Is number of knot strands an invariant? Question: Does the number of components in a knotwork depend on the particular planar embedding?


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*I've been investigating how to compute the number of components ("separate strands") in a Celtic knot based on the underlying planar graph structure. (See relationship between knots/links and planar graphs here).


*Apparently the calculation for general graphs is a little complicated; for example, the reference in this question points out that for a uniform $m\times n$ grid of squares, the number of components is $\mathrm{lcd}(m,n)$.


*It would satisfy me to find a formula for computing the number of components ("strands"), or a relationship between the number of strands and various graph properies such as its degree, spectrum, etc., even if those properties were hard to calculate.


*One approach I've taken is in terms of connected components: each separate strand  follows a particular trajectory, and the connected components of those trajectories correspond exactly to the strands. You can define the trajectory as a transition function mapping (some additional structure plus) each edge to its successor; this is a permutation on (structured) edges whose cycles are the components.


*The transition function can be encoded as its own, derived, directed graph (similar to a graph-encoded map), whose connected components are the components of the knotwork. From linear algebra, we know that the number of connected components can be recovered as the multiplicity of the zero eigenvalue of the adjacency matrix's Laplacian.
However, I know that the same graph $G$ can have multiple non-isomorphic planar embeddings (i.e. whose duals are non-isomorphic). So far in my experience, this has changed some of the knotting properties (such as number of twists in each component) but not the number of components:

My question is this:

Question: Does the number of components in a knotwork depend on the particular planar embedding? How do we prove it?

My intuition says that the number of components is an invariant, but I haven't been able to produce a counterexample or proof using my approach above.

Conjecture: If $G$ is a graph, then the corresponding knotwork has $c$ components, where
$$T_G(-1,-1) = (-1)^{|E(G)|}\cdot (-2)^{c - 1}$$
and $T_G$ is the Tutte polynomial, and $|E(G)|$ is the number of edges in the graph. (?)
 A: Let $D$ be the diagram of a link. For example, $D$ could be the diagram of the Celtic knot or link pictured in your post. Let $G$ be the checkerboard graph of $D$. The graph $G$ is the graph described in your first bullet point.
Answer: The number of components of $D$ is determined by the abstract graph $G$ and does not depend on how $G$ is embedded in the plane.
To the best of my knowledge, this was first proved by Michel Las Vergnas in 1979. He showed that the number of components of $D$ is determined by the Tutte polynomial evaluation $T_G(-1,-1)$. Since the Tutte polynomial doesn't depend on a particular embedding of $G$, the result follows. The reference for this paper is

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*Las Vergnas, Michel. On Eulerian partitions of graphs. Graph theory and combinatorics (Proc. Conf., Open Univ., Milton Keynes, 1978), pp. 62–75,
Res. Notes in Math., 34, Pitman, Boston, Mass.-London, 1979.

I could not easily find a copy of the above paper, so here's another way to get the solution, due to Dan Silver and Susan Williams (arXiv link). They define a matrix $Q_2(G)$ whose entries are in the field with two elements $\mathbb{F}_2$ as follows. Both the rows and columns of the matrix are indexed by the vertices $v_1,\dots,v_n$ of $G$. If $i\neq j$, then the $ij$ entry of $Q_2(G)$ is the number of edges between vertices $v_i$ and $v_j$ (taken$\mod 2$). The $ii$ entry of $Q_2(G)$ is the sum of the other entries in the row $i$ (again taken$\mod 2$). Equivalently, we could say the $ii$ entry in $Q_2(G)$ is the sum of the other entries in the column $i$.
In Theorem 1.1 of the linked paper, they prove that the number of components of $D$ equals the nullity of $Q_2(G)$. They note in Remark 1.2 that this implies the number of components of $D$ is independent of the plane embedding of $G$.
Edit: I don't have access to the Las Vergnas paper, but I can give another explanation of the result using the Tutte polynomial and the Jones polynomial.
Let $L$ be an alternating link, let $D$ be an alternating diagram of the link, and let $G$ be the checkerboard graph of $D$. Then the Tutte polynomial $T_G(x,y)$ of $G$ and the Jones polynomial $V_L(t)$ of $L$ are related as follows:
$$V_L(t) = f_D(t) T_G(-t,-t^{-1})$$
for the function $f_D(T)$ defined by
$$f_D(t) = (-1)^{w(D)}t^{\frac{1}{4}(|E| - 2(|V|-1)+3w(D))}$$
where $w(D)$ is the writhe of $D$, $|E|$ is the number of edges in $G$, and $|V|$ is the number of vertices of $D$. Notice that $|f_D(1)|=1$, and thus $|V_L(1)| = |T_G(-1,-1)|$.
The Jones polynomial satisfies the skein relation
$$(t^{\frac{1}{2}}-t^{-\frac{1}{2}})V_{L_0}(t) = t^{-1}V_{L_+}(t) - tV_{L_-}(t)$$
where $L_+,L_-,$ and $L_0$ are as below.

Setting $t=1$ in the above skein relation yields $V_{L_+}(1)=V_{L_-}(1)$. In other words the Jones polynomial evaluated at $t=1$ does not change under crossing changes, and thus $V_L(1)=V_{\bigcirc\sqcup\dots\sqcup\bigcirc}(1)$ where $\bigcirc\sqcup\dots\sqcup\bigcirc$ is the trivial link with the same number of components as $L$. The Jones polynomial of $\bigcirc\sqcup\dots\sqcup\bigcirc$ is $V_{\bigcirc\sqcup\dots\sqcup\bigcirc}(t) = (-t^{\frac{1}{2}}-t^{-\frac{1}{2}})^{m-1}$ where $m$ is the number of components of $\bigcirc\sqcup\dots\sqcup\bigcirc$. Thus
$$|T_G(-1,-1)|=|V_L(1)|=|V_{\bigcirc\sqcup\dots\sqcup\bigcirc}(1)| = 2^{m-1}.$$
The above case handles when $L$ is alternating. If $L$ is non-alternating, then proceed as follows. Let $D$ be any diagram of $L$. Define $D_{\text{alt}}$ to be a diagram with the same shadow as $D$ but whose crossings are changed to be alternating, and define $L_{\text{alt}}$ to be the link whose diagram is $D_{\text{alt}}$. Note that $D$ and $D_{\text{alt}}$ have the same checkerboard graph $G$. The above argument implies that $|T_G(-1,-1)|=2^{m-1}$ where $m$ is the number of components of $L_{\text{alt}}$. Since $L_{\text{alt}}$ and $L$ have the same number of components, the result follows for $L$ as well.
