**Proposition 3.13.** Let $T = \alpha_1 \cdots \alpha_n$ be a closed A-trail with $n \ge 2$, $\rho_T \subseteq \rho$, and such that assumptions (i) and (ii) hold. Let $p$ be an A-path of A-length $d$ such that $\mathfrak{d}_\beta(\bar{p}) = 0$ if $\beta \in Q_A \setminus \{\alpha_1, \dots, \alpha_n\}$ and which is not an A-subpath of a power of $T$. Then $p \notin \rho$.

*Proof:* The proof is very similar to that of Proposition 2.11, replacing paths with A-paths and using Consequence 3.5 in the proof that $d > k$. $\square$

*Remark 3.14.* We keep the assumptions and notation of Proposition 3.13. Set $T_i = \alpha_i \cdots \alpha_n \alpha_1 \cdots \alpha_{i-1}$. Then for any $j \ge 1$, we have

$$ \bar{T}_i^j \bar{\alpha}_k \neq 0 \iff k = i - 1, $$

$$ \bar{\alpha}_k \bar{T}_i^j \neq 0 \iff k = i. $$

**Lemma 3.15.** Let $T = \alpha_1 \cdots \alpha_n$ be a closed A-trail that satisfies assumptions (i) and (ii) and set $z_j = \sum_{i=1}^n \bar{T}_i^j$ with $nj = \delta(u)$ for some $u \ge 1$. Then $z_j \in Z$ if, and only if, $T$ satisfies Condition 3.6(2).

Moreover, if $T$ does not satisfy Condition 3.6(2), then no element in $B$ that is homogeneous with respect to $|\cdot|$ and $\mathfrak{d}$ (when viewed in $\mathfrak{h}\Lambda$) and that is a linear combination of non-trivial cycles lying on $T$ is in $Z$.

*Proof:* The proof is very similar to that of Lemma 2.13, replacing paths with A-paths, again using Remark 3.11, replacing the $d$-covering property with Consequence 3.3 and Proposition 2.11 with Proposition 3.13. Note also that, for the proof of the last part, testing commutation with paths in $B_2$ gives $s = n$ and $\lambda_k = \lambda_{k+d}$ for all $k$ and hence the result if $|z|$ is odd; and if $|z|$ is even, we must use the fact that $z$ commutes with elements in $B_3$ in a similar way to obtain, in addition, that $\lambda_k = \lambda_{k+d+1}$ for all $k$ and hence that $\lambda_k = \lambda_{k+1}$ for all $k$. $\square$

*Proof of Theorem 3.10:* We note first that if $\text{gldim } \Lambda$ is finite, then $\Lambda$ satisfies (Fg) and Condition 3.6 is empty. The implication (C2) ⇒ (C1) is Theorem 3.9. Again, the implication (C3) ⇒ (C4) is clear, and if in addition $K$ is algebraically closed, then the implication (C1) ⇒ (C3) follows from [7]. It remains to prove that (C4) implies (C2). The proof is similar to that of Theorem 2.8, again replacing paths with A-paths (we need not assume that the integers $k$ are even). $\square$

*Remark 3.16.* Suppose that $K$ is algebraically closed. We have now extended the equivalence between (C1) and (C3), already known for Koszul algebras from [7], as well as $d$-Koszul monomial algebras by Theorem 2.8, to $(D, A)$-stacked monomial algebras with $D \neq 2A$ whenever $A > 1$.

In particular, we can extend Corollary 2.15 to $(D, A)$-stacked monomial algebras.

**Corollary 3.17.** Let $\Lambda$ be a $(D, A)$-stacked monomial algebra over an algebraically closed field with $D \neq 2A$ whenever $A > 1$. Assume that $E(\Lambda)$ is a finitely generated $Z_{\text{gr}}(E(\Lambda))$-module. Then the algebra $Z_{\text{gr}}(E(\Lambda))$ is Noetherian.

# Appendix A. The Ext algebra of a $(D, A)$-stacked monomial algebra

Leader and Snashall gave in [17] a presentation of the Yoneda algebra $E(\Lambda)$ of a $(D, A)$-stacked monomial algebra by quiver and relations. However, in our proof of Theorem 2.8 that (C4) implies (C2) for $d$-Koszul monomial algebras, we used the description from [10, Sections 8 and 9] of $E(\Lambda)$ as an algebra contained, as a graded
vector space, in the Koszul dual $^1\Lambda$. In this appendix, we generalise this description to (D, A)-stacked monomial algebras.

Throughout this section, $\Lambda = KQ/I$ is a (D, A)-stacked monomial algebra with $D = dA$ and $d \ge 2$, where $I$ is an ideal generated by a set $\rho$ of A-paths of A-length $d = \frac{D}{A}$. We view $\Lambda$ as a quotient of the tensor algebra: $\Lambda = T_{\Lambda_0}(\Lambda_1)/I$.

All tensor products are taken over $\Lambda_0$ and we write $\overline{\otimes}$ for $\otimes_{\Lambda_0}$. The subspace $S = I \cap (\Lambda_1^{\overline{\otimes}D}) = \text{span}(\rho)$ of $T = T_{\Lambda_0}(\Lambda_1)$ is a $\Lambda_0$-$\Lambda_0$-submodule of $\Lambda_1^{\overline{\otimes}D}$; it is finite-dimensional over $K$. For an element $x \in T$, write $\bar{x}$ for its image in $\Lambda$. Note that for $0 \le i < D$ we have $\Lambda_i = \Lambda_1^{\overline{\otimes}i}$.

**A.1. Generalised Koszul complex of S.** Define spaces $H_{\delta(n)} \subseteq T_{\Lambda_0}^n(\Lambda_1)$ as follows:

$$H_0 = \Lambda_0, H_1 = \Lambda_1 \text{ and, for } n \ge 2, H_{\delta(n)} = \bigcap_{i+j=\delta(n)-d} (\Lambda_A^{\overline{\otimes}i} \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}).$$

For $n \ge 0$, let $\mathbf{P}^n$ be the right $\Lambda$-module defined by $\mathbf{P}^n = H_{\delta(n)} \overline{\otimes} \Lambda$; it is projective.

We have $H_{\delta(1)} = \Lambda_1 = H_{\delta(0)} \overline{\otimes} \Lambda_1$, $H_{\delta(2)} = S \subseteq \Lambda_1^{\overline{\otimes}D} = H_{\delta(1)} \overline{\otimes} \Lambda_1^{\overline{\otimes}D-1}$, and, for any $n \ge 3$, $H_{\delta(n)} \subseteq H_{\delta(n-1)} \overline{\otimes} \Lambda_A^{\overline{\otimes}(\delta(n)-\delta(n-1))}$. Indeed, for any $k \ge 1$,

$$H_{\delta(2k+2)} = H_{(k+1)d} = \bigcap_{j=0}^{kd} (\Lambda_A^{\overline{\otimes}(kd-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \subseteq \bigcap_{j=d-1}^{kd} (\Lambda_A^{\overline{\otimes}(kd-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \\ = \bigcap_{j=0}^{(k-1)d+1} (\Lambda_A^{\overline{\otimes}((k-1)d+1-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \overline{\otimes} (\Lambda_A^{\overline{\otimes}(d-1)}) = H_{\delta(2k+1)} \overline{\otimes} (\Lambda_A^{\overline{\otimes}(d-1)});$$

$$H_{\delta(2k+1)} = H_{kd+1} = \bigcap_{j=0}^{kd+1} (\Lambda_A^{\overline{\otimes}((k-1)d+1-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \subseteq \bigcap_{j=A}^{kd+1} (\Lambda_A^{\overline{\otimes}((k-1)d+1-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \\ = \bigcap_{j=0}^{kd} (\Lambda_A^{\overline{\otimes}((k-1)d-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \overline{\otimes} \Lambda_A = H_{\delta(2k)} \overline{\otimes} \Lambda_A.$$

It follows that the maps $F^1: \Lambda_1 \overline{\otimes} \Lambda \to \Lambda_0 \overline{\otimes} \Lambda \cong \Lambda$, $F^2: \Lambda_A^{\overline{\otimes}d} \overline{\otimes} \Lambda \to \Lambda_1 \overline{\otimes} \Lambda$, and, for $n \ge 3$, $F^n: \Lambda_A^{\overline{\otimes}\delta(n)} \overline{\otimes} \Lambda \to \Lambda_A^{\overline{\otimes}\delta(n-1)} \overline{\otimes} \Lambda$ defined by

$$F^1(x_1 \overline{\otimes} \lambda) = x_1 \lambda, \\ F^2(x_1 \overline{\otimes} \cdots \overline{\otimes} x_D \overline{\otimes} \lambda) = x_1 \overline{\otimes} x_2 \cdots x_D \lambda, \\ F^n(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(n)} \overline{\otimes} \lambda) = y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(n-1)} \overline{\otimes} y_{\delta(n-1)+1} \cdots y_{\delta(n)} \lambda,$$

where $x_i \in \Lambda_1$ and $y_i \in \Lambda_A$ for all $i$, induce maps $b^n: \mathbf{P}^n \to \mathbf{P}^{n-1}$. More specifically, for all $k \ge 1$,

$$F^{2k+1}(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{kd+1} \overline{\otimes} \lambda) = y_1 \overline{\otimes} \cdots \overline{\otimes} y_{kd} \overline{\otimes} y_{kd+1} \lambda \text{ if } n = 2k+1 \text{ is odd;} \\ F^{2k+2}(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{(k+1)d} \overline{\otimes} \lambda) = y_1 \overline{\otimes} \cdots \overline{\otimes} y_{kd+1} \overline{\otimes} y_{kd+2} \cdots y_{(k+1)d} \lambda \text{ if } n = 2k+2 \text{ is even.}$$

Define also $b^0: \mathbf{P}^0 = \Lambda_0 \overline{\otimes} \Lambda \cong \Lambda \to \Lambda_0$, which identifies with the natural projection. Moreover, for $n \ge 3$ we have

$$H_{\delta(n+1)} \subseteq H_{\delta(n)} \overline{\otimes} \Lambda_A^{\overline{\otimes}(\delta(n+1)-\delta(n))} \\ \subseteq H_{\delta(n-1)} \overline{\otimes} \Lambda_A^{\overline{\otimes}(\delta(n)-\delta(n-1))} \overline{\otimes} \Lambda_A^{\overline{\otimes}(\delta(n+1)-\delta(n))} = H_{\delta(n-1)} \overline{\otimes} \Lambda_A^{\overline{\otimes}d}$$
vector space, in the Koszul dual $^1\Lambda$. In this appendix, we generalise this description to $(D, A)$-stacked monomial algebras.

Throughout this section, $\Lambda = KQ/I$ is a $(D, A)$-stacked monomial algebra with $D = dA$ and $d \ge 2$, where $I$ is an ideal generated by a set $\rho$ of $A$-paths of $A$-length $d = \frac{D}{A}$. We view $\Lambda$ as a quotient of the tensor algebra: $\Lambda = T_{\Lambda_0}(\Lambda_1)/I$.

All tensor products are taken over $\Lambda_0$ and we write $\overline{\otimes}$ for $\otimes_{\Lambda_0}$. The subspace $S = I \cap (\Lambda_1^{\overline{\otimes}D}) = \text{span}(\rho)$ of $T = T_{\Lambda_0}(\Lambda_1)$ is a $\Lambda_0$-$\Lambda_0$-submodule of $\Lambda_1^{\overline{\otimes}D}$; it is finite-dimensional over $K$. For an element $x \in T$, write $\bar{x}$ for its image in $\Lambda$. Note that for $0 \le i < D$ we have $\Lambda_i = \Lambda_1^{\overline{\otimes}i}$.

**A.1. Generalised Koszul complex of S.** Define spaces $H_{\delta(n)} \subseteq T_{\Lambda_0}^n(\Lambda_1)$ as follows:

$$H_0 = \Lambda_0, H_1 = \Lambda_1 \text{ and, for } n \ge 2, H_{\delta(n)} = \bigcap_{i+j=\delta(n)-d} (\Lambda_A^{\overline{\otimes}i} \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}).$$

For $n \ge 0$, let $\mathbf{P}^n$ be the right $\Lambda$-module defined by $\mathbf{P}^n = H_{\delta(n)} \overline{\otimes} \Lambda$; it is projective.

We have $H_{\delta(1)} = \Lambda_1 = H_{\delta(0)} \overline{\otimes} \Lambda_1$, $H_{\delta(2)} = S \subseteq \Lambda_1^{\overline{\otimes}D} = H_{\delta(1)} \overline{\otimes} \Lambda_1^{\overline{\otimes}D-1}$, and, for any $n \ge 3$, $H_{\delta(n)} \subseteq H_{\delta(n-1)} \overline{\otimes} \Lambda_A^{\overline{\otimes}(\delta(n)-\delta(n-1))}$. Indeed, for any $k \ge 1$,

$$H_{\delta(2k+2)} = H_{(k+1)d} = \bigcap_{j=0}^{kd} (\Lambda_A^{\overline{\otimes}(kd-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \subseteq \bigcap_{j=d-1}^{kd} (\Lambda_A^{\overline{\otimes}(kd-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \\ = \bigcap_{j=0}^{(k-1)d+1} (\Lambda_A^{\overline{\otimes}((k-1)d+1-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \overline{\otimes} (\Lambda_A^{\overline{\otimes}(d-1)}) = H_{\delta(2k+1)} \overline{\otimes} (\Lambda_A^{\overline{\otimes}(d-1)});$$

$$H_{\delta(2k+1)} = H_{kd+1} = \bigcap_{j=0}^{kd+1} (\Lambda_A^{\overline{\otimes}((k-1)d+1-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \subseteq \bigcap_{j=A}^{kd+1} (\Lambda_A^{\overline{\otimes}((k-1)d+1-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \\ = \bigcap_{j=0}^{kd} (\Lambda_A^{\overline{\otimes}((k-1)d-j)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}j}) \overline{\otimes} \Lambda_A = H_{\delta(2k)} \overline{\otimes} \Lambda_A.$$

It follows that the maps $F^1: \Lambda_1 \overline{\otimes} \Lambda \to \Lambda_0 \overline{\otimes} \Lambda \cong \Lambda$, $F^2: \Lambda_A^{\overline{\otimes}d} \overline{\otimes} \Lambda \to \Lambda_1 \overline{\otimes} \Lambda$, and, for $n \ge 3$, $F^n: \Lambda_A^{\overline{\otimes}\delta(n)} \overline{\otimes} \Lambda \to \Lambda_A^{\overline{\otimes}\delta(n-1)} \overline{\otimes} \Lambda$ defined by

$$F^1(x_1 \overline{\otimes} \lambda) = x_1 \lambda, \\ F^2(x_1 \overline{\otimes} \cdots \overline{\otimes} x_D \overline{\otimes} \lambda) = x_1 \overline{\otimes} x_2 \cdots x_D \lambda, \\ F^n(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(n)} \overline{\otimes} \lambda) = y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(n-1)} \overline{\otimes} y_{\delta(n-1)+1} \cdots y_{\delta(n)} \lambda,$$

where $x_i \in \Lambda_1$ and $y_i \in \Lambda_A$ for all $i$, induce maps $b^n: \mathbf{P}^n \to \mathbf{P}^{n-1}$. More specifically, for all $k \ge 1$,

$$F^{2k+1}(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{kd+1} \overline{\otimes} \lambda) = y_1 \overline{\otimes} \cdots \overline{\otimes} y_{kd} \overline{\otimes} y_{kd+1} \lambda \text{ if } n = 2k+1 \text{ is odd;} \\ F^{2k+2}(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{(k+1)d} \overline{\otimes} \lambda) = y_1 \overline{\otimes} \cdots \overline{\otimes} y_{kd+1} \overline{\otimes} y_{kd+2} \cdots y_{(k+1)d} \lambda \text{ if } n = 2k+2 \text{ is even.}$$

Define also $b^0: \mathbf{P}^0 = \Lambda_0 \overline{\otimes} \Lambda \cong \Lambda \to \Lambda_0$, which identifies with the natural projection. Moreover, for $n \ge 3$ we have

$$H_{\delta(n+1)} \subseteq H_{\delta(n)} \overline{\otimes} \Lambda_A^{\overline{\otimes}(\delta(n+1)-\delta(n))} \\ \subseteq H_{\delta(n-1)} \overline{\otimes} \Lambda_A^{\overline{\otimes}(\delta(n)-\delta(n-1))} \overline{\otimes} \Lambda_A^{\overline{\otimes}(\delta(n+1)-\delta(n))} = H_{\delta(n-1)} \overline{\otimes} \Lambda_A^{\overline{\otimes}d}$$
and $H_{\delta(n+1)} \subseteq \Lambda_A^{\overline{\otimes}\delta(n-1)} \overline{\otimes} S$, we have $H_{\delta(n+1)} \subseteq (\Lambda_A^{\overline{\otimes}\delta(n-1)} \overline{\otimes} S) \cap (H_{\delta(n-1)} \overline{\otimes} \Lambda_A^{\overline{\otimes}d}) = H_{\delta(n-1)} \overline{\otimes} S$ (all the spaces involved are finitely generated and projective over $\Lambda_0$), hence $b^n \circ b^{n+1} = 0$. It is easy to check that $b^n \circ b^{n+1} = 0$ when $n=1, 2$, or $3$.

Therefore we have a complex $(\mathbf{P}^n, b^n)$ of projective right $\Lambda$-modules.

**Theorem A.1.** Let $\Lambda = KQ/I$ be a monomial algebra with $I$ generated in degree $D = dA$ with $d \ge 2$. Then $\Lambda$ is $(D, A)$-stacked monomial if, and only if, $(\mathbf{P}^\bullet, b^\bullet)$ is a minimal projective right $\Lambda$-module resolution of $\Lambda_0$.

*Proof:* By construction, $\mathbf{P}^n$ is generated in degree $\delta(n)$. Therefore, if $(\mathbf{P}^\bullet, b^\bullet)$ is a minimal projective right $\Lambda$-module resolution of $\Lambda_0$, then $\Lambda$ is a $(D, A)$-stacked monomial algebra.

Conversely, assume that $\Lambda$ is a $(D, A)$-stacked monomial algebra. We already have a complex $(\mathbf{P}^\bullet, b^\bullet)$ of right $\Lambda$-modules such that $\mathbf{P}^n$ is generated in degree $\delta(n)$. The beginning $\mathbf{P}^1 \xrightarrow{b^1} \mathbf{P}^0 \xrightarrow{b^0} \Lambda_0 \to 0$ is exact.

We prove exactness at $\mathbf{P}^{2n+1}$ for $n \ge 1$ (the proof of exactness at $\mathbf{P}^{2n}$ and at $\mathbf{P}^1$ is similar, without the need for Consequence 3.3).

First note that $b^{2n+2}(\mathbf{P}^{2n+2})$ is generated in degree $\delta(2n+2) = (n+1)D$ in $\mathbf{P}^{2n+1}$.

Let $z = \sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} \lambda_i$ be an element in $\text{Ker } b^{2n+1}$ with $x_{j,i} \in \Lambda_A$ and $\lambda_i \in \Lambda$ for all $i, j$. Then $\sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{2,i} \overline{\otimes} x_{1,i} \lambda_i$ is in $\overline{\text{T}} \overline{\otimes} I$. It follows that $\lambda_i \in \bigoplus_{k \ge D-A} \Lambda_k$ and that $\text{Ker } b^{2n+1}$ is generated in degrees at least $(n+1)D$. Therefore $z$ can be rewritten as $z = \sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} y_{d-1,i} \cdots y_{1,i} \lambda'_i$ with $y_{j,i} \in \Lambda_A$ and $\lambda'_i \in \Lambda$ for all $i, j$, with the $y_{j,i}$ right uniform and $t(y_{1,i}) \ne t(y_{1,k})$ when $i \ne k$. Write $\lambda'_i = \sum_{l \ge 0} \lambda'_{i,l}$ with $\lambda'_{i,l} \in \Lambda_l$ for all $i, l$, and $\lambda'_{i,0} = t(y_{1,i})$. Then each of the $\sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} y_{d-1,i} \cdots y_{1,i} \lambda'_{i,l}$ is in $\text{Ker } b^{2n+1}$ so in particular $z' := \sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} y_{d-1,i} \cdots y_{1,i} \in \text{Ker } b^{2n+1}$.

Consider $z'' := \sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} y_{d-1,i} \overline{\otimes} \cdots \overline{\otimes} y_{1,i} \overline{\otimes} t(y_{1,i}) \in \Lambda_A^{\overline{\otimes}((n+1)d)} \overline{\otimes} \Lambda$. We show that $z'' \in \mathbf{P}^{2n+2}$; this will imply that $z' = b^{2n+2}(z'') \in \text{Im } b^{2n+2}$. Since $\Lambda$ is monomial and $b^{2n+1}(z') = 0$, we may assume that, for all $i, x_{1,i} y_{d-1,i} \cdots y_{1,i}$ is a path; since it is in $I$ and has degree $D$, it is in $\rho = \mathcal{R}^2$. By definition of $\mathbf{P}^{2n+1}$, we may assume that $z$ is written so that, for each $i$, $x_{d,i} \cdots x_{1,i}$ is a path in $\rho = \mathcal{R}^2$. The path $x_{1,i} y_{d-1,i} \cdots y_{1,i}$ properly overlaps $x_{d,i} \cdots x_{1,i}$, therefore using Consequence 3.3 it follows that for all $k$ with $1 \le k \le d-1$ we have $x_{k,i} \cdots x_{1,i} y_{d-1,i} \cdots y_{k,i} \in \rho$ and hence $z'' \in \bigcap_{k=1}^{d-1} (\Lambda_A^{\overline{\otimes}(nd-k+1)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}(k-1)}) \overline{\otimes} \Lambda$. Finally, using the fact that $z \in \mathbf{P}^{2n+1} = H_{nd+1} \overline{\otimes} \Lambda$, we get $z'' \in H_{(n+1)d} \overline{\otimes} \Lambda = \mathbf{P}^{2n+2}$. We have proved that $(\text{Ker } b^{2n+1})_{(n+1)d} \subseteq \text{Im } b^{2n+2}$.

Since $\text{Im } b^{2n+2}$ is generated in degree $(n+1)d$ and $\text{Ker } b^{2n+1}$ is generated in degrees at least $(n+1)d$, it follows that $\text{Ker } b^{2n+1} \subseteq \text{Im } b^{2n+2}$ and lastly that $\text{Ker } b^{2n+1} = \text{Im } b^{2n+2}$.

Finally, since $\text{Im } b^{n+1}$ is generated in degree $\delta(n+1)$ and $\mathbf{P}^n$ is generated in degree $\delta(n) < \delta(n+1)$, $\text{Im } b^{n+1} \subseteq \text{Im } \mathbf{P}^n$ for all $n$ and therefore the resolution is minimal. $\square$

## A.2. The Ext algebra.

**A.2.1. Some duality results.** We recall without proof some results stated in [2]. All modules are finitely generated $\Lambda_0$-$\Lambda_0$-bimodules. All claims are easily checked for bimodules that are free and finitely generated as left or as right $\Lambda_0$-modules, and follow for arbitrary finitely generated modules (since $\Lambda_0$ is semisimple, all the $\Lambda_0$-modules (left or right) are projective).
and $H_{\delta(n+1)} \subseteq \Lambda_A^{\overline{\otimes}\delta(n-1)} \overline{\otimes} S$, we have $H_{\delta(n+1)} \subseteq (\Lambda_A^{\overline{\otimes}\delta(n-1)} \overline{\otimes} S) \cap (H_{\delta(n-1)} \overline{\otimes} \Lambda_A^{\overline{\otimes}d}) = H_{\delta(n-1)} \overline{\otimes} S$ (all the spaces involved are finitely generated and projective over $\Lambda_0$), hence $b^n \circ b^{n+1} = 0$. It is easy to check that $b^n \circ b^{n+1} = 0$ when $n=1, 2$, or $3$.

Therefore we have a complex $(\mathbf{P}^n, b^n)$ of projective right $\Lambda$-modules.

**Theorem A.1.** Let $\Lambda = KQ/I$ be a monomial algebra with $I$ generated in degree $D = dA$ with $d \ge 2$. Then $\Lambda$ is $(D, A)$-stacked monomial if, and only if, $(\mathbf{P}^\bullet, b^\bullet)$ is a minimal projective right $\Lambda$-module resolution of $\Lambda_0$.

*Proof:* By construction, $\mathbf{P}^n$ is generated in degree $\delta(n)$. Therefore, if $(\mathbf{P}^\bullet, b^\bullet)$ is a minimal projective right $\Lambda$-module resolution of $\Lambda_0$, then $\Lambda$ is a $(D, A)$-stacked monomial algebra.

Conversely, assume that $\Lambda$ is a $(D, A)$-stacked monomial algebra. We already have a complex $(\mathbf{P}^\bullet, b^\bullet)$ of right $\Lambda$-modules such that $\mathbf{P}^n$ is generated in degree $\delta(n)$. The beginning $\mathbf{P}^1 \xrightarrow{b^1} \mathbf{P}^0 \xrightarrow{b^0} \Lambda_0 \to 0$ is exact.

We prove exactness at $\mathbf{P}^{2n+1}$ for $n \ge 1$ (the proof of exactness at $\mathbf{P}^{2n}$ and at $\mathbf{P}^1$ is similar, without the need for Consequence 3.3).

First note that $b^{2n+2}(\mathbf{P}^{2n+2})$ is generated in degree $\delta(2n+2) = (n+1)D$ in $\mathbf{P}^{2n+1}$.

Let $z = \sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} \lambda_i$ be an element in $\text{Ker } b^{2n+1}$ with $x_{j,i} \in \Lambda_A$ and $\lambda_i \in \Lambda$ for all $i, j$. Then $\sum_j x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{2,i} \overline{\otimes} x_{1,i} \lambda_i$ is in $\overline{\text{T}} \overline{\otimes} I$. It follows that $\lambda_i \in \bigoplus_{k \ge D-A} \Lambda_k$ and that $\text{Ker } b^{2n+1}$ is generated in degrees at least $(n+1)D$. Therefore $z$ can be rewritten as $z = \sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} y_{d-1,i} \cdots y_{1,i} \lambda'_i$ with $y_{j,i} \in \Lambda_A$ and $\lambda'_i \in \Lambda$ for all $i, j$, with the $y_{j,i}$ right uniform and $\mathfrak{t}(y_{1,i}) \ne \mathfrak{t}(y_{1,k})$ when $i \ne k$. Write $\lambda'_i = \sum_{l \ge 0} \lambda'_{i,l}$ with $\lambda'_{i,l} \in \Lambda_l$ for all $i, l$, and $\lambda'_{i,0} = \mathfrak{t}(y_{1,i})$. Then each of the $\sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} y_{d-1,i} \cdots y_{1,i} \lambda'_{i,l}$ is in $\text{Ker } b^{2n+1}$ so in particular $z' := \sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} y_{d-1,i} \cdots y_{1,i} \in \text{Ker } b^{2n+1}$.

Consider $z'' := \sum_i x_{nd+1,i} \overline{\otimes} \cdots \overline{\otimes} x_{1,i} \overline{\otimes} y_{d-1,i} \overline{\otimes} \cdots \overline{\otimes} y_{1,i} \overline{\otimes} \mathfrak{t}(y_{1,i}) \in \Lambda_A^{\overline{\otimes}((n+1)d)} \overline{\otimes} \Lambda$. We show that $z'' \in \mathbf{P}^{2n+2}$; this will imply that $z' = b^{2n+2}(z'') \in \text{Im } b^{2n+2}$. Since $\Lambda$ is monomial and $b^{2n+1}(z') = 0$, we may assume that, for all $i, x_{1,i} y_{d-1,i} \cdots y_{1,i}$ is a path; since it is in $I$ and has degree $D$, it is in $\rho = \mathcal{R}^2$. By definition of $\mathbf{P}^{2n+1}$, we may assume that $z$ is written so that, for each $i$, $x_{d,i} \cdots x_{1,i}$ is a path in $\rho = \mathcal{R}^2$. The path $x_{1,i} y_{d-1,i} \cdots y_{1,i}$ properly overlaps $x_{d,i} \cdots x_{1,i}$, therefore using Consequence 3.3 it follows that for all $k$ with $1 \le k \le d-1$ we have $x_{k,i} \cdots x_{1,i} y_{d-1,i} \cdots y_{k,i} \in \rho$ and hence $z'' \in \bigcap_{k=1}^{d-1} (\Lambda_A^{\overline{\otimes}(nd-k+1)}) \overline{\otimes} S \overline{\otimes} (\Lambda_A^{\overline{\otimes}(k-1)}) \overline{\otimes} \Lambda$. Finally, using the fact that $z \in \mathbf{P}^{2n+1} = H_{nd+1} \overline{\otimes} \Lambda$, we get $z'' \in H_{(n+1)d} \overline{\otimes} \Lambda = \mathbf{P}^{2n+2}$. We have proved that $(\text{Ker } b^{2n+1})_{(n+1)d} \subseteq \text{Im } b^{2n+2}$.

Since $\text{Im } b^{2n+2}$ is generated in degree $(n+1)d$ and $\text{Ker } b^{2n+1}$ is generated in degrees at least $(n+1)d$, it follows that $\text{Ker } b^{2n+1} \subseteq \text{Im } b^{2n+2}$ and lastly that $\text{Ker } b^{2n+1} = \text{Im } b^{2n+2}$.

Finally, since $\text{Im } b^{n+1}$ is generated in degree $\delta(n+1)$ and $\mathbf{P}^n$ is generated in degree $\delta(n) < \delta(n+1)$, $\text{Im } b^{n+1} \subseteq \mathfrak{r}\mathbf{P}^n$ for all $n$ and therefore the resolution is minimal. $\square$

## A.2. The Ext algebra.

**A.2.1. Some duality results.** We recall without proof some results stated in [2]. All modules are finitely generated $\Lambda_0$-$\Lambda_0$-bimodules. All claims are easily checked for bimodules that are free and finitely generated as left or as right $\Lambda_0$-modules, and follow for arbitrary finitely generated modules (since $\Lambda_0$ is semisimple, all the $\Lambda_0$-modules (left or right) are projective).
For any bimodules $V$ and $W$, define $V^*=\operatorname{Hom}_{\Lambda_0-}(V,\Lambda_0)$ and $^{*}W=\operatorname{Hom}_{-\Lambda_0}(W,\Lambda_0)$; they are $\Lambda_0$-$\Lambda_0$-bimodules, for the actions given for all $e, e'$ in $\Lambda_0, \alpha \in V^*, \beta \in ^*W$, and $v \in V, w \in W$, by:

$$(e\alpha e')(v) = \alpha(ve)e' \text{ and } (e\beta e')(w) = e\beta(e'w).$$

There are natural isomorphisms of $\Lambda_0$-$\Lambda_0$-bimodules $\Lambda_0^* \cong \Lambda_0 \cong {}^*\Lambda_0$ which we view as identifications.

There are also natural bimodule isomorphisms $V \cong {}^*(V^*)$ and $W \cong {}^*(W)^*$ of bimodules.

If $V_1 \subseteq V$ (respectively $W_1 \subseteq W$), define $V_1^{\perp} = \{\alpha \in V^* | \alpha(V_1) = 0\}$ (respectively $\perp W_1 = \{\beta \in {}^*W | \beta(W_1) = 0\}$). If $V_1$ (respectively $W_1$) is a sub-bimodule, then they are sub-bimodules of $V^*$ and $^*W$ respectively.

Fix a $\Lambda_0$-$\Lambda_0$-bimodule $V$. Then:

(i) if $W$ is another $\Lambda_0$-$\Lambda_0$-bimodule, then there is an isomorphism of $\Lambda_0$-$\Lambda_0$-bimodules $\varphi_{V,W}: {}^*W \otimes {}^*W \to {}^*(W \otimes V)$ given for all $\alpha \in {}^*V$, $\beta \in {}^*W$, $v \in V$, and $w \in W$ by $\varphi_{V,W}(\alpha \otimes \beta)(w \otimes v) = \alpha(\beta(w)v)$. There is a similar isomorphism with right duals, which sends $\alpha \otimes \beta$ to the map $w \otimes v \mapsto \beta(w\alpha(v))$;

(ii) if $U$ and $W$ are sub-bimodules of $V$, then $\perp(U+W) = \perp U \cap \perp W$ and $(U+W)^{\perp} = U^{\perp} \cap W^{\perp}$;

(iii) if $U$ is a sub-bimodule of $V$, then $(V/U)^* \cong U^{\perp}$ and $^{*}(V/U) \cong \perp U$;

(iv) if $W$ is a sub-bimodule of $V$, then for any idempotents $e_i, e_j$ with $(i,j) \in Q_0^2$ we have $e_i {}^*W e_j = {}^*(e_j W e_i)$;

(v) if $U$ is a sub-bimodule of $V$, then for all $i, j$ in $Q_0$ we have $\dim e_j^{\perp} U e_i = \dim e_i V e_j - \dim e_i U e_j = \dim e_j U^{\perp} e_i$;

(vi) if $U$ is a sub-bimodule of $V$, then under the identification of $V$ with $^{*}(V^*)$ we have $\perp(U^{\perp})$ and under the identification of $V$ with $(^*V)^*$ we have $(\perp U)^{\perp}$;

(vii) if $U$ is a sub-bimodule of $V$ and $W$ and $Z$ are $\Lambda_0$-$\Lambda_0$-bimodules, there are bimodule isomorphisms $\perp(W \otimes U \otimes Z) \cong {}^*Z \otimes {}^\perp U \otimes {}^*W$ and $(W \otimes U \otimes Z)^{\perp} \cong Z^{\perp} \otimes U^{\perp} \otimes W^*$.

**A.2.2. Description of the Ext algebra.** From the above, there is a natural isomorphism $^{*}(\Lambda_A^{\otimes i}) \cong {}^*(\Lambda_A)^{\otimes i}$, which we view as an identification. We also view $S$ as a subspace of $\Lambda_A^{\otimes d}$ (rather than $\Lambda_1^{\otimes D}$). We may then consider $^{\perp}S = \{f \in {}^*(\Lambda_A)^{\otimes d} | f(x) = 0 \text{ for all } x \in S\}$. The dual algebra of $\Lambda$ is then $^{\natural}\Lambda = \mathbb{T}_{\Lambda_0}({}^*(\Lambda_A))/(^{\perp}S)$. It is a graded $d$-homogeneous algebra since $^{\perp}S$ is contained in $(^*(\Lambda_A)^{\otimes d}$, therefore $^{\natural}\Lambda = \bigoplus_{n \ge 0} {}^{\natural}\Lambda_n$.

In terms of quivers, we have $\Lambda_0 = KQ_0$ and $\Lambda_A = KQ_A$, the vector space whose basis is the set $Q_A$ of paths of length $A$ in $Q$; moreover, $^*\Lambda_A \cong KQ_A^{\text{op}}$ using (iv).

Then $^{\natural}\Lambda$ is isomorphic to $K\Gamma/(^{\perp}\rho)$, where $\Gamma$ is the quiver with the same vertices as $Q$ and whose set of arrows is $\Gamma_1 = \{\bar{\alpha}: i \to j \mid \text{there exists } \alpha \in Q_A, \alpha: j \to i\}$ and where $^{\perp}\rho$ is the left orthogonal of the set $\rho$ viewed as a set of $A$-paths, for the bilinear form $K\Gamma_d \times K(Q_A)_d \to K$ defined on paths of length $d$ in $\Gamma$ and $A$-paths of $A$-length $d$ in $Q$ by $\langle\bar{\alpha}_d \cdots \bar{\alpha}_1, \beta_1 \cdots \beta_d\rangle = 1$ if $\alpha_1 \cdots \alpha_d = \beta_1 \cdots \beta_d$ and $0$ otherwise, where the $\alpha_i$ and $\beta_i$ are paths of length $A$.

The algebra $K\Gamma/(^{\perp}\rho)$ is monomial, and the ideal $(^{\perp}\rho)$ has a minimal generating set $\sigma$ given by all the paths $\bar{\alpha}_d \cdots \bar{\alpha}_1$ such that the $A$-path $\alpha_1 \cdots \alpha_d$ is not in $\rho$. In particular, if $\bar{\gamma} = \bar{\alpha}_r \cdots \bar{\alpha}_1$ is a path in $\Gamma$, with $\alpha_i \in Q_A$ for all $i$, then $\bar{\gamma} \notin (\sigma)$ if, and only if, for each $i$ with $1 \le i \le r-d+1$, the path $\alpha_i \cdots \alpha_{i+d-1}$ is in $\rho$ (we use the fact that $\Lambda$ is monomial here).

**Lemma A.2.** There is an isomorphism $(^{\natural}\Lambda_{\delta(n)})^* \cong H_{\delta(n)}$.
For any bimodules $V$ and $W$, define $V^*=\operatorname{Hom}_{\Lambda_0-}(V,\Lambda_0)$ and $^{*}W=\operatorname{Hom}_{-\Lambda_0}(W,\Lambda_0)$; they are $\Lambda_0$-$\Lambda_0$-bimodules, for the actions given for all $e, e'$ in $\Lambda_0, \alpha \in V^*, \beta \in ^*W$, and $v \in V, w \in W$, by:

$$(e\alpha e')(v) = \alpha(ve)e' \text{ and } (e\beta e')(w) = e\beta(e'w).$$

There are natural isomorphisms of $\Lambda_0$-$\Lambda_0$-bimodules $\Lambda_0^* \cong \Lambda_0 \cong {}^*\Lambda_0$ which we view as identifications.

There are also natural bimodule isomorphisms $V \cong {}^*(V^*)$ and $W \cong {}^*(W)^*$ of bimodules.

If $V_1 \subseteq V$ (respectively $W_1 \subseteq W$), define $V_1^{\perp} = \{\alpha \in V^* | \alpha(V_1) = 0\}$ (respectively $\perp W_1 = \{\beta \in {}^*W | \beta(W_1) = 0\}$). If $V_1$ (respectively $W_1$) is a sub-bimodule, then they are sub-bimodules of $V^*$ and $^*W$ respectively.

Fix a $\Lambda_0$-$\Lambda_0$-bimodule $V$. Then:

(i) if $W$ is another $\Lambda_0$-$\Lambda_0$-bimodule, then there is an isomorphism of $\Lambda_0$-$\Lambda_0$-bimodules $\varphi_{V,W}: {}^*W \otimes {}^*W \to {}^*(W \otimes V)$ given for all $\alpha \in {}^*V$, $\beta \in {}^*W$, $v \in V$, and $w \in W$ by $\varphi_{V,W}(\alpha \otimes \beta)(w \otimes v) = \alpha(\beta(w)v)$. There is a similar isomorphism with right duals, which sends $\alpha \otimes \beta$ to the map $w \otimes v \mapsto \beta(w\alpha(v))$;

(ii) if $U$ and $W$ are sub-bimodules of $V$, then $\perp(U+W) = \perp U \cap \perp W$ and $(U+W)^{\perp} = U^{\perp} \cap W^{\perp}$;

(iii) if $U$ is a sub-bimodule of $V$, then $(V/U)^* \cong U^{\perp}$ and $^{*}(V/U) \cong \perp U$;

(iv) if $W$ is a sub-bimodule of $V$, then for any idempotents $e_i, e_j$ with $(i,j) \in Q_0^2$ we have $e_i^*We_j = {}^*(e_jWe_i)$;

(v) if $U$ is a sub-bimodule of $V$, then for all $i, j$ in $Q_0$ we have $\dim e_j^{\perp}Ue_i = \dim e_iVe_j - \dim e_iUe_j = \dim e_jU^{\perp}e_i$;

(vi) if $U$ is a sub-bimodule of $V$, then under the identification of $V$ with $^{*}(V^*)$ we have $\perp(U^{\perp})$ and under the identification of $V$ with $(^*V)^*$ we have $(\perp U)^{\perp}$;

(vii) if $U$ is a sub-bimodule of $V$ and $W$ and $Z$ are $\Lambda_0$-$\Lambda_0$-bimodules, there are bimodule isomorphisms $\perp(W \otimes U \otimes Z) \cong {}^*Z \otimes {}^\perp U \otimes {}^*W$ and $(W \otimes U \otimes Z)^{\perp} \cong Z^{\perp} \otimes U^{\perp} \otimes W^*$.

**A.2.2. Description of the Ext algebra.** From the above, there is a natural isomorphism $^{*}(\Lambda_A^{\otimes i}) \cong {}^*(\Lambda_A)^{\otimes i}$, which we view as an identification. We also view $S$ as a subspace of $\Lambda_A^{\otimes d}$ (rather than $\Lambda_1^{\otimes d}$). We may then consider $^{\perp}S = \{f \in {}^*(\Lambda_A)^{\otimes d} | f(x) = 0 \text{ for all } x \in S\}$. The dual algebra of $\Lambda$ is then $^{\natural}\Lambda = \mathbb{T}_{\Lambda_0}({}^*(\Lambda_A)/({}^{\perp}S)}$. It is a graded $d$-homogeneous algebra since $^{\perp}S$ is contained in $(^*(\Lambda_A)^{\otimes d}$, therefore $^{\natural}\Lambda = \bigoplus_{n \ge 0} {}^{\natural}\Lambda_n$.

In terms of quivers, we have $\Lambda_0 = KQ_0$ and $\Lambda_A = KQ_A$, the vector space whose basis is the set $Q_A$ of paths of length $A$ in $Q$; moreover, $^*\Lambda_A \cong KQ_A^{\text{op}}$ using (iv).

Then $^{\natural}\Lambda$ is isomorphic to $K\Gamma/({}^{\perp}\rho)$, where $\Gamma$ is the quiver with the same vertices as $Q$ and whose set of arrows is $\Gamma_1 = \{\bar{\alpha}: i \to j \mid \text{there exists } \alpha \in Q_A, \alpha: j \to i\}$ and where $^{\perp}\rho$ is the left orthogonal of the set $\rho$ viewed as a set of $A$-paths, for the bilinear form $K\Gamma_d \times K(Q_A)_d \to K$ defined on paths of length $d$ in $\Gamma$ and $A$-paths of $A$-length $d$ in $Q$ by $\langle\bar{\alpha}_d \cdots \bar{\alpha}_1, \beta_1 \cdots \beta_d\rangle = 1$ if $\alpha_1 \cdots \alpha_d = \beta_1 \cdots \beta_d$ and $0$ otherwise, where the $\alpha_i$ and $\beta_i$ are paths of length $A$.

The algebra $K\Gamma/({}^{\perp}\rho)$ is monomial, and the ideal $(^{\perp}\rho)$ has a minimal generating set $\sigma$ given by all the paths $\bar{\alpha}_d \cdots \bar{\alpha}_1$ such that the $A$-path $\alpha_1 \cdots \alpha_d$ is not in $\rho$. In particular, if $\bar{\gamma} = \bar{\alpha}_r \cdots \bar{\alpha}_1$ is a path in $\Gamma$, with $\alpha_i \in Q_A$ for all $i$, then $\bar{\gamma} \notin (\sigma)$ if, and only if, for each $i$ with $1 \le i \le r-d+1$, the path $\alpha_i \cdots \alpha_{i+d-1}$ is in $\rho$ (we use the fact that $\Lambda$ is monomial here).

**Lemma A.2.** There is an isomorphism $(^{\natural}\Lambda_{\delta(n)})^* \cong H_{\delta(n)}$.
Proof: By definition, $^{\natural}\Lambda_{\delta(n)} = \frac{(*\Lambda_A)\overline{\otimes}^{\delta(n)}}{\sum_{i=0}^{\delta(n)-d} (*\Lambda_A)\overline{\otimes}^{(\delta(n)-d-i)}\overline{\otimes}^{\perp} S\overline{\otimes}(*\Lambda_A)\overline{\otimes}^i}$. Therefore we have

$$ \begin{align*} ({}^{\natural}\Lambda_{\delta(n)})^* &\cong \left( \sum_{i=0}^{\delta(n)-d} (^{*\Lambda_A)}\overline{\otimes}^{(\delta(n)-d-i)}\overline{\otimes}^{\perp} S\overline{\otimes} (^{*\Lambda_A)}\overline{\otimes}^i \right)^{\perp} \\ &= \bigcap_{i=0}^{\delta(n)-d} ((^{*\Lambda_A)}\overline{\otimes}^{(\delta(n)-d-i)}\overline{\otimes}^{\perp} S\overline{\otimes} (^{*\Lambda_A)}\overline{\otimes}^i)^{\perp} \\ &= \bigcap_{i=0}^{\delta(n)-d} (((^{*\Lambda_A)}\overline{\otimes}^i)^*\overline{\otimes}^{(\perp S)^{\perp}}\overline{\otimes}((^{*\Lambda_A)}\overline{\otimes}^{(\delta(n)-d-i)})^*) \\ &\cong \bigcap_{i=0}^{\delta(n)-d} (\Lambda_A^{\overline{\otimes}^i}\overline{\otimes} S\overline{\otimes}\Lambda_A^{\overline{\otimes}^{(\delta(n)-d-i)}}) \\ &= H_{\delta(n)}. \end{align*} $$

This isomorphism takes $x = \sum x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(n)} \in H_{\delta(n)}$ to the map $g_x: {}^{\natural}\Lambda_{\delta(n)} \to \Lambda_0$ defined by

$$ g_x(\overline{\gamma_{\delta(n)} \otimes \cdots \otimes \gamma_1}) = \sum \gamma_{\delta(n)} (\gamma_{\delta(n)-1} (\cdots (\gamma_2(\gamma_1(x_1)x_2)\cdots)x_{\delta(n)-1})x_{\delta(n)}), $$

where the $x_i$ are in $\Lambda_A$ and the $\gamma_i$ are in $^*\Lambda_A$.

□

**Lemma A.3.** There is an isomorphism $\psi: {}^{\natural}\Lambda_{\delta(n)} \to \operatorname{Hom}_{\Lambda}(\mathbf{P}^n, \Lambda_0)$ given by

$$ \psi(\overline{f})(x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(n)} \overline{\otimes} \lambda) = f_{\delta(n)}(f_{\delta(n-1)}(\cdots (f_1(x_1)) \cdots x_{\delta(n)-1}) x_{\delta(n)}) \overline{\lambda}, $$

where $\overline{f} = \overline{f_{\delta(n)} \overline{\otimes} \cdots \overline{\otimes} f_1} \in {}^{\natural}\Lambda_{\delta(n)}$ with $f_i \in {}^*\Lambda_1$ for all $i$, $x_i \in \Lambda_1$ for all $i$ and $\lambda \in \Lambda$.

Proof: The isomorphism $\psi$ is the composition of the following isomorphisms:

* $ {}^{\natural}\Lambda_{\delta(n)} \to \operatorname{Hom}_{-\Lambda_0}(({}^{\natural}\Lambda_{\delta(n)})^*, \Lambda_0)$, which sends $\overline{f}$ to the map $[g \mapsto g(\overline{f})]$;

* $ \operatorname{Hom}_{-\Lambda_0}(({}^{\natural}\Lambda_{\delta(n)})^*, \Lambda_0) \to \operatorname{Hom}_{-\Lambda_0}(H_{\delta(n)}, \Lambda_0)$, which sends a map $h$ to the map $[x \mapsto h(g_x)]$, where $g_x$ is as in the proof of Lemma A.2;

* $ \operatorname{Hom}_{-\Lambda_0}(H_{\delta(n)}, \Lambda_0) \to \operatorname{Hom}_{-\Lambda}(H_{\delta(n)}\overline{\otimes}\Lambda, \Lambda_0)$, which sends a map $k$ to the map $[x\overline{\otimes}\lambda \mapsto k(x)\overline{\lambda}]$.

Applying these isomorphisms to $\overline{f}$ gives the expression in the statement.

□

Let $B$ be the vector space $B = \bigoplus_{n \ge 0} B_n$, where $B_n = {}^{\natural}\Lambda_{\delta(n)}$. Define a multiplication on $B$ as follows: for $x \in B_n$ and $y \in B_m$, set

$$ x.y = \begin{cases} 0 & \text{if } n \text{ and } m \text{ are odd;} \\ 0 & \text{if } n \text{ or } m \text{ is equal to } 1 \text{ and } n \ge 1, m \ge 1; \\ xy & \text{in } {}^{\natural}\Lambda \text{ otherwise.} \end{cases} $$

The algebra $B$ is a graded $K$-algebra generated in degrees 0, 1, 2, and 3.

We want to prove that $E(\Lambda) \cong B$ when $A > 1$ and $D \ne 2A$. We first need a description of the Yoneda product.
Proof: By definition, $^{\natural}\Lambda_{\delta(n)} = \frac{(*\Lambda_A) \overline{\otimes}^{\delta(n)}}{\sum_{i=0}^{\delta(n)-d} (*\Lambda_A) \overline{\otimes}^{(\delta(n)-d-i)} \overline{\otimes}^{\perp} S \overline{\otimes} (*\Lambda_A) \overline{\otimes}^i}$. Therefore we have

$$ 
\begin{align*}
(^{\natural}\Lambda_{\delta(n)})^* &\cong \left( \sum_{i=0}^{\delta(n)-d} (^{*\Lambda_A}) \overline{\otimes}^{(\delta(n)-d-i)} \overline{\otimes}^{\perp} S \overline{\otimes} (^{*\Lambda_A}) \overline{\otimes}^i \right)^{\perp} \\
&= \bigcap_{i=0}^{\delta(n)-d} ((^{*\Lambda_A}) \overline{\otimes}^{(\delta(n)-d-i)} \overline{\otimes}^{\perp} S \overline{\otimes} (^{*\Lambda_A}) \overline{\otimes}^i)^{\perp} \\
&= \bigcap_{i=0}^{\delta(n)-d} (((^{*\Lambda_A}) \overline{\otimes}^i)^* \overline{\otimes}^{(\perp)S} \overline{\otimes}^{\perp} ((^{*\Lambda_A}) \overline{\otimes}^{(\delta(n)-d-i)})^*) \\
&\cong \bigcap_{i=0}^{\delta(n)-d} (\Lambda_A^{\overline{\otimes}^i} \overline{\otimes} S \overline{\otimes} \Lambda_A^{\overline{\otimes}^{(\delta(n)-d-i)}}) \\
&= H_{\delta(n)}.
\end{align*}
$$

This isomorphism takes $x = \sum x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(n)} \in H_{\delta(n)}$ to the map $g_x: {}^{\natural}\Lambda_{\delta(n)} \to \Lambda_0$ defined by

$$ g_x(\overline{\gamma_{\delta(n)} \otimes \cdots \otimes \gamma_1}) = \sum \gamma_{\delta(n)} (\gamma_{\delta(n)-1} (\cdots (\gamma_2(\gamma_1(x_1)x_2)\cdots)x_{\delta(n)-1})x_{\delta(n)}), $$

where the $x_i$ are in $\Lambda_A$ and the $\gamma_i$ are in $^*\Lambda_A$.

□

**Lemma A.3.** There is an isomorphism $\psi: {}^{\natural}\Lambda_{\delta(n)} \to \operatorname{Hom}_{\Lambda}(\mathbf{P}^n, \Lambda_0)$ given by

$$ \psi(\overline{f})(x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(n)} \overline{\otimes} \lambda) = f_{\delta(n)}(f_{\delta(n-1)}(\cdots(f_1(x_1))\cdots x_{\delta(n)-1})x_{\delta(n)}) \overline{\lambda}, $$

where $\overline{f} = \overline{f_{\delta(n)} \overline{\otimes} \cdots \overline{\otimes} f_1} \in {}^{\natural}\Lambda_{\delta(n)}$ with $f_i \in {}^*\Lambda_1$ for all $i$, $x_i \in \Lambda_1$ for all $i$ and $\lambda \in \Lambda$.

Proof: The isomorphism $\psi$ is the composition of the following isomorphisms:

* $^{\natural}\Lambda_{\delta(n)} \to \operatorname{Hom}_{-\Lambda_0}((^{\natural}\Lambda_{\delta(n)})^*, \Lambda_0)$, which sends $\overline{f}$ to the map $[g \mapsto g(\overline{f})]$;

* $\operatorname{Hom}_{-\Lambda_0}((^{\natural}\Lambda_{\delta(n)})^*, \Lambda_0) \to \operatorname{Hom}_{-\Lambda_0}(H_{\delta(n)}, \Lambda_0)$, which sends a map $h$ to the map $[x \mapsto h(g_x)]$, where $g_x$ is as in the proof of Lemma A.2;

* $\operatorname{Hom}_{-\Lambda_0}(H_{\delta(n)}, \Lambda_0) \to \operatorname{Hom}_{-\Lambda}(H_{\delta(n)} \overline{\otimes} \Lambda, \Lambda_0)$, which sends a map $k$ to the map $[x \overline{\otimes} \lambda \mapsto k(x) \overline{\lambda}]$.

Applying these isomorphisms to $\overline{f}$ gives the expression in the statement.

□

Let $B$ be the vector space $B = \bigoplus_{n \ge 0} B_n$, where $B_n = {}^{\natural}\Lambda_{\delta(n)}$. Define a multiplication on $B$ as follows: for $x \in B_n$ and $y \in B_m$, set

$$ x.y = \begin{cases} 0 & \text{if } n \text{ and } m \text{ are odd;} \\ 0 & \text{if } n \text{ or } m \text{ is equal to } 1 \text{ and } n \ge 1, m \ge 1; \\ xy & \text{in } {}^{\natural}\Lambda \text{ otherwise.} \end{cases} $$

The algebra $B$ is a graded $K$-algebra generated in degrees 0, 1, 2, and 3.

We want to prove that $E(\Lambda) \cong B$ when $A > 1$ and $D \ne 2A$. We first need a description of the Yoneda product.
**Proposition A.4.** Let $\Lambda$ be a $(D, A)$-stacked monomial algebra with $A > 1$, $D = dA$, and $d \ge 3$. The Yoneda product of $f_n \in \operatorname{Hom}_\Lambda(\mathbf{P}^n, \Lambda_0)$ and $f_m \in \operatorname{Hom}_\Lambda(\mathbf{P}^m, \Lambda_0)$ is given by

$$f_n f_m = \left\{ \begin{array}{ll} 0 & \text{if } n \text{ and } m \text{ are odd;} \\ 0 & \text{if } n \ge 1, m \ge 1, \text{ and } n = 1 \text{ or } m = 1; \\ \sum x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(n)+\delta(m)} \overline{\otimes} \lambda \mapsto f_n \left( \sum f_m (x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(m)} \overline{\otimes} 1) x_{\delta(m)+1} \right. \\ \left. \overline{\otimes} x_{\delta(m)+2} \overline{\otimes} \cdots \overline{\otimes} x_{\delta(m)+\delta(n)} \overline{\otimes} \lambda \right) & \text{otherwise,} \end{array} \right.$$

where the $x_i$ are all in $\Lambda_A$.

*Proof:* If $m$ and $n$ are odd or if $m \ge 1$ or $n \ge 1$ is equal to 1, then, under the assumption that $A > 1$ and $D \ne 2A$, the Yoneda products vanish by [17, Theorem 3.4]. We now assume that $m$ or $n$ is even and that $m \ne 1$ and $n \ne 1$.

Let $\sigma: \Lambda_0 \to \Lambda$ be the natural inclusion.

Consider $f_m : \mathbf{P}^m \to \Lambda_0$; it lifts to $f_m^0 = \sigma \circ f_m : \mathbf{P}^m \to \Lambda$. We now define further liftings $f_m^i : \mathbf{P}^{m+i} \to \mathbf{P}^i$ for $i \ge 1$ as follows:

$$f_m^1(x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(m+1)} \overline{\otimes} \lambda) = f_m^0(x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(m)} \overline{\otimes} 1) x_{\delta(m)+1} \overline{\otimes} x_{\delta(m)+2} \cdots x_{\delta(m+1)} \lambda;$$

$$f_m^i(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+i)} \overline{\otimes} \lambda) = f_m^0(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+i)} \overline{\otimes} \lambda \quad \text{if } m \text{ or } i \ge 2 \text{ is even;}$$

$$f_m^i(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+i)} \overline{\otimes} \lambda) = f_m^0(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+i-1)+1} \quad \text{if } m \text{ and } i \ge 2 \text{ are odd,}$$

where the $x_i$ are in $\Lambda_1$ and the $y_i$ are in $\Lambda_A$. The proof that $(f_m^i)_{i \ge 0}$ is a family of liftings of $f_m$, that is, $f^{i-1} \circ b^{i+m} = b^i \circ f_m^i$ for all $i \ge 1$, is tedious but straightforward.

Finally, if $n$ or $m$ is even and $n \ge 2$, $m \ge 2$, then

$$ \begin{align*} f_n f_m (y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda) &= f_n \circ f_m^n (y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+n)} \overline{\otimes} \lambda) \\ &= f_n (f_m^0 (y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} y_{\delta(m)+2} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda) \\ &= f_n (f_m (y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} y_{\delta(m)+2} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda). \quad \square \end{align*} $$

**Theorem A.5.** If $\Lambda$ is a $(D, A)$-stacked monomial algebra, with $A > 1$, $D = dA$, and $d \ge 3$, then $E(\Lambda)$ and $B$ are isomorphic as graded algebras. In particular, $\operatorname{Ext}_\Lambda^n(\Lambda_0, \Lambda_0)$ is isomorphic to $\ell \Lambda_{\delta(n)}$.

*Proof:* We use the isomorphisms in Lemma A.3 and the cup product described in Proposition A.4. If $\overline{f} = f_{\delta(n)} \overline{\otimes} \cdots \overline{\otimes} f_1 \in B_n$ and $\overline{g} = g_{\delta(m)} \overline{\otimes} \cdots \overline{\otimes} g_1 \in B_m$, where $m$ or $n$ is even and both are at least 2, then

$$ \begin{align*} \psi(\overline{f})\psi(\overline{g})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda) \\ &= \psi(\overline{f})(\psi(\overline{g})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+n)} \overline{\otimes} \lambda) \\ &= f_{\delta(n)}(f_{\delta(n)-1}(\dots (f_2(f_1(\psi(\overline{g})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1}) y_{\delta(m)+2}) \dots)) \\ & \qquad \cdot y_{\delta'(m+n)-1} y_{\delta(m+n)}) \overline{\lambda} \\ &= f_{\delta(n)}(f_{\delta(n)-1}(\dots (f_2(f_1(g_{\delta(m)})(\dots (g_1(y_1)) \dots y_{\delta(m)}) y_{\delta(m)+1}) y_{\delta(m)+2}) \dots)) \\ & \qquad \cdot y_{\delta(m+n)-1} y_{\delta(m+n)}) \overline{\lambda} \\ &= \psi(\overline{fg})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda) \\ &= \psi(\overline{f} \cdot \overline{g})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda), \end{align*} $$
**Proposition A.4.** Let $\Lambda$ be a $(D, A)$-stacked monomial algebra with $A > 1$, $D = dA$, and $d \ge 3$. The Yoneda product of $f_n \in \operatorname{Hom}_\Lambda(\mathbf{P}^n, \Lambda_0)$ and $f_m \in \operatorname{Hom}_\Lambda(\mathbf{P}^m, \Lambda_0)$ is given by

$$f_n f_m = \begin{cases} 0 & \text{if } n \text{ and } m \text{ are odd;} \\ 0 & \text{if } n \ge 1, m \ge 1, \text{ and } n = 1 \text{ or } m = 1; \\ \sum x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(n)+\delta(m)} \overline{\otimes} \lambda \mapsto f_n \left( \sum f_m (x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(m)} \overline{\otimes} 1) x_{\delta(m)+1} \right. \\ \qquad \left. \overline{\otimes} x_{\delta(m)+2} \overline{\otimes} \cdots \overline{\otimes} x_{\delta(m)+\delta(n)} \overline{\otimes} \lambda \right) & \text{otherwise,} \end{cases}$$

where the $x_i$ are all in $\Lambda_A$.

*Proof:* If $m$ and $n$ are odd or if $m \ge 1$ or $n \ge 1$ is equal to 1, then, under the assumption that $A > 1$ and $D \ne 2A$, the Yoneda products vanish by [17, Theorem 3.4]. We now assume that $m$ or $n$ is even and that $m \ne 1$ and $n \ne 1$.

Let $\sigma: \Lambda_0 \to \Lambda$ be the natural inclusion.

Consider $f_m: \mathbf{P}^m \to \Lambda_0$; it lifts to $f_m^0 = \sigma \circ f_m: \mathbf{P}^m \to \Lambda$. We now define further liftings $f_m^i: \mathbf{P}^{m+i} \to \mathbf{P}^i$ for $i \ge 1$ as follows:

$$f_m^1(x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(m+1)} \overline{\otimes} \lambda) = f_m^0(x_1 \overline{\otimes} \cdots \overline{\otimes} x_{\delta(m)} \overline{\otimes} 1) x_{\delta(m)+1} \overline{\otimes} x_{\delta(m)+2} \cdots x_{\delta(m+1)} \lambda;$$

$$f_m^i(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+i)} \overline{\otimes} \lambda) = f_m^0(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+i)} \overline{\otimes} \lambda \quad \text{if } m \text{ or } i \ge 2 \text{ is even;}$$

$$f_m^i(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+i)} \overline{\otimes} \lambda) = f_m^0(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+i-1)+1} \quad \text{if } m \text{ and } i \ge 2 \text{ are odd,}$$

where the $x_i$ are in $\Lambda_1$ and the $y_i$ are in $\Lambda_A$. The proof that $(f_m^i)_{i \ge 0}$ is a family of liftings of $f_m$, that is, $f^{i-1} \circ b^{i+m} = b^i \circ f_m^i$ for all $i \ge 1$, is tedious but straightforward.

Finally, if $n$ or $m$ is even and $n \ge 2$, $m \ge 2$, then

$$\begin{align*} f_n f_m (y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda) &= f_n \circ f_m^n (y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+n)} \overline{\otimes} \lambda) \\ &= f_n (f_m^0 (y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} y_{\delta(m)+2} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda) \\ &= f_n (f_m (y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} y_{\delta(m)+2} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda). \quad \square \end{align*}$$

**Theorem A.5.** If $\Lambda$ is a $(D, A)$-stacked monomial algebra, with $A > 1$, $D = dA$, and $d \ge 3$, then $E(\Lambda)$ and $B$ are isomorphic as graded algebras. In particular, $\operatorname{Ext}_\Lambda^n(\Lambda_0, \Lambda_0)$ is isomorphic to $\ell \Lambda_{\delta(n)}$.

*Proof:* We use the isomorphisms in Lemma A.3 and the cup product described in Proposition A.4. If $\bar{f} = \overline{f_{\delta(n)} \overline{\otimes} \cdots \overline{\otimes} f_1} \in B_n$ and $\bar{g} = \overline{g_{\delta(m)} \overline{\otimes} \cdots \overline{\otimes} g_1} \in B_m$, where $m$ or $n$ is even and both are at least 2, then

$$\begin{align*} \psi(\bar{f})\psi(\bar{g})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda) \\ &= \psi(\bar{f})(\psi(\bar{g})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1} \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m+n)} \overline{\otimes} \lambda) \\ &= f_{\delta(n)}(f_{\delta(n)-1}(\dots (f_2(f_1(\psi(\bar{g})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)} \overline{\otimes} 1) y_{\delta(m)+1}) y_{\delta(m)+2}) \dots)) \\ & \qquad \cdot y_{\delta'(m+n)-1} y_{\delta(m+n)}) \overline{\lambda} \\ &= f_{\delta(n)}(f_{\delta(n)-1}(\dots (f_2(f_1(g_{\delta(m)})(\dots (g_1(y_1))) y_{\delta(m)} y_{\delta(m)+1}) y_{\delta(m)+2}) \dots)) \\ & \qquad \cdot y_{\delta(m+n)-1} y_{\delta(m+n)}) \overline{\lambda} \\ &= \psi(\overline{fg})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda) \\ &= \psi(\bar{f} \cdot \bar{g})(y_1 \overline{\otimes} \cdots \overline{\otimes} y_{\delta(m)+\delta(n)} \overline{\otimes} \lambda), \end{align*}$$
therefore $\varphi(\overline{f} \cdot \overline{g}) = \psi(\overline{f})\psi(\overline{g})$ and we have proved that $\psi$ is an isomorphism of graded algebras. $\square$

*Remark A.6.* Recall from Subsection 1.2 that the $m$-th projective in a minimal projective right $\Lambda$-module resolution of $\Lambda_0$ is $L^m = \bigoplus_{R_i^m \in \mathbb{R}^m} t(R_i^m)\Lambda$. By Consequence 3.4, $R_i^m \in \mathbf{P}^m$. We then have an isomorphism $\mathbf{P}^m \to L^m$ which is determined by $R_i^m \mapsto t(R_i^m)$ for all $i$.

As we mentioned in Subsection 1.2, the authors of [14] also gave a basis of $E(\Lambda)$, namely the set $\{g_i^m \in \operatorname{Hom}_\Lambda(L^m, \Lambda_0) \mid R_i^m \in \mathbb{R}^m\}$, where $g_i^m(t(R_j^m)) = t(R_i^m)$ if $j=i$ and $0$ otherwise. The element $g_i^m$ corresponds to a map in $\operatorname{Hom}_\Lambda(\mathbf{P}^m, \Lambda_0)$ which we denote again by $g_i^m$ and that is defined by $g_i^m(R_j^m) = t(R_i^m)$ if $j=i$ and $0$ otherwise.

We have isomorphisms $KQ_A \cong K\Gamma_1$ and $KQ_A \cong {}^*\Lambda_A = {}^*(KQ_A)$. Combining them, we obtain an isomorphism which associates to $\bar{\alpha} \in \Gamma_1$ the linear form $f_\alpha$ on $KQ_A$ that sends $\beta \in Q_A$ to $t(\alpha)$ if $\beta = \alpha$ and to $0$ otherwise. This extends to an isomorphism between the algebras $K\Gamma/(\sigma)$ and $\Gamma^2\Lambda$ that sends a path $\bar{p}$ of length $n$ to the class of the linear map $f_p \in {}^*(\Lambda_A)^{\bar{\otimes}n}$ defined on $A$-paths by $f_p(q) = t(p)$ if $q=p$ and $0$ otherwise.

Now consider $R_i^m = \alpha_1 \cdots \alpha_{\delta(m)}$ with $\alpha \in Q_A$ for all $i$ and $\bar{R}_i^m = \bar{\alpha}_{\delta(m)} \cdots \bar{\alpha}_1$ in $K\Gamma/(\sigma)$. It is easy to check that $g_i^m = \psi(\bar{f}_{\alpha_{\delta(m)}} \bar{\otimes} \cdots \bar{\otimes} \bar{f}_{\alpha_1})$ so that it corresponds, via the isomorphism above, to $\bar{R}_i^m$.

Therefore we have a basis $B_B = \{\bar{R}_i^m \mid R_i^m \in \mathbb{R}^m\}$ of $B$.

## References

[1] M. J. BARDZELL, The alternating syzygy behavior of monomial algebras, *J. Algebra* **188**(1) (1997), 69–89. DOI: 10.1006/jabr.1996.6813.

[2] A. BEILINSON, V. GINZBURG, AND W. SOERGEL, Koszul duality patterns in representation theory, *J. Amer. Math. Soc.* **9**(2) (1996), 473–527. DOI: 10.1090/S0894-0347-96-00192-0.

[3] R. BERGER, Koszulity for nonquadratic algebras, *J. Algebra* **239**(2) (2001), 705–734. DOI: 10.1006/jabr.2000.8703.

[4] V. DOTSENKO, V. GÉLINAS, AND P. TAMAROFF, Finite generation for Hochschild cohomology of Gorenstein monomial algebras, *Selecta Math. (N.S.)* **29**(1) (2023), Paper no. 14, 45 pp. DOI: 10.1007/s00029-022-00817-8.

[5] K. ERDMANN, Algebras with non-periodic bounded modules, *J. Algebra* **475** (2017), 308–326. DOI: 10.1016/j.jalgebra.2016.05.004.

[6] K. ERDMANN, M. HOLLOWAY, R. TAILLEFER, N. SNASHALL, AND Ø. SOLBERG, Support varieties for selfinjective algebras, *K-Theory* **33**(1) (2004), 67–87. DOI: 10.1007/s10977-004-0838-7.

[7] K. ERDMANN AND Ø. SOLBERG, Radical cube zero weakly symmetric algebras and support varieties, *J. Pure Appl. Algebra* **215**(2) (2011), 185–200. DOI: 10.1016/j.jpaa.2010.04.012.

[8] T. FURUYA AND N. SNASHALL, Support varieties for modules over stacked monomial algebras, *Comm. Algebra* **39**(8) (2011), 2926–2942. DOI: 10.1080/00927872.2010.498395.

[9] E. L. GREEN, D. HAPPEL, AND D. ZACHARIA, Projective resolutions over Artin algebras with zero relations, *Illinois J. Math.* **29**(1) (1985), 180–190. DOI: 10.1215/ijm/1256045849.

[10] E. L. GREEN, E. N. MARCOS, R. MARTÍNEZ-VILLA, AND P. ZHANG, D-Koszul algebras, *J. Pure Appl. Algebra* **193**(1-3) (2004), 141–162. DOI: 10.1016/j.jpaa.2004.03.012.

[11] E. L. GREEN AND R. MARTÍNEZ-VILLA, Koszul and Yoneda algebras. II, in: *Algebras and Modules*, II (Geiranger, 1996), CMS Conf. Proc. **24**, Published by the American Mathematical Society, Providence, RI, 1998, pp. 227–244.

[12] E. L. GREEN AND N. SNASHALL, Finite generation of Ext for a generalization of D-Koszul algebras, *J. Algebra* **295**(2) (2006), 458–472. DOI: 10.1016/j.jalgebra.2005.10.026.

[13] E. L. GREEN AND N. SNASHALL, The Hochschild cohomology ring modulo nilpotence of a stacked monomial algebra, *Colloq. Math.* **105**(2) (2006), 233–258. DOI: 10.4064/cm105-2-6.

[14] E. L. GREEN AND D. ZACHARIA, The cohomology ring of a monomial algebra, *Manuscripta Math.* **85**(1) (1994), 11–23. DOI: 10.1007/BF02568180.

[15] R. Y. JAWAD, Cohomology and finiteness conditions for generalisations of Koszul algebras, Thesis (Ph.D.)-University of Leicester (2019).
therefore $\varphi(\overline{f} \cdot \overline{g}) = \psi(\overline{f})\psi(\overline{g})$ and we have proved that $\psi$ is an isomorphism of graded algebras. $\square$

*Remark A.6.* Recall from Subsection 1.2 that the *m*-th projective in a minimal projective right $\Lambda$-module resolution of $\Lambda_0$ is $L^m = \bigoplus_{R_i^m \in \mathbb{R}^m} t(R_i^m)\Lambda$. By Consequence 3.4, $R_i^m \in \mathbf{P}^m$. We then have an isomorphism $\mathbf{P}^m \to L^m$ which is determined by $R_i^m \mapsto t(R_i^m)$ for all $i$.

As we mentioned in Subsection 1.2, the authors of [14] also gave a basis of $E(\Lambda)$, namely the set $\{g_i^m \in \operatorname{Hom}_\Lambda(L^m, \Lambda_0) \mid R_i^m \in \mathbb{R}^m\}$, where $g_i^m(t(R_j^m)) = t(R_i^m)$ if $j=i$ and $0$ otherwise. The element $g_i^m$ corresponds to a map in $\operatorname{Hom}_\Lambda(\mathbf{P}^m, \Lambda_0)$ which we denote again by $g_i^m$ and that is defined by $g_i^m(R_j^m) = t(R_i^m)$ if $j=i$ and $0$ otherwise.

We have isomorphisms $KQ_A \cong K\Gamma_1$ and $KQ_A \cong {}^*\Lambda_A = {}^*(KQ_A)$. Combining them, we obtain an isomorphism which associates to $\bar{\alpha} \in \Gamma_1$ the linear form $f_\alpha$ on $KQ_A$ that sends $\beta \in Q_A$ to $t(\alpha)$ if $\beta = \alpha$ and to $0$ otherwise. This extends to an isomorphism between the algebras $K\Gamma/(\sigma)$ and $\Gamma^2\Lambda$ that sends a path $\bar{p}$ of length $n$ to the class of the linear map $f_p \in {}^*(\Lambda_A)^{\bar{\otimes}n}$ defined on $A$-paths by $f_p(q) = t(p)$ if $q=p$ and $0$ otherwise.

Now consider $R_i^m = \alpha_1 \cdots \alpha_{\delta(m)}$ with $\alpha \in Q_A$ for all $i$ and $\bar{R}_i^m = \bar{\alpha}_{\delta(m)} \cdots \bar{\alpha}_1$ in $K\Gamma/(\sigma)$. It is easy to check that $g_i^m = \psi(\bar{f}_{\alpha_{\delta(m)}} \bar{\otimes} \cdots \bar{\otimes} \bar{f}_{\alpha_1})$ so that it corresponds, via the isomorphism above, to $\bar{R}_i^m$.

Therefore we have a basis $B_B = \{\bar{R}_i^m \mid R_i^m \in \mathbb{R}^m\}$ of $B$.

## References

[1] M. J. BARDZELL, The alternating syzygy behavior of monomial algebras, *J. Algebra* **188**(1) (1997), 69–89. DOI: 10.1006/jabr.1996.6813.

[2] A. BEILINSON, V. GINZBURG, AND W. SOERGEL, Koszul duality patterns in representation theory, *J. Amer. Math. Soc.* **9**(2) (1996), 473–527. DOI: 10.1090/S0894-0347-96-00192-0.

[3] R. BERGER, Koszulity for nonquadratic algebras, *J. Algebra* **239**(2) (2001), 705–734. DOI: 10.1006/jabr.2000.8703.

[4] V. DOTSENKO, V. GÉLINAS, AND P. TAMAROFF, Finite generation for Hochschild cohomology of Gorenstein monomial algebras, *Selecta Math. (N.S.)* **29**(1) (2023), Paper no. 14, 45 pp. DOI: 10.1007/s00029-022-00817-8.

[5] K. ERDMANN, Algebras with non-periodic bounded modules, *J. Algebra* **475** (2017), 308–326. DOI: 10.1016/j.jalgebra.2016.05.004.

[6] K. ERDMANN, M. HOLLOWAY, R. TAILLEFER, N. SNASHALL, AND Ø. SOLBERG, Support varieties for selfinjective algebras, *K-Theory* **33**(1) (2004), 67–87. DOI: 10.1007/s10977-004-0838-7.

[7] K. ERDMANN AND Ø. SOLBERG, Radical cube zero weakly symmetric algebras and support varieties, *J. Pure Appl. Algebra* **215**(2) (2011), 185–200. DOI: 10.1016/j.jpaa.2010.04.012.

[8] T. FURUYA AND N. SNASHALL, Support varieties for modules over stacked monomial algebras, *Comm. Algebra* **39**(8) (2011), 2926–2942. DOI: 10.1080/00927872.2010.498395.

[9] E. L. GREEN, D. HAPPEL, AND D. ZACHARIA, Projective resolutions over Artin algebras with zero relations, *Illinois J. Math.* **29**(1) (1985), 180–190. DOI: 10.1215/ijm/1256045849.

[10] E. L. GREEN, E. N. MARCOS, R. MARTÍNEZ-VILLA, AND P. ZHANG, D-Koszul algebras, *J. Pure Appl. Algebra* **193**(1-3) (2004), 141–162. DOI: 10.1016/j.jpaa.2004.03.012.

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Ruaa Jawad

Technical Instructors Training Institute, Middle Technical University, Baghdad, Iraq

E-mail address: ruaa.yousuf.jawad@mtu.edu.iq

Nicole Snashall

School of Computing and Mathematical Sciences, University of Leicester, University Road, Leicester
LE1 7RH, United Kingdom

E-mail address: njs5@leicester.ac.uk

Rachel Taillefer

Université Clermont Auvergne, CNRS, LMBP, F-63000 Clermont-Ferrand, France

E-mail address: rachel.taillefer@uca.fr

Received on January 9, 2023.

Accepted on October 13, 2023.
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Ruaa Jawad

Technical Instructors Training Institute, Middle Technical University, Baghdad, Iraq

E-mail address: ruaa.yousuf.jawad@mtu.edu.iq

Nicole Snashall

School of Computing and Mathematical Sciences, University of Leicester, University Road, Leicester
LE1 7RH, United Kingdom

E-mail address: njs5@leicester.ac.uk

Rachel Taillefer

Université Clermont Auvergne, CNRS, LMBP, F-63000 Clermont-Ferrand, France

E-mail address: rachel.taillefer@uca.fr

Received on January 9, 2023.

Accepted on October 13, 2023.