Nakayama Algebras and Fuchsian Singularities

Abstract

This present paper is devoted to the study of a class of Nakayama algebras $N_n(r)$ given by the path algebra of the equioriented quiver $\mathbb {A}_n$ subject to the nilpotency degree r for each sequence of r consecutive arrows. We show that the Nakayama algebras $N_n(r)$ for certain pairs (n, r) can be realized as endomorphism algebras of tilting objects in the bounded derived category of coherent sheaves over a weighted projective line, or in its stable category of vector bundles. Moreover, we classify all the Nakayama algebras $N_n(r)$ of Fuchsian type, that is, derived equivalent to the bounded derived categories of extended canonical algebras. We also provide a new way to prove the classification result on Nakayama algebras of piecewise hereditary type, which have been done by Happel–Seidel before.

Appendix A: Extension-free property

Let $\mathcal {D}$ be a Hom-finite triangulated category with Serre functor ${\mathbb {S}}=\tau [1]$, where $\tau$ is the Auslander-Reiten translation and [1] denotes the suspension functor of $\mathcal {D}$. The Serre duality of $\mathcal {D}$ is given by the following natural isomorphism

$$\begin{aligned} {\textrm{Hom}}_{\mathcal {D}}(X,Y)\cong D \mathrm{Hom}_{\mathcal {D}}(Y, {\mathbb {S}}X), \end{aligned}$$

which is functorial in $X,Y\in \mathcal {D}$.

Lemma A.1

Let $X\in \mathcal {D}$ and $U=\bigoplus \limits _{k=0}^{m}{\mathbb {S}}^{k}X$ for some $m\ge 0$. Then U is extension-free in $\mathcal {D}$ if and only if $\mathrm{{Hom}}_{\mathcal {D}}(X, {\mathbb {S}}^{k}X[n])=0$ for any $1\le k\le m+1$ and any non-zero integer n.

Proof

By definition, U is extension-free if and only if for any $n\ne 0$ and $0\le i,j \le m$, $\mathrm{{Hom}}_{\mathcal {D}} ({\mathbb {S}}^{i}X, {\mathbb {S}}^{j}X[n])=0$, or equivalently, $\mathrm{{Hom}}_{\mathcal {D}} (X, {\mathbb {S}}^{k}X[n])=0$ for any $n\ne 0$ and $-m\le k \le m$. For $-m\le k\le 0$, we have $\mathrm{{Hom}}_{\mathcal {D}} (X, {\mathbb {S}}^{k}X[n])\cong D\mathrm{{Hom}}_{\mathcal {D}} (X, {\mathbb {S}}^{1-k}X[-n])$ by using Serre duality, where $1\le 1-k\le m+1$. This finishes the proof. $\square$

Lemma A.2

Let $X,Y\in \mathcal {D}$ and $U=\bigoplus \limits _{k=0}^{m}{\mathbb {S}}^{k}X$, $V=\bigoplus \limits _{k=0}^{r}{\mathbb {S}}^{k}Y$ for some $m,r\ge 0$. Assume both of U and V are extension-free. Then $U\oplus V$ is extension-free if and only if $\mathrm{{Hom}}_{\mathcal {D}}(X, {\mathbb {S}}^{k}Y[n])=0$ for any $-m\le k\le r+1$ and any non-zero integer n.

Proof

By definition, $U\oplus V$ is extension-free if and only if for any $n\ne 0$, $0\le i \le m$ and $0\le j \le r$, $\mathrm{{Hom}}_{\mathcal {D}} ({\mathbb {S}}^{i}X, {\mathbb {S}}^{j}Y[n])=0$ and $\mathrm{{Hom}}_{\mathcal {D}} ({\mathbb {S}}^{j}Y[n], {\mathbb {S}}^{i}X)=D\mathrm{{Hom}}_{\mathcal {D}} ({\mathbb {S}}^{i}X, {\mathbb {S}}^{j+1}Y[n])=0$. Equivalently, $\mathrm{{Hom}}_{\mathcal {D}} (X, {\mathbb {S}}^{k}Y[n])=0$ for any $n\ne 0$ and $-m\le k \le r+1$. This proves the result. $\square$

In the following, we will show that the objects $T_{(2,4,5)}$, $T_{(2,4,7)}$, $T_{(2,5,5)}$, $T_{(2,5,6)}$ constructed in Theorem 5.5 are extension-free in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}$ case by case.

Recall that for any weight type $(p_1,p_2,p_3)$, there is a surjective group homomorphism $\delta :\mathbb {L}(p_1,p_2,p_3)\rightarrow \mathbb {Z}$ given by $\delta (\vec {x}_i)=\frac{p}{p_i}$ for $1\le i\le 3$, where $p=\mathrm{l.c.m.}(p_1,p_2,p_3)$. We denote the $\delta$-datum $(\delta (\vec {c}); \delta (\vec {x}_1), \delta (\vec {x}_2), \delta (\vec {x}_3); \delta (\vec {\omega }))$ by $\delta (\vec {c}; \vec {x}_1, \vec {x}_2, \vec {x}_3; \vec {\omega })$ for convenience.

For a weighted projective line $\mathbb {X}$ of negative Euler characteristic, the associated Fuchsian singularity R is the restricted subalgebra $S|_{\mathbb {Z}\vec {\omega }}$ of the coordinate algebra of $\mathbb {X}$ defined in Eq. 2.1, i.e.,

$$\begin{aligned} R=\bigoplus _{n\ge 0}\textrm{Hom}_{}({\mathcal O}_{\mathbb {X}}, \tau ^n{\mathcal O}_{\mathbb {X}})=S|_{\mathbb {Z}\vec {\omega }}. \end{aligned}$$

1.1 A.1 The Weight Type (2,4,5)

Assume $\mathbb {X}$ has weight type (2,4,5). By easy calculation we get the $\delta$-datum $\delta (\vec {c}; \vec {x}_1, \vec {x}_2, \vec {x}_3; \vec {\omega })$$=(20;10,5,4;1)$.

According to [24], the semigroup $\{n\vec {\omega }| n\vec {\omega }\ge 0\}$ of $\mathbb {Z}\vec {\omega }$ has minimal generating system $\{4\vec {\omega }=\vec {x}_3; 10\vec {\omega }=2\vec {x}_2; 15\vec {\omega }=\vec {x}_1+\vec {x}_2\}$. Hence, the subalgebra $R=S|_{\mathbb {Z}\vec {\omega }}\cong k[x,y,z]/(f)$, where $x=x_3, y=x_2^2, z=x_1x_2$ and $f=z^2+y^3+x^5y$. Moreover, R is $\mathbb {Z}$-graded in the sense that $\deg (x,y,z;f)=(4,10,15;30)$. Thus the second suspension functor of $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,5)$ is given by degree shift: $[2]=(30\vec {\omega })=(\vec {c}+2\vec {x}_2)$.

Since $4\vec {\omega }=\vec {x}_3>0$, the set $\mathcal {S}$ defined in Eq. 5.1 has the form

$$\begin{aligned} \mathcal {S}=\{\vec {x}\, |\, 0\le \vec {x}\le n\vec {\omega }+\vec {c}, \text {\ for\ any \ } 2\le n\le 5\} =\{0, \vec {x}_2\}. \end{aligned}$$

Hence each line bundle belongs to the orbit $\tau ^{\mathbb {Z}}{\mathcal O}$ or $\tau ^{\mathbb {Z}}{\mathcal O}(\vec {x}_2)$ by Proposition 5.1.

By Proposition 5.3, the projective cover of ${\mathcal O}(\vec {x}_2)$, under $\tau ^{\mathbb {Z}}{\mathcal O}$-exact structure on $\textrm{vect}-\mathbb {X}$, is given by $\mathcal {P}({\mathcal O}(\vec {x}_2))={\mathcal O}\oplus {\mathcal O}(-5\vec {\omega })$, which fits into the following exact sequence:

(A.1)

Lemma A.3

In $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,5)$, we have ${\mathcal O}(\vec {x}_2)[1]={\mathcal O}(\vec {x}_2)(15\vec {\omega }).$

Proof

By Eq. A.1 we have ${\mathcal O}(\vec {x}_2)[-1]={\mathcal O}(-\vec {x}_1)={\mathcal O}(\vec {x}_2)(-15\vec {\omega })$. Then the result follows from $[2]=(30\vec {\omega })$. $\square$

Moreover, by Proposition 5.4, we have

Lemma A.4

In $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,5)$, we have

$$\begin{aligned} \underline{\textrm{Hom}}({\mathcal O}(\vec {x}_2-\vec {x}), {\mathcal O}(\vec {x}_2))\ne 0 \quad \text {if\ and\ only\ if}\quad 0\le \vec {x}\le 4\vec {x}_3. \end{aligned}$$

(A.2)

Proposition A.5

$T_{(2,4,5)}$ is extension-free in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,5)$.

Proof

Recall that $T_{(2,4,5)}=\bigoplus \limits _{k=0}^{10}L(k\vec {x}_3)$, where $L={\mathcal O}(\vec {x}_2)$. For construction we assume $T_{(2,4,5)}$ is not extension-free, then there exist some integers $0\le a, b \le 10$ and $0\ne m\in \mathbb {Z}$, such that

$$\begin{aligned} \underline{\textrm{Hom}}(L(a\vec {x}_{3}), L(b\vec {x}_{3})[m])\ne 0. \end{aligned}$$

Note that $\vec {x}_3=4\vec {\omega }$ and $L[1]=L(15\vec {\omega })$. According to Eq. A.2, we have

$$\begin{aligned} (b-a)\vec {x}_{3}+15m\vec {\omega }=k\vec {x}_{3} \text {for\ some\ } 0\le k\le 4. \end{aligned}$$

(A.3)

It follows that $4(b-a)+15m=4k$ since $\vec {x}_3=4\vec {\omega }$. Hence 4|m, say, $m=4r$ for some integer $r\ne 0$. Then we have $0\le k=b-a+15r\le 4$, a contradiction to the assumption $0\le a, b \le 10$. We are done. $\square$

1.2 A.2 The Weight Type (2,4,7)

Assume $\mathbb {X}$ has weight type (2,4,7). Then the $\delta$-datum is given by $\delta (\vec {c}; \vec {x}_1, \vec {x}_2, \vec {x}_3; \vec {\omega })=(28;14,7,4;3)$.

According to [24], the semigroup $\{n\vec {\omega }| n\vec {\omega }\ge 0\}$ of $\mathbb {Z}\vec {\omega }$ has minimal generating system $\{4\vec {\omega }=3\vec {x}_3; 6\vec {\omega }=2\vec {x}_2+\vec {x}_3; 7\vec {\omega }=\vec {x}_1+\vec {x}_2\}$. Hence, the subalgebra $R=S|_{\mathbb {Z}\vec {\omega }}\cong k[x,y,z]/(f)$, where $x=x_3^3, y=x_2^2x_3, z=x_1x_2$ and $f=y^3+x^3y+xz^2$. Moreover, R is $\mathbb {Z}$-graded in the sense that $\deg (x,y,z;f)=(4,6,7;18)$. Thus the second suspension functor of $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,7)$ is given by degree shift: $[2]=(18\vec {\omega })=(\vec {c}+2\vec {x}_2+3\vec {x}_3)$.

Since $4\vec {\omega }=3\vec {x}_3>0$, the set $\mathcal {S}$ defined in Eq. 5.1 has the form

$$\begin{aligned} \mathcal {S}=\{\vec {x}\, |\, 0\le \vec {x}\le n\vec {\omega }+\vec {c}, \text {\ for\ any \ } 2\le n\le 5\} =\{\vec {x}\,|\, 0\le \vec {x}\le \vec {x}_2+2\vec {x}_3\}. \end{aligned}$$

Hence each line bundle belongs to the orbit $\tau ^{\mathbb {Z}}{\mathcal O}(\vec {x})$ for some $0\le \vec {x}\le \vec {x}_2+2\vec {x}_3$.

By Proposition 5.3, the projective covers of ${\mathcal O}(\vec {x}_2)$ and ${\mathcal O}(\vec {x}_2+\vec {x}_3)$, under $\tau ^{\mathbb {Z}}{\mathcal O}$-exact structure on $\textrm{vect}-\mathbb {X}$, are given by $\mathcal {P}({\mathcal O}(\vec {x}_2))={\mathcal O}\oplus {\mathcal O}(-5\vec {\omega })$ and $\mathcal {P}({\mathcal O}(\vec {x}_2+\vec {x}_3))={\mathcal O}\oplus {\mathcal O}(-\vec {\omega })$, which fit into the following exact sequences:

(A.4)

(A.5)

Lemma A.6

The following statements hold in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,7)$.

• (1) ${\mathcal O}(\vec {x}_2)[1]={\mathcal O}(\vec {x}_2+\vec {x}_3)(7\vec {\omega });$

• (2) ${\mathcal O}(\vec {x}_2+\vec {x}_3)[1]={\mathcal O}(\vec {x}_2)(11\vec {\omega }).$

Proof

Since we have exact sequences (A.4) and (A.5) for projective covers, the following hold in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,7)$:

• ${\mathcal O}(\vec {x}_2)[-1]={\mathcal O}(-\vec {x}_1-2\vec {x}_3)={\mathcal O}(\vec {x}_2+\vec {x}_3)(-11\vec {\omega });$

• ${\mathcal O}(\vec {x}_2+\vec {x}_3)[-1]={\mathcal O}(-\vec {x}_1)={\mathcal O}(\vec {x}_2)(-7\vec {\omega })$.

Then the result follows by noting $[2]=(18\vec {\omega })$ in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,7)$. $\square$

Moreover, by Proposition 5.4, we have

Lemma A.7

For any $\vec {x}=\sum _{1\le i\le 3}l_i\vec {x}_i+l\vec {c}$ in normal form, the following hold in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,7)$.

• (1) $\underline{\textrm{Hom}}({\mathcal O}(\vec {x}_2-\vec {x}), {\mathcal O}(\vec {x}_2))\ne 0$ if and only if $0\le \vec {x}\le \vec {c}+\vec {x}_3$ and $l_2=0$;

• (2) $\underline{\textrm{Hom}}({\mathcal O}(\vec {x}_2+\vec {x}_3-\vec {x}), {\mathcal O}(\vec {x}_2+\vec {x}_3))\ne 0$ if and only if $0\le \vec {x}\le \vec {c}$ and $l_1=0$.

Proposition A.8

$T_{(2,4,7)}$ is extension-free in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,7)$.

Proof

Recall that $T_{(2,4,7)}=\big (\bigoplus \limits _{k=0}^{6}L(3k\vec {x}_3)\big )\oplus \big (\bigoplus \limits _{k=0}^{5}L((3k+1)\vec {x}_3)\big )$, where $L={\mathcal O}(\vec {x}_2)$. For contradiction we assume $T_{(2,4,7)}$ is not extension-free, then there exist some integers $0\le a,b\le 18$ satisfying $a,b\ne 3k+2$ for $0\le k\le 5$, and $0\ne m\in \mathbb {Z}$, such that

$$\begin{aligned} \underline{\textrm{Hom}}(L(a\vec {x}_{3}), L(b\vec {x}_{3})[m])\ne 0. \end{aligned}$$

Recall that $[2]=(18\vec {\omega })$ in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,4,7)$. We consider the following two cases.

Case 1: m is even, say, $m=2n$. We get

$$\begin{aligned} \underline{\textrm{Hom}}(L(a\vec {x}_{3}), L(b\vec {x}_{3})[m])=\underline{\textrm{Hom}}(L(a\vec {x}_{3}-18n\vec {\omega }), L(b\vec {x}_{3})). \end{aligned}$$

For $b=3k, 0\le k\le 6$, then according to Lemma A.7 (1), we have $0\le (b-a)\vec {x}_{3}+18n\vec {\omega }\le \vec {c}+\vec {x}_{3}$ and the coefficient of $\vec {x}_2$ in the normal form of $(b-a)\vec {x}_{3}+18n\vec {\omega }$ is zero. Hence 4|18n, i.e, n is even, say, $n=2n'$. Then $(b-a)\vec {x}_{3}+18n\vec {\omega }=(b-a+27n')\vec {x}_3$. Thus we get $0\le b-a+27n'\le 8$. Since $0\le a, b\le 18$, we get $n'=0$. It follows that $n=0$ and then $m=0$, a contradiction.

For $b=3k+1, 0\le k\le 5$, then according to Lemma A.7 (2), we have $0\le (b-a)\vec {x}_{3}+18n\vec {\omega }\le \vec {c}$. If n is even, say, $n=2n'$, then $0\le (b-a)\vec {x}_{3}+27n'\vec {x}_3 \le \vec {c}$, hence $0\le b-a+27n'\le 7$. Since $0\le a, b\le 18$, we get $n'=0$. It follows that $n=0$ and then $m=0$, a contradiction. If n is odd, say, $n=2n'-1$, then $0\le (b-a)\vec {x}_{3}+18n\vec {\omega }=(b-a+27n'-17)\vec {x}_3+2\vec {x}_2\le \vec {c}$. It follows that $b-a+27n'-17=0$. Hence $a\equiv 2(\textrm{mod}\ 3)$, a contradiction.

Case 2: m is odd, say, $m=2n+1$. By Serre duality we get

$$\begin{aligned} \underline{\textrm{Hom}}(L(a\vec {x}_{3}), L(b\vec {x}_{3})[2n+1])=D\underline{\textrm{Hom}}(L(b\vec {x}_{3}+18n\vec {\omega }-\vec {\omega }), L(a\vec {x}_{3})). \end{aligned}$$

According to Lemma A.7, by similar arguments as above we obtain that in the normal form of $(a-b)\vec {x}_3-18n\vec {\omega }+\vec {\omega }$, the coefficient of $\vec {x}_1$ or $\vec {x}_2$ is zero. It follows that $2|18n-1$, a contradiction.

This finishes the proof. $\square$

1.3 A.3 The Weight Type (2,5,5)

Assume $\mathbb {X}$ has weight type (2,5,5). Then the $\delta$-datum is given by $\delta (\vec {c}; \vec {x}_1, \vec {x}_2, \vec {x}_3; \vec {\omega })=(10;5,2,2;1)$.

According to [24], the semigroup $\{n\vec {\omega }| n\vec {\omega }\ge 0\}$ of $\mathbb {Z}\vec {\omega }$ has minimal generating system $\{4\vec {\omega }=\vec {x}_2+\vec {x}_3; 5\vec {\omega }=\vec {x}_1\}$, and the subalgebra $R=S|_{\mathbb {Z}\vec {\omega }}\cong k[x,y,z]/(f)$, where $x=x_2x_3, y=x_1, z=x_2^5$ and $f=z^2+y^2z+x^5$. Moreover, R is $\mathbb {Z}$-graded in the sense that $\deg (x,y,z;f)=(4,5,10;20)$. Thus the second suspension functor of $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,5)$ is given by degree shift: $[2]=(20\vec {\omega })=(2\vec {c})$.

Since $4\vec {\omega }=\vec {x}_2+\vec {x}_3>0$, the set $\mathcal {S}$ defined in Eq. 5.1 has the form

$$\begin{aligned} \mathcal {S}=\{\vec {x}\, |\, 0\le \vec {x}\le n\vec {\omega }+\vec {c}, \text {for\ any} \ 2\le n\le 5\} =\{0, \vec {x}_2, \vec {x}_3, 2\vec {x}_2, 2\vec {x}_3\}. \end{aligned}$$

Hence each line bundle belongs to the orbit $\tau ^{\mathbb {Z}}{\mathcal O}(j\vec {x}_i)$ for some $0\le j\le 2\le i\le 3$.

By Proposition 5.3, the projective covers of ${\mathcal O}(\vec {x}_3)$ and ${\mathcal O}(2\vec {x}_3)$, under $\tau ^{\mathbb {Z}}{\mathcal O}$-exact structure on $\textrm{vect}-\mathbb {X}$, are given by $\mathcal {P}({\mathcal O}(\vec {x}_3))={\mathcal O}\oplus {\mathcal O}(-6\vec {\omega })$ and $\mathcal {P}({\mathcal O}(2\vec {x}_3))={\mathcal O}\oplus {\mathcal O}(-2\vec {\omega })$, which fit into the following exact sequences:

(A.6)

(A.7)

Lemma A.9

The following statements hold in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,5)$.

• (1) ${\mathcal O}(\vec {x}_3)[1]={\mathcal O}(\vec {x}_2)(10\vec {\omega });$

• (2) ${\mathcal O}(2\vec {x}_3)[1]={\mathcal O}(2\vec {x}_2)(10\vec {\omega }).$

Proof

Since we have exact sequences (A.6) and (A.7) for projective covers, the following hold in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,5)$:

• ${\mathcal O}(\vec {x}_3)[-1]={\mathcal O}(-4\vec {x}_2)={\mathcal O}(\vec {x}_2)(-10\vec {\omega });$

• ${\mathcal O}(2\vec {x}_3)[-1]={\mathcal O}(-3\vec {x}_2)={\mathcal O}(2\vec {x}_2)(-10\vec {\omega })$.

Then the result follows by noting $[2]=(20\vec {\omega })$ in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,5)$. $\square$

Moreover, by Proposition 5.4, we have

Lemma A.10

The following statements hold in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,5)$.

• (1) $\underline{\textrm{Hom}}({\mathcal O}(\vec {x}_3-\vec {x}), {\mathcal O}(\vec {x}_3))\ne 0$ if and only if $0\le \vec {x}\le \vec {x}_1+3\vec {x}_2$;

• (2) $\underline{\textrm{Hom}}({\mathcal O}(2\vec {x}_3-\vec {x}), {\mathcal O}(2\vec {x}_3))\ne 0$ if and only if $0\le \vec {x}\le \vec {x}_1+2\vec {x}_2+\vec {x}_3.$

Proposition A.11

$T_{(2,5,5)}$ is extension-free in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,5)$.

Proof

Recall that $T_{(2,5,5)}=\bigoplus \limits _{a=1}^{4}\bigoplus \limits _{k=0}^{2}{\mathbb {S}}^k({\mathcal O}(a\vec {x}_3))$. For contradiction we assume $T_{(2,5,5)}$ is not extension-free. Then by Lemma A.2 and using Serre duality, there exist some integers $1\le a, b \le 4$, $1\le k\le 3$ and $0\ne m\in \mathbb {Z}$, such that

$$\begin{aligned} \underline{\textrm{Hom}}({\mathcal O}(a\vec {x}_{3}), {\mathbb {S}}^k({\mathcal O}(b\vec {x}_{3})[m]))\ne 0. \end{aligned}$$

Recall that $\vec {c}=10\vec {\omega }$ and $[2]=(2\vec {c})=(20\vec {\omega })$. Hence, by Lemma A.9 we have

$$\begin{aligned} {\mathbb {S}}^k({\mathcal O}(b\vec {x}_{3})[m])= \left\{ \begin{array}{lll} {\mathcal O}(b\vec {x}_2+k\vec {\omega })((k+m)\vec {c}), &{}&{} \;\text {if}\; k+m \text {\;is\; odd};\\ {\mathcal O}(b\vec {x}_3+k\vec {\omega })((k+m)\vec {c}), &{}&{} \;\text {if}\; k+m \;\text {is\; even}. \end{array} \right. \end{aligned}$$

By Lemma A.10 (and its symmetric version by exchanging $\vec {x}_2$ and $\vec {x}_3$), we have for $i=2,3$,

$$\begin{aligned} 0\le b\vec {x}_i-a\vec {x}_3+k\vec {\omega }+(k+m)\vec {c}\le \left\{ \begin{array}{lll} \vec {x}_1+3\vec {x}_2, &{}&{} \text {\;if\;} {\mathcal O}(b\vec {x}_i)\in \tau ^\mathbb {Z}({\mathcal O}(\vec {x}_3));\\ \vec {x}_1+3\vec {x}_3, &{}&{} \text {\;if\;} {\mathcal O}(b\vec {x}_i)\in \tau ^\mathbb {Z}({\mathcal O}(\vec {x}_2));\\ \vec {x}_1+2\vec {x}_2+\vec {x}_3, &{}&{} \text {\;if\;} {\mathcal O}(b\vec {x}_i)\in \tau ^\mathbb {Z}({\mathcal O}(2\vec {x}_3));\\ \vec {x}_1+\vec {x}_2+2\vec {x}_3, &{}&{} \text {\;if\;} {\mathcal O}(b\vec {x}_i)\in \tau ^\mathbb {Z}({\mathcal O}(2\vec {x}_2)).\\ \end{array} \right. \end{aligned}$$

(A.8)

It follows that $0\le 2(b-a)+k+10(k+m)\le 11$ by considering the degrees, which yields that $k+m=0$ or 1.

Case 1: $k+m=0$, then $2\le k\le 3$ since $m\ne 0$ and $(b-a)\vec {x}_3+k\vec {\omega }\ge 0$, which implies that $(b-a)\vec {x}_3\ge 2\vec {x}_3$, i.e., $b-a\ge 2$. Hence $b=3$ or 4. If $b=3$, then $a=1$. Observe that $3\vec {x}_3=2\vec {x}_2+2\vec {\omega }$, i.e., ${\mathcal O}(3\vec {x}_3)\in \tau ^\mathbb {Z}({\mathcal O}(2\vec {x}_2))$. Thus $2\vec {x}_3+k\vec {\omega }\le \vec {x}_1+\vec {x}_2+2\vec {x}_3$ by Eq. A.8, a contradiction to $2\le k\le 3$. If $b=4$, then $a=1$ or 2. Observe that $4\vec {x}_3=\vec {x}_2+6\vec {\omega }$. By Eq. A.8 we have $0\le (4-a)\vec {x}_3+k\vec {\omega }\le \vec {x}_1+3\vec {x}_3$, contradicting to $2\le k\le 3$.

Case 2: $k+m=1$, then $m\ne 0$ implies $2\le k\le 3$. Moreover, $0\le 2(b-a)+k+10\le 11$ implies that $b<a$.

• If $b=1$, then by Eq. A.8 we have $0\le \vec {x}_2-a\vec {x}_3+k\vec {\omega }+\vec {c}\le \vec {x}_1+3\vec {x}_3$. By considering the normal forms, we see that the coefficient of $\vec {x}_2$ for the middle term equals zero, which implies $k=1$, a contradiction.

• If $b=2$, then by Eq. A.8 we have $0\le 2\vec {x}_2-a\vec {x}_3+k\vec {\omega }+\vec {c}\le \vec {x}_1+\vec {x}_2+2\vec {x}_3$. By considering the coefficient of $\vec {x}_2$ in the normal form for the middle term, we get $k=2$. In this case, $2\vec {x}_2-a\vec {x}_3+2\vec {\omega }+\vec {c}=(8-a)\vec {x}_3\le \vec {x}_1+\vec {x}_2+2\vec {x}_3$, a contradiction.

• If $b=3$, then $a=3$ or 4. Since $3\vec {x}_2=2\vec {x}_3+2\vec {\omega }$, by Eq. A.8 we have $0\le 3\vec {x}_2-a\vec {x}_3+k\vec {\omega }+\vec {c}\le \vec {x}_1+2\vec {x}_2+\vec {x}_3$. By considering the coefficient of $\vec {x}_3$ in the normal form for the middle term, we get $k=2$. In this case, $3\vec {x}_2-a\vec {x}_3+2\vec {\omega }+\vec {c}=\vec {x}_2+(8-a)\vec {x}_3\le \vec {x}_1+2\vec {x}_2+\vec {x}_3$, a contradiction.

This finishes the proof. $\square$

1.4 A.4 The Weight Type (2,5,6)

Assume $\mathbb {X}$ has weight type (2,5,6). Then the $\delta$-datum is given by $\delta (\vec {c}; \vec {x}_1, \vec {x}_2, \vec {x}_3; \vec {\omega })=(30;15,6,5;4)$.

According to [24], $\{n\vec {\omega }| n\vec {\omega }\ge 0\}$ of $\mathbb {Z}\vec {\omega }$ has minimal generating system $\{4\vec {\omega }=\vec {x}_2+2\vec {x}_3; 5\vec {\omega }=\vec {x}_1+\vec {x}_3; 6\vec {\omega }=4\vec {x}_2\}$, and the subalgebra $R=S|_{\mathbb {Z}\vec {\omega }}\cong k[x,y,z]/(f)$, where $x=x_2x_3^2, y=x_1x_3, z=x_2^4$ and $f=xz^2+y^2z+x^4$. Moreover, R is $\mathbb {Z}$-graded in the sense that $\deg (x,y,z;f)=(4,5,6;16)$. Thus the second suspension functor of $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,6)$ is given by degree shift: $[2]=(16\vec {\omega })=(\vec {c}+4\vec {x}_2+2\vec {x}_3)$.

Since $4\vec {\omega }=\vec {x}_2+2\vec {x}_3>0$, the set $\mathcal {S}$ defined in Eq. 5.1 has the form

$$\begin{aligned} \mathcal {S}=\{\vec {x}\, |\, 0\le \vec {x}\le n\vec {\omega }+\vec {c}, \text {\ for\ any \ } 2\le n\le 5\} =\{\vec {x}\, |\, 0\le \vec {x}\le 3\vec {x}_3 \text {\ or } 0\le \vec {x}\le 2\vec {x}_2+\vec {x}_3 \}. \end{aligned}$$

Hence each line bundle belongs to the orbit $\tau ^{\mathbb {Z}}{\mathcal O}(\vec {x})$ for some $0\le \vec {x}\le 3\vec {x}_3$ or $0\le \vec {x}\le 2\vec {x}_2+\vec {x}_3$.

By Proposition 5.3, the projective covers of ${\mathcal O}(j\vec {x}_i)$ for $1\le j\le 2\le i\le 3$ are given by the middle terms of the following exact sequences respectively:

(A.9)

(A.10)

(A.11)

(A.12)

Lemma A.12

The following statements hold in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,6)$.

• (1) ${\mathcal O}(\vec {x}_2)[1]={\mathcal O}(2\vec {x}_3)(6\vec {\omega });$

• (2) ${\mathcal O}(\vec {x}_3)[1]={\mathcal O}(\vec {x}_2+\vec {x}_3)(5\vec {\omega });$

• (3) ${\mathcal O}(2\vec {x}_2)[1]={\mathcal O}(2\vec {x}_2)(8\vec {\omega });$

• (4) ${\mathcal O}(2\vec {x}_3)[1]={\mathcal O}(\vec {x}_2)(10\vec {\omega }).$

Proof

Since we have exact sequences (A.9)–(A.12) for projective covers, the following hold in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,6)$:

• ${\mathcal O}(\vec {x}_2)[-1]={\mathcal O}(-\vec {c})={\mathcal O}(2\vec {x}_3)(-10\vec {\omega });$

• ${\mathcal O}(\vec {x}_3)[-1]={\mathcal O}(-\vec {x}_1-3\vec {x}_2)={\mathcal O}(\vec {x}_2+\vec {x}_3)(-11\vec {\omega });$

• ${\mathcal O}(2\vec {x}_2)[-1]={\mathcal O}(-4\vec {x}_3)={\mathcal O}(2\vec {x}_2)(-8\vec {\omega });$

• ${\mathcal O}(2\vec {x}_3)[-1]={\mathcal O}(-3\vec {x}_2)={\mathcal O}(\vec {x}_2)(-6\vec {\omega }).$

Then the result follows by noting $[2]=(16\vec {\omega })$ in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,6)$. $\square$

Moreover, by Proposition 5.4, we have

Lemma A.13

For any $\vec {x}=\sum _{1\le i\le 3}l_i\vec {x}_i+l\vec {c}$ in normal form, the following hold in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,6)$.

• (1) $\underline{\textrm{Hom}}({\mathcal O}(\vec {x}_2-\vec {x}), {\mathcal O}(\vec {x}_2))\ne 0$ if and only if $0\le \vec {x}\le \vec {x}_1+5\vec {x}_3$;

• (2) $\underline{\textrm{Hom}}({\mathcal O}(\vec {x}_3-\vec {x}), {\mathcal O}(\vec {x}_3))\ne 0$ if and only if $0\le \vec {x}\le \vec {c}+2\vec {x}_2 \text {\ and\ } l_3=0$;

• (3) $\underline{\textrm{Hom}}({\mathcal O}(2\vec {x}_2-\vec {x}), {\mathcal O}(2\vec {x}_2))\ne 0$ if and only if $0\le \vec {x}\le \vec {x}_1+\vec {x}_2+3\vec {x}_3;$

• (4) $\underline{\textrm{Hom}}({\mathcal O}(2\vec {x}_3-\vec {x}), {\mathcal O}(2\vec {x}_3))\ne 0$ if and only if $0\le \vec {x}\le \vec {x}_1+2\vec {x}_2+\vec {x}_3.$

Proposition A.14

$T_{(2,5,6)}$ is extension-free in $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,6)$.

Proof

Recall that $T_{(2,5,6)}=\big (\bigoplus \limits _{k=0}^{3}{\mathbb {S}}^k({\mathcal O}(\vec {x}_3))\big )\oplus \big ( \bigoplus \limits _{k=0}^{2}{\mathbb {S}}^k({\mathcal O}(2\vec {x}_3)\oplus {\mathcal O}(4\vec {x}_3)\oplus {\mathcal O}(6\vec {x}_3))\big ).$ In $\underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,6)$ we have $[2]=(16\vec {\omega })$, ${\mathcal O}(4\vec {x}_3)={\mathcal O}(2\vec {x}_2)(2\vec {\omega })$ and ${\mathcal O}(6\vec {x}_3)={\mathcal O}(\vec {x}_2)(6\vec {\omega })$. For any objects $X,Y\in \underline{\textrm{vect}}^{\mathbb {Z}}-\mathbb {X}(2,5,6)$, by Serre duality we have

$$\begin{aligned} \underline{\textrm{Hom}}(X, Y[2n+1])=D\underline{\textrm{Hom}}(Y[2n], \tau X)=D\underline{\textrm{Hom}}(Y, X((1-16n)\vec {\omega })). \end{aligned}$$

Denote by $V_0=\bigoplus \limits _{k=0}^{3}{\mathbb {S}}^{k}({\mathcal O}(\vec {x}_3))$ and $V_i=\bigoplus \limits _{k=0}^{2}{\mathbb {S}}^{k}({\mathcal O}(2i\vec {x}_3))$ for $1\le i\le 3$. Then $T=V_0\oplus V_1\oplus V_2\oplus V_3$. We consider the following steps.

• (i) $V_0$ is extension-free. For contradiction, by Lemma A.1 there exist $m\ne 0$ and $1\le k\le 4$, such that $\underline{\textrm{Hom}}({\mathcal O}(\vec {x}_3), \tau ^{k}{\mathcal O}(\vec {x}_3)[k+m])\ne 0$. By Lemma A.13 we have

  • $\underline{\textrm{Hom}}({\mathcal O}(\vec {x}_3), {\mathcal O}(\vec {x}_3+n\vec {\omega }))\ne 0$ if and only if $n=0$ or 6.

  If $k+m=2n$ for some n, then $k+16n=0$ or 6, it follows that $k=n=0$, a contradiction; if $k+m=2n+1$ for some n, then $1-(k+16n)=0$ or 6, it follows that $n=0$ and $k=1$, and then $m=0$, a contradiction.

• (ii) $V_i$ is extension-free for any $1\le i\le 3$. For contradiction, by Lemma A.1 there exist $m\ne 0$ and $1\le k\le 3$, such that $\underline{\textrm{Hom}}({\mathcal O}(2i\vec {x}_3), \tau ^{k}{\mathcal O}(2i\vec {x}_3)[k+m])\ne 0$. By Lemma A.13 we have

  • $\underline{\textrm{Hom}}({\mathcal O}(2\vec {x}_3), {\mathcal O}(2\vec {x}_3+n\vec {\omega }))\ne 0$ if and only if $n=0$ or 5;

  • $\underline{\textrm{Hom}}({\mathcal O}(4\vec {x}_3), {\mathcal O}(4\vec {x}_3+n\vec {\omega }))\ne 0$ if and only if $n=0,4,5$ or 9;

  • $\underline{\textrm{Hom}}({\mathcal O}(6\vec {x}_3), {\mathcal O}(6\vec {x}_3+n\vec {\omega }))\ne 0$ if and only if $n=0$ or 5.

  If $k+m=2n$ for some n, then $k+16n\in \{0,4,5,9\}$, it follows that $k=n=0$, a contradiction; if $k+m=2n+1$ for some n, then $1-(k+16n)\in \{0,4,5,9\}$, it follows that $n=0, k=1$, and then $m=0$, a contradiction.

• (iii) $V_0\oplus V_i$ is extension-free for any $1\le i\le 3$. For contradiction, by Lemma A.2 there exist $m\ne 0$ and $-3\le k\le 3$, such that $\underline{\textrm{Hom}}({\mathcal O}(\vec {x}_3), \tau ^{k}{\mathcal O}(2i\vec {x}_3)[k+m])\ne 0$. By Lemma A.13 we have

  • $\underline{\textrm{Hom}}({\mathcal O}(\vec {x}_3), {\mathcal O}(2i\vec {x}_3+n\vec {\omega }))\ne 0$ if and only if $n=0$;

  • $\underline{\textrm{Hom}}({\mathcal O}(2i\vec {x}_3), {\mathcal O}(\vec {x}_3+n\vec {\omega }))\ne 0$ if and only if $n=4i+1$.

  If $k+m=2n$ for some n, then $k+16n=0$, it follows that $k=n=0$, then $m=0$, a contradiction; if $k+m=2n+1$ for some n, then $1-(k+16n)=4i+1$, it follows that $n=0$, then $k=-4i$, a contradiction.

• (iv) $V_i\oplus V_j$ is extension-free for any $1\le i<j\le 3$. For contradiction, by Lemma A.2 there exist $m\ne 0$ and $-2\le k\le 3$, such that $\underline{\textrm{Hom}}({\mathcal O}(2i\vec {x}_3), \tau ^{k}{\mathcal O}(2j\vec {x}_3)[k+m])\ne 0$. By Lemma A.13 we have

  • $\underline{\textrm{Hom}}({\mathcal O}(2i\vec {x}_3), {\mathcal O}(2j\vec {x}_3+n\vec {\omega }))\ne 0$ if and only if $n=0$ or 5;

  • $\underline{\textrm{Hom}}({\mathcal O}(2j\vec {x}_3), {\mathcal O}(2i\vec {x}_3+n\vec {\omega }))\ne 0$ if and only if $n=4(j-i)$ or $4(j-i)+5$.

  If $k+m=2n$ for some n, then $k+16n=0$ or 5, it follows that $k=n=0$, then $m=0$, a contradiction; if $k+m=2n+1$ for some n, then $1-(k+16n)=4(j-i)$ or $4(j-i)+5$, it follows that $n=0$ and $k=1-4(j-i)$ or $-4-4(j-i)$, a contradiction.

Then the proof is finished. $\square$