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 show that a null–homologous transverse knot $K$ in the complement of an overtwisted disk in a contact 3–manifold is the boundary of a Legendrian ribbon if and only if it possesses a Seifert surface $S$ such that the self–linking number of $K$ with respect to $S$ satisfies ${\operatorname{sl}}(K,S)=-\chi(S)$. In particular, every null–homologous topological knot type in an overtwisted contact manifold can be represented by the boundary of a Legendrian ribbon. Finally, we show that a contact structure is tight if and only if every Legendrian ribbon minimizes genus in its relative homology class.'
address:
- 'S. Baader, Dept. Mathematik der ETH Z[ü]{}rich, R[ä]{}mistr. 101, 8092 Z[ü]{}rich, Switzerland'
- 'K. Cieliebak, Mathematisches Institut der LMU München, Theresienstr. 39, 80333 München, Germany'
- 'T. Vogel, Mathematisches Institut der LMU München, Theresienstr. 39, 80333 München, Germany'
author:
- 'S. Baader, K. Cieliebak, T. Vogel'
title: Legendrian ribbons in overtwisted contact structures
---

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

In this note, $(M,\xi)$ will always denote a cooriented contact 3–manifold, i.e. $\xi=\ker\lambda$ for a 1–form $\lambda$ on $M$ satisfying $\lambda\wedge d\lambda\neq 0$. An oriented knot $K \subset M$ is called [*(positively) transverse*]{} if $\lambda(\dot\gamma)>0$ for a positive parametrization $\gamma:S^1\to K$. We will always assume that $K$ is null–homologous. Then $K$ possesses a [*Seifert surface*]{}, i.e. an embedded oriented connected surface $\Sigma\subset M$ with boundary ${\partial}\Sigma=K$. Given $K$ and $\Sigma$, we choose a nowhere vanishing section $v$ of $\xi{\big|_{\Sigma}}$ and use $v$ to push $K$ away from itself. The resulting knot is denoted by $K'$.

The [*self–linking number*]{} ${\operatorname{sl}}(K,\Sigma)$ is defined as the algebraic intersection number of $K'$ and $\Sigma$.

The self–linking number ${\operatorname{sl}}(K,\Sigma)$ depends only on $K$ and $[\Sigma]\in H_2(M,K)$ and is independent of the choice of $v$. Moreover it is always odd (since it has the same parity as the Euler characteristic of $\Sigma$). When it is clear which Seifert surface we use we will simply write ${\operatorname{sl}}(K)$. Obviously, isotopic transverse knots have the same self–linking number (with respect to Seifert surfaces carried along with the isotopy).

A contact structure $\xi$ on a manifold $M$ is called *overtwisted* if $M$ contains an overtwisted disc, i.e. an embedded disc whose boundary is a Legendrian unknot with Thurston–Bennequin number zero (see Figure 1). Otherwise the contact structure $\xi$ is called *tight*.

![](overtwisted)

The following result was first proved for the standard contact structure on ${\mathbb{ R}}^3$ by D. Bennequin in [@Be] and then generalized to tight contact structures by Y. Eliashberg in [@El].

\[t:bennequin\] If $(M,\xi)$ is a tight contact 3–manifold and $K$ is a transverse knot with Seifert surface $\Sigma$, then $$\label{e:tb ungl}
{\operatorname{sl}}(K,\Sigma)\le-\chi(\Sigma).$$

A [*Legendrian graph*]{} in a contact 3–manifold $(M,\xi)$ is a trivalent embedded graph $\Gamma\subset M$ such that all edges are tangent to $\xi$.

To a Legendrian graph one can associate a transverse link type as follows. Choose a surface $\Sigma_\Gamma$ containing $\Gamma$ with smooth boundary such that $\Sigma_\Gamma$ is tangent to $\xi$ at every point of $\Gamma$.

We use the orientation of the contact structure to orient $\Sigma_\Gamma$. Then for $\Sigma_\Gamma$ sufficiently small (i.e. after replacing $\Sigma_\Gamma$ by a sufficiently small neighbourhood of $\Gamma$ in $\Sigma_\Gamma$), its boundary $\partial \Sigma_\Gamma$ is a link all of whose components are positive transverse knots. The isotopy class of the resulting transverse link depends only on $\Gamma$. We refer to $\Sigma_\Gamma$ as a [*Legendrian ribbon*]{}.

Legendrian Ribbons and overtwisted discs
========================================

The following theorem is an analogue of a result explained in [@Dy] for overtwisted knots. Before stating it, recall that positive transverse knots are obtained from knots tangent to $\xi$ by choosing an oriented framing of $\xi$ such that one of the components of the framing is tangent to the knot and pushing the knot in the direction opposite to the second component of the framing. As described for example in in [@Et], a positive transverse knot $\gamma$ can be stabilized to a knot $\hat{\gamma}$ with ${\operatorname{sl}}(\hat{\gamma})={\operatorname{sl}}(\gamma)-2$.

\[t:class\] Let $(M,\xi)$ be a contact 3–manifold and $D\subset M$ an overtwisted disc. If $L,K$ are two null–homologous transverse knots lying in the complement of $D$ such that $K,L$ represent the same topological knot type and ${\operatorname{sl}}(K)={\operatorname{sl}}(L)$ (with respect to Seifert surfaces carried along with the isotopy), then $L$ and $K$ are isotopic as transverse knots.

Any two transverse knots $K,L$ representing the same topological knot types become transversely isotopic after sufficiently many stabilizations; this fact can be shown as the analogous statement for Legendrian knots, cf. [@Et]. (At this point the contact structures are allowed to be tight or overtwisted.)

Of course, every stabilization changes the self–linking number of the knots and therefore this procedure by itself does not produce a transverse isotopy between the original knots $K,L$. However, when $K,L$ lie in the complement of a fixed overtwisted disc $D$, then one can neutralize each stabilization by pulling a segment of the knot over $D$. Indeed, let $\partial D$ be oriented in such a way that ${\operatorname{rot}}(\partial D)=1$, then its positive transversal push–off $\partial D^+$ has self–linking number one, according to the general formula $${\operatorname{sl}}(\gamma^+)={\operatorname{tb}}(\gamma)+{\operatorname{rot}}(\gamma),$$ which holds for all null–homologous Legendrian knots $\gamma$ in a 3–manifold (see [@Be]). Here ${\operatorname{tb}}$ and ${\operatorname{rot}}$ denote the Thurston–Bennequin number and the rotation number, respectively.

Given two transversal knots $\gamma, \gamma'$ in $M$ which lie in disjoint balls one can define the connected sum such that the self linking number of the resulting knot $\gamma\#\gamma'$ satisfies $${\operatorname{sl}}(\gamma\#\gamma')={\operatorname{sl}}(\gamma)+{\operatorname{sl}}(\gamma')+1.$$ Let $D_{ot}$ be an overtwisted disc and $\gamma'$ the positive push off of the boundary $\partial D_{ot}$ with ${\operatorname{sl}}(\gamma')=+1$. If $\gamma$ is a positive transverse knot which is disjoint from $D_{ot}$, then ${\operatorname{sl}}(\gamma\#\gamma')={\operatorname{sl}}(\gamma)+2$ and $\gamma\#\gamma'=\gamma$ as topological knot types. We call $\gamma\#\gamma'$ the destabilization of $\gamma$. A stabilization of $\gamma\#\gamma'$ yields a knot which is isotopic to $\gamma$ as positive transverse knot. The isotopy is obtained by a push off of the isotopy constructed in Lemma 4.7 in [@Dy] (where a similar situation for Legendrian knots is considered).

In order to construct a transverse isotopy between $K$ and $L$ we follow a procedure used in [@Dy] in the context of Legendrian knots. We first isotope a segment of $K$ such that it coincides with a segment $\sigma$ of $L$. Then we stabilize both knots sufficiently often on the complements of $\sigma$ and we destabilize $\sigma$ using $D$ in order to undo the stabilizations.

If this is done sufficiently many times, then the complements of $\sigma$ become transversely isotopic while $K$ and $L$ still coincide along $\sigma$. Because we have never changed the transverse knot types this shows that $K,L$ are isotopic transverse knots.

The following lemma shows that transverse knots obtained from Legendrian ribbons realize equality in the Thurston–Bennequin inequality . Y. Kanda has given examples of topological knot types for which the equality in cannot be realized by any transverse representative, cf. [@Ka]. In particular, such knots cannot be boundaries of Legendrian ribbons.

\[l:euler\] Let $K=\partial\Sigma_\Gamma$ be the boundary of a Legendrian ribbon $\Sigma_\Gamma$. Then ${\operatorname{sl}}(K,\Sigma_\Gamma)=-\chi(\Sigma_\Gamma)$.

Let $w$ be a vector field along $K={\partial}\Sigma_\Gamma$ which is tangent to $\Sigma_\Gamma$ and points outwards. As $\Sigma_\Gamma$ is almost tangent to $\xi$, $w$ projects to a nonvanishing section $v$ of $\xi$ along $K$. The self–linking number of $K$ can be described as the obstruction to the existence of a nowhere vanishing extension over $\Sigma_\Gamma$ of $v$ as a section of $\xi$, or equivalently, of $w$ as a section of $T\Sigma_\Gamma$. But the latter obstruction equals minus the Euler characteristic, hence ${\operatorname{sl}}(K,\Sigma_\Gamma)=-\chi(\Sigma_\Gamma)$.

\[t:ribbon\] Let $K$ be a transverse knot in an overtwisted contact 3–manifold $(M,\xi)$ with Seifert surface $\Sigma$. Assume that $M\setminus K$ is still overtwisted and that ${\operatorname{sl}}(K,\Sigma)=-\chi(\Sigma)$. Then there is a Legendrian graph $\Gamma$ with $\partial\Sigma_\Gamma=K$. Moreover, $[\Sigma_\Gamma]=[\Sigma]\in H_2(M,K)$ and $\chi(\Sigma_\Gamma)=\chi(\Sigma)$.

We fix an overtwisted disc $D$ in the complement of $K$. Without loss of generality we may assume that $D$ and $\Sigma$ do not intersect. The following constructions can be carried out in the complement of $D$.

Let $G\subset \Sigma$ be a trivalent graph such that $G$ is a deformation retract of $\Sigma$. After a $C^0$–small isotopy of $\Sigma$ we may assume that $\Sigma$ is tangent to $\xi$ at the vertices of $G$ such that the orientations of $\Sigma$ and of $\xi$ coincide at the vertices. Moreover, we may assume that the edges of $G$ are tangent to $\xi$ near the vertices.

Next, we smoothly isotope each edge to a Legendrian curve contained in a small neighbourhood of the original edge, fixing it near the vertices, and pull along $\Sigma$ with the isotopy. After doing this, $\Sigma$ and $\xi$ need not induce the same framing of the edges. However, the framings can be arranged to agree for each edge by either stabilizing the edge sufficiently often or by sliding the edge sufficiently often over $D$, cf. [@Dy]. These operations correspond to taking a connected sum with a Legendrian unknot with Thurston–Bennequin number $-1$ or $+1$, respectively. The latter only exist in the presence of an overtwisted disc. An example of a Legendrian ribbon whose core curve is a Legendrian unknot with Thurston–Bennequin number $+1$ is shown in Figure 2. The two ‘horizontal’ parts of the ribbon lie on the boundary of two parallel overtwisted discs there.

![](tbone)

We denote the resulting Legendrian graph by $\Gamma$ and the resulting transverse knot by $L={\partial}\Sigma_\Gamma$.

By construction, $K$ and $L$ are equivalent as topological knots, ${\operatorname{sl}}(K)=-\chi(\Sigma)=-\chi(\Sigma_\Gamma)={\operatorname{sl}}(L)$ by Lemma \[l:euler\], and the complement of $K\cup L$ is still overtwisted. So by [Theorem \[t:class\]]{} the knots $K$ and $L$ are transversely isotopic. This isotopy can be realized by an ambient contact isotopy (see [@Et]), so after pulling along $\Sigma_\Gamma$ with this isotopy we may assume ${\partial}\Sigma_\Gamma=K$. By construction we have $[\Sigma_\Gamma]=[\Sigma]\in H_2(M,K)$ and $\chi(\Sigma_\Gamma)=\chi(\Sigma)$, which finishes the proof of the theorem.

Denote by $g(S)$ the genus of a surface $S$.

Let $(M,\xi)$ be an overtwisted contact manifold. Let $K\subset M$ be a null–homologous transverse knot in the complement of an overtwisted disc and $S$ be a Seifert surface for $K$ which minimizes the genus in its relative homology class.

Then $K$ is the bondary of a Legendrian ribbon representing the class $[S]\in H_2(M,K)$ if and only if ${\operatorname{sl}}(K,S)\geq 2g(S)-1$. In particular, every null–homologous topological knot type in an overtwisted contact $3$–manifold can be represented by the boundary of a Legendrian ribbon.

Suppose first that ${\operatorname{sl}}(K,S)\geq 2g(S)-1$. By taking connected sums with null–homologous tori we can increase the genus of $S$ by any positive integer, without changing its relative homology class. Due to the assumptions, this allows us to find a Seifert surface $\Sigma$ for $K$, homologous to $S$, with ${\operatorname{sl}}(K,\Sigma)=2g(\Sigma)-1=-\chi(\Sigma)$. So by Theorem \[t:ribbon\], $K$ is the boundary of a Legendrian ribbon representing the class $[S]\in H_2(M,K)$.

Conversely, if $K={\partial}\Sigma_\Gamma$ for a Legendrian graph $\Gamma$ with $[\Sigma_\Gamma]=[S]\in H_2(M,K)$, then Lemma \[l:euler\] yields ${\operatorname{sl}}(K,S)=-\chi(\Sigma_\Gamma)=2 g(\Sigma_\Gamma)-1\geq 2g(S)-1$ because $S$ minimizes the genus in its relative homology class.

The last statement holds because any topological knot type in an overtwisted contact 3–manifold can be realized by a transverse knot $K$ in the complement of an overtwisted disk, and by repeated destabilization we can arrange ${\operatorname{sl}}(K,S)\geq 2g(S)-1$.

As it is shown in [@BaM], the situation in tight contact structures is quite different: A transverse knot bounding a Legendrian ribbon is quasipositive in the sense of Rudolph [@Ru]. Quasipositivity is quite a strong condition, as it implies chirality. For example, the figure–8 knot is not quasipositive since it is achiral, i.e. topologically equivalent to its mirror image. A classification of quasipositive knots up to 10 crossings is given in [@Ba].

To conclude this note, we give a characterization of tightness in terms of Legendrian ribbons.

A contact structure $\xi$ on a 3–manifold $M$ is tight if and only if every Legendrian ribbon has minimal genus among all embedded surfaces with the same boundary and in the same relative homology class.

Assume first that $\xi$ is tight and $\Sigma_\Gamma$ is a Legendrian ribbon with boundary ${\partial}\Sigma_\Gamma=K$. By Lemma \[l:euler\] we have ${\operatorname{sl}}(K,\Sigma_\Gamma)=-\chi(\Sigma_\Gamma)$, and by Theorem \[t:bennequin\] every other Seifert surface $\Sigma$ for $K$ homologous to $\Sigma_\Gamma$ satisfies ${\operatorname{sl}}(K,\Sigma)\leq-\chi(\Sigma)$, so $g(\Sigma)\geq g(\Sigma_\Gamma)$.

Conversely, if $\xi$ is overtwisted, then we construct a Legendrian ribbon $\Sigma_\Gamma$ of genus one whose boundary $K={\partial}\Sigma_\Gamma$ is the topological unknot, as shown in Figure 3.

![](unknot)

Since $\Sigma_\Gamma$ is contained in the neighbourhood of an overtwisted disc, i.e. in a ball, it is homologous rel $K$ to a disk and therefore not genus minimizing.

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