You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'We explain an algorithm for finding a boundary link Seifert matrix for a given multivariable Alexander polynomial. The algorithm depends on several choices and therefore makes it possible to find non-equivalent Seifert matrices for a given Alexander polynomial.'
address: 'Rice University, Houston, Texas, 77005-1892'
author:
- Stefan Friedl
title: Algorithm for finding boundary link Seifert matrices 
---

Introduction
============

Algebraic statement
-------------------

We call $A=(A_{ij})_{i,j=1,\dots,m}$ a (boundary link) Seifert matrix if $A$ is a matrix with entries $A_{ij}$ which are $(n_i \times n_j)$–matrices over $\Z$ such that $A_{ij}=A_{ji}^t$ for $i \ne j$ and $\det(A_{ii}-A_{ii}^t)=1$ (for more details cf. [@L77], [@K87]) Note that the $n_i$ are necessarily even numbers. Set $$T:=\diag(\underbrace{t_1,\dots,t_1}_{n_1},
\dots,\underbrace{t_m,\dots,t_m}_{n_m}),$$ then define the Alexander polynomial of $A$ to be $$\Delta(A):=\det(T)^{-\frac{1}{2}}\det(TA-A^t) \in \L_m:=\Z[t_1^{\pm 1},\dots,t_m^{\pm 1}].$$ This polynomial has the following well-known properties which can easily be verified from the definitions. $$\ba{rcl} \Delta(A)(1,\dots,1)&=&1, \\
 \Delta(A)(t_1,\dots,t_m)&=&\Delta(A)(t_1^{-1},\dots,t_m^{-1}). \ea$$

Now assume that $\Delta$ is a polynomial with the above properties. The goal of this paper is to give an algorithm how for finding a Seifert matrix $A$ in terms of the coefficients of $\Delta$ such that $\Delta(A)=\Delta$. In the case $m=1$, i.e. the case of Seifert matrices for knots, an algorithm has been found by Seifert (cf. [@S34], [@BZ85]).

Topological motivation
----------------------

We quickly recall how boundary link Seifert matrices appear in link theory. An $m$-link $L=L_1\cup\dots \cup L_m \subset S^{4q+3}$ is a smooth embedding of $m$ disjoint oriented $(4q+1)$-spheres. A boundary link is a link which has $m$ disjoint Seifert manifolds, i.e. there exist $m$ disjoint oriented $(4q+2)$-submanifolds $F_1,\dots,F_m \subset S^{4q+3}$ such that $\partial(F_i)=L_i, i=1,\dots,m$. One of the main tools for studying boundary links is the Seifert form $$\ba{rcl}
H_{2q+1}(F) \times H_{2q+1}(F) &\to & \Z \\
(a,b) &\mapsto &\lk(a,b_+),
\ea$$ where $b_+$ means that we push a representative of $b$ into $S^{4q+3}\sm F$ along the positive normal direction of $F$. More precisely, we can find an orientation preserving embedding $\iota:F\times [-1,1] \to S^{4q+3}$ and we define $a_+=\iota(a,+1)$ and $a_-=\iota(a,-1)$.

Now pick bases $l_{i,1},\dots,l_{i,n_i}$ for $H_{2q+1}(F_i), i=1,\dots,m$, then $$l_{1,1},\dots,l_{1,n_1},\dots,l_{m,1},\dots,l_{m,n_m}$$ form a basis for $H_{2q+1}(F)=H_{2q+1}(F_1)\oplus \dots \oplus H_{2q+1}(F_m)$. Representing the Seifert form with respect to this basis we get a boundary link Seifert matrix (cf. [@L77], [@K87 p. 670]).

We also need the notion of an $F_m$–link, this is a link with a map $\pi_1(S^{4q+3}\sm L)\to
F_m$, where $F_m$ denotes the free group on $m$ generators, which sends meridians to conjugates of the generators of $F_m$. A Thom argument shows that there is a one-to-one correspondence between isotopy classes of $F_m$–links and isotopy classes of boundary links with Seifert manifolds. It turns out that it is easier to study $F_m$–links, for example the addition of $F_m$–links is well-defined for $q\geq 1$. Boundary links and $F_m$–links are the best understood links, they have been studied thoroughly and many of the classifying results for higher dimensional knots can be done similarly in the context of such links (cf. [@L77], [@K87], [@D86]).

If $L$ is a boundary link with $m$ components then denote by $\ti{X}$ the universal abelian cover of $S^{4q+3}\sm L$, i.e. the cover induced by $\pi_1(S^{4q+3}\sm L)\to H_1(S^{4q+3}\sm L)=\Z^m$. Note that $H_*(\ti{X})$ has a natural $\Z[\Z^m]=\L_m$-module structure.

Let $L \subset S^{4q+3}$ be a boundary link with $m$ components, and $A$ a Seifert matrix of size $(n_1,\dots,n_m)$ for a Seifert manifold $F=F_1\cup \dots\cup F_m$. Then there exists a short exact sequence $$0\to \L_m^n/(AT-A^t)\L_m^n\to  H_{2q+1}(\ti{X}) \to  P\to 0,$$ where $n=\sum_{i=1}^m n_i$ and $P$ is some torsion free $\L_m$-module. Furthermore $P=0$ if $q>0$.

We will give a quick outline of the proof which follows well-known arguments in the knot case (cf. [@L66], [@R90]).

Let $Y=S^{4q+3}\sm F$. We can view $\ti{X}$ as the result of gluing $\Z^m$ copies of $Y$ together along $\Z^m$ copies of $F_1,\dots,F_m$. Consider the resulting Mayer-Vietoris sequence $$\ba{rcl}
 \dots \to H_{i}(F)\otimes \L_m& \to& H_{i}(Y)\otimes \L_m \to H_{i}(\ti{X}) \to \dots \\
       a_j \otimes p &\mapsto & (a_{j,+}t_j-a_{j,-})\otimes p,
\ea$$ where $a_j \in H_i(F_j)$. Note that for $i\in \{1,\dots,4q+1\}$ we have $H_i(Y)\cong H^{4q+2-i}(F)\cong H_i(F,L)\cong H_i(F)$ by Alexander duality, Poincaré duality and a long exact sequence argument. Pick a basis for $H_i(F)$ which gives $A$ as a Seifert matrix for $L$, then give $H_i(Y)$ the corresponding basis. An argument as in Rolfsen (cf. [@R90]) shows that the map $H_{2q+1}(F)\otimes \L_m
\to H_{2q+1}(Y)\otimes \L_m$ is given by $v\mapsto (AT-A^t)v$.

If $q=0$ then the sequence becomes $$\dots \to H_{1}(F)\otimes \L_m \to H_{1}(F)\otimes \L_m \to H_{1}(\ti{X}) \to
 H_{0}(F)\otimes \L_m \to H_{0}(Y)\otimes \L_m \to$$ It is clear that $\ker\{H_{0}(F)\otimes \L_m \to H_{0}(Y)\otimes \L_m\}$ is $\L_m$-torsion free since $H_0(Y)$ is $\L_m$-torsion free (cf. also [@S81b]).

Now consider the case $q>0$. We are done once we show that $H_{2q}(F)\otimes \L_m \to
H_{2q}(Y)\otimes \L_m\cong H_{2q}(F)\otimes \L_m$ is injective. Picking a basis for $H_{2q}(F)$ and giving $H_{2q}(Y)$ the corresponding basis, then we can represent this map by a matrix $B(t_1,\dots,t_m)$. We will prove that $B(1,\dots,1)$ is in fact the identity matrix, in particular $\det(B)\ne 0$. This concludes the proof of the proposition. Note that $B(1,\dots,1)$ represents the map $H_{2q}(F) \to H_{2q}(Y)$ given by $a\mapsto a_+-a_-$. Recall that the isomorphism $f:H_i(Y) \to H^{2q+2-i}(F)$ is induced by the linking pairing, in particular for $\s\in C_{2q+2-i}(F)$ $$f(a_+-a_-)(\s)=\lk(a_+-a_-,\s)=(a\times [-1,1])\cdot \s=a \cdot \s.$$ Thus under the Poincaré duality map $f(a_+-a_-)$ gets sent to $a$.

From the theory of fitting ideals for presentation matrices (cf. [@S81b]) it follows that $\det(AT-A^t)$ is a well-defined invariant for a boundary link $L$ up to multiplication by a unit in $\L_m$. It is easy to see that $\det(T)^{-\frac{1}{2}}\det(AT-A^t)$ is a well-defined invariant for boundary links, it is called the Alexander polynomial of $L$.

Gutierrez [@G74 p. 34] showed that any polynomial $\Delta(t_1,\dots,t_m)$ with the properties $$\ba{rcl} \Delta(1,\dots,1)&=&1 \\
 \Delta(t_1,\dots,t_m)&=&\Delta(t_1^{-1},\dots,t_m^{-1}) \ea$$ is the Alexander polynomial of a boundary link in dimension 1, in particular there exists a boundary link Seifert matrix $A$ with $\Delta(A)=\Delta$. But it is difficult to find an explicit boundary link Seifert matrix, which would be important to compute further invariants.

Farber [@F92] and Garoufalidis and Levine [@GL02] defined non-commutative invariants for boundary links which can be viewed as generalizations of the Alexander polynomial of a knot. Farber also proves a realization theorem.

$S$-equivalence class of Seifert matrices
-----------------------------------------

In the following we will call a matrix $P$ block diagonal if it commutes with $T$, equivalently if $P=P_1\oplus \dots \oplus P_m$ where $P_i$ is a $(n_i\times n_i)$-matrix.

The $S$-equivalence of Seifert matrices is the equivalence relation generated by the following two equivalences (for more details cf. [@L77], [@K85])

$A \sim PAP^t$ where $P$ is a block diagonal matrix over $\Z$ with $\det(P)=1$.

$A$ is equivalent to any row or column enlargement or reduction of $A$.

\[propko\] Any two Seifert matrices for an $F_m$–link are $S$-equivalent. Furthermore any Seifert matrix is the Seifert matrix of an $F_m$–link.

There exists a similar but more complicated proposition for boundary links (cf. [@K85]). It turns out that Seifert matrices for boundary links are related by $S$-equivalence and an action by (cf. also [@K87]) $$A_m:=\{\varphi:F_m\to F_m | \varphi(x_i)=\l_i x_i\l_i^{-1} \mbox { for some }\l_i\in F_m\}/\mbox{inner
automorphism}.$$ The groups $A_1,A_2$ are trivial [@K84], it follows that boundary link matrices with 2 components which are are related by $S$-equivalence and an action by $A_m$ are in fact $S$-equivalent.

It is easy to see that if $A_1,A_2$ are $S$-equivalent, then $\Delta(A_1)=\Delta(A_2)$, this shows again that the Alexander polynomial is an invariant for any $F_m$–link.

We call a Seifert matrix irreducible if no row or column reductions are possible.

\[propprops\]

A Seifert matrix of size $(n_1,\dots,n_m)$ is irreducible if and only if $$\rk(A_{i1} \dots A_{im})=n_i,
  \quad \rk\left(\ba{c} A_{1i}\\ \vdots \\A_{mi}\ea \right)=n_i$$ for all $i=1,\dots,m$. Put differently, a Seifert matrix is irreducible if and only if the block columns and block rows have maximal rank.

If $A_1,A_2$ are $S$-equivalent minimal Seifert matrices then $A_1=PA_2P^t$ where $P$ is a block diagonal matrix over $\Q$ with $\det(P)\ne 0$.

We will use this proposition to show that certain Seifert matrices are not $S$-equivalent.

The statement of the proposition is well-known in the case $m=1$ (cf. [@T73]). The first part of the proposition is fairly straight forward to show, whereas the second part is more difficult to prove. Using ideas of Farber [@F91] one can rewrite the proof of Trotter in the general case, but this requires many details, which we will omit here.

Statement of results
====================

Algebra
-------

For $v_1,\dots,v_l \in \Z$ and $\e_2,\dots,\e_l \in \{-1,+1\}$ define matrices $B_i:=B_i(v_1,\dots,v_i,\eps_2,\dots,\eps_i)$ inductively as follows. $$B_1 :=\bp v_1 & 0 \\-1 & 1 \ep \quad
B_i := \bp &&&&&&& v_i & 0 \\
&&&&&&& 0 & 1 \\&&&&&&& v_i & 1 \\&&&&B_{i-1}&&& 0 & 1 \\
&&&&&&& \vdots \\
&&&&&&& v_i & 1 \\&&&&&&& 0 & 1 \\
v_i &0&v_i&0&\dots &v_i &0&v_i&z_n \\
0&1&1&1 &&1&1&1-z_n&1 \ep,$$ where $z_i:=\frac{1}{2}(1+\e_i)$. Furthermore let $$Y_l:=\diag(y_1,y_1,y_2,y_2,\dots,y_l,y_l).$$

\[mainprop\] Set $v_{l+1}=0$, then $$\det(Y_lB_l-Y_l^{-1}B_l^t)=1-2v_1+ \sum_{j=1}^l (v_j-v_{j+1})
\left(y_1^2\prod_{i=2}^j y_i^{2\e_i}+y_1^{-2}\prod_{i=2}^j y_i^{-2\e_i}\right).$$

The proof will be given in Section \[sectionproof\].

Explanation of the algorithm
----------------------------

Let $\Delta \in \Z[t_1,\dots,t_m]$ be a polynomial with the following properties $$\ba{rcl} \Delta(1,\dots,1)&=&1, \\
 \Delta(t_1,\dots,t_m)&=&\Delta(t_1^{-1},\dots,t_m^{-1}). \ea$$ Then using the usual multiindex notation we can uniquely write $$\Delta(t_1,\dots,t_m)=\sum_{\a \in \Z^m}c_{\a}(t^{\a}+t^{-\a})
+1-\sum_{\a \in \Z^m}2c_{\a}, \quad c_{\a}\in \Z,$$ where $c_{\a}=0$ for all but finitely many $\a$ and $c_{(0,\dots,0)}=0$.

Denote the $\a$ with $c_{\a}\ne 0$ by $\a_1,\dots,\a_r$. Pick a map $p:\{0,\dots,l\}\to \Z^m$ with the following properties.

$p(0)=(0,\dots,0)$,

$|p(t)-p(t-1)|=1$ for all $t=1,\dots,l$,

for each $i=1,\dots,r$ there exists a $t_i\in \{1,\dots,l\}$ such that $p(t_i)=\a_i$. It is easy to see that such a map always exists. Denote the $i^{th}$ unit vector in $\Z^m$ by $e_i$, the second condition says that $p(t)=p(t-1)+\e_te_{s_t}$ for unique $\e_t \in \{-1,+1\},
s_t\in \{1,\dots,m\}$.

Now define $w_{t_i}=c_{p(t_i)}=c_{\a_i}$ for $i=1,\dots,r$ and $w_j=0$ otherwise. Let $v_i:=\sum_{j=i}^{l}w_j, j=1,\dots,l$. From Proposition \[mainprop\] it follows now immediately that for $B=B(v_1,\dots,v_l,\e_2,\dots,\e_l)$ and $Y:=\diag(y_{s_1},\dots,y_{s_l})$ we get $$\ba{rcl}
\det(YB-Y^{-1}B^t)&=& 1-2v_1+\sum_{j=1}^l (v_j-v_{j+1}) \left(y_{s_1}^2\prod_{i=2}^j
y_{s_i}^{2\e_i} +y_{s_1}^{-2}\prod_{i=2}^j y_{s_i}^{-2\e_i}\right).\ea$$ Using multiindex notation $y=(y_1,\dots,y_m)$ we can rewrite this as $$\sum_{j=1}^l w_j(y^{p(j)}+y^{-p(j)})
+1-\sum_{j=1}^l 2w_j = \sum_{\a \in \Z^m}c_{\a}(y^{2\a}+y^{-2\a}) +1-\sum_{\a \in \Z^m}2c_{\a},$$ in particular for $\ti{T}:=\diag(t_{s_1},\dots,t_{s_l})$ we get $$\det(\ti{T})^{\frac{1}{2}}\det(\ti{T}B-B^t)
=\sum_{\a \in \Z^m}c_{\a}(t^{\a}+t^{-\a}) +1-\sum_{\a \in \Z^m}2c_{\a}=\Delta.$$ We can find a permutation matrix $P$ such that $$P\ti{T}P^{-1}=\diag(\underbrace{t_1,\dots,t_1}_{n_1},
\dots,\underbrace{t_m,\dots,t_m}_{n_m})=:T$$ for some $n_1,\dots,n_m$. In fact we can and will assume that $P$ is of form $$P(v_{1,1},v_{1,2},v_{2,1},v_{2,2},\dots,v_{l,1},v_{l,2})
=P(v_{\s(1),1},v_{\s(1),2},v_{\s(2),1},v_{\s(2),2},
\dots,v_{\s(l),1},v_{\s(l),2})$$ for some permutation $\s \in S_l$, i.e. $P$ permutes pairs of coordinates. Note that $P^t=P^{-1}$ and $\det(P)=1$.

The matrix $A=PBP^{-1}$ is a boundary link Seifert matrix of size $(n_1,\dots,n_m)$ and $\Delta(A)=\Delta$.

Note that $B-B^t$ and hence $A-A^t$ is a block sum of $2 \times 2$ matrices of the form $\bp 0 & \pm 1 \\ \mp 1 &0 \ep$, in particular $A$ is a Seifert matrix of size $(n_1,\dots,n_m)$, furthermore $$\ba{rcl} \Delta(A)
&=&\det({T})^{-\frac{1}{2}}\det(TA-A^t)
=\det({T})^{-\frac{1}{2}}\det(PTP^{-1}PAP^{-1}-PA^tP^{-1})=\\
&=&\det(\ti{T})^{-\frac{1}{2}}\det(\ti{T}B-B^t)=\Delta. \ea$$

Using Proposition \[propko\] we get the following corollary.

Any $\Delta$ with $\Delta(1,\dots,1)=1$ and $\Delta(t_1^{-1},\dots,t_m^{-1})=\Delta(t_1,\dots,t_m)$ is the Alexander polynomial of a boundary link.

It is clear that $A$ depends on the map $p$, for example $A$ is a $(2l \times 2l)$–matrix, i.e. $p$ determines the size of $A$. We will see in the next section that different paths can in fact give non $S$-equivalent matrices.

Example
-------

### Minimality of matrices

Let $\Delta=c_{1,0}(t_1+t_1^{-1})+c_{1,1}(t_1t_2+t_1^{-1}t_2^{-1})+c_{0,1}(t_2^2+t_2^{-2})-17$, then $\a_1=(1,0), \a_2=(1,1), \a_3=(0,1)$. The map $p(0):=(0,0), p(1):=(1,0), p(2):=(1,1), p(3):=(0,1)$ satisfies the conditions on $p$. In this case $$\ba{rclrclrcl}
t_1&=&1,& t_2&=&2,&t_3&=&3,\\
s_1&=&1,& s_2&=&2,&s_3&=&1,\\
\e_1&=&1,& \e_2&=&1,&\e_3&=&-1,\\
w_1&=&c_{1,0},& w_2&=&c_{1,1},&w_3&=&c_{0,1},\\
v_1&=&c_{1,0}+c_{1,1}+c_{0,1},& v_2&=&c_{1,1}+c_{0,1},&v_3&=&c_{0,1}.
\ea$$ Then $$B=\bp v_1&0\hspace{0.2cm}&v_2&0\hspace{0.2cm}&v_3&0 \\
-1&1\hspace{0.2cm}&0&1\hspace{0.2cm}&0&1
\\[0.2cm]v_2&0\hspace{0.2cm}&v_2&1\hspace{0.2cm}&v_3&1
\\0&1\hspace{0.2cm}&0&1\hspace{0.2cm}&0&1
\\[0.2cm]v_3&0\hspace{0.2cm}&v_3&0\hspace{0.2cm}&v_3&0
\\0&1\hspace{0.2cm}&1&1\hspace{0.2cm}&1&1
\ep \mbox{ and }
A=\bp v_1&0\hspace{0.2cm}&v_3&0\hspace{0.2cm}&v_2&0 \\
-1&1\hspace{0.2cm}&0&1\hspace{0.2cm}&0&1
\\[0.2cm]v_3&0\hspace{0.2cm}&v_3&0\hspace{0.2cm}&v_3&0
\\0&1\hspace{0.2cm}&1&1\hspace{0.2cm}&1&1
\\[0.2cm]v_2&0\hspace{0.2cm}&v_3&1\hspace{0.2cm}&v_2&1
\\0&1\hspace{0.2cm}&0&1\hspace{0.2cm}&0&1
\ep,$$ where we chose $ \s= \left(\ba{ccc} 1 &2&3 \\ 1&3&2\ea \right)$.

Using Proposition \[propprops\] it is easy to see that $A$ forms an irreducible Seifert matrix of size $(2,1)$.

Consider $$A=\bp w_1+w_3 &0\hspace{0.2cm} &-w_3&0 \\ -1&1\hspace{0.2cm}&0&1 \\[0.2cm]
-w_3&0\hspace{0.2cm}&w_2+w_3&0 \\ 0&1\hspace{0.2cm}&-1&1 \ep$$ then $$\Delta(A)=w_1(t_1+t_1^{-1})+w_2(t_2+t_2^{-1})+w_3(t_1t_2+t_1^{-1}t_2^{-1})+1-2(w_1+w_2+w_3).$$ This shows that the algorithm does in general not produce a Seifert matrix of minimal size for a given Alexander polynomial.

### Uniqueness of result

A straight forward argument shows that for a knot Alexander polynomial $\Delta(t)$ different choices of maps $p$ will produce $S$-equivalent matrices. This is no longer true in the case $m>1$.

Consider $\Delta=w(t_1t_2+t_1^{-1}t_2^{-1})+1-2w, w\ne 0$. If we take maps $p_1,p_2$ with $p_1(0)=(0,0), p_1(1)=(1,0)$ and $p_1(2)=(1,1)$ and $p_2(0)=(0,0), p_2(1)=(0,1)$ and $p_2(2)=(1,1)$ then applying the algorithm we will get identical matrices $B$ but we have to use different permutations: $$\s_1 =\bp 1 & 2 \\ 1&2 \ep, \quad \s_1 =\bp 1 & 2 \\ 2&1 \ep.$$ We get Seifert matrices$$A_1=\bp w&1\hspace{0.2cm}&w&0 \\ 0&1\hspace{0.2cm}&1&1\\[0.2cm]
w&1\hspace{0.2cm}&0&1\\ 0&1\hspace{0.2cm}&0&1 \ep
\mbox{ and }
A_2=\bp 0&1\hspace{0.2cm}&w&1 \\ 0&1\hspace{0.2cm}&0&1\\[0.2cm]
w&0\hspace{0.2cm}&w&1\\ 1&1\hspace{0.2cm}&0&1 \ep.$$ Both matrices are minimal, but not block congruent, since $\det(A_{1,11})=w, \det(A_{2,11})=0$ Hence by Proposition \[propprops\] $A_1$ and $A_2$ are not $S$-equivalent.

Recall that any boundary link Seifert matrix corresponds to an $F_m$–link, we therefore can construct non-isotopic $F_m$–links with identical Alexander polynomials. I do not know whether the matrices are $S_m$-equivalent, in particular whether the corresponding boundary links are isotopic.

Using signature invariants one can show that these matrices are in fact not even matrix cobordant (for a definition cf. [@K87]), i.e. one can show that the corresponding $F_m$–links are in fact not even $F_m$–cobordant.

Proof of Proposition \[mainprop\] {#sectionproof}
=================================

Proof of a special case of Proposition \[mainprop\]
---------------------------------------------------

In this section we will consider the case $\eps_2=\dots=\eps_l=1$. We have to show that $$\det(Y_lB_l-Y_l^{-1}B_l^t)=1-2v_1+ \sum_{j=1}^l (v_j-v_{j+1})
\left(\prod_{i=1}^j y_i^2+\prod_{i=1}^j y_i^{-2}\right).$$ We will show how to compute the determinant, but we will give the matrices only for the case $l=4$ to simplify the notation.

Consider $Y_lB_l-Y_l^{-1}B_l^t$: [$$\bp
v_1(y_1-y_1^{-1})\hspace{-0.3cm}&y_1^{-1}
&v_2(y_1-y_1^{-1})\hspace{-0.3cm}&0&v_3(y_1-y_1^{-1})\hspace{-0.3cm}&0&v_4(y_1-y_1^{-1})&0 \\
-y_1\hspace{-0.3cm}&y_1-y_1^{-1}&0\hspace{-0.3cm}&y_1-y_1^{-1}&0\hspace{-0.3cm}&y_1-y_1^{-1}&0&y_1-y_1^{-1}
\\[0.2cm]
v_2(y_2-y_2^{-1})\hspace{-0.3cm}&0&v_2(y_2-y_2^{-1})\hspace{-0.3cm}&y_2&v_3(y_2-y_2^{-1})\hspace{-0.3cm}&y_2-y_2^{
-1}
&v_4(y_2-y_2^{-1})&y_2-y_2^{-1} \\
0\hspace{-0.3cm}&y_2-y_2^{-1}&-y_2^{-1}\hspace{-0.3cm}&y_2-y_2^{-1}&0\hspace{-0.3cm}&y_2-y_2^{-1}&0&y_2-y_2^{-1}
\\[0.2cm]
v_3(y_3-y_3^{-1})\hspace{-0.3cm}&0&v_3(y_3-y_3^{-1})\hspace{-0.3cm}&0&v_3(y_3-y_3^{-1})\hspace{-0.3cm}&y_3&v_4(y_3
-y_3^{-1})&y_3-y_3^{-1} \\
0\hspace{-0.3cm}&y_3-y_3^{-1}&y_3-y_3^{-1}\hspace{-0.3cm}&y_3-y_3^{-1}&-y_3^{-1}\hspace{-0.3cm}&y_3-y_3^{-1}&0&y_3
-y_3^{-1} \\[0.2cm]
v_4(y_4-y_4^{-1})\hspace{-0.3cm}&0&v_4(y_4-y_4^{-1})\hspace{-0.3cm}&0&v_4(y_4-y_4^{-1})\hspace{-0.3cm}&0&v_4(y_4-y
_4^{-1})&y_4 \\
0\hspace{-0.3cm}&y_4-y_4^{-1}&y_4-y_4^{-1}\hspace{-0.3cm}&y_4-y_4^{-1}&y_4-y_4^{-1}\hspace{-0.3cm}&y_4-y_4^{-1}&-y
_4^{-1}&y_4-y_4^{-1} \\
\ep.$$ ]{} We will first simplify the matrix to make the computation of the determinant easier. For $i=2,\dots,l$ multiply the second row by $\frac{y_i-y_i^{-1}}{y_1-y_1^{-1}}$ and subtract the result from the $2i$–th row, we get [$$\bp
v_1(y_1-y_1^{-1})\hspace{-0.3cm}&y_1^{-1}
&v_2(y_1-y_1^{-1})\hspace{-0.3cm}&0&v_3(y_1-y_1^{-1})\hspace{-0.3cm}&0&v_4(y_1-y_1^{-1})&0 \\
-y_1\hspace{-0.3cm}&y_1-y_1^{-1}&0\hspace{-0.3cm}&y_1-y_1^{-1}&0\hspace{-0.3cm}&y_1-y_1^{-1}&0&y_1-y_1^{-1}
\\[0.2cm]
v_2(y_2-y_2^{-1})\hspace{-0.3cm}&0&v_2(y_2-y_2^{-1})\hspace{-0.3cm}&y_2&v_3(y_2-y_2^{-1})\hspace{-0.3cm}&y_2-y_2^{
-1}&v_4(y_2-y_2^{-1})&y_2-y_2^{-1} \\
y_1\frac{y_2-y_2^{-1}}{y_1-y_1^{-1}}\hspace{-0.3cm} &0&-y_2^{-1}\hspace{-0.3cm}&0&0\hspace{-0.3cm}&0&0&0 \\[0.2cm]
v_3(y_3-y_3^{-1})\hspace{-0.3cm}&0&v_3(y_3-y_3^{-1})\hspace{-0.3cm}&0&v_3(y_3-y_3^{-1})\hspace{-0.3cm}&y_3&v_4(y_3
-y_3^{-1})&y_3-y_3^{-1} \\
y_1\frac{y_3-y_3^{-1}}{y_1-y_1^{-1}}\hspace{-0.3cm}&0

&y_3-y_3^{-1}\hspace{-0.3cm}&0&-y_3^{-1}\hspace{-0.3cm}&0&0&0 \\[0.2cm]
v_4(y_4-y_4^{-1})\hspace{-0.3cm}&0&v_4(y_4-y_4^{-1})\hspace{-0.3cm}&0&v_4(y_4-y_4^{-1})&0&v_4(y_4-y_4^{-1})\hspace
{-0.3cm}&y_4 \\y_1\frac{y_4-y_4^{-1}}{y_1-y_1^{-1}}\hspace{-0.3cm}&0&y_4-y_4^{-1}\hspace{-0.3cm}&0
&y_4-y_4^{-1}\hspace{-0.3cm}&0&-y_4^{-1}&0 \\
\ep.$$ ]{} For $i=1,\dots,l-1$ subtract the $(2i+1)$–st column from the $(2i-1)$–st column and for $i=l-1,\dots,1$ subtract the $2i$–th column from the $(2i+2)$–nd column, we get [$$\bp w_1(y_1-y_1^{-1})\hspace{-0.3cm}&y_1^{-1}&w_2(y_1-y_1^{-1})\hspace{-0.3cm}&-y_1^{-1}&
w_3(y_1-y_1^{-1})&0&w_4(y_1-y_1^{-1})&0 \\
-y_1\hspace{-0.3cm}&y_1-y_1^{-1}&0\hspace{-0.3cm}&0&0\hspace{-0.3cm}&0&0&0\\[0.2cm]
0\hspace{-0.3cm}&0&w_2(y_2-y_2^{-1})\hspace{-0.3cm}&y_2&w_3(y_2-y_2^{-1})\hspace{-0.3cm}&-y_2^{-1}&w_4(y_2-y_2^{-1
})&0\\
y_1\frac{y_2-y_2^{-1}}{y_1-y_1^{-1}}+y_2^{-1}\hspace{-0.3cm}&0&-y_2^{-1}\hspace{-0.3cm}&0&0\hspace{-0.3cm}&0&0&0\\[0.2cm]
0\hspace{-0.3cm}&0&0\hspace{-0.3cm}&0&w_3(y_3-y_3^{-1})\hspace{-0.3cm}
&y_3&w_4(y_3-y_3^{-1})\hspace{-0.3cm}&-y_3^{-1}\\
y_1\frac{y_3-y_3^{-1}}{y_1-y_1^{-1}}-(y_3-y_3^{-1})\hspace{-0.3cm}&0&y_3\hspace{-0.3cm}&0&-y_3^{-1}\hspace{-0.3cm}
&0&0&0\\[0.2cm]
0\hspace{-0.3cm}&0&0\hspace{-0.3cm}&0&0\hspace{-0.3cm}&0&w_4(y_4-y_4^{-1})&y_4\\
y_1\frac{y_4-y_4^{-1}}{y_1-y_1^{-1}}-(y_4-y_4^{-1})\hspace{-0.3cm}&0&0\hspace{-0.3cm}&0&y_4\hspace{-0.3cm}&0&-y_4^
{-1}&0 \ep,$$ ]{} where $w_i:=v_i-v_{i+1}, i=1,\dots,l-1$, recall that $v_{l+1}=0$ hence $w_l:=v_l$. For $i=2,\dots,l$ multiply the $(2i-1)$–st row by $\frac{y_{i-1}-y_{i-1}^{-1}}{y_i-y_i^{-1}}$ and subtract the result from the $(2i-3)$–rd row, furthermore for $i=2,\dots,l-1$ multiply the $2i$–th row by $y_iy_{i+1}$ and subtract the result from the $(2i+2)$–nd row. An induction argument shows that the result is a matrix $D_l$ which is inductively defined as follows. $$\ba{rcl}
D_1&=&\bp w_1(y_1-y_1^{-1})&y_1^{-1}\\-y_1&y_1-y_1^{-1}\ep \\[0.5cm]
D_2&=&\bp D_1 &&0&\frac{y_1^{-1}y_2^{-1}-y_1y_2}{y_2-y_2^{-1}} \\
&&0&0\\
0&0&w_2(y_2-y_2^{-1})&y_2\\
\frac{-y_1^{-1}y_2^{-1}+y_1y_2}{y_1-y_1^{-1}}&0&-y_2^{-1}&0 \ep
\ea$$ and for $n=3,\dots,l$ $$D_n =\bp
&&&&& 0&0 \\
&&&&& \vdots &\vdots \\
&&&&& 0&0 \\
&&D_{n-1}&&& 0&\frac{y_{n-1}^{-1}(y_{n-2}-y_{n-2}^{-1})}{y_{n-1}-y_{n-1}^{-1}}\\
&&&&& 0&0 \\
&&&&& 0&\frac{y_{n-1}^{-1}y_n^{-1}-y_{n-1}y_n}{y_{n}-y_{n}^{-1}}\\
&&&&& 0&0\\
0&0&0&\dots&0&w_n(y_n-y_n^{-1})&y_n \\
\frac{-y_1^{-1}y_n^{-1}+y_1y_2^2\cdot \dots \cdot y_{n-1}^2y_n}
{y_1-y_1^{-1}}&0&0&\dots&0&-y_n^{-1}&0 \ep.$$ Note that $\det(D_l)=\det(Y_lB_l-Y_l^{-1}B_l^t)$, we will now compute $\det(D_l)$. For $n=2,\dots,l$ we denote by $D_n'$ respectively $D_n''$ the matrix obtained from $D_n$ by deleting the first column and the $(2n-3)$–rd respectively $(2n-1)$–st row. Define $${\det}_n:=\det(D_n), \quad {\det}_n ':=\det(D_n '), \quad {\det}_n'':=\det(D_n'').$$ Using the last row to compute $\det(D_n)$ we get $$\hspace{-0.5cm}
\ba{rcl}{\det}_n\hspace{-0.35cm}&=&\hspace{-0.35cm}{\det}_{n\hspace{-0.05cm}-\hspace{-0.05cm}1}
\hspace{-0.1cm}-\hspace{-0.1cm}w_n\hspace{-0.05cm}(y_{n}
\hspace{-0.1cm}-\hspace{-0.1cm}y_{n}^{\hspace{-0.1cm}-1}\hspace{-0.05cm})
\hspace{-0.1cm}\frac{-y_1^ {\hspace{-0.1cm}-1} y_n^{\hspace{-0.1cm}-1}+y_1y_2^2\cdot \dots \cdot
y_{n-1}^2y_n}{y_1-y_1^{\hspace{-0.1cm}-1}} \hspace{-0.15cm}\left(
\frac{y_{n\hspace{-0.05cm}-\hspace{-0.05cm}1}^{\hspace{-0.1cm}-1}
(y_{n\hspace{-0.05cm}-\hspace{-0.05cm}2}-y_{n\hspace{-0.05cm}
-\hspace{-0.05cm}2}^{\hspace{-0.1cm}-1})}
{y_{n\hspace{-0.05cm}-\hspace{-0.05cm}1}-y_{n\hspace{-0.05cm}-
\hspace{-0.05cm}1}^{\hspace{-0.1cm}-1}} {\det}_{n\hspace{-0.05cm}-\hspace{-0.05cm}1}
'\hspace{-0.1cm}+\hspace{-0.1cm}\frac{y_{n\hspace{-0.05cm}-
\hspace{-0.05cm}1}^{-1}y_n^{\hspace{-0.1cm}-1}-y_{n-1}y_n}
{y_n-y_n^{\hspace{-0.1cm}-1}}{\det}_{n\hspace{-0.05cm}-\hspace{-0.05cm}1} ''
\hspace{-0.1cm}\right).
\ea$$

We make the following easy observations: $$\ba{rcl}
{\det}_n '&=&{\det}_{n-1}'',\\
{\det}_n''&=&-y_n^{-1}\left( \frac{y_{n-1}^{-1}(y_{n-2}-y_{n-2}^{-1})}{y_{n-1}-y_{n-1}^{-1}}
{\det}_{n-1} '+\frac{y_{n-1}^{-1}y_n^{-1}-y_{n-1}y_n}{y_n-y_n^{-1}}{\det}_{n-1} '' \right).
\ea$$ It follows that $${\det}_n={\det}_{n-1}-w_n(y_{n}-y_{n}^{-1})
\frac{-y_1^{-1}y_n^{-1}+y_1y_2^2\cdot \dots \cdot y_{n-1}^{2}y_n}{y_1-y_1^{-1}} y_n{\det}_n''.$$ Recall that we have to show that $$\ba{rcl}\det_n&=&1-2v_1+ \sum_{j=1}^l (v_j-v_{j+1})
\left(\prod_{i=1}^j y_i^2+\prod_{i=1}^j y_i^{-2}\right)\\
&=&1-2\sum_{j=1}^lw_j+ \sum_{j=1}^l w_j \left(\prod_{i=1}^j y_i^2+\prod_{i=1}^j y_i^{-2}\right).
\ea$$ The proof of the special case of Proposition \[mainprop\] is complete once we show that $$\ba{rcl}
{\det}_1\hspace{-0.1cm}&=\hspace{-0.1cm}&w_1(y_1^2+y_1^{-2})+1-2w_1\\
\frac{-y_1^{-1}y_n^{-1}+y_1y_2^2\cdot \dots \cdot y_{n-1}^{2}y_n}{y_1-y_1^{-1}}
y_n(y_n\hspace{-0.1cm}-\hspace{-0.1cm}y_n^{-1}){\det}_n''\hspace{-0.1cm}&=\hspace{-0.1cm}&y_1^2\cdot
\dots \cdot y_n^2+ y_1^{-2}\cdot \dots \cdot y_n^{-2}\hspace{-0.1cm}-\hspace{-0.1cm}2 \mbox{ for
} n\hspace{-0.1cm}=\hspace{-0.1cm}2,\dots,l.
\ea$$ The first equality follows from a simple computation. We now prove the second equality by induction on $n$. For $n=1,2$ this follows again from a direct computation. Now assume that the statement is true for all $k<n$, then using the above results we get $$\ba{rl}
&
\frac{-y_1^{-1}y_n^{-1}+y_1y_2^2\cdot \dots \cdot y_{n-1}^{-1}}{y_1-y_1^{-1}}
y_n(y_n-y_n^{-1}){\det}_n''=\\
=&\frac{-y_1^{-1}y_n^{-1}+y_1y_2^2\cdot \dots \cdot y_{n-1}^{-1}}{y_1-y_1^{-1}}
(y_n-y_n^{-1}){\det}_n'' \left(\frac{-y_{n-1}(y_{n-2}-y_{n-2}^{-1})}{y_{n-1}-y_{n-1}^{-1}}
{\det}_{n-2}''-\frac{y_{n-1}^{-1}-y_{n-1}y_n}{y_n-y_n^{-1}}{\det}_{n-1}'' \right).\ea$$ Using the induction hypothesis we get an expression in the five variables $y_1,y_2^2\cdot \dots \cdot
y_{n-3}^2,y_{n-2},y_{n-1},y_n$ which can be computed to equal $ y_1^2\cdot \dots \cdot y_n^2+
y_1^{-2}\cdot \dots \cdot y_n^{-2}-2$.

Proof of Proposition \[mainprop\] {#proof-of-proposition-mainprop}
---------------------------------

Let $\e_2,\dots,\e_l \in \{-1,+1\}$. Let $\varphi:\Z[y_1^{\pm 1},\dots,y_l^{\pm 1}] \to
\Z[y_1^{\pm 1},\dots,y_l^{\pm 1}]$ be the ring homomorphism induced by $\varphi(y_1)=y_1$ and $\varphi(y_i)=y_i^{\e_i}, i=2,\dots,l$, denote the induced map on $M_{2m \times 2m}(\Z[y_1^{\pm 1},\dots,y_l^{\pm 1}])$ by $\varphi$ as well. Write $B(\e_2,\dots,\e_l)$ for $B(v_1,\dots,v_l,\e_2,\dots,\e_l)$.

We see that if we multiply the $(2i-1)$–st and the $2i$–th row of $Y_lB(\eps_2,\dots,\eps_l)-Y_l^{-1}B(\eps_2,\dots,\eps_l)^t$ by $\eps_i$, $i=2,\dots,l$, then we get $\varphi(Y_lB(1,\dots,1)-Y_l^{-1}B(1,\dots,1)^t)$, in particular the determinants are the same, i.e. $$\ba{rl}
&\det(Y_lB(\eps_2,\dots,\eps_l)-Y_l^{-1}B(\eps_2,\dots,\eps_l)^t)\\
=&\varphi(\det(Y_lB(1,\dots,1)-Y_l^{-1}B(1,\dots,1)^t))\\
=&\varphi\left(1-2v_1+ \sum_{j=1}^l (v_j-v_{j+1})
\left(\prod_{i=1}^j y_i^2+\prod_{i=1}^j y_i^{-2}\right)\right)\\
=&1-2v_1+ \sum_{j=1}^l (v_j-v_{j+1}) \left(y_1^2\prod_{i=2}^j y_i^{2\e_i}+y_1^{-2}\prod_{i=2}^j
y_i^{-2\e_i}\right).
\ea$$ This proves Proposition \[mainprop\].

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