# Minimal Genus Seifert surface of $11$ crossing alternating knots

Neetal Neel

## Abstract

Kakimizu complexes have been found for several classes of links. O.Kakimizu
found the Kakimizu complexes of knots with crossing number less than or equal
to 10. Hatcher and Thurston found the 0-skeleton of the Kakimizu complex of
2-bridge links. Sakuma later generalized the result for special arborescent
links and found the Kakimizu complexes for the same. Jessica Banks gave a
complete proof of results announced by Hirasawa and Sakuma, describing
explicitly the Kakimizu complexes of non-split, prime special alternating
links. The Kakimizu complexes of prime, non-split alternating links have
finite number of vertices. In this paper we compute the Kakimizu complexes for
all $11$ crossing prime, alternating knots, explicitly describing each of
them. For most knots and links we use known algorithms. The rest of the
Kakimizu complexes were found by using Murasugi sums and sutured manifold
theory developed by Gabai, Scharlemann, Kakimizu and others.

##  1 Introduction

Let $K$(or $L$) be an oriented knot (or link) in $S^{3}.$ A Seifert surface on
$K$ (or $L$) is a compact, connected, orientable surface with $K$(or $L$) as
the boundary of the surface compatible with the orientation of $K$(or $L$) and
which does not contain any closed component. Let $E(L)$ be the link complement
defined by $S^{3}-(N(L))^{\circ}$, where $N(L)$ is a regular neighborhood of
$L$ homeomorphic to $L\times D^{2}$. $L$ will be used to denote a link (with
one component links considered as knots). $K$ will be used to specifically
denote a knot.

We will call $S$ to be a Seifert surface on $L$ to denote $S$ a Seifert
surface with no closed component and $L$ as the boundary of $S$ contained in
$S^{3}$. We will abuse notation to denote $S$, a Seifert surface on $L$ lying
on $E(L),$ i.e $S\cap E(L)$ with $\partial(S\cap E(L))$ is a parallel copy of
$L$ lying in the boundary tori of the regular neighborhood of the link $L$,
i.e $\partial N(L).$ It will be clear from the context otherwise will be
elaborated.

In 1992, Kakimizu introduced Kakimizu complex of an oriented knot. Kakimizu
complex aims to explore the configuration of Seifert surfaces on links $L$ in
the link complement, $E(L).$ More specifically, it aims to answer how disjoint
representatives of the set of isotopy classes of incompressible (or minimal
genus) Seifert surfaces on a link $L$ is situated in the knot/link complement,
expressed with the help of a complex.

In this paper, we will consider oriented, non-split links, since $E(L)$ for
non-split links are irreducible. Moreover we consider minimal genus Seifert
surfaces $S$ on a link $L$ so that $S$ is an incompressible surface. Note that
incompressible surfaces may not be a minimal genus Seifert surface but minimal
genus Seifert surfaces are incompressible. We consider incompressible surfaces
because given any surface, adding a compressible handle increases the genus
and represents a new surface on the given link. This would create an infinite
family of surfaces, hence an infinite complex on the given link $L$.

Kakimizu has computed the Kakimizu complex of all knots upto $10$ crossings.
In 2018, Hass, Tsvietkova and Thompson announced the result, that the number
of isotopy classes of Seifert surfaces of a given genus is bounded by a
polynomial depending only on the genus of the Seifert surface. This shows that
the Kakimizu complex is a finite complex for alternating links. Cromwell in
1989 showed that every homogenenous link $L$ (a superset of the set of
alternating links) is a $*-$product of special alternating links, $L_{i}$.
This is equivalent to, given a reduced alternating diagram $D$ of a
homogeneous link $L$ and applying Seifert’s algorithm on $D,$ we obtain a
Seifert surface $S$ which is the Murasugi sum of Seifert surfaces $S_{i}$ on
$L_{i}.$ This served as a motivation to try to find an algorithm for computing
the Kakimizu complex of homogeneous links $L.$ Now this problem turns out to
be very difficult owing to the fact, that the sutured manifold theory is
developed for plumbing but is absent for a general Murasugi $2n-$gon disk.
Moreover the Kakimizu complex depends on the embedding site of the Murasugi
disks in the corresponding Seifert surfaces on the special alternating links
$L_{i},$ In this paper, we compute the Kakimizu complex of all alternating
$11$ crossing knots using the methods known for special classes of links. Most
of the links either fall in one of the classes or could be found using
Theorems of Kakimizu and Gabai.

There are a few 11 crossing links whose Kakimizu complex had to be found using
sutured manifold theory. I found a new type of surface, called the parallel
surface, which is a priori not a plumbing.

I would like to thank Jennifer Schultens, my advisor, for the constant and
unwavering help throughout the work. I would also like to thank Andrew Alameda
for the valuable conversations we had.

##  2 Preliminaries

###  2.1 Sutured Manifold Theory

This section is mostly taken from the paper by Kakimizu, _Classification of
the incompressible spanning surfaces for prime knots of 10 or less crossings._
For a detailed reading please refer to [1].

A _sutured manifold $(M,\gamma)$_ is a compact oriented $3$-manifold $M$
together with a union of finitely many pairwise disjoint annuli
$\gamma\subset\partial M.$ For each component of $\gamma,$ which is an
annulus, there exists a core circle which is called a _suture_. The set of
sutures are denoted by $s(\gamma).$ Moreover $R(\gamma)$ defined as $\partial
M-\gamma^{\circ}$ is oriented and is coherent with respect to $s(\gamma).$ Let
$R_{+}(\gamma)$ (or $R_{-}(\gamma)$) be the components of $R(\gamma)$ whose
normal vectors point out of $M$ (respectively point into $M.$) The _knot
complement_ of a link $L,$ is $M=S^{3}-N(L),$ a compact, oriented 3-manifold
and $\partial M=\partial N(L)\cong L\times\mathbb{S}^{1}.$ The sutures,
$s(\gamma),$ generally are parallel to the components of the link $L.$

Let $(M,\gamma)$ be a sutured 3-manifold. A $\gamma$-surface $S$ is a properly
embedded compact oriented surface in $M$ such that $\partial
S\subset(\gamma)^{\circ}$ and is isotopic to $s(\gamma),$ with no closed
component. A $\gamma$-surface $S$ is said to be _parallel_ if there exists an
isotopy

$e:(S,\partial S)\times I\to(M,\gamma)$

with $e_{0}=\text{ identity}$ and $e_{1}(S)\subset R(\gamma).$ An _essential_
$\gamma$-surface is a $\gamma$-surface which is incompressible and not
parallel to a surface in $R(\gamma).$ Note that if $S$ is a $\gamma$-surface
in $(M,\gamma),$ $\partial M$ is connected and $S$ is parallel to a surface in
$R(\gamma),$ then $S$ is isotopic to $R_{+}(\gamma)$ or $R_{-}(\gamma).$

A $\gamma$-isotopy of $(M,\gamma)$ is a map $h:I\times M\to M$ such that
$h_{0}=$identity, with the property $h_{t}|_{R(\gamma)}=$identity, and
$h_{t}|_{\gamma}=$identity for all $0\leq t\leq 1$. We say two minimal genus
Seifert surfaces on a link $L$, $S$ and $S^{\prime},$ are isotopic to each
other if there is an ambient isotopy

$e:S^{3}\times I\to S^{3}$

such that $e_{0}=$identity map and $e_{1}(S)=S^{\prime}$ with
$e_{t}|_{L}=$identity. If $S$ and $S^{\prime}$ are isotopic then there is a
$\gamma$-isotopy $h:I\times M\to M$ of $(M,\gamma)=(S^{3}-N(L),\partial N(L))$
such that $h_{1}(\tilde{S})=\tilde{S^{\prime}}$ where
$\tilde{S},\tilde{S^{\prime}}$ denote $S$ and $S^{\prime}$ restricted to the
link complement ($S^{3}-N(L))$ respectively. We will think of a minimal genus
Seifert surface $S$ on $L$ as a $\gamma$-surface on
$(M,\gamma)=(S^{3}-N(L),\partial N(L))$ with the boundary of the surface in
$\partial N(L)\cap S$, a parallel copy of $L$ in $\partial N(L).$

Two major examples of sutured manifolds are product sutured manifolds and the
complementary sutured manifolds.

Let $F$ be a compact, oriented $2$-manifold. Then $(M,\gamma),$ homeomorphic
to $(F\times I,\partial F\times I),$ is called a product sutured manifold. If
$S$ (on $L$) is a minimal genus Seifert surface in the link complement then a
regular neighborhood,

$\bigg{(}N(S),(S\cap\partial N(L))\times
I\bigg{)}\subset\bigg{(}S^{3}-N(L),\partial N(L)\bigg{)}$

is a product sutured manifold since $N(S)\cong S\times I$ and $S\cap\partial
N(L))\cong L.$

Let $S$ be a minimal genus Seifert surface (on $L$) in the link complement. In
this paper, we are considering only links that are oriented and non-split. Let
$E(L)$ denote the link complement and $(N(S),\partial S\times I)$ be the
product sutured manifold.

A sutured manifold $(M_{S},\gamma)$ defined by

$(M_{S},\gamma)=\bigg{(}\overline{E(L)-N(S)},\hskip
5.69046pt\overline{\partial E(L)-((S\cap\partial N(L))\times I)}\bigg{)}$

is called the complementary sutured manifold. Given a minimal genus spanning
surface $S$ (on $L$) we look for essential surfaces in the complementary
sutured manifold of $S$, $(M_{S},\gamma).$ Any essential $\gamma$-surface
$S^{\prime}$ in $(M_{S},\gamma)$ is disjoint from $S$ and is not isotopic to
$S.$

###  2.2 Product Decomposition

Let $(M,\gamma)$ be a sutured manifold. Let $D\subset M$ be a disk properly
embedded in $M$ such that $\partial D$ intersects $s(\gamma)$ in exactly at
$2$ points. $D$ is said to be a _product disk_ in $(M,\gamma).$ An operation
on $(M,\gamma)$ called _product decomposition_ yields us another sutured
manifold $(M^{\prime},\gamma^{\prime}).$ Each component of
$R(\gamma^{\prime})$ is incompressible if and only if every component of
$R(\gamma)$ is incompressible. We denote this operation as

$(M,\gamma)\xrightarrow{D}(M^{\prime},\gamma^{\prime})$

Let $(M,\gamma)$ be a sutured manifold. Note that the complementary sutured
manifold of a non-split link is irreducible. Moreover a $3$-submanifold of an
irreducible $3$-manifold is irreducible.

Let $S\subset M$ be a $\gamma$-surface. By a $\gamma$-isotopy we can isotope
$S$ such that $\partial S=s(\gamma)$ and $S$ intersects transversally
$\partial D$ in an arc connecting the $2$ points of intersection of
$s(\gamma)$ and $D.$ After product decomposition, cut through the arc, and we
get a new $\gamma^{\prime}$-surface $S_{D}$ in $(M^{\prime},\gamma^{\prime}).$

This is from [[1] , Lemma 1.3]

######  Theorem.

Let $(M,\gamma)\xrightarrow{D}(M^{\prime},\gamma^{\prime})$ be a product
decomposition. Suppose that $M$ is irreducible and $\partial M^{\prime}$ is
connected. Then for each essential $\gamma$-surface $S\subset M$, the
$\gamma^{\prime}$ -surface $S_{D}$ is essential. Moreover if $S,S^{\prime}$
are $\gamma-$isotopic to each other then $S_{D},S_{D}^{\prime}$ are
$\gamma^{\prime}$-isotopic as well.

Let $S_{(M,\gamma)}$ be the isotopy classes of essential surfaces
$S\subset(M,\gamma).$ Two essential surfaces $S,S^{\prime}$ are in the same
isotopy class if they are $\gamma$-isotopic to each other. Let
$(M,\gamma)\xrightarrow{D}(M^{\prime},\gamma^{\prime})$ be product
decomposition. Consider the map:

$S_{(M,\gamma)}\xrightarrow{\widetilde{D}}S_{(M^{\prime},\gamma^{\prime})}\text{
given by }\widetilde{D}(S^{\prime})=S_{D}^{\prime}.$

The previous theorem indicates that the above map is well defined.

This is [[1] ,Lemma 1.4]

######  Theorem.

Suppose that $M$ is irreducible and $\partial M^{\prime}$ is connected. Then
the map:
$S_{(M,\gamma)}\xrightarrow{\widetilde{D}}S_{(M^{\prime},\gamma^{\prime})}$ is
a bijection.

For fibred links $L$ with fibre $S,$ the complementary sutured manifold
$(M_{S},\gamma)$ is a product sutured manifold homeomorphic to $(S\times
I,L\times I).$

Hence the next theorem is useful in this context:

######  Theorem.

Let $(M,\gamma)\xrightarrow{D}(M^{\prime},\gamma^{\prime})$ be a product
decomposition. Let $M$ be irreducible and that $(M^{\prime},\gamma^{\prime})$
has two components, $(M_{1},\gamma_{1})$ and $(M_{2},\gamma_{2}).$ Let
$(M_{2},\gamma_{2})$ be a product sutured manifold and $\partial M_{1}$ be
connected. Then we have a bijection:

$S_{(M,\gamma)}\xrightarrow{\widetilde{\bar{D}}}S_{(M_{2},\gamma_{2})}$

is a bijection.

Finally a fundamental theorem proved by Kakimizu using Haken and Hempel and
Waldhausen’s theorems. [2],

######  Theorem.

Let $X$ be a connected Haken $3$-manifold such that $\partial X$ is a union of
incompressible tori. Let $Y$ be a compact irreducible $3$-submanifold of $X$
(possibly disconnected) such that each component of $\partial Y$ is a properly
embedded incompressible surface in $X$. Let $F$ and $F^{\prime}$ be two
properly embedded orientable incompressible surfaces in $X$ (possibly
disconnected) which satisfy the following properties $(1)–(4)$. Then there is
an isotopy $h_{t}$ of X keeping $Y$ fixed so that $h_{0}=id$ and
$h_{1}(F)=F^{\prime}.$

  1. (1)

$F\cup F^{\prime}\subset X-Y.$

  2. (2)

Each component of $\partial X$ contains at most one component of $\partial F$
and $F$ has no closed components.

  3. (3)

There is a homotopy $f:F\times I\to X$ such that $f_{0}=id$ and $f_{1}:F\to
F^{\prime}$ is a homeomorphism and $f(\partial F\times I)\subset\partial X,$

  4. (4)

There is no component of $F$ which is parallel to a component of $\partial Y.$

The proof is in [1] .

An application of this theorem is as follows. Let $X$ (from the theorem) be
the link complement $E(L)$ for a non-split link $L.$ Then $X$ is an
irreducible Haken manifold with $\partial X$ are incompressible tori with the
core circles are link components. Let $S$ be an essential surface on $L.$
Consider the complementary sutured manifold $Y=(M_{S},\gamma).$ Let
$S^{\prime},S^{\prime\prime}$ be two essential embedded $\gamma$-surfaces in
$(M_{S},\gamma)$ such that $S^{\prime}$ and $S^{\prime\prime}$ are isotopic in
$X.$ Then it satisfies all the conditions $(1)-(4),$ the theorem implies that
there is a $\gamma$-sutured isotopy of $S^{\prime}$ and $S^{\prime\prime}$ in
$(M,\gamma).$

###  2.3 Murasugi sums and Plumbings

An oriented surface $S$ is said to be a Murasugi sum of $S_{1}$ and $S_{2}$ if
there exists a splitting sphere $S^{2}\subset S^{3}$ splitting $S^{3}$ into
two balls $V_{1}$ and $V_{2}$ ($V_{1}\cup V_{2}=S^{3}$, $V_{1}\cap
V_{2}=S^{2}$ ) with

$S_{1}\subset V_{1},S_{2}\subset V_{2};\hskip 28.45274ptS_{1}\cup
S_{2}=S,S_{1}\cap S_{2}=S\cap(S^{2}=\partial V_{i})=D$

where $D$ is an embedded $2n$-gon on $S$. The disk $D$ is said to be the
$2n$-Murasugi disk for the Murasugi sum $S=S_{1}\cup_{D}S_{2}.$

Let $\partial S=L$ and $\partial S_{i}=L_{i}.$ Then $S\cap E(L)\subset E(L)$
is also said to be the Murasugi sum of $S_{1}\cap E(L_{1})$ and $S_{2}\cap
E(L_{2}).$ Also $L$ is said to be the Murasugi sum of $L_{1}$ and $L_{2}.$ Let
$S^{c}=S_{1}\cup_{D^{c}}S_{2}$ with $D^{c}=(S^{2}=\partial V_{1})-D.$ This is
obtained by $S^{c}=(S-D)\cup D^{c}.$ Note that $S^{c}$ is also the Murasugi
sum of $S_{1}$ and $S_{2}.$ This surface is called the dual of $S.$ By an
isotopy of $S^{3},$ keeping the link $L$ fixed, we can isotope $S^{c}$ such
that $S\cap S^{c}=\phi.$ For $n=2,$ the Murasugi disk is a rectangle and
$4$-Murasugi sum is called plumbing.

Gabai showed the following 2 properties for Murasugi sum operation. Assume $S$
on link $L$ be the Murasugi sum of $S_{1}$ and $S_{2}$ on $L_{1}$ and $L_{2}$
respectively.

  * •

$S$ is a minimal genus Seifert surface on $L$ if and only if $S_{1}$ and
$S_{2}$ are both minimal genus Seifert surfaces on $L_{1}$ and $L_{2}$
respectively.

  * •

$L$ is a fibred link with $S$ being the fibre if and only if $L_{1}$ and
$L_{2}$ are fibred links with $S_{1}$ and $S_{2}$ being the fibres for the
respective fibred links.

##  3 Definition of Kakimizu Complex

Let $L$ be a non-split oriented link. Seifert’s algorithm applied on a reduced
alternating diagram $D$ of $L,$ yields a Seifert surface $S$ on $L.$ Let
$E(L)$ denote the link complement. $E(L)$ is an irreducible $3$-manifold since
$L$ is non split. Consider the surface $S\cap E(L)\subset E(L).$ This is a
properly embedded compact surface in $E(L)$ with the boundary, $\partial(S\cap
E(L))\subset\partial E(L)$ homeomorphic to $L.$ Let $S$ be a Seifert surface
in $E(L)$ which is compressible in $E(L).$ Then we can use the compression
disks to compress the surface until the resultant surface is incompressible.
Since every minimal genus Seifert surface on a link $L$ is incompressible, we
consider minimal genus Seifert surface for the rest of the paper. This
establishes that every oriented non-split link $L$ admits a minimal genus
Seifert surface.

Let $S,S^{\prime}$ be Seifert surfaces on a link $L.$ $S$ and $S^{\prime}$ are
said to be isotopic if there exists an isotopy fixing the link $L,$ taking $S$
to $S^{\prime}.$ Let $e$ be the isotopy $e:I\times S^{3}\to S^{3}$ with
$e_{0}=id$ ; $e_{1}(S)=S^{\prime}$ and $e_{t}|_{L}=id$ for all $0\leq t\leq
1.$ Given an isotopy $e,$ of $S^{3}$ taking $S$ to $S^{\prime}$ with the link
$L$ fixed, there exists an isotopy $e^{\prime}$ of $E(L)$ such that the
surface $S\cap E(L)$ is taken to $S^{\prime}\cap E(L)$ with
$e^{\prime}_{t}(\partial E(L))=\partial E(L)$ for all $0\leq t\leq 1.$

The $0$-skeleton of the Kakimizu complex of a link $L$ are isotopy classes of
minimal genus Seifert surfaces on the link $L.$ The $1$-skeleton of the
Kakimizu complex of $L$ are edges between two vertices $v$ and $v^{\prime}$ if
there exists representative surfaces $S$ and $S^{\prime}$ respectively such
that $S$ and $S^{\prime}$ are disjoint in $E(L).$ Kakimizu complex is a flag
complex. If there are $n+1$ vertices such that every pair of vertex has an
edge then there is an $n$ simplex on the corresponding $n+1$ vertices.

##  4 Computing the Kakimizu Complex of knots

###  4.1 Kakimizu complex of fibred knots

In 1972, W.Whitten showed in [3] that for a fibred link $L$ there is a unique
incompressible surface for a fibred link $L$, unique in the sense of the
definition of Kakimizu complex. Hence $MS(L)=\\{[S]\\}$ which implies that the
Kakimizu complex of a fibred knot is a point.

We derive the set of fibred knots from KnotInfo [4].

There is an algorithm to check if a homogeneous link $L$ (that includes the
class of alternating links) is fibred or not. In 1989, Cromwell [5] showed
that a homogeneous link $L$ is a *-product of special alternating links,
$L_{i}$. Given an oriented diagram $D$ of a homogeneous link $L$, applying
Seifert’s algorithm to the diagram $D$ yields a Seifert surface $S$ which is a
Murasugi sum of Seifert surfaces $S_{i}$ on special alternating links $L_{i}$.
Gabai [6] showed that the Murasugi sum of two links, $L_{1}$ and $L_{2}$ is
fibred if and only if both $L_{1},$ and $L_{2}$ are fibred.

Each homogeneous link $L$ is a Murasugi sum of special alternating link,
Jessica Banks [7] has given an algorithm to detect fibredness of a special
alternating link: Let $D$ be a reduced diagram of a special alternating link
$L.$

Let $S(D)$ be the partition of $S^{2}$ into black and white regions by
applying Seifert’s algorithm and each Seifert disk constitutes a black region
in $S^{2}$. Let $G(D)$ be the planar graph with a vertex in each white region
and an edge through each crossing. A special alternating link $L$ is fibred if
given a reduced diagram $D$ of $L,$ we can reduce the graph $G(D)$ to a single
vertex using the following moves:

  * •

Delete a loop.

  * •

Contract an edge if one of the endpoint is a vertex of valence 2.

If every summand of a Murasugi sum of links is fibred then the Murasugi sum of
links is a fibred link.

The list of fibred links for $11$ crossing alternating links is as follows:

$11_{3},11_{5},11_{7},11_{9},11_{14},11_{15},11_{17},11_{19},11_{22},11_{24},11_{25},11_{26},11_{28},11_{33},11_{34},11_{35},11_{40},11_{42},$
$11_{44},11_{47},11_{51},11_{53},11_{55},11_{57},11_{62},11_{66},11_{68},11_{71},11_{72},11_{73},11_{74},11_{76},11_{79},11_{80},11_{81},$
$11_{82},11_{83},11_{86},11_{88},11_{92},11_{96},11_{99},11_{106},11_{108},11_{109},11_{112},11_{113},11_{121},11_{125},11_{126},11_{127},$
$11_{128},11_{129},11_{131},11_{139},11_{142},11_{146},11_{147},11_{151},11_{156},11_{157},11_{158},11_{159},11_{160},11_{162},11_{163},$
$11_{164},11_{170},11_{171},11_{174},11_{175},11_{176},11_{177},11_{179},11_{180},11_{182},11_{184},11_{189},11_{194},11_{196},11_{203},$
$11_{206},11_{209},11_{215},11_{216},11_{217},11_{218},11_{221},11_{223},11_{228},11_{231},11_{232},11_{233},11_{239},11_{248},11_{250},$
$11_{251},11_{252},11_{253},11_{254},11_{255},11_{257},11_{259},11_{261},11_{264},11_{266},11_{267},11_{268},11_{269},11_{274},11_{277},$
$11_{281},11_{282},11_{284},11_{286},11_{287},11_{288},11_{289},11_{293},11_{300},11_{301},11_{302},11_{305},11_{306},11_{308},11_{314},$
$11_{315},11_{316},11_{326},11_{330},11_{332},11_{346},11_{348},11_{350},11_{351},11_{367}.$

The Kakimizu complex of fibred links is a single vertex.

$T$

###  4.2 Kakimizu Complex of special alternating knots

In 2012, Jessica Banks [7] gave an explicit algorithm to find the Kakimizu
complex of a prime, non-split, oriented, special alternating link $L$. A
special alternating link $L$ admits a reduced, oriented, alternating diagram
$D$, with the property that every Seifert circle is innermost in itself, when
Seifert’s algorithm is applied on $D.$

In 1991, Thistlewait and Menasco [8] proved the flyping conjecture. _Given any
two reduced alternating diagrams $D_{1}$ and $D_{2}$ of an oriented, prime
alternating link $L$ then $D_{1}$ can be transformed to $D_{2}$ by applying a
sequence of flypes._

_Every Seifert surface on a non-split, prime, oriented, special alternating
link $L$ is isotopic to a Seifert surface obtained by applying Seifert’s
algorithm on a reduced, oriented, prime special alternating diagram $D$ of L._

Hirasawa-Sakuma [9], in 1990 announced the algorithm to find the Kakimizu
complex for a special alternating link. Jessica Banks [7] and Joshua Greene
[10] proved the result about the characterization of Seifert surfaces on a
special alternating knot independently.

To compute the Kakimizu complex of a link $L,$ it suffices to compute the $0$
and the $1$ skeleton of the complex. Since the Kakimizu complex is a flag
complex, the $1$ skeleton completely describes the Kakimizu complex of the
link $L$.

_Kakimizu complex of a non-split, oriented link is connected._ Proved
independently by Thompson and Scharlemann [11] in 1988 and Kakimizu [1] in
1990, It led to the conjecture that the Kakimizu complex of a non-split, prime
link $L$ is contractible. This has been shown to be true for different classes
of links with Jessica Banks [7] proving the result announced by Hirasawa-
Sakuma for special alternating links.This was finally proved by Przytycki and
Schultens [12] in 2010.

The Kakimizu complex of special alternating links is given explicitly by
_Jessica Banks_ in [7]. We explain the algorithm here briefly.

Since the Kakimizu complex of a link $L$ is connected, it suffices to find the
largest simplices containing a generic vertex $v$ of the Kakimizu complex of a
special alternating link $L$.

Let $L$ be a special alternating link. Let $v$ be a generic vertex of the
Kakimizu complex of $L$ and let $S$ be a representative surface (in it’s
isotopy class) of $v.$ Since $L$ is a special alternating link, $S$ is
isotopic to a surface which can be realized by applying Seifert’s algorithm on
a reduced, non-split, prime, oriented, alternating diagram $D.$

Any generic vertex $v$ of a non-split, special, alternating link $L$ is given
by prescribing a reduced, alternating, oriented diagram $D$ of the link $L$
and applying Seifert’s algorithm on $D$ to obtain a representative surface
$S.$

The algorithm to find the largest complex containing the vertex $v$ is
described as follows. Let $L$ be a prime, special alternating link $L.$ Given
$v,$ let $D$ be the reduced alternating diagram of $L$ such that applying
Seifert’s algorithm on $D$ yields a representative Seifert surface $S$ of $v.$

  * •

Consider the Seifert graph associated to the oriented diagram $D.$
$\theta$-graph is obtained from the Seifert graph by the following process:

    * –

Give a weight 1 to every edge in the Seifert graph.

    * –

Identify the pair of edges with same end vertices such that they bound a bigon
region in between them in $D.$ [We consider the diagram $D$ to be embedded in
$S^{2}$]. Replace each bigon region with one edge between the two vertex, with
the new edge weight being the sum of the two weights of the previous two
edges. Continue till we get rid of all bigon regions.

    * –

Add edges to a pair of vertices in the resulting graph with weight 0 only if
there is another edge between the two vertices and the addition of this new
edge doesn’t create a bigon region in the graph.

    * –

Consider the sub graph of the resulting graph that contains edges such that
the endpoint vertices bound multiple edges.

    * –

We call this weighted graph, the $\theta$-graph of $S.$

  * •

A $\theta$\- graph of a diagram $D$ of $L,$ embedded in $S^{2}$ divides
$S^{2}$ into regions. The boundary of a region $r$ has edges of the
$\theta$-graph. We can assign a signature to each boundary edge depending on
the orientation of the diagram $D.$

  * •

We fix an order of the edges of the $\theta$-graph and denote the surface $S$
as a tuple $(e_{1},e_{2},\dots e_{n})$ with $e_{i}$ being the weight on the
$i$th edge according to the weight.

  * •

To get the maximal simplices containing the generic vertex $v,$ we need to
apply all the regions of the $\theta$-graph.

  * •

Applying a region to the $\theta$-graph yields a non-isotopic Seifert surface
which is disjoint from $S.$ Applying a region amounts to applying all the
flypes associated to the weights and signature of the boundary edges of the
region.

  * •

Let $r$ be a region. Let $(w_{1},w_{2},\dots,w_{k})$ be the boundary edges of
the region with the signature of the edges(w.r.t the region $r$) be
$(s_{1},s_{2},\dots,s_{k});$ $s_{i}=\pm 1.$

  * •

After applying the region $r,$ we get a new surface, with edge weights being
$w_{i}\to w_{i}+s_{i}$ and for non-boundary edges $e_{i}\neq w_{j},$ $e_{i}\to
e_{i}.$

  * •

Apply all the regions (each region once). This implies we have applied every
flype twice hence we will get back the same surface we started with.

  * •

A cycle (with every region of the $\theta$-graph applied) constitutes a
largest simplex containing $v.$

Since the Kakimizu complex is connected, we can obtain the Kakimizu complex of
a link $L$ by choosing a generic vertex $v$, computing the largest simplices
for each vertex (there is finitely many vertices in the Kakimizu complex) and
we obtain the Kakimizu complex of the link $L$.

$K=11_{237}$ is a special alternating knot and the Kakimizu complex is
computed here to demonstrate the algorithm. $(0,1,0)$ denotes a generic vertex
in the Kakimizu complex. In this example, by symmetry if we compute the
largest simplices containing a vertex, we will obtain the same $2$-simplex.
The Kakimizu complex of $K$ is the $2$-simplex, containing the $3$ vertices
$(0,1,0);(0,0,1);(1,0,0).$

Figure 1: The Kakimizu complex of $11_{237}.$ The knot $11_{237}$ is a special
alternating knot.

The list of special alternating links for $11$ crossing alternating links
(with the Kakimizu complex) is as follows:

  * •

$K=11_{43},11_{123},11_{124},11_{200},11_{227},11_{240},11_{241},11_{244},11_{245},11_{263},11_{291},11_{292},$
$11_{298},11_{299},11_{318},11_{319},11_{320},11_{329},11_{338},11_{354},11_{361}$

are spanned by a unique minimal genus Seifert surface up to isotopy.

The Kakimizu complex of the knots are a single vertex.

$T$

  * •

$K=11_{94}.$ The $\theta$-graph contains 2 regions. The Kakimizu complex is

$T_{1}$$T_{2}$

  * •

$K=11_{237}.$ The $\theta$-graph contains $3$ regions.

$T_{3}$$T_{2}$$T_{1}$$T_{2}$$T_{1}$$T_{3}$

  * •

$K=11_{340}.$ The $\theta$-graph contains $2$ regions. The Kakimizu complex
is:

$T_{1}$$T_{2}$

###  4.3 Kakimizu Complex of 2 bridge knots

Let $L$ be a $2$ bridge knot. Reidemeister [13], Schubert and Seifert [14]
classified all $2$ bridge knots. Every $2$ bridge knot is associated to a
reduced rational number $p/q.$ Two $2$ bridge knots with the associated bridge
index $p/q$ and $p^{\prime}/q^{\prime}$ are equivalent if and only if
$q=q^{\prime}$ and $p=p^{\pm 1}(mod$ $q).$ For our purpose, given an index for
a 2-bridge knot,

$\text{ Given }0\leq p/q<1\text{ we consider }:\begin{cases}p/q-1,&\text{if
}p,q\text{ both odd }\\\ p/q,&\text{otherwise}\end{cases}$

They represent the same respective knot but the rational number $p/q$ has at
least one of the entry (p,q) is even.

_Every rational number $p/q$ with one even entry p or q, yields a continued
fraction expansion with all even integers._

Example:

Let $K=11_{13}$ is a 2-bridge knot with the index $28/61.$

The continued fraction expansion of $28/61$ is given by:

$\mathrm{28/61}=\cfrac{1}{2-\cfrac{1}{-6-\cfrac{1}{-2-\cfrac{1}{2}}}}$

Hence according to the continued fraction expansion we can denote

$28/61=[2,-6,-2,2].$

The knot could be isotoped to a 2-bridge diagram (with respect to the height
function) such that it is ordered as

The $MS(L)$ (the 0-skeleton) of the Kakimizu complex can be computed from the
results of Hatcher and Thurston [15] . They showed that every minimal genus
Seifert surface on a $2$ bridge link is given by the inner or outer plumbing
of full twisted $n$ bands, the numbers determined from the continued fraction
expansion of the 2-bridge index of the knot.

Example:

Any surface on the knot $K=11_{13}$ with bridge index $28/61$ is given by

$S=B_{2}\cup_{D_{1}}B_{6}\cup_{D_{2}}B_{2}\cup_{D_{3}}B_{2}$ with $B_{i}$
being bands with $i$ twists and $D_{i}$ are the plumbing disks. Every minimal
genus Seifert surface on $K$ is given by inner or outer plumbings on the 3
plumbing disks. In our example since $B_{2}$ is the Hopf band, it’s fibred,
and $B_{6}$ is the unique incompressible, non-fibred minimal genus Seifert
surface on the $3$ full twisted link, the link is not fibred and since
$IS(\partial B_{6})=[B_{6}],$ hence deplumbing each Hopf band mounts to
product decomposition, and invoking the bijection from Boileau and Gabai’s
result with $S=S_{1}\cup_{D}F$, where $\partial F$ is a fibred link with fibre
$F$, [$L_{i}=\partial S_{i}$]

$\phi:IS(L,S)\to IS(L_{1},S_{1}=B_{6}\cup_{D_{2}}B_{2}\cup_{D_{3}}B_{2})\to
IS(L_{2},S_{2}=B_{6}\cup_{D_{2}}B_{2})$

$\to IS(L_{3},S_{3}=B_{6})=\Phi.$

This implies $IS(L,S)=\phi$ so $S$ is the unique spanning surface on $L.$

Sakuma[] has computed the Kakimizu complex of special arborescent links,
generalizing the Kakimizu complex of $2$ bridge links. We are going to give
the algorithm to find the Kakimizu complex for $2$ bridge links.

The $0-$ skeleton of a $2$ bridge link has already been found by Hatcher and
Thurston. To find the Kakimizu complex of the link $L,$ we will find the
largest complex containing a generic surface $S.$

Let $L$ be a $2$ bridge link with a bridge index $p/q.$ Let us consider the
continued fraction of $p/q(p/q-1,\text{ if p,q both odd})$ such that each
entry is even. Let $p/q=[e_{1},e_{2},\dots,e_{k}],$ with each $e_{i}$ being
even.

To find $MS(L),$ the 0-skeleton of the Kakimizu complex of a $2-$ bridge link,
denote $0$ as the inner plumbing and $1$ as the outer plumbing.

Any minimal genus Seifert surface
$S=S_{1}\cup_{D_{1}}S_{2}\cup_{D_{2}}\dots\cup_{D_{k-1}}S_{k}$ with
$S_{i}=B_{e_{i}}$ are the $e_{i}$ twisted bands and $D_{i}$ being the plumbing
disk, either inner or outer with respect to a standard 2-bridge diagram.

$B_{2n}$ is a band on the $2n$ twisted link. $n=1$ represents the Hopf link
with $B_{2}$ being the Hopf band. The Hopf link is fibred with fibre $B_{2}.$
For $n\geq 2,$ $B_{2n}$ is a unique minimal genus Seifert surface on the $2n$
twisted link.

Each surface in $MS(L)$ can be represented by a $k-1$ tuple of $0$ or $1.$

For example, let $L=[4,6,4].$ Then we have 2 plumbing disk, hence each surface
is a 2 tuple. The surfaces are denoted by,

$(0,0);(0,1);(1,0);(1,1).$

In case there is a Hopf band at the end of the chain, then

$(0,\dots)=(1,\dots)(\text{ if Hopf band is attached at the start of the
chain})$

or

$(\dots,0)=(\dots,1)(\text{ if Hopf band is attached at the end of the
chain}).$

If the Hopf band is attached in the middle between the kth and $k+1$th
plumbing disk,

$(\dots,s_{k},s_{k+1},\dots)$

then

$(\dots,0,0,\dots)=(\dots,1,1,\dots).$

This relationship holds for every Hopf band in the chain.

The algorithm to find the maximal simplex containing a generic vertex $v$ is
given:

  * •

For a $2$-bridge link, let $p/q$ $(p/q-1$ for p,q both odd) be the 2-bridge
index. Let $L=[e_{1},e_{2},\dots,e_{n}]$ be the even continued fraction
expansion of $L.$ Let $v$ be a generic vertex of the Kakimizu complex and let
$S$ be the surface denoted by $v=(s_{1},s_{2},\dots,s_{n-1}).$

  * •

To find a maximal simplex, we need to start from $v$ and apply the _component
surfaces_ to find a maximal cycle. Each such maximal cycle represents the
maximal simplices containing $v.$

  * •

Let $L=[e_{1},e_{2},\dots,e_{n}],$ with $v=(s_{1},s_{2},\dots s_{n-1}).$ This
implies

$S=B_{e_{1}}\cup_{D_{1}}B_{e_{2}}\cup_{D_{2}}B_{e_{3}}\dots\cup_{D_{n-1}}B_{e_{n}}.$

along with the fact that $D_{i}$ is an inner plumbing if $s_{i}=0$ and $D_{i}$
is an outer plumbing if $s_{i}=1.$

  * •

Given a surface $S,$ denoted by $(s_{1},s_{2},\dots,s_{n-1})$, we can apply a
component surface $B_{e_{k}}$ $(1<k<n)$ only if $s_{k}=s_{k+1}.$ If the
condition is satisfied then applying $B_{e_{k}}$ on $S$ yields us $S^{\prime}$
denoted by

$(s_{1},\dots,s_{k-1},(s_{k}+1)(\text{ mod }2),(s_{k+1}+1)(\text{ mod
}2),s_{k+2},\dots s_{n}).$

  * •

The two end component surfaces could be applied at any instant. If we apply
$B_{e_{1}}$, on a surface $S,$ denoted by $(s_{1},s_{2},\dots,s_{n-1}),$ it
yields us $S^{\prime}$ denoted by

$(s_{1}+1(\text{ mod }2),s_{2},\dots,s_{n-1}).$

Similarly, if we apply $B_{e_{n}}$, on a surface $S,$ denoted by
$(s_{1},s_{2},\dots,s_{n-1}),$ it yields us $S^{\prime}$ denoted by

$(s_{1},s_{2},\dots,s_{n-2},(s_{n-1}+1)(\text{ mod }2)).$

  * •

Given a generic vertex $v,$ let $S$ be the surface denoted by
$(s_{1},s_{2},\dots,s_{n}).$

Consider a chain of surfaces, starting from $S$ where each component surface
$B_{e_{k}},$ $1\leq k\leq n$ is applied exactly once (in some order). Note
that the end of the chain should exactly be $S.$ This is because any entry
$s_{i},$ is affected when $B_{e_{i}}$ and $B_{e_{i+1}}$ were applied. For any
other $B_{e_{k}},$ $s_{i}$ remains unaffected. Applying twice, we get

$\bigg{(}(s_{i}+1)(\text{ mod }2)+1\bigg{)}(\text{ mod }2)=s_{i}.$

Hence we get a maximal cycle of surfaces starting from $S,$ containing $n$
distinct surfaces. The $n$-vertices in the Kakimizu complex span a maximal
simplex containing $v.$

  * •

The Kakimizu complex of a non-split link is connected. Given a generic vertex
$v,$ we are able to find the maximal simplices containing $v.$ Since there are
finitely many vertex in the $0$-skeleton in the Kakimizu complex, applying
this on every vertex yields us the Kakimizu complex.

An example to illustrate the algorithm for $2$-bridge knots

Figure 2: The Kakimizu complex of $11_{192}.$ The knot $11_{192}$ is a
2-bridge knot.

The list of Kakimizu complex of $2$-bridge knots of $11$ crossings are as
follows:

  * •

$K=11_{13},11_{59},11_{65},11_{75},11_{77},11_{84},11_{85},11_{89},11_{90},11_{91},11_{93},11_{110},11_{111},$
$11_{117},11_{120},11_{140},11_{144},11_{178},11_{183},11_{185},11_{188},11_{190},11_{193},11_{195},11_{204},11_{205},$
$11_{207},11_{208},11_{211},11_{220},11_{224},11_{225},11_{230},11_{234},11_{242},11_{246},11_{247},11_{307},11_{309},$
$11_{334},11_{339},11_{342},11_{355},11_{358},11_{364}.$

are spanned by a unique minimal genus Seifert surface up to isotopy.

The Kakimizu complex of the knots are a single vertex.

$T$

  * •

The list of $2$-bridge knots with non-trivial Kakimizu complex.

Knots | 2-bridge index | Even continued fraction expansion | Kakimizu complex
---|---|---|---
$11_{95}$ | $33/73=-40/73$ | $[-2,-6,-4,-2]$ | $T_{1}$$T_{2}$
$11_{98}$ | $18/47$ | $[4,-4,-2,2]$ | $T_{1}$$T_{2}$
$11_{119}$ | $64/109$ |  $[2,-4,-4,2]$, | $T_{1}$$T_{2}$
$11_{145}$ | $22/83$ |  $[4,4,-2,2]$, | $T_{1}$$T_{2}$
$11_{154}$ | $37/67=-30/67$ | $[-2,4,-4,-2]$ | $T_{1}$$T_{2}$
$11_{166}$ | $45/59=-14/59$ | $[-4,4,-2,-2]$ | $T_{1}$$T_{2}$
$11_{186}$ | $39/95=-56/95$ | $[-2,-4,-2,-2,-4,-2]$ | $T_{1}$$T_{2}$$T_{2}$$T_{3}$$T_{3}$$T_{4}$
$11_{191}$ | $19/83=-64/83$ |  $[-2,-2,-2,-4,-4,-2]$, | $T_{1}$$T_{2}$
$11_{192}$ | $45/59=-14/59$ | $[-4,4,-2,-2]$ | $T_{3}$$T_{2}$$T_{1}$$T_{2}$$T_{1}$$T_{3}$$T_{3}$$T_{2}$$T_{4}$$T_{2}$$T_{4}$$T_{3}$
Knots | 2-bridge index | Continued fraction expansion | Kakimizu complex
---|---|---|---
$11_{210}$ | $16/73$ |  $[4,-2,-4,-2]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$
$11_{226}$ | $20/71$ |  $[4,2,-4,2]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$
$11_{229}$ | $55/71=-16/71$ |  $[-4,2,-4,-2]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$
$11_{235}$ | $49/71=-22/71$ |  $[-4,-2,-2,-2,-4,-2]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$$T_{3}$$T_{4}$$T_{4}$$T_{5}$
$11_{236}$ | $29/99=-70/99$ |  $[-2,-2,-4,-2,-4,-2]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$
$11_{238}$ | $53/65=-12/65$ |  $[-6,-2,-4,-2]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$
$11_{243}$ | $49/69=-20/69$ |  $[-4,-2,-6,-2]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$
$11_{311}$ | $61/79=-18/79$ |  $[-4,2,-2,-4]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$$T_{3}$$T_{4}$
$11_{333}$ | $14/65$ |  $[4,-2,-2,4]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$$T_{3}$$T_{4}$
$11_{335}$ | $17/75=-58/75$ |  $[-2,-2,-2,-4,-2,-4]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$
$11_{336}$ | $11/59=-48/59$ |  $[-2,-2,-2,-2,-4,-4]$, | $T_{1}$$T_{2}$
$11_{337}$ | $63/89=-16/89$ |  $[-6,-2,4,2]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$
$11_{343}$ | $27/31=-4/31$ |  $[-8,-4]$, | $T_{1}$$T_{2}$
$11_{356}$ | $55/79=-24/79$ |  $[-4,-2,-2,-4,-2,-2]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$$T_{3}$$T_{4}$
$11_{357}$ | $27/91=-64/91$ |  $[-2,-2,-4,-4,-2,-2]$, | $T_{1}$$T_{2}$
$11_{359}$ | $43/53=-10/79$ |  $[-6,-2,-2,-4]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$$T_{3}$$T_{4}$
$11_{360}$ | $47/57=-10/59$ |  $[-6,-4,-2,-2]$, | $T_{1}$$T_{2}$
$11_{363}$ | $29/35=-6/35$ |  $[-6,-6]$, | $T_{1}$$T_{2}$
$11_{365}$ | $35/51=-16/51$ |  $[-4,-2,-2,-2,-2,-4]$, | $T_{1}$$T_{2}$$T_{2}$$T_{3}$$T_{3}$$T_{4}$$T_{4}$$T_{5}$$T_{5}$$T_{6}$

###  4.4 General Algorithm

Let $D$ be a reduced oriented alternating diagram of an 11 crossing prime,
alternating knot $K.$

  * •

Firstly we identify (from the table [4]) if the given $11$ crossing
alternating link is a $2$-bridge knot. We compute the Kakimizu complex using
Hatcher and Thurston’s [15] result to find $MS(L).$ We use the generalised
result to find the $1$-skeleton of the Kakimizu complex of $2$-bridge knots.

  * •

If it is not a $2$-bridge knot, we can apply the Seifert’s algorithm on $D.$
We get a minimal genus Seifert surface $S$ with $\partial S=K.$

We know that the Kakimizu complex is a flag complex. Hence finding the graph
containing the $0$ and $1$ skeleton of the Kakimizu complex suffices to find
the whole complex.

Secondly, the Kakimizu complex is connected, implies that starting from a
minimal genus Seifert surface $S$ on the knot $K,$ we need to obtain all
vertices adjacent to $S$ and repeat the process for every new vertex found.

Let L be a non-split prime alternating link with $n>0$ crossings. It has been
shown by Hass,Thompson and Tsvietkova that for each fixed $g$ the number of
genus $g-$ Seifert surfaces for $L$ is bounded by an explicitly given
polynomial in $n.$

Hence we know that the process in the previous part ends, in other words the
Kakimizu complex for a link $L$ is finite.

  * •

Any alternating link $K$ with a diagram $D$ can be written as a $*-$ product
of special alternating links $K=*(L_{i})_{i}$. When we apply Seifert’s
algorithm on an oriented alternating reduced diagram $D$ of $K$ then we get
the surface $S$ as a Murasugi sums of spanning surfaces $S_{i}$ on special
alternating links $L_{i}$.

  * •

If $K$ is a special alternating link, then we use Jessica Banks’ result to
find the Kakimizu complex of $K.$

  * •

Identify all the fibred pieces (if any) plumbed on a surface $S_{1}$ to obtain
$S.$

  * •

If $S_{1}$ is non-fibred and has a unique Seifert surface, and is Murasugi
summed with fibred surfaces, then the Kakimizu complex of $K$ is a single
vertex.

Let $L$ be a non-split, prime link and let $S$ be a minimal genus Seifert
surface on $L.$ $IS(L,S)$ is the set of all isotopy classes of surfaces which
can be made disjoint from $S$ in the link complement. (With respect to
Kakimizu complex, surfaces(upto isotopy) in $IS(L,S)$ represents vertices in
the Kakimizu complex which share an edge with $v_{[S]}$).

Boileau and Gabai showed that,

_Let $L$ be a non-split oriented link and $S$ a connected minimal genus
Seifert surface for $L$. Suppose that $S$ is a Murasugi sum of $S_{1}$ and
$S_{2}$, where $S_{i}$ is a spanning surface for an oriented link $L_{i}$
$(i=1,2)$. Suppose further that $L_{2}$ is a fibred link with fibre $S_{2}$.
Then $L_{1}$ is non-split, and $S_{1}$ is connected and minimal genus.
Moreover there is a bijection_

$\phi:IS(L,S)\to IS(L_{1},S_{1}).$

Let

$S=S_{1}\cup_{D_{1}}F_{1}\cup_{D_{2}}F_{2}\cup\dots\cup_{D_{n}}F_{n}$

where $F_{i}$ are fibred surfaces on $\partial F_{i}$ and each
$S_{1}\cup_{D_{i}}F_{i}$ is a plumbing with $D_{i}$ being the plumbing disks
in the surface $S_{i}.$ Note that all the plumbing disks $D_{i}$ are disjoint.
Let $\partial S_{1}=L_{1}$. Then $L_{1}$ is non-split and $S_{1}$ is
connected.

Assume

$IS(L_{1})=[S_{1}].$

Then

$IS(L)=[S].$

###### Proof.

$IS(L_{1},S_{1})=\Phi.$

Consider

$S_{1}^{1}=S_{1}\cup_{D_{1}}F_{1}.$

We know that $F_{1}$ is a fibred surface for $L_{1}$. From the theorem by
Boileau and Gabai, we have

$IS(\partial S_{1}^{1},S_{1}^{1})=\Phi.$

Define

$S_{1}^{k}=S_{1}\cup_{D_{1}}F_{1}\cup_{D_{2}}F_{2}\cup\dots\cup_{D_{k}}F_{k}$

for every $1\leq k\leq n$

Repeating the same process as above we have,

$IS(\partial S_{1}^{k},S_{1}^{k})=\Phi.$

For $k=n$ we have

$IS(L,S)=\Phi\implies IS(L)=[S].$

∎

The list of the Kakimizu complexes of $11$ crossing alternating knots $K$,
that are Murasugi sums of links $L_{1}$ with a unique incompressible surface
and $L_{2}$ that are fibred:

  * •

The following knots $K$ spans an unique minimal genus Seifert surface up to
isotopy.

$K=11_{1},11_{2},11_{4},11_{6},11_{8},11_{10},11_{11},11_{12},11_{16},11_{18},11_{20},11_{21},11_{23},11_{27},11_{29},11_{30},11_{31},$
$11_{32},11_{36},11_{37},11_{38},11_{39},11_{41},11_{46},11_{48},11_{49},11_{50},11_{52},11_{54},11_{56},11_{58},11_{60},11_{63},11_{64},$
$11_{67},11_{69},11_{70},11_{78},11_{87},11_{97},11_{100},11_{101},11_{102},11_{104},11_{105},11_{107},11_{114},11_{115},11_{116},$
$11_{118},11_{122},11_{130},11_{132},11_{133},11_{134},11_{135},11_{136},11_{137},11_{138},11_{141},11_{143},11_{148},11_{149},11_{150},$
$11_{152},11_{153},11_{155},11_{161},11_{165},11_{167},11_{168},11_{169},11_{172},11_{173},11_{181},11_{187},11_{197},11_{198},11_{199},$
$11_{202},11_{212},11_{213},11_{214},11_{219},11_{222},11_{249},11_{256},11_{258},11_{260},11_{262},11_{265},11_{270},11_{271},11_{272},$
$11_{273},11_{275},11_{276},11_{278},11_{279},11_{283},11_{285},11_{290},11_{294},11_{295},11_{296},11_{297},11_{303},11_{304},11_{312},$
$11_{313},11_{317},11_{321},11_{322},11_{323},11_{324},11_{327},11_{328},11_{331},11_{344},11_{345},11_{347},11_{349},11_{352}.$

The Kakimizu complex of the knot $K$ is:

$T$

  * •

Let $K$ be an $11$ crossing alternating knot. Let $D$ be a reduced, prime,
alternating, oriented diagram of $L.$ Let $S$ be a Seifert surface obtained by
applying Seifert’s algorithm on $D.$

Let $S=S_{1}\cup_{D_{0}}S_{2}$ such that $S_{2}$ be a fibred surface on a
fibred link $L_{2}.$ We have,

$IS(L,S)\cong IS(L_{1},S_{1}).$ Let us call this bijective map

$f:IS(L,S)\to IS(L_{1},S_{1}).$

Assume that for every $T\in IS(L_{1},S_{1}),$ $f^{-1}(T)=T\cup_{E}S_{2}.$
Assume that this condition holds for every representative surface of a generic
vertex $S$ on $L.$ Then $IS(L)\cong IS(L_{1})$ and the Kakimizu complexes are
identical.

###### Proof.

Let $D$ be a reduced, alternating, oriented diagram of the knot $K.$ Let $S$
be the Seifert surface obtained by applying Seifert’s algorithm on $D.$

Let $v$ be a generic vertex in $MS(L)=IS(L).$ Since the Kakimizu complex of a
prime, non-split link is connected, there exists a path,
$[[S],v_{1},v_{2},\dots,v_{k}=v]$ in the Kakimizu complex.

Since $S=S_{1}\cup_{D_{0}}S_{2}$ with $S_{2}$ fibred, we have

$f_{1}:IS(L,S)\to IS(L_{1},S_{1});\text{ with }S=S_{1}\cup S_{2}.$

For every $v\in IS(L_{1},S_{1})$, there exists a representative surface $T$
such that $v=[T]$ and $f_{1}^{-1}([T])=[T\cup S_{2}].$

$v_{1}\in IS(L,S),$ $f_{1}(v_{1})\in IS(L_{1},S_{1}).$ There exists a
representative surface $T_{1}$ on $L_{1}$ such that

$f_{1}(v_{1})=[T_{1}]\text{ and }v_{1}=[T_{1}\cup S_{2}].$

Define a map $f:IS(L)\to IS(L_{1})$ with $f([S])=[S_{1}],$ and
$f(v_{1})=[T_{1}].$

$f_{2}:IS(L,T_{1}\cup S_{2})\to IS(L_{1},T_{1}).$

$v_{2}\in IS(L,T_{1}\cup S_{2}),$ $f_{2}(v_{2})\in IS(L_{1},S_{1}).$ There
exists a representative surface $T_{2}$ on $L_{1}$ such that

$f_{2}(v_{2})=[T_{2}]\text{ and }v_{2}=[T_{2}\cup S_{2}].$

$f(v_{2})=[T_{2}].$

Continuing the process, we obtain $f(v).$

Hence we define a one-one correspondence between $IS(L)$ and $IS(L_{1})$ and
so the Kakimizu complexes of $L$ are identical to $L_{1}.$

∎

The list of the Kakimizu complexes of $11$ crossing alternating knots $K$,
that are Murasugi sums of links $L_{1}$ with minimal genus Seifert surfaces
and $L_{2}$ that is fibred with fibre $S_{2}$ with the condition that every
surface on $K$ is a Murasugi sum of a surface on $L_{1}$ and $S_{2}.$

$K=11_{61}.$

$IS(K)=\\{[S_{1}],[S_{2}]\\}$ and the Kakimizu complex is: $[S]$$[S^{c}]$

### Kakimizu complex of plumbings of links with unique spanning surfaces

The following section is adopted from [1] and proved by Kakimizu, in 1992.

Let $(M,\gamma)$ be a sutured manifold. A marking $A$ is a properly embedded
arc in $R(\gamma).$ A sutured manifold with a prescribed marking
$(M,\gamma,A)$ is called a _marked sutured manifold_. If there is a product
disk $D$ in $(M,\gamma)$ with $A$ as an edge of $\partial D,$ then the sutured
manifold $(M,\gamma)$ with the opposite edge $B$ is also a marked sutured
manifold $(M,\gamma,B).$

_Let $(M,\gamma,A)$ be a marked sutured manifold. Suppose that $M$ is
irreducible and each component of $R(\gamma)$ is incompressible . If there is
a product disk with $A$ as an edge, then the ambient isotopy types of product
disks are unique._

For our purpose, if $L$ is non-split then $E(L)$ and complementary sutured
manifold, $(M,\gamma)$ for a Seifert surface $S$ are irreducible 3-manifolds.
If $S$ is incompressible then $R(\gamma)$ for the complementary sutured
manifold $(M,\gamma)$ is incompressible. Let $S$ be a plumbing of $S_{1}$ and
$S_{2}.$ Consider the complementary sutured manifolds $(M_{1},\gamma_{1})$ and
$(M_{2},\gamma_{2})$ for $S_{1}$ and $S_{2}$ respectively. Let $D_{0}$ be the
plumbing disk. $D_{0}\subset S_{1}$ is an embedded disk. Consider the core
curve of the disk $D_{0}$ in $S_{1}.$ Since $M=E(L)-N(S),$ push out the core
curve on the side where $S_{2}$ is attached. The push off of the core curve is
denoted by the marking $A_{1}\subset R(\gamma_{1}).$ Hence we get a marked
sutured manifold $(M_{1},\gamma_{1},A_{1}).$ Similarly we obtain a marked
sutured manifold for $S_{2}$ denoted by $(M_{2},\gamma_{2},A_{2}).$

######  Theorem.

Let $L$ be a non-split, prime, alternating link and let $D$ be a reduced
alternating, oriented diagram $D$ of $L.$ Let $S$ be the Seifert surface
obtained by applying Seifert’s algorithm on the diagram $D.$ Let $S$ be a
plumbing of $S_{1}$ and $S_{2}$ which are unique minimal genus Seifert
surfaces for links $L_{1}$ amd $L_{2}$ respectively. Assume that $S_{1}$ and
$S_{2}$ are not fibred. Let $(M_{i},\gamma_{i},A_{i})$ and
$(M_{i},\gamma_{i},A_{i}^{\prime})$ $(i=1,2)$ be the marked sutured manifolds
for $S=S_{1}\cup_{D_{0}}S_{2}$ and $S^{c}=S_{1}\cup_{D_{0}^{c}}S_{2},$ the
dual of $S$ respectively. Then

  * •

$MS(L)=\\{[S],[S^{c}]\\}$ and the Kakimizu complex is

$[S]$$[S^{c}]$

provided it satisfies the condition:

    * –

there is no product disk with $A_{1}$ or $A_{1}^{\prime}$ in $M_{1}.$

    * –

there is no product disk with $A_{2}$ or $A_{2}^{\prime}$ in $M_{2}.$

  * •

Let us assume that there is a product disk in $(M_{1},\gamma_{1},A_{1})$ with
$A_{1}$ as an edge. Let $B_{1}$ be the opposite mark of the product disk. Let
$S=S_{1}\cup_{D_{0}}S_{2}.$ Suppose $T=S_{1}\cup_{E_{0}}T_{2}$ is the plumbing
with respect to the marking $B_{1}.$ Note that $S=T.$

$MS(L)=\\{[S],[S^{c}],[T^{c}]\\}$ and the Kakimizu complex is

$[S^{c}]$$[S]$$[S]$$[T^{c}]$

provided it satisfies the condition:

    * –

there is no product disk with $A_{1}^{\prime}$ or $B_{1}^{\prime}$ in $M_{1}$

    * –

there is no product disk with $A_{2}$ or $A_{2}^{\prime}$ in $M_{2}.$

The list of the Kakimizu complexes of $11$ crossing alternating knots $K$,
that spans a surface $S$ that are plumbings of links with unique
incompressible surfaces :

  * •

$K=11_{45}.$ Let $S=S_{1}\cup_{D_{0}}S_{2}$ on links $L_{1}$ and $L_{2}.$

The link $L_{1}$ is a special alternating link with a unique minimal genus
Seifert surface $S_{1}$. $S_{2}$ is the $4$-half-twisted band, a unique
minimal genus Seifert surface. There are no product disks with the markings as
an edge. Hence the Kakimizu complex of $K$ is:

$[S]$$[S^{c}]$

  * •

$K=11_{280}.$

$S=S_{1}\cup_{D_{0}}S_{2}$ with $L_{1}=(2,2,2)$ and $L_{2}=(3,1,1)$ Pretzel
knots. Both $S_{1},S_{2}$ are unique minimal genus spanning surfaces for
$L_{1},L_{2}$ respectively. They Kakimizu complex is:

$[S]$$[S^{c}]$

  * •

$K=11_{325}.$ $S=S_{1}\cup_{D_{0}}S_{2}$ with $S_{1}$, a 4-half twisted band
and $S_{2}$ is a unique spanning surface on a special alternating link
$L_{2}.$ There is a product disk with the marking as an edge. The Kakimizu
complex is:

$T_{1}$$T_{2}$$T_{2}$$T_{3}$

Example:

Figure 3: Knot $K$ = $11_{2}$ with a Seifert Surface on $K$ Figure 4: Deplumb
the Hopf band to obtain a special alternating link $L_{1}$. There is a unique
Seifert surface on $L_{1}.$

## Special case: $11_{103}$

Let $D$ be a reduced oriented alternating diagram of an $11$ crossing prime
alternating knot $K.$ We consider $K=11_{103}.$

  * •

Apply Seifert’s algorithm on the diagram $D$ of $K.$ We obtain a surface
$T_{1}.$

$T_{1}=S_{1}(7_{4})\cup_{D_{1}}H_{1}\cup_{D_{2}}H_{2}$

where $S_{1}(7_{4}),S_{2}(7_{4})$ are the two distinct surfaces (up to
isotopy) as shown in the diagram and $H_{i}$ are Hopf bands plumbed onto
$S_{1}(7_{4}).$

Boileau and Gabai [16] showed that,

_Let $L$ be a non-split oriented link and $S$ a connected minimal genus
Seifert surface for $L$. Suppose that $S$ is a Murasugi sum of $S_{1}$ and
$S_{2}$, where $S_{i}$ is a spanning surface for an oriented link $L_{i}$
$(i=1,2)$. Suppose further that $L_{2}$ is a fibred link with fibre $S_{2}$.
Then $L_{1}$ is non-split, and $S_{1}$ is connected and minimal genus.
Moreover there is a bijection_

$\phi:IS(L,S)\to IS(L_{1},S_{1})$

Hence we have:

$IS(11_{103},T_{1})\xrightarrow{\phi_{1}}IS(L_{0},S_{1}(7_{4})\cup_{D_{1}}H_{1})\xrightarrow{\phi_{2}}IS(7_{4},S_{1}(7_{4}))=[S_{2}(7_{4})].$

where $\phi_{1}$ and $\phi_{2}$ are isomorphisms. Hence we call

$IS(11_{103},T_{1})=[T_{2}]=\phi_{1}^{-1}\circ\phi_{2}^{-1}([S_{2}(7_{4})]).$

This gives us the following Lemma.

######  Lemma.

$IS(11_{103},T_{1})=[T_{2}]$

_where $T_{2}$ is a non-plumbed disjoint surface in $E(L),$ not isotopic to
$T_{1}.$ _

$T_{1}=S_{1}(7_{4})\cup_{D_{1}}H_{1}\cup_{D_{2}}H_{2}.$

Since $H_{1}$ and $H_{2}$ are Hopf Bands, hence there are product disks
passing through the Hopf bands.

Product decomposition on $H_{1}$ and $H_{2}$ leads us to the complementary
sutured manifold of $S_{1}(7_{4}).$ $IS(7_{4},S_{1}(7_{4}))=[S_{2}(7_{4})].$
Inverting the operation of product decomposition on $H_{1}$ and $H_{2}$ yields
us $T_{2}$ in the complementary sutured manifold of $T_{1}.$

Figure 5:  Figure 6:  Figure 7:

Figure 8:  Figure 9:

## Finding $IS(L,T_{2})$

######  Theorem.

Let $L=11_{103},$ and let $D$ be an oriented reduced alternating diagram of
$L.$ Let $T_{1}$ be the surface obtained by applying Seifert’s algorithm to
$D.$ Let $IS(L,T_{1})=T_{2}.$

Then

$IS(L,T_{2})=T_{1}.$

From the previous Lemma,

$IS(L,T_{1})=T_{2}$

was obtained by applying 2 product decomposition to the surface $T_{1}.$

Let $H_{1}$ and $H_{2}$ be the Hopf bands plumbed to the surface
$S_{1}(7_{4}).$ The surfaces $T_{1}$ and $T_{2}$ are parallel along $H_{1}.$
[picture]

$L=7_{4}\cup_{D_{1}}\partial(H_{1})\cup_{D_{2}}\partial(H_{2}),$

Consider the boundary of the surface

$\tilde{L}=\partial(S_{1}(7_{4})\cup_{D_{2}}(H_{2}))$

From the previous lemma, we have:

$IS(L,T_{1})=[T_{2}].$

If we deplumb $H_{1}$ from $T_{i}$, we denote the new surfaces (isotopy class)
on $\tilde{L}$ as $S_{i}(\tilde{L}).$

$\phi_{1}:IS(L,T_{1})\to
IS(\tilde{L},(S_{1}(7_{4}))\cup_{D_{2}}H_{2})=IS(\tilde{L},(S_{1}(\widetilde{L}))$

and the map $\phi_{1}$ is a bijection.

$[S_{2}(\tilde{L})]:=\phi_{1}([T_{2}]).$

and let $S_{2}(\tilde{L})$ be a surface in the isotopy class which is disjoint
from $S_{1}(\tilde{L}).$

We will use the Lemma for

$\tilde{\phi}:IS(L,T_{2})\to IS(\tilde{L},S_{2}(\tilde{L}))$

is a bijection as well since we obtain $S_{2}(\widetilde{L})$ by deplumbing
the Hopf band $H_{1}$ (equivalent to product decomposition on $H_{1}$)

_We need to prove:_

$IS(\tilde{L},S_{2}(\tilde{L}))=S_{1}(\tilde{L})=S_{1}(7_{4})\cup_{D_{1}}H_{2}.$

###### Proof.

Let $[F]\in IS(\tilde{L},S_{2}(\tilde{L}))$ and $[F]\neq[S_{1}(\tilde{L})].$

Note that if $F\cap S_{1}(\tilde{L})=\phi$ this implies either

  * •

$[F]=[S_{1}(\tilde{L})]$ (which contradicts the assumption on
$[F]\neq[S_{1}(\tilde{L})]$); or

  * •

$[F]\in IS(\tilde{L},S_{1}(\tilde{L}))=[S_{2}(\tilde{L})].$ (which contradicts
the assumption that $[F]\in IS(\tilde{L},S_{2}(\tilde{L}))$)

Hence $F$ intersects $S_{1}(\tilde{L})$ essentially

Figure 10:

Consider the two balls $V$ and $W$ such that

$V\cup W=S^{3},$

with

$V\cap(S_{2}(\tilde{L})\cup D)=H_{1}\text{ and }W\cap(S_{2}(\tilde{L})\cup
D)=S_{2}(7_{4}),$

where $D$ is 4-gon disk, boundary of which is a rectangle with two of the
opposite sides are $\partial V\cap S_{2}(\tilde{L}),$ and
$S_{2}(\widetilde{L})$ is an embedded surface in $E(\widetilde{L}).$

[In the following pictures we assume the ball at the top is $V$ and the one at
the bottom is $W.$]

Let $F$ be as above. $F\cap\partial V$ consists of two arcs since $F$ is a
minimal genus Seifert surface and we assume that $F$ intersects $\partial V$
transversely. There are three cases as shown:

Figure 11:

We want to show that if $\tilde{L}$ is the boundary of the surface
$S_{1}(7_{4})\cup_{D}H_{2},$ then
$IS(\tilde{L},S_{2}(\tilde{L}))=S_{1}(7_{4})\cup H_{2}$

#### Case 1:

Consider $(M,\gamma)$ to be the complementary sutured manifold of
$S_{2}(\widetilde{L}).$ Let $(M_{1},\gamma_{1})$ (resp. $(M_{2},\gamma_{2})$)
be the complimentary sutured manifold for $(H_{2},\partial H_{2})$ (resp
$(S_{2}(7_{4}),7_{4})$). ($F,\partial F)\subset(M,\gamma)$ is a sutured
surface in the complementary sutured manifold of $S_{2}(\widetilde{L}).$ Let
$D_{1}$ and $D_{1}^{c}$ be the 4-gon’s whose boundary is $((F\cap
S^{2})\cup(S_{2}(\tilde{L})\cap S^{2}))\cap(M,\gamma).$ [Let $D_{1}$ be the
disk in the interior side of the diagram].

$(\widetilde{M_{1}},\widetilde{\gamma_{1}})=(M\cap V_{1}-N(D_{1}),(\gamma\cap
V_{1})\cup N(F\cap S^{2}))\cong(M_{1},\gamma_{1}).$ $F_{1}=F\cap V_{1}\text{
is a }\widetilde{\gamma_{1}}\text{ surface in }\widetilde{M_{1}}\text{ along
with }\partial F_{1}\cong\partial H_{2}.$

Since $H_{2}$ is a Hopf band, we know that $\partial H_{2}$ spans a unique
Seifert surface and $H_{2}$ is the fibred surface.

We isotope $F_{1}$ so that in $V_{1}$ it’s parallel to $H_{1}.$ Let
$e_{1}:[0,1]\times F_{1}\to\widetilde{M_{1}}$ be the isotopy with
$e_{1}(0)=Id_{F_{1}}$ and $e_{1}(1)(F_{1})=R_{(+)}(\widetilde{\gamma_{1}}),$ a
parallel copy of $H_{2}.$ Since $H_{2}$ is fibred , we have an isotopy
$e_{2}:[0,1]\times F_{1}\to\widetilde{M_{1}}$ with $e_{1}(0)=Id_{F_{1}}$ and
$e_{1}(1)(F_{1})=R_{(-)}(\widetilde{\gamma_{1}}),$ a parallel copy of $H_{2}.$

Note that $(\widetilde{M_{1}},\widetilde{\gamma_{1}})\cong(M_{1},\gamma_{1}).$
$H_{2}$ is a unique fibred surface on $\partial H_{2},$ and $F_{1}$ is a
sutured surface on $(\widetilde{M_{1}}).$ This implies that $F_{1}$ is
isotopic to $R_{+}(\widetilde{\gamma_{1}})$ or
$R_{-}(\widetilde{\gamma_{1}}).$

Without loss of generality, let us assume that $F_{1}$ is isotopic to
$R_{+}(\widetilde{\gamma_{1}}).$ Isotope the surface $F$ to $\widetilde{F}$
(consider an ambient sutured isotopy $e:I\times F\to M$ such that
$e(0)=\text{Identity }$ and $e(1)(F)=\widetilde{F}$ such that
$e(1)(F_{1})=R_{+}(\gamma_{1})$).

Hence we have $\widetilde{F}_{1}=R_{+}(\widetilde{\gamma_{1}})$

Consider $I\times D_{1}\subset\widetilde{M_{1}}$ with $\\{0\\}\times
D_{1}=D_{1}\subset\partial V$ and the vertices of $\partial(\\{t\\}\times
D_{1})\in\widetilde{\gamma_{1}}$

$D_{1}^{\prime}=\\{1\\}\times D_{1}\in\widetilde{M_{1}}$ is an isotopic copy
of $D$ in $\widetilde{M_{1}}.$

(If $F_{1}$ is isotopic to $R_{-}(\widetilde{\gamma_{1}}),$ then isotope the
surface $F$ to $\widetilde{F}$ via an ambient sutured isotopy $e:I\times F\to
M$ such that $e_{0}:F\to F\text{ is }\text{ the identity map }$ and
$e_{1}(F)=\widetilde{F}$ with $e_{1}(F_{1})=R_{-}(\gamma_{1})$).

We have $\widetilde{F}_{1}=R_{-}(\widetilde{\gamma_{1}})$

Consider $I\times(D_{1}^{c}=\partial V-D_{1}^{\circ})\subset\widetilde{M_{1}}$
with $\\{0\\}\times D_{1}^{c}=D_{1}^{c}\subset\partial V$ and the vertices of
$\partial(\\{t\\}\times D_{1}^{c})\in\widetilde{\gamma_{1}}$)

Let $V_{1}=V-{(\partial V\times[0,1))}$ and $W_{1}=W\cup{(\partial
V\times[0,1])}$ with

$V_{1}\cap W_{1}=\partial V_{1}=\partial
V\times\\{1\\}.[D_{1}\times\\{t\\}\subset\partial V\times\\{t\\}.]$

Since $\widetilde{F}\cap V=\widetilde{F_{1}}$ is parallel to $H_{2},$
$\widetilde{F}$ intersects the sphere $\partial V_{1}$ in two arcs
perpendicular to $\widetilde{F_{1}}\cap V.$

We have $[F]=[\widetilde{F}]\in IS(\tilde{L},S_{2}(\tilde{L}))$which means
$\widetilde{F}$ be a surface disjoint from $S_{2}(\tilde{L}).$ and we know
from the previous assumptions that $[\widetilde{F}]\neq[S_{1}(\tilde{L})].$

Consider the two balls $V_{1}$ and $W_{1}.$

$V_{1}\cap S_{2}(\tilde{L})\cup D^{\prime}=H_{1}\text{ and }W_{1}\cap
S_{2}(\tilde{L})\cup D^{\prime}=S_{2}(7_{4}),$

where $D=\\{1\\}\times D$ is the 4-gon disk, boundary of which is a rectangle
with two of the opposite sides are $\partial V_{1}\cap S_{2}(\tilde{L}).$

Since $\widetilde{F}$ intersects $D^{\prime}$( or $D^{\prime c}$) in
perpendicular arcs to $\widetilde{F}\cap D$( or $\widetilde{F}\cap D^{c}$) ,
Case 1 reduces to Case 2( or Case 3).

#### Case 2 (Similarly for Case 3):

$F\subset(M,\gamma)$ is a sutured surface in the complementary sutured
manifold of $T_{2}.$ Case 2 implies that $F\cap S^{2}$ is a pair of arcs
parallel to a copy of $T^{2}\cap S^{2}$ in $R_{+}(\gamma).$ [Case 3 indicates
the same in $R_{-}(\gamma).$]

$S^{2}$ separates $S^{3}$ in two balls $V_{1}$ and $V_{2}$ respectively
[$V_{1}$ denotes the ball at the top in the diagram.]

Filling $V_{1}$ to a point, we isotope $S^{3}$ to $V_{2},$ adjoining a point,
a new $S^{3}.$ We have the link $L=7_{4}$ and we have a Seifert surface
$F_{2}=F\cap V_{2}$ with boundary link $L=7_{4}.$ Let $r_{1}$ and $r_{2}$ be 2
arcs in $S^{2}$ perpendicular to the arcs $F_{2}\cap S^{2}.$ Let the 4-gon in
$S^{2}$ bounded by $r_{1},r_{2},F\cap S^{2}$ be $D_{2}.$

(For Case 3: the 4-gon is bounded by the same arcs. $D_{2}^{c}$ satisfies the
same condition, in the diagram it is the complement (in $S^{2}$) of the 4-gon
in Case 2).

$(\widetilde{M_{2}},\widetilde{\gamma_{2}})=(M\cap V_{2}-N(D_{2}),(\gamma\cap
V_{2})\cup N(F\cap S^{2})).$

Let $(M_{1},\gamma_{1})$ (resp. $(M_{2},\gamma_{2})$) be the complementary
sutured manifold for $(H_{2},\partial H_{2})$ (resp $(S_{2}(7_{4}),7_{4})$).
As mentioned before $F_{2}=F\cap V_{2}$ is a $\widetilde{\gamma_{2}}$ surface
in $\widetilde{M_{2}}$ along with $\partial F_{2}=7_{4}.$

Note that:

$(\widetilde{M_{2}},\widetilde{\gamma_{2}})\cong(M_{2},\gamma_{2}).$

This implies $F_{2}$ in $\widetilde{M_{2}}$ is isotopic to $S_{1}(7_{4})$ or
$S_{2}(7_{4}).$

#### Case a)

$F_{2}$ in $\widetilde{M_{2}}$ is isotopic to $S_{1}(7_{4}).$ Let
$S_{1}(\widetilde{L})$ be the standard embedded surface in $(M,\gamma)$ such
that $S_{1}(\widetilde{L})$ restricted to $V_{1}$ is in $R_{+}({\gamma}).$ Let
$S_{1}=(S_{1}(\widetilde{L})\cap V_{2})\cup D=S_{1}(7_{4})$ (where $D$ is the
plumbing disk) be the standard embedded surface in $\widetilde{M_{2}}.$ Since
$F_{2}$ is isotopic to $S_{1}(7_{4})$ in $\widetilde{M_{2}}$ and both the
surfaces are disjoint from $T_{2}$, we have an isotopy $e:I\times
F_{2}\to\widetilde{M_{2}}$ such that $e(0,F_{2})=F_{2}$ and $e(1,F_{2})$ is a
parallel copy of $S_{1}(7_{4})$ in $\widetilde{M_{2}}.$

A priori we have the assumption that $F$ intersects $S_{1}(\widetilde{L})$
essentially. We would like to show that we can isotope $F$ in $E(L)$ such that
$F$ is disjoint from $S_{1}(\widetilde{L}).$ That would prove the fact that
$F$ is isotopic to $S_{1}(\widetilde{L})=S_{1}(7_{4})\cup_{D}H_{2}.$ where
$\cup_{D}$ denotes the plumbing of the two surfaces.

Let $S_{1}(\widetilde{L})$ be the embedded surface. As before, $V_{1}$ and
$V_{2}$ denote the balls partitioning $S^{3}$ with $V_{1}$ being the ball at
the top and $V_{2}$ at the bottom of the diagram with the plumbing disk
$D\subset\partial V_{1}$. $F_{2}$ in $E(L)$ is isotopic to $S_{1}(7_{4}).$ Let
$(M,\gamma)$ be the complementary sutured manifold of $S_{1}(\widetilde{L}).$
There is an isotopy $e:I\times F_{2}\to M_{2}$ such that $e_{1}(F_{2})$ is
parallel to $R_{+}(\gamma)$ or $R_{-}{(\gamma)}$ and $e_{1}(F_{2})$ does not
intersect $S_{1}(\widetilde{L}).$ Without loss of generality, we can assume
$e_{1}(F_{2})=R_{+}(\gamma).$

Let $N(S^{2})=S^{2}\times[-1,1]$ with $S^{2}\times\\{0\\}=\partial V_{1}$ and
$S^{2}\times\\{-1\\}\in V_{2}$ be a regular neighborhood of the Murasugi
sphere $S^{2}.$

Let $V_{\text{down}}=V_{2}-(N(S^{2})^{\circ})$ and $V_{up}=\overline{V_{1}\cup
N(S^{2})}$ Let’s consider 2 arcs $\alpha$ perpendicular to
$S^{2}\times\\{-1\\}\cap e_{1}(F_{2})$ in $S^{2}\times\\{-1\\}$. Let $R$ be a
rectangular disk with the boundary being $\alpha$ and $S^{2}\times\\{-1\\}\cap
e_{1}(F_{2}).$ $F_{2}^{\prime}=(F\cap V_{down})\cup R$ is a sutured surface in
$(M_{2}^{\prime}=V_{\text{down}}\cap M,\gamma_{2}^{\prime})$ with the boundary
link of $F_{2}^{\prime}$ being $7_{4}.$ Then $F_{2}^{\prime}$ is parallel to
$R_{+}(\gamma_{2}^{\prime})$ since $F_{2}$ is isotopic to $R_{+}(\gamma),$
isotope $F$ by an ambient isotopy such that
$F_{2}^{\prime}=R_{+}(\gamma_{2}^{\prime}).$

Consider $(M_{1}^{\prime}=M\cap V_{\text{up}},\gamma_{1}^{\prime})$ with
$F_{1}^{\prime}=F\cap V_{up}.$ being a sutured surface isotopic to the Hopf
band . Hence we have an isotopy $e^{1^{\prime}}:I\times F_{1}^{\prime}\to
E(L)$ such that $e^{\prime}(1)$ is parallel to $R_{+}(\gamma_{1}).$ (since
Hopf band is fibred). Isotope $F$ such that $F\cap(V_{1}\cup
S^{2}\times[-1/2,1])\subset R_{+}(\gamma)$ and by isotopy extension we can
demand $F_{2}^{\prime}=R_{+}(\gamma_{2}^{\prime}).$ Hence, the surface $F$
intersects $S_{1}(\widetilde{L})$ $\in V_{2}$ in $S^{2}\times(-1/2,-1).$ Since
$F$ is a minimal genus Seifert surface, there is a product region between $F$
and $S_{1}(\widetilde{L}).$ Hence $F$ can be made disjoint from
$S_{1}(\widetilde{L})$ which implies that $F$ is isotopic to
$S_{1}(\widetilde{L}).$

#### Case b)

$F_{2}$ is isotopic to $S_{2}(7_{4}).$ In this case we know that $F_{2}$ is
parallel to $R_{+}(\widetilde{\gamma_{2}})$ or
$R_{-}(\widetilde{\gamma_{2}}).$ WLOG let’s assume $F_{2}$ is isotopic to
$R_{+}(\widetilde{\gamma_{2}}).$ Hence there is an sutured isotopy
$e:F_{2}\times I\to\widetilde{M_{2}}$ such that $e_{0}(F_{2})=F_{2}$ and
$e_{1}(F_{2})=R_{+}(\widetilde{\gamma_{2}})).$ This implies that there is a
product region between $F_{2}$ and $R_{+}(\widetilde{\gamma_{2}}).$

Consider a thin product, $S^{2}\times I\subset V_{2}$ such that
$S_{2}\times\\{0\\}=\partial V_{2}$ and $S_{2}\times\\{1\\}\subset V_{2}.$

Let

$V_{up}={V_{1}\cup(S^{2}\times I)};\hskip
28.45274ptV_{down}=\overline{V_{2}-(S^{2}\times[0,1])}.$

This implies

$V_{up}$ and $V_{down}$ are balls with $\partial V_{up}=\partial
V_{down}=S^{2}\times\\{1\\}.$

We move $F$ by a sutured isotopy such that $\widetilde{M_{1}}$ is fixed and
its restriction to $F_{2}$ is $e|_{F_{2}\times[0,r]}$ , $r$ close to $1.$

Let $(M^{*},\gamma^{*})=(M\cup V_{\text{down}},(\partial
V_{up}-M)\cup\gamma\cap V_{up})).$ Consider $F^{*}$ be the sutured surface
obtained by adding two rectangles along the arcs $F\cap\partial V_{up}.$ We
know that $F^{*}$ is a sutured surface on $(M^{*},\gamma^{*})$ homeomorphic to
$(M_{1},\gamma_{1}.)$ Since Hopf band is the unique surface spanned by the
core of $\gamma_{1}=\text{Hopf Link}$, $F^{*}$ is isotopic to Hopf band, (let
$e^{*}$ be the isotopy $e^{*}:F^{*}\times I\to M^{*}$). There are 2
possibilities $e^{*}_{1}(F^{*})=R_{+}(\gamma^{*})$ or $R_{-}(\gamma^{*}).$
Since the Hopf band is fibred, we have an isotopy from $R_{+}(\gamma^{*})$ to
$R_{-}(\gamma^{*}).$ We consider the isotopy
$e^{*}_{1}(F^{*})=R_{+}(\gamma^{*}).$ [If apriori, $F^{*}$ is isotopic to
$R_{-}(\gamma^{*})$, then we can compose with the isotopy arising from the
fibredness of the Hopf link].

So we have two embeddings,

$e((F\cap V_{\text{down}})\times[0,1])\subset M\cap
V_{\text{down}};e^{*}((F\cap V_{up})\times[0,1])\subset M\cap V_{up}$

and

$e|_{F\cap\partial V_{up}\times[0,1]}=e^{*}|_{F\cap\partial
V_{up}\times[0,1]}.$

hence we can connect these two embeddings to get a resulting embedding
$\widetilde{e}$ such that $\widetilde{e}_{0}=id$ and
$\widetilde{e}_{(}1)(F)=R_{(\gamma)},$ which proves that $F$ is isotopic to
$\widetilde{S_{2}}$ contradicting the fact that $F\in
IS(\widetilde{L},\widetilde{S_{2}}).$

This concludes that $IS(\widetilde{L},\widetilde{S_{2}})=\widetilde{S_{1}}.$

and

$IS(L,T_{2})=T_{1}.$

Hence $IS(11_{103})=\\{T_{1},T_{2}\\}$

$T_{1}$$T_{2}$

∎

## Special Case $11_{201}$

Figure 12:  Figure 13:

Let $D$ be an oriented alternating diagram of $11_{201}$ and let $S$ be the
surface obtained by applying Seifert’s algorithm on $D.$ $S=S_{1}\cup S_{2}$
where $S_{2}$ is the Hopf band. Since the Hopf band is a fibred surface on the
Hopf link, hence we have,

$IS(L,S)\cong IS(L_{1},S_{1}).$

$S_{1}$ is a plumbing of $2$ surfaces

$S_{1}=T_{1}\cup_{D}T_{2},$ two unique Seifert surfaces on special alternating
links $\tilde{L_{1}},$ and $\tilde{L_{2}}$ respectively.

If $(M,\gamma)$ be the complementary sutured manifold for $S_{1}$ and
$(M_{1},\gamma_{1},A_{1})$ along with $(M_{2},\gamma_{2},A_{2})$ be the marked
complementary sutured manifolds for $T_{1}$ and $T_{2}$ respectively, then
$M_{2}$ contains a product disk with $A_{2}$ as an edge. By the theorem of
Kakimizu (the plumbed surface $S_{1}$ satisfies the rest of the condition and
$S_{1}=\tilde{S}_{1}$) we know,

$IS(L_{1})=\leavevmode\hbox to78.06pt{\vbox
to23.65pt{\pgfpicture\makeatletter\hbox{\hskip 10.86494pt\lower-2.82861pt\hbox
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}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{}{{}}{} {}{}{}{}{{{}{}}}
{{{}{}{{}}{}}}{{{}}}{{{{}}{{}}}}{{}}{{{ }}}{}{}{{{}{}}}
{{{}{}{{}}{}}}{{{}}}{{{{}}{{}}}}{{}}{{{ }}}\pgfsys@beginscope\pgfsys@invoke{
}\color[rgb]{0.5,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.5,0.5,0}\pgfsys@color@cmyk@stroke{0}{0}{1}{.5}\pgfsys@invoke{
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}{}{}{}{}{{}}{}{}{}{{}}\pgfsys@moveto{4.6799pt}{0.0pt}\pgfsys@lineto{23.77286pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
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\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-7.53194pt}{9.06161pt}\pgfsys@invoke{
}\hbox{{\definecolor[named]{.}{rgb}{0.5,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.5,0.5,0}\pgfsys@color@cmyk@stroke{0}{0}{1}{.5}\pgfsys@invoke{
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}\hbox{{\definecolor[named]{.}{rgb}{0.5,0.5,0}\color[rgb]{0.5,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.5,0.5,0}\pgfsys@color@cmyk@stroke{0}{0}{1}{.5}\pgfsys@color@cmyk@fill{0}{0}{1}{.5}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}$[S_{1}^{c}]$}}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@color@rgb@fill{0}{0}{1}\pgfsys@invoke{ }\hbox{}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{20.92082pt}{9.06161pt}\pgfsys@invoke{
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\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{20.92082pt}{9.06161pt}\pgfsys@invoke{
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}\hbox{{\definecolor[named]{.}{rgb}{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}\pgfsys@color@rgb@stroke{0}{1}{0}\pgfsys@invoke{
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}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}}\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{
{}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{49.94995pt}{9.06161pt}\pgfsys@invoke{
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}\pgfsys@color@cmyk@fill{0}{0}{1}{.5}\pgfsys@invoke{
}\hbox{{\definecolor[named]{.}{rgb}{0.5,0.5,0}\color[rgb]{0.5,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.5,0.5,0}\pgfsys@color@cmyk@stroke{0}{0}{1}{.5}\pgfsys@color@cmyk@fill{0}{0}{1}{.5}{\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}$[\tilde{S}_{1}^{c}]$}}}
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}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}$

Therefore $IS(L_{1},S_{1})=\\{[S_{1}^{c}],[S_{2}^{c}]\\}.$

$IS(L,S)\cong IS(L_{1},S_{1}).$

This implies that $IS(L,S)=\\{{R_{1}},{R_{2}}\\}.$ Via the congruence, one of
the surface, say $R_{1}=S_{1}^{c}\cup_{D}S_{2}.$ The other surface is a
parallel surface, parallel near the plumbing disk with the Hopf band. [It is
similar to the surface $T_{2}$ found in the previous case.]

Since $R_{1}=S_{1}^{c}\cup_{D}S_{2}$ and $S_{2}$ is a Hopf band,
$IS(L,R_{1})\cong IS(L_{1},S_{1}^{c})=\\{S_{1}\\}.$ Hence
$IS(L,R_{1})=S_{1}\cup S_{2}.$

$S_{2}$ is a Hopf band and $R_{2}$ is a parallel surface with $S_{2}$ and
$L_{1}$

The same proof as finding $IS(L,T_{2})$ in the previous section applies and we
have

$IS(L,R_{2})=S.$

Hence the Kakimizu complex of $K=11_{201}$ is:

$[R_{1}]$$[S]$$[S]$$[R_{2}]$

Figure 14:  Figure 15:  Figure 16:

##  5 Further Questions

The Kakimizu complexes of the class of $11$ crossing prime alternating knots
is tabulated.

  * •

The most immediate question is to find the data of Kakimizu complexes of 12
crossing prime, alternating knots to gain some structural information of the
Kakimizu complex. Moreover, it seems plausible to extract from the current
proofs, the Kakimizu complex of a special class of alternating links such that
given an oriented diagram $D$ of $L,$ the Seifert surface obtained by applying
Seifert’s algorithm is a set of plumbings of Seifert surfaces on special
alternating links. Work is in progress on these 2 projects.

  * •

The general goal is to find an algorithm to compute the Kakimizu complexes of
every prime, non-split, oriented, alternating links.

If possible, the goal is to extend to all hyperbolic links.

  * •

Predict or provide more structure to the Kakimizu complex of a link. A general
question is, given a graph $K,$ can $K$ be realized as a $1$ skeleton of the
Kakimuzi complex of a link $L.$

## References

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