Line graphs of triangle-free graphs
Andrés Aranda
January 18, 2015
W ORKING D RAFT
Abstract
We present some results on the structure of line graphs of triangle-free graphs and discuss connections with homogeneity.
1
Notation
In this document, K3 f denotes the family of all triangle-free graphs, and L(K3 f ) is the family of all graphs
G such that G is the line graph of some H ∈ K3 f . We denote the line graph of G by G∗ . We use R to denote
0
a graph relation, with superscripts (RG , RG ) when necessary; edges are often denoted by their endpoints as
vv 0 . Throughout, RG (v) = {w ∈ G : RG (v, w)}, with the superscripts dropped whenever possible.
We call any graph isomorphic to K1,3 a claw, and we refer to the graph on four vertices with five edges
as K4− .
2
Line graphs of triangle-free graphs
The following results are due, respectively, to Hassler Whitney [Whi32] and Lowell Wayne Beineke [Bei68]:
Theorem 2.1. Let G and H be two connected graphs, neither of which is isomorphic to a claw. Then G is
isomorphic to H if and only if G∗ and H ∗ are isomorphic.
Theorem 2.2. A connected graph is a line graph if and only if it has no induced subgraphs isomorphic to
one of the following:
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Remark 2.3. H.A. Jung strengthened Whitney’s Theorem 2.1 in [Jun66], proving that an “edge isomorphism” (i.e., isomorphism of the line graphs) induces a “vertex isomorphism” (isomorphism of the graphs),
under the assumption that there are no isolated vertices in the original graphs.
The following proposition is an almost immediate corollary of Theorems 2.2 and 2.1.
Proposition 2.4. A finite graph G is the line graph of a triangle-free graph if and only if G does not embed
a claw or K4− .
Proof. First suppose that G is the line graph of a triangle-free graph. It follows directly from Theorem 2.2
that G does not embed a claw. Now observe that K4− is the line graph of a triangle with an edge sharing an
endpoint with the triangle; therefore, by Whitney’s theorem, the line graph of a triangle-free graph does not
embed a K4− .
Now suppose that G does not embed a claw or a K4− . All of Beineke’s forbidden structures which are
not a claw have K4− as an induced substructure. Therefore, G is the line graph of some H. Now, if there
are any connected components of H isomorphic to a triangle, substitute them with a claw and call the result
H 0 . Clearly, the line graph of H 0 is isomorphic to G, and we claim that H 0 is triangle-free. Suppose that
there is a triangle in H 0 ; because of the way we constructed H 0 from H, any such triangle does not form a
connected component, so there is at least one edge with an endpoint in the triangle. The line graph would in
that case embed a K4− , contradiction.
Observation 2.5. Let G be a K4− -free graph and v, v 0 ∈ G. If the graph induced by G on R(v) ∩ R(v 0 )
contains an edge ww0 and a non-edge u, u0 , then v = v 0 .
Proof. Suppose for a contradiction that v 6= v 0 . Then either vv 0 is an edge of G or v, v 0 is a non-edge of
G. In the former case, since R(v) ∩ R(v 0 ) embeds a non-edge u, u0 we reach a contradiction because the
structure induced by G on vv 0 uu0 is a K4− . And if v, v 0 do not form an edge, then we reach a contradiction
by the same argument using the edge ww0 . The conclusion follows.
Definition 2.6. Let G be a graph and e = vv 0 an edge in G. We define C(e) as the set {w ∈ G :
R(w, v) ∧ R(w, v 0 )} ∪ {v, v 0 }. We call C(e) the closure of e.
Observation 2.7. If G is a K4− -free graph and e = vv 0 is an edge in G, then C(e) is a clique.
Proof. The conclusion follows trivially if C(e) = {v, v 0 } or |C(e)| = 3, so we may assume that there exist
distinct vertices c, c0 in C(e) \ {v, v 0 }. We know by the definition of C(e) that cvv 0 and c0 vv 0 form triangles;
it follows by K4− -freeness that R(c, c0 ) holds.
Definition 2.8. Two edges vv 0 and ww0 in a graph G are said to be parallel if none of vw, v 0 w, vw0 , v 0 w0 is
an edge in G.
The following proposition gives us a local picture of line graphs of triangle-free graphs.
2
Proposition 2.9. Let G ∈ L(K3 f ) and suppose that for some v ∈ G, R(v) contains parallel edges. Then
any maximal collection {e1 , . . . , ek } of edges from G, all of whose endpoints are contained in R(v) and
such that the ei are pairwise parallel, is of size 2. Moreover, R(v) ∪ {v} = C(e1 ) ∪ C(e2 ) for any parallel
edges e1 , e2 with endpoints in R(v) and there are no edges between C(e1 ) \ {v} and C(e2 ) \ {v} (i.e., for
all x ∈ C(e1 ) \ {v} and y ∈ C(e2 ) \ {v} we have ¬R(x, y)).
Proof. The first assertion follows directly from claw-freeness. To prove R(v) ∪ {v} = C(e1 ) ∪ C(e2 ),
we proceed as follows: from the definition of C(e) we know that v ∈ C(e1 ) ∩ C(e2 ), so take any w ∈
R(v), w 6= v. It follows from claw-freeness that v is in an edge with at least one of the endpoints of e1 or e2 ;
let us assume without loss of generality that R(w, a) holds, where a is an endpoint of e1 . Since the graph is
K4− -free, it follows that there is an edge from a to the other endpoint of e1 , and therefore v ∈ C(e1 ); this
proves R(v) ∪ {v} ⊂ C(e1 ) ∪ C(e2 ).
Now suppose that x ∈ C(e1 ) ∪ C(e2 ) is distinct from v. Without loss, suppose x ∈ C(e1 ). From
K4− -freeness, it follows that R(x, v) holds, so C(e1 ) ∪ C(e2 ) ⊂ R(v) ∪ {v}. Therefore, R(v) ∪ {v} =
C(e1 ) ∪ C(e2 ). Notice that in the situation described where e1 , e2 are parallel edges in R(v), we have
C(e1 ) ∩ C(e2 ) = {v}: we already know that v ∈ C(e1 ) ∩ C(e2 ), so suppose we have y 6= x in the
intersection. If ¬R(x, v), then x, y, and the endpoints of e1 form a K4− ; and if R(x, y) then x, y and one
endpoint from each of e1 , e2 form a K4− , a contradiction in any case.
Finally, suppose for a contradiction that R(x, y) holds for some x ∈ C(e1 ) and y ∈ C(e2 ). By our
assumption that R(v) contains parallel edges, there is an endpoint a of e1 which is distinct from x. It
follows that x, a, y, v form a K4− , contradicting K4− -freeness.
Observation 2.10. Let G be a K4− -free graph, and let e = vv 0 be an edge in G. Then C(e) is the greatest
clique in G containing v, v 0 .
Proof. Let K denote the greatest clique in G containing v, v 0 . It follows from Observation 2.7 that C(e) ⊂
K. Now take any k ∈ K. Since K is a clique and contains v, v 0 , it follows that R(k, v) and R(k, v 0 ) hold in
G. Therefore, k ∈ C(e).
Observation 2.11. Let G be a K4− -free graph and e, e0 be edges in G. Then the following are equivalent:
1. C(e) = C(e0 ),
2. The endpoints of e, e0 span a K4 ,
3. The endpoints of e and e0 belong to the same maximal clique K in G (so C(e) = K).
Proof. Let v, w and v 0 , w0 denote the endpoints of e and e0 , respectively. If C(e) = C(e0 ), then it follows
by Observation 2.7 that vv 0 ww0 span a K4 ; this proves the implication 1→2.
Now suppose that the endpoints of e, e0 span a K4 . The maximal clique in G containing e is, by Observation 2.10, C(e); if for some vertex u ∈ C(e0 ) we have C(e0 ) 6∈ C(e), then u, v 0 , w0 , v span a K4− ,
contradicting the hypothesis of K4− -freeness. This proves the implication 2→3.
Finally, suppose that the endpoints of e and e0 belong to the same maximal clique K in G. Again,
Observation 2.10 implies that K = C(e) = C(e0 ), and 3→1.
Proposition 2.12. Let G ∈ L(K3 f ) be a finite graph withS
n edges. Then there exist edges e1 , . . . , em , with
m ≤ n, such that every edge and every vertex of G are in m
i=1 C(ei ). Assuming that we choose a minimal
m, equality is reached when each connected component of G is either 2-regular or a path.
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Proof. It suffices to prove the result for connected G. We know that for each edge e in G, C(e) is the
maximal clique in G containing e; if C(e) contains 3 or more vertices, then any other edge e0 with endpoints
in C(e) satisfies C(e) = C(e0 ). There is an equivalence relation E on the edges from G, E(e, e0 ) holds iff
C(e) = C(e0 ); any transversal to E satisfies the condition in the proposition.
Now we prove the second assertion. If G has n edges and n is the minimal number of edges needed to
cover G as in the proposition, then G is triangle-free. As G is claw-free, this means that each vertex is of
degree at most 2. If all vertices are of degree 2, then G is 2-regular and we are done. So suppose that there
is some vertex of degree 1. The graph has to be cycle-free in this case because it contains no triangles and
no vertices of degree 3 or greater. Therefore, G is a path.
Proposition 2.13. A graph G is the line graph of some triangle-free graph if and only if
1. Each vertex x of G either belongs to a unique maximal clique of G or is the only element in the
intersection of two maximal cliques of G, and
2. For any edge xy of G, R(x) ∩ R(y) is either empty or a clique.
Proof. We have already proved that if H is a triangle-free graph then its line graph satisfies both conditions.
To prove the converse, it suffices to prove that any G satisfying 1 and 2 is K4− - and K1,3 -free. Suppose for
a contradiction that the structure induced by G on the vertices a, b, c, d is a K1,3 centered on a. Then the
edges ab, ac, ad belong to distinct maximal cliques of G, contradicting condition 1. And if the structure on
a, b, c, d is isomorphic to K4− with a, b not forming an edge, then R(c) ∩ R(d) is not a clique, contradicting
condition 2.
Definition 2.14.
1. We will use E to denote the equivalence relation defined in the proof of Proposition 2.12.
2. Let G ∈ L(K3 f ); we say that the set of edgesS{e1 , . . . , em } from G is a minimal spanning set of
edges if every vertex and every edge of G are in m
i=1 C(ei ) and the ei are pairwise E-inequivalent in
G.
3. A weighted graph is a structure (V (G), E(G), w), where (V (G), E(G)) is a simple loopless graph
and w : V (G) → N ∪ {∞}.
4. The block graph of G ∈ L(K3 f ) is the graph with vertex set {C(ei ) : 1 ≤ i ≤ m}, where
{e1 , . . . , em } is a minimal spanning set of edges for G, and edge relation R(C(ei ), C(ej )) that holds
if C(ei ) ∩ C(ej ) 6= ∅. We denote the block graph of G by B(G).
5. The weighted block graph of G ∈ L(K3 f ) is the weighted graph (V (B(G)), E(V (B(G))), wG ),
where the weight function is defined by w(C(ei )) = |C(ei )|. We denote the weighted block graph of
G by (B(G), wG ).
Proposition 2.15. Let G ∈ L(K3 f ). Then B(G) ∈ K3 f .
Proof. Suppose for a contradiction that c1 , c2 , c3 form a triangle in B(G). This means that there exist edges
e1 , e2 , e3 with distinct closures, such that C(e1 ) ∩ C(e2 ) 6= ∅, C(e1 ) ∩ C(e3 ) 6= ∅, C(e3 ) ∩ C(e2 ) 6= ∅.
From K4− -freeness, it follows that each of these intersections is of size 1, so let v1 , v2 , v3 be the elements
of those intersections. By claw-freeness, the vi are distinct. But then v1 v2 v3 is a triangle, and therefore
C(e1 ) = C(e2 ) = C(e3 ), contradiction.
4
Now we wish to establish more connections between triangle-free graphs, their line graphs, and the
block graphs of their line graphs.
Proposition 2.16. Every finite triangle-free graph G is the block graph of some G̃ ∈ L(K3 f ).
Proof. Consider a connected finite triangle-free graph G. We wish to construct a graph G̃ whose block
graph is G. Enumerate the vertices of G as g1 , . . . , gn in such a way that gi+1 forms an edge with some gj ,
j ≤ i. Build a vertex set V0 as follows: for each vertex gi of degree 1, let h0i , hji , where R(gi , gj ) holds in
G, be a pair of vertices in V0 . Now let V1 = V0 ∪ {hji : i 6= j, dG (gi ) ≥ 2, G |= R(gi , gj )}.
At this point we have n distinct vertices in V1 for each vertex of degree n ≥ 2 in G and two vertices in
V1 for each vertex of degree 1 in G. Define an equivalence relation ∼ on V1 : hji ∼ hlk if (i, j) = (k, l) or
(i, j) = (l, k). Notice that the vertices in V1 with superindex 0 belong to ∼-equivalence classes of size one
and no class contains more than two elements. Let V (G̃) be the set of equivalence ∼-equivalence classes of
V1 .
Now we define a graph relation R on V (G̃). Declare RG̃ (hji / ∼, hki / ∼) if the equivalence classes are
distinct and there exist representatives in the classes with the same subindex. Call this graph G̃.
It follows directly from the construction of G̃ and Proposition 2.13 that G̃ is the line graph of a trianglefree graph.
To prove that B(G̃) ∼
= G, notice first that the number of maximal cliques in G̃ equals the number of
vertices in G, since we added max{2, d(g)} vertices for each vertex g ∈ G and then formed cliques with
them. Enumerate the maximal cliques of G̃ as ki , where i is the common subindex of representatives in the
clique. The function f : B(G̃) → G given by f (ki ) = gi is an isomorphism.
The isomorphism type of G ∈ L(K3 f ) is determined by its weighted block graph:
Proposition 2.17. Any weighted triangle-free graph G in which 2 ≤ wG (v) and wG (v) ≥ d(v) is the block
graph of a uniquely determined (up to isomorphism) G ∈ L(K3 f ).
Proof. This is proved using the same procedure as in the proof of Proposition 2.16, but substituting the
vertices of G with cliques of size wG (v) instead of cliques of size max{2, d(v)}.
Proposition 2.18. Let G, G0 ∈ L(K3 f ). If G embeds into G0 , then there exists an injective homomorphism
of graphs j : B(G) → B(G)0 such that wG (v) ≤ wG0 j(v) for all v ∈ B(G).
Proof. Suppose that G embeds into G0 , and let i : G → G0 be the embedding; we can think of G as the
induced subgraph of G0 on the set of vertices i(G). In this situation, calculating the closures of edges in
either graph corresponds to taking the appropriate intersections: if e is an edge with endpoints in i(G), then
0
C G (e) = C G (e) ∩ i(G). The result is a direct consequence of the following claims:
Claim 2.19. Let e, e0 be distinct edges of G0 with endpoints in i(G). Then C G (e) 6= C G (e0 ) if and only if
0
0
C G (e) 6= C G (e0 ).
Proof. This is a consequence of Observation 2.11 and the fact that i is an embedding when e, e0 are disjoint
edges: their endpoints will form a K4 in G0 if and only if they form a K4 in G.
0
0
If e and e0 have one vertex u in common, then C G (e) 6= C G (e0 ) if and only if ¬RG (v, w), where
0
uv = e and uw = e0 , if and only if ¬RG (v, w), since i is an embedding.
Claim 2.20. Let e, e0 be distinct edges of G0 with endpoints in i(G). If C G (e) and C G (e0 ) form an edge in
0
0
B(G), then C G (e) and C G (e0 ) form an edge in B(G).
0
0
Proof. It is clear that if C G (e) and C G (e0 ) form an edge, then so do C G (e) and C G (e0 ), since G is an
induced substructure of G.
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A homomorphism j from B(G) into B(G0 ) can be described as follows: let EG , EG0 be the equivalence
relations on the edge sets of G, G0 that holds for e, e0 if they have endpoints in G (G0 ) and C G (e) = C G (e0 )
0
0
(C G (e) = C G (e0 )). Recall that a vertex v ∈ B(G) corresponds to an EG -class. Let j(v) be the EG0 class
of any edge e in the EG -class corresponding to v. By the claims, this map is injective and preserves the edge
relation in B(G).
Proposition 2.21. Let G, G0 ∈ L(K3 f ). If i : B(G) → B(G0 ) is an embedding of graphs and for all
v ∈ B(G) we have wG (v) ≤ wG0 (i(v)), then G embeds into G0 .
Proof. We will assume without loss of generality that B(G) is an induced substructure of B(G0 ). Let
0
{e1 , . . . , em } be a minimal spanning set of edges of
SnG , such that {e1 /E, . . . , en /E} is the vertex set
of B(G). We will select a set of vertices V from i=1 C(ei ) such that, with the edges induced by G,
B(V ) ∼
= B(G) and has the same weight function.
For i 6= j, 1 ≤ i, j ≤ n, let pij be the unique element of C(ei ) ∩ C(ej ) if such an element exist;
otherwise, leave pij undefined. Let V0 be the set {pij : 1 ≤ i, j ≤ n}. From each C(ei ) select any
wG − |{pij : j S6= i, 1 ≤ j ≤ n}| vertices, and accumulate them in a set T (i). Let V be the graph with
vertex set V0 ∪ ni=1 T (i) and the edge relation induced by G. It is easy to verify that the block graph of V
is isomorphic to B(G).
Observation 2.22. The relation Q on L(K3 f ) defined by Q(G, G0 ) if B(G) ∼
= B(G0 ) is a nontrivial proper
equivalence relation; each class G/Q contains a unique element SG which embeds into every other element
of G/Q.
Proof. It is clear that Q is a nontrivial proper equivalence relation.
The second assertion follows from Proposition 2.17: SG is characterised by wG (v) = max{2, d(v)} for
each vertex in B(SG ).
3
Homogeneity and entomology
In this section, we will argue that we cannot use Covington’s theory of homogenizable structures to find
a finite relational language in which the line graph of the universal homogeneous triangle-free graph is
homogeneous. The problem is that L(K3 f ) does not have local failure of amalgamation, i.e., there exist
infinitely many pairwise non-embeddable amalgamation problems that cannot be solved in L(K3 f ).
Recall some definitions from Covington [Cov89], [Cov90].
Let C be a class of L-structures and α : A → B, β : A → C be L-embeddings between structures in C.
We denote by D[α, β] the diagram
7B
α
A
β
'
C
We say that a diagram D[α : A → B, β : A → C] of C-stuctures and L-embeddings can be amalgamated in
C if there exists a D ∈ C and embeddings α0 : B → D and β 0 : C → D such that α0 ◦ α = β 0 ◦ β. A class
C has the Amalgamation Property if every D in C can be amalgamated in C.
Definition 3.1. Let L be a finite relational language. Let Σ be a set of universal L-sentences and let C
be the class of models of Σ. Assume that C satisfies the Joint Embedding Property. Then we say that C is
homogenizable if there is a finite relational language L0 ⊃ L and a countable set Σ0 ⊃ Σ of universal axioms
in L0 such that:
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i The class C 0 of models of Σ0 satisfies AP and JEP;
ii Every C-structure has an L0 -extension to a C 0 -structure, and the L-reduct of each member of C 0 is in
C;
iii If Γ0 is the unique countable homogeneous L0 -structure with age Cf (all finite structures embeddable
in some model of Σ) and Γ is its L-reduct, then AutL (Γ) = AutL0 (Γ0 )
Definition 3.2. Let α : A → B, β : A → C, α0 : A0 → B0 and β0 : A0 → C0 be embeddings between
C-structures and let D = D[α, β] and D0 = D[α0 , β0 ]. Then (, η) is an embedding of α0 in α if the
following diagram commutes:
AO
/C
O
α
η
A0
α0
/ C0
We say that D0 is a subdiagram of D if there are embeddings : A0 → A, η : C0 → C, and ζ : B0 → B
such that (, η) embeds α0 in α and (, ζ) embeds β0 in β as in the following diagram:
CO g
η
7 BO
β
α
AO
ζ
7 B0
C0 g
α0
A0
β0
Definition 3.3. Let C be a class of structures in which there are finitely many diagrams Dj = D[ξj,0 : Xj →
Yj,0 ; ξj,1 : Xj → Yj,1 ], 0 ≤ j ≤ p such that any diagram D in C fails to be amalgamated in C if and only if
one of Dj is embeddable in D. Then we say that C has Local Failure of Amalgamation (LFA).
In [Cov90], Jacinta Covington proves the following theorem:
Theorem 3.4. Let L be a finite relational language, and Σ a set of universal L-sentences. Let C be the class
of models of Σ. Assume that Cf , the collection of finite elements of C, satisfies the JEP and that it has LFA.
Then C is homogenizable.
3.1
Infinitely many obstructions to amalgamation
There are some obvious obstructions to the Amalgamation Property in L(K3 f ), such as:
D1 •c
•αa1
and
•βa1
•αa2
•βa2
•αa3
•βa3
•αa4
•βa4
7
•b
(1)
D2 •c
D3 •c
•αa1
•βa1
•b
(2)
•αa1
•b1
(3)
•αa2
•αa2
•b2
•αa3
•αa3
•αa4
•αa4
•αa5
•αa5
and
•αa2
•βa2
•αa3
•βa3
•αa4
•βa4
•αa1
and
•αa1
•b1
•αa2
•αa2
•b2
•αa3
•αa3
•αa4
•αa4
•αa5
•αa5
D4 •c1
•αa1
• c2
and
(4)
Of these four diagrams, D1 and D2 have to do with the fact that the line graphs of triangle-free graphs
are K4− - and claw-free. The fourth one is related to the fact that any solution to the amalgamation problem
would necessarily identify c1 and b1 , by Observation 2.5, but c2 and b2 cannot be identified, since they
have different neighbourhoods in A; by claw-freeness, and by K4− -freeness, a non-edge is forced between
them. The third one makes use of the fact, stated in Proposition 2.15, that the block graphs of line graphs
of triangle-free graphs are triangle-free, and of the fact that in any solution to this problem, by Observation
2.5 would identify c and b1 . Diagram D3 is crucial, since there are infinitely many ways to express that
content with amalgamation problems which are mutually non-embeddable, all of whose subdiagrams have
a solution in L(K3 f ).
Definition 3.5. Let T N (n)1 denote an amalgamation problem equivalent to
1
The name T N is from Thasus neocalifornicus. The first few of these amalgamation problems look like them!
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•b1
•a1
• c1
•b2
•a2
• c2
•b3
•a3
• c3
..
.
•a4
..
.
•bn−1
..
.
•cn−1
•bn
•an+1
• cn
•an+2
•cn+1
•an+3
•an+4
In this problem, C = {a1 , . . . , an+4 } ∪ {c1 , . . . , cn } and B = {a1 , . . . , an+4 } ∪ {b1 , . . . , bn+1 }.
Proposition 3.6. Any amalgamation problem having T N (n) (n ≥ 1) as a subdiagram does not have a
solution in L(K3 f ).
Proof. We will first illustrate the argument with T N (1) and T N (2), and then generalise.
The diagram T N (1) is D3 from the beginning of this subsection. It follows from Observation 2.5 that
any solution to D3 will identify c and b1 , say in a vertex p. But this conflicts with the structures B and C,
since C has an edge from C to a5 , and B has a nonedge in that same position.
T N (2) is the following problem:
•b1
•a1
• c1
•b2
•a2
• c2
•a3
• c3
•a4
•a5
•a6
Again, we are forced to identify b1 and c1 . Now b2 and c2 are forced to correspond in any solution to
distinct vertices, because there are witnesses in A (namely, a5 , a6 ) saying that their neighbourhoods differ.
By K4− -freeness, this forces an edge from c2 to b3 , which in turn forces an edge c2 a4 , and again this means
that if we want to embed both B and C over A in a common graph G, then that graph will contain some
K4− .
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In the general case, we know from Observation 2.5 that any solution to the problem of amalgamating B
and C over A will identify c1 with b1 . There is no problem with the pairs bi , ci for i = 2, . . . , n − 1: we
can either identify them, or add them to the “solution” as distinct vertices; this is immaterial. But bn , bn+1 ,
and cn are forced to correspond to distinct vertices in any solution to T N (n) because there are witnesses
in A signaling that their neighbourhoods are distinct. If we don’t add any more edges at this point, then
an+3 , an+4 , bn+1 and cn form a K4− , so we are forced to add the edge cn bn+1 , which produces another K4− ,
this time at the vertices cn bn bn+1 an+2 , so we are foced to add cn an+2 , and amalgamation fails.
Proposition 3.7. If n < m and n, m ≥ 1, then T N (n) is not a subdiagram of T N (m).
Proof. Let An denote the structure on a1 , . . . , an+4 , Bn the structure on the A0 and the bi in T N (n), and
let αn denote an embedding (inclusion) An → Bn . Define Am , Bm , αm similarly in T N (m).
An embedding : An → Am takes edges to edges; let us assume without loss of generality that fixes a1 , a2 , an+3 , an+4 (the argument depends on the graph structure, not on the particular labelling). For
a function η : Bn → Bm to complete an embedding (, η) of αn into αm , the embedding η would have to
preserve the graph distance from a1 to an+4 in Bn , since we cannot skip (that is, leave out of the image of
η) any of the bi as they have witnesses in A that imply that they are distinct vertices. No function η satisfies
this condition.
Proposition 3.8. Fix n ≥ 1. Any subdiagram of T N (n) has a solution in L(K3 f ).
Proof. It suffices to show that omitting any vertex from the amalgamation problem results in a problem
with a solution in L(K3 f ). Continuing with our entomological analogy, in the problem T N (n), labelled as
in Definition 3.5, we call {a1 , a2 , a3 , b1 , c1 , b2 , c2 } the head of T N (n); {a4 , . . . , an } ∪ {c3 , . . . , cn−1 } ∪
{b3 , . . . , bn−1 } is the thorax of T N (n), and the remaining vertices are the abdomen of T N (n).
Removing a vertex from the head: If we remove b1 or b2 , the problem is solved by adding an edge
cn bn+1 , completing the clique in the abdomen. If we remove b2 or c2 , then all we need to do is identify
c1 with b1 and complete the clique in the abdomen. If we delete a1 or a2 , then the structure resulting from
adding the edge cn bn+1 solves the amalgamation problem in L(K3 f ).
Removing a vertex from the thorax: If we remove a4 , a solution to the amalgamation problem is identifying b1 and c1 , adding the edge c2 b2 and completing the clique in the abdomen.
Removing a5 is a bit more delicate: we are forced to identify bi with ci , for i = 1, 2, 3, do not identify
anything else (this forms a C4 with vertices p3 , b4 , c4 , a6 , where p3 is the vertex corresponding to b3 and
c3 ), and complete the clique in the abdomen. The problems resulting from deleting any of a6 , . . . , an are all
solved similarly.
Removing a vertex from B or C in this region results in easily solvable problems: simply identify bi and
ci up to the index preceding that of the deleted vertex, and complete the clique in the abdomen.
Removing a vertex from the abdomen: If we remove an+4 , the resulting problem can be solved by
identifying bi with ci for i = 1, . . . , n − 1, add the edge bn pn−1 to complete a clique, and leaving everything
else unchanged. The same procedure solves the subdiagram omitting an+3 .
If we delete an+2 , the solution is: identify bi and ci for i = 1, . . . , n − 1, add the edges bn cn and bn cn+1
to complete two cliques. The same recipe yields a solution to the problem that results from removing an+1 .
If we remove cn+1 , then all we need to do is identify bi with ci for i = 1, . . . , n. Doing exactly the
same, but up to n − 1, solves the problem if we delete bn or cn .
4
A language for Γ∗3 .
Let Γ3 denote the universal homogeneous triangle-free graph and DiamR (G) denote the R-diameter of the
graph G. We have:
10
Proposition 4.1. B(Γ∗3 ) ∼
= Γ3
Proof. It follows from Proposition 2.15 that B(Γ3 ∗) is a triangle-free graph. We know by Proposition 2.16
that Age(B(Γ∗3 )) = Age(Γ∗3 ). It suffices to prove that B(Γ∗3 ) satisfies the extension axioms of Γ3 .
Let A and B be finite disjoint subsets of B(Γ∗3 ), each without any edges. We wish to prove that there
exists a vertex v ∈ B(Γ∗3 ) such that R(v, a) for all a ∈ A and ¬R(v, b) for all b ∈ B. To prove this, we
will chase the meaning of such a vertex v back to Γ3 . There is a vertex v ∈ B(Γ∗3 ) satisfying the extension
axiom for A, B if and only if there is a maximal clique V in Γ∗3 sharing one vertex with each of the infinite
cliques (corresponding to elements) in A and disjoint from each of the cliques in B. A maximal clique in Γ∗3
corresponds to a maximal star in Γ3 , and we can associate each such star to its central vertex. Therefore, V
exists in Γ∗3 if and only if the corresponding extension axiom in Γ3 is satisfied. The conclusion follows.
Observation 4.2. DiamR (Γ∗3 ) = 3.
Proof. Take any v 6= v 0 in R(Γ∗3 ). If ¬R(v, v 0 ), then v and v 0 belong to distinct maximal cliques Cv , Cv0 in
R(Γ3 ∗). By Proposition 4.1, either Cv ∩ Cv0 6= ∅, in which case dR (v, v) = 2, or there exist a maximal
clique C such that C ∩ Cv 6= ∅ and C ∩ Cv0 6= ∅, and dR (v, v 0 ) = 3.
Proposition 4.3. Suppose that G, H are countable {K1,3 , K4− }-free graphs such that B(G) ∼
= B(H) ∼
= Γ3 .
∼
Then G = H.
Proof. It is easy to establish a back-and-forth system between G and H using the axioms of the universal
homogeneous triangle-free graph and the fact (also easy to prove) that in any countable {K1,3 , K4− }-free K
graph such that B(K) ∼
= Γ3 satisfies |C(e)| = ω for any edge e.
It follows that Γ∗3 is characterised as the unique (up to isomorphism) countable graph not embedding
K1,3 or K4− such that for all distinct edges e1 , e2 , e3 we have C(e1 ) = C(e2 ) = C(e3 ) ∨ (C(e1 ) ∩ C(e2 ) =
∅ ∨ C(e3 ) ∩ C(e2 ) = ∅ ∨ C(e1 ) ∩ C(e3 ) = ∅) and satisfying the axiom scheme
\
\
ϕn,m = ∀e1 , . . . , en , e01 , . . . e0m (
C(ei ) ∩ C(ej ) = ∅ → ∃e0 (
C(e0 ) ∩ C(ei ) 6= ∅∧
1≤i<j≤n
∧
\
C(e0 ) ∩
C(e0i )
1≤i≤n
= ∅))
1≤i≤m
Remark 4.4. It follows from Bednarek’s extension of Whitney’s theorem (see [Bed85]) to infinite graphs
that Γ3 is the unique graph (up to isomorphism) whose line graph is Γ∗3 .
Question 1. Suppose that H is a connected countably infinite K1,3 - and K4 -free graph such that all vertices
have infinite R-neighbourhoods. Is it true that L(B(H)) ∼
= H and B(L(H)) ∼
= H? Is it true for countably
infinite K1,3 - and K4 -free regular graphs?
From this we see that if we wish to know the type in Γ∗3 of a finite tuple of vertices over ∅ we need
to specify how the maximal cliques they belong to are related. We need to add three binary relations,
11
corresponding to the following situations:
@
@
@
@
u S(a, b) @u
@
@
@
@
@
@
u Q(a, b)@u
@
@
@
@
@
@
@
@u U (a, b)
u
@
@
@
@
@
In the drawings, the bullet points represent vertices and the lines represent the two maximal cliques to
which each vertex belongs. These predicates correspond to the relations
S(a, b) ↔ ¬R(a, b) ∧ ∃x(R(a, x) ∧ R(x, b) ∧ ∃y∃z(R(a, y) ∧ R(b, z) ∧ C(ay) ∩ C(bz) = ∅))
Q(a, b) ↔ ¬R(a, b) ∧ ∃x∃y(R(a, x) ∧ R(b, x) ∧ R(a, y) ∧ R(by))
U (a, b) ↔ ∀x∀y(R(a, x) ∧ R(b, y) → C(ax) ∩ C(by) = ∅)
Where C(xy) denotes the closure of the edge xy. The predicate U corresponds to those vertices which are
at R-distance 3 in the graph; S and Q are the two ways for two vertices to be at R-distance 2 in Γ∗3 .
We can interpret an incidence structure in Γ∗3 , where two points (vertices) x, y are collinear if R(x, y)
holds, and lines are the sets C(xy) for collinear x, y. This incidence structure satisfies:
1. Lines are infinite.
2. Every point lies on exactly two lines.
3. The lines do not form triangles.
4. Two distinct lines have at most one point in common.
Remark 4.5. This incidence structure satisfies all the conditions of a pseudoplane. Therefore, by the main
theorem in [Tho98], we cannot expect this to be homogenizable in a binary language.
The next result can be proved by inspection:
Observation 4.6. Of all the possible triangles in the language {R, S, U, Q}, only RU Q and RRU, do not
embed into Γ∗3 .
The predicate S is unlike the other two in the sense that if S(a, b) holds then over the parameter set
{a, b} the elements of the line through a that contains the intersection have a type different from that of
the other clique through a (likewise for b). This implies that any finite structure embedding a triangle with
at least two S-edges will embed in at least two different ways into Γ∗3 . Things can get quite messy. For
12
example, there are five embeddings of the triangle SSS into Γ∗3 :
@
@
@
@
@
@
@
@u
u
@
@u
u
@
@
@
@
@u
u
@
@
u
@
Q
@
Q
u QQ @ u
Q
Q
Q
Q
u
Q
u
Y
YY
Yu
YY
Y
YY
u
H
HH
H
Y
Q
Q
Qu
HH
H
HH
H
u
We could add ternary predicates for each of these embeddings (and for the embeddings of other triangles
with two S sides), but we prefer to add only two ternary predicates t1 , t2 :
t1 (a, b; c) : ∃x1 , x2 (x1 6= x2 ∧ S x1 (a, c) ∧ S x2 (b, c) ∧ K3 (c, x1 , x2 ))
t2 (a, b; c) : ∃x1 , x2 (S x1 (a, c) ∧ S x1 (b, c) ∧ S c (x1 , x2 ))
In these definitions, S x (u, v) means that x witnesses the relation S, so R(x, u) ∧ R(x, v). We will use
similar notation for Q (with two witnesses). By K3 (u, v, w) we mean R(u, v) ∧ R(u, w) ∧ R(v, w). Notice
that t1 and t2 make no assertions regarding a, b: they are open-ended. For mnemonics, one can think that
13
the ti say that a and b are in S-relations with c and a, b are on the same side of c (t1 ), or on opposite sides of
c (t2 ).
Is that all we neet to know? The predicates ti suffice to separate the orbits of almost all triangles with
two S sides not realising Q, but we still need to add a predicate L for a triangle SSR:
L(a, b; c) : ∃x(S x (ac) ∧ S x (bc) ∧ K3 (abx))
But even this is not enough. Consider a structure on four vertices a, b, c, d satisfying
Q(a, b), S(a, c), U (b, c), S(c, d), U (a, d), S(b, d), t2 (a, d; c), t2 (b, c; d)
There are two ways to realise this structure, namely
@
@
@
@
@
u
@
@u
@
@
u
H
HH
H
HH
H
Hu
u
H
H
HH
H H
H
H
u
HH
H
HH
H
HH
H
@u
HH
H
H
H
H
Hu
As in the case of the relations ti , here what we need to know is, relative to a Q-edge, whether the S-edges
are realised by vertices in opposite or contiguous sides of the quadrangle Q. A similar problem arises when
we consider the interaction of Q-edges and R-edges. To address this problem, we introduce two quaternary
relations; as with the ti , these relations imply an atomic diagram for their parameters.
W1 (a, b, c, d) :∃x1 , x2 , x3 , x4 (Qx1 x2 (a, b) ∧ Lx3 (a, x1 ; c) ∧ Lx4 (b, x1 ; d))
W2 (a, b, c, d) :∃x1 , x2 , x3 , x4 (Qx1 x2 (a, b) ∧ Lx3 (a, x1 ; c) ∧ Lx4 (b, x2 ; d))
Notice that these two predicates imply enough information to isolate the orbits of the two structures above,
and also to distinguish between the two possible embeddings of the triangle SSQ that satisfy t2 (a, b; c)
(namely, Wi (a, b, c, c)).We wish to prove that Γ∗3 eliminates quantifiers in the language
L = {R, S, U, Q, t1 , t2 , W1 , W2 , L}
Definition 4.7. We will say that an L-structure A ∈ Age(Γ∗3 ) includes witnesses to S (respectively, Q) if for
some a, a0 ∈ A with A |= S(a, a0 ) (Q(a, a0 )) there exists b ∈ A with R(b, a) ∧ R(b, a0 ) (similarly for Q).
Otherwise, we say that A does not contain witnesses for S. In the case of Q, since we need two witnesses,
it could be the case that for some Q-edges we have only one of the two witnesses in A. A Q-edge ab is fully
witnessed if there are two distinct elements c, c0 ∈ A such that R(a, c) ∧ R(b, c) ∧ R(a, c0 ) ∧ R(b, c0 ).
Definition 4.8. Given a nonempty subset X of {R, S, Q}, we say that an L-structure A is X-connected if
for all a, a0 ∈ A there exists a path in A from a to a0 such that each of the edges of the path is of one of the
types in X.
14
Definition 4.9. We say that a (simple, undirected, loopless) K4− , K1,3 -free graph B satisfies an L-structure
A if there exists an injective map i : A → B such that
RA (a, a0 ) → RB (i(a), i(a0 ))
S A (a, a0 ) → ¬RB (i(a), i(a0 )) ∧ ∃x(RB (i(a), x) ∧ RB (x, i(a0 ))∧
∧ ∀y(RB (i(a), x) ∧ RB (x, i(a0 )) → x = y))
QA (a, a0 ) → ¬RB (i(a), i(a0 )) ∧ ∃x, y ∈ B(x 6= y ∧ ¬RB (x, y) ∧ RB (i(a), x)
∧ RB (x, i(a0 )) ∧ RB (i(a0 ), y) ∧ RB (y, i(a)))
U A (a, a0 ) → ∀x(RB (x, i(a)) ↔ ¬RB (x, i(a0 ))
The link between these two definitions is that A ∈ Age(Γ∗3 ) is fully witnessed iff its reduct to the
language {R} satisfies A. If in addition to being fully witnessed A is of least size, then A is isomorphic to
its algebraic closure in Γ∗3 (we argue below that the isomorphism type of the algebraic closure of A in Γ∗3
does not depend on the embedding A → Γ∗3 ).
Given a finite substructure A of Γ∗3 , we can use Γ∗3 to find a graph B that satisfies the L-structure of A:
all we need to do is calculate acl(A) and consider its reduct to {R}, which we denote by acl(A) R .
Observation 4.10. An isomorphism between two finite substructures A, B of Γ∗3 can be extended to an
isomorphism between acl(A) and acl(B).
Proof. Given two finite substructures A, B in Γ∗3 , if they are isomorphic it is clear that we can extend the
isomorphism between them to their algebraic closures (the closures will be isomorphic by the homogeneity
of Γ3 ): let {vi : i ∈ ω} be an enumeration of the vertices of Γ3 , and enumerate Γ∗3 as {{i, j} : i, j ∈
ω, Γ3 |= R(vi , vj )}, so that R(x, y) holds in Γ∗3 iff the pairs of natural numbers x, y are distinct and have
nonempty intersection.In the case of S, it is clear that we should send the unique witness of that edge
in A to the unique witness of the corresponding edge in B. And if Q({x, y}, {a, b}) and this edge is
partially witnessed in A, then send the other witness to the witness of the corresponding pair in acl(B)
that is not in B. If Q({x, y}, {a, b}) is not witnessed in A, and the isomorphism A ∼
= B sends {x, y} 7→
{z, w}, {a, b} 7→ {α, β}, then we know that the vertices {vx , v, y, va , vb } form a 4-cycle in Γ3 , as do the
vertices {vz , vw , vα , vβ }, so there is an automorphism of Γ3 taking one cycle to the other and sending the
edge {vx , vy } to the edge {vz , vw }. This automorphism induces an isomorphism on the closure of the
Q-edge in Γ∗3 , which we can use to extend the isomorphism A ∼
= B.
Let A, B be two isomorphic finite substructures Γ∗3 . By the argument in the preceding Observation, their
algebraic closures are isomorphic, and if we consider the reducts of acl(A) and acl(B) to the language {R},
then the isomorphism between acl(A) and acl(B) is an isomorphism of line graphs. By Jung’s strengthening
of Whitney’s Theorem 2.1, (seeSRemark 2.3), fˆ induces an isomorphism g of the induced finite subgraphs
on the vertices vi such that i ∈ aclA. By the homogeneity of Γ3 , g can be extended to an automorphism
σ of Γ3 , which induces an automorphism of Γ∗3 extending fˆ and therefore f .
The argument in the preceding paragraphs indicates that if we want to prove the homogeneity of Γ∗3
in the language L, it suffices to prove that we can determine the isomorphism type of acl(A) from the
relational information contained in A (i.e., without embedding it first into Γ∗3 ). Essentially, all we need to
know is to which side (i.e., cliques from R(a)) of a vertex a the other elements attach, and it is more or less
clear that our predicates indicate exactly that. Given the size of the language and the complexity of some
of its predicates, the task of proving that we can determine the isomorphism type of acl(A) is laborious and
unenlightening; we present a simplified version of the argument next.
15
Given a finite substructure A of Γ∗3 , consider its reduct to {R}, A R . This is a K4− , K1,3 -free graph,
but it need not be connected, and in general will not satisfy A in the sense of Definition 4.9; indeed, it will
satisfy A exactly when A is algebraically closed in Γ∗3 .
Remark 4.11. A finite substructure A of Γ∗3 in the language L is {R, S, Q}-connected if and only if the
R-reduct of acl(A) R is connected.
Each connected components of A R is of one of four types: it could be a non-trivial clique, an isolated
vertex, connected graph lacking some witnesses, or it could include some S- or Q-edges and contain all the
witnesses to the relations in A between its vertices. We will show that we can reduce this variety to a case in
which each connected component of A R is either a non-trivial clique or a fully witnessed portion of A R .
First, we show how to insert witnesses to an R-connected structure of the third type.
Observe that acl(A) ∼
= acl(B) iff their reducts to {R} are isomorphic.
We want to calculate the isomorphism type of the acl(A) from the relational information contained in
A. It follows from the homogeneity of Γ3 and the arguments above that the isomorphism type of acl(A) not
be depend on the embedding of A into Γ∗3 . Therefore, the isomorphism type of acl(A) depends only on A.
Given a line graph G, we can always find a set of two-element sets G0 of natural numbers such that
G ∼
= G0 when G0 is equipped with the relation x ∼ y if x 6= y and x ∩ y 6= ∅. We will follow this
convention from this point on. In the case of line graphs of triangle-free graphs, we can use an enumeration
of the blocks of G B1 , . . . , Bm to name those elements that belong to two cliques (i.e., if x ∈ Bi ∩ Bj , then
associate x with the set {i, j}). Enumerate the vertices in A that belong to only one (possibly trivial) clique
in A as c1 , . . . , ck , and associate with cj the pair {i, m + j}, where cj ∈ Bi . We can also name the vertices
of any A ∈ Age(Γ∗3 ) by the same process applied to by considering A R . We will assume that all finite
substructures of Γ∗3 are given as pairs of natural numbers.
Definition 4.12. Given an L-structure A ∈ AgeΓ∗3 , considered as a set of pairs of natural numbers, we will
say that a set of pairs of natural numbers X solves A, or is a solution to A, if A ∪ X equipped with the
relation R(x, y) that holds iff x 6= y and x ∩ y 6= ∅ satisfies A. A solution X is minimal if for all x ∈ X
we have that A ∪ (X \ {x}) does not satisfy A.
Remark 4.13. It follows from Observation 4.10 and the fact that the algebraic closure of a finite subset of
an ω-categorical structure is finite that all A ∈ AgeΓ∗3 can be solved by a finite set. Also, minimal solutions
to A correspond to finding the elements of the algebraic structure of A.
Observation 4.14. Let A ∈ Age(Γ∗3 ) and suppose that for some a = {x, y}, b = {α, β} ∈ A we have
an unwitnessed edge S(a, b) and c = {x, z} ∈ R(a), d={α, γ} ∈ R(b). Then there is a unique minimal
solution that will satisfy all the relations holding in the set a, b, c, d.
Proof. First note that ¬R({a, b}, {c, d}) implies that a, b, c, d are four distinct natural numbers (corresponding to four distinct vertices in Γ∗3 ). We have the following implications in Γ∗3 :
S({x, y}, {α, β}} → {x, α} ∈ Γ∗3 ∨ {x, β} ∈ Γ∗3 ∨ {y, α} ∈ Γ∗3 ∨ {y, β} ∈ Γ∗3
U ({u, v}, {ε, θ}) → {ε, u} ∈
/ Γ∗3 ∧ {θ, u} ∈
/ Γ∗3 ∧ {ε, v} ∈
/ Γ∗3 ∧ {θ, v} ∈
/ Γ∗3
R({u, v}, {ε, θ}) → u = ε ∨ u = θ ∨ v = ε ∨ v = θ
and exactly one of the four options in each disjunction holds. We need to prove that any structure on the
four vertices gives us enough information to decide which one is true.
16
There are 25 possible structures on four vertices satisfying the hypotheses of this observation. But the
method to find the solution is more or less uniform, and is based on the following obvious implications:
L({x, y}, {x, z}; {α, γ}) → {α, x} ∈ Γ∗3 ∨ {γ, x} ∈ Γ∗3
S({x, y}, {α, β}) ∧ Q({x, y}, {α, γ}) → {α, x} ∈ Γ∗3 ∨ {α, y} ∈ Γ∗3
t1 ({x, y}, {x, z}; {α, β}) → ({α, y} ∈ Γ∗3 ∧ {α, z} ∈ Γ∗3 ) ∨ ({β, y} ∈ Γ∗3 ∧ {β, z} ∈ Γ∗3 )
t2 ({x, y}, {x, z}; {α, β}) → ({α, z} ∈ Γ∗3 ∧ {β, y} ∈ Γ∗3 ) ∨ ({β, z} ∈ Γ∗3 ∧ {α, y} ∈ Γ∗3 )
Again, only one atom of each disjunction is true. Of course, we can translate these implications to assertions
that make no reference to Γ∗3 , simply saying that a line graph can be extended by adding such a pair to a line
graph that satisfies the relations.
To illustrate how these implications give a solution to a configuration satisfying the hypotheses of this
observation, we solve one explicitly. Suppose we have the structure
S({x, y}, {α, β})
S({α, β}, {x, z})
U ({x, y}, {α, γ})
S({x, z}, {α, γ})
L({x, y}, {x, z}; {α, β})
t2 ({α, β}, {α, γ}; {x, z})
Then U ({x, y}, {α, γ}) implies that we cannot add any of {α, x}, {γ, x}, {α, y}, {γ, y}, while
L({x, y}, {x, z}; {α, β}) tells us that exactly one of {α, x}, {β, x} is in the solution. From this we already
know that we need to add {β, x}. Finally, t2 ({α, β}, {α, γ}; {x, z}) implies that if we add {β, x}, then
we also need to insert {γ, z}. Therefore, inserting the vertices {β, x}, {γ, z} will satisfy all the relations
(equivalently, those vertices are in Γ∗3 ). The other 24 cases are proved in a similar manner.
If an edge Q(a, b) is only partially witnessed, and we have a = {x, y}, b = {α, β}, and the witness is,
say, {x, α}, then the only way to complete the quadrangle is to insert a vertex {y, β}. So partially witnessed
Q-edges are easy to complete.
Observation 4.15. Let A ∈ Age(Γ∗3 ) and suppose that for some a = {x, y}, b = {α, β} ∈ A we have
an unwitnessed edge Q(a, b) and c = {x, z} ∈ R(a), d={α, γ} ∈ R(b). Then there is a unique minimal
solution that satisfies all the relations holding in the set a, b, c, d.
Proof. The same as before. There are eleven possible configurations on the vertices a, b, c, d, each of which
implies a unique possible solution. For example, given the structure
Q({x, y}, {α, β})
Q({α, γ}, {x, z})
S({x, y}, {α, γ})
S({x, z}, {α, β})
we get the following disjunctions (all of which are “exclusive or”):
({α, x}, {β, y} ∈ Γ∗3 ) ∨ ({β, x}, {α, y} ∈ Γ∗3 )
({α, x}, {γ, z} ∈ Γ∗3 ) ∨ ({γ, x}, {α, z} ∈ Γ∗3 )
{α, x} ∈ Γ∗3 ∨ {γ, x} ∈ Γ∗3 ∨ {α, y} ∈ Γ∗3 ∨ {γ, y} ∈ Γ∗3
{α, x} ∈ Γ∗3 ∨ {β, x} ∈ Γ∗3 ∨ {α, z} ∈ Γ∗3 ∨ {β, z} ∈ Γ∗3
So if we insert {α, x}, then we should also have {β, y}, {γ, z}. One can easily check that this is indeed a
solution. On the other hand, if we choose to add {β, x} then we would be forced by the first formula to add
{α, y} as well; but this is inconsistent with the third formula above. Similar arguments prove that the other
ten cases also imply a unique solution.
17
Proposition 4.16. Let A ∈ Age(Γ∗3 ) be such that for every a ∈ A we have R(a) 6= ∅. Then we can
reconstruct a K4− , K1,3 -free graph that realizes A.
Proof. The result follows from Propositions 4.14 and 4.15, since there is a unique solution to each unwitnessed edge.
Remark 4.17. It follows from Propositions 4.14 and 4.15 that given a finite A with some R-edges, the
substructure induced on A0 = {a ∈ A : R(a) 6= ∅} contains enough information to reconstruct the line
graph represented by A0 (which will be an induced subgraph of the graph represented by A). This is not
true of general induced substructures of A, as we may need some additional information to decide which
vertices to introduce.
After Proposition 4.16, we still have two cases to solve, namely the case of an R-free structure and the
case where some, but not all, vertices have nonempty R-neighbourhood. First, we reduce the former case to
the latter:
Observation 4.18. Suppose that A ∈ Age(Γ∗3 ) is R-free and is not a U -clique. Then we can insert vertices
so that some vertices in A have nonempty R-neighbourhood.
Proof. This is a consequence of common sense, but we include a proof for completeness. Since A is not
a U -clique, there exist a, b ∈ A such that S(a, b) ∨ Q(a, b) holds. Suppose that we have Q(a, b) and A is
RSQ-connected.
Let us write a = {x, y} and b = {α, β}. We claim that we can make an arbitrary choice for this edge
i.e., insert the vertex {x, α} (so the challenge is to make consistent subsequent choices). To prove this,
suppose that we have a minimal solution Y for A which includes the vertex {x, β}. We wish to prove that
we can find a set X of the same size as such that A ∪ X ∼
= Y ∪ B. Set
X = {{α, γ} : {β, γ} ∈ Y } ∪ {{β, γ} : {α, γ} ∈ Y } ∪ (Y \ {{c, d} ∈ Y : {α, β} ∩ {c, d} = ∅}
It is easy to verify that A ∪ X is isomorphic to B ∪ Y .
And if Q is not realised in A, then A is an S-clique. Choose any two vertices {x, y}, {α, β}. Then one
of {α, x}, {α, y}, {β, x}, {β, y} is in a solution Y of A, without loss, let us assume {α, x} ∈ Y . If we insert
arbitrarily a different witness {c, d} to this S-edge, then if {c, d} ∩ {α, x} =
6 ∅, we can form a new solution
of the same size by the same argument as before. And if we choose to insert {β, y}, then set
X ={{β, y}} ∪ {{y, c} : {x, c} ∈ Y, c 6= α} ∪ {{x, d} : {y, d} ∈ Y }∪
∪ {{β, γ} : {α, γ} ∈ Y, γ 6= x} ∪ {{α, δ} : {β, δ} ∈ y, δ}
Again, this set is a solution of A.
Proposition 4.19. Let A ∈ Age(Γ∗3 ). Then we can reconstruct a K4− , K1,3 -free graph that realizes A.
Proof. We may assume without loss that A is RSQ-connected. Consider A0 = {a ∈ A : R(a) 6= ∅},
and let A1 = A \ A0 ; we may assume that A0 is not empty by Observation 4.18. By Proposition 4.16 and
Remark 4.17, we may assume that A1 is fully witnessed.
Take {α, β} = a ∈ A1 . We wish to show that we can insert new vertices so that all the elements of
A0 ∪ {a} have nonempty R-neighbourhood. We have a few cases:
1. Suppose that there exist {x, y} = b, {x, w} ∈ A0 such that L({x, y}, {x, w}; {α, β}) holds. This
tells us that the witness to S(a, b) is either {x, α} or {x, β}. The same argument as in Observation
4.18 proves that any choice can be extended to a full solution of A.
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2. If there is no pair as in the preceding case, but S({α, β}) ∩ A0 6= ∅, then it must be the case that
there exist {x, y}, {x, w} ∈ A0 such that S({α, β}, {x, y}) ∧ ¬L({x, y}, {y, z}; {α, β}) (this is true
because every element of A0 is in a nontrivial R-clique), which tells us that the witness to this S-edge
is either {α, x} or {β, x}. Again, either choice can be extended to a full solution of A.
3. Otherwise, S({α, β}) ∩ A0 = ∅. If Q({α, β}) ∩ A0 = ∅, then by RSQ-connectedness, there exists
another vertex which is connected by S or by Q to A0 ; solve one of those vertices first. So we may
assume that {x, y} ∈ Q({α, β}) ∩ A0 . If {u, v} ∈ S({α, β}), then insert an arbitrary vertex {u, α}.
The choice at this point is immaterial since {u, α} is in A1 , but it will determine via the predicates
W1 , W2 which of {α, x} or {β, x} to insert (some predicate W will hold because R({x, y}) 6= ∅).
Otherwise, the choice of witnesses for this edge is immaterial.
Therefore we can extend the set of elements with nonempty R-neighbourhood one at a time and solve the
resulting structures by the method sketched in Proposition 4.16.
Theorem 4.20. Γ∗3 is homogeneous in the language L0 = {R, S, Q, t1 , W1 , L}
Proof. It follows from the discussion preceding Remark 4.11 and Proposition 4.19 that Γ∗3 is homogeneous
in L. Notice that we can define U (a, b) ↔ ¬R(a, b) ∧ ¬S(a, b) ∧ ¬Q(a, b), t2 (a, b; c) ↔ S(c, a) ∧ S(c, b) ∧
¬t1 (a, b; c)∧¬L(a, b; c) and W2 (a, b, c, d) ↔ Q(a, b)∧S(a, c)∧S(b, d)∧¬W1 (a, b, c, d). These predicates
are therefore not essential, and the theory of Γ∗3 also eliminates quantifiers in the language L0 .
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