ . Hence, $r^{2}p=8V_{0}$ .

When $r$ is even and $p$ is odd, let $r=2x$ .

	$\displaystyle\nu^{+rel}$	$\displaystyle=-\dfrac{1}{2}+\dfrac{2xp-\|\sigma\|_{1}}{2}$
		$\displaystyle=(x-1)p+\dfrac{2p-\|\sigma\|_{1}-1}{2}$

	$\displaystyle V_{\tfrac{r}{2}}$	$\displaystyle=V^{rel}_{0}$
		$\displaystyle=V^{\sigma}_{(x-1)p+\tfrac{2p-\|\sigma\|_{1}-1}{2}}$
		$\displaystyle=\dfrac{(x-1)^{2}p+(x-1)\|\sigma\|_{1}}{2}+(x-1)\dfrac{2p-\|\sigma\|_{1}-1}{2}+V^{\sigma}_{\tfrac{2p-\|\sigma\|_{1}-1}{2}}$
		$\displaystyle=\dfrac{\left(\tfrac{r}{2}-1\right)^{2}p+\left(\tfrac{r}{2}-1\right)\|\sigma\|_{1}}{2}+\left(\dfrac{r}{2}-1\right)\dfrac{2p-\|\sigma\|_{1}-1}{2}+\dfrac{p-1}{2}$

which simplifies to $V_{\tfrac{r}{2}}=\dfrac{r^{2}p}{8}-\dfrac{r}{4}$ .

Hence, $V_{0}\geq V_{\tfrac{r}{2}}=\dfrac{r^{2}p}{8}-\dfrac{r}{4}$ and $V_{0}\leq V_{\tfrac{r}{2}}+\tfrac{r}{2}=\dfrac{r^{2}p}{8}+\dfrac{r}{4}$ . This implies

8V_{0}-2r\leq r^{2}p\leq 8V_{0}+2r

When $r$ is odd, let $r=2x+1$ .

	$\displaystyle\nu^{+rel}$	$\displaystyle=\dfrac{(2x+1)p-\|\sigma\|_{1}}{2}$
		$\displaystyle=xp+\dfrac{p-\|\sigma\|_{1}}{2}$

	$\displaystyle V_{0}$	$\displaystyle=V^{rel}_{0}$
		$\displaystyle=V^{\sigma}_{xp+\tfrac{p-\|\sigma\|_{1}}{2}}$
		$\displaystyle=\dfrac{(x^{2}p+x\|\sigma\|_{1}}{2}+x\dfrac{p-\|\sigma\|_{1}}{2}+V^{\sigma}_{\tfrac{p-\|\sigma\|_{1}}{2}}$
		$\displaystyle=\dfrac{\left(\tfrac{r-1}{2}\right)^{2}p+\left(\tfrac{r-1}{2}\right)\|\sigma\|_{1}}{2}+\left(\dfrac{r-1}{2}\right)\dfrac{2p-\|\sigma\|_{1}}{2}+V^{\sigma}_{\tfrac{p-\|\sigma\|_{1}}{2}}$

which simplifies to $V_{0}=\tfrac{r^{2}p-p}{8}+V^{\sigma}_{\tfrac{p-|\sigma|_{1}}{2}}$ . Hence,

\dfrac{r^{2}p-p}{8}+\dfrac{p-odd(\sigma)}{8}\leq V_{0}\leq\dfrac{r^{2}p-p}{8}+\dfrac{p+3odd(\sigma)}{8}

and $\tfrac{r^{2}p-p}{8}+\tfrac{p-odd(\sigma)}{8}=V_{0}$ if and only if the even entries of $\sigma$ can be partitioned into two sets with equal sum. Result follows from rearranging this inequality. ∎

7. Linear changemakers

Greene [9] showed that $\sigma$ must take 1 of the 33 forms, including 26 small families which have a simple structure and 7 large families that involves a process called “weight expansion”. In this section, we work on 1 of the 26 small families. We believe that a similar approach can be used on all small families. Unfortunately, we do not see how does this generalizes to the large families.

The small family structure that we investigate in the section has changemakers in the form

\sigma=\begin{pmatrix}4s+3,&2s+1,&s+1,&s,&\smash{\underbrace{\begin{matrix}1,&\dots&,&1\end{matrix}}_{s\text{ copies of }1\text{'s}}}\end{pmatrix}

where $s\geq 2$ .

According to Greene’s work [9], this family corresponds to the $j\leq-3$ case in Berge Type X in Rasmussen’s table [16, §6.2].

At the end of this section we prove Theorem 1.11, which states that any knot in $S^{3}$ has at most 7 lensbordant surgeries with an associated changemaker coming from this family.

First we use the following lemma to convert that into a statement about the number of possible $\sigma$ instead of the number of possible surgeries.

Lemma 7.1.

Given a knot $K$ in $S^{3}$ and a changemaker vector $\sigma$ , if $\sigma\neq(1)$ or $\nu^{+}(K)$ is not a triangular number, there is at most one lensbordant surgeries on $K$ associated with $\sigma$ . When $\sigma=(1)$ and $\nu^{+}(K)$ is a triangular number, there are at most two lensbordant surgeries on $K$ associated with $\sigma$ .

Remark: When $\sigma=(1)$ , being a lensbordant sugery associated with $\sigma$ means that the surgery bounds a rational homology ball.

Proof of Lemma 7.1.

Suppose if the $r_{1}^{2}p$ -surgery and the $r_{2}^{2}p$ -surgery on $K$ are both lensbordant associated with $\sigma$ . Let $r_{2}\geq r_{1}$ .

Theorem 1.8 states that

2\nu^{+}+2(r-1)\geq r^{2}p-r|\sigma|_{1}\geq 2\nu^{+}

Hence,

r_{1}^{2}p-r_{1}|\sigma|_{1}\geq r_{2}^{2}p-r_{2}|\sigma|_{1}-2(r_{2}-1)

which can be rearranged into

0\geq(r_{2}-r_{1})\left((r_{2}-r_{1})p-|\sigma|_{1}\right)+(2r_{1}p-2)(r_{2}-r_{1}-1)+2r_{1}(p-1)

Therefore, either $r_{2}=r_{1}$ or ( $r_{2}=r_{1}+1$ and $p=1$ ).

When $p=1$ , we know that

2\nu^{+}+2(r-1)\geq r^{2}-r\geq 2\nu^{+}

which can be rearranged into

\dfrac{r(r-1)}{2}\geq\nu^{+}\geq\dfrac{(r-1)(r-2)}{2}

Hence, we can determine the value of $r$ unless $\nu^{+}$ is a triangular number, in which case there can be two values of $r$ . ∎

Remark: The case with two values of $r$ can indeed happen. Both the $p^{2}$ -surgery and the $(p+1)^{2}$ -surgery on the torus knot $T_{p,(p+1)}$ bound a smooth rational homology ball [2, Th. 1.4].

Given Lemma 7.1, we now focus on the number of $\sigma$ ’s instead of the number of surgeries.

The majority of the content in this section is for proving the following

Theorem 7.2.

Let $s\geq 5$ and $r\geq 2$ . Let $K$ be a knot in $S^{3}$ and there is a $(r,p)$ -lensbordant surgery on $K$ associated with a changemaker vector in the form

\sigma=\begin{pmatrix}4s+3,&2s+1,&s+1,&s,&\smash{\underbrace{\begin{matrix}1,&\dots&,&1\end{matrix}}_{s\text{ copies of }1\text{'s}}}\end{pmatrix}

Let $a$ be the number of elements in the set $\{i\geq 0\mid 296\leq V_{i}(K)\leq 300\}$ .

Let $b$ be the biggest value satisfying the property that there are at least $\dfrac{a}{2}$ elements in the set $\{i\geq 0\mid V_{i}(K)=b\}$ .

Then

s=\left\lfloor\sqrt{\dfrac{b}{11}}\right\rfloor\text{ or }\left\lfloor\sqrt{\dfrac{b}{11}}\right\rfloor+1\text{ or }\left\lfloor\sqrt{\dfrac{b}{11}}\right\rfloor+2

From this point onwards, until we finish proving Theorem 7.2, we let

\sigma=\begin{pmatrix}4s+3,&2s+1,&s+1,&s,&\smash{\underbrace{\begin{matrix}1,&\dots&,&1\end{matrix}}_{s\text{ copies of }1\text{'s}}}\end{pmatrix}

and we assume that $s\geq 5$ and $r\geq 2$ .

The following lemma converts information about $V^{rel}_{i}$ ’s into information about $V_{i}$ ’s.

Lemma 7.3.

For any $m_{1}>m_{2}\geq 0$ ,

(T^{rel}_{m_{1}}-T^{rel}_{m_{2}}+1)r>|\{i\geq 0\mid m_{1}\geq V_{i}>m_{2}\}|>(T^{rel}_{m_{1}}-T^{rel}_{m_{2}}-1)r

Proof.

$T^{rel}_{m_{1}}-T^{rel}_{m_{2}}$ is the number of relevant coefficients having values that are both $\leq m_{1}$ and $>m_{2}$ . Let $T$ be maximal such that $V_{T}$ is relevant and $>m_{2}$ . Let $t$ be minimal such that $V_{t}$ is relevant and $\leq m_{1}$ . Since relevant coefficients having indices $r$ apart from each other,

T-t=(T^{rel}_{m_{1}}-T^{rel}_{m_{2}}-1)r

Hence, by monotonicity of the $V$ coefficients, we conclude that

|\{i\geq 0\mid m_{1}\geq V_{i}>m_{2}\}|>(T^{rel}_{m_{1}}-T^{rel}_{m_{2}}-1)r

Also, since $V_{T+r}\leq m_{2}$ , and also if $t\geq r$ then $V_{t-r}>m_{1}$ ,

	$\displaystyle\|\{i\geq 0\mid m_{1}\geq V_{i}>m_{2}\}\|$	$\displaystyle\leq(T+r-1)-(t-r+1)+1$
		$\displaystyle=T-t+2r-1$
		$\displaystyle=(T^{rel}_{m_{1}}-T^{rel}_{m_{2}}+1)r-1$

∎

Now we calculate some values of $T^{\sigma}$ .

Lemma 7.4.

T^{\sigma}_{295}=110s+75

Proof.

With 295 coins, when $s\geq 15$ , an integral rational greedy algorithm up to $\tfrac{5}{s}$ coin per point gives

\alpha=(20,10,5,5,0,\dots)

which implies

T^{\sigma}_{295}=110s+75

When $14\geq s\geq 5$ , rational greedy algorithm up to $\tfrac{21}{4s+3}$ coin per point gives

\alpha=(20+\dfrac{5}{21},10,5,4,0,\dots)

which implies

T^{\sigma,\mathbb{Q}}_{295}=110s+75+\dfrac{15-s}{21}<110s+76

Hence,

T^{\sigma}_{295}\leq 110s+75

By considering $\alpha=(20,10,5,5,0,\dots)$ , we conclude that

T^{\sigma}_{295}=110s+75

∎

Lemma 7.5.

T^{\sigma}_{300}=110s+80

Proof.

By considering $\alpha=(20,10,5,5,1,1,1,1,1,0,\dots)$ , we know that

T^{\sigma}_{300}\geq 110s+80

Hence, it suffices to show that

T^{\sigma}_{300}\leq 110s+80

Let $m=\tfrac{1}{2}\sum\limits_{j}\alpha_{j}(\alpha_{j}+1)$ . Let $\alpha=(\alpha_{0},\dots,\alpha_{n})$ . Then

\alpha\cdot\sigma=\alpha_{0}(4s+3)+\alpha_{1}(2s+1)+\alpha_{2}(s+1)+\alpha_{3}s+(\alpha_{4}+\dots+\alpha_{n})

Let

S=\{\alpha\in S_{\leq 300}\mid\alpha\cdot\sigma=T^{\sigma}_{300}\text{ and }\alpha_{3}\geq 5\}

First of all, we argue that if the set $S$ is non-empty, then $T^{\sigma}_{300}=110s+80$ .

If $S$ is non-empty, pick an $\alpha$ in $S$ with minimal $\alpha_{3}$ . We know that $\alpha_{2}\geq 5$ because otherwise we can swap $\alpha_{2},\alpha_{3}$ to increase $\alpha\cdot\sigma$ without increasing $m$ . So, $\alpha_{1}\geq 10$ because otherwise we can decrease $\alpha_{2},\alpha_{3}$ by 1 and increase $\alpha_{1}$ by 1, which does not increase $m$ and keep $\alpha\cdot\sigma$ unchanged while decreasing $\alpha_{3}$ . So, $\alpha_{0}\geq 20$ because otherwise we can decrease $\alpha_{1},\alpha_{2},\alpha_{3}$ by 1 and increase $\alpha_{0}$ by 1 to increase $\alpha\cdot\sigma$ without increasing $m$ . Notice that

\dfrac{20\times 21}{2}+\dfrac{10\times 11}{2}+\dfrac{5\times 6}{2}+\dfrac{5\times 6}{2}=295

Therefore $\alpha$ must be $(20,10,5,5,1,1,1,1,1)$ because otherwise $m$ will be bigger than 300. This gives

T^{\sigma}_{300}=110s+80

Now we consider the case when $S$ is empty. In this case, all optimal $\alpha$ ’s have $\alpha_{3}\leq 4$ . Among these $\alpha$ ’s, pick one with maximal $\alpha_{3}$ and satisfies

\alpha_{4}\geq\alpha_{5}\geq\dots

We know that at most $4$ values of $\alpha_{4},\alpha_{5},\dots$ can be non-zero, because otherwise we can decrease $(\alpha_{3}+1)$ of those values by 1 and increase $\alpha_{3}$ by 1, which increases $\alpha_{3}$ but does not increase $m$ and does not decrease $\alpha\cdot\sigma$ .

Let $\sigma_{1}=(4,2,1,1,0,\dots)$ and $\sigma_{2}=(19,9,5,4,1,1,1,1,0,\dots)$ . We have

\alpha\cdot\sigma=(s-4)\alpha\cdot\sigma_{1}+\alpha\cdot\sigma_{2}

First we investigate $\sigma_{1}$ . It has $p_{1}=22$ and $|\sigma_{1}|_{1}=8$ . Lemma 6.15 implies

V^{\sigma_{1}}_{110}=295\text{ and }V^{\sigma_{1}}_{111}=301

Therefore, $T^{\sigma_{1}}_{300}=110$ .

Now we investigate $\sigma_{2}$ . It has $p_{2}=487$ and $|\sigma_{2}|_{1}=41$ . Since $T^{\sigma_{2}}_{2}=28$ and $T^{\sigma_{2}}_{3}=38$ , we know that $V^{\sigma_{2}}_{33}=V^{\sigma_{2}}_{34}=3$ . Lemma 6.15 implies

V^{\sigma_{2}}_{520}=300\text{ and }V^{\sigma_{1}}_{521}=301

Therefore, $T^{\sigma_{2}}_{300}=520$ .

Hence,

	$\displaystyle\alpha\cdot\sigma$	$\displaystyle=(s-4)\alpha\cdot\sigma_{1}+\alpha\cdot\sigma_{2}$
		$\displaystyle\leq 110(s-4)+520$
		$\displaystyle=110s+80$

∎

Lemma 7.6.

T^{\sigma}_{11s^{2}-33s+24}\leq 22s^{2}-22s-28

Proof.

Let $\alpha\in S_{\leq 11s^{2}-33s+24}$ be such that $\alpha\cdot\sigma=T^{\sigma}_{11s^{2}-33s+24}$ .

Let $\sigma_{1}=(4,2,1,1,0,\dots)$ and $\sigma_{2}=(4s-5,2s-3,s-1,s-2,1,\dots,1)$ . We have

\alpha\cdot\sigma=2\alpha\cdot\sigma_{1}+\alpha\cdot\sigma_{2}

First we investigate $\sigma_{1}$ . It has $p_{1}=22$ and $|\sigma_{1}|_{1}=8$ . Since $T^{\sigma_{1}}_{1}=4$ and $T^{\sigma_{1}}_{2}=6$ and $T^{\sigma_{1}}_{3}=8$ , we know that $V^{\sigma_{1}}_{6}=2$ and $V^{\sigma_{1}}_{7}=3$ . Lemma 6.15 implies

V^{\sigma_{1}}_{22(s-2)+6}=11s^{2}-34s+26\text{ and }V^{\sigma_{1}}_{22(s-2)+7}=11s^{2}-33s+25

Therefore,

T^{\sigma_{1}}_{11s^{2}-33s+24}=22(s-2)+6=22s-38

Now we investigate $\sigma_{2}$ . Rational greedy algorithm up to $\tfrac{4s-6}{4s-5}$ coin per point gives

\alpha=(4s-7+\dfrac{4s-7}{4s-6},2s-4,s-2,s-3,0,\dots)

which implies

T^{\sigma_{2},\mathbb{Q}}_{11s^{2}-33s+24}=22s^{2}-66s+48+\dfrac{4s-7}{4s-6}

which implies

T^{\sigma_{2}}_{11s^{2}-33s+24}\leq 22s^{2}-66s+48

Hence,

	$\displaystyle\alpha\cdot\sigma$	$\displaystyle=2\alpha\cdot\sigma_{1}+\alpha\cdot\sigma_{2}$
		$\displaystyle\leq 2(22s-38)+22s^{2}-66s+48$
		$\displaystyle=22s^{2}-22s-28$

∎

Lemma 7.7.

T^{\sigma}_{11s^{2}-33s+25}\geq 22s^{2}-22s-24

Proof.

This follows from observing that when $\alpha=(4s-6,2s-4,s-2,s-3)$ , we have

\alpha\in S_{11s^{2}-33s+25}\text{ and }\alpha\cdot\sigma=22s^{2}-22s-24

∎

Now we can prove Theorem 7.2, which states that when $s\geq 5$ and $r\geq 2$ , if we let $a$ be the number of elements in the set $\{i\geq 0\mid 296\leq V_{i}\leq 300\}$ and let $b$ be the biggest value satisfying the property that there are at least $\dfrac{a}{2}$ elements in the set $\{i\geq 0\mid V_{i}=b\}$ , then

s=\left\lfloor\sqrt{\dfrac{b}{11}}\right\rfloor\text{ or }\left\lfloor\sqrt{\dfrac{b}{11}}\right\rfloor+1\text{ or }\left\lfloor\sqrt{\dfrac{b}{11}}\right\rfloor+2

Proof of Theorem 7.2.

First of all, note that $p=22s^{2}+31s+11$ and $|\sigma|_{1}=9s+5$ . So we have

T^{rel}_{V_{0}^{rel}}=\nu^{+rel}\geq-\dfrac{1}{2}+\dfrac{1}{2}(2p-|\sigma|_{1})=22s^{2}+\dfrac{53}{2}s+8

Hence, when $r\geq 2$ ,

T^{\sigma}_{V_{0}^{rel}}\geq 22s^{2}+\dfrac{53}{2}s+8

Notice that when $s\geq 5$ , $22s^{2}+\dfrac{53}{2}s+8$ is strictly bigger than $110s+80$ and $22s^{2}-22s-28$ . Lemma 7.5 and Lemma 7.6 implies that $300$ and $11s^{2}-33s+24$ are strictly less than $V_{0}^{rel}$ . This implies $295$ , $300$ , $11s^{2}-33s+24$ , and $11s^{2}-33s+25$ are all less than or equal to $V_{0}^{rel}$ . Hence, in the statements of Lemma 7.4-7.7, $T^{\sigma}$ can be replaced by $T^{rel}$ .

Together with Lemma 7.3, we conclude that

6r>a>4r

and

|\{i\geq 0\mid V_{i}={11s^{2}-33s+25}\}|>3r

Hence, we know that $b\geq 11s^{2}-33s+25$ .

To get an upper bound of $b$ in terms of $s$ , notice that since

|\{i\geq 0\mid V_{i}=b\}|\geq\dfrac{a}{2}>2r

Lemma 7.3 implies that $T^{rel}_{b}-T^{rel}_{b-1}\geq 2$ . Lemma 6.11 implies that

b\leq\dfrac{p-|\sigma|_{1}}{2}=11s^{2}+11s+3

Result follows from

11(s-2)^{2}<11s^{2}-33s+25\leq b\leq 11s^{2}+11s+3<11(s+1)^{2}

∎

We conclude this section by proving Theorem 1.11, which states that any knot in $S^{3}$ has at most 7 lensbordant surgeries with an associated changemaker coming from this family.

Proof of Theorem 1.11.

Fix a knot in $S^{3}$ .

When $r=1$ , $T^{rel}_{1}=4s+3$ is the number of $V_{i}$ ’s with value 1. Therefore, there is at most 1 possible value of $s$ .

When $r\geq 2$ , Theorem 7.2 implies that there are at most 3 possible values of $s\geq 5$ .

Hence, together with the possibilities $s=2,3,4$ , there are at most 7 possible values of $s$ . Result follows from Lemma 7.1. ∎

8. Knots in the Poincaré homology sphere

In this section, we work on knots in the Poincaré homology sphere (denoted as $\mathcal{P}$ ) instead of in $S^{3}$ . We use Caudell’s work [6] to generalize our work in Section 3 of this paper to knots in $\mathcal{P}$ .

We make the following definitions that can be found in [6, §1.4]:

Let $a\leq b$ be integers that are either both odd or both even. We define $PI(a,b)$ to be the set of integers between $a$ and $b$ (including $a,b$ ) that have the same parity (odd or even) as $a,b$ .

For every negative-definite unimodular lattice $L$ ,

m(L):=\max\{\langle\mathfrak{c},\mathfrak{c}\rangle\mid\mathfrak{c}\in Char(L)\}

short(L):=\{\mathfrak{c}\mid\mathfrak{c}\in Char(L)\text{ and }\langle\mathfrak{c},\mathfrak{c}\rangle=m(L)\}

Short(L):=\{\mathfrak{c}\mid\mathfrak{c}\in Char(L)\text{ and }\langle\mathfrak{c},\mathfrak{c}\rangle=m(L)-8\}

For every vector $v\in L$ ,

c(v):=\max\{\langle\mathfrak{c},v\rangle\mid\mathfrak{c}\in short(L)\}

C(v):=\max\{\langle\mathfrak{c},v\rangle\mid\mathfrak{c}\in Short(L)\}

Definition 8.1.

[6, Def 1.6] A vector $\tau=(s,\sigma)\in-E_{8}\oplus-\mathbb{Z}^{n-7}$ is an $E_{8}$ -changemaker if

PI(-c(\tau),c(\tau))=\{\langle\mathfrak{c},\tau\rangle\mid\mathfrak{c}\in short(-E_{8}\oplus-\mathbb{Z}^{n-7})\}

and

PI(c(\tau)+2,C(\tau))\subset\{\langle\mathfrak{c},\tau\rangle\mid\mathfrak{c}\in Short(-E_{8}\oplus-\mathbb{Z}^{n-7})\}

Caudell noted that $short(-E_{8}\oplus-\mathbb{Z}^{n-7})=\{0\}\oplus\{\pm 1\}^{n-7}$ , and therefore the first condition of Definition 8.1 is equivalent to saying that $\sigma$ satisfies the changemaker condition (as in Definition 3.1, but the entries of $\sigma$ are allowed to be 0).

Note that the definition of changemaker in [6] allows the entries to be 0. That is different from Definition 3.1 which comes from [9].

One important result by Caudell [6, Th. 1.19] is that if the canonical negative definite plumbing 4-manifold of $L(p,q)$ embeds in $-E_{8}\oplus-\mathbb{Z}^{n-7}$ as an orthogonal complement of an $E_{8}$ -changemaker, then $L(p,q)$ is a positive integer surgery on one of the knots described by Tange [18] in $\mathcal{P}$ . In this section, we use this to generalize the content of Section 3 to L-space knots in $\mathcal{P}$ .

The aim of this section is to prove

Theorem 1.12.

Suppose $r^{2}p$ -surgery on a knot $K\subset\mathcal{P}$ produces an L-space that is smoothly $\mathbb{Z}_{2}$ -homology cobordant to a reduced $L(p,q)$ . Let

i(K,r,p)=\begin{cases}2g(K)+r\text{ if }p\text{ is odd}\\ 2g(K)\quad\ \ \text{ if }p\text{ is even}\end{cases}

If $r^{2}p\geq i(K,r,p)$ , then $L(p,q)$ must be a positive integer surgery on a knot in $\mathcal{P}$ .

If $r^{2}p<i(K,r,p)$ , then $L(p,q)$ must be a positive integer surgery on a knot in $S^{3}$ .

We use a similar set up as Section 3. In this section, we let $K$ be an L-space knot in $K\subset\mathcal{P}$ . $L(p,q)$ is reduced and is smoothly $Z_{2}$ -homology cobordant to the $r^{2}p$ -surgery on $K$ which is denoted as $K_{r^{2}p}$ . Let $W$ be the cobordism. Let $P$ be the canonical negative definite plumbed 4-manifold with boundary $L(p,q)$ . Let $W_{r^{2}p}$ be the 4-manifold obtained by attaching a $r^{2}p$ -framed 2-handle along $K\subset\mathcal{P}\subset\partial(\mathcal{P}\times[0,1])$ . Let $i(K,r,p)$ be as described in Theorem 1.12. (sorry for using $\mathcal{P}$ and $P$ to denote different things, which came from using notations from both [6] and [1])

Since $L(p,q)$ is smoothly $Z_{2}$ -homology cobordant to $K_{r^{2}p}$ , it must be smoothly rational homology cobordant to $K_{r^{2}p}$ , and we also know that $|H_{1}(W)|$ is odd.

We also define the surface $\Sigma$ and label the $\text{spin}^{\text{c}}$ structures on $K_{r^{2}p}$ in the same way as in Section 3.

Let $X=P\cup_{L(p,q)}W$ and $Z=X\cup_{K_{r^{2}p}}W_{-r^{2}p}$ .

Figure 3. The set up of Section 8.

Since $\mathcal{P}$ is an integer homology 3-sphere, Lemma 3.2-3.8 and Definition 3.9 still applies. To use notations similar to [6], we let $\tau$ (instead of $\sigma$ ) to be a generator of the free part of $H_{2}(W\cup-W_{r^{2}p})$ such that

[\Sigma]=r\tau+torsion

In Section 3, the first time we used the information that $K\subset S^{3}$ was in the discussion between Lemma 3.8 and Definition 3.9 when we introduced the $V$ coefficients and how they relate the $d$ -invariants. So, we start again from there.

Instead of using those $V_{i}$ coefficients, we use the torsion coefficients $t_{i}$ defined as follows. Consider the symmetrized Alexander polynomial of $K$

\Delta_{K}(T)=\sum\limits_{j}a_{j}T^{j}

The torsion coefficients are defined as

t_{i}(K):=\sum\limits_{j\geq 1}ja_{|i|+j}

Some properties of torsion coefficients of L-space knots in $\mathcal{P}$ addressed in [18] and also mentioned in [5]. For $i\geq 0$ , the $t_{i}$ coefficients form a monotonic decreasing sequence of non-negative integers. Also, for any positive integer $k$ , whenever $|i|\leq\tfrac{k}{2}$ ,

(11)

2-2t_{i}(K)=d(K_{k},i)-d(U_{k},i)

where $K_{k}$ is the $k$ -surgery on $K\subset\mathcal{P}$ , and $U_{k}$ is the $k$ -surgery on the unknot in $S^{3}$ (which is $L(k,k-1)$ ). The $i$ behind denotes the $\text{spin}^{\text{c}}$ structure with label $i$ .

Equation (11) is almost the same as what we saw in Section 3, but with an extra 2 on the left hand side coming from the fact that $d(\mathcal{P})=2$ .

Lemma 8.2.

Let $i$ be relevant. Then

8-8t_{i}=\max_{\begin{subarray}{c}\mathfrak{c}\in Char(Z)\\ \langle\mathfrak{c},[\Sigma]\rangle+r^{2}p=2i\end{subarray}}(\mathfrak{c}^{2}+(n+1))

Proof.

Everything in the proof of Lemma 3.10 still applies, with $t_{i}$ replacing $V_{i}$ , and an extra 8 on the left hand side coming from the extra 2 on the left hand side of equation (11). ∎

Recall that in this section, $|H_{1}(W)|$ is odd. Consider the Mayer-Vietoris sequence

\rightarrow H_{1}(P\cup-W_{-r^{2}p})\oplus H_{1}(W)\rightarrow H_{1}(Z)\rightarrow H_{0}(L(p,q)\cup K_{r^{2}p})\xrightarrow{injective}

Since $H_{1}(P\cup-W_{-r^{2}p})=0$ , we know that $|H_{1}(Z)|$ divides $|H_{1}(W)|$ . Hence, $H_{1}(Z)$ has no 2-torsion.

Lemma 8.3.

If $r^{2}p\geq i(K,r,p)$ , the intersection form of $Z$ is $-E_{8}\oplus-\mathbb{Z}^{n-7}$ .

If $r^{2}p<i(K,r,p)$ , the intersection form of $Z$ is $-\mathbb{Z}^{n+1}$ .

Proof.

According to Scaduto’s work [17, Cor. 1.4], if a smooth compact oriented negative definite 4-manifold with $\mathcal{P}$ boundary and $b_{2}=n+1$ has no 2-torsion in its homology, then its intersection form is either $-\mathbb{Z}^{n+1}$ or $-E_{8}\oplus-\mathbb{Z}^{n-7}$ .

Since $K$ is an L-space knot in $\mathcal{P}$ , by combining [20, Prop. 3.7], [13, Prop. 3.5], and [16, Prop. 3.1], we know that the genus $g(K)$ coincides with the leading degree of the symmetrized Alexander polynomial.

By comparing the definition of $i(K,r,p)$ and Definition 3.9, and consider the fact that relevant coefficients are monotonic decreasing, we see that $r^{2}p\geq i(K,r,p)$ if and only if there exists some relevant $i$ such that $t_{i}=0$ .

Hence, it suffices to show that when there exists some relevant $i$ such that $t_{i}=0$ , the intersection form of $Z$ cannot be $-\mathbb{Z}^{n+1}$ , and show that when the intersection form of $Z$ is $-E_{8}\oplus-\mathbb{Z}^{n-7}$ , there exists some relevant $i$ such that $t_{i}=0$ .

Suppose there exists some relevant $i$ such that $t_{i}=0$ .

Similar to the argument in [6, §3], observe that for any characteristic vector $\mathfrak{c}$ in $-\mathbb{Z}^{n+1}$ , $\mathfrak{c}^{2}$ is at most $-(n+1)$ . Hence, it is impossible to obtain

8=\mathfrak{c}^{2}+(n+1)

Therefore, we can conclude that the intersection form of $Z$ cannot be $-\mathbb{Z}^{n+1}$ .

Suppose the intersection form of $Z$ is $-E_{8}\oplus-\mathbb{Z}^{n-7}$ .

Let $\tau=s+\sigma$ , with $s\in-E_{8}$ and $\sigma\in-\mathbb{Z}^{n+1}$ . By choosing a suitable orthonormal basis, we can assume that all entries of $\sigma$ are non-negative.

By considering $\mathfrak{c}=(0,(1,\dots,1))\in short(-E_{8}\oplus-\mathbb{Z}^{n-7})$ , Lemma 8.2 implies that

t_{\tfrac{1}{2}r(rp-|\sigma|_{1})}=0

Note that

p=\langle\tau,\tau\rangle=-\langle s,s\rangle-\langle\sigma,\sigma\rangle

Since $E_{8}$ is an even lattice, $\langle s,s\rangle$ is always even. Hence, $p$ has the same parity (odd or even) as $|\sigma|_{1}$ . Therefore, $t_{\tfrac{1}{2}r(rp-|\sigma|_{1})}$ is relevant. ∎

Lemma 8.4.

If the intersection form of $Z$ is $-\mathbb{Z}^{n+1}$ , then $L(p,q)$ is a positive integer surgery on a knot in $S^{3}$ .

Proof.

This is implied by the argument in Section 3, with $t_{i}$ replacing $V_{i}$ , $\tau$ replacing $\sigma$ , Lemma 8.2 replacing Lemma 3.10, and in the proof of Lemma 3.11 we consider the relevant coefficients that equals to 1 instead of relevant coefficients that equals to 0. ∎

From this point onwards, until we start proving Theorem 1.12, we assume that the intersection form of $Z$ is $-E_{8}\oplus-\mathbb{Z}^{n-7}$ .

Let $\tau=s+\sigma$ , with $s\in-E_{8}$ and $\sigma\in-\mathbb{Z}^{n-7}$ . By choosing a suitable orthonormal basis, we can assume that all entries of $\sigma$ are non-negative.

We now prove that $\tau$ is an $E_{8}$ -changemaker. The definition of $E_{8}$ -changemaker consists of two conditions. We prove them one by one using arguments similar to the ones in [6, §3].

Lemma 8.5.

$\sigma$ is a changemaker (with entries allowed to be 0)

Proof.

We follow a similar argument as in the proof of Lemma 3.11.

When $\mathfrak{c}=(0,(1,\dots,1))\in short(-E_{8}\oplus-\mathbb{Z}^{n-7})$ , we have $\mathfrak{c}^{2}+(n+1)=8$ and

\langle\mathfrak{c},[\Sigma]\rangle+r^{2}p=2\left(\dfrac{r^{2}p-r|\sigma|_{1}}{2}\right)

By Lemma 8.2, we conclude that

t_{\tfrac{1}{2}r(rp-|\sigma|_{1})}=0

By Lemma 8.2, we know that for all relevant $i\geq\tfrac{1}{2}r(rp-|\sigma|_{1})$ , there exists some $\mathfrak{c}\in Char(-E_{8}\oplus-\mathbb{Z}^{n-7})$ such that $\mathfrak{c}^{2}=-(n-7)$ and $\langle\mathfrak{c},[\Sigma]\rangle+r^{2}p=2i$ . Since $short(-E_{8}\oplus-\mathbb{Z}^{n-7})=\{0\}\oplus\{\pm 1\}^{n-7}$ , we know that such a $\mathfrak{c}$ must be in $\{0\}\oplus\{\pm 1\}^{n-7}$ . The rest of the proof of Lemma 3.11 applies. ∎

Lemma 8.6.

If $\langle s,s\rangle\leq-4$ , then

PI(c(\tau)+2,C(\tau))\subset\{\langle\mathfrak{c},\tau\rangle\mid\mathfrak{c}\in Short(-E_{8}\oplus-\mathbb{Z}^{n-7})\}

Proof.

From [6, Lemma 3.4], we know that if $\langle s,s\rangle\leq-4$ , then for all $\mathfrak{c}\in Short(-E_{8}\oplus-\mathbb{Z}^{n-7})$ , we have $|\langle\mathfrak{c},\tau\rangle|\leq|\langle\tau,\tau\rangle|=p$ . Hence, for all $\mathfrak{c}\in Short(-E_{8}\oplus-\mathbb{Z}^{n-7})$ , we have

(12)

\langle\mathfrak{c},\tau\rangle+rp\geq(r-1)p

Let $i_{0}$ be the smallest relevant index such that $t_{i_{0}}=0$ . Let $i_{1}$ is the smallest relevant index relevant index such that $t_{i_{1}}\leq 1$ .

Case 1: $r$ is odd or $p$ is even
Let $i^{\prime}:=\tfrac{i}{r}$ . Lemma 8.2 can be rewritten as:

For all $0\leq i^{\prime}\leq\tfrac{rp}{2}$ ,

8-8t_{ri^{\prime}}=\max_{\begin{subarray}{c}\mathfrak{c}\in Char(Z)\\ \langle\mathfrak{c},\tau\rangle+rp=2i^{\prime}\end{subarray}}(\mathfrak{c}^{2}+(n+1))

Hence, when $0\leq i^{\prime}\leq\tfrac{rp}{2}$ ,

t_{ri^{\prime}}=0\Leftrightarrow\exists\mathfrak{c}\in short\text{ such that }\langle\mathfrak{c},\tau\rangle+rp=2i^{\prime}

and

t_{ri^{\prime}}=1\Leftrightarrow t_{ri^{\prime}}\neq 0\text{ and }\exists\mathfrak{c}\in Short\text{ such that }\langle\mathfrak{c},\tau\rangle+rp=2i^{\prime}

By (12), we know that

i_{1}=\dfrac{r}{2}\left(rp-C(\tau)\right)\text{ and }i_{0}=\dfrac{r}{2}\left(rp-c(\tau)\right)

Let $j\in PI(c(\tau)+2,C(\tau))$ . We want to show that $j\in\{\langle\mathfrak{c},\tau\rangle\mid\mathfrak{c}\in Short(-E_{8}\oplus-\mathbb{Z}^{n-7})\}$ . Let

i=\dfrac{r}{2}\left(rp-j\right)

Since $j$ has the same parity (odd or even) as $C(\tau)$ , $i$ differs from $i_{1}$ by a multiple of $r$ . So, $i$ is relevant. Since $i_{1}\leq i<i_{0}$ , we must have $t_{i}=1$ . Hence, there exists some $\mathfrak{c}\in Short$ such that

\langle\mathfrak{c},\tau\rangle+rp=\dfrac{2i}{r}

This implies $\langle-\mathfrak{c},\tau\rangle=j$ .

Case 2: $r$ is even and $p$ is odd
Let $i^{\prime}:=\tfrac{i}{r}-\tfrac{1}{2}$ . Lemma 8.2 can be rewritten as:

For all $0\leq i^{\prime}\leq\tfrac{rp}{2}-1$ ,

8-8t_{ri^{\prime}}=\max_{\begin{subarray}{c}\mathfrak{c}\in Char(Z)\\ \langle\mathfrak{c},\tau\rangle+rp=2i^{\prime}+1\end{subarray}}(\mathfrak{c}^{2}+(n+1))

Since $r$ is even, (12) implies that for all $\mathfrak{c}\in Short$ , we have

\langle\mathfrak{c},\tau\rangle+rp-1\geq 0

The rest of the proof of Case 1 applies. ∎

Lemma 8.7.

$\tau$ is an $E_{8}$ -changemaker.

Proof.

According to the arguments in the proofs of [6, Prop. 3.6] and [6, Prop. 4.2], $\tau$ is an $E_{8}$ -changemaker when $\langle s,s\rangle=0,-2$ . Since $E_{8}$ is an even lattice, the only other possibility is $\langle s,s\rangle\leq-4$ , which Lemma 8.6 says that $\tau$ is an $E_{8}$ -changemaker. Hence, in all cases, $\tau$ is an $E_{8}$ -changemaker. ∎

Now we can prove Theorem 1.12.

Proof of Theorem 1.12.

By Lemma 3.3, Lemma 8.7, and [6, Th. 1.19], we conclude that when the intersection form of $Z$ is $-E_{8}\oplus-\mathbb{Z}^{n-7}$ , $L(p,q)$ is a positive integer surgery on a knot in the Poincaré homology sphere described by Tange. The rest of the proof of Theorem 1.12 follows from Lemma 8.3 and Lemma 8.4. ∎

References

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	$\displaystyle V_{0}$	$\displaystyle=V^{rel}_{0}$
		$\displaystyle=V^{\sigma}_{(x-1)p+\tfrac{2p-\|\sigma\|_{1}}{2}}$
		$\displaystyle=\dfrac{(x-1)^{2}p+(x-1)\|\sigma\|_{1}}{2}+(x-1)\dfrac{2p-\|\sigma\|_{1}}{2}+V^{\sigma}_{\tfrac{2p-\|\sigma\|_{1}}{2}}$
		$\displaystyle=\dfrac{\left(\tfrac{r}{2}-1\right)^{2}p+\left(\tfrac{r}{2}-1\right)\|\sigma\|_{1}}{2}+\left(\dfrac{r}{2}-1\right)\dfrac{2p-\|\sigma\|_{1}}{2}+\dfrac{p}{2}$

	$\displaystyle V_{\tfrac{r}{2}}$	$\displaystyle=V^{rel}_{0}$
		$\displaystyle=V^{\sigma}_{(x-1)p+\tfrac{2p-\|\sigma\|_{1}-1}{2}}$
		$\displaystyle=\dfrac{(x-1)^{2}p+(x-1)\|\sigma\|_{1}}{2}+(x-1)\dfrac{2p-\|\sigma\|_{1}-1}{2}+V^{\sigma}_{\tfrac{2p-\|\sigma\|_{1}-1}{2}}$
		$\displaystyle=\dfrac{\left(\tfrac{r}{2}-1\right)^{2}p+\left(\tfrac{r}{2}-1\right)\|\sigma\|_{1}}{2}+\left(\dfrac{r}{2}-1\right)\dfrac{2p-\|\sigma\|_{1}-1}{2}+\dfrac{p-1}{2}$

	$\displaystyle V_{0}$	$\displaystyle=V^{rel}_{0}$
		$\displaystyle=V^{\sigma}_{xp+\tfrac{p-\|\sigma\|_{1}}{2}}$
		$\displaystyle=\dfrac{(x^{2}p+x\|\sigma\|_{1}}{2}+x\dfrac{p-\|\sigma\|_{1}}{2}+V^{\sigma}_{\tfrac{p-\|\sigma\|_{1}}{2}}$
		$\displaystyle=\dfrac{\left(\tfrac{r-1}{2}\right)^{2}p+\left(\tfrac{r-1}{2}\right)\|\sigma\|_{1}}{2}+\left(\dfrac{r-1}{2}\right)\dfrac{2p-\|\sigma\|_{1}}{2}+V^{\sigma}_{\tfrac{p-\|\sigma\|_{1}}{2}}$

Integer surgeries rational homology cobordant to lens spaces

Abstract.

1. Introduction

Definition 1.1.

Conjecture 1.2.

Theorem 1.3.

Theorem 1.4.

Theorem 1.5.

Definition 1.6.

Theorem 1.7.

Theorem 1.8.

Corollary 1.9.

Theorem 1.10.

Theorem 1.11.

Theorem 1.12.

2. Analyzing the Aceto-Celoria-Park map

Lemma 2.1.

Proof.

Lemma 2.2.

Proof.

Lemma 2.3.

Proof.

Lemma 2.4.

Proof.

Corollary 2.5.

Proof.

3. Proof of Theorem 1.5

Theorem 1.5.

Definition 3.1.

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.

Lemma 3.4.

Proof.

Lemma 3.5.

Proof.

Lemma 3.6.

Proof.

Lemma 3.7.

Proof.

Lemma 3.8.

Proof of Lemma 3.8.

Definition 3.9.

Lemma 3.10.

Proof of Lemma 3.10.

Lemma 3.11.

Proof.

Proof of Theorem 1.5.

4. Number of possible surgeries when fixing rr

Definition 4.1.

Lemma 4.2.

Proof of Lemma 4.2.

Theorem 1.7.

Lemma 4.3.

Proof.

Lemma 4.4.

Proof.

Theorem 1.4.

Proof of Theorem 1.4.

Lemma 4.5.

Proof.

Lemma 4.6.

Proof.

Proof of statement (2) and (3) of Theorem 1.7.

5. Some bounds on the surgery slope r2​pr^{2}p

Theorem 1.8.

Proof.

Corollary 1.9.

Proof.

6. Structure of relevant coefficients

Theorem 6.1.

Definition 6.2.

Definition 6.3.

Definition 6.4.

Lemma 6.5.

Proof.

Corollary 6.6.

Proof.

Lemma 6.7.

4. Number of possible surgeries when fixing $r$

5. Some bounds on the surgery slope $r^{2}p$