Proof The first part of item 1 is [Reference Agarwala and Marin-Amat5, Equation (9)] and surrounding discussion, and the second part is standard matroid theory. Item 2 is [Reference Agarwala and Marin-Amat5, Lemma 3.27]. Item 3 is a direct consequence of [Reference Agarwala and Marin-Amat5, Theorem 3.20] and the fact that $F(P_1)^c = V(P_1^c)$ , and item 5 is given in [Reference Agarwala and Marin-Amat5, Lemma 3.28].

To prove item 4, we need to show that $F(P)$ is maximally dependent. If $F(P) = [n]$ then this is automatic, so suppose not and let $v \in [n] \setminus F(P)$ . In other words, $v \in V(P^c)$ and so v supports some propagator $q \not \in P$ . Let $S \subseteq F(P)$ be an independent set of maximal size. Then $\text {Prop}(S) \subseteq P$ (from the definition of $F(P)$ ) and no subset of S supports fewer propagators than the number of vertices it contains (by Theorem 2.6). Since v supports a new propagator $q \not \in P$ , the set $S \cup \{v\} \subseteq F(P) \cup \{v\}$ also satisfies this independence condition.

Since S is independent, it follows from the first point in this lemma that $|S| = \text {rk} (S) = \text {rk} F(P)$ . Similarly,

$$ \begin{align*}|S \cup \{v\}| = |S|+1 = \text{rk} (S \cup \{v\}) \leq \text{rk} (F(P) \cup \{v\}) \leq \text{rk} F(P) + 1.\end{align*} $$

Thus $\text {rk} (F(P) \cup \{v\}) = \text {rk} F(P) + 1$ , as required. ▪

Proof Since the vertices of $\tau (W)$ are labeled by the edges of W, which are cyclically ordered, this gives an ordering to the vertices and the outer face of $\tau (W)$ is a cycle. Since W does not admit any propagators of the form $p = (i, i+1)$ due to its admissibility, there is exactly one edge connecting any two adjacent edges of $\tau (W)$ . Similarly, there does not exist two propagators $p,q$ such that both p and q start at edge i and end at edge j. Therefore, no two vertices of $\tau (W)$ can be connected by more than one edge, and so $\tau (W)$ is a simple graph. Finally, the embedding of $\tau (W)$ is induced from the embedding of the graph W, and since W is weakly admissible we know that no pair of propagators can cross. Therefore, it is a planar embedding. ▪

Proof We begin by giving an algorithm for the decomposition, then prove its uniqueness. Let $W = (\mathcal {P}, [n])$ , with $|\mathcal {P}| = k$ .

By splitting a vertex v we will mean replacing v by new vertices $v_1, v_2,\ldots , v_{\text {deg}(v)}$ such that each $v_i$ has exactly one neighbor and the union of the $v_i$ is the neighborhoodFootnote ⁵ of v.

Let $T(W)$ be the dual graph of $\tau (W)$ with the vertex corresponding to the infinite face split. Since $\tau (W)$ is an embedded graph (with a fixed distinguished embedding) by Lemma 3.1, $T(W)$ is an uniquely defined graph.

Furthermore $T(W)$ is a tree because it is connected, has $n+k+1$ vertices ( $k+1$ from the internal faces of $\tau (W)$ and n from the outer face) and $n+k$ edges (since $\tau (W)$ has $n+k$ edges). Additionally, since $\tau (W)$ is simple, $T(W)$ has no vertices of degree $2$ .

Now split every vertex of $T(W)$ which has degree $>3$ ; see Figure 1 for an example of this. That is, we split $T(W)$ at every vertex corresponding to a nontriangle face of $\tau (W)$ .

Figure 1:

The polygon dissection $\tau (W_3)$ overlaid with its dual graph $T(W_3)$ (left diagram), and the forest obtained by splitting at the only vertex of degree $>3$ (right diagram).

We claim that the connected components of $T(W)$ correspond to the decomposition of $\tau (W)$ into maximal triangulated pieces and edges originally in the polygon of $\tau (W)$ . To see this, let f be the forest thus obtained. The vertices of f either have degree 1 or 3. Trees of f with no trivalent vertices correspond to either edges in the polygon of $\tau (W)$ (if they were originally leaves of $T(W)$ ) or to maximal trivial triangulated pieces. Splitting at all the faces that are not triangles ensures maximality of the decomposition. If the splitting were not maximal, then one could add a triangle to a connected component of the splitting, but this would imply that that splitting happened at a degree $3$ vertex.

To see uniqueness, consider a different maximal decomposition of $\tau (W)$ . This induces a splitting on $T(W)$ , where each connected component of the new decomposition corresponds to a subtree. Call this forest $f'$ . Since $f' \neq f$ , there are two trees t and $t'$ in f and $f'$ respectively that are distinct, but share at least one edge of $T(W)$ . Since $f'$ is also maximal, $t'$ is not a subtree of t. Therefore, the edges of $t'$ can be found in at least two trees in the forest f. In particular, there is a nonleaf vertex v in $t'$ that corresponds to a split vertex of $T(W)$ in the original decomposition. This implies that v has degree greater than $3$ in $T(W)$ , and thus the corresponding face of $\tau (W)$ is not a triangle. In other words, the piece of $\tau (W)$ corresponding to $f'$ is not a triangulation.▪

Proof Consider any two distinct maximal triangulated pieces of $\tau (W)$ . Using notation as in the previous proof, these two pieces correspond to subtrees of $T(W)$ and intersect, at most, at a vertex in the interior of $\tau (W)$ . Since the subtrees corresponding to the maximal triangulated pieces are edge disjoint, and the edges of $T(W)$ correspond to the edges of $\tau (W)$ , this forces the maximal triangulated pieces to be edge disjoint as well. ▪

Proof Let us first record a few elementary facts about polygon triangulations (that is, about triangulations with all vertices on the outer face). If such a triangulation has n vertices then it has n edges on the polygon (that is, on the outer face) and $n-3$ edges which are not. No planar simple graph with the same vertices and the same outer face can have more edges than a triangulation, and every such simple graph with $n-3$ edges off the outer face is a triangulation.Footnote ⁶

Since W is weakly admissible, by Lemma 3.1 $\tau (W)$ is a simple graph. Let t be a triangulated piece of the decomposition of $\tau (W)$ given in Lemma 3.5. Note that t cannot be equal to $\tau (W)$ by the definition of admissible diagrams.

If t has two vertices then t corresponds to a propagator that connects two nonadjacent edges. Therefore, the trivial triangulation is a trivial exact subdiagram.

Now suppose that t has $m>2$ vertices. We count how many edges of t are not on the outer face of $\tau (W)$ . These are exactly the edges of t defined by propagators of W. Consider the intersection of t with the outer face of $\tau (W)$ : this is a possibly disconnected subgraph of the polygon of $\tau (W)$ and this subgraph has m vertices. Call this subgraph S, and let j be the number of connected components of S. To join the components of S into the outer face of t, t must have j edges in its outer face which are not in the outer face of $\tau (W)$ . Furthermore, t has $m-3$ edges not in its outer face and so also not in the outer face of $\tau (W)$ . Thus, there are $m-3+j$ edges of t not in the outer face of $\tau (W)$ .

Each of these $m-3+j$ edges corresponds to a propagator in W. Call this set of propagators P. Next, we count the size of $V(P)$ . Each of the m vertices in the outer face of t corresponds to an edge of W. These m edges define j connected components of the outer polygon of W. Thus, the set $V(P)$ has $m+j$ vertices. In other words,

$$ \begin{align*}|V(P)| = m+j = |P| +3\;.\end{align*} $$

Thus, the subdiagram $(P,V(P))$ defined by t is exact.

Conversely, suppose we have an exact subdiagram $(P, V(P))$ of $W = (\mathcal {P}, [n])$ supported on $|V(P)| = |P|+3$ vertices, and let t be the subgraph of $\tau (W)$ corresponding to $(P,V(P))$ .

Suppose $|P|=1$ , and let p be the single propagator belonging to P. The exactness condition on $(P, V(P))$ says that the four supporting vertices of p are distinct. If the support of p is four consecutive vertices, then $V(p)$ defines three consecutive boundary edges of W, so t is a single triangle, hence a triangulated piece. If the support of p is not four consecutive vertices, then the vertices which are the ends of t are separated by at least two vertices along the cycle. This implies that t is a trivial triangulated piece.

Now suppose $|P|>1$ . Let $j=|P|$ , $m= j+3=|V(P)|$ , and suppose that the set $V(P)$ defines c disjoint cyclic intervals of $[n]$ . Then t has $m-c$ vertices. If t were a triangulation, then t would have $j -c$ internal edges, so we calculate the number of internal edges of t.

The graph t has j edges that come from propagators, and $m - 2c$ edges that come from the boundary polygon of $\tau (W)$ . Since t has $m-c$ vertices, it has $m-c$ external edges, of which c come from propagators. Therefore, of the j edges of t that come from propagators, $j-c$ are internal to the connected component. Therefore, t is a triangulated piece. ▪

Proof Combining Lemmas 3.5 and 3.8 yields the unique decomposition into maximal exact subdiagrams, and Corollary 3.6 ensures that no propagator appears in more than one subdiagram in this decomposition. Since the diagonal edges of $\tau (W)$ correspond to the propagators of W, the decomposition of $\tau (W)$ induces a partition of $\mathcal {P}$ . ▪

Proof This follows from the definition of equivalent Wilson loop diagrams (Definition 2.7), the definition of retriangulations (Definition 3.4), and Lemma 3.8. ▪

Proof (1) If F is an independent flat, then $C = \emptyset $ and the statement is trivially true. Now suppose that F is a dependent set, so C is nonempty.

Let $v \in C$ . Clearly $\text {Prop}(v) \subseteq \text {Prop}(C)$ , and hence $F(\textrm {Prop}(v)) \subseteq F(\textrm {Prop}(C))$ (this is easily verified by applying the definition and simplifying out all of the complements). Since $v \in F(\textrm {Prop}(v))$ by the definition of propagator flat, it follows that $v \in F(\textrm {Prop}(C))$ . Thus, $C \subseteq F(\textrm {Prop}(C))$ .

Now suppose there exists some $w \in F(\text {Prop}(C)) \setminus C$ . Let B be an independent subset of C of maximal rank. We first show that $B \cup \{w\}$ is a dependent set in $M(W)$ . By Lemma 2.7, we have

(3.1) $$ \begin{align} |\text{Prop}(C)| = \text{rk} (C) = |B| = \text{rk} (B) .\end{align} $$

Note that $\text {rk} (B) \leq |\text {Prop}(B)|$ , and $B \subseteq C$ implies that $\textrm {Prop}(B) \subseteq \textrm {Prop}(C)$ . Combining these facts with equation (3.1), it follows that $\textrm {Prop}(B) = \textrm {Prop}(C)$ . Combining this with the fact that $w \in F(\text {Prop}(C))$ , we have $\text {Prop}(B\cup \{w\}) \subseteq \text {Prop}(B)$ . We can now say something about $\text {rk} (B\cup \{w\})$ : by Lemma 2.7, we have

$$ \begin{align*} \text{rk} (B \cup \{w\}) \leq \min\{|B\cup \{w\}|, |\text{Prop}(B\cup \{w\})|\} \leq \min\{|B|+1,|\text{Prop}(B)|\}, \end{align*} $$

which becomes

$$ \begin{align*}\text{rk} (B\cup \{w\}) \leq \min\{|B|+1,|B|\} = |B| < |B \cup\{w\}|\end{align*} $$

via equation (3.1). Therefore, $B\cup \{w\}$ is a dependent set, which implies that $B\cup \{w\}$ contains a circuit D. In particular, we must have $w \in D$ (since B was an independent set), and since we have already shown that $C \subseteq F(\text {Prop}(C))$ , we also have the chain of inclusions

$$ \begin{align*}D \subseteq B\cup \{w\} \subseteq F(\text{Prop}(C)) \subseteq F.\end{align*} $$

Since C was the union of all circuits in F, we finally obtain $w \in C$ , which is a contradiction. This completes the proof of (1).

For part (2), first note that $F \setminus C$ is automatically independent as it contains no circuits.

Note that since $F(\emptyset )$ has rank $0$ , $F(\emptyset )$ is contained in every flat of W. Therefore, if $F(\emptyset ) \neq \emptyset $ , then $F\setminus C$ cannot be a flat.

Now suppose that $F(\emptyset ) = \emptyset $ . We must show that $F\setminus C$ is a flat. For any $e \not \in F$ we certainly have $\text {rk} ((F\setminus C)\cup \{e\}) = \text {rk} (F\setminus C) +1$ , since F is a flat. Now let $e \in C$ , and suppose that $\text {rk} ((F\setminus C)\cup \{e\}) = \text {rk} (F\setminus C)$ . This implies that $(F\setminus C)\cup \{e\}$ is dependent, and hence contains a circuit. Since $F(\emptyset ) = \emptyset $ this circuit must contain at least two elements: one of these elements is e, but any others must come from $F \setminus C$ . This contradicts the fact that C was the union of all circuits in F.

Thus, $\text {rk} ((F\setminus C)\cup \{e\}) = \text {rk} (F\setminus C) + 1$ for any $e \not \in F\setminus C$ , and hence $F\setminus C$ is a flat. ▪

Proof By Lemma 3.11 the set $C = F(\text {Prop}(C))$ is a propagator flat, and $S := F \setminus C$ is independent. ▪

Proof First note that if $(P,V(P))$ is exact then the admissibility of W guarantees that $V(P) \neq [n]$ , so $F(P^c) = V(P)^c$ is not empty.

We know from Lemma 2.7 that $F(P^c)$ is always a flat, so for item 1 we only need to show that $F(P^c)$ is dependent.

Since W is admissible, we have $n \geq |\mathcal {P}| + 4$ . Rewriting this inequality as

$$ \begin{align*}|V(P)| + |F(P^c)| \geq |P| + |P^c| +4,\end{align*} $$

and combining it with the fact that $|V(P)| = |P| + 3$ (from the exactness of $(P,V(P))$ ), we obtain

(3.2) $$ \begin{align}|F(P^c)|> |P^c| \;.\end{align} $$

Equation (3.2) implies that $F(P^c)$ supports fewer propagators than the number of vertices it contains, i.e., $F(P^c)$ is not an independent set.

For item 2, suppose that $(P, V(P))$ is a maximal exact subdiagram of W. By Lemma 3.11 item 1 and Corollary 3.12 we can decompose $F(P^c)$ as

(3.3) $$ \begin{align}V(P)^c = F(P^c) = C \sqcup S = F(\text{Prop}(C)) \sqcup S,\end{align} $$

where C is the largest cyclic flat contained in $F(P^c)$ and S is an independent set. We show that $\text {Prop}(C) = P^c$ , thus forcing S to be empty and implying that $F(P^c) = C$ , i.e., $F(P^c)$ is a cyclic flat. Note that C must be nonempty since $F(P^c)$ is nonempty and dependent (by item 1).

By similar arguments to Lemma 3.11, the fact that S is independent and that C is the union of all circuits in $F(P^c)$ implies that every element of S is independent of C. Further, we know that $\text {rk} (C) = |\textrm {Prop}(C)|$ from Lemma 2.7(2), and $\textrm {rk} (S) = |S|$ from Lemma 2.7 item 1 and the independence of S. Combining these gives

(3.4) $$ \begin{align} \text{rk} \big(F(P^c)\big) = \textrm{rk} (C) + \textrm{rk} (S) = |\textrm{Prop}(C)| + |S| \leq |\textrm{Prop}(F(P^c))|\;, \end{align} $$

where the inequality follows from the bound on ranks of vertex sets in Lemma 2.7 item 1. For ease of future notation, write $Q = \text {Prop}(C)^c$ .

From Remark 2.2 it follows that $\text {Prop} (F(P^c)) \subseteq P^c$ , and so equation (3.4) implies

$$ \begin{align*} |Q^c| + |S| \leq |P^c| \;.\end{align*} $$

Rearranging and adding $|P|$ to both sides gives

(3.5) $$ \begin{align} |P| + |S| \leq |P| + |P^c| - |Q^c| = |Q| \;.\end{align} $$

Meanwhile, by taking complements of equation (3.3) it follows that

$$ \begin{align*} V(P) \sqcup S = C^c = F(\text{Prop}(C))^c = V(Q)\;. \end{align*} $$

Combining this with equation (3.5) gives

(3.6) $$ \begin{align} |V(Q)| = | V(P) | + |S| = | P | + 3 + |S| \leq |Q| +3 \;. \end{align} $$

Since W is admissible this cannot be a strict inequality, so we have $|V(Q)| = |Q| +3$ and the subdiagram $(Q, V(Q))$ is exact.

Since $\text {Prop}(C) \subseteq \textrm {Prop}(F(P^c)) \subseteq P^c$ , it follows that $P \subseteq \textrm {Prop}(C)^c = Q$ . In other words, $(P, V(P))$ is a subdiagram of $(Q, V(Q))$ . Since $(P, V(P))$ is a maximal exact subdiagram by assumption, we must have $P=Q$ . Recalling that $Q = \textrm {Prop}(C)^c$ , we finally obtain $\text {Prop} (C) = P^c$ and hence $S = \emptyset $ . This completes the proof of item 2.

For item 3, first note that when $(P,V(P))$ is maximal exact, item 2 of this Lemma combined with Lemma 2.7 gives that

$$ \begin{align*}\text{rk} (F(P^c)) = |\text{Prop}(F(P^c))| = |P^c|.\end{align*} $$

To prove the result in the general case, let $(R,V(R))$ be an exact subdiagram of W that is not maximal and let $(P, V(P))$ be the maximal exact subdiagram containing it. That is, $R \subsetneq P$ .

Since $R \subsetneq P$ we can write

(3.7) $$ \begin{align}V(P) = V(R) \sqcup S, \quad \text{ where } S := V(P) \setminus V(R).\end{align} $$

Since P and R both define exact subdiagrams, it follows that $|V(P)| = |V(R)| + |P \setminus R|$ , and hence that S is a vertex set of size $|P \setminus R|$ . Furthermore, by taking complements of both sides of equation (3.7), we see that $F(R^c) = F(P^c) \sqcup S$ .

We will show that S is an independent set in the subdiagram $(P, V(P))$ , and thus in W. This will imply that for any maximal independent set $T \subseteq F(P^c)$ , the set $T \sqcup S$ is independent in $F(R^c)$ , and hence that

$$ \begin{align*} \text{rk} (F(R^c)) \geq |T| + |S| = \text{rk} (F(P^c)) + |S| = |P^c| + |P\setminus R| = |R^c|.\end{align*} $$

The reverse inequality $\text {rk} (F(R^c)) \leq |R^c|$ always holds (see Lemma 2.7), so if S is independent in $(P,V(P))$ we obtain $\text {rk} (F(R^c)) = |R^c|$ as desired.

It remains to show that S is an independent set in the subdiagram $(P, V(P))$ . By way of contradiction suppose otherwise. Then by Theorem 2.6 there is a subset $U \subseteq S$ such that $|\text {Prop}(U)| < |U|$ . Then the remaining propagators $P \setminus \text {Prop}(U)$ are supported entirely on $V(P) \setminus U$ , i.e.,

$$ \begin{align*} V(P \setminus \text{Prop}(U)) \subseteq V(P) \setminus U\; .\end{align*} $$

Comparing sizes of these sets, we get

$$ \begin{align*} |V(P \setminus \text{Prop}(U))| \leq |V(P)| - |U| & = |P| +3 -|U| \\ & < |P| +3 -|\text{Prop}(U)| = |P \setminus \text{Prop}(U)| +3 ,\end{align*} $$

which violates admissibility. Therefore, S must be independent in $(P, V(P))$ after all, and (as described above) it follows that $\text {rk} (F(R^c)) = |R^c|$ . ▪

Proof This follows from Lemma 3.14. Lemma 3.13 shows that the supports of exact subdiagrams satisfy the conditions of this lemma. ▪

Proof It follows directly from the definitions that a matroid of rank r is uniform if and only if all circuits have rank r. We therefore focus on the circuits of $M(W')$ .

We prove the following claim: $W'$ is exact if and only if $V(P)$ contains no circuits C with $\text {rk} (C)< |P|$ in $(P, V(P))$ . Since $\text {rk} (M(W'))$ is bounded above by $|P|$ , this will give that $M(W')$ is uniform.

First suppose $C \subseteq V(P)$ is a circuit of rank $m < |P|$ . Observe that $\text {Prop}_{W'}(C)\subseteq P$ by definition, where the subscript to $\text {Prop}$ specifies the diagram we are working in. By Lemma 2.7 we know that ${|\text {Prop}_{W'}(C)| = m}$ . Thus, the set $P \setminus \text {Prop}_{W'}(C)$ must be nonempty, and we can consider the subdiagram $W":= (P\setminus \text {Prop}_{W'}(C),V(P\setminus \text {Prop}_{W'}(C))$ . By the density condition on subdiagrams of admissible diagrams, we have

$$ \begin{align*}|V(P\setminus\text{Prop}_{W'}(C))| \geq |P\setminus\text{Prop}_{W'}(C)| + 3.\end{align*} $$

It is easy to verify that $V(P\setminus \text {Prop}_{W'}(C)) \subseteq V(P)\setminus C$ . Since $C \subseteq V(P)$ and $\textrm {Prop}_{W'}(C) \subseteq P$ , it follows from the previous inequality that

$$ \begin{align*}|V(P)| - (m+1) \geq |V(P\setminus\text{Prop}_{W'}(C))| \geq |P| - m + 3.\end{align*} $$

Simplifying, we obtain $|V(P)| \geq |P| + 4$ , i.e., $(P,V(P))$ is not an exact diagram.

Conversely, suppose that $W'$ is not exact and for a contradiction suppose also that $M(W')$ is uniform of rank $|P|$ . Take $p \in P$ . Then $|V(P)\setminus V(p)| = |V(P)| - 4 \geq |P|$ by nonexactness. By uniformity there must be an independent set of size $|P|$ in $V(P)\setminus V(p)$ . However, this is impossible because the submatrix of $C(W')$ corresponding to this independent set has $|P|$ rows but the one corresponding to p is all $0$ , so it cannot be full rank. ▪

Proof Since $W'$ is weakly admissible, by Theorem 2.6 the associated matroid $M(W')$ can be realized by an element of the totally non-negative Grassmannian $\mathbb {G}_{\mathbb {R}, \geq 0}(|P|,|V(P)|)$ . However, since $W'$ is an exact subdiagram its corresponding matroid $M(W')$ is uniform, so all $|P| \times |P|$ minors of $C(W')$ must be nonzero and $M(W')$ can be represented by a matrix with all maximal minors strictly positive. Since the unique top dimensional cell of $\mathbb {G}_{\mathbb {R}, \geq 0}(|P|, |V(P)|)$ is defined by precisely those points in $\mathbb {G}_{\mathbb {R}, \geq 0}(|P|, |V(P)|)$ with the property that all Plücker coordinates are strictly greater than $0$ , this completes the proof. ▪

Proof One direction has been proved in [Reference Agarwala and Marin-Amat5, Theorem 1.18], but we give a different proof here to be consistent with the method of this paper.

For the if direction, it suffices to prove the result for two equivalent diagrams that differ in exactly one exact subdiagram. Let W and $W'$ be equivalent Wilson loop diagrams which can be decomposed as

$$ \begin{align*}W = (P \sqcup R, [n]) \qquad \text{and} \qquad W' = (P \sqcup R', [n]),\end{align*} $$

where $(R,V(R))$ and $(R',V(R'))$ are different maximal exact subdiagrams of W and $W'$ respectively, and $V(R) = V(R')$ .

By Theorem 3.17, $M(R, V(R))$ and $M(R',V(R'))$ are both uniform matroids of rank $|R| = |R'|$ . In particular, this means that any subset of $V(R)$ is independent in both $M(W)$ and $M(W')$ .

Now consider the propagator flat $F(P)$ , which is a cyclic flat of rank $|P|$ in both $M(W)$ and $M(W')$ by Lemma 3.13. Let $B \subseteq F(P)$ be an independent set of rank $|P|$ . Adding any element of $V(R)$ to B increases its rank in both matroids since $F(P)$ is a flat and $F(P) \cap V(R) = \emptyset $ , so we can lift any independent set in $F(P)$ to an independent set of both $M(W)$ and $M(W')$ by adjoining any independent set in $V(R)$ . In particular, $B\sqcup U$ is a basis in both $M(W)$ and $M(W')$ for any $U \subseteq V(R)$ with $|U| = |R|$ .

Finally, note that any basis of $M(W)$ or $M(W')$ must be of this form, since $[n] = F(P) \sqcup V(R)$ and $\text {rk} F(P) + \textrm {rk} V(R) = |P| + |R| = n$ . Thus, both matroids have the same set of bases, proving that they are the same.

For the converse, suppose that $M(W) = M(W') = M$ . This immediately implies that $|\mathcal {P}| = |\mathcal {P}'|$ , so to demonstrate the equivalence of W and $W'$ it suffices to show that for any maximal exact subdiagram $(P,V(P))$ in W, there is a corresponding maximal exact subdiagram $(P',V(P'))$ in $W'$ with $V(P) = V(P')$ .

So let $(P,V(P))$ be a maximal exact diagram in W. By Lemma 3.13, this implies that $F:=F(P^c)$ is a cyclic flat in M. Viewing M as the matroid of $W'$ , Lemma 3.11 implies that $F = F(\text {Prop}_{W'}(F))$ (recall that “cyclic” simply means “a union of circuits,” so the union of all circuits in F is F itself). Define $Q^c = \text {Prop}_{W'}(F)$ , so that $F(P^c) = F = F(Q^c)$ , and therefore, $V(P) = V(Q)$ .

By Lemma 3.13 we know that $\text {rk} (F) = |P^c|$ . However, by Lemma 2.7 we also know that the rank of a cycle is equal to the number of propagators supported on that cycle, so $\textrm {rk} (F) = |\textrm {Prop}_{W'}(F)| = |Q^c|$ as well. Since $(P,V(P))$ is exact and we have shown that $V(P) = V(Q)$ and $|P| = |Q|$ , it follows that $|V(Q)| = |Q| + 3$ , i.e., $(Q,V(Q))$ is exact in $W'$ .

However, $(Q,V(Q))$ is not a priori maximal exact, so let $(P',V(P'))$ be the maximal exact subdiagram in $W'$ that contains $(Q,V(Q))$ as a subdiagram (possibly trivially). Applying the argument above to $(P',V(P'))$ , we obtain an exact subdiagram $(R,V(R))$ in W with $V(R) = V(P')$ and $|R| = |P'|$ . Now we have

$$ \begin{align*}V(P) = V(Q) \subseteq V(P') = V(R),\end{align*} $$

which implies that $(P,V(P)) = (R,V(R))$ by the maximality of $(P,V(P))$ , and hence $(Q,V(Q)) = (P',V(P'))$ is maximal exact in $W'$ .▪

Proof Let W and X be two equivalent Wilson loop diagrams. From Corollary 3.10, we know that $\tau (W)$ and $\tau (X)$ differ from each other only by retriangulations of their triangulated pieces.

All edges of $A_n$ are polygon dissections which are only one diagonal away from being triangulations. Such dissections consist of triangles and exactly one quadrilateral. Consider the edges of $A_n$ bounding the faces corresponding to $\tau (W)$ and $\tau (X)$ . These edges have their one quadrilateral within the nontriangulated parts of $\tau (W)$ or $\tau (X)$ , but these nontriangulated parts are the same in $\tau (W)$ and $\tau (X)$ , so we get a bijection between the edges of $A_n$ bounding the face of $\tau (W)$ and the edges of $A_n$ bounding the face of $\tau (X)$ which matches edges with the same quadrilateral.

Consider now an edge in $A_n$ whose quadrilateral is defined by the vertices of $\tau (W)$ or $\tau (X)$ , which we label $x_i$ , $x_j$ , $x_k$ , and $x_l$ in that order (see Figure 3 below). Let $t_1$ and $t_2$ be the triangulations corresponding to the two ends of the edge where $t_1$ has a diagonal from $x_j$ to $x_l$ and $t_2$ has a diagonal from $x_i$ to $x_k$ . From the construction of the secondary polytope $A_n$ in Definition 4.2, we see that $s_{t_1}$ and $s_{t_2}$ only differ at coordinates i, j, k, and l: all of the triangles incident to any of these vertices other than the one coming from triangulating $x_i, x_j, x_k, x_l$ are the same in both $t_1$ and $t_2$ .

Figure 3:

The shaded regions are triangulated regions which are the same in $t_1$ and $t_2$ . Consider $s_{t_1}$ and $s_{t_2}$ at coordinate i. In $s_{t_1}$ this coordinate is given by the area of those shaded triangles incident to $x_i$ (denoted by $b_i$ in the proof) along with $a(i,j,l)$ . In $s_{t_2}$ this coordinate is given by $b_i$ along with $a(i,l,k) + a(i,j,k) = a$ .

Let a be the area of the quadrilateral and let $a(i,j,k)$ be the area of the triangle with corners $x_i$ , $x_j$ and $x_k$ (and similarly for other indices), and let $b_i$ denote the sum of the areas of triangles incident to vertex $x_i$ which are common to both triangulations. Then $s_{t_1}$ on coordinates i, j, k, l is

$$ \begin{align*}(a(i,j,l) + b_i, a+b_j, a(k,l,j)+b_k, a+b_l),\end{align*} $$

and $s_{t_2}$ on these coordinates is

$$ \begin{align*}(a + b_i, a(j,k,i)+b_j, a+b_k, a(l,i,k)+b_l).\end{align*} $$

This is illustrated in Figure 3.

Thus a vector parallel to this edge is the vector which is

(4.1) $$ \begin{align} \begin{gathered} (a(i,j,l) - a, a-a(j,k,i), a(k,l,j)-a, a-a(l,i,k)) \\= (-a(j,k,l), a(i,k,l), -a(i,j,l), a(i,j,k))\end{gathered}\end{align} $$

on i, j, k, l, and $0$ in all other coordinates.

Returning to $\tau (W)$ and $\tau (X)$ , since the direction vector of an edge in $A_n$ depends only on its quadrilateral and the bijection between the edges bounding $\tau (W)$ and the edges bounding $\tau (X)$ preserves the quadrilateral, we have the same direction vectors bounding $\tau (W)$ as $\tau (X)$ . Furthermore, the bijection also maintains the relative configuration of the edges and so the edges corresponding to $\tau (W)$ and $\tau (X)$ are parallel. ▪

Proof Since $V_{>3}(D_1) = V_{>3}(D_2)$ , the vertices of the nontriangle faces of $D_1$ and $D_2$ are the same, but $D_1$ and $D_2$ do not have the same set of nontriangle faces. Suppose that no diagonal of $D_1$ crosses through a nontriangle face of $D_2$ . Then all nontriangle faces of $D_2$ must lie strictly within faces of $D_1$ . Since the nontriangle faces of the two dissections are different at least one of these containments must be proper, and so some diagonal bounding one of the nontriangle faces of $D_2$ must cross through a nontriangle face of $D_2$ . Therefore, swapping $D_1$ and $D_2$ if necessary, there must be some diagonal d of $D_1$ that crosses through a nontriangle face f of $D_2$ . If f has degree at least 5 then it must contain four vertices in the configuration described in the statement.

Suppose no f of degree at least 5 crossed by a diagonal of the other dissection occurs in either $D_1$ or $D_2$ . Then all the nontriangle faces of $D_2$ which are crossed by a diagonal of $D_1$ must be quadrilaterals and likewise for $D_1$ . This means that any nontriangle and nonquadrilateral faces must be common between $D_1$ and $D_2$ . These common faces are irrelevant for the statement as none of them can be the face containing $x_i, x_j, x_k, x_l$ , and so by triangulating any common nontriangle faces, we may assume that $D_1$ and $D_2$ have only quadrilateral and triangle faces.

Assume for a contradiction that no quadrilateral as in the statement exists, and let d continue to be a diagonal in $D_1$ crossing a nontriangle face f of $D_2$ . Then, f must have two vertices strictly on each side of d. We name the vertices on one side of d as $x_i$ and $x_j$ , and those on the other side as $x_k$ , $x_l$ in that order; see Figure 5. Let g be the face in $D_1$ that is incident to d on the $x_i, x_j$ side of d.

Figure 5:

The three possible configurations in the final part of the proof of Lemma 4.2. The thin lines come from the polygon dissection with f as a face while the thick lines come from the polygon dissection with d as an edge.

We first show that none of the vertices defining g can lie in the cyclic interval $[x_i,x_j]$ . Indeed, if g is a triangle then we immediately see that its third vertex cannot lie in the interval $[x_i,x_j]$ , as we could obtain a configuration as described in the statement by replacing d with one of the other edges of g.

Similarly, if g is a quadrilateral then by the same argument none of its vertices can lie in the interval $(x_i, x_j)$ , nor can we have exactly one of $x_i$ or $x_j$ being a vertex of g for the same reason. If $x_i$ and $x_j$ are both vertices of g (so g is the quadrilateral defined by $x_i$ , $x_j$ , and the two endpoints of d) then we still obtain a configuration as described in the lemma statement by swapping the roles of f and g: now the four vertices would be the ones defining g, and the diagonal would be one of the edges of f. (For instance d can now be the edge connecting $x_l$ to $x_i$ or $x_j$ to $x_k$ .) Thus, we can conclude that none of the vertices of the face g are in the interval $[x_i, x_j]$ .

There are now two possibilities for g: either it has one vertex in the interval $(x_j, x_k)$ and the rest in $(x_l, x_i)$ (or vice versa), or it has two vertices in $(x_j, x_k)$ and two in $(x_l, x_i)$ . In the first case, g must be a triangle since if g was a quadrilateral then its four vertices provide the desired configuration (with the edge of f that connects $x_j$ and $x_k$ providing the desired diagonal). In the second case, g has four vertices and is therefore a quadrilateral.

These cases are illustrated in Figure 5. If g is a triangle then let $d'$ be the other diagonal across f (leftmost two cases in the figure), while if g is a quadrilateral let $d'$ be the other diagonal that crosses f. In all cases, note that $d'$ is strictly closer to $[x_i,x_j]$ than d was, since we chose g to be the face incident to d on the $x_i,x_j$ side.

Now $d'$ is also a diagonal of $D_1$ that crosses the nontriangle face f of $D_2$ , and we can repeat the above argument with $d'$ instead of d. This process can be repeated indefinitely but the polygon is finite, yielding a contradiction to our assumption that no quadrilateral with the configuration described in the statement exists. ▪

Proof Let W and X be inequivalent. If $\tau (W)$ and $\tau (X)$ do not have the same number of diagonals (hence their faces in $A_n$ do not have the same dimension) then they are not parallel by definition. Now suppose that $\tau (W)$ and $\tau (X)$ do have the same number of diagonals and hence their faces in $A_n$ have the same dimension. Let the face associated to $\tau (W)$ in $A_n$ be $f_W$ and the face associated to $\tau (X)$ be $f_X$ .

As observed in the proof of Proposition 4.1, any nonzero vector in $\mathbb {R}^n$ parallel to an edge of $A_n$ has nonzero entries only in the coordinates corresponding to the four vertices bounding the quadrilateral of the dissection associated with the edge. Thus, if $V_{>3}(\tau (W)) \neq V_{>3}(\tau (X))$ (recall the notation defined before Lemma 4.2) then the span of the vectors parallel to their edges has different support and so they cannot be parallel. So, we can assume that $V_{>3}(\tau (W)) = V_{>3}(\tau (X))$ .

Since $\tau (W)$ and $\tau (X)$ do not have the same set of nontriangle faces, there must be some diagonal d of $\tau (W)$ that crosses through a nontriangle face f of $\tau (X)$ . It will be most convenient to consider faces in terms of their normal vectors. We make use of the construction discussed on page 8 of [Reference Ceballos, Santos and Ziegler11]: namely, pick one side of d to be positive and define the ith coordinate of the vector $\omega ^+_d$ to be

$$ \begin{align*} \omega^+_d(i) = \begin{cases} \text{dist}(x_i, d) & \text{if}\ x\ \text{is on the positive side of}\ d,\\ 0 & \text{otherwise.} \end{cases} \end{align*} $$

We have from [Reference Ceballos, Santos and Ziegler11] (see the proof of Proposition 3.5 therein) that $\omega ^+_d$ is a normal vector for any face with d as a diagonal. However, we can also check this fact in a completely elementary way using the shoelace formula for the area of a triangle. The shoelace formula says that if $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ are the coordinates of the corners of a triangle in counterclockwise order then the area of the triangle is

$$ \begin{align*}\textstyle\frac{1}{2}(x_1y_2 - x_1y_3 + x_2y_3 - x_2y_1 + x_3y_1 - x_3y_2).\end{align*} $$

We now check that $\omega ^+_d$ is a normal vector for any face associated to a dissection including d, which we accomplish by showing it is normal to every edge bounding the face. To do so, choose the coordinate system for the polygon so that d is along the x-axis. Let $(a_i, h_i)$ , $i\in \{1,2,3,4\}$ be the coordinates of the vertices of a quadrilateral in counterclockwise order and which corresponds to an edge bounding the face. Since the dissection includes d, either all four points are on the negative side of d and so the dot product of $\omega ^+_d$ with the direction vector for the quadrilateral is immediately $0$ , or, by the formula in equation (4.1) and the shoelace formula, the dot product of $\omega ^+_d$ with the vector associated to the quadrilateral is

$$ \begin{align*} & \textstyle\frac{1}{2}h_1(a_2h_3-a_2h_4+a_3h_4-a_3h_2+a_4h_2-a_4h_3) \\ & \quad - \textstyle\frac{1}{2} h_2(a_1h_3-a_1h_4+a_3h_4-a_3h_1+a_4h_1-a_4h_3) \\ & \quad + \textstyle\frac{1}{2}h_3(a_1h_2-a_1h_4+a_2h_4-a_2h_1+a_4h_1-a_4h_2) \\ & \quad - \textstyle\frac{1}{2}h_4(a_1h_2-a_1h_3+a_2h_3-a_2h_1+a_3h_1-a_3h_2), \end{align*} $$

which also evaluates to 0.

Now, $\tau (W)$ and $\tau (X)$ satisfy the hypotheses of Lemma 4.2, so we can assume that f and d have the property that we can choose a side of d to be the positive side in such a way that exactly one vertex of f is strictly on the positive side. Then $\omega ^+_d$ and the direction vector of d have common support exactly on that one vertex and so their dot product is nonzero. Since the direction vector of d is contained in the direction of $\tau (X)$ but is not orthogonal to $\omega ^+_d$ , while $\omega ^+_d$ is normal to the face of $\tau (W)$ , it follows that the faces associated to $\tau (W)$ and $\tau (X)$ are not parallel, as desired. ▪