Proof. Due to the existence of right $\operatorname {add} U[-1]$ -approximations of R, the assertion (1) implies $R\in \operatorname {rigid-}(U[-1]\ast U)$ . Similarly, the assertion (2) implies $R\in \operatorname {rigid-}(U[-1]\ast U)$ , too. So it suffices to show (1) and (2). We only prove (1) since (2) can be proved dually.

Since $\iota _0$ is a right $\operatorname {add} U[-1]$ -approximation, we have $U^0_R\in \operatorname {add} U$ . Since $U\in R\ast R[1]$ and $R\ast R[1]$ is closed under taking direct summands, we have $U^0_R\in R\ast R[1]$ . Hence, $U^1_R\in R\ast U^0_R\subseteq R\ast R[1]$ . Applying $\operatorname {Hom}(U[-1],-)$ to the triangle (2.2), we get a long exact sequence in $\mathcal {D}$

$$\begin{align*}\begin{array}{rl} &\operatorname{Hom}(U[-1],U^0_R[-1])\xrightarrow{\operatorname{Hom}(U[-1],\iota_0)}\operatorname{Hom}(U[-1],R)\\ \xrightarrow{\operatorname{Hom}(U[-1],\iota_1)}&\operatorname{Hom}(U[-1],U^1_R) \to \operatorname{Hom}(U[-1],U^0_R). \end{array}\end{align*}$$

Since U is rigid, the last term $\operatorname {Hom}(U[-1],U^0_R)=0$ . Since $\iota _0$ is a right $\operatorname {add} U[-1]$ -approximation of R, the morphism $\operatorname {Hom}(U[-1],\iota _0)$ is surjective. So $\operatorname {Hom}(U[-1],U^1_R)=0$ , which implies $\operatorname {Hom}(U^1_R,U[1])=0$ by the 2-Calabi-Yau property. In particular, $\operatorname {Hom}(U^1_R,U^0_R[1])=0$ .

For any $f\in \operatorname {Hom}(U^1_R,U^1_R[1])$ , consider the following diagram.

Since $\iota [1]\circ f\in \operatorname {Hom}(U^1_R,U^0_R[1])=0$ , there exists $f_1\in \operatorname {Hom}(U^1_R,R[1])$ such that $f=\iota _1[1]\circ f_1$ . Since $f_1\circ \iota _1\in \operatorname {Hom}(R,R[1])=0$ , there exists $f_2\in \operatorname {Hom}(U^0_R,R[1])$ such that $f_1=f_2\circ \iota $ . Since $\iota _0[1]$ is a right $\operatorname {add} U$ -approximation of $R[1]$ , there exists $f_3\in \operatorname {Hom}(U^0_R,U^0_R)$ such that $f_2=(\iota _0[1])f_3$ . Hence, we have $f=\iota _1[1]\circ \iota _0[1]\circ f_3\circ \iota $ , which is zero since $\iota _1[1]\circ \iota _0[1]=0$ . So $U^1_R$ is rigid, and hence $\operatorname {Hom}((U^1_R\,{\oplus}\, U),(U^1_R\,{\oplus}\, U)[1])=0$ . Since U is maximal rigid with respect to $R\ast R[1]$ , we have $U^1_R\in \operatorname {add} U$ .

Proof. The proof of [Reference Dehy and Keller21, Theorem 2.4] also works here.

Proof. Let $U=\,{\oplus}\, ^m_{i=1}U_i$ be a basic rigid object in $R\ast R[1]$ with $U_i,1\leq i\leq m$ , indecomposable. By Proposition 2.6, we have

$$\begin{align*}|U|=\operatorname{rank}\{\operatorname{ind}_RU_j|1\leq j\leq m\}\leq\operatorname{rank}\{\operatorname{ind}_R R_i|1\leq i\leq n\}=|R|,\end{align*}$$

where $\operatorname {rank} X$ denotes the rank of a set $X\subseteq K_0^{\mbox {sp}}(R)$ .

Let U be a maximal rigid object with respect to $R\ast R[1]$ . On the one hand, U is rigid in $R\ast R[1]$ , which implies $|U|\leq |R|$ . On the other hand, by Lemma 2.4, we have $R\in \operatorname {rigid-}(U[-1]\ast U)$ , which implies $|R|\leq |U|$ . Hence, we have $|R|=|U|$ . Conversely, let $U\in \operatorname {rigid-}(R\ast R[1])$ with $|U|=|R|$ . For any $N\in R\ast R[1]$ such that $U\,{\oplus}\, N$ is rigid, we have $|U\,{\oplus}\, N|\leq |R|$ . Then $|U|=|U\,{\oplus}\, N|$ , which implies $N\in \operatorname {add} U$ . So U is maximal rigid with respect to $R\ast R[1]$ .

Proof. For any $U\in \operatorname {max}\operatorname {rigid-}(R\ast R[1])$ , by Lemma 2.4, we have $R\in \operatorname {rigid-}(U[-1]\ast U)$ , and by Proposition 2.7, we have $|U|=|R|$ . Then $|R|=|U[-1]|$ . So by Proposition 2.7 again, we have $R\in \operatorname {max}\operatorname {rigid-}(U[-1]\ast U)$ . The opposite implication can be obtained by switching R with $U[-1]$ .

Proof. Let $N=U\setminus Y$ . If $[\operatorname {ind}_{U[-1]}R:Y[-1]]= 0$ then by the definition of index, R admits an $N[-1]$ -presentation. So by Proposition 2.7, we have $|R|\leq |N|<|U|$ . Since $R\in \operatorname {max}\operatorname {rigid-}(U[-1]\ast U)$ , again by Proposition 2.7, we have $|R|=|U|$ , a contradiction.

Proof. Let $[\operatorname {ind}_{(Y\,{\oplus}\, N)[-1]}R:Y[-1]]=t$ and $[\operatorname {ind}_{(Y^{\prime }\,{\oplus}\, N)[-1]}R:Y^{\prime }[-1]]=s.$ Assume conversely $ts\geq 0$ . By Lemma 2.9, we have $t\neq 0$ and $s\neq 0$ . So either $t<0$ and $s<0$ , or and . We only make a contradiction for the case $t<0$ and $s<0$ since the other case is similar. Let $U=N\,{\oplus}\, Y$ and $U^{\prime }=N\,{\oplus}\, Y^{\prime }$ . Consider the following diagram

where the first (resp. second) row is a minimal $U[-1]$ -presentation (resp. $U^{\prime }[-1]$ -presentation) of R. Since $t<0$ and $s<0$ , by Proposition 2.2, both $U_R^0$ and ${U^{\prime }}^0_R$ belong to $\operatorname {add} N$ , and Y and $Y^{\prime }$ are direct summands of $U^1_R$ and ${U^{\prime }}^1_R$ , respectively. So both $\delta $ and $\delta ^{\prime }$ are minimal right $\operatorname {add} N[-1]$ -approximations of R. Hence, there exists an isomorphism $\phi :U_R^0[-1]\to {U^{\prime }}^0_R[-1]$ such that the middle square of the above diagram commutes. It follows that there exists an isomorphism $\psi :U_R^1\to {U^{\prime }}_R^1$ , and hence, $Y\cong Y^{\prime }$ , a contradiction.

Proof. We only prove the equivalences between (1), (2) and (3), since the equivalences between (1’), (2’) and (3’) can be proved dually.

$(1)\Rightarrow (2)$ . Take a minimal $U[-1]$ -presentation of R:

$$\begin{align*}U^1_R[-1]\longrightarrow U^0_R[-1]\xrightarrow{\iota_0} R\xrightarrow{\iota_1} U^1_R.\end{align*}$$

Since $\iota _1$ is a left $\operatorname {add} U$ -approximation of R, for any morphism $g\in \operatorname {Hom}(R,Y)$ , there exists $h\in \operatorname {Hom}(U^1_R,Y)$ such that $g=h\circ \iota _1$ . Since , by Proposition 2.2, we have $U^1_R\in \operatorname {add} N$ . Since $\alpha _Y$ is a right $\operatorname {add} N$ -approximation of Y, there exists $h^{\prime }\in \operatorname {Hom}(U^R_1,N_0)$ such that $h=\alpha _Y\circ h^{\prime }$ . Then we have $g=\alpha _Y\circ h^{\prime }\circ \iota _1$ , which implies $\operatorname {Hom}(R,\alpha _Y)$ is surjective.

$(2)\Rightarrow (3)$ . Let

$$\begin{align*}R^1_Y\to R^0_Y\overset{\iota^0_Y}{\longrightarrow}Y\overset{\iota^1_Y}{\longrightarrow} R^1_Y[1]\end{align*}$$

be a minimal R-presentation of Y. Since $\operatorname {Hom}(R,\alpha _Y)$ is surjective, there exists $\iota ^{\prime }_0\in \operatorname {Hom}(R^0_Y,N_0)$ such that $\iota ^0_Y=\alpha _Y\circ \iota ^{\prime }_0$ . So by the octahedral axiom, we have the following commutative diagram of triangles.

Since $M\in N_0\ast R^0_Y[1]\subseteq R\ast R[1]$ , we have $Z\in R^1_Y\ast M\subseteq R\ast R[1]$ . Thus, by Lemma 2.13, we have $\mu ^+_Y(U)=N\,{\oplus}\, Z\in \operatorname {max}\operatorname {rigid-}(R\ast R[1])$ .

Applying $\operatorname {Hom}(R,-)$ to the right exchange triangle (2.4), we have the following exact sequence:

$$\begin{align*}\operatorname{Hom}(R,N_0)\xrightarrow{\operatorname{Hom}(R,\alpha_Y)}\operatorname{Hom}(R,Y)\longrightarrow\operatorname{Hom}(R,Z[1])\xrightarrow{\operatorname{Hom}(R,\alpha_N[1])}\operatorname{Hom}(R,N_0[1]).\end{align*}$$

Since $\operatorname {Hom}(R,\alpha _Y)$ is surjective, we have $\operatorname {Hom}(R,\alpha _N[1])$ is injective. So by the 2-Calabi-Yau property, the morphism $\operatorname {Hom}(\alpha _N,R[1])$ is surjective. Hence, any $g\in \operatorname {Hom}(Z[-1],R)$ factors through $N_0[-1]$ . So any right $\operatorname {add} N[-1]$ -approximation of R is also a right $\operatorname {add}\mu ^+_Y(U)[-1]$ -approximation of R. By Lemma 2.9, it follows that $[\operatorname {ind}_{\mu ^+_Y(U)[-1]}R:Z[-1]]<0$ .

$(3)\Rightarrow (1)$ . Since $Z\not \cong Y$ , this implication follows from Lemma 2.11 directly.

Proof. By Proposition 2.7, there is a completion X of N. By Lemma 2.9, we have $[\operatorname {ind}_{(N\,{\oplus}\, X)[-1]}R:X[-1]]\neq 0$ . If , we take $Y=X$ and $Y^{\prime }=\mu ^+_X(N\,{\oplus}\, X)\setminus N$ . Then by Proposition 2.14, the triangle (2.4) becomes the required one. If $[\operatorname {ind}_{(N\,{\oplus}\, X)[-1]}R:X[-1]]<0$ , we take $Y^{\prime }=X$ and $Y=\mu ^-_{X}(N\,{\oplus}\, X)\setminus N$ . Then by Proposition 2.14, the triangle (2.5) becomes the required one.

Proof. Let $Y^{\prime }$ be another completion of N. If $\mu ^+_Y(U)\in R\ast R[1]$ , then it is maximal rigid with respect to $R\ast R[1]$ by Lemma 2.13. So by Corollary 2.16, we have $\mu ^+_Y(U)=N\,{\oplus}\, Y^{\prime }$ and . Then by Theorem 2.15, there is a triangle

$$ \begin{align*}Y^{\prime}\to E\to Y\to Y^{\prime}[1],\end{align*} $$

with $E\in \operatorname {add} N$ and such that there is an exact sequence

$$ \begin{align*}\operatorname{Hom}(R,Y^{\prime})\to\operatorname{Hom}(R,E)\to\operatorname{Hom}(R,Y)\to 0.\end{align*} $$

In particular, we have $\operatorname {Hom}(R,U)\in \operatorname {Fac}\operatorname {Hom}(R,N)\subseteq \operatorname {Fac}\operatorname {Hom}(R,\mu ^+_Y(U))$ . By definition, we have , which implies $\Phi (\mu ^+_Y(U))=\mu ^+_{\Phi (Y)}(\Phi (U))$ , as required. The case $\mu ^-_Y(U)\in R\ast R[1]$ can be proved dually.

Proof. Let $U=G\,{\oplus}\, X$ be a basic object in $R\ast R[1]$ . By [Reference Iyama and Yoshino35, Lemma 4.8], U is rigid in $\mathcal {D}$ if and only if X is rigid in ${\mathcal {D}_G}$ . Since an indecomposable object in $R\ast R[1]$ is isomorphic to zero in ${\mathcal {D}_G}$ if and only if it is isomorphic to a direct summand of G, we have $|U|=|R|$ in $\mathcal {D}$ if and only if $|X|=|R\setminus G|$ in ${\mathcal {D}_G}$ . So by Proposition 2.7, we have that $U\in \operatorname {max}\operatorname {rigid}_G\operatorname {-}(R\ast _{\mathcal {D}} R[1])$ if and only if $X\in \operatorname {max}\operatorname {rigid-}(R\ast _{\mathcal {D}_G} R{\langle }1{\rangle })$ . Thus, we get the bijection $\Psi $ .

By Theorem 2.23, a minimal U-presentation

$$\begin{align*}R\xrightarrow{f} U^1_R\xrightarrow{g} U^0_R\to R[1]\end{align*}$$

of $R[1]$ in $\mathcal {D}$ gives rise to a minimal X-presentation

$$\begin{align*}R\xrightarrow{\overline{f}} U^1_R\xrightarrow{\overline{g}} U^0_R\to R{\langle}1{\rangle}\end{align*}$$

of $R{\langle }1{\rangle }$ in ${\mathcal {D}_G}$ . So for any indecomposable summand Y of X, we have

$$ \begin{align*}[\operatorname{ind}_{U[-1]}R:Y[-1]]=[\operatorname{ind}_{U{\langle}-1{\rangle}}R:Y{\langle}-1{\rangle}].\end{align*} $$

Then for any $\varepsilon \in \{+,-\}$ , by Proposition 2.14, we have $\mu ^{\varepsilon }_Y(U)\in \operatorname {max}\operatorname {rigid}_G\operatorname {-}(R\ast _{\mathcal {D}} R[1])$ if and only if $\mu ^{\varepsilon }_{\Psi (Y)}(\Psi (U))\in \operatorname {max}\operatorname {rigid-}(R\ast _{\mathcal {D}_G} R{\langle }1{\rangle })$ . Then the last assertion follows from [Reference Iyama and Yoshino35, Proposition 4.4 (2)].

Construction 3.8. Let $\mathbf {R}$ be a partial tagged triangulation. We construct a partial ideal triangulation $\mathbf {R}^\circ $ by

1. (1) for $p\in \mathbf {P}_{\mathbf {R}}$ with $\kappa _{\mathbf {R}}(p)\neq 0$ , removing the taggings of tagged arcs in $\mathbf {R}$ at p, and

2. (2) for $p\in \mathbf {P}_{\mathbf {R}}$ with $\kappa _{\mathbf {R}}(p)=0$ , replacing the two tagged arcs in $\mathbf {R}$ incident to p by a self-folded triangle enclosing p, as shown in Figure 1 (from right to left).

Note that when $\mathbf {R}=\mathbf {T}$ is a tagged triangulation, $\mathbf {T}^\circ $ is an ideal triangulation. For any tagged arc $\gamma \in \mathbf {R}$ , we denote by $\gamma ^{\circ _{\mathbf {R}}}$ the arc in $\mathbf {R}^\circ $ corresponding to $\gamma $ . Usually, we simply denote $\gamma ^{\circ _{\mathbf {R}}}$ by $\gamma ^{\circ }$ , when there is no confusion arising.

Construction 3.9. Let $\mathbf {R}$ be a partial tagged triangulation. For any tagged arc $\delta $ , we construct a tagged arc $\delta ^{\mathbf {R}}$ , which is obtained from $\delta $ by changing taggings at $p\in \mathbf {P}_{\mathbf {R}}$ with $\kappa _{\mathbf {R}}(p)<0$ .

Proof. Starting from the quiver $Q^{\mathbf {T}}$ , there is a cluster algebra $\mathcal {A}$ [Reference Fomin and Zelevinsky27]. By [Reference Fomin, Shapiro and Thurston25, Theorem 7.11], there is a bijection x from the set $\operatorname {A}^\times (\mathbf {S})$ of tagged arcs on $\mathbf {S}$ to the set of cluster variables of $\mathcal {A}$ , such that x induces a bijection from the set of tagged triangulations to the set of clusters of $\mathcal {A}$ , which sends $\mathbf {T}$ to the initial cluster and commutes with flips of tagged triangulations and mutations of clusters.

For the 2-Calabi-Yau category $\mathcal {C}(\mathbf {S})$ with cluster tilting object $T^{\mathbf {T}}$ , by [Reference Palu47] and Theorem 3.15 (4), there is a bijection $cc$ (called cluster character) from the set of indecomposable objects in $\mathcal {C}(\mathbf {S})$ to the set of cluster variables of $\mathcal {A}$ , such that $cc$ induces a bijection from the set of basic cluster tilting objects to the set of reachable clusters of $\mathcal {A}$ , which sends $T^{\mathbf {T}}$ to the initial cluster and commutes with mutations.

Then by Theorem 3.15, we have the following commutative diagram of bijections:

So for any tagged arc $\delta $ , $x(\delta )=cc(X(\delta ))$ . By [Reference Reading50, Proposition 5.2] (cf. also [Reference Labardini-Fragoso41, Theorem 7.1]), the g-vector of $x(\delta )$ is $-b_{\mathbf {T}}(e(\delta ))$ . However, by [Reference Plamondon48, Proposition 3.6], the g-vector of $cc(X(\delta ))$ is $\operatorname {ind}_{T^{\mathbf {T}}}X(\delta )$ . Thus, we get (3.2).

For (3.3), we have

$$ \begin{align*} \operatorname{ind}_{X(\mathbf{T})[-1]}X(\delta)&{}=\operatorname{ind}_{X(\mathbf{T})}X(\delta)[1]\\&=\operatorname{ind}_{X(\mathbf{T})}X(\rho(\delta)){}\\ &=-b_{\mathbf{T}}(e(\rho(\delta)))\\&=-b_{\mathbf{T}}(e^{op}(\delta)), \end{align*} $$

where the second equality holds by Theorem 3.15 (1), the third one is (3.2), and the last one is due to Remark 3.12.

Proof. By Proposition 3.16, $b_{\gamma ,\mathbf {T}}(e(\delta ))=0$ if and only if $[\operatorname {ind}_{X(\mathbf {T})}X(\delta ):X(\gamma )]=0$ . Then by Remark 2.5, we get this assertion.

Proof. By Theorem 3.15 and the definition of index, we have

$$ \begin{align*}[\operatorname{ind}_{X(\mathbf{T})}X(\delta):X(\gamma)]=\begin{cases}1&\mathrm{if}\ \gamma=\delta,\\ 0&\mathrm{otherwise,}\end{cases}\end{align*} $$

and

$$ \begin{align*}[\operatorname{ind}_{X(\mathbf{T})[-1]}X(\delta):X(\gamma)[-1]]=\begin{cases}-1&\mathrm{if}\ \gamma=\delta,\\ 0&\mathrm{otherwise.}\end{cases}\end{align*} $$

Then by Proposition 3.16, the required formulas follow.

Proof. We only show the equivalences between statements for the $\mathbf {R}$ -standard case, since the $\mathbf {R}$ -co-standard case can be shown dually. By Corollary 3.17, (4) is equivalent to (3). By definition, (1) is equivalent to that $\delta ^{\mathbf {R}}$ is $\mathbf {R}^\circ $ -standard, (2) is equivalent to that $e(\delta ^{\mathbf {R}})$ shears $\mathbf {R}^\circ $ . By Remark 3.10, (3) is equivalent to that for some/any tagged triangulation $\mathbf {T}$ satisfying the condition in Remark 3.10 and $\mathbf {R}\subset \mathbf {T}$ , we have $b_{\gamma ^\circ ,\mathbf {T}^\circ }(e(\delta ^{\mathbf {R}}))=0$ for any $\gamma \in \mathbf {T}\setminus \mathbf {R}$ . So we may assume that $\mathbf {R}$ is a partial ideal triangulation.

‘(1) $\Rightarrow $ (2)’: We list all the possible cases of arc segments of an $\mathbf {R}$ -standard tagged arc $\delta $ and its elementary laminate $e(\delta )$ in Figure 6, where the pictures in the first row are for $\delta \in \mathbf {R}^\times $ and $\delta ^\circ $ not a side of a self-folded triangle, those in the second row are for $\delta \in \mathbf {R}^\times $ and $\delta ^\circ $ a side of a self-folded triangle, those in the third row are for $\delta $ adjoint to $\mathbf {R}^\times $ but not in $\mathbf {R}^\times $ , those in the fourth row are for the arc segments in the first two pictures of Figure 4, and those in the last row are for the arc segments in the last two pictures of Figure 4. In each case, any segment of $e(\delta )$ cuts out an angle between two arcs in $\mathbf {R}\cup B$ . Hence, (2) holds.

Figure 6

The elementary laminate of a standard tagged arc.

‘(2) $\Rightarrow $ (3)’: By (2), $e(\delta )$ is divided by $\mathbf {R}$ into segments, each of which cuts out an angle between two arcs in $\mathbf {R}\cup B$ . Let $\mathbf {T}$ be an ideal triangulation containing $\mathbf {R}$ , $\gamma \in \mathbf {T}\setminus \mathbf {R}$ and q an intersection between $\gamma $ and a segment $\eta $ of $e(\delta )$ . Then $\gamma $ divides the angle cut by $\eta $ into two parts. Hence, $b_{q,\gamma ,\mathbf {T}}(e(\delta ))=0$ (cf. the third picture in Figure 3). Thus, we have $b_{\gamma ,\mathbf {T}}(e(\delta ))=0$ .

‘(3) $\Rightarrow $ (1)’: If $\delta $ has no interior intersections with any arc in $\mathbf {R}$ , then we have the following three subcases.

1. (i) $\operatorname {Int}(\delta ,\gamma ^\times )=0$ for any $\gamma \in \mathbf {R}$ . Then there is a tagged triangulation $\mathbf {T}_1$ containing $\mathbf {R}^\times $ and $\delta $ . By Corollary 3.18, we have $b_{\delta ,\mathbf {T}_1}(e(\delta ))=-1$ . Hence by (3), we have $\delta \in \mathbf {R}^\times $ , which implies $\delta $ is $\mathbf {R}$ -standard.

2. (ii) Exactly one endpoint of $\delta $ , say $\delta (0)$ , is a tagged intersection with some arc in $\mathbf {R}^\times $ . Then $\delta (0)\neq \delta (1)$ . If $\kappa _\delta (0)=1$ , then there is an arc $\epsilon \in \mathbf {R}$ such that $\epsilon ^\times (0)=\delta (0)$ and $\kappa _{\epsilon ^\times }(0)=-1$ . Then $\epsilon $ is a loop enclosing $\delta (0)$ . Note that $\delta $ has no interior intersections with any arc in $\mathbf {R}$ . So $\delta $ is in the once-punctured monogon enclosed by $\epsilon $ . Thus, $\delta $ is homotopic to $\epsilon $ . Then by the definition of tagged intersections, $\delta (1)$ is also a tagged intersection between $\delta $ and $\epsilon ^\times $ , a contradiction with the setting of subsection (ii). Hence, $\kappa _\delta (0)=-1$ . Let $\delta _0$ be the tagged arc homotopic to $\delta $ and whose tagging at each tagged end is $1$ . Since $\delta _0$ has zero (if $\kappa _\delta (1)=1$ ) or two (if $\kappa _\delta (1)=-1$ ) tagged intersections with $\delta $ , there are arcs in $\mathbf {R}$ which have $\delta (0)$ as an endpoint and is not homotopic to $\delta _0$ . Take $\gamma $ to be the first arc among them next to $\delta _0$ in the anticlockwise order around $\delta (0)$ . Let $\gamma '$ be the arc or the boundary segment such that $\Delta =\{\gamma ^{\prime },\delta _0,\gamma \}$ is a (possibly self-folded) triangle with the angle from $\delta _0$ to $\gamma $ in the anticlockwise order as an inner angle. Since $\delta _0$ has no interior intersections with arcs in $\mathbf {R}$ , neither does $\gamma ^{\prime }$ . Let $\mathbf {T}_2$ be an ideal triangulation containing $\mathbf {R}$ , $\delta _0$ and $\gamma ^{\prime }$ . If $\Delta $ is self-folded, since $\gamma $ is not homotopic to $\delta $ and $\delta $ is a tagged arc (which implies that $\delta $ is not the non-folded side of a self-fold triangle), we have that $\gamma ^{\prime }$ and $\delta _0$ are folded, and $\gamma $ is the non-folded side (cf. the last picture in Figure 7). So $\delta (1)$ is a tagged intersection with either $\gamma ^\times $ (if $\kappa _\delta (1)=1$ ) or ${\gamma ^{\prime }}^\times =\delta _0$ (if $\kappa _\delta (1)=-1$ ). But since $\gamma \in \mathbf {R}$ , by Definition 3.2, we have $\gamma ^{\prime }\in \mathbf {R}$ , a contradiction with the setting of subsection (ii). Hence, the triangle $\Delta $ is not self-folded. In particular, none of $\gamma $ , $\delta _0$ and $\gamma ^{\prime }$ is the folded side of a self-folded triangle in $\mathbf {T}_2$ .

   Figure 7

   No interior intersections but one tagged intersection.

   

   1. (a) If $\gamma ^{\prime }$ is a boundary segment (see the first picture in Figure 7), by Definition 4.2 (2) (see the first picture in Figure 4), $\delta $ is $\mathbf {R}$ -standard.

   2. (b) If $\gamma ^{\prime }$ is not a boundary segment and either $\delta (1)\in \mathbf {M}$ or $\delta (1)\in \mathbf {P}$ with $\kappa _\delta (1)=1$ (see the second and the third pictures in Figure 7), by definition, $b_{q,\gamma ^{\prime },\mathbf {T}_2}(e(\delta ))\neq 0$ . Then by Remark 3.14, we have $b_{\gamma ^{\prime },\mathbf {T}_2}(e(\delta ))\neq 0$ , which implies $\gamma ^{\prime }\in \mathbf {R}$ by (3). Then by Definition 4.2 (2) (see the first picture in Figure 4), $\delta $ is $\mathbf {R}$ -standard.

   3. (c) If $\delta (1)\in \mathbf {P}$ and $\kappa _\delta (1)=-1$ (see the fourth picture in Figure 7), we have $\delta =\rho (\delta _0)$ . So by Corollary 3.18, $b_{\delta _0,\mathbf {T}_2}(e(\delta ))=-1$ . Then by (3), $\delta _0\in \mathbf {R}$ , a contradiction with the setting of subsection (ii), since $\delta (1)$ is a tagged intersection between $\delta $ and $\delta ^\times _0\in \mathbf {R}^\times $ .

3. (iii) Both endpoints of $\delta $ are tagged intersections between $\delta $ and some arcs in $\mathbf {R}^\times $ . Let $\delta _0$ be the underlying arc of $\delta $ . Note that $\delta _0$ has no intersections with $\mathbf {R}$ . Let $\mathbf {T}_3$ be an ideal triangulation containing $\mathbf {R}$ and $\delta _0$ . Then $b_{\delta _0,\mathbf {T}_3}(e(\delta ))\neq 0$ by Corollary 3.18, which by (3) implies $\delta _0\in \mathbf {R}$ . So $\delta $ is adjoint to $\delta ^\times _0\in \mathbf {R}^\times $ and hence is $\mathbf {R}$ -standard.

If $\delta $ has interior intersections with some arc in $\mathbf {R}$ , then $\delta $ is divided by $\mathbf {R}$ into at least two arc segments. Let $\gamma $ be the first arc in $\mathbf {T}\cup B$ next to $\delta $ around $\delta (0)$ in the anticlockwise order if $\kappa _\delta (0)=-1$ or in the clockwise order otherwise. If $\gamma \in \mathbf {T}$ , we have $b_{\gamma ,\mathbf {T}}(e(\delta ))\neq 0$ as shown in the pictures of the last row in Figure 6. Then by (3), we have $\gamma \in \mathbf {R}$ . So the end arc segment of $\delta $ is as shown in the last two pictures in Figure 4. For any arc segment $\eta $ of $\delta $ between two arcs $\gamma _1,\gamma _2\in \mathbf {R}$ , let $\gamma ^{\prime }_0=\gamma _1,\gamma ^{\prime }_1,\cdots ,\gamma ^{\prime }_{s-1},\gamma ^{\prime }_s=\gamma _2$ be the arcs in $\mathbf {T}$ which $\eta $ crosses in order. By (3), we have $b_{\gamma ^{\prime }_i,\mathbf {T}}(e(\delta ))=0$ for any $1\leq i\leq s-1$ . Hence, $\eta $ cuts out an angle bounded by $\gamma _1$ and $\gamma _2$ (cf. the third picture in Figure 3). So $\delta $ is $\mathbf {R}$ -standard.

Proof. By the equivalence between (1) and (4) in Proposition 4.4, we get the required bijections. The last assertion then follows from Proposition 3.16 and the equality (3.1).

Proof. This follows directly from Theorem 4.7 and Proposition 2.7.

Proof. By Theorem 4.7, $\mathbf {U}\in D(\mathbf {R})$ if and only if $X(\mathbf {U})\in \operatorname {max}\operatorname {rigid-}(X(\mathbf {R})\ast X(\mathbf {R})[1])$ , and $\mathbf {R}\in D^{op}(\mathbf {U})$ if and only if $X(\mathbf {R})\in \operatorname {max}\operatorname {rigid-}(X(\mathbf {U})[-1]\ast X(\mathbf {U}))$ . So the statement follows by Corollary 2.8.

Proof. By Corollary 4.9, $\mathbf {R}$ is a $\mathbf {U}$ -co-dissection. In particular, any $\gamma \in \mathbf {R}$ is $\mathbf {U}$ -co-standard. Then for any $l\in \mathbf {U}$ , by (4.4), we have

$$ \begin{align*}\sum_{\gamma\in\mathbf{R}}b_{l,\mathbf{U}}(e^{op}(\gamma))=-[\operatorname{ind}_{X(\mathbf{U})[-1]}X(\mathbf{R}):X(l)[-1]],\end{align*} $$

which is not zero by Lemma 2.9. Hence, there exists $\gamma \in \mathbf {R}$ such that $b_{l,\mathbf {U}}(e^{op}(\gamma ))\neq 0$ . The last assertion then follows by Remark 2.17.

Proof. For any $\gamma \in \operatorname {A}^\times _{\mathbf {R}\operatorname {-st}}\setminus \rho (\mathbf {R})$ , define $M(\gamma )=\operatorname {Hom}_{\mathcal {C}(\mathbf {S})}(X(\mathbf {R}),X(\gamma ))$ . Then by Theorem 4.7 and Theorem 2.21, M is a bijection from $\operatorname {A}^\times _{\mathbf {R}\operatorname {-st}}\setminus \rho (\mathbf {R})$ to $\operatorname {ind} \operatorname {\tau -rigid}\operatorname {mod}\Lambda $ .

Let $X_i=X(\gamma _i)$ and $M_i=M(\gamma _i)$ , $i=1,2$ . Then we have

$$\begin{align*}\begin{aligned} \operatorname{Hom}_{\mathcal{C}(\mathbf{S})}(X_1,X_2[1])&{}=\operatorname{Hom}_{\mathcal{C}(\mathbf{S})/R[1]}(X_1,X_2[1])\,{\oplus}\,[R[1]](X_1,X_2[1])\\ &\cong\operatorname{Hom}_{\mathcal{C}(\mathbf{S})/R[1]}(X_1,X_2[1])\,{\oplus}\, D\operatorname{Hom}_{\mathcal{C}(\mathbf{S})/R[1]}(X_2,X_1[1])\\ &\cong\operatorname{Hom}_{\Lambda_R}(M_1,\tau M_2)\,{\oplus}\, D\operatorname{Hom}_{\Lambda_R}(M_2,\tau M_1), \end{aligned}\end{align*}$$

where the first isomorphism is due to [Reference Chang, Zhang and Zhu18, Lemma 2.3] and the last one is due to Theorem 2.21. By Theorem 3.15, we have $\operatorname {Int}(\gamma _1,\gamma _2)=\dim _{\mathbf {k}}\operatorname {Hom}_{\mathcal {C}(\mathbf {S})}(X_1,X_2[1])$ . Hence, we get the required formula.