Reflexive graphs with Near-Unanimity but no Semilattice Polymorphisms

We show that every generator, in a certain set of generators for the variety of reflexive near unanimity graphs, admits a semilattice polymorphism. We then find a retract of a product of such graphs (paths, in fact) that has no semilattice polymorphism. This verifies for reflexive graphs that the variety of graphs with semilattice polymorpisms does not contain the variety of graphs with near-unanimity, or even 3-ary near-unanimity polymorphisms. Mathematics Subject Classifications: 05C75, 08B05


Introduction
For relational structures such as graphs, the existence of relation preserving operations, or polymorphisms, satisfying various identities has been of great interest recently due to its relation to the complexity of the decision problem of homomorphism to the given structure.We refer the reader to [6] for a general discussion of such topics, to [9] for a discussion of the results on general digraphs, or to [5] and [8] for more concise discussion directly related to the present paper.
In this paper we look at near-unanimity (NU), and semilattice (SL) polymorphisms on reflexive graphs.For context also talk of totally symmetric idempotent (TSI) polymorphsims.The necessary definitions of these are given in the next section.
It is a trivial fact that any structure with an SL polymorphism has a TSI polymorphsim, and it is known, see [11], that the converse is not generally true.Moreover, there are structures admitting SL (and so TSI) polymorphsims, but not NU polymorphsims, and vice versa.
When one restricts ones scope to reflexive graphs though, things change.It is known, see for example [13], that any graph having an NU polymorphism also has TSI polymorphisms of all arities.Moreover it was shown in [5] that any reflexive graph with a NU polymorphism has a symmetric NU polymorphism.It is natural to ask if the existence of an NU polymorphism on a reflexive graph might imply the existence of an SL polymorphism, or vice-versa.Indeed, it was asked in [11] and again in [9] if there are posets (another subfamily of reflexive digraphs) that are NU but not SL.For the rest of the paper, all graph are reflexive and symmetric.
The class, NU, of graphs admitting NU polymorphisms has been well studied; see, for example, [1], [3], [5], [10], and [12].In [3] for example, it was verified that the class NU is a variety, i.e., is closed under products and retractions, as is the class k-NU of graphs admitting k-ary NU polymorphisms, for all k ≥ 3. The variety (k + 1)-NU contains k-NU for all k.It was also shown that every chordal graph is in k-NU for some k, but also that for every k there chordal graphs in (k + 1)-NU but not k-NU.In [5] an explicit description of the generators of the variety k-NU was given for all k ≥ 3.
The class SL of graphs admitting SL polymorphisms, on the other hand, has not been so extensively studied.It has only been looked at recently in [8] and in a more specialised context in [14].In [8], we showed that SL contains the class of chordal graphs.We also verified that the class SL is not closed under retraction, so though it is closed under products, it is not a variety.This shows that it is different from the classes TSI and NU.We also found graphs in SL that are not in NU, and asked, as [9] did for posets, whether or not every reflexive graph in NU must also be in SL.In this paper, we answer this question in the negative.
In Proposition 3.2, we observe that the generators of the variety k-NU, found in [5], all have SL polymorphisms.This is, of course, a first step towards showing that k-NU is in SL.Our second result however, Theorem 4.1, answering the question above, shows that this is not true.We find a retracton of a product of paths (the generators of 3-NU) which does not have SL.This shows that 3-NU, and so NU, is not contained in SL.
Proposition 3.2 is not simply a ploy for building tension before surpising the reader with Theorem 4.1.It also yields some alternate proofs of known facts, which we discuss briefly now, and raises some questions that we talk of in Section 5.
It was observed in [4] that every NU structure H is the retraction of some universal structure U TSI (H) which can easily be shown to admit an SL polymorphism.So every NU graph is a retract of an SL graph.This follows also from Proposition 3.2.The graph U TSI (H) is large as its vertices all subsets of vertices of H; our result generally embeds an NU graph as a retract of a much smaller SL graph.
In [3] it was shown that there are chordal graphs in k-NU but not (k − 1)-NU for all k ≥ 4. As chordal graphs were shown to have SL polymorphisms in [8], it follows that there are SL graphs that are in k-NU but not (k − 1)-NU.Corollary 3.3 points out how this also follows from Proposition 3.2, but the examples it provides are far from chordal, and the proof is much different.
In Section 2 we introduce the required definitions.In Section 3 we introduce the generators of k-NU from [5] and prove Proposition 3.2.In Section 4 we prove Theorem 4.1.Finally, in Section 5 we ask some questions.

Basics
2.1.Semilattices.In this subsection we recall some standard definitions related to semilattices.
A semilattice on a set V can alternately be described as a an ordering ≤ such that for every pair u, v ∈ V there is a unique greatest lower bound denoted u ∧ v; or as a 2-ary function , symmetric, and associative.We thus use '∧' and '≤' interchangably, and may refer to either of them as a semilattice ordering.
It is well known, and easily verified, that the two defininition are related through the identity 2016/06/06 Preprint of submitted article

REFLEXIVE GRAPHS WITH NEAR-UNANIMITY BUT NO SEMILATTICE POLYMORPHISMS 3
As our semilattices are finite, the existence of a lower bound for every pair of elements extends by The width of an semilattice is the maximimum number of pairwise incomparible elements.An element v covers or is a cover of an element u if v ≥ u and there is no 2.2.Semilattice polymorphisms.We denote the adjacency of two vertices u and v of a graph by , on the vertex set V (G), which satisfies the following for all choices of u i , v i ∈ V (G).
A reflexive graph G is an SL graph if it admits a semilattice polymorphism, and SL is the class of all SL graphs.The (categorical) product of two graphs G 1 and G 2 is the graph As we mentioned above, it was verified in [8] that SL is closed under taking products by verifying the following standard fact; though it was shown that SL is not closed under retractions.
We will frequently use products of paths.Let P denote the path of length having vertex set [0, ], where a ∼ b if a − b ≤ 1. Figure 1 shows the product P = P 3 × P 3 of two 3-paths, and a typical retract R of P. Though the graph is reflexive, we have omitted all loops from the figure.We will do so on all figures.
Given a semilattice (V, ∧), a sub-semilattice consists of a subset In particular, an SL polymorphism of an graph G induces a sub-semilattice on any conservative set.So the subgraph of any SL graph induced by any conservative set is also an SL graph.It is well known that the i th distance neighbourhood N i (v) of a reflexive graph G, consisting of all vertices that are distance at most i from a vertex v, are conservative.It is also known that the intersection of conservative sets is conservative.
That is to say, we have the following.
The product P of 3-paths and a retract R. (Loops omitted.)As the only SL on a two element set is a totally ordered set, the following useful fact is immediate from the above fact.
Fact 2.3.If an edge (u, v) of an SL graph G is the intersection of distance neighbourhoods of vertices of G, then either u ≥ v or v ≥ u with respect to any compatible SL.

NU polymorphisms
There are many characterisations of the class NU.We introduce here the description from [5].Definition 2.4.Let T be a tree with k leaves and m edges e 1 , . . ., e m .Let U and D be the partition of its vertices into colour classes, and let U * and D * be the subsets of U and D respectively, of vertices of degree at least 2. Define a graph K 0 (T ) as follows: its vertices are the tuples (x 1 , . . ., x m ) such that (i) (ii) for each u ∈ U * , x i = 2 for at least one edge e i incident with u; and (iii) for each d ∈ D * , x i = 0 for exactly one edge e i incident with d.Tuples (x 1 , . . ., x m ) and (y 1 , . . ., y m ) are adjacent if x i − y i ≤ 1 for all i.
The following was a main result of [5] Theorem 2.5.A reflexive graph G admits a k-NU polymorphism if and only if it is a retract of the product of the graphs K 0 (T i ), for a finite family of trees T 1 , . . ., T d each having at most k − 1 leaves.

The generators of k-NU have SL
We define an SL on K 0 (T).Definition 3.1.Let T be a tree with k leaves and m edges e 1 , . . ., e m , having vertex partition U and D as in Definition 2.4.Choosing a root z of T orient the edges of T towards z, (so that any vertex of T has at most one incident edge oriented away from it).
Define ∧ = ∧ z ∶ K 0 (T ) × K 0 (T ) → K 0 (T ) as follows: for x = (x 1 , . . ., x m ) and y = (y 1 , . . ., y m ) ∈ K 0 (T ), let where z i is max(x i , y i ) if e i is oriented towards U , and is min(x i , y i ) if e i is oriented towards D. (See Figure 3 for an example.) Proof.To see that this function is onto K 0 (T) we must show for (x 1 , . . ., x m ) and (y 1 , . . ., y m ) in K 0 (T ) that (z 1 , . . ., z m ) = (x 1 , . . ., x m ) ∧ z (y 1 , . . ., y m ) is also in K 0 (T ).This requires showing that for any d ∈ D * , there is at least one each edge e i incident to d such that z i = 0; and for any u ∈ U * there is an incident edge e i such that z i = 2.We show the former, the proof of the latter is essentially the same.
To see that ∧ z it is a homomorphism, assume that x i ∼ x ′ i and y i ∼ y ′ i for all i, where x, x ′ , y and y ′ are in K 0 (T ).Then ) are clearly at most 1, we get that z i − z ′ i ≤ 1 for all i, and so (z 1 , . . ., z m ) ∼ (z ′ 1 , . . ., z ′ m ).The homomorphism is symmetric and idempotent and associative, as it is in each coordinate.
In [5] it was shown that for a tree T with k − 1 leaves, K 0 (T ) admits k-NU polymorphisms but not (k − 1)-NU polymorphisms.Thus we get the following.This is already known, it follows from [3] and [8]; but the examples one gets from these papers are chordal or products of chordal graphs.The above examples are far from this.
In [5] we defined retracts K(T ) of the K 0 (T ) which also served as generators of the variety k-NU.One can show that the SL-polymorphism defined above survives the retraction from K 0 (T ) to K(T ), so these smaller generators are also in SL.The proof of this is basic, but is too messy for what it gains us; we chose to omit if from the paper1 .

Retract of product of 3 paths without SL
In this section we prove the following theorem.
Theorem 4.1.There exists a reflexive graph admitting a 3-NU polymorphism, but no SL polymorphism.
Our proof is constructive.For a graph G with a compatible SL ∧, we say an edge u ∼ v is oriented u → v by ∧ if u > v in the SL.The following simple observation is key.
Lemma 4.2.Given an SL on a graph there can be no bad 2-path: an induced path of edges oriented away from each other: The main idea, is that in light of the above observation, very few different SL polymorphisms are possible on a product of paths P, and that they, upto some skewing, are much like the product of SLs on the component paths.With respect to the minimal vertex of an assumed SL, we can split the product of paths up into orthants, on which the SL is of a very restrictive form.If one of these orthants is big enough, we can 'kill' the SL on that orthant with a retraction.For a product of long enough paths, we will make several 'killing' retractions in such a way that wherever the minimun vertex is for an SL, one of the killing retractions is contained well inside an orthant, and so kills the SL on that orthant.
Before we get to the actually constructive proof, we set up for it by defining our 'killing' retractions.4.1.Setup for the proof.Let P = P 3 be a product of 3 paths of length , so vertices of P are triples v = (v 1 , v 2 , v 3 ) ∈ [0, ] 3 .For i ∈ [3], let e i denote the vector with a 1 in the i th coordinate, and 0 elsewhere, so that for a vertex v, v + e i is the vertex we get from it by increasing the i-coordinate by one.
A vertex v of P is an positive (or negative) i-boundary vertex if v i = (or v i = 0).An edge of P of the form {v, v + e i } is i-square.It is inner i-square if it does not contain any j-boundary vertices for j ≠ i.These will be important because of the following simple observation.Lemma 4.3.Any inner square edge of P is oriented.
Proof.Clearly we may assume our inner square edge is of the form {u, u + e 1 }.In this case, as it is an inner square edge, the set For a subgraph G of P, the square edges are consistantly oriented if for all i, there exists d i = ±1 such that all i-square edges of G are oriented v → v + d i e i .If d i = 1 they are positively oriented, if d i = −1 they are negatively oriented.
For a square edge {v, v + de i } let C(v; d, i) be the cone of vertices that are closer to v + de i than to v.That is, let The notation is chosen so that P ∖ C(v; d, i) is the graph we get from removing from P the 'cone above v in the positive i direction if d = +1 and in the negative i direction if d = −1'.This graph is in fact a retract of P, as one can easily check that the following map r, which 'pushes C(v; d, i) in the negative (positive if The retract R in Figure 1 can be viewed as is a 2-dimensional version of the construction P ∖ C(v; d, i).Specifically it would be P ∖ C((2, 1); −1, 1)) as we have removed the cone of vertices closer to (1, 1) than to (2, 1).
Figure 4 shows two different depictions of B = P 2 ∖ C((1, 1, 1); +1, 1).The first depiction shows the subgraphs induced by the 1-layers of P, the i th 1-layer being the set of vertices v such that v 1 = i.We have only shown the edges between 1-layers that involve the vertex (1, 0, 2).The second depiction, in which the i-coordinate increases towards you, is more suggestive of how B is as subgraph of a product of paths achieved by removing a cone.Many edges are hidden, but any 'unit cube' in this picture induces a clique of B.
The following will be our main obstruction to SL polymorphisms.
So we must show that at least one of these vertices are in R.
If none of them are, then as the subgraphs B(v) of P are distance at least 3 apart, all of these vertices are in C(v) for some single v ∈ V.
They, along with x + de 2 make up copy of P 2 × P 3 and all have the same 2coordinate x 2 + 1.The only way they can fit into a single 2 , and if one of x and x + e 1 is v.But then v ∈ R contradicts the fact that both x and x + e 1 are in R. ◇ Now, for a vertex v ∈ P let the i-line of v be the set of vertices that differ from it in at most the i th coordinate.
Observe that for any i-line L = L i (v) of inner square edges, which we know are oriented by the above claim, Lemma 4.2 gives us that L contains a center vertex c L towards which all edges must be oriented.Claim 4.7.Let u and v be adjacent, then for any i, the centers of L i (u) and L i (v) are adjacent.
Proof.Assume that the centers a = c Li(u) and b = c Li(v) are not adjacent.Then as they are in the i-lines of adjacent vertices, their i-coordinates must differ by more than one.We may assume that a i < b i − 1.Let a ′ be the vertex in which contradicts the fact that a ′ − e i ∼ b. ◇ The inner i-floor is graph induced by the set of centers of the inner i-lines.By the claim we have that it is a product of two paths.A simple consequence of the claim is that if x and x ′ are in the i-floor then (2) now implies that the inner 1-floor cannot intersect B(v) and B(v 1 between their 1 coordinates is at least 17 − 4 = 13.So we may assume that the 1-floor does not intersect B(v) for any v ∈ V + 1 .Similarily equation (2) gives us that the the inner 2-floor can intersect B(v) for at most three v ∈ V + 1 .We show that for any m ∈ {4, 9, 13}, it can intersect B(v) for at most one v with v 1 = 15 and v 3 = m.Indeed, if it were to contain x ∈ B(v) and 1 such that B(v) is not intersected by any i-floor.This means that its square edges are consistantly oriented, and in particular its 1-square edges are negatively oriented.This is the contradiction of Lemma 4.4 we were looking for.
As the graph R is a retract of a product of paths, it follows by [7] that it admits a 3-NU polymorphism, so this completes the proof of Theorem 4.1.

5.1.
Getting an NU from an SL.Given a meet-semilattice ≤, various subsets S of the vertices may also have least upper bounda ⋁ S. It was observed in [2] that if the k-ary function is well-defined, then it is near-unanimity.It was further shown that in this case the function has many strong properties such as symmetry and so-called superassociativity.
Let T be a polyad a tree that has exactly one vertex z of degree greater than 2. It was shown in [5] that K 0 (T ) can alternately be described as follows.
• y j = 1 if x j = 0 = x i for some i < j such that there is some v ∈ D * incident to both e i and e j .• y j = 1 if x j = 2 = x i for some i < j such that there is some v ∈ U * incident to both e i and e j .• y j = x j otherwise.
We now show that the SL homomorphism on K 0 (T ) from Definition survives the retraction r ∶ K 0 (T ) → K(T ) Theorem A.2. Let T be a tree with k leaves and z a vertex of T .Then where ∧ = ∧ z is from Definition A, and r is from Definition A.1, the composition is a semilattice polymorphism on K(T ).
Proof.That r ○ ∧ is a polymorphism is immediate as the composition of homomorphisms is a homomorphism.That it is symmetric and idempotent is immediate from the fact that ∧ is symmetric and idempotent, and that r is a retraction to (so the identity on) K(T ).What has to be shown is that r ○ ∧ is associative.We show that a ∶= r(r(x ∧ y) ∧ z) = r(x ∧ y ∧ z) =∶ b.
The fact that r(x ∧ r(y ∧ z)) = r(x ∧ y ∧ z) follows by the symmetry of the operation ∧, and so this yields associativity.As a i and b i are in {0, 1, 2} it is clearly enough to show for each i that a i = 0 if and only if b i = 0 and a i = 2 if and only if b i = 2. Further, for d ∈ D * (or similarily for u ∈ U * ), with incident edge e i , showing a i = 0 only if b i = 0 accomplishes the same thing, as we already know a i = 0 for exactly one e i incident to d, and b j = 0 for exactly one e j incident to d.
By the symmetry of the construction with respect to U and D, it is there for enough to show the following.
Claim A.3.For a leaf d ∈ D, where e i is the edge incident to d, a i = 0 if and only if b i = 0.For each vertex in d ∈ D * , and each edge e i incident to d, a i = 0 implies b i = 0. We complete our proof by proving the claim.For the first part, let d be a leaf in D and e i be the incident edge.As d is not in D * any change r makes to the i th coordinate is from 2 to 1.So a i = 0 if and only if min(x i , y i , z i ) = 0 if and only if b i = 0. Now, for the second statement of the lemma, fix d ∈ D * and let e 1 , . . ., e c be the edges adjacent to d in T .Let α be the unique index in [c] such that u α = 0. Similarily let β and γ be the indexes in [c] such that v β = w γ = 0.
For i ∈ [c], if e i is directed up, then a i = 0 only if α = β = γ = i; and in this case we clearly have that b i = 0.So assume that e i is directed down.
Then a i = 0 means that either (i) min{α, β, γ} = i, or (ii) e m is directed up where m = min{α, β, γ}, and i = min({α, β, γ} − {m}).In the first case, clearly b i = 0.In the second case, we have that α, β and γ are not all the same.So at least one of u m , v m and w m are non-zero.If it is u m or v m then (r(u ∧ v)) m is non-zero, (as e m is directed up); if it is w m then (r(u ∧ v) ∧ w) m

Figure 2 .Fact 2 . 2 .
Figure 2. The graph K 0 (T ) when T = K 1,3 , the reflexive claw.All vertices are shown, some edges are hidden: two vertices are adjacent if the are in the same unit cube.

Figure 3 .
Figure 3.The semilattice ∧ z on K 0 (T ) where T = K 1,3 and z is the leaf of the edge e 3 .The semilattice ∧ z is the transitive closure of the shown digraph on K 0 (T ).

Corollary 3 . 3 .
For all k ≥ 4 there are graphs in SL ∩(k − NU) that are not in (k − 1) − N U .

Figure 5 9 Figure 5 .
Figure 5 shows part of the graph R. The dimples in the face of the cube are the 54 different cones C(v) removed from P. Lemma 4.5.The retract R of P = P 317 defined above has no SL polymorphisms.Proof.For each v ∈ V let B(v) be the neighbourhood of v in R. It is easy to check that the vertices in V are far enough apart that each subgraph B(v) is isomorphic