Proof. If p and q are paths in a path-consistent dimer model Q with $h(p)=t(q)$ , then we may write $[p]=[f^{m_p}r_p]$ and $[q]=[f^{m_q}r_q]$ for some minimal paths $r_p$ and $r_q$ . Then $m_p$ is the c-value of p and $m_q$ is the c-value of q. Then, using the fact that $[f]$ is central, we calculate

$$\begin{align*}[qp]=[f^{m_q}r_qf^{m_p}r_p]=[f^{m_q+m_p}r_qr_p].\end{align*}$$

We have shown that $[f^{m_q+m_p}]$ may be factored out of $[qp]$ ; hence, the c-value of $[qp]$ is greater than or equal to $m_q+m_p$ .

Proof. Suppose p is a proper subpath of a face-path $f_v$ starting at $v:=t(p)$ . Let $p'$ be the subpath of $f_v$ such that $p'p=f_v$ . If p is not minimal, then by definition of path-consistency, $[p]=[rf^m]$ for some positive integer m and some minimal path r from v to $h(p)$ homotopic to p. Then $[p'p]=[p'rf^m]=[p'rf_v^m]$ . Moreover, r is homotopic to p, and hence, $p'r$ is homotopic to the face-path $p'p$ and hence is null-homotopic. Then by Remark 2.10, it has some positive c-value of $m'$ . By definition of path-consistency, $[p'r]=[f_v^{m'}]$ . It follows that

$$\begin{align*}[f_v]=[p'p]=[p'rf_v^m]=[f_v^{m'}f_v^m]=[f_v^{m'+m}],\end{align*}$$

which is a contradiction since $m+m'\geq 1+1=2$ but all face-paths trivially have a c-value of 1. It follows that p is minimal.

Proof. Suppose p is a proper subpath of a face-path of Q which is a cycle. By Lemma 2.11, the path p is minimal. Since Q is simply-connected, p is null-homotopic. The only minimal null-homotopic path from a vertex to itself is the constant path, so this is a contradiction.

Proof. Suppose Q is path-consistent. Choose vertices $\tilde v_1$ and $\tilde v_2$ of $\widetilde Q$ ; these correspond to vertices $v_1$ and $v_2$ of Q and induce a homotopy class C of paths between them. By Remark 2.13, the paths in $\widetilde Q$ from $\tilde v_1$ to $\tilde v_2$ correspond to the paths in Q from $v_1$ to $v_2$ in C. By path-consistency of Q, each such path in Q is equivalent to $p_{v_2v_1}^C$ composed with some power of a face-path, hence $\widetilde Q$ is path-consistent.

The other direction is similar.

Proof. This follows because $[p]=[q]$ in $A_Q$ if and only if $[\tilde p]=[\tilde q]$ in $A_{\widetilde Q}$ , where $\tilde p$ and $\tilde q$ are any lifts of p and q to $\widetilde Q$ with $t(\tilde p)=t(\tilde q)$ .

Proof. Let $p=\gamma _m\dots \gamma _1$ be a product of arrows. For each $\gamma _i$ , let $R_{\gamma _i}$ be a return path of $\gamma _i$ . The path $R_{\gamma _1}\dots R_{\gamma _m}\gamma _m\dots \gamma _1$ is equivalent to $f^m_{t(p)}$ and begins with p.

Proof. By Lemma 2.16 and Proposition 2.14, it suffices to show the result on the universal cover $\widetilde Q$ .

Suppose that $\widetilde Q$ is path-consistent. We prove that $A_{\widetilde Q}$ is a cancellation algebra. Accordingly, take paths $p,q,a$ of ${\widetilde Q}$ with $h(a)=t(p)=t(q)$ and $h(p)=h(q)$ . We show that $[pa]=[qa]\implies [p]=[q]$ . The case of left composition is symmetric. By path-consistency, we may write $[p]=[rf^{m_p}]$ and $[q]=[rf^{m_q}]$ , where r is a minimal path from $t(p)$ to $h(p)$ , necessarily homotopic to p and q by simple connectedness. Given $[pa]=[qa]$ , we have $[f^{m_p}ra]=[f^{m_q}ra]$ . Then $m_p=m_q$ by path-consistency. We have shown that

$$\begin{align*}[q]=[rf^{m_q}]=[rf^{m_p}]=[p],\end{align*}$$

and the proof of this direction is complete.

Suppose now that $A_{\widetilde Q}$ is a cancellation algebra. We first show that only a finite number of face-paths may be factored out of any path p, and that this number is bounded by the number of arrows in p. Suppose to the contrary that there is some path p of Q with m arrows such that we may write $[p]=[p'f^{m'}]$ for some $m'>m$ . By Lemma 2.17, $[p'f^{m'}]=[lp]$ for some nonconstant cycle l at $h(p)$ . Applying the cancellation property to the equation $[p]=[p'f^{m'}]=[lp]$ gives that l is equivalent to the constant path, which is a contradiction. This shows that only a finite number of face-paths may be factored out of any path of ${\widetilde Q}$ .

Then any path p of ${\widetilde Q}$ is equivalent to $rf^m$ for a minimal path r and a nonnegative integer m. Suppose $[p]=[rf^m]=[r'f^{m'}]$ for some nonnegative integers m and $m'$ and minimal paths r and $r'$ . Without loss of generality, suppose $m\leq m'$ . By the cancellation property, $[r]=[r'f^{m'-m}]$ . Then if $m'>m$ , we have factored a face-path out of r, contradicting minimality of r; hence, $m'=m$ and $[r]=[r']$ .

Then if ${\widetilde Q}$ is not path-consistent, there must be minimal paths p and q between the same vertices which are not equivalent. Take m which is greater than the length of p and the length of q. By Lemma 2.17, $[pf^m]=[pq'q]$ for some path $q'$ from $h(q)$ to $t(q)$ . Suppose we have shown that $pq'$ is equivalent to $f^{m'}$ for some $m'$ . Then

$$\begin{align*}[pf^m]=[pq'q]=[qf^{m'}].\end{align*}$$

By the cancellation property, we get either $[p]=[qf^{m'-m}]$ or $[q]=[pf^{m-m'}]$ . Since p and q are minimal, we have $m=m'$ and $[p]=[q]$ , contradicting our initial assumption. The proof is then complete if we show that any cycle is equivalent to a composition of face-paths. We do so now.

Suppose to the contrary and take a simple cycle $l=\gamma _{s'}\dots \gamma _1$ which is not equivalent to a composition of face-paths and such that every simple cycle inside the disk bounded by l is equivalent to a composition of face-paths. Note that if $\widetilde {S(Q)}$ is a sphere, then choose one of the two regions that l bounds as being the disk. For any arrow $\gamma _i\in l$ , let $R_{\gamma _i}$ be the return path of $\gamma _i$ inside of the disk bounded by l. As in Lemma 2.17, set $l':=R_1\dots R_{s'}$ . Then $l'l=R_1\dots R_{s'}\gamma _{{s'}}\dots \gamma _1$ is equivalent to $f_{t(l)}^{s'}$ . Moreover, $l'$ is a cycle lying in the area bounded by l. If $l'$ is a simple cycle strictly contained in l, then $l'$ is equivalent to a composition of face-paths by choice of l. If $l'$ is not a simple cycle, then one-by-one we remove simple proper subcycles of $l'$ , each of which is strictly contained in the area defined by l and hence is equivalent to a composition of face-paths. Then we replace them with a composition of face-paths until we get $[l']=[f_{t(p)}^{s}]$ for some s.

Either way, $l'$ is equivalent to some composition of face-paths $f_{t(p)}^s$ . Then $[f_{t(p)}^{s'}]=[l'l]=[f_{t(p)}^sl]$ . Since $l'$ is a subpath of $l'l$ , we must have that ${s'}\geq s$ . Then the cancellation property gives $[f_{t(p)}^{{s'}-s}]=[l]$ and l is equivalent to a composition of face-paths, contradicting the choice of l. This completes the proof that all cycles are equivalent to a composition of face-paths and yields the theorem.

Proof. If $F=F_\alpha ^{cc}$ or $F=F_\alpha ^{cl}$ , then $R_\alpha ^{cc}$ winds around F for some positive angle $0\leq \theta \leq 2\pi $ , while $R_\alpha ^{cl}$ winds around F for an angle of $\theta -2\pi $ . Left-morphing at $\alpha $ switches the former for the latter, leading to a net decrease of $2\pi $ radians. Then $\text {Wind}(m_\alpha (p)q,F)=\text {Wind}(qp,F)-1$ in this case. If F is any other face, then $R_\alpha ^{cl}$ and $R_\alpha ^{cc}$ do not wind differently around F and the winding number does not change.

Proof. Parts (1) and (3) follow from the definitions. We prove (2). Certainly, if $\omega _\alpha (p)$ is constant, then it contains no arrow of $R_\alpha ^{cc}$ . However, if $\omega _\alpha (p)$ is nonconstant, then in the notation of Definition 3.2, we must have $a<b$ . Then some arrow of $R_\alpha ^{cc}$ is contained in $R'$ and hence is contained in $\omega _\alpha (p)=q"R'q'$ .

Proof. To see (1), it suffices to reduce to the case where p and q share only their start and end vertices. In this case, they bound one disk $q^{-1}p$ and a left-morph of p at $\beta $ results in a path contained in the area enclosed by $q^{-1}p$ .

We now prove (2). We show that $\text {Wind}(\omega _\alpha (p),F)\geq \text {Wind}(p,F)$ for any face F. It follows that $\omega _\alpha (p)$ is also an elementary counter-clockwise cycle. Since any face F is in the interior of p (respectively $\omega _\alpha (p)$ ) if and only if $\text {Wind}(p,F)>0$ (respectively $\text {Wind}(\omega _\alpha (p),F)>0$ ), it further follows that if F is in the interior of p, then F is in the interior of $\omega _\alpha (p)$ and the statement is proven.

Let p be an elementary counter-clockwise cycle in $\widetilde Q$ and let $\alpha $ be a right-morphable arrow for p such that $m_\alpha (p)$ has no proper counter-clockwise subcycle. By Lemma 2.20, $\text {Wind}(m_\alpha (p),F)\geq \text {Wind}(p,F)$ . The path $\omega _\alpha (p)$ is obtained from $m_\alpha (p)$ by deleting some number of proper elementary subcycles. All of these are clockwise and hence have a winding number less than or equal to zero around F, by assumption. It follows that their deletion can only increase the winding number around F, and we have

$$\begin{align*}\text{Wind}(\omega_\alpha(p),F)\geq\text{Wind}(m_\alpha(p),F)\geq\text{Wind}(p,F).\end{align*}$$

If F is enclosed within p, then $\text {Wind}(p,F)=1$ , and the above inequality forces $\text {Wind}(\omega _\alpha (p),F)=1$ . It follows that F is enclosed within $\omega _\alpha (p)$ . This ends the proof.

Proof. Any subpath of $\omega _\alpha (p)$ which is not a subpath of p must contain some arrow $\gamma $ of $R_\alpha ^{cc}$ . Since $\omega _\alpha (p)$ does not contain $\alpha $ , the only counter-clockwise return path which $\gamma $ could be a part of is $R_\alpha ^{cc}$ . This shows that $\omega _\alpha (p)$ contains no counter-clockwise return paths which are not in p, other than possibly $R_\alpha ^{cc}$ . The result follows.

Proof. Let $b=\beta _s\dots \beta _1$ be a crossing-creating left-chain of $\omega _\alpha (p)$ . Let j be maximal such that $b':=\beta _j\dots \beta _1$ is a left-chain of p as well as $\omega _\alpha (p)$ . If $b'$ is crossing-creating for p, then we are done, so suppose $b'$ is not a crossing-creating chain of p.

Suppose first that b is a left-chain of p, and hence that $j=s$ and $b'=b$ . Since b creates a crossing of $\omega _\alpha (p)$ but not of p, there must be a proper subcycle of $m_{b}(\omega _\alpha (p))$ starting at a vertex v of $\omega _\alpha (p)$ which is not a vertex of p. Then left-morphing $m_{\beta _{s-1}\dots \beta _{1}}(p)$ at $\beta _s$ must add v to p. By Lemma 3.10 (2), the area bounded by p is contained in the area bounded by $\omega _\alpha (p)$ . In particular, vertices of $\omega _\alpha (p)$ which are not vertices of p are not in the area bounded by p. Since left-morphing p at b does not create any crossings, $p=\omega _b(\omega _b(p))$ is obtained by applying the (cycle-removing) right-chain b to $\omega _b(p)$ . Then Lemma 3.10 (2) applied to $\omega _b(p)$ shows that $\omega _b(p)$ is strictly contained in the area bounded by p. Since v is not in the area bounded by p, the vertex v cannot be in $\omega _b(p)$ . This is a contradiction.

It follows that b is not a left-chain of p and hence that $j<s$ . We claim that $R_\alpha ^{cl}\in m_{\beta _{j'}\dots \beta _{1}}(p)$ and that no arrow of $F_\alpha ^{cl}$ is in $m_{\beta _{j'}\dots \beta _{1}}(\omega _\alpha (p))$ for any $1\leq j'\leq j$ . Since $R_\alpha ^{cl}\in p$ and $\beta _1$ does not create a crossing of p, we must have $\beta _1\not \in F_\alpha ^{cl}$ . Since no arrow of $F_\alpha ^{cl}$ is in $\omega _\alpha (p)$ and $R_{\beta _1}^{cc}\in \omega _\alpha (p)$ , for any $\alpha '\in F_\alpha ^{cl}$ we must have $\beta _1\not \in F_{\alpha '}^{cc}$ . It follows that $R_\alpha ^{cl}$ is in $m_{\beta _1}(p)$ and that no arrow of $F_\alpha ^{cl}$ is in $m_{\beta _1}(\omega _\alpha (p))$ . We repeat this argument to see that $R_\alpha ^{cl}\in m_{\beta {j'}\dots \beta _{1}}(p)$ and that no arrow of $F_\alpha ^{cl}$ is in $m_{\beta {j'}\dots \beta _{1}}(\omega _\alpha (p))$ for any $1\leq j'\leq j$ , proving the claim.

First, we suppose that $m_\alpha (p)$ is not elementary and hence that $\omega _\alpha (p)$ does not contain all of $R_\alpha ^{cc}$ . In particular, either the arrow $\alpha '$ following $\alpha $ in $R_\alpha ^{cc}$ or the arrow $\alpha "$ preceding $\alpha $ in $R_\alpha ^{cc}$ (or both) are absent from $\omega _\alpha (p)$ . Suppose that $\alpha '$ is not in $\omega _\alpha (p)$ ; the $\alpha "$ case is the same. Since $b'$ is a left-chain of p and $R_\alpha ^{cl}$ is in $m_{\beta _{j'}\dots \beta _{1}}(p)$ by the claim above, the arrows $\alpha '$ and $\alpha $ may not be added by any morph in $b'$ without creating a crossing in p, which would contradict our choice of $b'$ . Hence, neither $\alpha '$ nor $\alpha $ is in $m_{\beta _j\dots \beta _1}(\omega _\alpha (p))$ . In particular, no return path of an arrow in $F_\alpha ^{cc}$ is a subpath of $m_{\beta _j\dots \beta _1}(\omega _\alpha (p))$ . Since $\beta _{j+1}$ is a left-morphable arrow for $m_{b'}(\omega _\alpha (p))$ but not of $m_{b'}(p)$ , it must be the case that $R_{\beta _{j+1}}^{cc}$ is a subpath of $m_{b'}(\omega _\alpha (p))$ but not of $m_{b'}(p)$ . Since the only such subpaths contain some arrow of $R_\alpha ^{cc}$ , we must have that $\beta _{j+1}\in F_\alpha ^{cc}$ . This contradicts the fact that no return path in $F_\alpha ^{cc}$ is a subpath of $m_{\beta _j\dots \beta _1}(\omega _\alpha (p))$ .

However, suppose that $m_\alpha (p)$ is elementary, and hence that $\omega _\alpha (p)=m_\alpha (p)$ . As above, the arrow $\beta _{j+1}$ must be in $F_\alpha ^{cc}$ . If $\beta _{j+1}\neq \alpha $ , then $\alpha \in m_{\beta _j\dots \beta _1}(\omega _\alpha (p))$ , contradicting the claim in the third paragraph, so $\beta _{j+1}=\alpha $ . Consider the chain $\beta _{j+1}\beta _j\dots \beta _1\alpha $ of p. The left-morph $\beta _{j+1}$ cancels out the right-morph $\alpha $ in this chain; hence, the left-chain $\beta _{j+2}\beta _j\dots \beta _1$ of p is equivalent to $\beta _{j+2}\dots \beta _2\beta _1\alpha $ . Then $\beta _s\dots \beta _{j+2}\beta _j\dots \beta _1$ is a crossing-creating left-chain of p. This completes the proof.

Proof. The proof is the same as that of Lemma 3.13, with l taking the place of $\omega _\alpha (p)$ .

Proof. Any arrow added by right-morphing at $\alpha $ is only a part of one counter-clockwise face, which is $F_\alpha ^{cc}$ . Since p is elementary, $\alpha $ is not in p and hence is not in $m_\alpha (p)$ . The result follows.

Proof. We induct on the c-value of p. For the base case, let p be an elementary counter-clockwise cycle with a minimal c-value among elementary counter-clockwise cycles. By path-consistency, p must be equivalent to a composition of face-paths and hence must have a crossing-creating chain. If p has a good crossing-creating chain $a=\alpha _r\dots \alpha _1$ such that $\alpha _r$ is a right-morph, then $\omega _a(p)$ is an elementary counter-clockwise cycle by Lemma 3.10 (2), and necessarily has a lower c-value than p, contradicting our choice of p. Suppose p has a good crossing-creating chain $a=\alpha _r\dots \alpha _1$ such that $\alpha _r$ is a left-morph. If a is a left-chain, then there is nothing to show; otherwise, let j be maximal such that $\alpha _j$ is a right-morph. Applying Lemma 3.13 to $m_{\alpha _1\dots \alpha _{j-1}}(p)$ shows that there is a crossing-creating left-chain of $m_{\alpha _1\dots \alpha _{j-1}}(p)$ . Repeating this process for each right-morph of a gives that there is a crossing-creating left-chain of p. If p has a bad crossing-creating chain $a=\alpha _r\dots \alpha _1$ , then $\alpha _r$ is a right-morph and $m_a(p)$ contains a proper subpath l which is a simple counter-clockwise cycle. By Lemma 3.15, l is not a face-path and hence is elementary. This contradicts our choice of p, since l has a strictly lower c-value than p. This completes the proof of the base case.

Now suppose that p is an elementary counter-clockwise cycle which does not have a minimal c-value. There must be a crossing-creating chain $c=\gamma _r\dots \gamma _1$ of p. We first show that there must be a crossing-creating chain $ab$ of p where b is a left-chain of p and a is a (possibly empty) right-chain of $m_b(p)$ . Suppose c is not already of this form and let j be minimal such that $\gamma _{j-1}$ is a right-morph in c and $\gamma _j$ is a left-morph in c.

If $\gamma _{j-1}$ and $\gamma _j$ are the same arrow, then $\gamma _j$ cannot create a crossing, and the term $\gamma _{j}\gamma _{j-1}$ may be removed from c to get an equivalent chain $\gamma _r\dots \gamma _{j+1}\gamma _{j-2}\dots \gamma _1$ . If $\gamma _{j-1}$ and $\gamma _j$ are not the same arrow, then by Lemma 3.11, $\gamma _j$ is a left-morphable arrow for $m_{\gamma _{j-2}\dots \gamma _1}(p)=\omega _{\gamma _{j-2}\dots \gamma _1}(p)$ . If this left-morph creates a crossing, then let $c_1:=\gamma _j\gamma _{j-2}\dots \gamma _1$ . Otherwise, note that $\gamma _j\gamma _{j-1}$ and $\gamma _{j-1}\gamma _{j}$ are equivalent chains of $m_{\gamma _{j-2}\dots \gamma _1}(p)$ ; hence,

$$\begin{align*}c_1:=\gamma_r\dots\gamma_{j+1}\gamma_{j-1}\gamma_j\gamma_{j-2}\dots\gamma_1\end{align*}$$

is a crossing-creating chain of p equivalent to c. We apply the above logic repeatedly to move all left-morphs in c to the front. We end up with some crossing-creating chain $c_m=ab$ , where b is a left-chain of p and a is a right-chain of $m_b(p)$ . If a is trivial, then b is a crossing-creating left-chain and we are done. Otherwise, let $a=\alpha _r\dots \alpha _1$ . Lemma 3.14 or Lemma 3.13 (depending on whether $m_{ba}(p)$ contains a proper counter-clockwise subcycle) along with the induction hypothesis shows that $m_{\alpha _{r-1}\dots \alpha _1b}(p)$ has a crossing-creating left-chain. We now repeatedly apply Lemma 3.13 to see that $m_b(p)$ has a crossing-creating left-chain $b'$ . Then $b'b$ is a crossing-creating left-chain of p.

Proof. We prove by induction on the c-value of p. The base case when p is minimal is trivial. Suppose the result has been shown for paths with a strictly lower c-value than that of some non-minimal path p. We first show that p has a good crossing-creating chain. Since p is not minimal, there is some crossing-creating chain $a=\alpha _r\dots \alpha _1$ of p. If $\alpha _r$ is a left-morph or if $m_a(p)$ contains no proper counter-clockwise subcycle, then a is a good crossing-creating chain of p. If not, then $m_a(p)$ contains a proper counter-clockwise simple subcycle l, which is not a face-path by Lemma 3.15. By Proposition 3.16, l has a crossing-creating left-chain. By Lemma 3.14, $m_{\alpha _{r-1}\dots \alpha _{1}}(p)$ has a crossing-creating left-chain $a'$ . Then $a'\alpha _{r-1}\dots \alpha _1$ is a good crossing-creating chain of p. Then we may suppose that p has a good crossing-creating chain a. Since $\omega _a(p)$ is an elementary path with a lower c-value than p, by the induction hypothesis there is a good cycle-removing chain $a"$ from $\omega _a(p)$ to a minimal path. Then $a"a$ is a good cycle-removing chain from p to a minimal path.

Proof. Suppose to the contrary that $m_\alpha (r)$ enters the interior of $q^{-1}p$ . Since p is leftmost and q is rightmost, it must be the case that $\alpha $ is a right-morph at an arrow of p or a left-morph at an arrow of q. Suppose the former is true; the latter case is symmetric. The segment $r'$ of r from $t(p)$ to $h(\alpha )$ either winds right or left around $q^{-1}p$ . See Figure 6. In either case, as can be seen from Figure 6, the path $R_\alpha ^{cl}r'$ may not be completed to a path from $t(p)$ to $h(p)$ without entering the interior of $q^{-1}p$ or causing a self-intersection.

Proof. Let $\widetilde Q$ be path-consistent and simply connected dimer model that is not on a sphere. Take a pair $(p,q)$ of paths such that $q^{-1}p$ is an elementary counter-clockwise cycle-walk. We show that p has a left-morphable arrow or q has a right-morphable arrow. If p is a cycle and q is trivial, then Proposition 3.16 shows the desired result. If p is trivial and q is a cycle, the dual of Proposition 3.16 shows the desired result. We may now assume that p and q are not cycles. Suppose for the sake of contradiction that $(p,q)$ is an irreducible pair.

Choose a face F in the interior of $q^{-1}p$ . By Proposition 3.17, there is a good cycle-removing chain $a=\alpha _r\dots \alpha _1$ from p to some minimal path $\omega _a(p)$ . By repeated application of Lemma 3.18, no intermediate path of this chain enters the interior of $q^{-1}p$ . In particular, no arrow $\alpha _i$ is an arrow of F. Then Lemma 2.20 tells us that $\text {Wind}\big (q^{-1}m_{\alpha _i}(\omega _{\alpha _{i-1}\dots \alpha _{1}}(p)),F\big )=\text {Wind}\big (q^{-1}\omega _{\alpha _{i-1}\dots \alpha _{1}}(p),F\big )$ for each i. There may be some clockwise cycles removed from $m_{\alpha _i}(\omega _{\alpha _{i-1}\dots \alpha _{1}}(p))$ to get $\omega _{\alpha _i\dots \alpha _1}(p)$ , but since a is good, no counter-clockwise cycles are removed. Hence,

$$ \begin{align*} \text{Wind}\big(q^{-1}\omega_{\alpha_i\dots\alpha_1}(p),F\big) &\geq\text{Wind}\big(q^{-1}m_{\alpha_i}(\omega_{\alpha_{i-1}\dots\alpha_1}(p)),F\big)\\ &=\text{Wind}(\omega_{\alpha_{i-1}\dots\alpha_1}(p),F). \end{align*} $$

By applying this result for each i, we see that

$$\begin{align*}\text{Wind}(q^{-1}\omega_a(p),F)\geq\text{Wind}(q^{-1}p,F)=1.\end{align*}$$

Dually, there is a cycle-removing chain $b=\beta _s\dots \beta _1$ of q removing only counter-clockwise cycles such that $\omega _b(q)$ is minimal and

$$\begin{align*}\text{Wind}(q^{-1}\omega_b(q),F)\leq\text{Wind}(q^{-1}q,F)=0.\end{align*}$$

By path-consistency, $\omega _a(p)$ and $\omega _b(q)$ are equivalent. Then there is a chain $c=\gamma _t\dots \gamma _1$ such that $m_c(\omega _a(p))=\omega _b(q)$ . As above, Lemma 3.18 shows that no arrow $\gamma _i$ of c is an arrow of F; hence, repeated application of Lemma 2.20 gives that $1\leq \text {Wind}(q^{-1}\omega _a(p),F)=\text {Wind}(q^{-1}\omega _b(p),F)\leq 0$ , a contradiction.

Proof. Given a path-consistent dimer model Q, Proposition 2.14 implies that $\widetilde Q$ is path-consistent. By Proposition 4.7, it suffices to show that $\widetilde Q$ is strand-consistent. We will do this by showing that self-intersecting strands, interior cycles and bad lenses in the strand diagram give rise to irreducible pairs, which cannot occur in strand-consistent models by Theorem 3.19, since $\widetilde Q$ is not a sphere. See Figure 10.

First, suppose there is some self-intersecting strand z of $D_{\widetilde Q}$ . Recall from the discussion following Definition 4.8 that intersections of the strand z correspond to multiple incidences of an arrow in its associated zigzag path in ${\widetilde Q}$ . Then there is some segment $C=\gamma _0\gamma _t\dots \gamma _1\gamma _0$ of the zigzag path associated to z such that $\gamma _i\neq \gamma _j$ for $i\neq j$ . Suppose that the segment of the strand z corresponding to $\gamma _t\dots \gamma _0$ is a clockwise cycle; the counter-clockwise case is symmetric. See the left of Figure 10. Let $v_1:=h(\gamma _0)$ and $v_2:=t(\gamma _0)$ . Since the cycle runs clockwise, $v_1$ and $v_2$ are both zag vertices of z. Let $z':=\gamma _t\dots \gamma _1$ and let p be the elementary right return path of $z'$ . Then p is a leftmost path from $v_2$ to $v_1$ , and $\gamma _0^{-1}p$ is an elementary counter-clockwise cycle-path winding counter-clockwise around $z'$ ; hence, $(p,\gamma _0)$ is an irreducible pair. This contradicts Theorem 3.19.

Now suppose there is a strand z of $D_{\widetilde Q}$ which is an interior cycle. By the above, we may suppose that z contains no self-intersections. Then we may realize z as a path $z':=\gamma _t\dots \gamma _0$ of ${\widetilde Q}$ such that $t(\gamma _t)=h(\gamma _0)$ is a zag vertex of z and $\gamma _i\neq \gamma _j$ for $i\neq j$ . See the middle of Figure 10. As above, we assume that $z'$ winds clockwise. Let p be the elementary right return path of $z'$ . Then p winds counter-clockwise around $z'$ and thus is a nontrivial counter-clockwise path which is leftmost, contradicting Theorem 3.19.

Now, suppose there is a bad lens in $D_{\widetilde Q}$ . Accordingly, we may take subpaths of zigzag paths $z':=\beta \gamma _t\dots \gamma _1\alpha $ and $w':=\beta \delta _s\dots \delta _1\alpha $ such that $\gamma _i\neq \delta _j$ for $i\neq j$ . By the above, the strands have no self-intersections, and hence, $z'$ and $w'$ have no repeated arrows. Suppose without loss of generality that $z'$ is to the left of $w'$ . Then $t(\alpha )$ and $h(\beta )$ are zig vertices of w and zag vertices of z. See the right of Figure 10. Let p be the right elementary return path of $z'$ and let q be the left elementary return path of $w'$ .

Let $p_0:=p$ and $q_0:=q$ . Choose a face F in the interior of the bad lens. Then $p_0$ is a leftmost elementary path and $q_0$ is a rightmost elementary path such that $\text {Wind}(q_0^{-1}p_0,F)=1>0$ , since $q^{-1}p$ winds counter-clockwise around the lens. If $q_0^{-1}p_0$ is simple, then $(p_0,q_0)$ is an irreducible pair and we are done. If not, then $q_0^{-1}p_0$ has some simple proper subcycle-walk l. The paths $p_0$ and $q_0$ are elementary, so l must be of the form $(q^{\prime }_0)^{-1}p^{\prime }_0$ , where $p^{\prime }_0,q^{\prime }_0,p_1,q_1$ are paths such that $p_0=p^{\prime }_0p_1$ and $q_0=q_1q^{\prime }_0$ . If $(q^{\prime }_0)^{-1}p^{\prime }_0$ is counter-clockwise, then $(p^{\prime }_0,q^{\prime }_0)$ forms an irreducible pair, since any subpath of p is leftmost and any subpath of q is rightmost. If not, then the removal of $(q^{\prime }_0)^{-1}p^{\prime }_0$ from $q_0^{-1}p_0$ may only increase the winding number around F; hence, $\text {Wind}(q_1^{-1}p_1,F)\geq \text {Wind}(q_0^{-1}p_0,F)>0$ . We now start the process over with $p_1$ and $q_1$ in place of $p_0$ and $q_0$ . This process must eventually terminate when some $(p^{\prime }_i,q^{\prime }_i)$ forms an irreducible pair, contradicting Theorem 3.19.

Proof. By Theorem 2.18, Q is path-consistent if and only if $A_Q$ is cancellative. By Lemma 2.16, this is true if and only if $A_{\widetilde Q}$ is cancellative. Hence, it suffices to show that $A_{\widetilde Q}$ is a cancellation algebra. Accordingly, suppose $p,q,a$ are paths of $\widetilde Q$ such that $h(a)=t(p)=t(q)$ and $h(p)=h(q)$ and $[pa]=[qa]$ . Then there is a finite sequence of morphs taking $pa$ to $qa$ in $\widetilde Q$ . Let $Q'$ be a disk submodel of $\widetilde Q$ containing every intermediate path in this sequence. Since $D_{Q'}$ is a restriction of $D_{\widetilde Q}$ , which has no bad configurations, the former also has no bad configurations. By [Reference Çanakç, King and Pressland13, Proposition 2.15], $Q'$ is path-consistent. By Theorem 2.18, $A_{Q'}$ is cancellative; hence, $[p]=[q]$ in $Q'$ . Then there is a sequence of morphs taking p to q in $Q'$ ; this is necessarily a sequence of morphs in $\widetilde Q$ , so $[p]=[q]$ in $\widetilde Q$ .

It may similarly be shown that if $p,q,b$ are paths of $\widetilde Q$ such that $t(b)=h(p)=h(q)$ and $t(p)=t(q)$ , then $[bp]=[bq]$ implies $[p]=[q]$ . This completes the proof that $A_{\widetilde Q}$ is cancellation, and thus that Q is path-consistent.

Proof. Path-consistency implies strand-consistency by Theorem 4.10. Strand-consistency implies path-consistency by Proposition 4.12. Path-consistency is equivalent to cancellativity by Theorem 2.18.

Proof. We will use the dual definition of a perfect matching. We must then show that there is a set $\mathcal N$ of edges of the plabic graph $\mathcal G$ of $\widetilde Q$ such that every vertex of $\mathcal G$ is incident to exactly one edge of $\mathcal N$ .

We first claim that it suffices to show that any collection of m internal black vertices is connected to at least m white vertices and that any collection of m internal white vertices is connected to at least m black vertices. Suppose this is true. By applying Theorem 6.5 to the internal black vertices, we see that there is a matching from the set of internal black vertices into the white vertices. Symmetrically, we get a matching from the set of internal white vertices into the black vertices. Then Theorem 6.6 shows that there exists a perfect matching. This ends the proof of the claim.

We show that any collection of m internal white vertices is connected to at least m black vertices. The remaining case is symmetric. Take a set S of m internal white vertices of $\mathcal G_{\widetilde Q}$ . These correspond to m internal faces of $\widetilde Q$ . Let $Q'$ be a disk submodel of $\widetilde Q$ containing all faces of S. Such a disk submodel must exist since $\widetilde Q$ is simply connected. Since $D_{\widetilde Q}$ has no bad configurations and $D_{Q'}$ is a restriction of $D_{\widetilde Q}$ , the latter also has no bad configurations. By [Reference Çanakç, King and Pressland13, Proposition 2.15], $Q'$ is a path-consistent dimer model. By [Reference Çanakç, King and Pressland13, Corollary 4.6], $\mathcal G_{Q'}$ has a perfect matching. In particular, by Hall’s marriage theorem (Theorem 6.5), the set S of white vertices considered as vertices of $\mathcal G_{Q'}$ has at least m neighbors in $\mathcal G_{Q'}$ ; hence, S has at least m neighbors in $\mathcal G_{\widetilde Q}$ . This completes the proof.

Proof. Let C be a collection of all perfect matchings on Q. Since Q is finite, C has finite cardinality. Given a path p of Q, we give p the grading

$$\begin{align*}G(p)=\sum_{\mathcal M\in C}\mathcal M(p),\end{align*}$$

where $\mathcal M(p)$ is the number of arrows of p which are in $\mathcal M$ . Note that, for any perfect matching $\mathcal M$ , the quantity $\mathcal M(p)$ is unchanged by applying a basic morph to p. This means that the quantity $G(p)$ is a well-defined number of the equivalence class of p. It is clear that if p, q, and $qp$ are paths, then $G(p)+G(q)=G(qp)$ . It follows that G gives a positive $\mathbb Z$ -grading on A through which every arrow is given a positive degree. The second statement follows because the degree of any face-path is equal to the number of perfect matchings on Q.

Proof. Since Q is strongly consistent, we may fix a positive $\mathbb Z$ -grading G of the completed dimer algebra $\widehat A_Q$ as in Lemma 6.11. We first show that an arbitrary $x\in \widehat A_Q$ is of the desired form. By definition, an element $x\in \widehat A_Q$ is of the form

(3) $$ \begin{align}x=\sum_{p\text{ a path of }Q}a_p[p],\end{align} $$

for some coefficients $a_p\in \mathbb C$ . Define $G_{\min }:=\text {min}\{G(\alpha )\ :\ \alpha \in Q_1\}$ .

Let p be any path of Q. Let C be the homotopy class of p. By path-consistency, write $[p]=[r_Cf^{m}]$ for some value m and a minimal path $r_C$ in homotopy class C. Let $M_{C,m}$ be an integer greater than $\frac {G(p)}{G_{\min }}$ . Then any path of Q with at least $M_{C,m}$ arrows must have a grading greater than $G(p)$ ; hence, every path equivalent to p must have less than $M_{C,m}$ arrows. Such paths are finite in number, and hence, the path equivalence class of p is finite. Then we may define $a_{C,m}:=\sum _{q\text { equivalent to }p}a_q$ . The desired

$$\begin{align*}x=\sum_{v,w\in Q_0}\sum_{C:v\to w}\sum_{m\geq0}a_{C,m}[r_Cf^m]\end{align*}$$

now follows from rearranging the terms of equation (3).

Now suppose that some $a_{C',m'}$ is nonzero. Define a real number $G'>G(r_C'f^{m'})$ . We show that the graded part of x below $G'$ must be nonzero, and hence that x itself is nonzero. Define for any homotopy class C and $m\geq 0$

$$\begin{align*}a^{\prime}_{C,m}:=\begin{cases} a_{C,m}&G(r_Cf^m)<G' \\ 0&\text{else.} \end{cases}\end{align*}$$

Then the sum

(4) $$ \begin{align}x_{G'}:=\sum_{v,w\in Q_0}\sum_{C:v\to w}\sum_{m\geq0}a^{\prime}_{C,m}[r_Cf^m]\end{align} $$

gives the graded part of x below $G'$ .

Set $M'>\frac {G'}{G_{\min }}$ . Then any path of Q with grading less than $G'$ has less than $M'$ arrows. So, the sum (4) has a finite number of nonzero summands. Then the sum (4) is in the noncompleted path algebra of Q. Since $a_{C',m'}$ is nonzero and in this sum, then $x_{G'}$ is not in the ideal $I_Q$ of the noncompleted dimer algebra. Moreover, since every path with at least $M'$ arrows has a grading greater than $G'$ , the sum $x_{G'}$ cannot be in the completion of $I_Q$ with respect to the arrow ideal. This shows that $x_{G'}$ is nonzero in the completed dimer algebra; hence, x is nonzero in the completed dimer algebra.