Weakly Cohen-Macaulay Posets and a Class of Algebras

# Weakly Cohen-Macaulay posets and a class of finite-dimensional graded quadratic algebras

Tyler Kloefkorn Department of Mathematics
University of Arizona
Tucson, AZ 85721
April 1, 2016
###### Abstract.

To a finite ranked poset we associate a finite-dimensional graded quadratic algebra . Assuming satisfies a combinatorial condition known as uniform, is related to a well-known algebra, the splitting algebra . First introduced by Gelfand, Retakh, Serconek, and Wilson, splitting algebras originated from the problem of factoring non-commuting polynomials. Given a finite ranked poset , we ask: Is Koszul? The Koszulity of is related to a combinatorial topology property of called Cohen-Macaulay. Kloefkorn and Shelton proved that if is a finite ranked cyclic poset, then is Cohen-Macaulay if and only if is uniform and is Koszul. We define a new generalization of Cohen-Macaulay, weakly Cohen-Macaulay, and we note that this new class includes posets with disconnected open subintervals. We prove: if is a finite ranked cyclic poset, then is weakly Cohen-Macaulay if and only if is Koszul.

###### Key words and phrases:
Koszul algebra, splitting algebra, Cohen-Macaulay poset
###### 2010 Mathematics Subject Classification:
Primary: 16S37, Secondary: 05E15

## 1. Introduction

We fix a field . Let denote a finite ranked poset with unique minimal element and order . If , we write if covers .

###### Definition 1.1.

The algebra is the graded quadratic -algebra with degree one generators for all and relations

 rx∑x→yry=0andrxrw=0 whenever x↛w.

The algebra has a relatively complex history – we give a brief summary. In [4], Gelfand, Retakh, Serconek and Wilson associate to a connected graded -algebra , which is called the splitting algebra of ; splitting algebras are related to the problem of factoring non-commuting polynomials. Retakh, Serconek and Wilson later showed that if satisfies a combinatorial condition called uniform, then an associated graded algebra of the splitting algebra is quadratic, and it follows that is quadratic (c.f. [10]). These authors then asked a standard question in homological algebra: given a finite uniform ranked , is Koszul?

###### Definition 1.2.

A connected graded -algebra is Koszul if the trivial right -module admits a linear projective resolution.

We note that Koszul algebras are quadratic and that there are many equivalent definitions (c.f. [8]).

If is Koszul, one can use the Hilbert series condition of numerical Koszulity to extract combinatorial data from the algebra. Recent work in the area of splitting algebras often focuses on calculating Hilbert series (c.f. [11] and [12]).

Given a specific , it is difficult to determine if is Koszul. In fact, preliminary literature incorrectly asserted that is Koszul for all uniform . We thus pass to a related question. Following [10] and assuming is uniform, we filter by rank in . We denote the associated graded algebra by . Finally, we study the quadratic dual of , which is denoted by . Applying standard techniques, we know that if is Koszul, then so is . We then ask: given a finite uniform ranked , is Koszul? We note in [11] and subsequent papers by the same authors, is denoted by .

If is uniform, then . The notation is from [2]; Cassidy, Phan and Shelton assume is uniform and denote with . We emphasize: for Definition 1.1, need not be uniform.

We are interested in the algebra and Koszulity of , even if is not uniform and we draw no conclusions about splitting algebras. In [2], Cassidy, Phan and Shelton show that there exists a non-Koszul . Also, if stems from a geometric object, then the Koszulity of gives important combinatorial and topological data; for example, the authors show that if is the intersection poset of a regular CW complex, then is Koszul. Sadofsky and Shelton study posets associated to regular CW complexes in [14] – they show Koszulity for is a topological invariant.

Kloefkorn and Shelton found connections between and a combinatorial topology property known as Cohen-Macaulay. We say is cyclic if it has a unique maximal element. If in , then denotes the order complex of .

###### Definition 1.3.

A finite ranked cyclic poset is Cohen-Macaulay relative to if for all in , for all .

A finite ranked cyclic poset is Cohen-Macaulay if every open interval is, as we say, a cohomology bouquet of spheres (or CBS). The Cohen-Macaulay property has been studied extensively; for a survey of all things Cohen-Macaulay, see [1].

It is worth noting that both the Cohen-Macaulay and Koszul properties are relative to the field .

In [5], Kloefkorn and Shelton proved the following theorem.

###### Theorem 1.4.

Let be finite ranked cyclic poset. Then is Cohen-Macaulay if and only if is uniform and is Koszul.

In this paper we define a new generalization of the Cohen-Macaulay property: weakly Cohen-Macaulay. The main theorem is as follows.

###### Theorem 1.5.

Let be finite ranked cyclic poset. Then is weakly Cohen-Macaulay if and only if is Koszul.

This paper is organized as follows. Section 2 gives important background information. Sections 3 through 5 build the necessary machinery for our main theorem. Section 6 gives the definition of weakly Cohen-Macaulay and the statement of our main theorem. Finally, Section 7 gives important examples and remarks.

## 2. Definitions and preliminaries

###### Definition 2.1.

Let be a poset with unique minimal element and strict order . We say is ranked if for all , any two maximal chains in have the same length. The length of such a maximal chain is then referred to as the rank of and written . Set . Let be the elements of of rank .

(1) is pure of rank if for every maximal element of .

(2) If is pure, then is the poset adjoined with a unique maximal element.

(3) If is pure, then is the poset .

(4) denotes the interval in .

(5) is cyclic if for some .

(6) For any , .

(7) For each we set .

For any , we say that covers , written , if the closed interval has order 2, or equivalently . This makes into a directed graph that is often referred to as a layered graph.

The above definition of ranked poset is taken from [4], a fundamental paper in the area of splitting algebras. We note that this definition differs from the traditional one wherein every maximal chain has the same length (c.f. [16]).

From now forward, denotes a finite ranked poset and we will assume that has unique minimal element (without explicitly stating so).

We recall the definition of uniform from [4].

###### Definition 2.2.

For and , write if there exists and extend to an equivalence relation on . We say is uniform if, for every , has a unique equivalence class.

The order complex of a finite poset is a standard tool in combinatorial topology and we refer the reader to [5] for basic definitions. We denote the order complex of by . The order complex of is a simplicial complex and its geometric realization, or total space is denoted . All cohomology groups are all calculated with coefficients in our base field, .

Recall from Definition 1.3 that a finite ranked cyclic poset is Cohen-Macaulay relative to a field if the order complex of any open subinterval has non-zero reduced cohomology only in the degree equal to its dimension. Before giving an equivalent definition of Cohen-Macaulay, we recall notation from [11].

###### Definition 2.3.

For any and we set

 Γa,i={wrkΓ(a)−i∩(∗,a)

From [5], we note: is a subposet of and the dimension of is . Also, , and .

From [15], we now give an equivalent definition of uniform.

###### Proposition 2.4.

Let be a finite ranked poset. Then is uniform if and only if for all of rank at least three, is connected as a graph.

Similarly, we give an equivalent definition of Cohen-Macaulay. This statement and its proof are slightly different from those in [15].

###### Proposition 2.5.

Let be a finite ranked cyclic poset. Then is Cohen-Macaulay if and only if for all and all , .

###### Proof.

Assume is Cohen-Macaulay and let . By Theorem 1.4, is Koszul, from which it follows that is Koszul. Applying Theorem 2.7 from [5], we see for all . Theorem 4.2 together with Lemma 6.3 from [5] tell us that for all , which completes the proof of the forward direction.

For the reverse direction, we proceed by induction on the rank of . For a poset of rank one, there is nothing to show; we assume has rank with .

By the inductive hypothesis, is Cohen-Macaulay for all in . Let . We observed that is a closed subspace of . We obtain the standard long exact sequence

 ⋯→~Hn(Δ(Γb,m−1))→~Hn+1(Δ(Γb,m),Δ(Γb,m−1))→~Hn+1(Δ(Γb,m))→⋯.

Then, by assumption, for all . Similar to Lemma 6.4 from [5], we get

 Hn(Δ(Γb,m),Δ(Γb,m−1))=⨁a∈Sb(m−1)~Hn−1(Δ((a,b))).

This implies

 ⨁a∈Sb(m−1)~Hj(Δ((a,b)))=0

for all . This implies is Cohen-Macaulay. ∎

We now turn our attention to form Definition 1.1. Cassidy, Phan, and Shelton proved the following lemma in [2]. This lemma is a very powerful tool; we will use it repeatedly and without further comment. We note that need not be uniform.

###### Lemma 2.6 ([2], (3.1)).

Let be a finite ranked poset. Then

 (RΓ)+=⨁x∈Γ+rxRΓ.

We adopt the following notation and conventions from [5].

###### Definition 2.7 ([5], (2.5)).

Let be a finite ranked poset.

(1) . Also let denote the function given by left (but never right) multiplication by .

(2) For all , set .

We note that the space denotes the span of such monomials with and and the degree of such a monomial is . Since , for each we have a cochain complex:

 ⋯RΓ(n−1,k)dΓ−→RΓ(n,k)dΓ−→RΓ(n+1,k)⋯.

## 3. The Right Annihilator Condition

Throughout the remainder of this paper denotes an arbitrary finite ranked poset with unique minimal element and rank .

We remind the reader of the following fact from Section 3 in [2]: for a uniform , the Koszulity of is equivalent to a condition on right annihilators of certain elements of . In this section, we establish a more general result. We first need some notation and some new combinatorial objects.

###### Definition 3.1.

Let . Define the linearly generated right annihilator of , .

###### Definition 3.2.

Let and . Then

 rW:=∑z∈Wrz.

For all we write . Suppose . We write if for all

 q=∑z∈Γ+qzrz,(qz∈F)

implies .

We note defines an equivalence relation on . Equivalence classes will be denoted by and thus

 r[z0]W=∑z∈[z0]Wrz.
###### Proposition 3.3.

Let and be given as in Definition 3.2. Then

 LΓ(rW)=⨁[z0]Wr[z0]WRΓ.
###### Proof.

Suppose

 rW⋅∑z∈Γ+qzrz=0.

Then

 ∑z∈Γ+qzrz=∑[z0]Wq[z0]Wr[z0]W∈⨁[z0]Wr[z0]WRΓ.

We also need to show for all . If , then and . Now assume . We partition into two sets: and is the compliment of in . Then we compute

 rW⋅r[z0]W=(rW1+rW2)r[z0]W=rW1r[z0]W=0

and this completes our proof. ∎

###### Definition 3.4.

Let and be given as in Definition 3.2. Let . We define

 AΓ(rW):=⨁[z0]W⊆Ur[z0]WRΓ,BΓ(rW):=⨁z∈Γ(n−1)∖UrzRΓ

and

 HΓ(n−1):=⨁z∈Γ+∖Γ(n−1)rzRΓ.

We observe:

 LΓ(rW)=AΓ(rW)⊕BΓ(rW)⊕HΓ(n−1).
###### Definition 3.5.

Assume . We define

 Tm+1(Γ)={Γ(m+1)}={{x}}

and recursively define for

 Ti(Γ)={[z0]W⊂Γ+|W∈Ti+1 and there % exists s∈W such that s→z0}.

Then define

 T(Γ)=m+1⋃i=1Ti(Γ).

For an arbitrary finite ranked poset, we define

 T(Γ)=⋃x∈Γ+T(Γx).
###### Remark 3.6.

By (3.3) from [2], we see that if is uniform, then consists of sets of the form where and .

The following lemma is (3.3) of [2], without the uniform hypothesis.

###### Theorem 3.7.

Let be a finite ranked poset. The algebra is Koszul if and only if for all , .

###### Proof.

For the forward direction, assume is Koszul. Then the right -modules and are Koszul. Let , so there is and with . By (reverse) induction on , we will simultaneously prove and is a Koszul module.

If , then . By definition of , . is a direct summand of by Lemma 2.6, thus it is a Koszul module.

We now assume . Then there exists , and such that and . By induction, . We then have a short exact sequence

 0→LΓ(rU)→RΓ→rURΓ→0.

Again, by induction, is Koszul. Thus is a Koszul module. The module is a direct summand of and therefore, is a Koszul module. Now is linearly generated and . This completes the proof of the forward direction.

We now prove the reverse direction. Let be the set of right ideals in of the form

 I=m⨁i=1rWiRΓ

where are pairwise disjoint. We include the zero ideal in and observe . To prove that is Koszul, we show that is a Koszul filtration (c.f. [7]).

Let

 0≠I=m⨁i=1rWiRΓ∈F.

Then set

 J=m−1⨁i=1rWiRΓ

and observe and . Also, the right ideal (the conductor of into ) is equal to . Since , . We conclude is a Koszul filtration, which completes our proof. ∎

The following lemma is (3.5) from [2], again, without the uniform hypothesis.

###### Theorem 3.8.

Let be a finite ranked poset. Then is Koszul if and only if is Koszul for all .

###### Proof.

Let and . By Lemma 2.6,

 rannRΓ(rW)∩RΓx=rannRΓx(rW).

The theorem then follows from Theorem 3.7. ∎

###### Corollary 3.9.

Let be a finite ranked poset with rank less than or equal to three. Then is Koszul.

###### Remark 3.10.

We often use the following fact, which is Theorem 3.7 together with Theorem 3.8. Suppose is a finite ranked cyclic poset with . Set and assume is Koszul. Then is Koszul if and only if for all , .

We thus have a generalized right annihilator condition, but we wish to refine it. In order to do so, we first define additional notation for .

###### Definition 3.11.

Let and . We set

 ΓW={a∈Γ:a≤s for some s∈W}=⋃s∈W[∗,s].

Also, for , we set

 Γ(W,k)=ΓW∩Γ≥n−k.

We say is linked if is connected as a graph. We say is maximally linked relative to if is maximal amongst linked subsets of .

The collection of all maximally linked subsets of forms a partition of .

###### Lemma 3.12.

Let and be given as in Definition 3.11. Then if and only if for all maximally linked subsets , .

###### Proof.

Due to the observation immediately following Definition 3.4, it is enough to prove the lemma for . We therefore assume . We enumerate the maximally linked subsets of : . For all , set .

For the forward direction, suppose towards a contradiction that for some , there exists . Then with

Clearly , thus . Also, is nonzero and ; if not, then . We compute

 rWA1=(rWk+rΓ(n)∖Wk)A1=rWkA1=0.

Therefore, . Also, we see since . Thus , which is a contradiction.

For the converse, we assume for all . Carefully applying Definition 3.4, we compute

 rannRΓ(rW) =p⋂i=1rannRΓ(rWi) =p⋂i=1LΓ(rWi) =p⨁i=1AΓ(rWi)⊕HΓ(n−1) =⨁[z0]Wr[z0]WRΓ =LΓ(rW),

which completes our proof. ∎

###### Lemma 3.13.

Let and be given as in Definition 3.11. Assume is linked. Then if and only if for all , .

###### Proof.

Again, it is enough to prove the claim for . We set . Then if and only if

 rannRΓ(rW)=rΓ(n−1)RΓ⊕HΓ(n−1).

This equality holds if and only if {diagram} R_Γ(n-3, k) & \rTo^r_Γ(n-1) ⋅ & R_Γ(n-2, k) & \rTo^r_W ⋅ & R_Γ(n-1, k) is exact for all . ∎

###### Definition 3.14.

Assume . We define

 Mn(Γ)=⋃W∈Tn(Γ){W′:W′ is % maximally linked relative to W}

and

 M(Γ)=m+1⋃n=1Mn(Γ).
###### Remark 3.15.

Using Remark 3.6, we see that if is uniform then for all .

We combine the definition of with Lemmas 3.12 and 3.13 and Remark 3.10 for the following theorem about the Koszulity of .

###### Theorem 3.16.

Let be a finite ranked poset and assume . Set and assume is Koszul. The following are equivalent:

1. is Koszul;

2. for all , ;

3. for all , and , .

We end this section with an example that motivates the definition of weakly Cohen-Macaulay and our main theorem.

###### Example 3.17.

Let be the ranked poset shown in Figure 2.

This poset was first introduced by Cassidy and was studied extensively in [6]. By inspection, we see is uniform. Also, by inspection, we know is not Cohen-Macaulay; is homotopic to . By Theorem 1.4, is not Koszul.

Let be the dual poset of , as shown in Figure 2. Since is disconnected as a graph, is not uniform and is not Cohen-Macaulay. Also, is homotopic to .

We claim is Koszul. By Theorem 3.16, it suffices to show that is linearly generated. We see . Then, by inspection,

 rannRΘ∗(rm+rn)=(rs+rp+rq)RΘ∗⊕HΘ∗(2).

## 4. A Spectral Sequence Associated to Δ(Γ∖{∗})

Let . We start by defining an unusual filtration on . Let be an increasing filtration on given by

 FpCn(Y)={(α0,…,αn)∈Cn(Y):rkΓ(αn)≥(m+1)−p}.

We note: consists of all co-chains of that emanate from the top layers of . Also, observe

 0=F−1C∙(Y)⊂F0C∙(Y)⊂⋯⊂Fm−1C∙(Y)⊂FmC∙(Y)=C∙(Y).

Let denote the associated cohomology spectral sequence with

It is bounded since if , , or . The differentials are maps induced by the differential on .

###### Proposition 4.1.

Let be a ranked poset. Let be the spectral sequence of the filtration on . For all and , . Also .

###### Proof.

The equality holds because is the zero map on . The convergence statement follows from Theorem 5.5.1 of [17]. ∎

The page of our spectral sequence can be written in terms of reduced cohomology of open intervals of .

###### Proposition 4.2.

Let be a finite ranked poset. Let be the spectral sequence of the filtration on . For all and ,

 E1p,q=⨁rkΓ(x)=(m+1)−p~Hp+q−1(Δ((∗,x))).
###### Proof.

We fix and identify two chain complexes: and

 ⎛⎝⨁rkΓ(x)=(m+1)−p~C∙(Δ((∗,x))),⨁rkΓ(x)=(m+1)−pdΔ((∗,x))⎞⎠.

For convenience of notation, elements of will be denoted by sums of chains of the form . Define a map

 f:E0p,q=FpCp+q(Y)/Fp−1Cp+q(Y)→⨁rkΓ(x)=(m+1)−p~Cp+q−1(Δ((∗,x)))

via and if

 (α0,…,αp+q)↦(α0,…,αp+q−1)αp+q,

and extend linearly. It is easy to see that is bijective. We claim is a co-chain map. Let and note . We compute

 ⨁rkΓ(x)=(m+1)−pdΔ((∗,x))(f(α0,…,αp+q)) =⨁rkΓ(x)=(m+1)−pdΔ((∗,x))((α0,…,αp+q−1)αp+q) =dΔ((∗,αp+q))(α0,…,αp+q−1)

and on the other hand

 f(d0E(α0,…,αp+q)) =f(∑x<α0(x,α0,…,αp+q)) =∑x<α0(x,α0,…,αp+q−1)αp+q +p+q−1∑i=0(−1)i+1∑αi

It follows that is a co-chain isomorphism. ∎

We now make an observation related to Theorem 1.4.

###### Corollary 4.3.

Let be a finite ranked cyclic poset and let be the spectral sequence of the filtration on . Assume is Cohen-Macaulay. Then unless . Moreover, .

The first page of our spectral sequence is important for our main theorem, so we will introduce some convenient notation.

###### Definition 4.4.

For , define

 Sn(Γ+)=⨁rkΓ(x)=n+1~Hn−1(Δ((∗,x))).

For , we denote elements in with linear combinations of equivalence classes of the form . This notation requires: . Elements in will be denoted by scalar multiples of .

Define by extending linearly from the formula

 [1]x↦⨁x←y[x]y

and for

 [x1←⋯←xn]x↦⨁x←y(−1)n[x1←⋯←xn←x]y.

It is worth noting that we will later study the spaces and , which is consistent with our above notation.

###### Proposition 4.5.

The map is well-defined.

###### Proof.

If , the map is clearly well-defined. Assume . It is sufficient to observe the inclusion

 dSn(Γ+)(im(dn−2Δ((∗,x)):Cn−2(Δ((∗,x)))→Cn−1(Δ((∗,x))))) ⊂im(dn−1Δ((∗,z)):Cn−1(Δ((∗,z)))→Cn(Δ((∗,z)))).

for all . ∎

###### Proposition 4.6.

is a co-chain complex.

###### Proof.

Let . Then

 dS1(Γ+)∘dS0(Γ+)([1]x) =dS1(Γ+)(⨁x←y[x]y) =⨁x

since

 d0Δ((∗,z))(x)=−∑x←y←z(x,y).

for all .

Assume . Let