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Below I will describe a link invariant, denoted by me as $inv(L)$. Has anyone encountered this invariant in the literature? If so, what is its name? Also, any references to papers or books that mention it would be fantastic.

Let $D$ be a diagram for a link $L$. Let $inv(D)$ denote the minimum number $k$ of components $C_i$ necessary to add to the diagram $D$ such that

  1. each added component $C_i$ has no self-crossings, and
  2. the resulting link $D\cup\left(\bigcup_{i=1}^{k} C_i\right)$ is an alternating diagram.

Define $inv(L)=\min\{inv(D)\}$ where $D$ is a diagram of $L$.

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    $\begingroup$ Out of curiosity, is this guaranteed to be anything other than 0? Or 1? $\endgroup$ – zibadawa timmy Nov 7 '13 at 1:05
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    $\begingroup$ @zibadawatimmy that's a good question. I've only started to look at this invariant, but I have an idea of how to construct links L where inv(L) is arbitrarily large. This would involve finding a computable lower bound for inv(L) and showing that the lower bound can be arbitrarily large. $\endgroup$ – Adam Lowrance Nov 7 '13 at 1:42
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    $\begingroup$ Are you sure that such an number always (or even) exists? Assuming you aren't allowing virtual links. $\endgroup$ – Robin Oct 26 '14 at 22:07
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It has been ages since I posted this question, but I have the answer now. The invariant described is $$\operatorname{inv}(L) = \begin{cases} 0 & \text{if $L$ is alternating,}\\ 1 & \text{if $L$ is nonalternating.} \end{cases}$$ In other words, the name of the invariant is "alternating" vs. "nonalternating."

In order to prove this, we proceed in two steps. The first step is to show that $\operatorname{inv}(D)$ is defined for each diagram, that is, we show there exists a collection of curves $\{C_1,\dots, C_k\}$ (drawn in red) with no self crossings such that $D\cup\left(\bigcup_{i=1}^k C_i\right)$ is alternating. The second step is to show how to modify the set of red curves $\{C_1,\dots,C_k\}$ to another valid set of red curves $\{C_1',\dots, C_{k-1}'\}$, where valid means that $D\cup\left(\bigcup_{i=1}^{k-1} C_i'\right)$ is alternating.

Step 1. Consider the diagram $D$ as a $4$-regular graph whose vertices are the crossings. Decorate the ends of each edge of the graph to indicate over/under crossings. If both ends of an edge are over-crossings or if both ends of an edge are under-crossings, then we call that edge nonalternating. If one end of the edge is an over-crossing and the other is an under-crossing, then the edge is alternating. Around any face of $D$ there is an even number of nonalternating edges. Moreover, if we ignore the alternating edges on the face, then the nonalternating edges go back and forth between over-crossing edges and under-crossing edges.

Mark each nonalternating edge with a red point in its midpoint. Inside of each face, draw a red arc connecting a marked point on an under-crossing nonalternating edge with the marked point on the adjacent over-crossing nonalternating edge in the clockwise direction. Performing this operation for each face yields a collection of simple closed curves $\{C_1,\dots, C_k\}$ which intersect $D$ transversely at midpoints of nonalternating edges. Choose crossing information between the red curves and the curves of $D$ so that if the marked point comes from an over-crossing nonalternating edge, then the red curve goes over the curve of $D$, and if the marked point comes from an under-crossing nonalternating edge, then the red curve goes under the curve of $D$. The resulting diagram $D\cup\left(\bigcup_{i=1}^k C_i\right)$ is alternating. See the figure below for a picture of this process.

enter image description here

Step 2. In this step, we assume that $k\geq 2$ and give an algorithm to reduce the number of red curves by one, while keeping the overall link diagram $D\cup \{\text{red curves}\}$ alternating.

The first part of this algorithm is to make two arcs of distinct red curves lie in the same face of $D$. This can be accomplished by performing a finger move of a red curve past a curve of $D$. Since we are allowed to choose crossing information between red curves and curves of $D$, this operation can be performed while keeping $D\cup \{\text{red curves}\}$ alternating. See the figure below.

enter image description here

After a suitable number of finger moves, we have arcs of two distinct red curves, say $C_i$ and $C_j$, that lie in the same face of $D$. We perform a saddle move on those two arcs to surger curves $C_i$ and $C_j$ into one red curve. See the figure below.

enter image description here

Repeating this step $k-1$ times yields a single red curve $C$ such that $D\cup C$ is an alternating diagram.

Conclusion. If the link $L$ is alternating, then it has a diagram $D$ that is alternating, and the addition of $0$ curves to $D$ is an alternating diagram. Thus $\operatorname{inv}(L)=0$. If the link $L$ is nonalternating, then we can take any diagram $D$ of $L$ and perform the algorithm described above. In the end, we obtain an alternating diagram $D\cup C$ where $C$ is the red curve with no self crossings. Hence $\operatorname{inv}(L)=1$.

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