We prove that the bounded derived category of the lattice of order ideals of the product of two ordered chains is fractionally Calabi-Yau. We also show that these lattices are derived equivalent to higher Auslander algebras of type A. The proofs involve the study of intervals of the poset that have resolutions described with antichains having rigid properties. These two results combined corroborate a conjecture by Chapoton linking posets to Fukaya Seidel Categories.

§ INTRODUCTION

The notion of Calabi-Yau categories was introduced by Kontsevich in the late nineties <cit.>.
A triangulated category $\mathcal{T}$ with a Serre functor $\Serre$ is said to be fractionally Calabi-Yau if there exists $l$ and $d$ such that $\Serre^{l}$ is isomorphic as a functor to the suspension functor applied $d$ times. We say that $\mathcal{T}$ is $\frac{d}{l}$-Calabi-Yau.

When $\mathcal{T} = D^b(A)$, the bounded derived category of an algebra $A$ of finite global dimension over a field $\field$, we can take $\Serre = -\otimes_A^{\mathbbm{L}} DA$, the derived Nakayama functor. In that case, the Calabi-Yau property can be further relaxed. If $\phi$ is an automorphism of $A$ , then $D^b(A)$, is said to be twisted fractionally Calabi–Yau if $\Serre^{l}\simeq [d]\circ \phi^{*}$ where $\phi^*$ twists the action of $A$ on a module by $\phi$. We recover the previous definition when $\phi = id_A$. The following theorem makes it easier to detect twisted fractionally Calabi–Yau algebras.

Let $\Lambda$ be a finite dimensional $\field$-algebra of finite global dimension. The following conditions are equivalent.

* $\Lambda$ is twisted $\frac{d}{l}$-Calabi–Yau.

* $\Serre^{l}\Lambda\simeq \Lambda [d]$.

This leads to a question which is still far from being answered in general.

[<cit.>]Is every twisted fractionally Calabi-Yau algebra fractionally Calabi-Yau?

Further, the trivial extension algebra of a (twisted) fractionally Calabi–Yau finite dimensional algebras of finite global dimension is (twisted) periodic <cit.>. Hence Question <ref> is linked to the following conjecture of Erdmann and Skowroński <cit.>.

[<cit.>]Is every finite-dimensional twisted periodic algebra periodic?

In the case of finite posets with a unique maximal element or a unique minimal element, the answer to question <ref> is positive as per <cit.>. However, for a given incidence algebra the existence of an isomorphism
\[\Serre^l (A)\simeq A[d]\]
is still in general very hard to check<cit.><cit.>. In this article we provide a relaxation of <cit.> to help overcome that difficulty.

Let $L$ be a finite lattice, $d$ and $l$ integers and $(C_{\alpha})_{\alpha\in L}$ be a family of indecomposable modules with simple head $S_{\alpha}$ and a boolean resolution. If for all $\alpha\in L$ it holds that $\Serre^{l}(C_{\alpha})\simeq C_{\alpha}[d]$, then $L$ is $\frac{n}{m}$- fractionally Calabi–Yau.

This theorem does not provide with a recipe to find such appropriate families, but it suggests certain criteria which restrict the search for the perfect combinatorial candidates.

Fractionally Calabi-Yau posets are fascinating objects in part due to a hypothetical relation to product formulas due to Chapoton <cit.>. In combinatorics, many families of sets $(S_n)_{n\in \mathbb{N}}$ can be counted by product formulas $|S_n| = \Pi_{i=1}^{n}\frac{D-d_i}{d_i}$ where the sum of the numerator and denominator is constant and equal to $D$. Such families include the Catalan numbers, the number of alternating sign matrices, the West family and the Tamari intervals family. Chapoton's highly conjectural explanation is that there should exist a partial order on $S_n$ whose derived category is $\frac{C}{D}$-Calabi-Yau, where $C = \sum_i D-2d_i$. Moreover, the bounded derived category in question should be equivalent to a type of Fukaya-Seidel category constructed from the data of $D$ and the $d_i$ coefficients. The conjecture also provides predictions regarding the coxeter polynomial or the Milnor number of the surface some of which we can tested with a computer on examples. Some consequences of these conjectures have been proven since (<cit.>). The starting point of this paper was to prove another one of these resulting conjecture which was already studied in part in <cit.>.

Observe that the binomial ${m+n\choose m}$ can be written as

\begin{equation}
\frac{m+n}{1}\frac{m+n-1}{2}\cdot \dots\cdot\frac{m + 1}{n}
\end{equation}

where $D = m + n + 1$. This is probably the most natural example of product formula discussed above. The poset of order ideals of a product of total orders of length $m$ and $n$ has cardinality ${n+m \choose m}$. We write it $J_{m,n}$. Using our results on boolean antichain modules we are able to confirm Chapoton's prediction about the Calabi-Yau dimension of these posets.

The bounded derived category of $J_{m,n}$ is $\frac{mn}{m+n+1}$-Calabi-Yau.

As a corollary this gives a positive answer to the Chapoton-Yildirim conjecture on cominuscule posets of type A and B<cit.>.
The bounded derived category of cominuscule posets of type $A$, $B$, $D$ are fractionally Calabi-Yau. For types A and B, the denominator is $h + 1$ where h is a constant associated with the root system.

The key observation one needs for applying Theorem <ref> to cominuscule posets  is their classification into types $C_{I}$, $C_{II}$ or $C_{III}$ depicted below (<cit.>).

[step=0.5cm, rotate = 45] (0, 0) grid (1.5, 1.5);
[step=0.5cm, rotate = 45] (1.99, 0) grid (3, 1.5);
[step=0.5cm, rotate = 45] (0, 1.99) grid (1.5, 3);
[step=0.5cm, rotate = 45] (1.99, 1.99) grid (3, 3);
[step=0.5cm, rotate = 45, dotted] (0, 0) grid (3, 3);

[step=0.5cm, rotate = -45] (-3, 0) grid (-2, 1);
[step=0.5cm, rotate = -45, dotted] (-3, 0) grid (-1.5, 1.5);
[step=0.5cm, rotate = -45] (-1.5, 0) grid (-0.5, 0.5);
[step=0.5cm, rotate = -45] (-1.5, 0.5) grid (-1, 1);
[rotate = -45] (0,0) – (-0.5, 0);
[rotate = -45, step = 0.5cm] (-3, 1.49) grid (-2, 2);
[step = 0.5cm, rotate = -45] (-3, 2) grid (-2.5, 2.5);
[step = 0.5cm, rotate = -45](-3, 2.5) – (-3, 3);

[rotate = 45, step = 0.5] (0,0) – (0.5,0.5);
[rotate = 45, step = 0.5, dotted] (0.5,0.5) – (1, 1);
[rotate = 45, step = 0.5, dotted] (1.5,1.5) – (2,2);
[rotate = 45, step = 0.5] (2,2) – (2.5, 2.5);
[rotate = 45, step = 0.5] (0.99,0.99) grid (1.51, 1.51);

The three types of cominuscule posets

Interestingly, there is no one to one correspondance between this classification and the ADE classification of the root poset one started with. However cominuscule posets of type $A$ and $B$ all follow the pattern of  $C_{I}$. The corollary to Theorem <ref> follows from that. Type $D$ follows pattern $C_{III}$ which is proved using Ladkani's flip flop techniques <cit.>. Type $C$ follows pattern $C_{II}$ and seems to be different and is not a consequence of our work.

Our proof of Theorem <ref> gives a good understanding of the Serre functor for this category. However, one would like to have a more structural reason behind the Calabi-Yau property for posets. For us, a good reason why $J_{m,n}$ should be fractionally Calabi–Yau is our second main result which is the following derived equivalence.

The algebra of the poset $J_{m,n}$ is derived equivalent to the higher Auslander algebra $A_{m+1}^{n-1}$.

Higher auslander algebras were introduced by Iyama in <cit.> as part of a series of seminal articles on higher representation theory. Higher auslander algebras of type A were soon after described in <cit.> and are known to be fractionally Calabi-Yau.

Higher algebras of type $A$ are fractionally calabi Yau.

Hence with Theorem <ref> we have a new proof for an already known theorem. Previous proofs of this result have different flavours. The first one stemmed from symplectic geometry<cit.>, the second one, from the theory of infinity categories<cit.> and the most recent from an intricate study of the properties of a certain preprojective algebra and its Nakayama automorphism, linking it to the the Serre functor<cit.>. The proof exposed here is combinatorial and elementary. Of course knowing Theorems <ref> and <ref> also gives a proof of Theorem <ref>.

It is also satisfying to note that this derived equivalence ties back into the Chapoton conjectures as a partially wrapped Fukaya category can be associated to higher Auslander algebras of type $A$ <cit.>. A recent preprint <cit.> also links the higher Auslander algebras of type A to Fukaya Seidel Categories with the Milnor number predicted by Chapoton.

Acknowledegments I would like to thank my supervisor Baptiste Rognerud for introducing me to the subject as well as for all the discussion, guidance and careful reading of my work at every step of the way. I would like to thank my second supervisor, Bernhard Keller for  Figure (<ref>). This work will  also appear as part of my Phd thesis.

§.§ Notation

Generalities Let $\field$ be a field and $X$ a finite partially ordered set or poset. Define its incidence algebra $\IncAlg = \IncAlg_{\field}(X)$ over $\field$ to be the $\field$-vector space with basis pairs $(x, y)$ such that $x\leq y$ with multiplication defined by

\begin{equation*}
(x, y)(z, t) = \begin{cases}(x, t)& \text{if } y=z,\\
0 &\text{otherwise}.\end{cases}
\end{equation*}

For $x\in X$ we write $e_x = (x, x)$ the primitive idempotent. Then we have $1_\IncAlg = \sum_{x\in X} e_x$.
Throughout this work we consider finite dimensional left modules over $\IncAlg$. For every element $x\in X$ the associated simple module is $S_x \cong k$ with action $(y, t)\cdot 1_{\field} = 0$ unless $y=t=x$ in which case $e_x\cdot 1_{\field} = 1_{\field}$. Its projective cover $P_x = \mathcal{A}\cdot e_x$ has basis $\{(y, x)|y\leq x\}$.  Its injective hull is the injective indecomposable $I_x = (e_x \cdot\mathcal{A})^*$ and has basis $\{(x, y)^*| x\leq y\}$. Recall that morphisms between the projective indecomposables are characterised by

\begin{equation*}
\Hom_{\IncAlg} (P_x, P_y) = \Hom_{\IncAlg}(\mathcal{A} e_x, \mathcal{A} e_y)\cong\begin{cases}e_x \mathcal{A} e_y \cong \field&\text{if }x\leq y,\\0 &\text{otherwise.}\end{cases}
\end{equation*}

We denote the canonical inclusion as $\iota_x^y: P_x\hookrightarrow P_y$ whenever $x\leq y$ which is right multiplication by $(x, y)$. More generally for any left $\IncAlg$-module $M$, we have $\Hom_{\IncAlg}(P_x, M)\cong e_xM$. This isomorphism makes the following diagram commute

\begin{equation}\label{eq:sqIota}
\begin{tikzcd}
f\ar[d, mapsto]\ar[r, phantom, "\in" description]&\Hom_{\IncAlg} (P_x, M)\ar[d]&\ar[l, "\circ \iota_x^y"]\ar[d] \Hom_{\IncAlg} (P_y, M)\ar[r, phantom, "\ni" description]&g\ar[d, mapsto]\\
f(e_x)\ar[r, phantom, "\in" description, near start]&e_xM&\ar[l, "{(x,y)\cdot}"]e_yM\ar[r, phantom, "\ni" description, near end]&g(e_y)
\end{tikzcd}
\end{equation}

The total hom complex $\Hom^{\bullet}_{\IncAlg}(C, M)$ where $C$ is a chain complex $C = ((C_n)_n, (\partial_n))$ of $\IncAlg$ modules and $M$ is an $\IncAlg$-module, is the complex

\[\dots\to\Hom_{\IncAlg }(C_n, M) \xrightarrow{\partial_{n+1}^* }\Hom_{\IncAlg }(C_{n+1}, M) \to \dots\]

Note that we omit a conventional sign for the boundary map as it plays no role in our computations. This is a cochain complex as the functor $\Hom (-, M)$ is contravariant. Assuming that $C_n = \bigoplus_{x\in S_n} P_x$ with $S_n\subseteq X$ and taking its cohomology gives shifted morphisms in the homotopy category $\HomotopyCat(\IncAlg)$ <cit.>, which in turn are isomorphic to the shifted morphism in the derived category because the source is a perfect complexe:

\begin{equation}\label{eq:TotalHomHom}
\HH^{i}(\Hom^{\bullet}_{\IncAlg}(C, M)) \cong \Hom_{\HomotopyCat(\IncAlg)}(C, M[i])\overset{u}{\cong}\Hom_{D^b}(C, M[i]).
\end{equation}

Most computations will be carried out explicitly in the homotopy category. When needed the passage from one to the other will be discussed. Finally, using equation (<ref>) we have an isomorphism of cochain complexes

\begin{equation}\label{eq:totHomComp}
\begin{tikzcd}
\dots\ar[r]&\Hom_{\IncAlg}(\bigoplus_{x\in S_n} P_x, M)\ar[r, "\partial_{n+1}^* "]\ar[d]&\Hom_{\IncAlg}(\bigoplus_{x\in S_{n+1}} P_x, M)\ar[r]\ar[d]&\dots\\
\dots\ar[r]&\bigoplus_{x\in S_n}e_xM\ar[r]& \bigoplus_{x\in S_{n+1}}e_xM\ar[r]&\dots
\end{tikzcd}
\end{equation}

The boundary maps of the bottom complex are linear combinations of left multiplication by elements $(x, y)$ of the algebra with coefficients inherited from the top complex.

Antichain Modules
Let $(L, \wedge, \vee)$ be a finite lattice. We write $\hat{1}$ its greatest element and $\hat{0}$ its least one. Let $C$ be an antichain in $L$ i.e a subset C of L that consists of pairwise incomparable elements of $L$. We say an antichain $C$ is below an element $\alpha$ of $L$ if for all $c\in C$, we have $c\leq \alpha$, and when needed we record this information by the notation $C_{\alpha}$. Following <cit.> we associate to an antichain $C = \{c_1, \dots, c_r\}$ the submodule

$$N_C = \sum_{i=1}^{r} A\cdot(c_i,\hat{1})$$

of the projective indecomposable $P_{\hat{1}}$ generated by the antichain. It follows directly from the same proposition that there is a one to one correspondance between antichains and submodules of $P_{\hat{1}}$. The antichain module associated to $C$ is defined by

$$M_C := P_{\hat{1}}/N_C.$$

We will talk of antichain modules below $\alpha\in L$ by restricting to the sublattice $[\hat{0}, \alpha]$ of $L$. Then $\alpha$ is the greatest element of this lattice and there is a bijection between submodules of $P_\alpha$ and antichains below $\alpha$. The corresponding modules will be denoted $N_C^{\alpha}$ and $M_C^{\alpha}$.
As our main example consider $a\leq b$ in $L$. The maxima of the set of elements of $L$ which are strictly less than $b$ but not above $a$ form an antichain $C$ and the antichain module below $b$ associated to $C$ has support the interval $[a, b]$. The corresponding antichain module is usually called an interval module. In the rest of the paper we identify intervals with their interval modules.

Intervals are antichain modules.

With the convention of the previous paragraph, morphisms between interval modules follow a simple rule

\begin{equation}\label{eq:morphIntS}
\Hom_{\IncAlg}([a, b], [c, d]) = \begin{cases}\field &\text{if } a\leq c \leq b\leq d\\
0 &\text{otherwise.} \end{cases}
\end{equation}

By  <cit.>, for every antichain $C$ of cardinal $r$
of a lattice $L$ the associated antichain module $M_C$ has a projective resolution $\PRes_{C}$ of the form

$$0\to P_r \to \dots\to P_0 \to M_C \text{ where } P_0 = P_{\hat{1}} \text{ and } P_l = \bigoplus_{\substack{S\subseteq C\\|S|=l}} P_{\wedge S}\text{ for } 1\leq l\leq r. $$

Similarly, define a resolution $\PRes_C^{\alpha}$ for the antichain module $M_C^{\alpha}$ below $\alpha$ by replacing $P_{\hat{1}}$ by $P_{\alpha}$.
The boundary maps are defined by fixing an arbitrary total ordering of elements in $C$ and, in each degree, setting the following maps between the indecomposable summands of the source and target in each degree:

\begin{align}\label{eq:boundComp}\begin{split}
P_{\wedge  S} \quad &\to\quad\quad\quad P_{\wedge T}\\
\big(x, \wedge S\big)\quad &\mapsto \quad\begin{cases}(-1)^{|i|_S}\big(x,\wedge T\big) &\text{if }T \sqcup \{i\} = S, \\ \quad \quad 0&\text{otherwise}\end{cases}
\end{split}\end{align}

for each $S = \{i_1, \dots, i_k\}$ and $\big(\wedge T,\wedge S\big)\in P_{\wedge S}$ where $|i|_S = |\{j\in S | j \leq i\}|$.

§.§ Detailed outline

In section <ref> we introduce four properties on antichains in lattices: intersectivity, inclusivity, strength and booleanity. They are related as in figure (<ref>).

[12]l0.4[squarednode/.style=rectangle, draw, fill=gray!5, very thick, minimum size=5mm]
[squarednode] (intersective) intersective;
[squarednode] (inclusive)  [below = of intersective] inclusive;
[squarednode] (strong)  [below = of inclusive] strong;
[squarednode] (boolean) [right = of inclusive] boolean;
(-1.3, -3.7) rectangle  (1.3, .4) ;
at (1.15, -1.6) (box) ;
[implies-implies, double] (box) –node[above]def (boolean);
[implies-implies, double] (inclusive) –node[left] <ref> (strong);

Properties of Antichains

In Lemma (<ref>) we show that an antichain $C$ is boolean if and only if it spans a boolean sublattice in $L$ and $C$ is boolean. This justifies the terminology. Boolean antichains have a crucial property (Theorem <ref>): hom spaces from a boolean antichain module to an interval are at most one dimensional and are concentrated in one degree. This does not hold for antichains which are only strong or intersective. However, certain hom spaces can still be controlled well enough. More specifically Lemmas <ref> and <ref> describe the maps between a resolution of a strong antichain module and its truncations. Using these lemmas we prove Theorem <ref> which is the main technical result of this paper.

[step = 1cm](0,0) grid (1,5) node[above] $\PRes_{C}\qquad$;
[step = 1cm](3,0) grid (4,5) node[above] $\Serre \PRes_{C}\qquad$;;
[thick, gray] (-0.1, -0.1) rectangle (4.1, 3.1);
[thick, gray] (-0.2, -0.2) rectangle (4.2, 4.1);
[->, thick, gray] (1.3, 2.8) – (2.7, 2.8);
[->, thick, gray] (1.3, 3.8) – (2.7, 3.8);
in .5, 1.5, 2.5, 3.5[Circle[]->] (.5, ) – (3.5, );
[->, dashed] (2, 2.9) to [bend right] (2, 3.7);

Isomorphisms of graded objects out of homogeneous one

It is a categorification theorem that builds upon and broadens <cit.> which states that the Calabi-Yau property can be checked on the projective indecomposable modules in a finite poset with a least or greatest element. Our result implies that, for finite lattices, the property can be checked on any family of non zero strong antichain modules as long as it is sufficiently large. The proof consists in constructing isomorphisms $\Serre^{d}(P)\cong P[l]$ for projective indecomposable modules using the isomorphisms $\Serre^{d}(\PRes_C)\cong \PRes_C[l]$ on the family of antichain modules. We proceed by strong induction on the elements of the lattice and use the axiom TR3 and the 2 out of 3 Lemma <ref>. We use Lemmas <ref> and <ref> to ensure that required squares commute. They will also help to tackle the main difficulty of the proof: constructing an isomorphism between graded objects out of ismorphisms in each degrees. The fact that these objects come from strong antichains will be crucial and Lemmas <ref> and <ref> will be used again. This will take the form of a so called inner induction hypothesis. See Figure (<ref>) for an illustration.

Theorem <ref> will be applied in Section <ref> to the incidence algebra of the lattice of order ideals of the product of two linearly ordered sets. To discuss antichains it is convenient to see the elements of this lattice as paths in grids. We consider a family of antichains introduced in <cit.>. Its associated objects are written $(\PRes_{\alpha})_{\alpha\in J_{m,n}}$. After recalling the main arguments of the proof from <cit.> and adapting them to our convention, we will show that these antichains satisfy the conditions of Theorem <ref> and thus prove Theorem <ref>.

In section <ref> we describe morphisms between the objects $\PRes_{\alpha}$ and their shifts. The goal of Section <ref> is to prove Theorem <ref>. We call $\mathcal{Y}_{m,n}$ the full subcategory of $D^{b}(\IncAlg)$ who's objects are the antichain modules $\PRes_{\alpha}$ and their shifts. Knowing that their corresponding antichains are boolean  Theorem <ref> implies that each hom space is of dimension at most one. Proposition <ref> gives the following canonical factorisation of morphisms in $\mathcal{Y}_{m,n}$ into extensions and degree zero morphisms.

\begin{equation}\label{eq:IntroHomDecomp}
\begin{tikzcd}[column sep = tiny]
\PRes_{\alpha}\ar[rr]\ar[dr]&&\PRes_{\beta}[|J|]\\
\end{tikzcd}
\end{equation}

The extension is explicitly described in Proposition <ref> and further decomposed into elementary extensions in Lemma <ref>. The proof relies on computations in the homotopy category of complexes as the source of the morphism is identified with its projective resolution and the target is an interval.

It follows from Proposition <ref> that the degree one morphisms are parametrised by a subset of the original antichains themselves. It is also crucial to the proof that not all subsets of the antichains yield extensions. Definitions <ref> provides a characterisation of these subsets which we call allowed subsets.

The second component of the composition in equation (<ref>) is a shifted morphism between intervals of the form $[f(\alpha), \alpha]$. Morphisms between intervals are described by equation (<ref>) and comparison of partitions is done term wise. Most of the proofs amount thus to checking inequalities of the form:

\[f(\alpha)_i\leq f(\beta)_i \leq \alpha_i \leq \beta_i\]

for appropriate indices $i \leq m$. Lemma <ref> gives alternative characterisations of morphisms between the objects $\PRes_{\alpha}$. Paragraph <ref> highlights certain morphisms
$\PRes_{\alpha}\to\PRes_{p_J(\alpha)}$ using the new characterisations. Finally, Lemma <ref> provides a decomposition into elementary morphisms.

Next, Proposition <ref> describes the relations between these morphisms. It uses Lemma <ref> which identifies the relations using the same manipulation of morphisms between intervals and complexes as before. In the process, in Proposition <ref> we interpret the morphisms in a combinatorial setting that links $\mathcal{Y}_{m,n}$ to Higher Auslander algebras of type A. In Corollary <ref> we give three slightly different yet convenient presentations of the category $\mathcal{Y}_{m,n}$ with generators and relations. This leads us to prove Theorem <ref>. In Proposition <ref> and Lemma <ref> we extract a tilting object from the category $\mathcal{Y}_{m,n}$.

\begin{equation}\label{eq:tiltings}
T: = \bigoplus_{\alpha\in J_{m,n}} \PRes_{\alpha^{\star}}[i_{\alpha}]
\end{equation}

The integers $i_{\alpha}$ ensure that $T$ has no self extensions and are encoded using only the partitions $\alpha$. The fact that $\Thicc(T)$ generates the derived category can already be seen in the proof of <ref> as every projective is obtained as a succession of cones from the family of antichains.
One of the presentations of Corollary <ref> concludes the proof of Theorem <ref> while another shows that $\End(T)^{op}$ is isomorphic to the quadratic dual $(A_{n+1}^{m-1})^{\dag}$ of the higher Auslander algebra of type $A$. Thus $\End(T)^{op}$ is an intermediate object between the Auslander algebra and its quadratic dual.

§ ANTICHAIN MODULES AND THE CALABI-YAU PROPERTY

This section contains technical results about certain classes of antichain modules, their morphisms and extensions. The main one gives a way to study the fractionally Calabi-Yau property on lattices. Once the correct antichains are identified, the proof is formal. Some lemmas rely on observations about (co)simplicial sets<cit.> and their associated chain complexes.

§.§ Boolean antichains

Let $C$ be an antichain in a lattice $L$ and  $M^{\alpha}_C$ its associated antichain module below $\alpha\in L$.
Note that in degree $i$ of the projective resolution $\PRes^{\alpha}_C$ of $M^{\alpha}_C$ there are $\binom{r}{i}$ indecomposable components in the direct sum.  If one forgets the modules, the complex has the shape of the power set of $C$, however the indices of the modules in each degree are not necessarily in bijection with with the lattice $(\PRes(C), \subseteq, \cup, \cap)$ (see Figure (<ref>)).

(0,0)  – (1,1) – (1, 2) – (0,3)–(-1,2)
–(-1, 1)–(0,0)–(0,1)–(-1, 2);
(-1, 1) – (0,2) – (1,1);
(0,1) – (1, 2);
(0, 2) – (0,3);
[red] (-1, 2)node["$c_1$" color = black] circle (2pt);
[red] (0, 2)node["$c_2$" color = black, left] circle (2pt);
[red] (1, 2)node["$c_3$" color = black] circle (2pt);

*Boolean antichain

(0,0)  – (1,1) – (1, 2) – (0,3)–(-1,2)
–(-1, 1)–(0,0)–(0,1)–(1, 2);
(-1, 1) – (0,2) – (1,1);
(0,1) – (-1, 1.6);
(0, 2) – (0, 3);
[red] (-1, 2)node["$c_1$" color = black] circle (2pt);
[red] (0, 2)node["$c_2$" color = black] circle (2pt);
[red] (1, 2)node["$c_3$" color = black] circle (2pt);
(-1, 1.6)node[label=below:$\substack{(c_2\wedge c_1)\\ \vee (c_3\wedge c_1)}$] circle (2pt);
(-1, 2) circle (3pt);

*Strong, not intersective, antichain

(0,0) – (-1, 1) – (0,2) – (0,1) – (0,0) – (1,1) – (0,2);
[red] (-1, 1)node["$c_1$" color = black] circle (2pt);
[red] (0, 1)node["$c_2$  " color = black, left] circle (2pt);
[red] (1, 1)node["$c_3$" color = black] circle (2pt);
(0, 0)node["$c_1\wedge c_2 = c_1\wedge c_3$" below] circle (2pt);

*Intersective, not strong, antichain

(0,0) – (-1, 1) – (0,2) – (0.5,1.5) – (1,1) – (0,0) – (0,1) – (0.5, 1.5);
[red] (-1, 1)node["$c_1$" color = black] circle (2pt);
[red] (0, 1)node["$c_2$  " color = black, left] circle (2pt);
[red] (1, 1)node["$c_3$" color = black] circle (2pt);
(0, 0)node["$c_1\wedge c_2 = c_1\wedge c_3$" below] circle (2pt);
(0.5, 1.5)node["$\ \ c_2\vee c_3$" color = black, right];

*Antichain which is neither

Examples and non examples for key properties of antichains

To make this statements more precise, let us introduce three conditions on $C$ as an antichain of $[0, \alpha]$ for some $\alpha\in L$.

Inclusive antichainFor all subsets $S$ and $S'$ of  $C$, if $\wedge S \leq \wedge S'$ then $S'\subseteq S$.

Intersective antichainFor all subsets $S$ and $S'$ of  $C$, we have $(\wedge S) \vee (\wedge S') = \wedge(S\cap S')$.

Strong antichainFor all $S, S'$ subsets of $C$ of same cardinal, $\wedge S$ and $\wedge S'$ are incomparable i.e if $\wedge S \leq \wedge S'$ then $S = S'$.

Boolean antichain C is both Inclusive and intersective.

Note that intersectivity depends on the choice of a top element $\alpha$ whereas inclusivity and strength do not. The meet and join operations, will be computed in the interval $[\hat{0}, \alpha]$. Note also the following lemma.

An antichain is inclusive if and only if it is strong.

The inclusivity condition gives the strong antichain condition when the subsets $S$ and $S'$ have the same cardinal. To see the converse, assume that the antichain  $C$ is a strong antichain and let $S$ and $S'$ be two subsets of $C$ such that $\wedge S' \leq \wedge S$. Suppose at first that $|S| +n = |S'|$ with $n > 0$. Then there exists $s_1, \dots s_n \in S' \setminus S$. Set $S'' = S\sqcup \{s_1, \dots, s_n\}$.  Because the inequalities $\wedge S' \leq \wedge S $ and $\wedge S' \leq \wedge  \{a_1, \dots, a_n\}$ hold, we have

\begin{equation*}
\wedge S' \leq (\wedge S)\wedge (\wedge\{s_1, \dots, s_n\}) = \wedge S''.
\end{equation*}

Because $|S'|=|S''|$, the antichain is strong, $S' = S''$ hence $S\subseteq S'$. Next if $|S| = |S'| + n$, then take $s_1, \dots, s_n$ in $S\setminus S'$. Then we have

\[\wedge S'' = (\wedge S')\wedge (\wedge\{s_1, \dots, s_n\})\leq \wedge S\]

The antichain is strong so $S = S''$. Then $S' \subseteq S$ so $\wedge S' \geq \wedge S$ and thus $\wedge S' = \wedge S$. Using the first part of the proof we get $S = S'$. This contradicts the assumption on the integer $n$.

The strong antichain condition implies that for each $n$, the set
\[\{\wedge S | S\subseteq C \text{ with } |S| = n\}\]
is an antichain. This condition is strong enough for the main result (see <ref>). However it is not strong enough for  computing morphisms sets.

Denote $\langle C\rangle_{\vee, \wedge}^{\alpha}$ the lattice generated by the elements of $C$ and $\alpha$ in the sublattice $[0, \alpha]$ of $L$, equipped with the lattice operations of $L$. The following lemma motivates the terminology.

An antichain is boolean if and only if the map

\begin{align*}
(\mathcal{P}(C), \cap, \cup)&\xrightarrow{\phi} (\langle C\rangle_{\vee, \wedge}^{\alpha},  \wedge, \vee)\\
S\quad\quad &\mapsto \quad\quad \wedge S
\end{align*}

is a lattice anti-isomorphism.

Assume that the map $\phi$ is a lattice anti-isomorphism. Then $C_{\alpha}$ is intersective because $\phi$ sends $\cap$ to $\vee$. Now consider $S, S'\subseteq C$ such that $\wedge S \leq \wedge S'$. This is equivalent to the following equality

$$\wedge S = (\wedge S) \wedge (\wedge S').$$

The right hand side is equal to $\wedge (S\cup S')$. Because $\phi$ is a bijection, $S = S\cup S'$ meaning that $S'\subseteq S$. Thus $C$ is inclusive. Conversely assume $C$ is both inclusive and the intersective below $\alpha$. The fact that $\phi$ sends $\cup$ to $\wedge$ is true for any subset of a lattice. The intersection property makes $\phi$ send $\cap$ to $\vee$. To see that $\phi$ is injective, note that if $\wedge S = \wedge S'$ then the inclusion property forces $S = S'$. To see that the map is surjective, notice that the image of $\phi$, $\Ima(\phi) = \{\wedge S | S\subseteq C\}$ is a lattice, using the properties we just exhibited. Moreover, any sublattice of $[\hat{0}, \alpha]$ containing $C$ contains $\Ima(\phi)$. It is thus the sublattice of $[\hat{0}, \alpha]$ generated by  $C$ and $\alpha$, i.e, $\langle C\rangle_{\vee, \wedge}^{\alpha} = C\{\wedge S | S\subseteq C\}$ and $\phi$ is surjective.

§.§ Morphisms

In this subsection we fix an antichain $C$ below an element $\alpha$ of a lattice $L$ as well as an interval $I = [a, b]$ of $L$.
If $C$ is boolean more can be said about the total hom complex $\Hom^{\bullet}_{\IncAlg}(\PRes_C^{\alpha}, I)$. Recall the notation from equation (<ref>). Note that

\begin{equation}\label{eq:morphInt}
e_x\cdot I = \begin{cases}\field &\text{if } x\in I \\0&\text{otherwise}\end{cases}
\end{equation}

as a special case of equation (<ref>). Denote by $E$ the set $\{S\subseteq C | \wedge S\in I\}$. A projective module $P_{\wedge S}$ appearing in the projective resolution $\PRes_{C}^{\alpha}$ can contribute to the total hom if and only if $S \in E$. Note that $C$ is finite hence its set of subsets is finite. Assume that $E$ is not empty otherwise the total hom complex is zero. Because of the intersection property, if $S$ and $S'$ are in $E$ then

\[a\leq \wedge(S\cup S')\leq \wedge (S\cap S')\leq b\]

meaning that $S\cup S'$ and $S\cap S'$ are in $E$ as well. It follows that $E$ has a largest element $\cup_{S\in E} S = S_M$ as well as a least element $\cap_{S\in E} S = S_m$. Moreover, if $S \in [S_m, S_M]$ then
\[a\leq \wedge S_M\leq \wedge S \leq \wedge S_m\leq b\]
hence $S\in E$ and we have proved the following lemma
$E = [S_m, S_M]$.

To describe the total hom complex, $\Hom^{\bullet}_{\IncAlg}(\PRes_C^{\alpha}, I)$, we write $m = |S_m|$ and $M = |S_M|$. For each degree $m\leq i\leq M$  there are exactly $\binom{M-m}{i - m}$ subsets $S$ of $C$ with cardinal $i$ in $E$. Hence, by equations ((<ref> and <ref>)) the complex $\Hom^{\bullet}_{\IncAlg}(\PRes_C^{\alpha}, I[0])$ has shape:

\begin{equation}\label{eq:simplicialComp}
0\leftarrow \field^1 \leftarrow \dots \leftarrow \field^{\binom{M-n}{j}} \leftarrow \dots \leftarrow \field\leftarrow 0.
\end{equation}

This is precisely the shape of the simplicial module associated to the standard simplex. It remains to describe that the boundary map match the standard simplex as well. The map is post composition by the boundary map of $\PRes_C^{\alpha}$ in degree $i$. It sends a map defined by a vector $(f_S)_{|S|= i}$ to a vector $(g_{S'})_{|S'| = i+1}$ described by

\begin{equation*}
g_{S'} = \begin{cases}
0 &\text{if } S\not \subseteq S'\\
(-1)^{\epsilon(S, x)}\cdot f_S &\text{otherwise}\end{cases}
\end{equation*}

Where $|x|_{S} = |\{y\in S' | y\leq x\}|$. Indexing the vector elements by their complements and writing the basis vectors $e_J$ we get

\begin{equation*}
e_J \mapsto \sum_{x\in J} (-1)^{|x|_{J^{c}}} e_{J-\{x\}}
\end{equation*}

Like in the proof of  <cit.>, we have an isomorphism

\begin{equation}\label{eq:isoKoszulComplex}
\Hom^{\bullet}_{\IncAlg}(\PRes_C^{\alpha}, I)\cong (\field \xleftarrow{id} \field)^{\otimes (M-m)}[m]
\end{equation}

The left hand side is a Koszul complex over the base field $\field$. As a tensor product of acyclic complexes it is either acyclic or concentrated in one degree when the set $S_M\setminus S_m$ is empty using Künneth's formula <cit.> or <cit.>.

Let $C$ be an antichain of a lattice $L$ and let $I \subseteq L$ be an interval. Suppose the set $E = \{S\subseteq C | \wedge S\in I\}$ is an interval of the lattice $\PRes(C)$. Then there exists at most one integer $p$ such that $\Hom_{\HomotopyCat}(M_C, I[p])$ is non zero. When such an integer exists, the hom set is one dimensional.

Given the previous calculations and remarks, the result follows from (<ref>).

Moreover we know exactly that such a degree exists if and only if the set $E$ is a singleton i.e there exists a unique $S\subseteq C$ such that $\wedge S \in I$. In this case $p = |S|$. Combining this proposition with Lemma <ref> and the isomorphism $u$ of equation <ref> we have proved the following theorem.

Let $C$ be a boolean antichain of a lattice $L$. Let $I \subseteq L$ be an interval. There exists at most one integer $p$ such that $\Hom_{D^b}(M_C, I[p])$ is non zero. When such an integer exists, the hom space is one dimensional.

§.§ Truncations

Recall the stupid truncation $\sigma_{\geq i} R$ of a complexe $R = ((R_n)_{n\in \mathbb{Z}}, (\partial_n)_{n\in\mathbb{Z}})$ defined by

\begin{equation}
(\sigma_{\geq i} R)_n = \begin{cases} R_n &\text{ if } n \geq i\\ 0&\text{otherwise}
\end{cases}
\text{ with boundaries }
\partial'_n = \begin{cases} \partial_n &\text{ if} n > i\\ 0&\text{ otherwise. }
\end{cases}
\end{equation}

In this subsection we fix a strong antichain $C$. To make notation lighter we do not say if it is below some $\alpha$, though the two lemmas below hold in both $[\hat{0}, \alpha]$ and $L$.

With $C$ as above and notation from the previous subsections, for all $r = |C| \geq i\geq 0$ there is a bijection

\begin{align*}
\Phi : \End_{\IncAlg} ((\PRes_C)_{i-1}) &\to \Hom_{\HomotopyCat}(\sigma_{\geq i}\PRes_C, (\PRes_C)_{i-1}[i])\\
f\quad\quad & \mapsto \quad\quad f\circ \partial_i [i].
\end{align*}

If $i = 0$, $\PRes_{i-1} = 0$. So we assume $r \geq i > 0$. The antichain is strong so the indices of the indecomposable summands of $(\PRes_C)_{i-1} = \bigoplus_{|S| = i-1} P_{\wedge S}$ cannot be compared. Thus the endomorphisms of this module decompose as

\begin{equation}\label{eq:endP}f = \bigoplus_{|S| = i-1} \lambda_S\cdot id_{P_{\wedge S}}.\end{equation}

By projecting on the summands of the target, it suffices to show that there is a bijection

\begin{align*}
\Phi_S : \End (P) &\to \Hom_{\HomotopyCat}(\sigma_{\geq i}\PRes_C,P[i])  \\
f\quad\quad & \mapsto \quad\quad f\circ \pi_P\circ \partial_i [i]
\end{align*}

for all $P = P_{S}$ with $S\subseteq C$ of cardinal $i-1$. Write $S = \{s_1, \dots, s_{i-1}\}$.
The morphisms on the right hand side are of the form:

\begin{equation}\label{diag:trunc1}
\begin{tikzcd}
\bigoplus_{|S'| = i+1}P_{\wedge S'}\ar[r, "\partial_{i+1}"]\ar[d, "\dots\quad\quad"']& \bigoplus_{|S'| = i}P_{\wedge S'}\ar[r]		\ar[d, "\phi"]\ar[dl, red, "0"]&0\ar[dl, red, "0"]\\
0\ar[r]& P\ar[r]&0
\end{tikzcd}
\end{equation}

The red arrows represent potential homotopy maps. There cannot be a non zero homotopy so

\[ \Hom_{\HomotopyCat}(\sigma_{\geq i}R,P[i]) \cong  \Hom_{C}(\sigma_{\geq i}R,P[i]).\]

The relation $\partial^2 = 0 $ implies that $\Phi$ is well defined. An element of $\End(P)$ is either $0$ or an automorphism. Moreover, considering the specific form of the boundary maps of the complex $\PRes_C$ in equation (<ref>
), the projection of $\partial_i$ on the factor $P$ of its target is non zero. Hence if $f$ is non zero then $\Phi(f)$ is non zero and $\Phi$ is injective. It remains to see that $ \Hom_{\HomotopyCat}(\sigma_{\geq i}R,P[i])$ is one dimensional.

Notice that there is an isomorphism between $\Hom_{C}(\PRes_{C}, P[i])$ and  $\Hom_{C}(\sigma_{\geq i}\PRes_{C}, P[i])$ as any map $\phi$ as in equation (<ref>) also yields a map from the untruncated complex to $P[i]$ and vice versa. The map $\Phi_S$ is surjective if and only if every such map factors through $\partial_{i}$ i.e if and only if every element of  $\Hom_{C}(\PRes_{C}, P[i])$ is zero homotopic. Because the antichain is strong we have

\[E = \{S'\subset C| \wedge S'\leq \wedge S\} = \{S'\subseteq C | S \subseteq S' \}\]

where $E$ is the set of contributing subsets of $C$ in the total hom. The assumptions on $i$ and the cardinal of $C$ ensures that $E$ contains at least two elements. By Proposition <ref> we have

\[\Hom_{\HomotopyCat}(\PRes, P_{\wedge S}[i]) \cong 0.\]
This concludes the proof.

Let $C$ be as above. Then

\begin{equation*}
\dim \Hom_{\HomotopyCat}(\PRes_C, \sigma_{\geq 1} \PRes_C) \leq 1.
\end{equation*}

If $r = |C| = 0$, then $\sigma_{\geq 1}\PRes_C = 0$ and the space of morphisms in question is 0 dimensional. Assume that $r$ is strictly bigger than 1. The setting of the lemma can be illustrated by the following diagram

\begin{equation*}
\begin{tikzcd}
\dots\ar[r] & \bigoplus_{s, t\in C} P_{s\wedge t}\ar[d]\ar[r]& \bigoplus_{s\in C} P_S\ar[d]\ar[r]\ar[dl, red, "0"]  & P_{\hat{1}}\ar[d]\ar[dl, red, "0"] \\
\dots\ar[r] & \bigoplus_{s, t\in C} P_{s\wedge t}\ar[r] &\bigoplus_{s\in C} P_S\ar[r] & 0
\end{tikzcd}
\end{equation*}

The antichain $C$ is inclusive so the projective indecomposables in degree $i$ are associated to elements which are either bigger than the ones in degree $i+1$ or cannot be compared with them. Hence there are no non zero maps of degree $1$. We thus need to describe the maps between $\PRes_{C}$ and $\sigma_{\geq 1}\PRes_C$ in the category of complexes. For $k\in \llbracket 1, r \rrbracket$ both complexes have the same components
so morphism of complexes are determined by morphisms of modules

\[\phi_k \in \End\big( \bigoplus_{\substack{S\subseteq C,\\ |S| = k}}P_{\wedge S}\big)\]

satisfying the relation

\begin{equation}\label{eq:relMorph}
\phi_k\circ\partial_{k+1} = \partial_{k+1}\circ\phi_{k+1}.
\end{equation}

Since $C$ is a strong antichain, the elements $\wedge S$ with a fixed cardinal cannot be compared. Hence an  endomorphism $\phi_k$ of this module is of the form

\begin{equation}\label{eq:morphDecomp}
\phi_k = \bigoplus_{|S| = k} \lambda^k_S\cdot id_{P_{\wedge S}}
\end{equation}

with $\lambda_S^k\in\field$. The goal is to show that for $1\leq k\leq r$ we have $\phi_k = \lambda^r_{S}\cdot id_k$. In other words, for all $S$, $S'$ subsets of $C$, $\lambda^{|S|}_S = \lambda^{|S'|}_{S'}$. We proceed by downward induction on $k$ starting with $k = r$. In that case, we already have $\phi_r = \lambda_C^r id_{P_{\wedge C}}$ as there is only one projective indecomposable summand in degree $r$. Now take $1 \leq k <r$ and assume $\phi_{k+1} = \lambda_C^r\cdot id_{k+1}$. We now put together equations (<ref>), (<ref>) and (<ref>). On the one hand we have

\begin{equation*}\partial_{k+1}\circ \phi_{k+1} = \lambda^r_C \cdot \partial_{k+1}\end{equation*}

Evaluating at $e_{\wedge S}\in P_{\wedge S} \subseteq \bigoplus_{|S| = k+1} P_{\wedge S}$.

\begin{equation}
\partial_{k+1}\circ\phi_{k+1}(e_{\wedge S}) = \lambda^r_C\cdot \sum_{i\in S} (-1)^{|i|_S}\cdot e_{\wedge (S\setminus \{i\})}.
\end{equation}

On the other hand we have

\[\phi_k\circ\partial_{k+1}(e_{\wedge S}) = \sum_{s\in S} (-1)^{|s|_S}\lambda_{S\setminus\{s\}}^k\cdot e_{\wedge (S\setminus\{s\})}. \]

Because the $e_{\wedge (S\setminus \{s\})}$ are linearly independent we get

\[\lambda^k_{(S\setminus \{i\})} = \lambda^r_C\]

for all $S\subset C$ of cardinal $k+1$ and $s\in S$. Noticing that any subset of $C$ of cardinal $k < r$ can be expressed as $S\setminus\{i\}$ for some $S$ and some $i$, this finishes the proof.

§.§ Main result

Let $L$ be a finite lattice, let $d$ and $l$ be integers. Suppose there exists a family of antichains $(C_{\alpha})_{\alpha\in L}$ of L such that for all $\alpha\in L$, the following assumptions hold.

* The antichain $C_{\alpha}$ is below $\alpha$.

* The module $M^{\alpha}_C$ is non zero and there is an isomorphism

\begin{equation}\label{eq:CY}
\Serre^l M_{C_{\alpha}}^{\alpha}\cong M_{C_{\alpha}}^{\alpha}[d]
\end{equation}
in $D^b(\mathcal{A})$.

* The antichain $C_{\alpha}$ is strong.

Then $D^b(\IncAlg)$ is $\frac{d}{l}$-Calabi-Yau.

In practice, assumption (2) is the hardest to investigate. The proof relies on the following theorem.
<cit.> Let $X$ be a finite poset with a unique minimal or unique maximal element. If there are integers $d$ and $l$ such that $\Serre^l(P)\simeq P[d]$ for all projective indecomposable modules of $X$, then the category $D^b(\IncAlg_{\field}(X))$ is $\frac{d}{l}$-fractionally Calabi-Yau.

and this classical two-out-of three lemma from  <cit.>

Let $\mathcal{T}$ be a triangulated category with self equivalence $T$ and let

\begin{equation}
\begin{tikzcd}
\end{tikzcd}
\end{equation}

be a morphism between distinguished triangles. If two of the vertical morphisms are isomorphisms, then the third one is an isomorphism as well.

To simplify notation we now set $F = \Serre^{l}[-d]$. We will often refer to equation (<ref>) as the Calabi-Yau property for a certain complex of $\IncAlg$ modules.

By Theorem (<ref>), it suffices to prove that the Serre functor satisfies the Calabi-Yau property on the projective indecomposable modules. We proceed by induction on the elements $\alpha$ of the lattice $L$. For the initial step notice that, by the first assumption of the claim, the indecomposable projective module associated to the minimum of the poset $P_{\hat{0}}[0] \simeq M^{\hat{0}}_{C_{\hat{0}}}[0]$ satisfies equation (<ref>). Hence fix an element $\alpha > \hat{0}$ and assume as the induction hypothesis, that equation (<ref>) is true for all $\alpha' < \alpha$. We will refer to this as the outer induction hypothesis (OIH). Write $C = C_{\alpha}$ for the antichain indexed by $\alpha$ and let $R = \PRes_{C}^{\alpha}$ be its associated projective resolution. If $r = |C| = 0$, then $R$ is a projective module concentrated in degree $0$ and there is nothing to prove by the second assumption in <ref>.

Step1. So assume  $ r > 0$ and consider the distinguished triangle:

\begin{equation}\label{tikzcd:trig1}
\begin{tikzcd}
P_{\alpha}[0]\ar[r]&R\ar[r]&\sigma_{\geq 1}R\ar[r, "f"]& P_{\alpha}[1]
\end{tikzcd}\end{equation}

induced by the truncation short exact sequence <cit.>. A computation shows that the map $f$ is the boundary map $\partial_1$ in degree 1 and zero in all other degrees. With TR2 this triangle can be shifted to

\begin{equation}\label{tikzcd:trig2}
\begin{tikzcd}
R\ar[r]&\sigma_{\geq 1}R\ar[r]&P_{\alpha}[1]\ar[r]&R[1]
\end{tikzcd}\end{equation}

Because $F$ is triangulated, it sends this triangle to a triangle and we have the following diagram.

\begin{equation*}
\begin{tikzcd}
R\ar[r]\arrow{d}[anchor=center, rotate=-90,yshift=0.5ex]{\sim}&\sigma_{\geq 1}R\ar[r]\ar[d, dashed, "\exists ?" ]&P_{\alpha}[1]\ar[r]&R[1]\arrow{d}[anchor=center, rotate=-90,yshift=0.5ex]{\sim}\\
F(R)\ar[r]&F(\sigma_{\geq 1}R)\ar[r]&F(P_{\alpha}[1])\ar[r]&F(R)[1]
\end{tikzcd}
\end{equation*}

If we can show that there exists an isomorphism following the dashed arrow and that the left square can be chosen to commute, we can apply TR3 to complete the diagram followed by Lemma <ref> to finish the proof.

Step 2. To construct an isomorphism

\[\sigma_{\geq 1}R\xrightarrow{\sim}F(\sigma_{\geq 1 }R)\]

we show by downward induction on $i$ that

\[\sigma_{\geq i}R\xrightarrow{\sim}F(\sigma_{\geq i }R)\]

for $i \in \{1, \dots, r \}$. We refer to this as the inner induction hypothesis. Taking $i = r$, the initial step follows from the outer induction hypothesis as $P_{\wedge C}[r] \simeq \sigma_{\geq r} R$. Now, fix $1\leq i < r$ and assume the property is true for $i$. To show that it is also true for $i-1$, like (<ref>), consider the truncation triangle

\begin{equation*}
\begin{tikzcd}
\sigma_{\geq i}R\ar[r, "\partial_i{[i]}"]&R^{i-1}[i]\ar[r]&\sigma_{\geq i-1}R[1]\ar[r]&\sigma_{\geq i}R[1]
\end{tikzcd}
\end{equation*}

shifted using TR2. Again, we want to conclude by using TR3 and Lemma <ref> so it suffices to show that we can choose the vertical isomorphisms of the following square such that it commutes.

\begin{equation}\label{tikzcd:square}
\begin{tikzcd}
\sigma_{\geq i}R\ar[r, "\partial_i{[i]}"]\arrow{d}[anchor=center, rotate=-90,yshift=0.5ex]{\sim}& R^{i-1}[i]\arrow{d}[anchor=center, rotate=-90,yshift=0.5ex]{\sim}\\
F(\sigma_{\geq i}R)\ar[r, "\phi"]& F(R^{i-1} [i])
\end{tikzcd}
\end{equation}

By the inner induction hypothesis we can choose an isomorphism $f$ for the left vertical arrow and by the outer induction hypothesis we can choose an isomorphism $g$ for the right vertical arrow.

Step 3. The isomorphism $g$ can be rectified to make the square commutative. We compare the morphism $\partial_i$, with $g^{-1}\circ\phi\circ f$.
Recall the isomorphism $u$ between the morphism set in the homotopy and bounded derived category, provided that the source is a complexe of projectives. The isomorphism is functorial in the target and is defined by sending a class of morphisms up to homotopy $c$ to the morphism

\begin{equation*}
\begin{tikzcd}[row sep=small]
P\ar[ur, "c"]&&C\ar[ul,"\sim" labl, "id"']
\end{tikzcd}
\end{equation*}

in the derived category. This is the "fraction" $\frac{c}{1}$. In the homotopy category we want to compare the morphism $\partial_i$, which is already a morphism of complexes, with $u^{-1}(g^{-1}\circ\phi\circ f)$. By Lemma <ref> there exists $T$ an endomorphism of $\bigoplus_{|S| = i-1} P_{\wedge J}$ such that

\begin{equation}\label{eq:Tpartial}( T \circ \partial_i) [i]= u^{-1}( g^{-1}\circ\phi\circ f).\end{equation}

Applying $u$ again we get

\begin{equation*}
\frac{T[i]}{1} \circ \frac{\partial_i[i]}{1}= g^{-1}\circ\phi\circ f.\end{equation*}

From now on all the morphisms will be in the derived category but we omit the denominator of maps coming from the homotopy category as it is always equal to one.

We prove that the map $T$ is an isomorphism. It has form $\bigoplus_{S} \lambda_S\cdot id_{P_{\wedge S}}$ because the antichain $C$ is strong. It is enough to show that the projections of $T$ onto the indecomposable summands of its target are non zero. Consider the following diagram.

\begin{equation}\label{tikzcd:3.5squares}
\begin{tikzcd}[column sep = large]
&R^{i-1}[i]\ar[d, "T{[i]}"]\ar[r, "\pi_S{[i]}"]&P_{\wedge S}[i]\ar[d, "\textcolor{red}{\lambda_S}\cdot id{[i]}"]&\text{Shifted module maps}\ar[dll, red, shorten >=10ex, bend right = 10]\\
\sigma_{\geq i}R\ar[ur, "\partial_i{[i]}"]\ar[d, "f", "\text{IIH}"' blue]& R^{i-1}[i]\ar[d, "g", "\text{OIH}"' blue]\ar[r, "\pi_S{[i]}"]&P_{\wedge S}[i]\ar[d, "h\leadsto \tilde h", "\text{OIH}"' blue]&\dim\Hom (R^{i-1}[i], FP_{\wedge S}[i]) = 1\ar[dll, red, shorten >=8ex, bend right = 10]\\
F\sigma_{\geq i}R\ar[r, "F(\partial_i{[i]})"]\ar[uur, phantom, "\text{by } \ref{lem:HoCatHom_Bool1}" , blue]& FR^{i-1} [i]\ar[r, "F(\pi_S{[i]})"]&FP_{\wedge S}[i]&
\end{tikzcd}
\end{equation}

The pentagon on the left is commutative since

\[g\circ T[i]\circ \partial_i[i] = g\circ g^{-1}\circ F(\partial_i[i])\circ f =  F(\partial_i[i])\circ f.\]

The top right square illustrates equation (<ref>) so it commutes as well. For the bottom right square, the outer induction hypothesis gives  the isomorphism $h$. Next, $F$ is fully faithful so it induces an injective linear map on hom spaces. Hence $F(\pi_S[i])$ is non zero. Finally, $g$ is an isomorphism so the composition $F(\pi_{\wedge S}[i])\circ g$ is non zero too. Because the antichain is strong,

\[\dim \Hom_{\IncAlg}(R^{i-1}, P_{\wedge S} [i]) = 1\]

and because $P_{\wedge S}[i]$ and $ FP_{\wedge S} [i]$ are isomorphic, we also have

\[\dim \Hom_{\IncAlg}(R^{i-1}, FP_{\wedge S} [i]) = 1.\]

Hence it is possible to replace the map $h$ by $\tilde h = \lambda\cdot h$ for some $\lambda$ making it commute. Chasing around the diagram we thus compute

\begin{equation*}
\lambda_S\cdot \tilde{h} \circ \pi_S[i]\circ\partial[i] = F(\pi_S\circ\partial_i[i])\circ f.
\end{equation*}

By construction the projections of $\partial_i$ are all non zero and $\tilde{h}$ and $f$ are isomorphisms. Again, $F$ is an equivalence of categories so  the right hand side is non zero. Hence $\lambda_S\not = 0$ for all $S\subseteq C$ of cardinal $i$ and $T$ is an automorphism. Rearranging equation (<ref>) we get

\begin{equation*}
\partial_i = T^{-1}\circ g^{-1}\circ\phi\circ f.
\end{equation*}

So replacing $g$ by $g\circ T$ the square (<ref>) becomes commutative. Applying Lemma (<ref>) and the inner induction hypothesis we have
$$F(\sigma_{\geq 1}R)\cong \sigma_{\geq1}R.$$

Step 4. To complete the outer induction  choose such an isomorphism $g$, and an isomorphism $f: R \to F(R)$. Just like in step 3 above, we want to compare the map $p: R \to \sigma_{\geq 1}R$ with $g^{-1}\circ F(p)\circ f$ and, if needed, rectify the isomorphisms $g$ and $f$ to make the square commute. Lemma <ref> implies that $g^{-1}\circ F(p)\circ f = \lambda\cdot p $. Because $g$ and $f$ are isomorphisms and F is induces a linear bijection between hom spaces, $\lambda$ is non zero and we can replace $g$ by $\lambda\cdot g$. This concludes the proof.

§ A FAMILY OF FRACTIONALLY CALABI-YAU POSETS

The goal of this chapter is to prove Theorem <ref>. We use the combinatorial family of antichain modules introduced in <cit.> and show that it satisfies the conditions of Theorem <ref>. For the convenience of the reader and because it will be used heavily in the rest of the article, we recall Yildirim's combinatorial framework and proof, modifying the statements when needed to obtain results in the derived category.

§.§ Grids and their order ideals

Let $m$ and $n$ be positive integers and $G_{m, n}$ be the product of two total orders of size $m$ and $n$ respectively. Recall that an order ideal $I$ of a poset $P$ is a subset $I\subseteq P$ which is downward closed, i.e
\[\text{if }x\in I \text{ and } y\leq x \text{ then }y\in I.\]
Order ideals of a poset can be ordered by inclusion and form a distributive lattice when equipped with the union and the intersection of subsets. We denote by $J_{m,n}$ the lattice of order ideals of $G_{m,n}$.

BijectionsWe now present several descriptions of this poset. The first three play a crucial role in the proof of Theorem <ref> while the visualisations will be more helpful in the next section when describing hom spaces between certain objects. First, take an order ideal $I$ of $G_{m,n}$. Because $I$ is closed by below, counting the number of elements that belong to $I$ in each column, with increasing first value, gives a monotone sequence which completely determines the ideal. We thus obtain a bijection

\begin{equation}
J_{m,n} \cong \{(a_1, \dots, a_m)| a_i\in \llbracket 0, n\rrbracket \text{ and } a_1\leq \dots, \leq a_n\}
\end{equation}
with non decreasing sequences.
If the second set is equipped with term wise comparison this is an isomorphism of posets. We call these non decreasing sequences partitions. They can also be written as

\[(\lambda_1^{\mu_1}, \dots, \lambda_r^{\mu_r} )\]

with $\sum_i \mu_i = m$, where $\mu_i$ encodes the multiplicity of the value $\lambda_i$ and $\lambda_i \not = \lambda_j$ if $i\not = j$.
These partitions can classically be counted as follows: choosing a partition amounts to putting $m$ balls corresponding to coefficients $(a_1, \dots, a_m)$ into $n+1$ boxes, the possible values of the coefficients. Which also amounts to placing $n+1$ sticks among $m+n$ slots. This means there are exactly
$${m+n\choose m}$$
The second bijection that we introduce sends these non decreasing sequences of length $m$ to increasing sequences of length $m$. Let
\[\mathcal{Z} = \{-m, \dots, -1, 0, 1, \dots, n\}\]
be a set of representatives of $\mathbb{Z}/(m+n)\mathbb{Z}$. A configuration $C$ of $\mathcal{Z}$ is a subset of size $m$ of $\mathcal{Z}$. We write it $C = \{c_1< \dots < c_m\}$ is an increasing sequence naively using the order relation on $\mathbb{Z}$.  We write $C_{m,n}$ the set of configurations of length $m$ on $\mathcal{Z}$. Choosing a configuration amounts to picking $m$ distinct elements from a set of cardinal $m+n+1$ so the cardinality of $C_{m,n}$ is
\[{m+ n + 1\choose m}.\]

Given a partition $\alpha$ we can construct a configuration containing $\alpha$'s coefficients in its non negative side and encoding the multiplicities of $\alpha$ in its negative side.

\[x_i = \sum_{k = 1}^i \mu_i\]

to record the index of the last occurence of the $i^{th}$ coefficient. It will be called the ending index. We set $x_0 = 0$ as a convention. The index $x_{i-1} + 1$ is the first occurence of $\lambda_i$ and will be called the starting index. Think of the negative side as the indices of the elements of the sequence $\alpha$ but with a minus sign. Out of the many available options to encode the multiplicities, here are two that turn out to fit the problem perfectly:

* keep all negative elements except the opposite of the starting index of each coefficient. The resulting configuration is called the left configuration associated to $\alpha$, and we denote by it $L_{\alpha}$. The map sending $\alpha$ to $L_{\alpha}$ is denoted by $\phi_l$;

* keep all negative elements except the opposite of the ending index of each coefficient. The resulting configuration is called the right configuration associated to $\alpha$, and we denote it by $R_{\alpha}$. The map sending $\alpha$ to $R_{\alpha}$ is denoted $\phi_r$.

Take $n = 7$, $m = 5$ and consider the partition $a = (0, 2, 3, 7, 7)$. So we have $r = 4$ and

$$x_1 = 1, x_2 = 2, x_3 = 3 \text{ and } x_4 = 5.$$

The associated left and right configuration are respectively

$$\{-5< 0<2<3<7\} \text{ and } \{-4<0<2<3<7\}.$$

The maps $\phi_l$ and $\phi_r$ are injective.

Partitions are entirely determined by their coefficients and multiplicities. These can be recovered from the positive elements of a configurations and its negative gaps respectively.
They are not surjective as $\phi_l(\alpha)$ cannot contain $-1$ and $\phi_r(\alpha)$ cannot contain $-m$.

[step = 0.5cm, color = gray](0,0) grid (2,3);
[thick](.5, 0)–(.5, .5)–(1, .5)–(1, 1)–(1.5, 1)–(1.5, 3)–(2,3);
(.5,.5) circle (1pt);
(1,1) circle (1pt);
(1.5,3) circle (1pt);
(2,3) circle (1pt);

One can also think of partitions as paths in an $m\times n$ grid as depicted in the figure for the partition $(0, 2, 3, 7, 7)$. For a non decreasing sequence $(a_1, \dots, a_m)$ the path is obtained by putting a dot at height $a_i$ in the $i^{th}$ column from the left and then take the minimal path going through this dots. If $a_i = 0$ put no dot. To visualise configurations consider a table with $m+n+1$ columns and put a dot, or a bead in the columns corresponding to the elements of the configuration at hand. We call this the abacus associated to the configuration.

The abacus associated with the right configuration of the partition $a$ from example <ref> is as follows

-5   -4   -3   -2   -1   0   1   2   3   4   5   6   7

$\bullet$            $\bullet$      $\bullet$   $\bullet$            $\bullet$.

We used the map $\phi_r$.

Each bead of the negative side adds one to the multiplicity of one of the coefficients. If we are using the map $\phi_r$ to obtain the configuration, the bead sitting in the $-k^{th}$ column, is associated to the $k^{th}$ beads to its right. Indeed, gaps in the negative side indicate changes of coefficients. If the columns $-1$ to $-k+1$ are empty then exactly $(k-1)^{th}$ coefficients end before the $k^{th}$ element of the sequence and the claim holds. Any extra bead between $-1$ and $-k+1$ removes one such gap. If there are $s$ such beads in the negative side, the coefficient  we are looking for corresponds to the $(k-s)^{th}$ bead in the positive side. Note that $s$ is at most $k-1$, so the procedure always yields a bead in the positive side.

Illustrating the partial inverse of $\phi_r$

Figure (<ref>) illustrates this idea on the abacus of example <ref>. Its only negative bead can be connected to the coefficient $\lambda_4 = 7$. It is the only coefficient with multiplicity 2 in the partition $a$.

Similarly, if the map $\phi_l$ was used, negative beads can be associated to their positive counterparts by skipping $k-1$ beads to the left.

§.§ Yıldırım's theorem

The following key result is a categorified version of <cit.>.
Let $\alpha$ be a right enhanced partition. Then

$$\Serre^{m+n+1}(\PRes_{\alpha}) \cong \PRes_{\alpha}[mn].$$

First we describe the action of the Serre functor on the object $\PRes_{\alpha}$. Recall that it sends the projective indecomposable $P_{\alpha}$ to the injective indecomposable $I_{\alpha}$.

We call $\mathcal{I}_{\alpha}$ the image of the projective resolution $\PRes_{\alpha}$ under the Serre functor. Because $\PRes_{\alpha}$ is a complexe of projectives, we get $\mathcal{I}_{\alpha}$ by tensoring the components of  $P_{\alpha}$ by $\IncAlg^{*}$ which gives us

\begin{equation}\label{eq:InjRes}
\mathcal{I}_{\alpha} : 0\to I_{q_{S_{\alpha}}(\alpha)}\xrightarrow{\partial_r}\bigoplus_{\substack{J\subseteq S_{\alpha},\\|J| = r-1}} I_{q_J(\alpha)}
\to\dots\to\bigoplus_{\substack{J\subseteq S_{\alpha},\\ |J| = r-k}} I_{q_J(\alpha)} \overset{\partial_{r-k}}{\to}\dots\to I_{\alpha}\to 0.
\end{equation}

Just like for $\PRes_{\alpha}$, we can show that $\mathcal{I}_{\alpha}$ is an injective resolution. Its homology is concentrated in degree $|S_{\alpha}|$. To describe this module, define the map

\begin{equation}\label{eq:DefG}
g : (0^{\alpha_0}| \lambda_1^{\alpha_1}, \dots, \lambda_r^{\alpha_r}) \mapsto (\lambda_1^{\alpha_0+1}, 					\lambda_2^{\alpha_1}, \dots, \lambda_r^{\alpha_{r-1}}|n^{\alpha_r - 1})
\end{equation}

which is a counterpart to the map $f$, and set the following enhancement on the partition $q_{S_{\alpha}}(\alpha)$

\begin{equation}\label{eq:deltaEnhancement}
\begin{cases}
(0^{\alpha_1}|(\lambda_2 - 1)^{\alpha_2}, \dots, (\lambda_r-1)^{\alpha_r}, n^{\alpha_{r+1}}) & \text{if } \lambda_1 = 0\\
(|(\lambda_1 - 1)^{\alpha_1}, \dots, (\lambda_r-1)^{\alpha_r}, n^{\alpha_{r+1}}) & \text{otherwise.}
\end{cases}
\end{equation}

We denote $\delta$ the map sending $\alpha$ to this enhancement of $q_{S_{\alpha}}(\alpha)$. Then the support of the homology of $\mathcal{I}_{\alpha}$ in degree $|S_{\alpha}|$ is the interval

\begin{equation}\label{eq:IntIm}
[\delta(\alpha), g\circ \delta(\alpha)].
\end{equation}

The map delta amounts to representing the corresponding configuration in an abaci using $\phi_r$, shift all its beads to the left and interpret the configuration using $\phi_l$. The maps $f$ and $g$ are further related by the following lemma.

The function $f$ and $g$ are inverse of one another.

To see that it suffices to compute both composition

\begin{align*}
f\circ g : (\lambda_1^{\alpha_1}, \dots, \lambda_r^{\alpha_r} | n^{\alpha_{r+1}}) &\mapsto (0^{\alpha_1-1}|, 		\lambda_1^{\alpha_2}, \dots, \lambda_{r-1}^{\alpha_r}, \lambda_r^{\alpha_{r+1}+1})\\
&\mapsto (\lambda^{\alpha_1}, \lambda_2^{\alpha_2}, \dots, \lambda_r^{\alpha_r}|n^{\alpha_{r+1}+1 - 1})
\end{align*}

and, almost dually we have :

\begin{align*}
g\circ f : (0^{\alpha_0}| \lambda_1, \dots, \lambda_r^{\alpha_r}) &\mapsto (\lambda_1^{\alpha_0 +1}, 			\lambda_2^{\alpha_1}, \dots, \lambda_r^{\alpha_{r-1}} | n^{\alpha_r - 1})\\
&\mapsto (0^{\alpha_0 +1 -1}| \lambda_1, \dots, \lambda_{r-1}^{\alpha_{r-1}}, \lambda_r^{\alpha_r -1 + 1}).
\end{align*}

The key for Yildirim's theorem is the following proposition which mirrors the combinatorics of Proposition 4.2 in <cit.> but with slightly different combinatorial objects.

On the abacus, the map $f$ is a shift to the left of the negative beads or is simply $f(\alpha) = \phi^{-1}_l\circ \phi_r(\alpha)$. Conversely $g$ is the reinterpretation of the configuration using $\phi_r$ instead of $\phi_l$ i.e. $g(\alpha) = \phi^{-1}_r\circ \phi_l(\alpha)$.

Both $\alpha$ and $f(\alpha)$ have the same plain coefficients $\lambda_1, \dots, \lambda_r$ of $\alpha$ hence their respective abaci have the same positive beads regardless of the map used. To see that they coincide in the negative side notice that

\[x^{f(\alpha)}_{i-1}+1 = x_{i}^{\alpha}\]

for $i\in S_{\alpha}$. The result on $g$ follows from lemma <ref>

More importantly we can now show that Yildirim's family of intervals is stable under the Serre functor.
Write $\tilde{f} = g\circ \delta$.

Let $\alpha$ be a left enhanced partition. Then $S(\mathcal{P}_{\alpha}) \simeq \mathcal{P}_{\tilde{f}(\alpha)}[|S_{\alpha}|]$.

Recall that $\Serre(\mathcal{P}_{\alpha})$ is isomorphic to the interval $[ q_{S_{\alpha}}(\alpha), g(q_{S_{\alpha}}(\alpha))]$ shifted by $|S_{\alpha}|$. But because $f$ and $g$ are inverse of one another this is the interval

$$[f(\tilde{f}(\alpha)), \tilde{f}(\alpha)] $$

which itself is isomorphic to $\mathcal{P}_{\tilde{f}(\alpha)}$ by Equation <ref>. This gives us the result.

Hence describing the action of the Serre functor on the Yildirim modules amounts to describing the action of $\tilde{f}$ on abaci. This turns out to be quite simple.

The map $\tilde{f}$ acts on right abaci associated to right enhanced partitions as a shift to the left

This is clear form the description in terms of abaci of $g$ in Proposition <ref> and of $\delta$ below equation <ref>.

The previous Lemma implies that $\tilde{f}^{n+m+1}$ is the identity on configurations hence on partitions meaning that $S^{m+n+1}(\mathcal{P}_{\alpha})$ is isomorphic to $\mathcal{P}_{\alpha}$ with a certain shift. To compute that shift, recall that applying the Serre functor to $\mathcal{P}_{\alpha}$ shifts the resolution by $|S_{\alpha}|$ to the left so the total shift is:

$$\sum_{i=1}^{m+n+1} |S_{\tilde{f}^{i}(\alpha)}|.$$

But $|S_{\alpha'}|$ is the number of non zero beads in the positive side of the right abacus associated to $\alpha'$. When applying $\tilde{f}$ to $\alpha$ a total of $n+m+1$ times, each bead will be in the positive side exactly $n$ times. Because there are $m$ beads, this means the shift is $nm$.

Let us check that the family

$$\{\PRes_{\alpha} | \alpha \text{ enhanced } (m,n) \text{-partition} \}$$

satisfies the three conditions of Theorem <ref>. First, no matter the enhancement of a partition $\alpha$, the projective cover of the intervalle $[f(\alpha), \alpha]$ is the indecomposable projective $P_{\alpha}$. Theorem <ref> gives the second condition. The third condition follows from Proposition <ref> which concludes the proof.

§ TILTING TO HIGHER AUSLANDER ALGEBRAS OF TYPE $A$

The goal of this chapter is to prove Theorem (<ref>). We start by describing morphisms and extensions between the antichain modules $\PRes_{\alpha}$ introduced in the previous chapter. After discussing relations between the morphisms we construct a tilting complex between $J_{m,n}$ and $A_{n+1}^{m+1}$. In doing so we make explicite an isomorphism between the higher auslander algebra of type $A_s^d$ and the quadratic dual of $A_{d+2}^{s-2}$.

§.§ Describing Homspaces

Graph of the  $\mathcal{Y}_{2,2}$

We define the category $\mathcal{Y}_{m,n}$ to be the full subcategory of $D^b(J_{m,n})$ whose objects are the objects $\PRes_{\alpha}$ and all their shifts. The goal of this subsection is to describe morphisms between objects and to identify irreducible morphisms in the category $\mathcal{Y}_{m,n}$. These include morphisms of modules which are concentrated in degree zero as well as morphisms between shifted objects which are in fact extensions. Given a morphism in $\mathcal{Y}_{m,n}$, we will first factor it through an extension of the same degree but which we can easily describe using our antichains. This factorisation yields a degree zero map. We then decompose further these two components. Starting the process of our factorisation with the extension yields a form of asymmetry between morphism and extensions which is not integral to $\mathcal{Y}_{m,n}$. It is an artefact of this proof which will be smoothed out in the next subsection.

As a visual help and source of examples, see figure (<ref>) for the graph of $\mathcal{Y}_{2, 2}$. The labels on the arrows indicate the degree in which the morphisms are concentrated. Note that relations do not appear on the figure: many compositions of arrows are in fact zero.
Let $\alpha = (\lambda_1^{\mu_1}, \dots, \lambda_r^{\mu_r}|n^{\mu_{r+1}})$ and $\beta = (l_1^{m_1}, \dots, l_s^{m_s}|n^{m_{s+1}})$ be partitions. In this section we will consistently use the following conventions for the ending indices of the coefficients of $\alpha$ and $\beta$

\[x_i = \sum_{k\leq i} \mu_k \text{ and } y_j = \sum_{k\leq j} m_k\]

to distinguish data from the two partitions as well as the indices that we might be looking at any given moment. Recall from Theorem <ref> that there exists at most one integer $p$ such that

\begin{equation*}
\dim \Hom (\PRes_{\alpha}, \PRes_{\beta}[p]) \not= 0.
\end{equation*}

If it exists then $\dim \Hom (\PRes_{\alpha}, \PRes_{\beta}[p]) = 1$. It exists if and only if there exists a unique subset $J$ of $S_{\alpha}$ such that $q_J(\alpha) \in [f(\beta), \beta]$. In that case, $p = |J|$. When $p = 0$ we can give different characterisations of this property.

The following are equivalent

* There exists a nonzero morphism $\phi : \PRes_{\alpha}\to\PRes_{\beta}$.

* The inequalities $f(\alpha)\leq f(\beta)\leq \alpha\leq\beta$ hold.

* The partition $\alpha$ is in $[f(\beta), \beta]$ and for all non empty $J\subseteq S_{\alpha}$ $q_J(\alpha)\not\in [f(\beta),\beta]$.

* The partition $\alpha$ is in $[f(\beta), \beta]$ and $\{\lambda_i |i\in S_{\alpha}\} \subseteq \{l_j|j\in S_{\beta}\}$.

* For all $j\in S_{\beta}$ there exists $i\in S_{\alpha} \cup \{r+1\}$ such that $\lambda_i = l_j$ and $x_{i-1} < y_j \leq x_i < y_{j+1}$ and $\{\lambda_i |i\in S_{\alpha}\} \subseteq \{l_j|j\in S_{\beta}\}$.

(<ref>) $\Leftrightarrow$ (<ref>) is the characterisation of morphisms between intervals recalled in equation (<ref>).

(<ref>) $\Leftrightarrow$ (<ref>) follows from Theorem <ref>.

(<ref>) $\Leftrightarrow$ (<ref>) Reformulate (<ref>) using the ideas of the proof of Proposition <ref> which we reproduce here. Recall that the interval $[f(\beta), \beta]$ is the set of partitions

\[\{\gamma | \gamma\leq \beta \text{ and for all }i\in S_{\beta},  \gamma\not\leq q_j(\beta)\}.\]
Because the partition $q_j(\beta)$ differs from $\beta$ between indices $y_{j-1}+1$ and $y_j$, this is equivalent to there existing $k\in [y_{j-1}+1, y_j]$ such that $l_j -1 <a_k = b_k = l_j$. Equivalently, there exists $i\in S_{\alpha}$ such that $a_k = \lambda_i = l_j = b_k$. The equality $a_k = b_k$ is equivalent to saying there is an overlap of the occurrences of the value $l_j = \lambda_i$ in $\beta$ and $\alpha$ which translates into the interlacing  $x_{i-1} < y_j \leq x_i < y_{j+1}$.

To complete the proof we argue that (<ref>) $\Rightarrow$ (<ref>) $\Rightarrow$ (<ref>).

(<ref>) $\Rightarrow$ (<ref>)Assume that for all $i\in S_{\alpha}$, $q_i(\alpha)\not\in [f(\beta), \beta]$. Thus there exists $j\in S_{\beta}$ such that $(q_i(\alpha))_{y_j} < l_j$. Because $\alpha$ and $q_i(\alpha)$ differ only for indices between $x_{i-1}+1$ and $x_i$ we conclude that $\lambda_i = l_j$. The second inclusion follows directly from the characterisation of $[f(\beta),\beta]$ as detailed for the previous equivalence.

(<ref>) $\Rightarrow$ (<ref>)
Remember that in $f(\beta)$, the value $l_j$ runs from $y_j$ to $y_{j+1}-1$. Hence, the interlacing condition implies that $(f(\beta))_{x_i} = \lambda_i$ and $f(\beta)\in[f(\alpha), \alpha]$ because for all $i\in S_{\alpha}$, we have $q_i(\alpha)\not\leq f(\beta)$.

To describe some degree zero morphisms, we introduce new transformations on $J_{m,n}$.

Let $\alpha = (\lambda_1^{\mu_1}, \dots \lambda_r^{\mu_r}|n^{\mu_{r+1}})$ be an enhanced partition. Let $i$ be in $S_{\alpha}$. If $\mu_i > 1$ define the partition $p_i(\alpha) = (\lambda_1^{m_1}, \dots, \lambda_r^{m_s}| n^{m_{s+1}})$ with multiplicities

\begin{equation*}
m_j = \begin{cases}\mu_j - 1 & \text{if } j=i,\\
\mu_j + 1 &\text{if } j = i+1,\\
\mu_j &\text{otherwise.}
\end{cases}
\end{equation*}

There is a special case when $i = 1$, $\mu_1 = 1$ and $\lambda_1 = 0$. We still define the partition $p_1$ in that case by

\begin{equation*}
p_1(\alpha) = p_1(0^1, \lambda_2^{\mu_2}, \dots, \lambda_r^{\mu_r}) = (\lambda_2^{\mu_2+1}, \lambda_3^{\mu_3}, \dots, \lambda_r^{\mu_r}|n^{\mu_{r+1}}).
\end{equation*}

The resulting partition has one coefficient less than $\alpha$ which reduces by one the index the coefficients which remain in the partition.
Finally, when $\lambda_r < n$ and $\mu_{r+1} > 0$, we denote $p_{\star}(\alpha)$, the partition

\begin{equation*}
(\lambda_1^{\mu_1}, \dots \lambda_r^{\mu_r},n^{\mu_{r+1}}|)
\end{equation*}

which has one more mutable coefficient than $\alpha$. In these three cases we say that $p_i$ is well defined.

Please note the asymmetry: where $q_j$ changes the values of the partition, $p_i$ acts on its multiplicities.
[9]r0.5[scale = 1.5]
[step = 0.5cm, color = gray](0,0) grid (3.5,2);
[thick, dotted](0, 0)–(0, 0.5)–(0.5, 0.5);
[thick, dotted](2, .5)–(2, 1.5)–(2.5, 1.5);
[thick, dotted, red](3.5, 2)–(3, 2)–(3, 1.5);
[thick](0.5, 0)–(0.5, 0.5)–(2.5, 0.5)–(2.5, 1)–(2.5, 1.5)–(3.5, 1.5) – (3.5, 2);
[red, ->](0.44, 0.25)–(0.06, 0.25);
[red, ->] (2.44, 1.25)–(2.06, 1.25);
[red, ->] (3.44, 1.75)–(3.06, 1.75);
(0.5, 0.5) circle (1pt);
(1, .5) circle (1pt);
(1.5, .5) circle (1pt);
(2, .5) circle (1pt);
(2.5, 1.5) circle (1pt);
(3, 1.5) circle (1pt);
(3.5, 2) circle (1pt);
[red] (3.5, 2) circle (1pt);

$p_1$, $p_2$ and $p_3$ on $\alpha = (0, 2^4, 4^2|5)$

Using the previous proposition and this new definition we list some notable degree zero morphisms.
Consider a right enhanced partition $\alpha$. There exists a non zero morphism $\PRes_{\alpha}\to\PRes_{\beta}$ whenever $\beta= p_i(\alpha)$ and $i\in \{1, \dots, r\}\cup \{\star\}$ such that $p_i$ is well defined

We first consider the case where $i\not = \star$ and $\mu_i > 1$. Notice that if $p_i$ is well defined, then the ending indices of $\alpha$ and $p_i(\alpha)$ are the same except from $x_i^{\alpha} = x_i^{p_i(\alpha)} + 1$. Hence, knowing that $\mu_i > 1$, we have $\alpha_{x_j^{\alpha} -1} = \lambda_i = p_i(\alpha)_{x_j^{p_i(\alpha)}}$ for all $j\in S_{\alpha}$. It follows that $\alpha \in [f(p_i(\alpha)), p_i(\alpha)]$ using the characterisation from equation (<ref>). We also know from Proposition <ref> that $f(p_i(\alpha))$ and $p_i(\alpha)$ have the same values at indices $x_i^{p_i(\alpha)}$ from which we can deduce that and $f(p_i(\alpha)) \in [f(\alpha), \alpha]$ using the characterisation equation (<ref>) again. We conclude with item (<ref>) of Proposition <ref>. The other two cases can be treated very similarly and are left to the reader. See figure (<ref>) for a visualisation of the intervals.

For a pair $(\alpha, \beta)$ as in Lemma <ref> consider the composition of canonical maps

\[P_{\alpha }\xhookrightarrow{\iota_{\alpha}^{\beta}}P_{\beta} \twoheadrightarrow [f(\beta), \beta].\]

By Item <ref> of Lemma <ref>, for $j\in S_{\alpha}$ we have $q_j(\alpha)\not \in [f(\beta), \beta]$. Hence the map above sends generators of $N_C^{\alpha}$ to zero and this factors uniquely through $\PRes_{\alpha} \cong P_{\alpha} / N_{\alpha} \cong [f(\alpha), \alpha]$ providing us with one instance of non zero morphism which we denote $\prescript{0}{}{u_i}$. Because the hom space is one dimensional, any other non zero morphism is proportional to $\prescript{0}{}{u_i^{\alpha}}$.

0.45[scale = 2]
[step = .5cm, thin, lightgray] (0, 0) grid (3, 2);
[thick] (0, .5) – (1, .5) – (1, 1) – (2, 1) – (2, 2) – (3, 2);
[thick, dashed] (0, 0) – (.5, 0) – (.5, .5) – (1.5, .5) – (1.5, 1) – (3, 1) – (3, 2);
[thick, gray](0, 0.05) – (.55, 0.05) – (.55, .45) – (1.05, .45)– (1.05, 0.95) – (2.05, .95) – (2.05, 1.95) – (3, 1.95);
[thick, gray, dashed](0, 0.05) – (.55, .05) – (.55, .45) – (1.55, .45) – (1.55, .95) – (2.95, .95) – (2.95, 2);
[thick](3.02, 1.7)–(3.02, 2.1);
[thick, gray] (3.05, 1.7)–(3.05, 2.1);
[->] (0.1, 0.25)–(0.4, 0.25);

0.45[scale = 2]
[step = .5cm, thin, lightgray] (0, 0) grid (3, 2);
[thick] (0, .5) – (1, .5) – (1, 1) – (2, 1) – (2, 2) – (3, 2);
[thick, dashed] (0, 0) – (.5, 0) – (.5, .5) – (1.5, .5) – (1.5, 1) – (3, 1) – (3, 2);
[thick, gray](0, 0.45) – (1.55, .45)– (1.55, 0.95) – (2.05, .95) – (2.05, 1.95) – (3, 1.95);
[thick, gray, dashed](0, 0.05) – (1.05, .05) – (1.05, .45) – (1.55, .45) – (1.55, .95) – (2.95, .95) – (2.95, 2);
[thick](3.02, 1.7)–(3.02, 2.1);
[thick, gray] (3.05, 1.7)–(3.05, 2.1);
[->] (1.1, 0.75)–(1.4, 0.75);

0.45[scale = 2]
[step = .5cm, thin, lightgray] (0, 0) grid (3, 2);
[thick] (0, .5) – (1, .5) – (1, 1) – (2, 1) – (2, 2) – (3, 2);
[thick, dashed] (0, 0) – (.5, 0) – (.5, .5) – (1.5, .5) – (1.5, 1) – (2.5, 1) – (2.5, 2) – (3, 2);
[thick, gray](0, 0.45) – (1.05, .45)– (1.05, 0.95) – (2.05, .95) – (2.05, 1.95) – (3, 1.95);
[thick, gray, dashed](0, 0.05) – (.55, .05) – (.55, .45) – (1.55, .45) – (1.55, .95) – (2.95, .95) – (2.95, 2);
[thick] (2.7, 1.7)–(2.7, 2.1);
[thick, gray] (3.05, 1.7)–(3.05, 2.1);
[->] (2.75, 1.8)–(3.02, 1.8);

0.45[scale = 2]
[step = .5cm, thin, lightgray] (0, 0) grid (3, 2);
[thick] (0, .5) – (1, .5) – (1, 1) – (2, 1) – (2, 2) – (3, 2);
[thick, dashed] (0, 0) – (.5, 0) – (.5, .5) – (1.5, .5) – (1.5, 1) – (3, 1) – (3, 2);
[thick, gray](0, 0.45) – (1.05, .45)– (1.05, 0.95) – (2.05, .95) – (2.05, 1.95) – (3, 1.95);
[thick, gray, dashed](0, 0.05) – (.55, .05) – (.55, .45) – (1.55, .45) – (1.55, .95) – (3, .95);
[thick, gray] (1.9, 1.7)–(1.9, 2.1);
[thick](3.05, 1.7)–(3.05, 2.1);
[->, bend right] (3, 2.05) – (1.95, 2.05);

[thick] node [label=left:Legend:] (0, 0) – (0.5, 0) node [right]$\beta$;
[thick, dashed] (2.5, 0) – (3, 0) node [right]$f(\beta)$;
[thick, gray] (5, 0) – (5.5, 0) node [right, black]$\alpha$;
[thick, gray, dashed] (7.5, 0) – (8, 0) node [right, black]$f(\alpha)$;

Illustration of the interpolation condition for certain degree zero morphisms.

We will later see that degree zero morphisms occur from $\PRes_{\alpha}$ to $\PRes_{\beta}$ if and only if there exists a sequence $I = (i_1, \dots, i_k)$ of  $i\in \{1, \dots, r\}\cup \{\star\}$ satisfying  $\beta = p_{i_k}\circ\dots\circ p_{i_1}(\alpha)$ such that the intermediate transformations are well defined. In that case we write $\beta = p_I(\alpha)$. Moreover, the $\prescript{0}{}{u_i}$ will be shown to correspond to elementary morphisms. Because of the combinatorial nature of the argument we will deal with this later and focus first on the extensions. The following definition characterises the subsets $J$ of $S_{\alpha}$ that will play a role in describing extensions.

A subset $J$ of $S_{\alpha}$ is allowed when for all $i\in J$ such that $i-1\in S_{\alpha}\setminus J$, we have $\lambda_{i-1}< \lambda_{i}-1$.

Consider the partition $\alpha = (0, 1^2, 3, 4^2, 5 | 5)$ with $S_{\alpha} = \{2, 3, 4, 5\}$ and $\mu_{r+1} = 1$. Then the subset $J = \{2, 3, 4\}$ and $\{2\}$ are allowed while $\{4\}$ and $\{5\}$ are not.

The following lemma records some useful facts about allowed subsets and the combinatorics of the transformations $p_i$. The notation $J - 1$ refers to the set $\{j-1 |j\in J\}$.
Let $J$ be a subset of $S_{\alpha}$, let $h, j$ be elements of $S_{\alpha}$ and let  $i, k$ be in $\{1, \dots, r\}\cup \{\star\}$. The following assertions hold.

*  If $J$ is allowed, the partitions $\alpha$ and $q_J(\alpha)$ have the same multiplicities for non zero values.

*  If $J$ is allowed and $2\not \in J$ or $\lambda_2 > 1$, then for all $J\subseteq I\subseteq S_{\alpha}$ we have
\[q_I(\alpha) = q_{I\setminus J}(q_J(\alpha)).\]

*  If $J$ is allowed and $\lambda_2 = 1$ and $2\in J$ then $q_I(\alpha) = q_{(I\setminus J) -1 }(q_J(\alpha))$.

*  If $J$ is allowed, consider a pair of subsets $I\subseteq S_{\alpha}$ and $I'\subseteq S_{q_J(\alpha)}\setminus J$ of cardinal $k$ and $k-|J|$ respectively. If $2\not \in J$ or $\lambda_2 > 1$ assume $I'\not = I \setminus J$. Otherwise $I'\not = (I\setminus J) -1$. Then the partitions $q_I(\alpha)$ and $q_I'(q_J(\alpha))$cannot be compared.

*  If $p_i$ is well defined  and $i\not = 1$ or $\mu_1 > 1$ then $q_J(p_i(\alpha)) > q_J(\alpha)$ and for any $J' \subseteq S_{\beta}$ different from $J$ but of same cardinal, we have $q_{J'}(p_i(\alpha)) \not > q_J(\alpha)$.

*  If $i= 1$, $\mu_1 = 1$ and $\lambda_1 = 0$ then $q_{J-1}(p_i(\alpha)) > q_J(\alpha)$ and for any $J' \subseteq S_{\beta}$ different from $J-1$ but of same cardinal, we have $q_{J'}(p_i(\alpha)) \not > q_J(\alpha)$.

*  If both $h$ ans $j$ are allowed for $\alpha$ we have $q_hq_j(\alpha) = q_jq_h(\alpha)$.

*  If both $p_i$ ans $p_k$ are well defined for $\alpha$ we have $p_ip_k(\alpha) = p_kp_i(\alpha)$.

* From Definition <ref>, the partition $q_J(\alpha)$ can be written as
\[(\lambda'_1, \dots, \lambda'_1, \lambda'_2,\dots, \lambda'_2, \dots, \lambda'_r, \dots, \lambda'_r|n,\dots, n)\]
where the coefficient $\lambda'_i$ appears $\mu_i$ times. In general we do not know if the coefficients $\lambda'_1, \dots, \lambda'_r$ are all distinct. Because $J$ is allowed, for all $j\in J$ we either have $j-1\in J$ and $\lambda'_j = \lambda_{j} - 1 > \lambda'_{j-1} = \lambda_{j-1} - 1$, or, $j-1\not\in J$ in which case, if $j\not=1$, $\lambda'_j = \lambda_j -1 > \lambda'_{j-1} = \lambda_{j-1}$. For $i\not\in J$, then we have $\lambda'_i = \lambda_i > \lambda_{i-1}\geq \lambda'_{i-1}$ regardless of wether $i-1\in S_{\alpha}$ or not. Hence all the non zero values $\lambda'_i$ appear exactly $\mu_i$ times just like their counterpart in $\alpha$. When $\lambda_2 = 1$ and $2\in J$, then $0$ appears $\mu_1 + \mu_2$ times in $q_i(\alpha)$.

* Let $i$ be in $S_{\alpha}$ and $k\in [x_{i-1} +1 , x_i]$. Then

\[(q_{I\setminus J}(q_J(\alpha)))_k = \begin{cases}
\lambda_i - 1 &\text{if }i\in I\setminus J \text{ or } i\in J \text{ \emph{i.e} if } i\in I,\\
\lambda_i &\text{otherwise}\end{cases}.\]

This corresponds exactly to $q_I(\alpha)$.

* When $\lambda_2 = 1$ and $2\in J$, the first non zero coefficient has values $\lambda_3$. It is the second coefficient in $q_J(\alpha)$. More generally, when $J$ is allowed, the coefficient present between the indices $x_{i-1} + 1$ and $x_i$ is $\lambda_i$ or $\lambda_{i}-1$ but is the $(i-1)^{th}$ coefficient in $q_J(\alpha)$. Hence, the subset $I'$ of $S_{q_J(\alpha)}$ which satisfies $q_{I'}(q_J(\alpha)) = q_I(\alpha)$ is $(I\setminus J)-1$.

* The transformation $p_i$ is well defined and $i \not = 1$ or $\mu_i > 1$, so $\alpha$ and $p_i(\alpha)$ have the same coefficients $\lambda_1, \dots, \lambda_r$. Hence $q_J(\alpha)$ and $q_J(p_i(\alpha))$ have the same coefficients $\lambda'_1, \dots,  \lambda'_s$. They differ only at the index $x_i$ where $q_J(p_i(\alpha))_{x_i} = \lambda'_{i+1} > \lambda'_{i} = q_J(\alpha)_{x_i}$. Hence $q_J(p_i(\alpha)) \geq q_J(\alpha)$. Now take $J'\subseteq p_i(\alpha)$ of same cardinal as  $J$ but different from it and let $j$ be in $J'\setminus J$. The partitions $q_{J'}(p_i(\alpha))$ has value $\lambda_j - 1$ between indices $x'_{j-1} +1$ and $x'_j$, where the values $x'_{j-1} +1$ and $x'_j$ vary depending on wether $i = j-1$, $i = j$ or neither. It is easy to check that in all three cases we have $x'_{j-1}+1 \leq x_j \leq x'_j$. At the same time partition $q_J(\alpha)$ has value $\lambda_j$ at index $x_j$. Hence $q_{J'}(p_i(\alpha)) \not\geq q_J(\alpha)$.

The proofs of the remaining items are left for the reader hoping the ones we gave provide enough intuition on how the combinatorics of partitions will be discussed in the rest of the paper.

There is a nonzero morphism from $\PRes_{\alpha}$ to $\PRes_{q_J(\alpha)}[i]$ in the derived category if and only if $J$ is allowed and $|J| = i$

Clearly it is true that $q_J(\alpha)\in [f(q_J(\alpha)), q_J(\alpha)]$. By Theorem <ref> there is an extension of degree $|J|$ if and only if there is no other subset $J'$ such that $q_{J'}(\alpha)\in [f(q_J(\alpha)), q_J(\alpha)]$. So we want to show that this is equivalent to $J$ being allowed. Assume $J$ is allowed and take $J'\not = J$. Because the antichain is boolean it suffices to consider $J'\supset J$. Let $j$ be in $J'\setminus J$. Then

\[(q_{J'}(\alpha))_{x_j} = \lambda_j - 1 < \lambda_j = (q_J(\alpha))_{x_j}\]

where $x_i$ is the end of the coefficient $\lambda_i$ in $\alpha$. Because $J$ is allowed it is also the end of the coefficient $\lambda_i$ in $q_J(\alpha)$. By condition <ref> of Proposition <ref> the inequality above implies that $q_{J'}(\alpha)\not\in[f(q_J(\alpha)), q_J(\alpha)]$. Reciprocally, assume that $J$ is not allowed. Then there exists $j\in S_{\alpha}$ such that $j\in J$, $j-1\not \in J$ and $\lambda_{j-1} = \lambda_j- 1$. One can then check that $\alpha_{J\sqcup\{j-1\}} \in [f(q_J(\alpha)), q_J(\alpha)]$ using the characterisation for the support of the $\PRes_{\alpha}$ and conclude with Lemma <ref>. Figure <ref> provides an illustration for the arguments of the proof.

0.45[scale =2]
[step = .5cm, thin, lightgray] (0, 0) grid (3, 2);
[thick] (0, .5) – (1, .5) – (1, 1) – (2, 1) – (2, 2) – (3, 2);
[thick, gray](0, 0.45) – (1.05, .45)– (1.05, 0.95) – (2.05, .95) – (2.05, 1.5) – (3, 1.5);
[thick, lightgray](0, 0) – (1.1, 0)– (1.1, 0.9) – (2.1, .9) – (2.1, 1.45) – (3, 1.45);
[thick, gray, dashed](0, 0) – (.5, 0) – (.5, .45) – (1.5, .45) – (1.5, .95) – (3, .95) – (3, 1.5);
[thick](3.02, 1.7)–(3.02, 2.1);
[thick, gray] (3.05, 1.7)–(3.05, 2.1);
[->] (2.25, 1.9)–node[right]$J = \{3\}$(2.25, 1.6);
[->] (.8, .4)–node[right]$J' = \{1, 3\}$(.8, .1);
[red] (.4, 0.05) rectangle (.6, 0.55);

*$q_{\{1, 3\}}(\alpha)\not \in [f(\alpha_{\{3\}}), \alpha_{\{3\}}]$

0.45[scale =2]
[step = .5cm, thin, lightgray] (0, 0) grid (3, 2);
[thick] (0, .5) – (1, .5) – (1, 1) – (2, 1) – (2, 2) – (3, 2);
[very thick, lightgray] (0, 0) – (1, 0) – (1, .5) – (2.1, .5) – (2.1, 1.9) – (3, 1.9);
[thick, gray](0, 0.45) – (2.05, 0.45) – (2.05, .95) – (2.05, 1.95) – (3, 1.95);
[thick, gray, dashed](0, 0) – (1.5, 0) – (1.5, .45) – (3, .45) – (3, 1.95);
[thick](3.02, 1.7)–(3.02, 2.1);
[thick, gray] (3.05, 1.7)–(3.05, 2.1);
[->] (1.2, .9)–node[right]$J = \{2\}$(1.2, .6);
[->] (.2, .4)–node[right]$J' = \{1, 2\}$(.2, .1);
[red] (1.5, 0.5) circle (3pt);

*$q_{\{1, 2\}}(\alpha) \in [f(\alpha_{\{2\}}), \alpha_{\{2\}}]$

[thick] node [label=left:Legend:] (0, 0) – (0.5, 0) node [right]$\alpha$;
[very thick, lightgray] (2.5, 0) – (3, 0) node [right, black]$q_{J'}(\alpha)$;
[thick, gray] (5, 0) – (5.5, 0) node [right, black]$q_J(\alpha)$;
[thick, gray, dashed] (7.5, 0) – (8, 0) node [right, black]$f(q_J(\alpha))$;

Illustration of the proof of Fact <ref>

There exists a canonical realisation of these extensions defined as morphisms of complexes between the projective resolution associated to $\alpha$ and the interval module associated to $q_J(\alpha)$ concentrated in degree $|J|$. We take the morphisms of modules in degree $|J|$ to be the canonical projection of the factor $P_{q_J(\alpha)}$ onto $[f(q_J(\alpha)), q_J(\alpha)]$ and write the resulting morphism of complexes $u_J^{\alpha}$. Among the extensions we just exhibited we are particularly interested in the ones where $J = \{i\}\subset S_{\alpha}$ and write these $\prescript{1}{}{u_i^{\alpha}}$.

Up to homotopy there exists a unique lift of these morphisms along the projective resolution $\PRes_{q_J(\alpha)}[|J|]$.

Let  $\alpha$ be a left enhanced partition and let $J$ be an allowed subset of $S_{\alpha}$ of size $j$. The lift $\phi$ of $u^{\alpha}_J : \PRes_{\alpha} \to \PRes_{q_J(\alpha)}[|J|]$ has support $\bigoplus_{J\subseteq I} P_{q_I(\alpha)}$ in each degree. More precisely, in degree $j+k$, we have

\[\phi_{j+k} = \bigoplus_{\substack{J\subset I\\ |I| = j+k}} (-1)^{|I\setminus J|_J + |J|\cdot k}\cdot id_{P_{q_I(\alpha)}}\]

where $|I|_J = \sum_{i\in I}|i|_J$.

The morphism of complexes $\phi = (\phi_l)_l$ is concentrated in degrees greater than the cardinal $|J| =j$ of $J$. In each degree, because the antichains are strong, the morphism of module $\phi_{j+k}$ decomposes as linear combinations of the identity morphisms of the indecomposable projective summands appearing in both the source and the target. These are precisely the summands $P_{q_I(\alpha)}$ satisfying $J\subseteq I$. See Lemma <ref> for more details. It remains to determine the coefficients $\varepsilon(I)$ associated to each such summand. Assume we have computed the coefficient for degrees between $j$ and $j+k$. We want to compute $\varepsilon(I\sqcup\{i\})$ where $i$ is an element of $S_{\alpha}$ and $I$ is of cardinal $k$. First we assume that $\lambda_2 > 1$ or that $2\not \in J$. The relevant morphisms fit in the following commutative square.

\begin{equation}
\begin{tikzcd}[column sep = huge]
P_{q_{I\sqcup \{i\}}(\alpha)}\ar[r, "(-1)^{|i|_{I\sqcup \{i\}}}"]\ar[d, "\varepsilon(I\sqcup\{i\})"']&P_{q_I(\alpha)}\ar[d, "\varepsilon(I)"]\\
P_{q_{I\setminus J \sqcup \{i\}}(q_J(\alpha))}\ar[r, "(-1)^{j+|i|_{I\setminus J\sqcup \{i\}}}"]& P_{q_{I\setminus J}(q_J(\alpha))}
\end{tikzcd}
\end{equation}

Note that the bottom boundary map has a $(-1)^j$ sign because the complex is shifted by $[j]$. For the square to commute it must be that
\[ (-1)^{j + |i|_{I\setminus J\sqcup\{i\}}}\times \varepsilon(I\sqcup\{i\}) = (-1)^{|i|_{I\sqcup\{i\}}}\times \varepsilon(I).\]

Recall from equation (<ref>) that

\begin{equation*}
|i|_I = \{h\in I | h\leq i\} = |i|_{I\setminus J} + |i|_{J}.
\end{equation*}

Hence we get a recursive formula for the coefficient $\varepsilon(I\sqcup \{i\}) = (-1)^{j + |i|_J}\times \varepsilon(I)$.
By setting the degree $j$ sign to be $1$, we necessarily get the following formula for the degree $j+k$ maps

\[\phi_{j+k} = \bigoplus_{\substack{J\subset I\\ |I| = j+k}} (-1)^{|I\setminus J|_J + |J|\cdot k}\cdot id_{P_{q_I(\alpha)}}\]

where we introduce the notation $|K|_J = \sum_{i\in K}|i|_J$. The resulting coefficient is the product of the contribution of the previous signs which results in the two sums in the exponent. It is independent of the choice of $i$ made before. It is then clear that this indeed yields a morphism of complexes. Just like in Lemma <ref>, when $\lambda_2 = 1$ and $2\in J$ replace $I\setminus J$ by $I\setminus J - 1$ to have the corresponding subset of $S_{q_J(\alpha)}$. The index $2$ is the minimum of $J$ and of $I$. We can then check that we have

\[|i|_I = \{h\in I | h\leq i\} = |i|_{I\setminus J - 1} + |i|_{J}\]

which yields the same recursive formula and thus the same coefficients for the lift. In both cases there is no non trivial homotopy for the pair of complexes, so the explicit form we provided is in fact the only one and we shall refer to it several times in what follows. To conclude this proof we underline the fact that $\varepsilon(I)$ is indeed not zero for all subset $I$ of $S_{\alpha}$ containing $J$.

Similarly, it is possible to determine precisely the lift of the degree zero morphisms along the projective resolutions. We will do this at the end of this subsection.

Let $\alpha = (\lambda^{\mu_1}_1, \dots, \lambda_r^{\mu_r}|n^{\mu_{r+1}})$ and $\beta = (l_1^{m_1}, \dots, l_s^{m_s}| n ^{m_{s+1}})$ be two partitions and let $J$ be a subset of $S_{\alpha}$. The subset $J$ is the unique subset of $S_{\alpha}$ such that $q_J(\alpha)\in [f(\beta), \beta]$ if and only if there exists a morphism in degree zero from $\PRes_{q_J(\alpha)}$ to $\PRes_{\beta}$ and $J$ is allowed with the extra condition that if $\epsilon \in J$ and $\lambda_{\epsilon} = 1$, then the partition $\beta$ has its first value equal to zero and $y_1 > x_{\epsilon - 1}$. This means that the zeros in $\beta$ overlap with the first non zero value of $\alpha$.

Assume that $J$ is the unique subset of $S_{\alpha}$ satisfying $q_J(\alpha) \in [f(\beta), \beta]$. First we will prove that $J$ is allowed by contraposition. Assume that there exists $j\in J$ such that $j-1\not \in J$. If such an element $j$ does not exist then $J$ is automatically allowed. From these two assumptions it follows that $q_{J\sqcup\{j-1\}}(\alpha) \not \in [f(\beta), \beta]$ and $q_{J\setminus\{j\}}(\alpha) \not \in [f(\beta), \beta]$. Hence

* there exists $k$ such that $l_k = \lambda_{j}-1$ and $x_{j-1} <y_k \leq x_j$;

* there exists $k'$ such that $l_{k'} = \lambda_{j-1}$ and $x_{j-2} < y_{k'}\leq x_{j-1}$.

In particular, $k\not = k'$ and $\lambda_{j-1}\not = \lambda_j-1$. This proves that $J$ is allowed. Next we show that there is degree zero morphism. Note that the statement above about $k$ is true for any element $j$ of $J$ and the one about $k'$ for any $j'\not\in J$. It follows that

\[\{\lambda_i| i\in S_{\alpha}\}\subseteq \{l_j| j \in S_{\beta}\}.\]
By Proposition <ref> Item <ref> there is a non zero morphism from $\PRes_{q_J(\alpha)}$ to $\PRes_{\beta}$. Lastly we check the extra condition. We want to argue that if $\epsilon \in J$, $\lambda_{\epsilon} = 1$ and $l_1 > 0$ then $q_{J\setminus \{\epsilon\}}(\alpha)$ is also in $[f(\beta), \beta]$ as it satisfies all the conditions that define the interval. This would be a contradiction to $J$ being the unique subset of $S_{\alpha}$ such that $q_J(\alpha)\in [f(\beta), \beta]$. To see this, first notice that because there exists a nonzero morphism from $\PRes_{q_J(\alpha)}$ to $\PRes_{\beta}$, by Lemma <ref>, Item <ref>, the first non zero value of $\beta$, say $l_1$, is also the first non zero value of $q_J(\alpha)$. Moreover, in $\beta$, it runs beyond the index $x_{\epsilon}$. Hence, if the first non zero value of $\beta$ had started before the index $x_{\epsilon-1}$, we would still have $q_{J\setminus \{\epsilon\}}(\alpha)\leq \beta $. Moreover, $q_{J\setminus \{\epsilon\}}(\alpha)$ and $q_J(\alpha)$ have the same values after the index $x_{\epsilon}$ and the defining conditions of the interval $[f(\beta), \beta]$ are checked at indices $y_i$ which all appear after  $x_{\epsilon}$. Hence the first non zero value of $\beta$ starts before $x_{\epsilon -1} +1$.

Conversely, assume that $J$ is allowed, satisfies the extra condition and that there is a morphism from $\PRes_{q_J(\alpha)}\to\PRes_{\beta}$. Consider a subset $J'$ of $S_{\alpha}$ and suppose that $q_{J'}(\alpha)\in [f(\beta), \beta]$. We will show that $J' = J$. By <ref> for all non empty $J''\subseteq S_{q_J(\alpha)}$, the element $q_{J''}(q_J(\alpha))$ is not in $[f(\beta), \beta]$. Because the element

\[q_{J'}(\alpha)\wedge q_{J} (\alpha) = q_{J\cup J'}(\alpha) = q_{J'\setminus J} (q_J(\alpha))\]

is in the interval $[f(\beta), \beta]$ as well, we conclude that $J'\setminus J = \varnothing$ and $J' \subseteq J$ . To see that in fact $J = J'$, say there exists $j\in J\setminus J'$. Take $j$ to be minimal. Assume first that $j\not = 2$ or that $\lambda_2 > 1$.

Because $J$ is allowed, the partitions $q_J(\alpha)$ and $q_{J'}(\alpha)$ have the same non zero coefficients and the same multiplicities for non zero coefficients up until the index $x_{j-1}$. Moreover, the $j^{th}$ coefficient of $q_J(\alpha)$, $\lambda_j - 1$, runs from $x_{j-1}+1$ to $x_j$ and no further. The $j^{th}$ coefficient $\lambda_j$ of $q_{J'}(\alpha)$, runs from $x_{j-1}+1$ to $x_j$. This follows from $J$ being allowed, but we point out that it could extend further as the subset $J'$ needs not be allowed. By Lemma <ref> there exists $k\in S_{\beta}$ such that $x_{j-1} < y_k \leq x_j$ and $l_k = \lambda_j -1$. If we assume that $q_{J'}(\alpha)\in [f(\beta), \beta]$ then we would also have $l_k = \lambda_j$ which is impossible hence $J' = J$.

If $\lambda_2 = 1$ and $2\in J$, then our assumption on the zero value in $\beta$ implies that $2$ is necessarily in $J'$. Hence the previous argument concludes the proof.

Let $\phi : \PRes_{\alpha}\to \PRes_{\beta}[i]$ be a non zero morphism in $\mathcal{Y}_{m,n}$. Then there exists a unique subset $J$ of $S_{\alpha}$ such that $\phi$ factors through $\PRes_{q_J(\alpha)}[i]$ and $|J| = i$ completing the following commutative diagram.

\[
\begin{tikzcd}[column sep = tiny]
\PRes_{\alpha}\ar[rr]\ar[dr]&&\PRes_{\beta}[|J|]\\
\end{tikzcd}
\]

By Theorem <ref>, there exists a non zero morphism $\phi$ if and only if there exists a unique subset $J$ such that $q_J(\alpha)\in [f(\beta), \beta]$. Moreover any two morphisms between $\PRes_{\alpha}$ and $\PRes_{\beta}[i]$ differ by multiplication by a scalar. By Lemma <ref> the following morphisms exist and are unique up to multiplication by a scalar

$$\PRes_{\alpha}\to \PRes_{q_J(\alpha)}[|J|]\to\PRes_{\beta}[|J|].$$

It remains to check that their composition is non zero in the homotopy category. We represent the morphisms in the following diagram.

\[
\begin{tikzcd}
\PRes_{\alpha}:\ar[d]\ar[dd, bend right = 50, red, "?"]&\dots\ar[r]\ar[d]&\bigoplus_{|I| = |J|}P_{\alpha_I}\ar[d]\ar[r]&\bigoplus_{|I| = |J| - 1 }P_{\alpha_I}\ar[r]\ar[d]&\dots\ar[r]&P_{\alpha}\\
\PRes_{q_J(\alpha)}[|J|]\ar[d]:&\dots\ar[r]\ar[d]&P_{q_J(\alpha)}\ar[r]\ar[d]&0\ar[d]&&\\
\PRes_{\beta}:&0\ar[r]&{[f(\beta), \beta]}\ar[r]\ar[from = uu, bend right = 40, crossing over, red, "?"']\ar[from = uur, crossing over, red, "0"']&0&&
\end{tikzcd}
\]

The composition of the two morphisms of modules in degree $|J|$ is non zero given the support of the top map as described in Lemma <ref>. No homotopy can be constructed because by assumption if $I\not= J$, $q_I(\alpha)\not\in [f(\beta), \beta]$. By correcting one of the intermediate maps we can factor $\phi$ through $\PRes_{q_J(\alpha)}[i]$.

Next we decompose further the extensions $\PRes_{\alpha} \to \PRes_{q_J(\alpha)}[|J|]$ and the degree zero morphisms.
Let $\alpha$ be a partition and $J$ be an allowed subset of $S_{\alpha}$. Then the extension $u:\PRes_{\alpha}\to\PRes_{q_J(\alpha)}[k]$ discussed in Proposition <ref> decomposes as

\[\PRes_{\alpha}\xrightarrow{\prescript{1}{}{u_{j_1}}}\PRes_{q_{\{j_1\}}(\alpha)}[1]\xrightarrow{\prescript{1}{}{u_{j_2}[1]}}\dots\xrightarrow{\prescript{1}{}{u_{j_k}}[k-1]} \PRes_{q_J(\alpha)}[k],\]

where $J = \{j_1, \dots j_k\}$ with $j_t < j_{t+1}$.

First notice that the truncated set $J_i = \{j_1, \dots, j_i\}$ are all allowed. We proceed by induction on $k$. Assume the result holds for $k-1$. Then by Theorem <ref> it suffices to show that the composition

\[\PRes_{\alpha}\to\PRes_{q_{J_{k-1}}(\alpha)}[k-1] \to \PRes_{q_J(\alpha)}[k]\]

is non zero. Again we can draw a diagram to visualise the situation:

\[
\begin{tikzcd}
\PRes_{\alpha}:\ar[d]\ar[dd, bend left = 60, red, "?"]&[-35pt]\dots\ar[r]\ar[d]&\bigoplus_{|I| = k}P_{q_I(\alpha)}\ar[r]\ar[d]&[-25pt]\bigoplus_{|I| = k-1}P_{q_I(\alpha)}\ar[d]\ar[r]&[-30pt]\bigoplus_{|I| = k-2}P_{q_I(\alpha)}\ar[d]\ar[r]&[-20pt]\dots\\
\PRes_{q_{J_{k-1}}(\alpha)}[k-1]: \ar[d]&\dots\ar[d]\ar[r]&\bigoplus_{i\in S_{q_{J_{k-1}}(\alpha)}}P_{q_{i}(q_{J_{k-1}}(\alpha))}\ar[d]\ar[r]&P_{q_{J_{k-1}}(\alpha)}\ar[d]\ar[r]&0&\\
\PRes_{q_J(\alpha)}[k]:&0\ar[r]&{[f(q_J(\alpha)), q_J(\alpha)]}\ar[r]\ar[from = uu, bend right = 75, crossing over, red, "?"']\ar[from = uur, bend left = 10, crossing over, red, "0"]&0&&
\end{tikzcd}
\]

It is clear that $q_I(\alpha)\not\leq q_J(\alpha)$ when $|I| = k-1$. It might not be clear however that the composition of the two morphisms of module in degree $k$ is non zero. Because the bottom map has support $P_{q_J(\alpha)}$ we need to argue that the restriction of the top map to $P_{q_J(\alpha)}$ is non zero either. This is the case by Lemma <ref>. To do so, use the fact that  the top right square commutes and that the composition of modules morphisms

\[P_{q_J(\alpha)} \xrightarrow{\partial_k}\bigoplus_{|I| = k-1}P_{q_I(\alpha)}\xrightarrow{\pi_{J_{k-1}}}P_{q_{J_{k-1}}(\alpha)}\]

is non zero. Finally, we need to point out that this results in the canonical map $u_J$ and not just proportional to it meaning that no negative sign appears when composing the two maps. We comput the sign of the module map in degree $j$ with Lemma <ref>. It has support the indecomposable projective module $P_{q_J(\alpha)}$. Its sign is

\[(-1)^{|j_k|_{J_{k-1}} + 1\cdot(k-1)} = (-1)^{2\cdot(k-1)} = 1\]

Hence no sign appears.

A similar result holds for degree zero morphisms.

Let $\phi : \PRes_{\alpha}\to\PRes_{\beta}$ be a non zero morphism of modules. Then there exists a sequence $(d_1, \dots, d_r, d_{\star})$ such that for all $0\leq i< r $, we have $0\leq d_j< \mu_{i}$, if $\lambda_r = n$, we have $d_{\star} = 0$, otherwise $d_{\star}\in \{0, 1\}$ and

\[\beta = p_1^{d_1}\circ\dots\circ p_r^{d_r}\circ p_{\star}^{d_{\star}}(\alpha).\]

Moreover, the morphism factors through each of the objects associated to the intermediate partitions.

By Theorem <ref> and item <ref> of Proposition <ref> the inclusion

\[\{\lambda_i | i\in S_{\alpha}\}\subseteq \{l_j | j\in S_{\beta}\}\]
holds. In other words, for all $i\in S_{\alpha}$, there exists $j\in S_{\beta}$ such that $\lambda_i = l_j$. Because $\alpha\leq\beta$ the ending indices of these coefficients must interpolate: $x_{i-1}<y_j \leq x_i < y_{j+1}$. For $i\in S_{\alpha}$ we set $d_i = x_i-y_j$. If $l_1 = 0$, because $\alpha \leq \beta$ we also have $\lambda_1 = 0$ and $y_1\leq x_1$ so set $d_1 = x_1 - y_1$. Lastly, we set $d_{\star} = 1$ if $\lambda_r < n$, $\mu_{r+1} >0$ and $m_{s+1} = 0$. Otherwise, $d_{\star} = 0$. The condition $d_i< \mu_i$ holds because $x_{i-1} < y_{j}$ when $i\not= 1$. The resulting partition has the same coefficients as $\beta$. The multiplicities also match when $i\leq r$:

\[x_{i} - d_i - x_{i-1} - d_{i-1} = x_{i} - x_{i} + y_{j} - x_{i-1} +x_{i-1} - y_{j-1} = m_j.\]

To see that the map factors through the intermediate partitions use item <ref> of Proposition (<ref>):
let $k$ be in $S_{\alpha}$, $0\leq d < d_k$ and set $\gamma = p_k^d\circ p_{k+1}^{d_{k+1}}\circ\dots\circ p_{r}^{d_r}\circ p_{\star}^{d_{\star}}(\alpha)$ and assume the morphism $\PRes_{\alpha} \to \PRes_{\gamma}$ can be decomposed through all the intermediate links. By construction, $\alpha\in [f(p_k(\gamma)), \gamma]\cap [f(\gamma), \gamma]\cap [f(\alpha), \alpha]$ so the map induced by the composition

\[\PRes_{\alpha}\to \PRes_{\gamma}\xrightarrow{\prescript{0}{}{u_k^{\gamma}}}\PRes_{p_{k}(\gamma)}\]

is non zero on this vertex hence the composition itself is non zero. Recall that by Theorem <ref> the hom spaces are one dimensional it follows that the map $\phi$ can be factored into a sequence of $\prescript{0}{}{u_k^{\alpha'}}$ maps.

0.45[scale =2]
[step = .5cm, thin, lightgray] (0, 0) grid (3, 2);
[thick] (0, .5) – (1, .5) – (1, 1) – (2, 1) – (2, 2) – (3, 2);
[thick, lightgray](0, .45) – (1.5, .45) – (1.5, .95) – (2.05, 0.95) – (2.05, 1.95) – (3, 1.95);
[thick, dashed] (0, 0) – (.5, 0) – (.5, .5) – (1.5, .5) – (1.5, 1) – (3, 1) – (3, 2);
[thick](3.05, 1.7) – (3.05, 2.1);
[thick, gray] (1.95, 1.7) – (1.95, 2.1);
[->] (1.1, .65)–node[above]$d_1$(1.4, .65);
[->] (2.9, 2.05)–node[below]$d_r$(2.1, 2.05);

*$\alpha_{x_{i-1}} = \beta_{x_{i-1}}$

[thick] node [label=left:Legend:] (0, 0) – (0.5, 0) node [right]$\beta$;
[thick, dashed] (2, 0) – (2.5, 0) node [right, black]$f(\beta)$;
[thick, lightgray] (4, 0) – (4.5, 0) node [right, black]$\alpha$;
(1.5, -1) – node[]$d_0 = 0, d_1 = 1, d_2 = 0, d_3 = d_r = -2$ (1.5, -1);

Illustration of the proof of Lemma <ref>

Morphisms in the category $\mathcal{Y}_{m,n}$ can thus be decomposed into the morphisms of degree and the morphisms of degree zero we have exhibited. Most of them cannot be decomposed further.

Let $\alpha$ be a partition, $h\in S_{\alpha}$ allowed and $k\in \{1, \dots, r\}\cup \{\star\}$ such that $p_k$ is well defined. Assume that $h \not = \epsilon$ or that $ \lambda_2 > 1$. Then the morphisms $\prescript{1}{}{u_h}$ and $\prescript{0}{}{u_k}$ are irreducible.

The situation is summed up in the following diagrams.

\[\begin{tikzcd}
\PRes_{\alpha}\ar[rr, "\prescript{1}{}{u_h}"]\ar[dr]&&\PRes_{q_h(\alpha)}[1]&&\PRes_{\alpha}\ar[rr, "\prescript{0}{}{u_k}"]\ar[dr]&&\PRes_{p_k(\alpha)}\\
\end{tikzcd}\]

Consider an elementary morphism in degree zero $\prescript{0}{}{u_k}$ and assume it factors through an object $\PRes_{\beta}$ as $f\circ g$. The letters $\lambda_{i}$ and $x_{i}$ refer to $\alpha$, $\lambda'_{i}$ and $x'_{i}$ to $p_k(\alpha)$ and $l_{j}$ and $y_{j}$ to $\beta$. Then, by Lemma <ref> item <ref>, it holds that

\[ \{\lambda'_{i}| i \in S_{p_k(\alpha)}\}\subseteq \{l_j|j\in S_{\beta}\} \subseteq \{\lambda_{i}|i\in S_{\alpha}\}\]

and for all $j\in S_{\beta}$ there exists $i\in S_{\alpha}$ such that $x_{i-1}< y_{j}\leq x_i< y_{j+1}$ and $l_{j} = \lambda_{i}$.
The index $i$ in $S_{\alpha}$ associated to the value $\lambda_i$ in $\alpha$ corresponds to an index $i'\in S_{p_k(\alpha)}$ associated to the same value $\lambda'_{i'} = \lambda_i$. If $k = 1$, $\lambda_1 =0$ and $\mu_1 = 1$, then $i' = i - 1$, otherwise, $i = i'$. Note that it is possible that $i = r+1$ when $k = \star$. In that case, $i'$ can equally correspond to an enhanced or unenhanced coefficient.
In turn, for such an $i'$ there exists $j'$ such that $y_{j'-1}< x'_{i'}\leq y_{j'}< x'_{i'+1}$ and $\lambda'_{i'} = l_{j'}$.

It follows that $j = j'$, and that $x'_{i'}\leq y_{j}\leq x_{i}$. Hence, when $i\not = k$ we have $x'_{i'} = y_{j'} = x_{i}$. Otherwise, when $i = k \not = \star$, $y_j$ can be either $x_i$ or $x'_{i'} = x_i - 1$ which means that $\beta = \alpha$ or $\beta = p_k(\alpha)$ and  $\prescript{0}{}{u_k}$ is irreducible. When $k = \star$, It is easier to use Item <ref> of Lemma <ref>. We then conclude that $\alpha$, $p_k(\alpha)$ and $\beta$ have the same underlying plain partition and that $f(\alpha)\leq f(\beta)\leq f(p_k(\alpha))$. The plain partitions $f(\alpha)$ and $f(p_k(\alpha))$ differ only at the index $m$. Knowing already that $\beta$ has the same underlying plain partition as the other two we deduce that $f(\beta) = f(\alpha)$ or $f(\beta) = f(\alpha)$ which determines the enhancement of $\beta$.
Next consider an elementary extension $\prescript{1}{}{u_h}$ with $\alpha$ and $\beta$ as before. The object $\PRes_{\beta}$ is shifted by either $0$ or $1$. We consider first the case where it is shifted by $0$. On the one hand, $\alpha$ and $\beta$ satisfy item <ref> of Lemma <ref>. On the other hand by Theorem <ref> and Proposition <ref> there exists $f$ such that $q_f(\beta)$ and $q_i(\alpha)$ satisfy item <ref> of Lemma <ref>. Write $l'_j$ the coefficients of $q_f(\beta)$. Combining these observations we get that for all $i\in S_{\alpha}$ there exists $j$ such that $\lambda_{i} = l_{j}$ and $x_{i-1} < y_{j} \leq x_{i}<y_{j+1}$. For such a $j$ there exists $i'$ such that $l'_{j} = \lambda'_{i'}$ and $y_{j-1}<x_{i'}\leq y_{j} < x_{i'+1}$. Weather $i = h$ or not, we get that $i = i'$, $y_{j} = x_i$ and $\beta = \alpha$. The case where $\PRes_{\beta}$ is shifted by one is very similar: $\beta$ and $q_h(\alpha)$ satisfy condition <ref> of Lemma <ref> and by Proposition <ref> there exists a unique $f\in S_{\alpha}$ allowed such that $q_f(\alpha)$ and $\beta$ satisfy item <ref> of Lemma <ref>. Write $\lambda''_i$ the coefficients of $q_f(\alpha)$. Combining these observations we get that for all $i\in S_{\alpha}$ there exists $j$ such that $\lambda''_{i} = l_{j}$ and $x_{i-1} < y_{j} \leq x_{i}<y_{j+1}$. For such a $j$ there exists $i'$ such that $l_{j} = \lambda'_{i'}$ and $y_{j-1}<x_{i'}\leq y_{j} < x_{i'+1}$. Whether $i = h$ or not, we get that $i = i'$, $y_{j} = x_i$ and $\beta = q_{h}(\alpha)$.

If $\lambda_2 = 1$, then $\prescript{1}{}{u_2}$ decomposes into $\prescript{1}{}{u_1}\circ\prescript{0}{}{u_1^{\mu_1-1}}\circ\prescript{0}{}{u_0}$ where $\mu_1$ is the multiplicity of the value zero in the source partition. We give a concrete example below.

\[
\begin{tikzcd}[column sep = tiny]
\PRes_{(0011)}\ar[rr]\ar[dr]&&\PRes_{(0000)}[1]\\
\end{tikzcd}
\]

Before we move on to the description of the relations between these morphisms, we give an explicit lift of the canonical morphism $\prescript{0}{}{u_i}$.
Let $\phi : \PRes_{\alpha}\to \PRes_{\beta}$ be a non zero morphism. Then its lift along the projective resolutions  is made of monomorphism in each degree and all the signs are positive.

According to Lemma <ref> we can write $\beta$ as $p_{I}(\alpha)$ where $I = (i_1, \dots, i_k)$ is a sequence of elements of $\{1, \dots, r\}\cup \{\star\}$. Let $J$ be a subset of $S_{\alpha}$ of size $k$. If $1$ does not appear in $I$ or $\mu_1 > 1$ then $J$ can also be seen as a subset of $S_{\beta}$. Applying Item <ref> of Lemma <ref> repeatedly, the subset $J$ is then the only subset of $S_{\beta}$ satisfying $q_J(\alpha)\leq q_{J}(\beta)$.
If $1\in J$ and $\mu_1 = 1$, we apply Items <ref> and <ref> of Lemma <ref> to get that $I-1$ is indeed the only subset of size $k$ of $S_{\beta}$ for which $q_I(p_J(\beta)) > q_{I'}(\beta)$. Hence, when $1$ does not appear in $I$ or $\mu_1 > 1$ we have
\[\phi_k = \oplus_{J\subseteq S_{\alpha}}c_{J} \iota_{q_J(\alpha)}^{q_J(\beta)},\]

otherwise it is

\[\phi_k = \oplus_{J\subseteq S_{\alpha}}c_{J} \iota_{q_{J-1}(\alpha)}^{q_J(\beta)}.\]

It remains to see that $c_I = 1$ makes all the squares commute similarly to the proof of Lemma <ref>. This time the boundary maps are identical in the top and the bottom complexe hence no signs appear. As a conclusion we have an explicit lift of the canonical degree zero map which is a monomorphism in each degree.

§.§ Configurations and relations

We can now describe the elementary morphisms of subsection <ref> using configurations through the map $\phi_r$  from subsection <ref>. To do so we define a partial function on configurations, as in <cit.> for the construction of higher Auslander algebras of type A. If $R$ is a configuration, $k \in R$ and $k - 1 \not\in R$, we write

\begin{equation}\sigma_k^-(R) = (R\cup{\{k-1\}})\setminus\{k\}.\end{equation}

Note that if $k = -m$, then $\sigma^-_k(R) = (R\cup{\{n\}})-\{-m\}$ as $k$ represents an element of $\mathbb{Z}/(m+n+1)\mathbb{Z}$. From now on, we will also write the objects of the category $\mathcal{Y}_{m,n}$ as $\PRes_{R}$, implicitly identifying $\alpha$ with $R$ using the bijection $\phi_r$.

Recall the right abacus from example <ref> associated to the partition $a = (0, 2, 3, 7, 7)$.

-5   -4   -3   -2   -1   0   1   2   3   4   5   6   7

$\bullet$            $\bullet$      $\bullet$   $\bullet$            $\bullet$

Applying $\sigma^-_{-4}$ to the configuration associated to $a$ consists of taking the bead placed in $-4$ and sliding it down to the $-5$. We then get the following abacus,

-5   -4   -3   -2   -1   0   1   2   3   4   5   6   7

$\bullet$               $\bullet$      $\bullet$   $\bullet$            $\bullet$

which is associated to the partition $(0,1,2,3,7|\,7)$.

In terms of configurations, the elementary morphisms in degree zero, correspond to arrows

\[\PRes_R\to \PRes_{\sigma^-_k(R)}\text{ with } 0 \geq k \in R.\]

Next the elementary morphisms in degree 1 correspond to the following transformation on configurations

\[\PRes_R\to \PRes_{\sigma^-_k(R)}\text{ with } 0 < k\in R.\]

Note that, for a partition $\alpha$ with $\lambda_2 = 1$, the extension $\prescript{1}{}{u_2^{\alpha}}$ does not appear in the statement as the transformation $q_2$ of the partition does not correspond to a well defined transformation $\sigma_1^-$ of the corresponding abacus.
First we prove the statement about the degree zero elementary morphisms. Let $\alpha$ be a partition. Recall from Notation <ref> and Lemma <ref> that these morphisms were of the form

$$\PRes_{\alpha} \to \PRes_{p_i(\alpha)}$$
with $i\in \{1, \dots, r\}\cup \{\star\}$ and $\mu_i > 1$. First, taking $i\not = \star$ and using <ref>, we compute

\begin{equation}
x_j^{p_i(\alpha)} = \begin{cases} x_{i} - 1 &\text{ if } j = i\\
x_j &\text{ otherwise. }
\end{cases}\end{equation}
This implies that $R_{p_i(\alpha)} = (R_{\alpha}\cup\{-x_i + 1\}) \setminus\{-x_i\} = \sigma_{-x_i}^-(R_{\alpha})$. Now take $i = \star$, with $\lambda_r < n$ and $\mu_{r+1 > 0}$. The degree zero morphisms corresponding to the transformation $p_{\star}$ moves the enhancement bar all the way to the right. On the configuration this amounts to removing the coefficient $-m$ and adding $n$. Finally the degree 1 morphisms were
$$\PRes_{\alpha} \to \PRes_{q_{i}(\alpha)}[1]$$
where $\lambda_i -1> \lambda_{i-1}$. Here the configuration associated $\alpha_{\{i\}}$ is clearly the configuration associated to $\alpha$ where the $\lambda_i$ was replaced by $\lambda_i - 1$.
Reciprocally, every $k\in R$ such that $\sigma^{-}_k(R)$ is well defined, corresponds to either an elementary morphism or  an elementary extension through the inverse of $\phi_r$ (recall Figure <ref>). If $k \leq 0$, then $\sigma^{-}_k(R)$ corresponds to the transformation $p_i(\alpha)$ if $k -1$ is the $i^{th}$ gap in the negative side counting from the right. It is associated to $\prescript{0}{}{u_i}$. If $k > 0 $, then $\sigma^{-}_k(R)$ corresponds to the transformation $q_j(\alpha)$ if $k$ is the $j^{th}$ positive element of the configuration, in increasing order. It is associated to $\prescript{1}{}{u_j}$. This conclude the proof.

Note that configurations make the description of the elementary morphisms more homogeneous. This will also be the case for the relations between them. To simplify notations, the morphism, in degree zero or one, from $\PRes_{R}$ to $\PRes_{\sigma_k^-(R)}$ will be denoted $u_k^R$ since the degree of the morphism is encoded in the signe of $k$.
We will now express the relations between the morphisms and extensions in the language of configurations.

Let $R$ be a configuration, take $k, l\in R$ such that $k - 1, l - 1 \not\in R$. We have the following equalities in the category  $\mathcal{Y}_{m,n}$:

\begin{equation}\label{eq:rel}
\rho_{k, l}^R :  u_l^{\sigma_k^-(R)}\circ u_{k}^R = \varepsilon u^{\sigma^-_l(R)}_k \circ u_l^R \text{ and }z_k^R : u_{k-1}^{\sigma^-_{k}(R)}\circ u_{k}^R = 0
\end{equation}

where $\varepsilon = -1$ if $k$ and $l$ are positive and $1$ otherwise.

We distinguish several cases depending on the sign of the integers $k$ and $l$. First consider $\rho_{k, l}^R $ when both $k$ and $l$ are positive i.e the elementary morphisms are concentrated in degree 1. Then the morphisms in question can be summed up in the following diagram. We use partitions because the order relation is not clear on configurations. The elements $k$ and $l$ of $R$ can be uniquely associated to $i, j\in S_{\alpha}$

\begin{equation}\label{diag:proofRel1-1_1}
\begin{tikzcd}
\PRes_{\alpha}\ar[d]\ar[dd, bend right = 40, red, "?"']:&[-30pt] \dots\ar[d]\ar[r]&\bigoplus_{i', j'\in S_{\alpha}}P_{q_{\{i',j'\}}(\alpha)}\ar[d]\ar[r]&[-20pt]\bigoplus_{i'\in S_{\alpha}}P_{q_{i'}(\alpha)}\ar[d]\ar[r]&P_{\alpha}\ar[d]\ar[r]&[-10pt]0\\
\PRes_{q_i(\alpha)}\ar[d]:&\dots\ar[d]\ar[r]&\bigoplus_{j'\in S_{\alpha}}P_{q_{j'}(q_{i}(\alpha))}\ar[d]\ar[r]&P_{q_i(\alpha)}	\ar[d]	\ar[r]&0&\\
\PRes_{q_{\{i, j\}}(\alpha)}:& 0\ar[r]& \left [f(q_{\{i, j\}}(\alpha)), q_{\{i, j\}}(\alpha)\right] \ar[r]\ar[from = uu, bend left = 50, red, 			"?", 	crossing over]&0&&
\end{tikzcd}
\end{equation}

The resulting morphism of modules in degree two is the morphism from $\PRes_{\alpha}$ to $\PRes_{q_{\{i, j\}}(\alpha)}$ described in <ref> up to a sign because $i$ and $j$ are assumed to be allowed. This is symmetric in $i$ and $j$ hence the result. To be more specific, according to Lemma <ref>, signs appears in the upper middle square. The boundary map component going from $P_{q_{i, j}(\alpha)}$ to $P_{q_{\{i\}}(\alpha)}$ has sign $(-1)^{|j|_{\{i, j\}}}$. Hence the two compositions differ by a factor $-1$.

Next we consider $z_k^R$ when $k > 2$: the morphisms in question can be summed up by the following diagram using partitions.

\begin{equation}\label{diag:proofRel1-1_2}
\begin{tikzcd}
\PRes_{\alpha}\ar[d]\ar[dd, bend right = 40, red, "?"']:&[-30pt] \dots\ar[d]\ar[r]&\bigoplus_{i', j\in S_{\alpha}}P_{q_{\{i', j\}}(\alpha)}\ar[d]\ar[r]&[-20pt]\bigoplus_{i'\in S_{\alpha}}P_{q_{i'}(\alpha)}\ar[d]\ar[r]&P_{\alpha}\ar[d]\ar[r]&[-10pt]0\\
\PRes_{q_i(\alpha)}[1]\ar[d]:& \dots\ar[d]\ar[r]&\bigoplus_{j\in S_{q_i(\alpha)}}P_{q_j(q_{i}\alpha)}\ar[d]\ar[r]&P_{q_{i}(\alpha)}\ar[d]\ar[r]&0&\\
\PRes_{q_{i}(q_{i}(\alpha))}[2]:& 0\ar[r]& \left [f(q_{i}(q_{i}(\alpha))), q_{i}(q_{i}(\alpha))\right] \ar[r]\ar[from = uu, bend left = 50, red, "?", crossing over]&0&&
\end{tikzcd}
\end{equation}

There cannot be a morphism of modules in degree $2$ because the $q_{\{i', j\}}(\alpha)$ cannot be compared with $	q_{i}(q_{i}\alpha)$ when $j\not = i$.

We now take $k \leq 0$ and $l\leq 0$. For the $\rho^R_{k,l}$, notice that both branches are non zero when $l\neq k-1$ and conclude by saying that hom sets are one dimensional.

To see that $z_k$ is a relation recall from <ref> that there exists a morphism between $\PRes_{\alpha}$ and $\PRes_{\beta}$ if and only if $\alpha$ and $\beta$ have the same coefficients up to bars and zeros  and $\alpha_{y_j} = \beta_{y_j}$ for all $i$ in $S_{\beta}$. Let $\alpha$ be a partition, such that its corresponding right configuration $R$ contains $k$ but neither $k-1$ nor $k-2$. Say the negative coefficient $k$ is associated to the $i^{th}$ value of the partition. Figure (<ref>) illustrate the action of $\sigma_{k}^-$ and $\sigma_{k-1}^-$ on $R$ when $k> -m - 1$. On the right side, we see that the coefficients at $y_{i+1} = x_i + 2$ are different in $\alpha$ and $\beta$ meaning no arrow exists between the first and last partition. When $k = -m-1$ the argument can be summed up with a similar table which we leave to the reader. Finally, when $k = -m$, $z_k$ is indeed a relation because there is no extension from $\alpha$ to the partition $q_{r+1}(p_{\star}(\alpha))$. The partitions $\alpha$ and $p_{\star}(\alpha)$ have the same underlying plane partitions but not the same associated antichains and $q_{r+1}(p_{\star}(\alpha))$ does not appear in the antichain of $\alpha$.

[state] (abacus)

…    $k-2$     $k-1$     $k$    …

right of=abacus,
node distance= 6.3 cm,
anchor=center] (partition)

…   $x_i$     $x_i+1$     $x_i+2$    …

$\lambda_i$     $\lambda_i$     orange$\lambda_{i+1}$     $\lambda_{i+2}$

$\lambda_i$     $\lambda_{i+1}$     $\lambda_{i+1}$     $\lambda_{i+2}$

$\lambda_i$     $\lambda_{i+1}$     orange$\lambda_{i+2}$     $\lambda_{i+2}$

(abacus) edge[in = 183, out = 175, loop, distance=0.5cm]  node[anchor=east,left] $\sigma_{k}^-$ (abacus)
(abacus)  edge[out = 185, in = 193, loop , distance=0.5cm] node[anchor=east,left]$\sigma_{k1}^-$ (abacus)
(partition)  edge[out = 364, in = 358, loop, distance=0.5cm]  node[anchor=west,right] $p_i^{-1}$ (partition)
(partition) edge[out = 354, in = 348, loop, distance=0.5cm]   node[anchor=west,right] $p_{i+1}^{-1}$(partition);

relations between  elementary degree zero morphisms

Next we assume $k$ is positive and $l$ is non positive. We can check that both sides of the square yield the same result by looking at the diagrams <ref> and <ref> and discussing the the support of the morphisms in degree $1$.

\begin{equation}\label{diag:proofRel1-0_1}
\begin{tikzcd}
\PRes_{\alpha}\ar[d]\ar[dd, bend right = 50, red, "\neq 0", near start]:&[-15pt] \dots\ar[d]\ar[r]&\bigoplus_{i'\in S_{\alpha}}P_{q_{i'}(\alpha)}\ar[d]\ar[r]&P_{\alpha}\ar[d]\ar[r]&0\\
\PRes_{q_i(\alpha)}[1]\ar[d]:&\dots\ar[d]\ar[r]&P_{q_i(\alpha)}\ar[d]\ar[r]&0\ar[d]\\
\PRes_{p_j(q_i(\alpha))}[1]:& 0\ar[r]& \left [f(p_j(q_i(\alpha))), p_j(q_i(\alpha))\right] \ar[r]\ar[from = uu, bend right = 50, red,"\substack{\text{supported}\\ \text{ on }P_{q_i(\alpha)}}"', crossing over, near start]\ar[from = uur, red, "0", crossing over, near start]&0&
\end{tikzcd}
\end{equation}

\begin{equation}\label{diag:proofRel1-0_0}
\begin{tikzcd}
\PRes_{\alpha}\ar[d]\ar[dd, bend right = 40, red, "\neq 0", near start]:&[-15pt] \dots\ar[d]\ar[r]&\bigoplus_{i'\in S_{\alpha}}P_{q_{i'}(\alpha)}\ar[d]\ar[r]&P_{\alpha}\ar[d]\ar[r]&0\\
\PRes_{p_j(\alpha)}\ar[d]:&\dots\ar[d]\ar[r]&\bigoplus_{i'\in S_{\alpha}}P_{q_{i'}(p_j(\alpha))}\ar[d]\ar[r]&P_{p_j(\alpha)}	\ar[d]	\ar[r]&0\\
\PRes_{p_j(q_i(\alpha))}[1]:& 0\ar[r]& \left [f(p_j(q_i(\alpha))), p_j(q_i(\alpha))\right] \ar[r]\ar[from = uu, bend right = 70, red, 	"\substack{\text{supported }\\ \text{on }P_{q_i(\alpha)}}"', crossing over, near start]\ar[from = uur, red, "0", crossing over, near start]&0&
\end{tikzcd}
\end{equation}

The fact that some squares commute and other square anticommute is inconvenient for Theorem <ref>. However we will see that these signs can be corrected.

A subset $J$ of a configuration $R$ is allowed if $\sigma_J^-(R)$ is well defined. Note that if $J$ is in the positive side of $R$, $J$ corresponds to an allowed subset of $S_{\alpha}$.

Let $R_0, \dots, R_t$ be configurations such that for all $q\in\{0, \dots, t-1\}$, there exists $k_q\in R_q$ such that $R_{q+1} = \sigma^-_{k_q}(R_q)$. Then the morphism $f = u_{k_{p-1}}^{R_{p-1}}\circ \dots \circ u_{k_0}^{R_0}$ is non zero only if for all $q$, $k_q\in R_0$.

Assume that there exists $q$ such that $k_q\not\in R$. Then there exists $r< q$ with $k_r = k_q+1$. Take $q$ to be minimal and $r$ as close as possible to $q$. Take $s \in \{r, \dots, q\}$. Then, by the assumptions on $q$ and $r$, we either have $k_s > k_q+1$ or $k_s < k_q$. Whenever $s$ is such that $k_{s+1} < k_q < k_q+1 < k_{s}$, equation (<ref>) provides the equality

\[u_{k_{s+1}}\circ u_{k_s} = u_{k_s}\circ u_{k_{s+1}}.\]

Hence using equation (<ref>) we can rewrite the morphisms $f$ as

\[f = \epsilon \times ( \dots \circ u_{k_q}\circ u_{k_{s(q-1)}} \circ\dots\circ u_{k_{s(\tau+1)}} \circ u_{k_{s(\tau)}}\circ\dots\circ u_{k_{s(r+1)}} \circ u_{k_r} \circ\dots)\]

where $s$ is a permutation on the set $\{r+1, \dots , q-1\}$ and if $i > \tau$, $k_{s(i)} > k_q+1$ and $k_{s(i)} < k_q$ otherwise. Finally, we use equation (<ref>) again to get

\[f = \epsilon \times ( \dots \circ u_{k_{s(q-1)}} \circ\dots\circ u_{k_{s(\tau+1)}}\circ u_{k_q}\circ u_{k_r}  \circ u_{k_{s(\tau)}}\circ\dots\circ u_{k_{s(r+1)}} \circ\dots)\]

The central term is zero by Lemma <ref>. This concludes the the proof.

We now want to find a canonical way of decomposing morphisms into elementary ones. Knowing that elementary morphisms are indexed by certain elements of the partition $R$, we want to equip $R$ with a convenient order relation. Recall that when we defined configurations on $\mathcal{Z}$ we wrote them as increasing sequences, by equipping $\mathcal{Z}$ with the order relation on $\mathbb{Z}$. This ordering is not adapted to morphisms: for any partition $R$ containing $-m$ but not $n$, it is in general unclear weather the morphism $u^R_{-m}$ should come before or after $u^R_{k}$ when $k$ is positive. Let $R$ and $R'$ be configurations. Suppose that $R' = \sigma_{k_t}\circ\dots\circ\sigma_{k_1}(R)$, that for each $q\leq t$, $k_q\in R$ and the partition $R^q = \sigma_{k_q}\circ\dots\circ\sigma_{k_1}(\alpha)$ is well defined. We have the following composition of elementary morphisms

\[f = u_{k_t}^{R_{t-1}}\circ\dots\circ u_{k_1}^R\]

in $\mathcal{Y}_{m,n}$. Order the combinatorial data $K = (k_1, \dots, k_p)$ and by the same process the elements of $\mathcal{Z}$ as follows: let $k_{min}$ be the maximal element of $\mathcal{Z}$ for the naive order which  does not appear in K and is less than or equal to $1$. We consider the total order

\[k_{min} \prec_f k_{min} + 1 \prec_f\dots\prec_f n \prec_f -m \prec_f\dots \prec_f k_{min} - 1\]

Because we picked the combinatorial data so that the consecutive morphisms are well defined, if $k_{min} \not = 1$ then $k_{min}$ is not an element of $R$. Alternatively, if $k_{min} = 1$, either $1\in R$ and the first extension has datum $k > 2$ or $1\not\in R$ and $k> 1$. Hence, if $k$ is the minimum of $K$ with regard to $\prec_{f}$, then $k-1$ is not in $R$.
Consider two partitions $R$, $R'$ and a morphism $f$ as above with combinatorial data $k_1, \dots, k_t$ and order $\prec_f$. Suppose also that for all $1\leq q \leq t$, if $k_{q} \in R$. Then there exist a permutation $p\in \mathfrak{S}$ such that we have

\[k_{p(1)} \prec_f \dots \prec_f k_{p(t)}\]

\[f = \pm u_{k_{p(t)}}^{R_{p(t-1)}}\circ\dots\circ u_{k_{p(1)}}^{R}.\]

Recall from equation (<ref>) of Lemma <ref>, that for each $q$, if $k_q, k_{q+1} \in R^{q-2}$ while $k_{q} - 1\not\in R^{q-2}$ and $k_{q-1} - 1\not\in R^{q-2}$ then

\begin{equation*}u_{k_{q+1}}^{R^{q}}\circ u_{k_{q}}^{R^{q-1}} = \epsilon u^{\sigma_{k_{q+1}}(R^{q-1})}_{k_{q}}\circ u_{k_{q+1}}^{R^{q-1}}\end{equation*}

with the transformation $\sigma_{k_q}$ being well defined on $R^{q-2}$. We proceed by contraposition. Assume there exists $q < q' $ such that $k_{q'} = k_q - 1 $. We choose $q$ such that the value $k_q$ is minimal in $R$ with this property. It follows that $k_q - 1 \not\in R$, otherwise, there should be $q''$ such that $k_{q''} = k_q-1$ and $q''' > q''$ with $k_{q'''} = k_{q''} - 1$ contradicting the minimality of $k_q$. Similarly we can sow that $k_q = k_q'$ also contradicts the assumption. Hence when $k_q \prec_f k_{q-1}$ we do have $k_q, k_{q+1} \in R^{q-2}$ while $k_{q} - 1\not\in R^{q-2}$ and $k_{q-1} - 1\not\in R^{q-2}$. Using the bubble sort algorithm, the chain can be ordered up to a sign by applying relation $\rho_{k_q, k_{q-1}}$ (see equation (<ref>)) when $k_q \prec_f k_{q-1}$. Indeed the bubble sort algorithm <cit.> only swaps pairs of entries when they are not in the correct order. Moreover the assumptions of the lemma are maintained at each swap.

Let $R_0, \dots, R_t$ be configurations such that for all $q\in\{0, \dots, t-1\}$, there exists $k_q\in R_q$ such that $R_{q+1} = \sigma^-_{k_q}(R_q)$. Then the morphism $f = u_{k_{p-1}}^{R_{p-1}}\circ \dots \circ u_{k_0}^{R_0}$ is non zero if and only if for all $q$, $k_q\in R_0$.

Assume that $k_q\in R = R_0$ for all $0 \leq q\leq t-1$. By Lemma <ref> we can assume the elements of the chain are ordered with respect to the ordering $\prec_{f}$. Let $\tau$ be the maximal index such that $k_{\tau} \leq n$. First we show that $f_{ext} = u_{k_{\tau}}\circ\dots\circ u_{k_{1}}$ is a non zero extension. Because $k_{min}\not \in R $ or $k_{1} > 1$ (see remark before Lemma <ref>), the set $\{k_{1}, \dots, k_{\tau}\}$ is allowed as per Definition <ref> and see subsection <ref>. So $f_{ext}$ is non zero  in $\mathcal{Y}_{m,n}$ by Proposition <ref> and Lemma <ref>.
We now consider the morphism $f_{0} = u_{k_{t-1}}\circ\dots\circ u_{k_{\tau + 1}}$ which is concentrated in degree zero. To see that it is non zero we argue that its source and its target satisfy the condition of Lemma <ref> item <ref>. The source of the morphism is associated to configuration $R^{\tau} = \sigma^-_{k_{\tau}}\circ\dots\circ \sigma^-_{k_{1}}(R)$ and its target to $R' = \sigma^-_{k_{p+1}}\circ\dots\circ \sigma^-_{k_{\tau + 1}}(R_{\tau})$. Denote $\alpha, \alpha'$ the partition associated to $R, R'$. Thanks to how we chose the integer $\tau$ the two configurations identical in columns $k_{min}$ to $n-1$. It follows that, in terms of partitions we have $\{\lambda_i |i\in S_{\alpha_{\tau}}\} \subseteq \{l_j|j\in S_{\alpha'}\}$ completing the second part of Lemma <ref> item <ref>.
For the first part, let $j$ be an element of $S_{\alpha'}$ its associated value $l_j$ being an element of $R'$ and its ending index $y_j$ corresponding to the absence of $-y_j$ in $R'$. Because $R'$ is obtained from $R_{\tau}$ by moving beads of the abacus by one slot to the left, the gap in $-y_j$ was obtained by filling some gap $k\leq -y_j$ in $R_{\tau}$. The gap $k$ corresponds to an ending index $-x_i$ of the coefficient $\lambda_i$ in the partition $\alpha_{\tau}$. Moreover, $-y_i$ must be strictly less then the next gap to the right of $k$ in $R_{\tau}$ because it would require moving a beed from $-x_{i-1}$ to $-x_{i-1}-1$ and that $-x_{i-1}\not \in R_{\tau}$.
We need to show that $l_j = \lambda_i$. Because beads can be moved twice without contradicting the assumption that the $k_i's$ are in $R$ there are the same number of gaps in $R'$ and $R_{\tau}$ between column $-x_i$ and $k_{min}$. Moreover, recall that the $t^{th}$ gap in the negative side must correspond to the $t^{th}$ coefficient of the partition. There are three cases depending on whether zero is a value of the source configuration or the second. If both partitions contain the value zero, then all the gaps are in the negative side and $-x_i$ and $-y_j$ correspond to the same bead. If both partitions do not contain zero as well. Lastly, it is possible that $R_{\tau}$ contains $0$ but $R'$ does not. In that case, the $i^{th}$ coefficient in $\alpha_{\tau}$ is the $(i - 1)^{th}$ coefficient of $\alpha'$ and we again have $\lambda_i = l_j$. Hence there exists a non zero morphism from $\PRes_{R_{\tau}}$ to $\PRes_{R'}$. In turn this means that the map $f_0$ is non zero because of how composition of morphisms between intervals work.
The final step is to argue that $f = f_0\circ f_{ext}$ is non zero in $\mathcal{Y}_{m,n}$. Then we have showed that there exists $J\subseteq S_{\alpha}$ allowed such that there exists a non zero morphism from $\PRes_{q_J(\alpha)}$ to $\PRes_{\alpha'}$. Moreover, if $1 = \lambda_{\epsilon} \in R$ and $\epsilon\in J$, then $0$ is an element of $R'$ because the bead associated to $1$ can only be moved once. The same is true for all the beads associated to zero if it was an element of $R$. Hence if $\epsilon \in J$ and $\lambda_{\epsilon} = 1$ then the partition $\beta$ should have zerp for its first value i.e $l_1 = 0$ and $y_1 > x_{\epsilon - 1}$ and we can apply Lemma <ref> to conclude that there exists a morphism in the derived category from $\PRes_{\alpha}$ to $\PRes_{\alpha'}$. Moreover, by Theorem <ref>, the set $J$ that we have identified is the unique subset of $S_{\alpha}$ such that $q_J(\alpha)\in [f(\alpha', \alpha']$. By the proof of Proposition <ref> $f = f_0\circ f_{ext}$ is non zero. This concludes the proof.

By Lemmas <ref>, <ref> and <ref> all morphisms decompose into an extention postcomposed by a degree zero morphism. In turn these can be further decomposed into the elementary morphsims we have identified. The list we obtain is called the canonical list associated to a non zero morphism. Its ordering coinsides with $\prec_f$. Its positive elements are the elements of the unique subset $J$ of $S_{\alpha}$ such that $q_J(\alpha)\in [f(\beta), \beta]$ gives positive elements of the canonical list. Then, the inequalities between the ending indices described in item <ref> of Lemma <ref> gives the remainer of the canonical list by looking at the transformation associated to the integers $]-x_i, -y_j]$.

As a consequence we have the following proposition
The relations described in equation (<ref>) generate the relations between morphisms in the category  $\mathcal{Y}_{m,n}$.

Consider the category $C$ defined as follows:

* the objects of $C$ are pairs $(R, l)$ where $R$ is a configuration and $l$ is an integer;

* the morphisms are generated by arrows $(R, l)\to(\sigma_k^-(R), l')$ with $l' = l + 1$ if $k>0$ and $l' = l$ otherwise,

* with relations $\rho_{k,l}$ and $z_k$ identified in Lemma <ref>.

By the same lemma there is a well defined essentially surjective functor $F: C\to \mathcal{Y}_{m,n}$. To prove the current proposition we need to argue that it is an equivalence of categories. By Proposition <ref>, Lemma <ref> and Lemma <ref> combined with Proposition <ref> the functor induces surjective maps between the hom sets $\Hom_{C}(R, R')$ and $\Hom_{D^b(J_{m,n})}(\PRes_R, \PRes_{R'})$. It remains to see that this map is injective. Consider a element

\begin{equation}\label{eq:genrel}
\sum_{i=1}^q a_i\cdot f_i \text{ of } \Hom_{C}(R, R')
\end{equation}