An equality for the reduced homology related to the comparability graph of a poset

Question

$\DeclareMathOperator\width{width}$Let $P$ be a finite poset with $n$ elements (we can assume that $P$ is connected and has width at most $n-2$). The comparability graph $G_P=(V,E)$ associated to $P$ is by definition the finite graph with vertices $V=P$ and two elements $v, w \in V$ are adjacent if $v<w$ or $w<v$. A subset $S$ of $V$ is called a dominating set in $G$ if every vertex $v$ in $G$ belongs to $S$ or is adjacent to an element of $S$.

The dominance complex $D(G_P)$ is the simplicial complex consisting of the subsets of $V$ whose complements are dominating. We look at homology of such a simplicial complex over fields here (see Perkinson - Homology of Simplicial Complexes for a definition).

Question (Formulation 1): Is it true that the first degree in which the reduced homology is non-zero of $D(G_P)$ is equal to $n-1-\width(P)$?

(Here $\width(P)$ denotes the maximal cardinality of an antichain in the poset $P$.)

By theorem 1 in Matsushita - Dominance complex and vertex cover number, we should have that the first non-zero degree of the reduced homology is less than or equal to $n-1-\width(P)$. And the question asks whether we have equality in theorem 1 of the previous article in the case of graphs given as comparability graphs of finite posets.

Here is a more direct alternative formulation of the problem:

Let $P$ be a finite poset with $n$ elements (we can assume that $P$ is connected and the width of $P$ is at most $n-2$). For $p \in P$ set $J(p):=\{q \in P \mid p \nleq q \}$ and $I(p):= \{ q \in P \mid q \leq p \}$. For a subset $S \subseteq P$ set $J(S) := \bigcap\limits_{p \in S}{J(p)}$ and $I(S):= \bigcup\limits_{p \in S}{I(p)}$.

Then the simplicial complex $\Gamma(P)$ associated to $P$ is defined by the condition $S \in \Gamma(P)$ if and only if $J(S^c) \subseteq I(S^c)$.

Question (Formulation 2): Is it true that the first non-zero positive degree of the homology of $\Gamma(P)$ appears at $n-1-\width(P)$?

The question is tested with a computer and true for all posets with at most 10 elements (thus for nearly 3 million examples).

(Background: The original formulation of the question is Formulation 2 and the connection to the article Matsushita - Dominance complex and vertex cover number was noted by Hugh Thomas.)

Answer 1

I made a Sage program to check whether the question has a positive answer for a given graph.

I found no counterexample yet for comparability graphs of posets. Here is the program to test it for a randomly generated poset:

P = posets.RandomPoset(14, 0.3)
display(P)
G = P.comparability_graph();
display(G)
def test_homologygraph(G):
    U = Subsets(G)
    GGU=Set(G)
    T=[t for t in U if G.is_dominating(GGU.difference(t))==true]
    TT=[list(r) for r in T]
    S=SimplicialComplex(TT)
    W=S.homology(base_ring=GF(2))
    tt=len(W)
    F=[dimension(W[t]) for t in [0..tt-1]]
    FF=[r for r in [0..tt-1] if F[r]>0]
    uu=min(FF)
    return(uu+1)

def conjecturetestforperfectgraph(G):
    oo=test_homologygraph(G)
    oo2=G.vertex_cover(value_only=True)
    return(oo==oo2)

display(conjecturetestforperfectgraph(G))

But it gave a counterexample for perfect graphs (one with 6 elements). Note that this is not part of the question in this thread but was a suggested in the comments as a generalisation of the question.

def test_homologygraph(G):
    U = Subsets(G)
    GGU=Set(G)
    T=[t for t in U if G.is_dominating(GGU.difference(t))==true]
    TT=[list(r) for r in T]
    S=SimplicialComplex(TT)
    W=S.homology(base_ring=GF(2))
    tt=len(W)
    F=[dimension(W[t]) for t in [0..tt-1]]
    FF=[r for r in [0..tt-1] if F[r]>0]
    uu=min(FF)
    return(uu+1)

def conjecturetestforperfectgraph(G):
    oo=test_homologygraph(G)
    oo2=G.vertex_cover(value_only=True)
    return(oo==oo2)
n=6
U=graphs(n)
UU=[g for g in U if g.is_perfect()]
n=len(UU)
UU2=[UU[t] for t in [1..n-1]]
UU3=[g for g in UU2 if conjecturetestforperfectgraph(g)==false]
G=UU3[0]
display(G)
display(conjecturetestforperfectgraph(G))

edit: Here a (new) program to test the conjecture for posets with $n$ points:

n=7
def test_homologyposet(P):
    G = P.comparability_graph();
    U = Subsets(G)
    GGU=Set(G)
    T=[t for t in U if G.is_dominating(GGU.difference(t))==true]
    TT=[list(r) for r in T]
    S=SimplicialComplex(TT)
    W=S.homology(base_ring=GF(2))
    tt=len(W)
    F=[dimension(W[t]) for t in [0..tt-1]]
    FF=[r for r in [0..tt-1] if F[r]>0]
    uu=min(FF)
    return(uu+1)

def conjecturetestforposet(P):
    G= P.comparability_graph();
    oo=test_homologyposet(P)
    oo2=G.vertex_cover(value_only=True)
    return(oo==oo2)

PP=Posets(n)
PP1=[p for p in PP if len(p)>p.width()+1]
PP2=[p for p in PP1 if conjecturetestforposet(p)==true]
display(PP1==PP2) 

Here the example for the join of two diamonds with 5 elements:

P1=posets.DiamondPoset(5)
P2=posets.DiamondPoset(5)
U=P1.ordinal_sum(P2)
display(U)
def test_homologyposet(P):
    G = P.comparability_graph();
    U = Subsets(G)
    GGU=Set(G)
    T=[t for t in U if G.is_dominating(GGU.difference(t))==true]
    TT=[list(r) for r in T]
    S=SimplicialComplex(TT)
    W=S.homology(base_ring=GF(2))
    tt=len(W)
    F=[dimension(W[t]) for t in [0..tt-1]]
    FF=[r for r in [0..tt-1] if F[r]>0]
    uu=min(FF)
    return(uu+1)

def conjecturetestforposet(P):
    G= P.comparability_graph();
    oo=test_homologyposet(P)
    oo2=G.vertex_cover(value_only=True)
    return(oo==oo2)

display(conjecturetestforposet(U))

Here how to calculate the homology over a field K of a given poset P (here a random poset):

P = posets.RandomPoset(14, 0.3)
display(P)
G = P.comparability_graph();
display(G)

def posethomology(G,K):
    U = Subsets(G)
    GGU=Set(G)
    T=[t for t in U if G.is_dominating(GGU.difference(t))==true]
    TT=[list(r) for r in T]
    S=SimplicialComplex(TT)
    W=S.homology(base_ring=K)
    return(W)

display(posethomology(G,ZZ))

display(posethomology(G,QQ))

display(posethomology(G,GF(2)))

Some random examples revealed that it can happen that there is a unique non-zero degree in which the homology has dimension/rank more than one.

Answer 2

This is too long for a comment so I post it as a community wiki answer.

I recently saw a result in the book "Ordered sets" by Schröder and I wonder whether it can be applied when choosing the right point v: Let $K=(V,S)$ be a (finite always) simplicial complex. Then the link complex $(N(v) \setminus \{v \},Lk(v))$ of a point $v$ is the simplicial complex with $N(v):= \{ w \in V | \{v,w \} \in S \}$ and $Lk(v)= \{ b \in S | \{v\} \cup b \in S , \{v\} \cap b = \emptyset \}$.

Then theorem A.18 in the book states:

Theorem: Let $K=(V,S)$ be a simplicial complex, $q \geq 1$ and let $v \in V$ be such that $H_q((N(v) \setminus \{v \},Lk(v)))= \{0 \}$. Then the followng hold:

a) If $q=1$ and $H_0((N(v) \setminus \{v \},Lk(v)))= \mathbb{Z}$, then $H_1(K)=H_1(K [V \setminus \{v \}])$.

b) If $q \geq 2$ and $H_{q-1}((N(v) \setminus \{v \},Lk(v)))= \{0\}$, then $H_q(K)= H_q(K[V \setminus \{v \}])$.

Corollary A.19 then states :

Corollary: Let $K(V,S)$ be a simplicial complex and $v \in V$ such that the link complex $(N(v) \setminus \{v \},Lk(v))$ is acyclic. Then, for all $q \in \mathbb{N} \cup \{0\}$, we have $H_q(K)=H_q(K[V \setminus \{v\}])$.

Maybe someone has an idea for a good choice of $v$ to use induction or another argument that might at least work for certain posets to prove the statement for the comparability graphs (or the alexander dual equivalent statement as in the comment by Geva Yashfe). I will do some computer tests whether such a thing can work.