Intrinsically knotted graphs with linklessly embeddable simple minors

It has been an open question whether the deletion or contraction of an edge in an intrinsically knotted graph always yields an intrinsically linked graph. We present a new intrinsically knotted graph that shows the answer to both questions is no.


Introduction
A graph is intrinsically knotted (resp.intrinsically linked) if every embedding of it in S 3 contains a nontrivial knot (resp.2-component link).We abbreviate intrinsically knotted (resp.linked) as IK (resp.IL), and not intrinsically knotted (resp.linked) as nIK (resp.nIL).Robertson, Seymour, and Thomas [12] showed that every IK graph is IL.It is also known that coning one vertex over an IL graph yields an IK graph.(This is shown by combining [12] and the work of Foisy [4] and Sachs [13].)However, it has been difficult to make the relationship between IK and IL graphs stronger.For example, Adams [1] asked if deleting a vertex from an IK graph always yields an IL graph, but Foisy [5] provided a counterexample.Deleting a vertex from a graph also deletes all edges incident to that vertex, so it might seem more likely that deleting, or contracting, a single edge of an IK graph should leave it IL.Naimi, Pavelescu, and Schwartz [10] tried to show that this is the case when the edge belongs to a 3-cycle, but their proof contained an error (which we will describe in Section 6).They also asked if deleting or contracting an edge in an IK graph always yields an IL graph.We verify (using a computer program) that the answer to this question is yes for graphs of order at most 9, but we show that in general the answer is no.We present an IK graph G 11;35 of order 11 and size 35 with edges e and f such that neither G 11;35 e (edge deletion) nor G 11;35 =f (edge contraction) is IL.We argue that G 11;35 is a minimal-order example of an IK graph that yields a nIL graph by deleting one edge, and that ten is the smallest order for an IK graph that yields a nIL graph by contracting one edge.The graph G 11;35 is also a counterexample to the main result of [10].
Graphs that are IK but yield a nIL graph by deleting one vertex or edge or by contracting one edge are intriguing from the perspective of Colin de Verdière's graph invariant .This is an integer-valued graph invariant that is difficult to compute in general; its value is known only for certain classes of graphs with "nice" topological properties.For example, for any graph G, .G/ Ä 3 if and only if G is planar (see Colin de Verdière [2]), and .G/ Ä 4 if and only if G is nIL; see van der Holst, Lovász, and Schrijver [7].
An important open question is how to characterize graphs G with .G/ Ä 5.Even though many known minor-minimal IK (MMIK) graphs have -invariant 6, intrinsic knottedness is not the answer.A minor of a graph G is a graph obtained by contracting zero or more edges in a subgraph of G. We'll say an edge deletion minor (resp.edge contraction minor) of G is a graph obtained by deleting (resp.contracting) exactly one edge of G.Both are called simple minors of G.As we explain in Section 5, if an IK graph G has a nIL simple minor then .G/ D 5. Thus, our graph G 11;35 , together with other IK graphs obtained from it (as described in Section 5), join Foisy's graph as new examples of IK graphs with -invariant 5.These examples show that .G/ Ä 5 is not equivalent to G being nIK.
In the next section we describe the graph G 11;35 and we show it is IK and minor-minimal for that property in Sections 3 and 4, respectively.In Section 5 we make some observations about the Colin de Verdière invariant and prove that 10 is the least order for an IK graph with an edge-contraction minor that is IL.Section 6 goes over the error in [10], and we conclude with an appendix that provides edge lists for three graphs we discuss.
To complete this introduction, we provide several definitions.A graph G is n-apex if one can delete n vertices from G to obtain a planar graph; G is apex if it is 1-apex, and 0-apex is a synonym for planar.A graph G is minor minimal with respect to a property if G has that property but no minor of it has that property.The complete graph on n vertices is denoted by K n .V .G/ and E.G/ denote the vertex set and the edge set of G, respectively.A graph G is the clique sum of two subgraphs G 1 and The rY -move and Y r-move are defined as shown in Figure 1.The family of a graph G is the set of all graphs obtained from G by doing zero or more rY and Y r moves.The Petersen family of graphs is the family of the Petersen graph (which is also the family of K 6 ). Figure 2, right.This implies that H is an apex graph; thus it is nIL by [13].Similarly, the graph K shown in Figure 3, left, is nIL since deleting vertex 5 from K yields a maximal planar graph, as in Figure 3, right.
Notice that deleting the vertices 3, 4, 5, and 6 from both H and K produces connected subgraphs.So, by [9,Lemma 14], the clique sum of H and K over the K 4 induced by f3; 4; 5; 6g is a nIL graph, denoted by M and depicted in Figure 4.
The graph G 11;35 is obtained by adding the edge .2;11/ to the nIL graph M (see Figure 5).We prove in Section 3 that G 11;35 is IK.We have thus obtained an IK graph that has a nIL edge deletion minor.Further, since the edge .2;11/ is in a 3-cycle in G 11;35 , this also gives a counterexample to the main result of [10].Notice that contracting the edge .2;3/ in G 11;35 yields a graph that is a minor of M , and therefore nIL.Hence, G 11;35 also has a nIL edge contraction minor.The edge list of G 11;35 is given in the appendix.
Remark The edge .2;3/ in G 11;35 is triangular (ie it belongs to one or more triangles), so contracting it results in the deletion of parallel edges.One can ask whether contracting a nontriangular edge in an IK  Remark The graph G 11;35 is a minimal-order IK graph with a nIL edge deletion minor.To verify this, we took every maxnIL graph of order 10 (there are 107 of them [11]), and checked (with computer assistance) that adding one edge to it never yields an IK graph.However, 11 is not the smallest order of an IK graph that has a nIL edge contraction minor.The graph G 10;30 , depicted in Figure 6, is a minor-minimal IK graph of order 10.Contracting the edge .2;6/ gives the nIL minor in Figure 7, left.This graph is nIL since adding the edge .8;9/ produces a graph isomorphic to the clique sum, over the K 4 subgraph induced by f2; 3; 8; 9g, of K 5 and a subgraph isomorphic to H , introduced in Figure 2. By the following proposition, G 10;30 is a minimal-order IK graph with a nIL edge contraction minor.In Section 5, we show that G 10;30 has -invariant 5. Furthermore, according to our computer program, this graph is MMIK.
Proposition 2.1 Ten is the smallest order for an IK graph which admits a nIL edge contraction minor.
We defer the proof to Section 5. ) has odd linking number.The following lemma was proved by Foisy [4]; a more general version was proved independently by Taniyama and Yasuhara [14].
Lemma 3.1 Every double-linked D 4 contains a nontrivial knot.
We will also use the following (well known and easy to prove) lemma.Proof We shall prove that every embedding of G 10;26 has a double-linked D 4 minor.It then follows from Lemma 3.1 that G 10;26 is IK.For the remainder of this proof, we will say two disjoint simple closed curves ˛and ˇin S 3 are linked, or ˛links ˇ, if ˛[ ˇhas odd linking number.
In G 10;26 we select the subgraphs A, B, C , D, E, and F shown in Figure 10 (these are not induced subgraphs).All these subgraphs are either in the Petersen family of graphs or have minors in this family, and are therefore intrinsically linked: A contains a K 3;3;1 minor obtained by contracting the edge (4,6); B is isomorphic to K 4;4 ; C and F contain K 4;4 minors obtained by contracting the edges .8;9/ and .1;9/, respectively; D and E contain G 7 minors obtained by contracting the edges .6;7/ and .5;7/, respectively.
We organize the proof into several cases and subcases, according to which two cycles of each subgraph are linked.We start with the subgraph A. The vertices of G 10;26 can be partitioned into six equivalence classes up to symmetry: f1; 8g, f2; 3g, f4g, f5; 6g, f7; 10g, and f9g.All of these except vertex 9 are in A. This gives, up to symmetry, four different pairs of cycles in A: .A1/ .4;1; 5/ [ .2;7; 3; 10/; .A3/ .Since A is intrinsically linked, given any embedding of G    Case (A3) Assume .4;6; 7; 5/ [ .2;1; 3; 10/ is a nontrivial link of A. We identify a nontrivial link in D and show the existence of a double-linked D 4 for all cases except one.We then identify a nontrivial link in F and show the existence of a double-linked D 4 for all cases except one.If both exceptional cases occur at the same time, the existence of a double-linked D 4 is shown.

G 10;26 is MMIK
In this section we prove G 10;26 is MMIK by showing that each of its simple minors is nIK.The graph G 10;26 has ten vertices, labeled 1; 2; : : : ; 10. Due to the symmetries of the graph, the vertices can be partitioned into six equivalence classes: f1; 8g, f2; 3g, f4g, f5; 6g, f7; 10g, and f9g.Up to symmetry, G 10;26 has eleven types of edges.Representatives for each possible type of edge are listed in the first column of Table 1.For each such edge type, we constructed two graphs, one by deleting the edge and one by contracting the edge.The graph obtained by deleting the edge is 2-apex, since the removal of the two vertices listed in the second column gives a planar graph.There is one exception: the graph obtained by deleting the edge .4;5/ is not 2-apex.This graph is shown to be nIK in the next paragraph.The graph obtained by contracting the edge listed in the first column is 2-apex, since the removal of the two vertices listed in the third column gives a planar graph.When contracting an edge e, the new vertex inherits the smaller label among the endpoints of e, and all vertices not incident to e maintain their labels.
The graph G 0 obtained from G 10;26 by deleting the edge .4;5/ is not 2-apex.We show it is nIK.Denote by G 00 the graph obtained from G 0 through a rY -move on the triangle .1;5; 9/.Call the new vertex 11; Algebraic & Geometric Topology, Volume 24 (2024) see Figure 29.Delete vertices 2 and 11 of G 00 to obtain a planar graph.This proves G 00 is 2-apex, and thus nIK.Sachs [13] showed that the rY -move preserves intrinsic linking.Essentially the same argument shows that the rY -move also preserves intrinsic knotting.So the graph G 0 is nIK.

D IK graphs
In this section we describe what is known about graphs G with Colin de Verdière invariant 5. We begin with some basic observations.Let K 1 G denote the graph obtained by coning a vertex over G, ie we add a vertex a to G along with edge av for every v 2 V .G/.
Lemma 5.1 [7] Let G be a graph.
For v 2 V .G/, let G v denote the graph that results after deleting v and all its edges.Proof The proof is similar to that of the previous lemma.In particular .G/ 5.By definition, there is an edge e such that G e or G=e is nIL.Suppose first that G e is nIL.By Lemma 5.2, .G e/ Ä 4.
We can form a graph G 0 homeomorphic to G by adding a degree-two vertex between a and b, the vertices of e. Then G 0 is a subgraph of K 1 .G e/, and, using Lemma 5.1, .G 0 / Ä 5. Since G is a minor of G 0 , by Lemma 5.1, .G/ Ä 5.
Next, suppose G=e is nIL, so that .G=e/ Ä 4. We can again recognize G as a subgraph of K 1 .G=e/, which implies .G/ Ä 5. We remark that many of the known MMIK graphs have D 6.In [3], the authors provide a listing of 264 MMIK graphs, of which 105 are in the families of K 7 , K 3;3;1;1 , and E 9 C e. We will now verify that each of these three graphs has D 6.By Lemma 5.3, all 105 graphs have -invariant 6.As shown in [2], .K n / D n 1 when n > 1, so .K 7 / D 6.The graph K 3;3;1;1 is K 1 K 3;3;1 .Since K 3;3;1 is an obstruction for intrinsic linking [12], by Lemma 5.2, .K 3;3;1;1 / D .K 3;3;1 / C 1 6.On the other hand, K 3;3;1;1 is 3-apex, which, by Lemma 5.4, shows .K 3;3;1;1 / Ä 6.Since E 9 is in the K 7 family, by Lemma 5.  30, left, all the others are either in the K 7 family, the K 3;3;1;1 family, or the E 9 +e family, and thus have D 6.It follows that G must have order 9 and that G 9;28 is a subgraph of G.If contracting an edge e of G produces a nIL minor, then deleting either endpoint of e must also produce a nIL minor (subgraph).Since G 9;28 is a subgraph of G, deleting the same vertex must produce a nIL subgraph of G 9;28 .The graph G 9;28 is highly symmetric, having a rich automorphism group, and it is structured as two nonadjacent cones over the complement of a 7-cycle (the graph depicted in Figure 30, right).Up to isomorphism, there are only two induced subgraphs of order 8 inside G 9;28 : the graph obtained by deleting the vertex labeled 9, and the graph obtained by deleting the vertex labeled 7. Neither of these are nIL, since they both have a K 6 minor.For the first graph, contracting the edges .4;7/ and .2;6/ produces a complete minor on the 6 vertices.For the second graph, contracting the edges .4;9/ and .2;6/ also produces a complete minor on the 6 vertices.

Erratum
In this section we discuss an error in the proof of [10,Proposition 2].The proposition asserts that if a graph G has a paneled embedding, and an edge is added to G between two vertices a and b that have a common adjacent vertex v, then G C ab has a knotless embedding.
In the proof of Proposition 2, it is first shown that one can assume there is a path P ab G from a to b disjoint from v. Next, the proof claims that, in any paneled embedding of G, if D is a panel for the cycle P ab [ av [ vb in , then embedding the new edge ab in D yields a knotless embedding 0 of G C ab. Figure 31 shows a counterexample to this claim, and will be used to explain where the error in the proof of Proposition 2 lies.
It is not difficult to see that in Figure 31, left, every cycle in is paneled.In particular, the cycle acdbva bounds a panel D such that vc and vd lie below and above D, respectively, in the figure.If we embed the edge ab in D as in Figure 31, right, we see that the cycle abcvda is a trefoil, and hence 0 isn't a knotless embedding as claimed.
The error is specifically in the last few sentences of the penultimate paragraph in the proof, where it mentions a type 1 Reidemeister move on P 1 [ feg.The proof overlooks the possibility that P bv may prevent this Reidemeister move, as is the case in Figure 31, right (for reference, the paths acdb, adv, and bcv in Figure 31 represent the paths P ab , P av , and P bv , respectively, in the proof of Proposition 2).

Figure 29 :
Figure 29: Left: the graph G 00 obtained from G 10;26 by removing the edge .4;5/ followed by a rY -move on the triangle .1;5; 9/.Center: the graph G 000 obtained from G 00 by deleting vertices 2 and 11.Right: the planar embedding of G 000 .

Lemma 5 . 4 5 .Lemma 5 . 6
If G is n-apex for n 0, then .G/ Ä n C 3.Proof We use induction on n.If n D 0, the result follows from Lemma 5.2.Suppose G is .nC1/-apexandv 2 V is such that G v is n-apex.Then G is a subgraph of K 1 .G v/,and, by Lemma 5.1, .G/ Ä .G v/ C 1 Ä .nC 1/ C 3. Lemma 5.5 If G is IK and there is a vertex v such that G v is nIL , then .G/ D 5. Proof Robertson, Seymour, and Thomas [12] established that G being IK implies G is IL.By Lemma 5.2, .G/ 5 and .G v/ Ä 4. Since G is a subgraph of K 1 .G v/, using Lemma 5.1, .G/ Ä For e 2 E.G/, let G e denote the edge deletion minor and G=e the edge contraction minor of G.If G is IK and has a nIL simple minor, then .G/ D 5.

Figure 31 :
Figure 31: Left: every cycle in is paneled.Right: 0 contains a trefoil.Algebraic & Geometric Topology, Volume 24 (2024) graph can result in a nIL graph.The answer is yes: In G 11;35 , if we do a rY move on the triangle with vertices 2; 3, and 11, we obtain a new IK graph G 0 with a new vertex, denoted by x.Contracting the edge .x;3/ -which is nontriangular -in G 0 yields a graph isomorphic to G 11;35 .2;11/, which is nIL.

Table 1 :
Thomas W Mattman, Ramin Naimi, Andrei Pavelescu and Elena Pavelescu The graph obtained by deleting the edge in the first column becomes planar when deleting the two vertices in the second column.The graph obtained by contracting the edge in the first column becomes planar when deleting the two vertices in the second column.
[8].E 9 / D .K 7 / D 6.By Lemma 5.1, .E 9 C e/ .E Since G 10;26 is a minor of G 11;35 , we have .G 10;26 / Ä .G 11;35 / D 5. On the other hand, as we proved in Section 3, G 10;26 is IK, hence IL [12], and .G 10;26 / 5 by Lemma 5.2.By Lemma 5.3, graphs in the families of G 10;26 , G 10;30 , and G 11;35 also have D 5. Using computers, the G 10;26 family alone provides more than 600new examples of IK graphs with Colin de Verdière invariant 5. Assume there exists an IK graph G of order less than 10 which admits a nIL edge contraction minor.As such, by Lemma 5.6, .G/ D 5. Since is minor monotone (Lemma 5.1), any MMIK minor of G must have D 5. By work of Goldberg, Mattman, and Naimi[6], and Mattman, Morris, and Ryker[8], the MMIK graphs of order at most 9 are known.With the exception of G 9;28 , depicted in Figure [5] D 6. On the other hand, E 9 C e is 3-apex, so, byLemma 5.4, .E 9 C e/ Ä 6.By Lemma 5.3, all 110 graphs in the E 9 C e family have D 6 (not just the 33 that are MMIK).Note that these 110 graphs are all IK[6].In contrast, here we have introduced several new examples of IK graphs with D 5. Such examples were known previously.For example, Foisy[5]provided an example of an MMIK graph F that becomes nIL on deletion of a vertex.By Lemma 5.5, .F / D 5. By Lemma 5.6, .G 11;35 / D 5 as it is IK with both a nIL edge deletion minor as well a nIL edge contraction minor.Similarly, .G 10;30 / D 5 since it is IK with a nIL edge contraction minor.Finally, we argue that .G 10;26 / D 5.