Proof. Since the regions of D cover the disc, the underlying graph of $Q_D$ is connected. Moreover, every arrow a of $Q_D$ is contained in at least one fundamental cycle—exactly one if a is frozen, and exactly two otherwise. Thus every arrow has a path from its head to its tail, so $Q_D$ is even strongly connected.

Proof. First we consider the case that the untwisting move happens in the interior, as in the upper part of Figure 4. Then $Q_{D'}$ is obtained from $Q_D$ by removing two arrows a and b, which form a fundamental cycle, as in the left-hand side of Figure 5. Since these arrows are unfrozen, they are each contained in a second fundamental cycle, and we write these cycles as $ap$ and $bq$ for some paths p and q, respectively. Since these paths do not contain the arrows a or b, they are also paths in $Q_{D'}$ . Moreover, the relation $\partial _{a}{W_D}=0$ implies that $b=p$ in $A_D$ , and the relation $\partial _{b}{W_D}=0$ that $a=q$ in $A_D$ . Define a map $\varphi \colon \mathbb {K}\langle \hspace {-0.1em}\langle Q_D\rangle \hspace {-0.1em}\rangle \to \mathbb {K}\langle \hspace {-0.1em}\langle Q_{D'}\rangle \hspace {-0.1em}\rangle $ , fixing all vertex idempotents (the two quivers having the same vertex set) and all arrows different from a and b, and with $\varphi (a)=q$ and $\varphi (b)=p$ . This map induces a homomorphism $A_D\to A_{D'}$ inverse to that induced by the map $\mathbb {K}\langle \hspace {-0.1em}\langle Q_{D'}\rangle \hspace {-0.1em}\rangle \to \mathbb {K}\langle \hspace {-0.1em}\langle Q_D\rangle \hspace {-0.1em}\rangle $ arising from the inclusion map between arrow sets.

On the other hand, if the untwisting move takes place at the boundary, then $Q_{D'}$ is obtained from $Q_D$ by removing a frozen arrow a lying in a fundamental $2$ -cycle, and $F_{D'}$ is obtained from $F_D$ by replacing a by the other arrow b in this $2$ -cycle. As before, since b is unfrozen it lies in a second fundamental cycle $bq$ for some path q not involving a, and it follows from the relation $\partial _{b}{W_D}=0$ that $a=q$ in $A_D$ . Thus defining $\varphi $ as in the first case except for $\varphi (b)=b$ provides an isomorphism of dimer algebras in an analogous way.

Proof. A $2$ -cycle in $Q_D$ containing at least one unfrozen arrow corresponds to a possible untwisting move (see Figures 4 and 5) and so does not occur when D is reduced. A $2$ -cycle consisting only of frozen arrows would correspond to a pair of boundary regions meeting at two points on the boundary of the disc, which must then be the only boundary regions. But D has at least $3$ boundary regions, so this is also impossible.

Proof. As in [Reference Baur, King and Marsh2, §4], we may weight the arrows of $Q=Q_D$ by elements of the set $\left \{1,\dotsc ,n\right \}$ , labelling the boundary marked points of $\Sigma $ . An arrow a of Q is crossed by two strands of D, say that starting at marked point i from right to left and that starting at j from left to right. We then weight a by the (indicator function of) the cyclic interval $[i,j-1]$ (compare to [Reference Baur, King and Marsh2, Defn. 4.1]). A path in Q is weighted by the sum $\mathrm {wt}_p$ of weights of its arrows, and its total weight is defined to be $\sum _{i=1}^n\mathrm {wt}_p(i)$ , which is always at least $1$ if the path has nonzero length.

The proof of [Reference Baur, King and Marsh2, Cor. 4.4], stated for $(k,n)$ -Postnikov diagrams, remains valid in our more general setting to show that every fundamental cycle has constant weight $w(i)=1$ for all $1\leq i\leq n$ . If $p_+=p_-$ is one of the defining relations of $A_D$ , then there is an arrow $a\in Q_1$ such that both $a p_+$ and $a p_-$ are fundamental cycles, from which it follows that the weights of $p_+$ and $p_-$ agree. It follows that the weight, and hence the total weight, descends to a grading of A.

Now let $v,w\in Q_0$ . Since Q is strongly connected as in Proposition 2.6, there is some path from v to w in Q, and we choose p to be such a path with minimal total weight. If q is any other path from v to w, then [Reference Baur, King and Marsh2, Prop. 9.3] applies to show that there is a path $r\colon v\to w$ and nonnegative integers $N_p$ and $N_q$ such that

$$\begin{align*}p=t^{N_p}r,\qquad q=t^{N_q}r\end{align*}$$

in $A_D$ . As before, this proposition is stated only in the case that D is a $(k,n)$ -Postnikov diagram, but its proof is still valid under our weaker assumptions—the key property of D here is (P4).

Since the total weight of t is nonzero, and p has minimal total weight among paths from v to w, we must have $N_p=0$ and $p=r$ in $A_D$ . Thus $q=t^{N_q}p$ , and we see that $e_wAe_v$ is generated over Z by p. It is moreover freely generated since each element of $\{t^Np:N\geq 0\}$ has a different total weight, which implies that these elements are linearly independent in $A_D$ .

Proof. By Proposition 2.11, there are $m,n\geq 0$ such that $p=t^mp_{vu}$ and $q=t^np_{wv}$ in $A_D$ . Thus $qp=t^{m+n}p_{wv}p_{vu}=t^{m+n+\delta }p_{wu}$ for some $\delta \geq 0$ . But $qp$ is minimal, so by Proposition 2.11 again we must have $m+n+\delta =0$ , and so in particular $m=n=0$ . Hence both p and q are minimal paths.

Proof. The algebra $A_D$ may be graded by total weights of paths, as discussed in the proof of Proposition 2.11. As already mentioned in that proof, each arrow has positive total weight.

Proof. We first observe that if $p\colon u\to w$ is a minimal path in Q passing through the vertex v, then p includes our desired arrow as follows. Write $p=p_2ap_1$ for some $a\in \mathrm {H}_{v}$ . Then by Lemma 2.13, each subpath of p is also minimal, so in A we have $p_2=p_{wv}$ and $p_2a=p_{w,ta}$ , and hence a is our desired arrow.

Now for each $a\in \mathrm {H}_{v}$ , choose a minimal path $m_a\colon ta\to w$ . If any of these paths passes through v, then we find our desired arrow as above, so we assume the contrary. Since v is mutable, each $b\in \mathrm {T}_{v}$ is involved in two fundamental cycles $a_+pb$ and $a_-qb$ with $a_{\pm }\in \mathrm {H}_{v}$ , and we can write $p'=m_{a_+}p$ and $q'=m_{a_-}q$ . By our assumption on the $m_a$ , neither $p'$ nor $q'$ passes through v.

Using again that v is mutable, the union of fundamental cycles containing v is a disc with v in the interior, so there must be some b for which the paths $p'$ and $q'$ are two sides of a digon containing v (compare to the argument in [Reference Broomhead3, Lem. 6.17]). From now on, we assume that we are in this situation, shown in Figure 6.

Figure 6

Fundamental cycles $a_+pb$ and $a_-qb$ together with minimal paths $m_{a_{\pm }}\colon ta_{\pm }\to w$ such that $p'=m_{a_+}p$ and $q'=m_{a_-}q$ form a digon containing v. Solid arrows represent arrows in Q, whereas dashed arrows represent paths.

Observe that $p_{wv}a_+=t^{\delta }p_{w,ta_+}$ for some $\delta \geq 0$ by Proposition 2.11. If $\delta =0$ , then $a_+$ was our desired arrow, so assume $\delta>0$ . In a similar way, we find $m,n\geq 0$ such that $p_{w,ta_+}p=t^mp_{w,hb}$ and $p_{w,hb}b=t^np_{wv}$ . Since $pba_+$ is a fundamental cycle, we must have

$$\begin{align*}tp_{w,ta_+}=p_{w,ta_+}pba_+=t^mp_{w,hb}ba_+=t^{m+n}p_{wv}a_+=t^{m+n+\delta}p_{w,ta_+}.\end{align*}$$

By Proposition 2.11 again, we must have $m+n+\delta =1$ , so we conclude from positivity of $\delta $ that $m=n=0$ (and $\delta =1$ ). In particular, this means $p'=m_{a_+}p$ is a minimal path, representing $p_{w,hb}$ . Repeating the argument with $a_-$ and q replacing $a_+$ and p, we see that either $a_-$ is our desired arrow or $q'=m_{a_-}q$ is also a minimal path representing $p_{w,hb}$ .

In the latter case, $p'$ and $q'$ are paths from $hb$ to w defining the same element of A, namely $p_{w,hb}$ , and bounding a digon containing v. It is then a consequence of [Reference Broomhead3, Lem. 6.17] that there is a path $r\colon hb\to w$ passing through v and also defining $p_{w,hb}$ in A. In particular, r is a minimal path passing through v, so we may find our desired arrow as in the first paragraph of the proof.

Proof. Exactness of equation (3.1) and equation (3.2) are equivalent because of the existence of a grading as in Proposition 3.1, by an argument that is essentially due to Broomhead [Reference Broomhead3, Prop. 7.5]. The reader can find an explanation of how this argument extends to the case of frozen Jacobian algebras in [Reference Pressland42, §4]. The second equivalence is simply the observation that the complex in equation (3.2) is the direct sum of the complexes in equation (3.3).

Proof. As already noted in Remark 3.5, the complex in equation (3.1) can only fail to be exact in degrees $-2$ and $-3$ , so, by Proposition 3.6, we need only check that the complex in equation (3.3) is exact at the terms $A\mathbin {\otimes }\mathbb {K} Q_3^{\mathrm {m}}\mathbin {\otimes }S_{v}$ and $A\mathbin {\otimes }\mathbb {K} Q_2^{\mathrm {m}}\mathbin {\otimes }S_{v}$ for each $v\in Q_0$ .

We first deal with exactness at $A\mathbin {\otimes }\mathbb {K} Q_3^{\mathrm {m}}\mathbin {\otimes }S_{v}$ . This term is zero unless $v\in Q_0^{\mathrm {m}}$ , so we may additionally assume this. By the third isomorphism in equation (3.4), we then have $A\mathbin {\otimes }\mathbb {K} Q_3^{\mathrm {m}}\mathbin {\otimes }S_{v}\cong Ae_{v}$ , so let $x\in Ae_{v}$ . We calculate

$$\begin{align*}(\mu_3\otimes S_{v})(x)=\sum_{a\in\mathrm{H}_{v}^{\mathrm{m}}}xa\otimes[a],\end{align*}$$

using the notation in (3.5) for elements of $A\mathbin {\otimes }\mathbb {K} Q_2^{\mathrm {m}}\mathbin {\otimes }S_{v}\cong \bigoplus _{a\in \mathrm {H}_{v}^{\mathrm {m}}}Ae_{ta}$ . Thus $x\in \ker (\mu _3\otimes S_{v})$ if and only if $xa=0$ for all $a\in \mathrm {H}_{v}^{\mathrm {m}}$ . Note that since v is mutable, $\mathrm {H}_{v}^{\mathrm {m}}=\mathrm {H}_{v}\neq \varnothing $ .

Let $w\in Q_0$ , so $e_{w}x\in e_{w}Ae_{v}$ . Since A is thin by Proposition 2.11, we have $e_{w}Ae_{v}=Zp_{wv}$ , so we may write $e_{w}x=\sum _{n\in \mathbb {N}}z_nt^n\cdot p_{wv}$ for some sequence of scalars $z_n\in \mathbb {K}$ . Using thinness again, for any $a\in \mathrm {H}_{v}^{\mathrm {m}}$ , there is some $\delta _a\in \mathbb {N}$ such that $p_{wv}a=t^{\delta _a}p_{w,ta}$ . It then follows that

$$\begin{align*}e_{w}xa=\sum_{n\in\mathbb{N}}z_nt^n\cdot p_{wv}a=\sum_{n\in\mathbb{N}}z_nt^{n+\delta_a}\cdot p_{w,ta}.\end{align*}$$

Thus if $xa=0$ , we have $z_n=0$ for all n, and so $e_{w}x=0$ . Since w was chosen arbitrarily, we conclude that $x=0$ , and so the kernel of $\mu _3\otimes S_{v}$ is trivial. This establishes exactness of equation (3.3) at the term $A\mathbin {\otimes }\mathbb {K} Q_3^{\mathrm {m}}\mathbin {\otimes }S_{v}$ .

We now move to the term $A\mathbin {\otimes }\mathbb {K} Q_2^{\mathrm {m}}\mathbin {\otimes }S_{v}$ . Using the relevant isomorphisms in equation (3.4) and the notation of equation (3.5), a general element of this term is of the form $\phi =\sum _{a\in \mathrm {H}_{v}^{\mathrm {m}}}x_a\mathbin {\otimes }[a]$ for $x\in Ae_{ta}$ , and its image under $\mu _2\mathbin {\otimes }S_{v}$ is

$$\begin{align*}(\mu_2\mathbin{\otimes}S_{v})(\phi)=\sum_{b\in\mathrm{T}_{v}}\Bigg(\sum_{a\in\mathrm{H}_{v}^{\mathrm{m}}}x_a\partial^r_{b}{\partial_{a}{W}}\Bigg)\mathbin{\otimes}[b]=\sum_{b\in\mathrm{T}_{v}}\Bigg(\sum_{a\in\mathrm{H}_{v}^{\mathrm{m}}}x_a\partial^l_{a}{\partial_{b}{W}}\Bigg)\mathbin{\otimes}[b],\end{align*}$$

where the second equality follows from equation (3.8). The reader is warned that since $b\in \mathrm {T}_{v}$ may be frozen, the derivative $\partial _{b}{W}$ is not necessarily zero in A.

Now assume that $(\mu _2\mathbin {\otimes }S_{v})(\phi )=0$ or, equivalently by the above calculation, that

(3.9) $$ \begin{align} \sum_{a\in\mathrm{H}_{v}^{\mathrm{m}}}x_a\partial^l_{a}{\partial_{b}{W}}=0 \end{align} $$

for all $b\in \mathrm {T}_{v}$ . Picking $w\in Q_0$ and multiplying by $e_{w}$ on the left, we obtain elements $e_{w}x_a\in e_{w}Ae_{ta}$ for each $a\in \mathrm {H}_{v}^{\mathrm {m}}$ .

As in the first part of the proof, we will now use thinness of A to write the elements $e_{w}x_a$ as the product of a power series in t with $p_{w,ta}$ , but in order to keep the notation clean in the subsequent argument, we will do this in a slightly unusual way. First, note that, having fixed the vertex w, any path p determines $\delta _p\in \mathbb {N}$ with the property that $p_{w,hp}p=t^{\delta _p}p_{w,tp}$ . Since this equality is formulated in the algebra A, the quantity $\delta _p$ depends only on the class of p in the algebra.

Using this notation, we can write

(3.10) $$ \begin{align} e_{w}x_a=t^{\delta_a}\sum_{n\in\mathbb{Z}}z_n^{w,a}t^n\cdot p_{w,ta} \end{align} $$

for some scalars $z_n^{w,a}\in \mathbb {K}$ . Since the right-hand side of this expression may only involve nonnegative powers of t, we have $z_n^{w,a}=0$ whenever $n<-\delta _a$ .

The key step in the argument is to show, under our assumption that $(\mu _2\mathbin {\otimes }S_{v})(\phi )=0$ , that the coefficient $z_n^{w,a}$ depends only on n and w, and not on $a\in \mathrm {H}_{v}^{\mathrm {m}}$ . We consider the configuration

(3.11)

Here solid arrows represent arrows in $Q_1$ , dashed arrows represent paths in Q and dotted arrows represent linear combinations of paths. Our assumption is that $a_+$ and $a_-$ are unfrozen arrows, with head at v as depicted, which appear in the two fundamental cycles $a_+pb$ and $a_-qb$ containing the arrow b, also unfrozen, with tail at v. As indicated by the notation in equation (3.11), we will usually omit the letter a when $a_+$ or $a_-$ appears in subscripts or superscripts in the following argument—for example, this means $x_+:=x_{a_+}$ . Our aim is to show that $z_n^{w,+}=z_n^{w,-}$ for all $n\in \mathbb {Z}$ ; it will then follow from the way the quiver Q is embedded in the disc that $z_n^{w,a}$ is independent of $a\in \mathrm {H}_{v}^{\mathrm {m}}$ , since if $a,a'\in \mathrm {H}_{v}^{\mathrm {m}}$ , we can find sequences $a_0,\dotsc ,a_k$ and $b_1\cdots b_k$ such that $a_0=a$ , $a_k=a'$ , and for all $1\leq i\leq k$ the triple of arrows $(a_{i-1},b_i,a_i)$ has the same configuration as $(a_-,b,a_+)$ in equation (3.11) (or its mirror image), so that $z_n^{w,a_{i-1}}=z_n^{w,a_i}$ .

So consider the configuration shown in equation (3.11). We see from the figure that $\partial _{b}{W}=a_+p-a_-q$ . While it can happen that $a_+=a_-$ (if and only if v is a bivalent vertex of Q), in this case there is nothing to prove, so we may assume $a_+\ne a_-$ . Then for $a\in \mathrm {H}_{v}^{\mathrm {m}}$ , we have

$$\begin{align*}\partial^l_{a}{\partial_{b}{W}}=\begin{cases}p,&a=a_+,\\-q,&a=a_-,\\0,&\text{otherwise}.\end{cases}\end{align*}$$

Thus

$$\begin{align*}\sum_{a\in\mathrm{H}_{v}^{\mathrm{m}}}x_a\partial^l_{a}{\partial_{b}{W}}=x_+p-x_-q.\end{align*}$$

Since $\phi =\sum _{a\in \mathrm {H}_{v}^{\mathrm {m}}}x_a\mathbin {\otimes }[a]$ is in the kernel of $\mu _2\mathbin {\otimes }S_{v}$ , this quantity is zero by equation (3.9), so $x_+p=x_-q$ . It follows that

$$ \begin{align*} e_{w}x_+p&=t^{\delta_+}\sum_{n\in\mathbb{Z}}z_n^{w,+}t^n\cdot p_{w,ta_+}p=t^{\delta_++\delta_p}\sum_{n\in\mathbb{Z}}z_n^{w,+}t^n\cdot p_{w,hb},\\ e_{w}x_-q&=t^{\delta_-}\sum_{n\in\mathbb{Z}}z_n^{w,-}t^n\cdot p_{w,ta_-}q=t^{\delta_-+\delta_q}\sum_{n\in\mathbb{Z}}z_n^{w,-}t^n\cdot p_{w,hb}, \end{align*} $$

must also be equal. Since $a_+p=a_-q$ in A (because of the relation $\partial _{b}{W}$ ), we must have $\delta _++\delta _p=\delta _-+\delta _q$ , so we can conclude, for any $n\in \mathbb {Z}$ , that $z_n^{w,+}=z_n^{w,-}$ , and further that $z_n^{w,a}$ is independent of $a\in \mathrm {H}_{v}^{\mathrm {m}}$ , as required. From now on, we abbreviate $z_n^w:=z_n^{w,a}$ , and for any $a\in \mathrm {H}_{v}^{\mathrm {m}}$ , we have

(3.12) $$ \begin{align} e_{w}x_a=t^{\delta_a}\sum_{n\in\mathbb{Z}}z_n^wt^n\cdot p_{w,ta}. \end{align} $$

We now return to our main purpose, to show that the complex in equation (3.1) is exact. We recall that we are assuming that $\phi =\sum _{a\in \mathrm {H}_{v}^{\mathrm {m}}}x_a\mathbin {\otimes }[a]$ satisfies $(\mu _2\mathbin {\otimes }S_{v})(\phi )=0$ and wish to show that $\phi $ is in the image of $\mu _3\mathbin {\otimes }S_{v}$ .

First we deal with the case that v is a frozen vertex, in which case $\mu _3\mathbin {\otimes }S_{v}=0$ , and we wish to conclude that $x_a=0$ for all $a\in \mathrm {H}_{v}^{\mathrm {m}}$ . This is equivalent to showing that the complex numbers $z_n^w$ appearing in equation (3.12) are zero for all n and all w.

It follows from the construction of $(Q_D,F_D)$ that, since v is frozen, it is incident with a frozen arrow. If $\mathrm {H}_{v}^{\mathrm {m}}=\varnothing $ , then we have nothing to show, and otherwise there is an arrow $a\in \mathrm {H}_{v}^{\mathrm {m}}$ as in one of the following configurations

(3.13)

in which $apb$ and $cqb$ are fundamental cycles. As earlier in the paper, the bold arrows are frozen, and the dashed arrows represent paths. The vertices indicated by diamonds are frozen, like the vertex v, whereas the others may be mutable or frozen.

In the two configurations in equation (3.13), we have $\partial _{b}{W}=\pm ap$ and $\partial _{b}{W}=\pm (ap-cq)$ , respectively. Thus in either case we calculate for $a'\in \mathrm {H}_{v}^{\mathrm {m}}$ that

$$\begin{align*}\partial^l_{a'}{\partial_{b}{W}}=\begin{cases}\pm p,&a'=a,\\0,&\text{otherwise}.\end{cases}\end{align*}$$

Since $(\mu _2\mathbin {\otimes }S_{v})(\phi )=0$ , we deduce from equation (3.9) that

$$\begin{align*}x_ap=\pm\sum_{a'\in\mathrm{H}_{v}^{\mathrm{m}}}x_a\partial^l_{a'}\partial_{b}{W}=0.\end{align*}$$

Multiplying on the left by $e_{w}$ and using that A is thin, we see that

$$\begin{align*}0=e_{w}x_ap=t^{\delta_a}\sum_{n\in\mathbb{Z}}z_n^wt^n\cdot p_{w,ta}p=t^{\delta_a+\delta_p}\sum_{n\in\mathbb{Z}}z_n^wt^n\cdot{p_{w,hb}},\end{align*}$$

so $z_n^w=0$ for all n and w, as required.

Now we treat the case that v is a mutable vertex; in this case, $\mathrm {H}_{v}^{\mathrm {m}}=\mathrm {H}_{v}$ and we use the latter to keep the notation simpler. In this case, we want to construct $y\in Ae_{v}$ such that

$$\begin{align*}(\mu_3\mathbin{\otimes}S_{v})(y)=\sum_{a\in\mathrm{H}_{v}}ya\mathbin{\otimes}[a]=\phi.\end{align*}$$

Using the sequence of complex numbers $z_n^w$ appearing in equation (3.12), we write

$$\begin{align*}y_w=\sum_{n\in\mathbb{Z}}z_n^wt^n\cdot p_{wv}.\end{align*}$$

For this to be a well-defined element of $e_{w}Ae_{v}$ , we need $z_n^w=0$ for all $n<0$ . Choosing any $a\in \mathrm {H}_{v}$ , we have that $z_n^w=z_n^{w,a}$ is zero whenever $n<-\delta _a$ , as in equation (3.10). But by Lemma 3.2, there is some $a\in \mathrm {H}_{v}$ such that $\delta _a=0$ , so $y_w\in e_{w}Ae_{v}$ as required.

Now, letting $y=\sum _{w\in Q_0}y_w\in Ae_{v}\cong A\mathbin {\otimes }\mathbb {K} Q_3^{\mathrm {m}}\mathbin {\otimes }S_{v}$ , we can calculate

$$ \begin{align*} (\mu_3\mathbin{\otimes}S_{v})(y)=\sum_{w\in Q_0}\sum_{a\in\mathrm{H}_{v}}y_wa\mathbin{\otimes}[a]&=\sum_{w\in Q_0}\sum_{a\in\mathrm{H}_{v}}\Bigg(\sum_{n\in\mathbb{N}}z_n^wt^n\cdot p_{wv}a\Bigg)\mathbin{\otimes}[a]\\ &=\sum_{w\in Q_0}\sum_{a\in\mathrm{H}_{v}}\Bigg(\sum_{n\in\mathbb{N}}z_n^wt^{n+\delta_a}\cdot p_{w,ta}\Bigg)\mathbin{\otimes}[a]\\ &=\sum_{w\in Q_0}\sum_{a\in\mathrm{H}_{v}}e_{w}x_a\mathbin{\otimes}[a]=\sum_{a\in\mathrm{H}_{v}}x_a\mathbin{\otimes}[a]=\phi, \end{align*} $$

so we conclude that $\phi \in \operatorname {\mathrm {im}}(\mu _3\mathbin {\otimes }S_{v})$ , as required.

Proof. Since $A_D$ is thin, as in Proposition 2.11, it is finitely generated as a module over the commutative Noetherian ring $\mathbb {K}[\hspace {-0.1em}[t]\hspace {-0.1em}]$ and so is Noetherian.

By thinness again, $A_D$ has a basis $\{t^np_{wv}:v,w\in Q_0,n\in \mathbb {N}\}$ . To see that $A_D/\langle e\rangle $ is finite-dimensional, we show that all but finitely many of these basis vectors are zero in the quotient. Picking any $v,w\in Q_0$ , let c be any cycle at w passing through a boundary vertex, which exists since $Q_D$ is strongly connected as in Proposition 2.6. Then since $A_D$ is thin, we have $cp_{wv}=t^Np_{wv}$ for some $N\in \mathbb {N}$ . By construction, this element is zero in the quotient $A_D/\langle e\rangle $ , and thus $t^np_{wv}=0$ in this quotient algebra for all $n\geq N$ . Running over all pairs $v,w\in Q_0$ completes the proof.

Proof. This can be proved exactly as in [Reference Baur, King and Marsh2, Prop. 12.2]; while this proposition is stated only for $(k,n)$ -diagrams, the proof does not use the extra assumption on the strand permutation.

Proof. Again combining [Reference Postnikov41, Thm. 13.4] and [Reference Postnikov41, Thm. 17.1], any two diagrams determining the same positroid are connected by a sequence of geometric exchanges, so the result follows from Proposition 4.6.

Proof. Note that the quivers Q and $\mu _kQ$ have the same set of frozen vertices, so the statement makes sense. It is straightforward to check that $A_0/\langle e_{i}\rangle $ and $A/\langle e_{i}\rangle $ are, like $A_0$ and A, frozen Jacobian algebras of ice quivers with potential related by mutation at k. (The required ice quivers with potential are obtained from those of $A_0$ and A by removing the vertex i and all incident arrows from the quiver, and removing from the potential all terms given by cycles through i.) That mutations preserve finite dimensionality of frozen Jacobian algebras follows via a straightforward adaptation of the proof of [Reference Derksen, Weyman and Zelevinsky15, Cor. 6.6] for ordinary Jacobian algebras.

Proof. Since $e_{i}$ corresponds to projection onto an indecomposable projective summand $P_i$ of T, we have $e_{i}Ae_{i}=\operatorname {\mathrm {End}}_{B}(P_i)^{\mathrm {op}}$ . Since $P_i$ is an indecomposable summand of every cluster-tilting object of $\operatorname {\mathrm {GP}}(B)$ , including $T_D$ , this endomorphism algebra is isomorphic to Z by Proposition 2.11 and Theorem 4.5.

Proof. We use a result of Psaroudakis [Reference Psaroudakis45, Prop. 4.15], implying that it is enough to check that the image, denoted by $\mathsf {F}(P)$ in [Reference Psaroudakis45], of the counit

$$\begin{align*}\varepsilon\colon\ell(e_{i}P)=Ae_{i}\mathbin{\otimes}_Ze_{i}P\to P\end{align*}$$

is projective for all projective $P\in \operatorname {\mathrm {mod}}{A}$ . We may write $P=\operatorname {\mathrm {Hom}}_{B}(T,T')$ for $T'\in \operatorname {\mathrm {add}}{T}\subseteq \operatorname {\mathrm {GP}}(B)\subseteq \operatorname {\mathrm {CM}}(B)$ so that $e_{i}P=\operatorname {\mathrm {Hom}}_B(P_i,T')$ is a free Z-module. Thus $Ae_{i}\mathbin {\otimes }_Ze_{i}P\cong (Ae_{i})^n$ , where n is the rank of the Z-module $e_{i}P$ , is projective. We claim that the counit $\varepsilon $ is injective so that its image is isomorphic to the projective module $(Ae_{i})^n$ .

To see this, observe that $Ae_{i}\mathbin {\otimes }_Ze_{i}P\cong (Ae_{i})^n$ and P are Cohen–Macaulay A-modules—that is, they are free and finitely generated over Z—so this property is inherited by the kernel K of $\varepsilon $ . On the other hand, $e_{i}\varepsilon \colon e_{i}Ae_{i}\mathbin {\otimes }_Z e_{i}P\to e_{i}P$ is an isomorphism by Proposition 5.2, so $e_{i}K=0$ . Thus K is a finitely generated $A/\langle e_{i}\rangle $ -module and hence finite-dimensional by Lemma 5.1. Since the only finite-dimensional Cohen–Macaulay A-module is $0$ , we have the result.

Having established that $\iota $ is a homological embedding, it is then immediate that $\operatorname {\mathrm {gldim}}{A/\langle e_{i}\rangle }\leq \operatorname {\mathrm {gldim}}{A}$ . That $\operatorname {\mathrm {gldim}}{A}\leq 3$ is a consequence of [Reference Pressland43, Prop. 3.7], since A is the endomorphism algebra of the ( $2$ -)cluster-tilting object $T\in \operatorname {\mathrm {GP}}(B)$ , and $\operatorname {\mathrm {GP}}(B)\subseteq \operatorname {\mathrm {mod}}(B)$ is the category of third syzygy modules since B has Gorenstein dimension at most $3$ by Theorem 4.5(1).

Proof. Since $(Q,F,W)$ is reduced, Q is the Gabriel quiver of J [Reference Pressland44, Rem. 3.4]. That Q has no loops then follows from the no-loops theorem [Reference Igusa24, Reference Lenzing35], without using that J is a frozen Jacobian algebra. However, this latter assumption means that for each $i\in Q_0$ , there is an exact sequence

$$\begin{align*}\bigoplus_{b\in\mathrm{H}_{i}^{\mathrm{m}}}Je_{tb}\to\bigoplus_{a\in\mathrm{T}_{i}}Je_{ha}\to Je_{i}\to S_i\to 0\end{align*}$$

for $S_i=Je_{i}/\operatorname {rad}(J)e_{i}$ , the simple J-module supported at vertex i, and this sequence is the start of a projective resolution of $S_i$ [Reference Buan, Iyama, Reiten and Smith6, Prop. 3.3(b)]. Since Q has no loops, $tb\ne i$ for all $b\in \mathrm {H}_{i}^{\mathrm {m}}$ , so

$$\begin{align*}\operatorname{\mathrm{Hom}}_J\Bigg(\bigoplus_{b\in\mathrm{H}_{i}^{\mathrm{m}}}Je_{tb},S_i\Bigg)=0.\end{align*}$$

It follows that $\operatorname {\mathrm {Ext}}^2_J(S_i,S_i)$ , which is a subquotient of this space, is also zero. That Q has no $2$ -cycles then follows from [Reference Geiß, Leclerc and Schröer20, Prop. 3.11].

Proof. Let i be a frozen vertex and $e_{i}\in A$ the corresponding idempotent. As in the proof of Lemma 5.1, the algebra $A/\langle e_{i}\rangle $ is again a frozen Jacobian algebra. It is finite-dimensional by the same lemma and has finite global dimension by Lemma 5.3. Thus by Proposition 5.4, its quiver has no $2$ -cycles. Since this quiver is obtained from $Q'$ by deleting the vertex i and all incident arrows, there are no $2$ -cycles in $Q'$ incident with i. Since D has at least $3$ boundary regions, $Q'$ has at least three frozen vertices, and by repeating this process we rule out $2$ -cycles in $Q'$ altogether.

Proof. The results of this section show that the desired properties of T are preserved by mutations. Since they hold in the case $T=T_D$ by Theorem 4.5—reducedness of D implying that $Q_D$ has no loops or $2$ -cycles—we obtain the result by induction.

Proof. This is proved in exactly the same way as [Reference Fu and Keller17, Lem. 2.2] (in particular, using the assumption that $\mathbb {K}$ is algebraically closed), noting that $Q(\mathcal {T}\kern1.5pt)$ has no loops by Definition 6.2(3) and the assumption that $\mathcal {T}$ is reachable.

Proof. The argument given for [Reference Fu and Keller17, Thm. 3.3] applies equally well here; while Fu and Keller assume that $\mathcal {E}$ is Hom-finite, our assumption that $A=\operatorname {\mathrm {End}}_{\mathcal {E}}(T)^{\mathrm {op}}$ is Noetherian is sufficient for their proof to go through. (For example, it is still the case that any finitely generated A-module, such as a finite-dimensional $\underline {A}$ -module, is finitely presented.) Indeed, our additional assumption that $\operatorname {\mathrm {gldim}}{A}<\infty $ even allows some parts of the proof in [Reference Fu and Keller17] to be simplified.

Proof. This is proved exactly as [Reference Fu and Keller17, Thm. 5.4], despite our weakening of the definition of a cluster structure. Indeed, the relevant maps are shown to be injective in [Reference Fu and Keller17, Lem. 4.2, Prop. 4.3] without assuming that $\mathcal {E}$ has a cluster structure in either sense. The proof of surjectivity (and that the maps are well-defined: that is, that they take values in the appropriate sets) uses an inductive argument. We have already seen in Remark 6.7 that $\varphi ^{T}$ takes the indecomposable summands of T to the initial cluster variables $x_i$ of $\mathscr {A}_{Q}$ (and the indecomposable projective summands to the frozen variables, by the definition of the ice quiver structure on $Q(T)=Q$ ). The conclusion then follows by using compatibility between mutations of reachable cluster-tilting objects and mutations of clusters in $\mathscr {A}_{Q}$ . For example, the quivers of endomorphism algebras transform correctly by parts (3) and (4) of Definition 6.2, and in the notation of Definition 6.2(2), we have

$$\begin{align*}\varphi^{T}(X)\varphi^{T}(X^*)=\varphi^{T}(U^+)+\varphi^{T}(U^-)\end{align*}$$

by Lemma 6.4 and Theorem 6.8, corresponding to an exchange relation in $\mathscr {A}_{Q}$ by Lemma 6.4 again. The inductive nature of the argument means we do not require any properties of nonreachable cluster-tilting objects, since these do not appear.

Proof. The category $\operatorname {\mathrm {GP}}(B_D)$ is a stably $2$ -Calabi–Yau Frobenius category, and $T_D\in \operatorname {\mathrm {GP}}(B_D)$ a cluster-tilting object, by Theorem 4.5. It is idempotent complete and admits a fully faithful embedding to a Krull–Schmidt category by [Reference Çanakçı, King and Pressland11, Prop. 3.6] (see Proposition 7.1 below), so it is Krull–Schmidt. Moreover, $(\operatorname {\mathrm {GP}}(B_D),T_D)$ has a cluster structure as follows. Conditions (1) and (2) are automatic, as discussed in Definition 6.2. Condition (3) holds by Theorem 5.7, and as a consequence, condition (4) holds by [Reference Pressland44, Thm. 5.15]. Since $Q(T_D)$ is isomorphic to $Q_D$ by Theorem 4.5 again, $(\operatorname {\mathrm {GP}}(B_D),T_D)$ is a Frobenius $2$ -Calabi–Yau realisation of $\mathscr {A}_{D}$ , and the required bijections then follow from Theorem 6.10.

Proof. The map is well-defined since $\rho $ is exact and injective since $\rho $ is fully faithful; these conclusions hold even under the weaker assumption $Y\in \operatorname {\mathrm {CM}}(B)$ . For surjectivity, let $P\to X$ be a projective cover of X, and consider the commutative diagram

(7.1)

in which the lower row is an extension in $\operatorname {\mathrm {CM}}(C)$ and the right-hand square is a pull-back. Since $\rho $ is fully faithful, to show that the the lower row comes from an extension of B-modules, it suffices to show that E is in the essential image of $\rho $ .

Note that $\rho B$ is rigid, as follows. By [Reference Çanakçı, King and Pressland11, Prop. 8.2], we have $\rho B\cong \bigoplus _{I\in \mathcal {I}}M_I$ . Since the elements of $\mathcal {I}$ are weakly separated [Reference Oh, Postnikov and Speyer39, Lem. 4.5], the required rigidity follows from [Reference Jensen, King and Su28, Prop. 5.6]. As a consequence, if $Y=B$ , then the upper row of equation (7.1) splits, since $P\in \operatorname {\mathrm {add}}(B)$ . It follows that E is a factor of $\rho B\oplus \rho P$ , so it lies in the essential image of $\rho $ by [Reference Curtis12, §37, Ex. 10]; here we use the additional properties of $\rho $ from Proposition 7.1 to see that $\rho B\subseteq C$ and $\rho B[t^{-1}]=C[t^{-1}]$ . Thus

(7.2) $$ \begin{align} \operatorname{\mathrm{Ext}}^1_B(X,B)\stackrel{\sim}{\to}\operatorname{\mathrm{Ext}}^1_C(\rho X,\rho B). \end{align} $$

Now, for general $Y\in \operatorname {\mathrm {GP}}(B)$ , the upper row of equation (7.1) is an element of

$$\begin{align*}\operatorname{\mathrm{Ext}}^1_C(\rho P,\rho Y)=\mathrm{D}\operatorname{\mathrm{Ext}}^1_C(\rho Y,\rho P)=\mathrm{D}\operatorname{\mathrm{Ext}}^1_B(Y,P)=0,\end{align*}$$

where the first equality is the $2$ -Calabi–Yau property of $\operatorname {\mathrm {CM}}(C)$ , the second follows from equation (7.2) since $P\in \operatorname {\mathrm {add}}{B}$ , and the last uses that $P\in \operatorname {\mathrm {add}}(B)$ and $Y\in \operatorname {\mathrm {GP}}(B)$ . Thus this upper row once again splits, and we conclude as before that E is in the essential image of $\rho $ .

Proof. First, observe that $X\mapsto \Psi (\rho X)|_{\widehat {\Pi }^\circ }$ defines a cluster character $\operatorname {\mathrm {GP}}(B)\to \mathbb {C}[\widehat {\Pi }^\circ ]$ since $\Psi $ is a cluster character on $\operatorname {\mathrm {CM}}(C)$ ; this uses both Proposition 7.1 and, to see that the restricted map still satisfies the multiplication formula from Definition 6.5(3), Proposition 7.2.

Thus it suffices to check the required equality on the indecomposable summands of a single reachable cluster-tilting object, which we may take to be the initial object T. For each $j\in Q_0$ , the summand $T_j=eAe_j$ satisfies

$$\begin{align*}\Psi(\rho T_j)|_{\widehat{\Pi}^\circ}=\Psi(M_{I_j})|_{\widehat{\Pi}^\circ}=\Delta_{I_j}|_{\widehat{\Pi}^\circ}=\Phi(T_j)\end{align*}$$

by Propositions 7.3 and 7.4.

Proof. By Proposition 7.4, we have $\Delta _{I}=\Psi (M_I)=\Psi (\rho M)$ . Thus by Proposition 7.5, it suffices to check that M is reachable—it is rigid by Proposition 7.2 (see also the proof of [Reference Çanakçı, King and Pressland11, Lem. 10.4]).

Consider a geometric exchange at an alternating region j of D (corresponding to a mutation at the vertex j of Q), producing a new diagram $D'$ . By considering the exchange sequences as in Definition 6.2(2), we may see that the mutation $T_j'$ of the direct summand $T_k$ of T satisfies $\rho (T_j')=M_{I_j'}$ . Since $\rho $ is fully faithful, $T_j'$ is also the unique such object in $\operatorname {\mathrm {GP}}(B)$ up to isomorphism. Inductively, it follows that $M\in \operatorname {\mathrm {GP}}(B)$ with $\rho (M)\cong M_I$ is reachable as long as I appears as a label in a Postnikov diagram related to D by a sequence of geometric exchanges. By [Reference Oh, Postnikov and Speyer39, Thm. 6.6], such subsets I are precisely those appearing in some maximal weakly separated collection $\mathcal {C}$ with $\mathcal {I}\subseteq \mathcal {C}\subseteq \mathcal {P}$ : that is, they are those $I\in \mathcal {P}$ , which are weakly separated from every element of $\mathcal {I}$ .

So assume $M\in \operatorname {\mathrm {GP}}(B)$ has $\rho (M)\cong M_I$ . Since $M\in \operatorname {\mathrm {CM}}(B)$ , it follows from [Reference Çanakçı, King and Pressland11, Prop. 8.6] that $I\in \mathcal {P}_D$ . Since M is even Gorenstein projective, it follows from Proposition 7.2 that

$$\begin{align*}0=\operatorname{\mathrm{Ext}}^1_{B}(M,T_j)=\operatorname{\mathrm{Ext}}^1_C(M_I,M_{I_j})\end{align*}$$

for $T_j=eA_De_j$ with j a frozen vertex, these being precisely the indecomposable summands of B. By [Reference Jensen, King and Su28, Prop. 5.6], this means precisely that I and $I_j$ are weakly separated. Since $\mathcal {I}$ is the set of labels $I_j$ of frozen vertices j, we have the result.