aa r X i v : . [ m a t h . G T ] O c t GRAPHS OF 20 EDGES ARE –APEX, HENCE UNKNOTTED THOMAS W. MATTMAN
Abstract.
A graph is 2–apex if it is planar after the deletion of at mosttwo vertices. Such graphs are not intrinsically knotted, IK. We investigatethe converse, does not IK imply 2–apex? We determine the simplest possiblecounterexample, a graph on nine vertices and 21 edges that is neither IK nor2–apex. In the process, we show that every graph of 20 or fewer edges is2–apex. This provides a new proof that an IK graph must have at least 21edges. We also classify IK graphs on nine vertices and 21 edges and find nonew examples of minor minimal IK graphs in this set. Introduction
We say that a graph is intrinsically knotted or IK if every tame embedding of thegraph in R contains a non-trivially knotted cycle. Blain, Bowlin, et al. [BBFFHL]and Ozawa and Tsutsumi [OT] independently discovered an important criterion forintrinsic knotting. Let H ∗ K denote the join of the graph H and the completegraph on two vertices, K . Proposition 1.1 ([BBFFHL],[OT]) . A graph of the form H ∗ K is IK if and onlyif H is non–planar. A graph is called l –apex if it becomes planar after the deletion of at most l vertices (and their edges). The proposition shows that 2–apex graphs are not IK.It’s known that many non IK graphs are 2–apex. As part of their proof thatintrinsic knotting requires 21 edges, Johnson, Kidwell, and Michael [JKM] showedthat every triangle-free graph on 20 or fewer edges is 2–apex and, therefore, notknotted. In the current paper, we show Theorem 1.2.
All graphs on 20 or fewer edges are –apex. This amounts to a new proof that
Corollary 1.3.
An IK graph has at least 21 edges.
Moreover, we also show
Proposition 1.4.
Every non IK graph on eight or fewer vertices is –apex. This suggests the following:
Question 1.5.
Is every non IK graph –apex? We answer the question in the negative by giving an example of a graph, E ,having nine vertices and 21 edges that is neither IK nor 2–apex. (We thank RaminNaimi [N] for providing an unknotted embedding of E , which appears as Figure 8 Mathematics Subject Classification.
Primary 05C10, Secondary 57M15.
Key words and phrases. spatial graphs, intrinsic knotting, apex graphs. in Section 3.) Further, we show that no graph on fewer than 21 edges, no graph onfewer than nine vertices, and no other graph on 21 edges and nine vertices has thisproperty. In this sense, E is the simplest possible counterexample to our Question.The notation E is meant to suggest that this graph is a “cousin” to the set of14 graphs derived from K by triangle–Y moves (see [KS]). Indeed, E arises froma Y–triangle move on the graph F in the K family. Although intrinsic knottingis preserved under triangle–Y moves [MRS], it is not, in general, preserved underY–triangle moves. For example, although F is derived from K by triangle–Ymoves and, therefore, intrinsically knotted, the graph E , obtained by a Y–trianglemove on F , has an unknotted embedding.Our analysis includes a classification of IK and 2–apex graphs on nine verticesand at most 21 edges. Such a graph is 2–apex unless it is E , or, up to addition ofdegree zero vertices, one of four graphs derived from K by triangle–Y moves [KS].(Here | G | denotes the number of vertices in the graph G and k G k is the number ofedges.) Proposition 1.6.
Let G be a graph with | G | = 9 and k G k ≤ . If G is not –apex,then G is either E or else one of the following IK graphs: K ⊔ K ⊔ K , H ⊔ K , F , or H . The knotted graphs are exactly those four descendants of K : Proposition 1.7.
Let G be a graph with | G | = 9 and k G k ≤ . Then G is IK iffit is K ⊔ K ⊔ K , H ⊔ K , F , or H . In particular, we find that there are no new minor minimal IK graphs in the setof graphs of nine vertices and 21 edges.We remark that a result of Sachs [S] suggests a similar analysis of 1–apex graphs.A graph is intrinsically linked (IL) if every tame embedding includes a pair of non-trivally linked cycles.
Proposition 1.8 (Sachs) . A graph of the form H ∗ K is intrinsically linked if andonly if H is non–planar. It follows that 1–apex graphs are not IL and one can ask about the converse. Acomputer search suggests that the simplest counterexample (a graph that is neitherIL nor 1–apex) in terms of vertex count is a graph on eight vertices and 21 edgeswhose complement is the disjoint union of K and a six cycle. B¨ohme also gavethis example as graph J in [B]. In terms of the number of edges, the disjoint unionof two K , ’s is a counterexample of eighteen edges. It’s straightforward to verify,using methods similar to those presented in this paper, that a counterexample musthave at least eight vertices and at least 15 edges. Beyond these observations, weleave open the Question 1.9.
What is the simplest example of a graph that is neither IL nor –apex? The paper is organized into two sections following this introduction. In the firstwe prove Theorem 1.2. In the second we prove Propositions 1.4, 1.6, and 1.7.2.
Graphs on twenty edges
In this section we will prove Theorem 1.2, a graph of twenty or fewer edges is2–apex. We will use induction and break the argument down as a series of six
RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 3 propositions that, in turn, treat graphs with eight or fewer vertices, nine vertices,ten vertices, eleven vertices, twelve vertices, and thirteen or more vertices. Follow-ing a first subsection where we introduce some useful definitions and lemmas, wedevote one subsection to each of the six propositions.2.1.
Definitions and Lemmas.
In this subsection we introduce several definitionsand three lemmas. The first lemma and the definitions that precede it are based onthe observation that, in terms of topological properties such as planarity, 2–apex,or IK, vertices of degree less than three can be ignored.Let N ( c ) denote the neighbourhood of the vertex c . Definition 2.1.
Let c be a degree two vertex of graph G . Let N ( c ) = { d, e } . Smoothing c means replacing the vertex c and edges cd and ce with the edge de toobtain a new (multi)graph G ′ . If de was already an edge of G , we can remove oneof the copies of de to form the simple graph G ′′ . We will say G ′′ is obtained from G by smoothing and simplifying at c . We will use δ ( G ) to denote the minimal degree of G , i.e., the least degree amongthe vertices of G . Definition 2.2.
Let G be a graph. The multigraph H is the topological simpli-fication of G if δ ( H ) ≥ and H is obtained from G by a sequence of the followingthree moves: delete a degree zero vertex; delete a degree one vertex and its edge;and smooth a degree two vertex. Definition 2.3.
Graphs G and G are topologically equivalent if their topo-logical simplifications are isomorphic. The following lemma demonstrates that in our induction it will be enough toconsider graphs of minimal degree at least three, δ ( G ) ≥
3. For a a vertex of graph G , let G − a denote the induced subgraph on the vertices other than a : V ( G ) \ { a } .Similarly, G − a, b and G − a, b, c will denote induced subgraphs on V ( G ) \ { a, b } and V ( G ) \ { a, b, c } . Lemma 2.4.
Suppose that every graph with n > vertices and at most e edges is –apex. Then the same is true for every graph with n + 1 vertices, at most e + 1 (respectively, e ) edges, and a vertex of degree one or two (respectively, zero).Proof. Let G have n + 1 vertices and e ′ edges where n > e ′ ≤ e + 1.If G has a degree zero vertex, c , we assume further that e ′ ≤ e . In this case,deleting c results in a 2–apex graph G − c , i.e., there are vertices a and b such that G − a, b, c is planar. This implies G − a, b is also planar so that G is 2–apex.If G has a vertex c of degree one, we may delete it (and its edge) to obtain agraph, G − c on n vertices with e ′ − G − c is 2–apex,so that G − a, b, c is planar for an appropriate choice of a and b . This means G − a, b is also planar so that G is 2–apex.If G has a vertex c of degree two, smooth and simplify that vertex to obtain thegraph G ′ on n vertices and e ′ − e ′ − a, b in V ( G ′ ) such that G ′ − a, b is planar. Since V ( G ′ ) = V ( G ) \ { c } , a and b arealso vertices in G . Notice that G − a, b is again planar so that G is 2–apex. (cid:3) In showing that all graphs of 20 or fewer edges are 2–apex, we will frequentlyinvestigate a graph G of 20 edges and delete two vertices to obtain G ′ = G − a, b THOMAS W. MATTMAN which we may assume to be non–planar. By the previous lemma, we can assume G has no vertices of degree less than three (i.e., δ ( G ) ≥ δ ( G ′ ) ≥ G − a, b of this form.In the proof we will make use of the Euler characteristic χ ( G ) = | G | − k G k ,where | G | is the number of vertices and k G k the number of edges. Lemma 2.5.
Let G be a non–planar graph on n vertices, where n ≥ , with δ ( G ) ≥ . Then G has at least n + 3 − ⌊ ( n − / ⌋ edges.Proof. First observe that if G is connected, G will have at least n + 3 edges. Indeed,by Kuratowski’s theorem, G must have K or K , as a minor. If there is a K , minor, then we can construct K , from G by a sequence of edge deletions andcontractions. Since both G and K , are connected, we can arrange for the sequenceto pass through a sequence of connected graphs. We will delete any multiple edgesthat result from an edge contraction so that the intermediate graphs are also simple.To complete the argument notice that an edge deletion or contraction can onlyincrease the Euler characteristic. As χ ( K , ) = −
3, we conclude that χ ( G ) ≤ − k G k ≥ n + 3. If, instead G has a K minor, then, since χ ( K ) = −
5, asimilar argument shows that k G k ≥ n + 5 > n + 3.If G is not planar, then it must have a connected component G ′ for which χ ( G ′ ) ≤ −
3. Additional components will increase χ ( G ) only if they are trees, i.e., χ ( G ) ≤ − T where T denotes the number of tree components of G . If G ′ has atleast six vertices, then, as a tree component requires at least two vertices (recall that δ ( G ) ≥ T ≤ ⌊ ( n − / ⌋ . Thus k G k ≥ n +3 − T ≥ n +3 −⌊ ( n − / ⌋ , asrequired. If G ′ doesn’t have six vertices, then G ′ = K and χ ( G ′ ) = −
5. In this case,a similar argument shows that k G k ≥ n + 5 − ⌊ ( n − / ⌋ > n + 3 − ⌊ ( n − / ⌋ . (cid:3) Table 1.
A count of non–planar graphs with δ ( G ) ≥
1. Columnsare labelled by the number of edges and rows by the number ofvertices.
Remark 2.6.
Table 1 gives the number of graphs satisfying the hypotheses of thelemma. Moreover, using the reasoning outlined in the proof of the lemma, we cancharacterise such a non–planar graph G according to the number of vertices asfollows.If G has six vertices and nine edges, then G = K , . If | G | = 6 and k G k = 10 ,then G = K , with one additional edge.If G has seven vertices and ten edges, it is one of the two graphs illustrated inFigure 1 obtained from K , by splitting a vertex. As for | G | = 7 and k G k = 11 , RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 5
Figure 1.
Two non–planar graphs with seven vertices and ten edges.
Figure 2.
The nine non–planar graphs with seven vertices andeleven edges. there are nine such graphs obtained by splitting a vertex of a non–planar graph onsix vertices or else by adding an edge to a graph on ten edges, see Figure 2.
THOMAS W. MATTMAN
Figure 3.
Non–planar graphs with eight vertices and eleven edges.
The disjoint union K , ⊔ K is the only graph G with eight vertices and tenedges. The 11 graphs G with | G | = 8 and k G k = 11 are illustrated in Figure 3.Two of the three graphs with | G | = 9 and k G k = 11 are formed by the union of K with the two graphs having seven vertices and ten edges. The third is the unionof K , and the tree of two edges.The unique graph with | G | = 10 and k G k = 11 is K , ⊔ K ⊔ K . Of the 15graphs with | G | = 10 and k G k = 12 , 11 are formed by the union of K with one of RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 7 the non–planar graphs on eight vertices and eleven edges, two are the union of thetree of two edges with a non–planar graph on seven vertices and ten edges, and theremaining two are formed by the union of K , with the two trees of three edges.The graphs with | G | = 11 and k G k = 12 are formed by the union of K with anon–planar graph on nine vertices and 11 edges. If G has vertices and edges,then, it is either K ⊔ K ⊔ K ⊔ K , or else it has exactly one tree component, therest of the graph having a K , minor. Almost all of the graphs mentioned in the remark have K , minors. The fol-lowing definition seeks to take advantage of this. Definition 2.7.
Let G be a graph with vertex v and let v , v , v and w , w , w denote the vertices in the two parts of K , . The pair ( G ; v ) is a generalised K , if the induced subgraph G − v is topologically equivalent to K , − v . It followsthat the vertices of G − v can be partitioned into five disjoint sets V , V , W , W ,and W , where each of these five sets induces a tree as a subgraph of G − v , suchthat when each of these trees is contracted down to a single vertex, the tree on V i becomes the vertex v i in K , − v and similarly for the W i . When there is a choiceof partitions, a partition of a generalised K , will be one for which V and V areminimal. We next observe that when G − a, b is a generalised K , this will have implica-tions for N ( a ) and N ( b ), under the assumption that G is not 2–apex. Lemma 2.8.
Suppose that G is not –apex and that ( G − a, b ; c ) is a generalised K , . Then N ( a ) and N ( b ) each include at least one vertex from each of W , W ,and W .Proof. Let V , V , W , W , and W be the partition of the vertices of G − a, b, c asin the definition of a generalised K , .Suppose a has no neighbour in W . Note that by contracting the subgraphs of G − b, c induced by V , V , W , W , and W , we obtain a (multi)graph ( K , − v )+ a formed by adding a vertex a to K , − v . As a is not adjacent to w , it follows that( K , − v ) + a has a planar embedding. Now, reversing the contractions performedearlier, this results in a planar embedding of G − b, c , a contradiction.Therefore, a has a neighbour in W . Similarly, b also has a neighbour in W ,and both a and b have neighbours in W and W . (cid:3) Remark 2.9.
Lemma 2.8 also applies (with obvious modifications) when G − a, b has a generalised K , component with the remaining components being trees. Eight or fewer vertices.
We are now in a position to prove Theorem 1.2.We begin with graphs of eight or fewer vertices.
Remark 2.10.
In what follows, we will often make use of the following strategy.To argue that a graph G is –apex, proceed by contradiction. Assume G is not –apex. This means that every subgraph of the form G − a, b is non–planar. Usingthis assumption we eventually deduce that a particular G − a, b is planar. Althoughwe won’t always say it explicitly, in demonstrating a planar G − a, b , we have infact derived a contradiction that shows that G is –apex. Proposition 2.11.
A graph G with | G | ≤ and k G k ≤ is –apex. THOMAS W. MATTMAN
Proof.
We can assume | G | ≥ G is planar and a fortiori | G | ≤
7, then G is a proper subgraph of K . So, with an appropriate choice ofvertices a and b , G − a, b is a proper subgraph of K and therefore planar. Thus, G is 2–apex.So, we may assume | G | = 8 and we will also take k G k = 20. We will investigateinduced subgraphs G − a, b formed by deleting two vertices a and b . Notice that a and b may be chosen so that k G − a, b k ≤
10. Indeed, the maximum degree of G is at most seven, while the pigeonhole principle implies the maximum degree is atleast five: 5 ≤ ∆( G ) ≤
7. By Lemma 2.4, the minimum degree is at least three: δ ( G ) ≥
3. Since k G k = 20, the sum of the vertex degrees is 40 and it follows thatthere are vertices a and b such that G − a, b has at most ten edges.Assume G is not 2–apex. Then for each pair of vertices a and b , G − a, b is notplanar. By Lemma 2.5 such a non–planar G − a, b has at least nine edges. Thus,it will suffice to consider the cases where G has a non–planar subgraph G − a, b ofnine or ten vertices. We may assume d ( a ) ≥ d ( b ).Suppose first that G − a, b is non–planar and has nine edges. By Remark 2.6, G − a, b = K , . Let v , v , v be the vertices in one part of K , and w , w , w those in the other. Since k G k = 20, k G − a, b k = 9, and d ( a ) ≥ d ( b ), then d ( a ) isseven or six. In either case, k N ( a ) ∩ N ( b ) ∩ V ( G − a, b ) k ≥
3, so we can assume v and v , say, are in the intersection. If d ( a ) = 7, it follows that G − a, v is planarand G is 2–apex. If d ( a ) = 6, by Lemma 2.8, { w , w , w } ⊂ N ( b ). But then, since k N ( a ) ∩ V ( G − a, b ) k ≥
5, we can assume aw ∈ E ( G ) (i.e., aw is an edge of G )and it follows that G − a, w is planar whence G is 2–apex.Next suppose G − a, b is non–planar and has ten edges. That is, by Remark 2.6, G − a, b is K , with an extra edge. Again, v i and w i , ( i = 1 , ,
3) will denote thevertices in the two parts of K , and let v v be the additional edge.Suppose first that d ( a ) = 5. This implies d ( b ) = 5, ab E ( G ), and there are fouror five elements in N ( a ) ∩ N ( b ). If five, then G has K , as an induced subgraphafter deleting two vertices, a case we considered earlier. So, we can assume thereare four vertices in the intersection, including at least one of the vertices v , v , v ,call it v and at least one w i vertex, say w . Then, G − v, w is planar and G is2–apex.So, we can assume d ( a ) >
5. By Lemma 2.8, { w , w , w } ⊂ N ( b ). In that case,without loss of generality, aw ∈ E ( G ). Then G − a, w is planar and G is 2–apex.This completes the argument when G − a, b has ten edges.We have shown that when | G | = 8 and k G k = 20, G is 2–apex. It follows thatgraphs having | G | = 8 and k G k ≤
20 are also 2–apex. (cid:3)
Nine vertices.
In this subsection we prove Theorem 1.2 in the case of graphsof nine vertices. We begin with a lemma.
Lemma 2.12.
Let G be a graph with | G | = 9 , k G k = 20 , δ ( G ) ≥ , ∆( G ) = 5 ,and such that all degree five vertices are mutually adjacent. Then G is –apex.Proof. The degree bounds imply that G has four, five, or six degree five vertices.If G has six degree five vertices, then, as they are mutually adjacent, G has a K component. This implies the other component, on three vertices, has at most threeedges and the graph has at most 18 edges in total, which is a contradiction. So, infact, G cannot have six degree five vertices. RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 9 If G has five degree five vertices, then the induced subgraph on the other fourvertices has five edges, so it is K − e ( K with a single edge deleted). Let c bea degree four vertex that has degree two in the induced subgraph K − e and let a and b be the two degree five neighbours of c . Then G − a, b is planar and G is2–apex. Figure 4.
The six graphs of six edges on five vertices.If G has four degree five vertices, then the induced subgraph on the other fivevertices has six edges, so it is one of the six graphs in Figure 4. For graphs i, ii,and iii, the argument is similar to the previous case. That is, let c be a degree fourvertex of G that has degree two in the induced subgraph and let a and b be thedegree five neighbours of c . Then G − a, b is planar and G is 2–apex. For graph iv,if a and b are any of the degree five vertices, G − a, b is planar and G is 2–apex.For graphs v and vi, the argument is a little more involved, but, again, there arevertices a and b such that G − a, b is planar and G is 2–apex. (cid:3) We are now ready to prove Theorem 1.2 in the case of nine vertices.
Proposition 2.13.
A graph G with | G | = 9 and k G k ≤ is –apex.Proof. First, we’ll assume k G k = 20. Then, 5 ≤ ∆( G ) ≤ δ ( G ) ≥
3. If ∆( G ) >
5, by appropriate choice of vertices a and b , G − a, b has atmost ten edges. This is also true when ∆( G ) = 5 unless all degree 5 vertices aremutually adjacent. As Lemma 2.12 treats that case, we may assume that there isa G − a, b of at most ten edges. Moreover, we’ll take d ( a ) ≥ d ( b ).Assuming G is not 2–apex, then that G − a, b is non–planar. By Remark 2.6, G − a, b is one of the two graphs in Figure 1. Suppose first that it is the graph atleft in the figure. As u has degree three or more in G , both a and b are adjacent to u . By Lemma 2.8, { w , w , w } ⊂ N ( b ). Without loss of generality, we can assume aw ∈ E ( G ). Then G − a, w is planar and G is 2–apex.Suppose, then, that G − a, b is the graph at right in Figure 1. By Lemma 2.8, { w , w } ⊂ N ( a ) ∩ N ( b ) and at least one of w or u is a neighbour of each a and b .Now, as G is not 2–apex, G − w , w is non–planar and it is also a graph on sevenvertices and ten edges with either u or w of degree at least four. In other words, G − w , w is the graph on the left of Figure 1, a case we considered earlier.We have shown that if k G k = 20, then G is 2–apex. It follows that the same istrue for graphs with k G k ≤ (cid:3) Ten vertices.
In this subsection we prove Theorem 1.2 for graphs of tenvertices. We begin with a lemma that treats the case of a graph of degree four.
Lemma 2.14.
Suppose G is a graph with | G | = 10 , k G k = 20 , and such that everyvertex has degree four. Then G is –apex.Proof. We can assume that G has at least three vertices a , b , and c that are pairwisenon–adjacent for otherwise G must be K ⊔ K and is 2–apex. Then ∆( G − a, b ) = 4as c will retain its full degree in G − a, b . Also, δ ( G − a, b ) = 2; since c N ( a ) ∪ N ( b ), a and b must share at least one neighbour in the remaining seven vertices. Thiswill become a degree two vertex in G − a, b .Now, G − a, b is a graph on eight vertices and 12 edges with at least one degreetwo vertex. Smoothing that vertex, we arrive at G ′ , a multigraph on seven verticesand 11 edges that we can take to be non–planar (otherwise G is 2–apex). In otherwords, G ′ is either one of the graphs in Figure 2 or else one of the two graphs inFigure 1 with an edge doubled. Moreover, ∆( G ′ ) = 4 and δ ( G ′ ) ≥
2. Examiningthese candidates for G ′ , we see that G − a, b has degree sequence { , , , , , , , } or { , , , , , , , } . Figure 5.
The six non–planar graphs with degree sequence { , , , , , , , } .The six non–planar graphs with degree sequence { , , , , , , } (see Figure 5)are obtained by either doubling an edge at u in the graph on the right of Figure 1or else by adding a degree two vertex to graph v of Figure 2. If G − a, b is one ofthe graphs ii, iii, or iv in Figure 5, then we argue that G is 2–apex as follows. Byapplying Lemma 2.8 to ( G − a, b ; v ), we find { w , w } ⊂ N ( a ) ∩ N ( b ). But then d ( w ) = d ( w ) = 5, contradicting our hypothesis that all vertices have degree four.A similar argument (using ( G − a, b ; w ) and v , v in place of w , w ) applies when G − a, b is graph i. For graphs v and vi, the same approach shows that at least one RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 11 of w and w has degree five. The contradiction shows that G is 2–apex in case G − a, b has degree sequence { , , , , , , } .So, we may assume G − a, b has degree sequence { , , , , , , , } . Then G − a, b is either obtained by doubling an edge of the graph at right in Figure 1 or else byadding a degree two vertex to graph iii, iv, vi, or viii of Figure 2.Suppose first that G − a, b comes from doubling an edge of the right graph ofFigure 1 (and adding a degree two vertex to one of the two edges in the double).Up to symmetry, the doubled edge is either v w or else v w . In either case,( G − a, b ; w ) is a generalised K , , whence v ∈ N ( a ) ∩ N ( b ). But then d ( v ) = 5in contradiction to our hypotheses. So G is 2–apex in this case.Finally, to complete the proof, suppose G − a, b is graph iii, iv, vi, or viii ofFigure 1. The strategy here is similar to the previous case. We identify a degreefour vertex, c , of G − a, b , ( c is v , except for graph viii in which case c is v ) andobserve that ( G − a, b ; c ) is a generalised K , . We then find a vertex x (either w or w depending on the placement of the degree two vertex) that must lie in N ( a ) ∩ N ( b ). Consequently d ( x ) = 5, a contradiction. The contradiction showsthat G is 2–apex. (cid:3) We can now prove Theorem 1.2 for graphs of ten vertices.
Proposition 2.15.
A graph G with | G | = 10 and k G k ≤ is –apex.Proof. Suppose | G | = 10 and k G k = 20. Then 9 ≥ ∆( G ) ≥
4. By Lemma 2.4, wecan take δ ( G ) ≥ G − a, b is non–planar, it has at least tenedges. So, we may assume ∆( G ) ≤ a and b sothat k G − a, b k <
10 whence G is 2–apex.If ∆( G ) = 7, then G is 2–apex unless every subgraph G − a, b has at least tenedges. So, we can assume G has degree sequence { , , , , , , , , , } with eachof the degree four vertices adjacent to the vertex, a , of degree seven. For almost allchoices of b , k G − a, b k = 10 so that, by Remark 2.6, G − a, b = K , ⊔ K . Then G − a, b has two degree one vertices which must arise from degree three verticesof G from which two edges have been deleted. This implies a is adjacent to atleast two degree three vertices in G . This is a contradiction as N ( a ) includes onlyone degree three vertex, the remaining six vertices being those of degree four. Thecontradiction shows that G is 2–apex in case ∆( G ) = 7.If ∆( G ) = 4, then, in fact every vertex of G has degree four. This case is treatedin Lemma 2.14. Thus, the remainder of this proof treats the case where ∆( G ) = 6or 5. Then there are vertices a and b such that k G − a, b k ≤
11. By Remark 2.6we may assume G − a, b is either K , ⊔ K or else one of the graphs in Figure 3.Further, we will assume ∆( G ) = d ( a ) ≥ d ( b ).Suppose G − a, b is K , ⊔ K and let v , v , v and w , w , w be the vertices inthe two parts of K , while u , u will denote the vertices of K . By Remark 2.9,( K , ; v ) shows { w , w , w } ⊂ N ( b ). Similarly, ( K , ; w ) implies { v , v , v } ⊂ N ( b ). Finally, as u and u have degree one in G − a, b , both must be adjacent to b in G . This implies d ( b ) ≥ G ) ≤ G is 2–apex in case it has a subgraph of the form K , ⊔ K .We may now assume that k G − a, b k = 11 and that for any other pair a ′ , b ′ , k G − a ′ , b ′ k ≥
11. This allows us to dismiss the case where ∆( G ) = d ( a ) = 6.Indeed, the condition k G − a ′ , b ′ k ≥
11 then implies that the other vertices of G have degree at most four and each degree four vertex is adjacent to a . But then G would have degree sequence { , , , , , , , , , } and there are too many degreefour vertices for them all to be adjacent to a . The contradiction shows that G is2–apex in this case.Suppose then that ∆( G ) = 5, δ ( G ) ≥
3, and that for every choice of a ′ and b ′ , k G − a ′ , b ′ k ≥
11. Further, let a and b be vertices such that k G − a, b k = 11. Then G − a, b is one of the graphs in Figure 3 and we can assume that d ( a ) = 5. Thefollowing argument applies to all but the last two graphs in the figure.By Lemma 2.8 (or Remark 2.9), { w , w } ⊂ N ( a ) ∩ N ( b ). However, either this isalready a contradiction because w or w now has degree greater than ∆( G ) = 5, orelse, d ( w ) = d ( w ) = 5. In the latter case, as w w E ( G ) then k G − w , w k = 10,contradicting our assumption that k G − a ′ , b ′ k ≥
11. The contradiction shows that G is 2–apex.Similar considerations show that if G − a, b is graph x or xi of Figure 3, then,again, G must be 2–apex. This completes the argument in the case that k G k = 20.We have shown that if k G k = 20, then G is 2–apex. It follows that the same istrue for graphs with k G k ≤ (cid:3) Eleven vertices.
In this subsection, we prove Theorem 1.2 for graphs of 11vertices. We begin with a lemma that handles the case where ∆( G ) = 4. Lemma 2.16.
Let G have | G | = 11 , k G k = 20 , and ∆( G ) = 4 . Then G is –apex.Proof. By Lemma 2.4, we can take δ ( G ) ≥ G has degree sequence { , , , , , , , , , , } . Let a and b be two non–adjacent vertices of degreefour. Then G − a, b has nine vertices and 12 edges. Since k G − a, b k = 12 and δ ( G − a, b ) ≥
1, we see that G − a, b has at least two vertices of degree less thantwo. Deleting or smoothing those two, we arrive at a multigraph G ′ with seven ver-tices and ten edges. We can assume G ′ is non–planar as otherwise G − a, b is planarand G is 2–apex. Thus G ′ is either one of the two graphs in Figure 1, K , ⊔ C where C is a loop on a single vertex, K ⊔ K ⊔ K , or else the union of K and K , with an extra edge. We will consider these five possibilities in turn.If G ′ is K ⊔ K ⊔ K , then G − a, b = K ⊔ K ⊔ K . In order to bring the fourdegree one vertices of G − a, b up to degree three in G , each must be adjacent toboth a and b . Then the induced subgraph on a , b , and the vertices of the two K ’sis planar so that G is not only 2–apex, it’s actually 1–apex.Suppose next that G ′ is the union of K and K , with an extra edge. Let v , v , v and w , w , w be the vertices in the two parts of K , . Without loss ofgenerality, the extra edge of K , is either v w (doubling an existing edge) or else v v . By Remark 2.9, a and b both have neighbours in the three sets W , W , and W . Moreover, at least one of these three sets consists of a single vertex w . Butthen d ( w ) = 5, a contradiction. The contradiction shows that G is 2–apex in thiscase. If G ′ = K , ⊔ C or G ′ is the graph at the left of Figure 1, the same argumentapplies and we conclude G is 2–apex.Now, if G ′ is the graph at the right of Figure 1, then u is a degree two vertexnear w (so that W includes at least those two vertices) and the additional twodegree one and two vertices might lie near w and w so that in the generalised K , , ( G ′ ; v ), none of the W i ’s is a single vertex. For example, G − a, b may begraph i of Figure 6 below. Actually, we can conclude that G − a, b must be graphi. For otherwise, examining ( G − a, b ; v ) in turn for all choices of vertex v , we will RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 13 discover at least one v i or w i vertex, call it w , that must lie in N ( a ) ∩ N ( b ) whichleads to the contradiction that d ( w ) = 5.Thus, we are left to consider the case where G − a, b is graph i of Figure 6 below.Each of the three vertices u , u , and u is adjacent to at least one of a and b as the u i ’s must have degree at least three in G . Without loss of generality, we can assume u and u are neighbours of a . Also, N ( a ) must include at least one vertex from thesix v i and w i vertices. Up to symmetry, this gives two cases: { u , u , v } ⊂ N ( a )and { u , u , v } ⊂ N ( a ).Suppose first that { u , u , v } ⊂ N ( a ). Then in the generalised K , , ( G − a, b ; v ), W = { w , u } and W ∩ N ( a ) = ∅ . But, if aw ∈ E ( G ), then G − b, w is planar. So we can assume that N ( a ) = { u , u , u , v } . Note that v N ( b ) forotherwise d ( v ) = 5, contradicting our assumption about the maximum degree of G .Also, we’ve assumed that ab E ( G ). Then G − u , u is planar unless bu ∈ E ( G ).Similarly, G − u , u and G − u , u show that we can assume u , u ∈ N ( b ). Now,up to symmetry, we can assume that the fourth vertex of N ( b ) is either v , w , or w , so we consider those three cases. If N ( b ) = { u , u , u , v } then G − u , v isplanar and G is 2–apex. If N ( b ) = { u , u , u , w } then G − u , v is planar and G is 2–apex. If N ( b ) = { u , u , u , w } then G − u , v is planar and G is 2–apex.The argument in the case that { u , u , v } ⊂ N ( a ) is similar. (cid:3) Having treated the case where ∆( G ) = 4, we are ready to prove Theorem 1.2 forgraphs of 11 vertices. Proposition 2.17.
A graph G with | G | = 11 and k G k ≤ is –apex.Proof. Suppose | G | = 11 and k G k = 20. Then 10 ≥ ∆( G ) ≥
4. By Lemma 2.4, wecan take δ ( G ) ≥ G − a, b is non–planar, it has at least 11edges. So, we may assume ∆( G ) ≤ a and b sothat k G − a, b k <
11 whence G is 2–apex. Lemma 2.16 deals with graphs having∆( G ) = 4 and we treat the case of ∆( G ) = 6 in the following paragraph.Suppose ∆( G ) = 6 and let a be a vertex of maximum degree. If G is not 2–apex, then, to meet the requirement that k G − a, b k ≥
11 for every choice of b , theremaining vertices have degree three or four with all degree four vertices adjacentto a . It follows that G has degree sequence { , , , , , , , , , , } . Then a is adjacent to exactly two of the degree three vertices, call them c and d . Thus N ( c ) ∪ N ( d ) consists of at most four other vertices beside a . Let b be a vertexnot in N ( c ) ∪ N ( d ). Then G − a, b has 11 edges and no degree one vertex. ByRemark 2.6, G − a, b is planar and G is 2–apex.So, for the remainder of the proof, we assume ∆( G ) = 5. If G is not 2–apex,then, the condition k G − a, b k ≥
11 implies all degree five vertices are mutuallyadjacent. Moreover, either there are vertices a and b with d ( a ) = d ( b ) = 5 and k G − a, b k = 11, or else G has degree sequence { , , , , , , , , , , } .Suppose, first, that k G − a, b k = 11 with d ( a ) = d ( b ) = 5. Assuming G is not2–apex, by Remark 2.6, G − a, b is one of three graphs. If G − a, b is the union of thegraph at the left of Figure 1 and K , then a must be adjacent to each of the threedegree one vertices of G − a, b as otherwise they will have degree at most two in G .By Remark 2.9, { w , w , w } ⊂ N ( a ) which implies d ( a ) ≥
6, a contradiction. So G is 2–apex in this case. If G − a, b is either the union of the graph at the right ofthe figure and K or else the union of K , and a tree on three vertices, again, a must be adjacent to the two degree one vertices in the tree. But, by Remark 2.9, { v , v , w , w } ⊂ N ( a ). This again gives the contradiction d ( a ) ≥
6, which showsthat G is 2–apex in this case as well.Thus, we can assume that G has degree sequence { , , , , , , , , , , } . Fur-ther, we can assume all the degree four vertices are adjacent to a , the vertex ofdegree five. For otherwise, let b be a degree four vertex not adjacent to a . Then k G − a, b k = 11 so it is one of the three graphs mentioned in Remark 2.6, each ofwhich has two degree one vertices. As b is adjacent to all the degree one vertex, ithas at most two neighbours in { v , v , v , w , w , w } . That would imply G − a, b isplanar, a contradiction.So, let a be adjacent to all the degree four vertices. Then G − a has all vertices ofdegree three and, for any vertex b , G − a, b has degree sequence { , , , , , , , , } .Smoothing one of the degree two vertices, we have the multigraph G ′ with | G ′ | = 8and k G ′ k = 11. If G is not 2–apex, then G ′ is non–planar and, by Remark 2.6, iseither K , ⊔ K with one edge doubled or else it is one of the graphs of Figure 3with an additional degree two vertex. Then G − a, b is either K , ⊔ C , where C isthe cycle of three vertices, or else G − a, b is K , with the addition of three degreetwo vertices. However, if G − a, b is K , ⊔ C we deduce that G − a is K , ⊔ K .Let v be one of the vertices of K , , then G − a, v is planar and G is 2–apex. Figure 6.
Three non–planar graphs with degree sequence { , , , , , , , , } .So, we can assume G − a, b is K , with the addition of three degree two vertices.Let v , v , v and w , w , w denote the vertices in the two parts of K , as wellas the corresponding vertices in G − a, b . Suppose the degree two vertices are allon the edges, v w , v w , and v w of K , . Then G − a, v is planar so that G is 2–apex. Thus, we can assume G − a, b is one of the three graphs in Figure 6.Now, if G − a, b is graph ii or iii, then G − a, w is planar and G is 2–apex. So, theremainder of the proof treats the case of graph i.Assume then that G − a, b is graph i of Figure 6 and that G is not 2–apex.Further, let ab E ( G ). Since G − u , u is non–planar, then au ∈ E ( G ) and byremoving the pairs u , u and u , u in turn, we see that we can assume that a is adjacent to u , u , and u . Then a is adjacent to exactly two vertices of K , ,without loss of generality, either v , v ; v , w ; or v , w . Let us examine these threesubcases in turn. If N ( a ) = { u , u , u , v , v } , then G − u , v is planar and G is2–apex. If N ( a ) = { u , u , u , v , w } , then G − u , v is planar and G is 2–apex.If N ( a ) = { u , u , u , v , w } , then G − u , w is planar and G is 2–apex. Thiscompletes the argument in case G − a, b is graph i of Figure 6 and with it the caseof a graph G of twenty edges. RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 15
We have shown that if k G k = 20, then G is 2–apex. It follows that the same istrue for graphs with k G k ≤ (cid:3) Twelve vertices.
In this subsection we prove Theorem 1.2 in the case of agraph of 12 vertices.
Proposition 2.18.
A graph G with | G | = 12 and k G k ≤ is –apex.Proof. Suppose | G | = 12 and k G k = 20. Then 11 ≥ ∆( G ) ≥
4. By Lemma 2.4, wecan take δ ( G ) ≥ G − a, b is non–planar, it has at least 11edges. So, we may assume ∆( G ) ≤ a and b sothat k G − a, b k <
11 whence G is 2–apex.In fact, we can assume ∆( G ) ≤
5. Indeed, suppose instead ∆( G ) = 6 with a avertex of maximum degree. As there are only twenty edges in all, there must be adegree three vertex b not adjacent to a . Then k G − a, b k = 11. If G is not 2–apex,then, by Remark 2.6, G − a, b = K , ⊔ K ⊔ K . However, as d ( b ) = 3, G − a, b canhave at most three degree one vertices. The contradiction shows that G is 2–apexwhen ∆( G ) = 6.Let ∆( G ) = 5 and suppose that G has two degree five vertices a and b . Assuming G is not 2–apex, then G − a, b is non–planar. By Remark 2.6, a and b are adjacentand G − a, b = K , ⊔ K ⊔ K . It follows that each of a and b is adjacent to each ofthe four degree one vertices in G − a, b as these vertices come to have degree threein G . In particular, the induced subgraph on a , b , and the vertices of the two K ’sis planar. If v is a vertex in the K , component of G − a, b , then G − v is planarso that G is 1–apex and, therefore, also 2–apex.So, we can assume G has exactly one degree five vertex a . It follows that G hasexactly two degree four vertices with the remaining vertices of degree three. Wecan assume that both degree four vertices are adjacent to a as otherwise a similarargument to that of the last paragraph shows that G is 1–apex. Let b be one ofthe degree four vertices. Then k G − a, b k = 12. Assuming G is not 2–apex, then G − a, b is non–planar and therefore one of the 15 graphs described in Remark 2.6.However, as a is adjacent to the two degree four vertices, we see that ∆( G − a, b ) = 3which leaves seven candidate graphs: the union of K with graph viii, ix, x, or xi ofFigure 3; the union of the tree on two edges with the graph to the right in Figure 1;or K , union a tree on three edges. (There are two such trees.) We will considereach possibility in turn.If G − a, b is K ⊔ H where H is graph ix, x, or xi of Figure 3, then we deducethat a is adjacent to one of the degree three vertices of H , call it v , as that is theonly way to produce a second degree four vertex in G (besides b ). We claim that G − a, v is planar. Indeed, k G − a, v k = 12. But G − a, v is connected, so it is notone of the non–planar graphs described in Remark 2.6. As G − a, v is planar, G is2–apex.If G − a, b is K ⊔ H where H is graph viii of Figure 3, again, a is adjacent toa degree three vertex of H . If that vertex is one of the six v i or w i vertices, theargument proceeds as above. So assume instead a is adjacent to the seventh degreethree vertex. In this case G − v is planar so G is 1–apex, hence 2–apex.If G − a, b is the union of the right graph of Figure 1, call it H , with a tree T of two edges, we again conclude that if a is adjacent to v , a degree three vertex, of H then G − a, v is planar whence G is 2–apex. The only other way to produce adegree four vertex for G is if a and b are both adjacent to all three vertices of T . However, in this case we find that the subgraph induced by a , b , and the verticesof the tree is planar so that G is 1–apex and, therefore, also 2–apex.Similar arguments apply when G − a, b is the union of K , and the tree P , thepath of three edges: either a is adjacent to a vertex v of K , , which means that G − a, v is planar, or else the graph induced by a , b and P is planar so that G is,in fact, 1–apex, hence 2–apex. As for K , ⊔ S , where S is the star of three edges,again G − a, v is planar where v is the vertex of K , adjacent to a if there is suchand otherwise v is an arbitrary vertex of K , . This completes the argument when∆( G ) = 5.Finally, suppose ∆( G ) = 4. Then there are four degree four vertices with theremaining vertices of degree three. If there are non–adjacent degree four vertices a and b , then k G − a, b k = 12 and the analysis is much as the one just completedin the ∆( G ) = 5 case. That is, we can assume G − a, b is one of the fifteen graphsdescribed in Remark 2.6 with the additional condition that ∆( G − a, b ) ≤ a , b , c , and d , are mutually adjacent. Then c and d become two adjacent degreetwo vertices in G − a, b . Smoothing these we arrive at G ′ where | G ′ | = 8 and k G ′ k = 11. We can assume that G ′ is non–planar (otherwise G − a, b is planar and G is 2–apex) so that it is one of the graphs of Figure 3, K , ⊔ K with an edgedoubled, or else the union of one of the graphs of Figure 1 with C , a loop on onevertex. In addition, ∆( G ′ ) = 3, which leaves six possibilities: graph viii, ix, x, orxi of Figure 3, K , ⊔ C , where C is the cycle on two vertices, or else the union of C and the graph at the right of Figure 1. We’ll consider these in turn.If G ′ is graph viii, ix, x, or xi of Figure 3, let xy be the edge of G ′ that contained c and d before smoothing. That is, x and y are the vertices in G − a, b such that xc , cd , and dy is a path. Then G − x, y is planar and G is 2–apex.If G ′ is K , ⊔ C , then G − a, b is K , ⊔ C with c and d two of the vertices inthe 4–cycle C . Then G − a, v is planar where v is a vertex of K , . Finally, if G ′ is the union of C and the right graph of Figure 1, call it R , then G − a, b is C ⊔ R where c and d are two of the vertices in the 3–cycle C . It follows that G − v isplanar so that G is 1–apex, hence 2–apex.This completes the case where ∆( G ) = 4, and with it the proof for k G k = 20.As usual, since all graphs with k G k = 20 are 2–apex, the same is true for graphswith k G k ≤ (cid:3) Thirteen or more vertices.
In this subsection, we complete the proof ofTheorem 1.2 by examining graphs with 13 or more vertices.
Proposition 2.19.
A graph G with | G | ≥ and k G k ≤ is –apex.Proof. Suppose | G | = 13 and k G k = 20. By Lemma 2.4, we can assume δ ( G ) ≥ G has a single vertex a of degree four with all other vertices of degree three.Let b be a vertex that is not adjacent to a so that k G − a, b k = 13. Assume G isnot 2–apex. Then G − a, b is non–planar. Now, ∆( G − a, b ) = 3, so G − a, b has no K component. By Remark 2.6, G − a, b has exactly one tree component T , withthe rest of the graph G ′ = G − a, b \ T having a K , minor. As δ ( G − a, b ) ≥ ≤ | T | ≤ | T | = 2, then T is K and G ′ = G − a, b \ T is a non-planar graph on ninevertices with 12 edges. As ∆( G ′ ) = 3 and δ ( G ′ ) ≥ G ′ has a vertex of degreetwo. By smoothing that vertex, we have either a multigraph obtained by doubling RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 17 an edge of the graph K , ⊔ K or else one of the graphs of Figure 3. Moreover, as∆( G ′ ) = 3, of the graphs in the figure, only viii, ix, x, and xi are possibilities.Suppose then that, after smoothing and simplifying, G ′ is K , ⊔ K . Then, as∆( G ′ ) = 3, the doubled edge is that of the K and G ′ = K , ⊔ C , where C denotes the cycle on three vertices. Thus, G − a, b = K , ⊔ C ⊔ K . Let c be oneof the vertices in the K , component. Then G − a, c is planar and G is 2–apex.If, after smoothing a degree two vertex, G ′ becomes graph viii, ix, x, or xi ofFigure 3, then G − a, v is planar and G is 2–apex.Next suppose | T | = 3. As, | G ′ | = 8, k G ′ k = 11, and ∆( G ′ ) = 3, we concludethat G ′ is graph viii, ix, x, or xi of Figure 3. Whichever it is, G − a, v will be aplanar subgraph of G so that G is 2–apex.Similarly, if | T | = 4, then | G ′ | = 7, k G ′ k = 10. As ∆( G ′ ) = 3, we conclude that G ′ is the graph to the right of Figure 1. Then G − a, v is planar and G is 2–apex.Finally, if | T | = 5, then | G ′ | = 6 and k G ′ k = 9 so that G ′ is K , . Again, G − a, v is planar and G is 2–apex.We have shown that a graph with | G | = 13 and k G k = 20 is 2–apex. It followsthat the same is true for graphs having | G | = 13 and k G k ≤ | G | ≥
14 and k G k = 20. If δ ( G ) ≥
3, then the degree sum is atleast 3 ×
14 = 42 >
40, a contradiction. So, we may assume δ ( G ) < G is 2–apex by Lemma 2.4. It follows that any graph of 14 or more vertices withfewer than 20 edges is also 2–apex. (cid:3) Graphs on twenty-one edges
In this section we prove Propositions 1.4 (in the first subsection) and Proposi-tions 1.6 and 1.7 (in the second subsection).3.1.
Eight or fewer vertices.
In this subsection we prove Proposition 1.4, a non-IK graph of eight or fewer vertices is 2–apex. This implies that for these graphs2–apex is equivalent to not IK and the classification of 2–apex graphs on eight orfewer vertices follows from the IK classification due to [BBFFHL] and [CMOPRW].
Proposition 1.4.
Every non IK graph on eight or fewer vertices is –apex.Proof. By Proposition 2.11, the theorem holds if k G k ≤
20, so we may assume that k G k ≥
21. The only graph with 21 edges and fewer than eight vertices is K , whichis IK. So we may assume | G | = 8. Figure 7.
Complements of the non IK graphs G and G .Knotting of graphs on eight vertices was classified independently by [CMOPRW]and [BBFFHL]. Using the classification, the non IK graphs with 21 or more edgesare all subgraphs of two graphs on 25 edges, G and G , whose complements appearin Figure 7. Each of these two graphs has at least two vertices of degree seven and,for both graphs, deleting two such vertices leaves a planar subgraph of K . Thus,both G and G are 2–apex and the same is true of any subgraph of G and G . (cid:3) Nine vertices.
In this subsection we prove Propositions 1.6 and 1.7, whichclassify the graphs of nine vertices and at most 21 edges with respect to 2–apexand IK.We begin with Proposition 1.6: among these graphs, all but E (see Figure 8)and four graphs derived from K by triangle–Y moves ( K ⊔ K ⊔ K , H ⊔ K , F , and H , see [KS]) are 2–apex. We first present four lemmas that show this isthe case when there is a subgraph G − a, b of the form shown in Figure 2. The firstlemma shows that we can assume δ ( G − a, b ) ≥
2. The next three treat the fivegraphs (iii, iv, v, vi and viii) of Figure 2 that meet this condition.
Lemma 3.1.
Let G be a graph with | G | = 9 , k G k = 21 and δ ( G ) ≥ . Supposethat, for each pair of vertices a ′ and b ′ , k G − a ′ , b ′ k ≥ with equality for at leastone pair a , b . Then, a and b can be chosen so that one of the following two holds. • The vertices a and b have degrees six and five, respectively, G has one of thefollowing degree sequences: { , , , , , , , , } , { , , , , , , , , } , or { , , , , , , , , } , and a is adjacent to each degree five vertex (including b ). • The vertices a and b both have degree five, G has one of the following degreesequences: { , , , , , , , , } or { , , , , , , , , } , and a and b arenot neighbours.Moreover, a and b can be chosen so that δ ( G − a, b ) ≥ .Proof. We can assume ∆( G ) = d ( a ) ≥ d ( b ). Since k G − a ′ , b ′ k ≥
11 for every pairof vertices a ′ , b ′ , we must have d ( a ) = 6 or d ( a ) = 5.If d ( a ) = 6, the condition k G − a ′ , b ′ k ≥
11 implies that there is exactly onedegree six vertex, a , and every degree five vertex is adjacent to a . As k G k = 21,the degree sum is 42 and, therefore, there are only three possibilities for the degreesequence. In particular, there is always a vertex of degree five b which is adjacentto a so that k G − a, b k = 11.Similarly, if d ( a ) = 5, then the condition k G k = 21 leaves two possible degreesequences. There must be two degree five vertices a and b that are not adjacent sothat k G − a, b k = 11. This is clear for the degree sequence with seven degree fivevertices. In the case of six degree five vertices, if they were all mutually adjacent,they would constitute a K component of 15 edges. The other component hasthree vertices and, at most, three edges. In total, G would have at most 18 edges,a contradiction.Finally, we argue that it is always possible to choose a and b so that δ ( G − a, b ) ≥
2. Indeed, this is obvious when δ ( G ) ≥ a and b can reduce the degreeof the other vertices by at most two. For the sequence { , , , , , , , , } , thedegree six vertex a is adjacent to each degree five vertex and is, therefore, notadjacent to either of the degree three vertices. Hence in G − a, b these degree threevertices have degree at least two and δ ( G − a, b ) ≥
2. For { , , , , , , , , } ,the degree three vertex is adjacent to at most three of the degree five vertices. Bychoosing b as one of the other degree five vertices, we will have δ ( G − a, b ) ≥ { , , , , , , , , } , the degree three vertex is adjacent to at mostthree of the degree five vertices, call them v v , and v . We can find degree fivevertices a and b that are not adjacent and not both neighbours of the degree threevertex (so that δ ( G − a, b ) ≥ v , v , v , and v are all mutually adjacent and also all adjacent to v , v , RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 19 and v . But this is not possible, e.g., d ( v ) = 5, so it cannot have all of the otherdegree five vertices as neighbours. (cid:3) Lemma 3.2.
Let G be a graph with | G | = 9 , k G k = 21 , and δ ( G ) ≥ . Supposethat there are vertices a and b such that G − a, b is graph iii of Figure 2 and forany pair of vertices a ′ and b ′ , k G − a ′ , b ′ k ≥ . If G is not –apex, then G is H .Proof. By Lemma 3.1, d ( a ) = 6 or 5. Also, since u has degree three or more in G ,at least one of a and b is adjacent to u .Assuming G is not 2–apex, by Lemma 2.8, { w , w , w } ⊂ N ( a ) ∩ N ( b ). Then G − u, w is planar (and G is 2–apex) in the case d ( a ) = 5.So, we can assume d ( a ) = 6 and we are in the first case of Lemma 3.1. As above { w , w , w } ⊂ N ( a ) ∩ N ( b ). Then, since G − a, w is non–planar, we deduce that N ( b ) = { a, w , w , w , u } . Finally, since G − w , w is non–planar, v and v arealso neighbours of a , i.e., N ( a ) = { b, v , v , w , w , w } . But these choices of N ( b )and N ( a ) result in the graph H . So, if G is not 2–apex, it is H . (cid:3) Lemma 3.3.
Let G be a graph with | G | = 9 , k G k = 21 , and δ ( G ) ≥ . Supposethat there are vertices a and b such that G − a, b is graph iv, v, or vi of Figure 2and for any pair of vertices a ′ and b ′ , k G − a ′ , b ′ k ≥ . Then G is –apex.Proof. By Lemma 3.1, d ( a ) = 6 or 5. If d ( a ) = 5, note that G − a, b, v , w is acycle. By placing a inside the cycle and b outside, G − v , w is planar and G is2–apex. So, we may assume d ( a ) = 6 and we are in the first case of Lemma 3.1.Assume G is not 2–apex and apply Lemma 2.8 to ( G − a, b ; v ), for which W = { u, w } and W i = { w i } , i = 2 ,
3. If G is graph iv or v, then G − w , w is planarand G is 2–apex. So, let G be graph vi. Then, since G − w , w is non–planar,either v or v , say v , is a neighbour of b . But, then the degree of v in G is at leastfive. We deduce that av ∈ E ( G ), for otherwise, d ( v ) = 5 and, by Lemma 3.1, v is adjacent to a , a contradiction. However, as a is adjacent to v , d ( v ) = 6which again contradicts Lemma 3.1 as a is the unique vertex of degree six. Thecontradiction shows that G is 2–apex. (cid:3) Lemma 3.4.
Let G be a graph with | G | = 9 , k G k = 21 , and δ ( G ) ≥ . Supposethat there are vertices a and b such that G − a, b is graph viii of Figure 2 and forany pair of vertices a ′ and b ′ , k G − a ′ , b ′ k ≥ . If G is not –apex, then G is E .Proof. By Lemma 3.1, d ( a ) = 6 or 5. Also, since u has degree three or more in G ,at least one of a and b is adjacent to u .Assume G is not 2–apex and apply Lemma 2.8 using W = { u, w } and W i = { w i } , i = 2 ,
3, to see that { w , w } ⊂ N ( a ) ∩ N ( b ) and that N ( a ) and N ( b ) bothintersect W . Similarly ( G − a, b ; w ) shows { v , v } ⊂ N ( a ) ∩ N ( b ) and both a and b have a neighbour in V = { u, v } . If d ( a ) = 6, then | N ( b ) ∩ ( V ( G ) \ { a, b } ) | = 4,which contradicts what we already know about N ( b ). So, it must be that d ( a ) = 5,from which it follows that N ( a ) = N ( b ) = { u, v , v , w , w } and that G = E . (cid:3) Having treated graphs containing an induced subgraph as in Figure 2, we areready to prove Proposition 1.6.
Proposition 1.6.
Let G be a graph with | G | = 9 and k G k ≤ . If G is not –apex,then G is either E or else one of the following IK graphs: K ⊔ K ⊔ K , H ⊔ K , F , or H . Figure 8.
An unknotted embedding of the graph E . Proof.
By Theorem 1.2, we can assume k G k = 21.As in the proof of Lemma 2.4, δ ( G ) ≥ G has a vertex of degree lowerthan three whose deletion (or smoothing in the case of a degree two vertex) resultsin a graph that is not 2–apex. As all graphs of 20 edges are 2–apex, this is possibleonly in the case that G has a degree zero vertex; deleting that vertex must result in agraph on eight vertices with 21 edges that is not 2–apex. By Proposition 1.4 such agraph is IK and, by the classification of knotting of eight vertex graphs, we concludethat G is either the union of K with two degree zero vertices, K ⊔ K ⊔ K , orelse G is H ⊔ K , where H is the graph obtained by a single triangle–Y move on K (see [KS]).In other words, so long as G = K ⊔ K ⊔ K and G = H ⊔ K , we can assume δ ( G ) ≥
3. Also, 5 ≤ ∆( G ) ≤
8. Now, if ∆( G ) = 5, there are at least six degree fivevertices and, therefore, there must be a pair of non–adjacent degree five vertices.Thus, whatever the maximum degree ∆( G ), by appropriate choice of vertices a and b , we may assume G − a, b has at most 11 edges.If ∆( G ) = 8, then G is 2–apex. Indeed, if a has degree eight, then, as k G k = 21,there is a vertex b with d ( b ) ≥
5. This means G − a, b has at most nine edgesand, by Lemma 2.5, is planar. So we’ll assume G − a, b has at most 11 edges andthat 7 ≥ d ( a ) ≥ d ( b ) . Assume G is not 2–apex; then G − a, b is non–planar. ByRemark 2.6, G − a, b is one of the two graphs in Figure 1 or one of the nine inFigure 2.Suppose first that G − a, b is the graph at left in Figure 1. Since k G − a, b k = 10,we can assume that d ( a ) = 7 or 6 and d ( b ) ≤
6. As u has degree three in G , both a and b are adjacent to u . By Lemma 2.8, { w , w , w } ⊂ N ( b ). Without loss ofgenerality, we can assume aw ∈ E ( G ). Then G − a, w is planar and G is 2–apex.If G − a, b is the graph at right in Figure 1, then, as u has degree three or morein G , at least one of a and b is a neighbour of u . Again, k G − a, b k = 10 so d ( a ) = 7or 6 and d ( b ) ≤
6. Applying Lemma 2.8 with W = { u, w } and W i = { w i } , i = 2 ,
3, we see that N ( a ) and N ( b ) each include at least one vertex from each RAPHS OF 20 EDGES ARE 2–APEX, HENCE UNKNOTTED 21 W i . Similarly, ( G − a, b ; w ) shows N ( a ) and N ( b ) each include at least one vertexfrom each of V = { u, v } and V i = { v i } . i = 2 ,
3. In particular, we conclude that { v , v , w , w } ⊂ N ( a ) ∩ N ( b ). Then G − w , w is a non–planar graph on sevenvertices and eleven edges, i.e., one of the graphs in Figure 2.In particular, if d ( a ) = 7, then the degree of a in G − w , w is five. The onlygraph of Figure 2 with a degree five vertex is i. However, in that graph, the degreefive vertex is adjacent to a degree one vertex which is not a possibility for a . So,we conclude G − w , w is planar and G is 2–apex if d ( a ) = 7.Thus, we can assume d ( a ) = 6 and d ( b ) = 5 or 6. If d ( b ) = 5, then the discussionabove shows that N ( b ) = { u, v , v , w , w } and G − v , w is planar (so that G is 2–apex). If d ( b ) = 6, then ab ∈ E ( G ) which implies N ( a ) ∩ N ( b ) = { u, v , v , w , w } and G = F .We can now assume that there is a pair of vertices a and b such that k G − a, b k =11 and G − a, b is one of the nine graphs in Figure 2. Moreover, we can also positthat for any pair of vertices a ′ , b ′ , k G − a ′ , b ′ k ≥
11, for otherwise the subgraph hasten vertices (in order to ensure it is non–planar, see Lemma 2.5), which is the casewe just treated above. Lemma 3.1 describes the possible degrees for such a graph.In particular, δ ( G − a, b ) ≥ G is not 2–apex, G − a, b must be graph iii,iv, v, vi, or viii in Figure 2. Lemmas 3.2 through 3.4 show that in those cases, if G is not 2–apex, then G is E or H . This completes the proof. (cid:3) Finally, we prove Proposition 1.7.
Proposition 1.7.
Let G be a graph with | G | = 9 and k G k ≤ . Then G is IK iffit is K ⊔ K ⊔ K , H ⊔ K , F , or H .Proof. It follows from [KS] that, if G is one of the four listed graphs, then it is IK.By Proposition 2.13, G is 2–apex and, therefore, not IK when k G k ≤
20. In case k G k = 21, Proposition 1.6 shows G is 2–apex and not IK unless G is one of thefour listed graphs or E . However, Figure 8 is an unknotted embedding of E . So,if G is IK, it must be one of the four listed graphs. (cid:3) Remark 3.7.
It is straightforward to verify that Figure 8 is an unknotted embed-ding. For example, here’s a strategy for making such a verification. Number thecrossings as shown. It is easy to check that there are 16 possible crossing combina-tions for a cycle in this graph: 1236, 134, 1457, 1459, 1479, 16789, 234689, 23479,236789, 25689, 2578, 345689, 3457, 3578, 36789, and 4578. That is, any cycle inthe graph will have crossings that are a subset of one of those 16 sets. For example,a cycle that includes crossings 1, 2, 3, and 6 must include the edges ab and bc andtherefore, cannot have the edge bd required for crossing 4. Indeed, a cycle that in-cludes 1, 2, 3, and 6 can have none of the other five crossings. To show that thereare no knots, consider cycles that correspond to each subset of the 16 sets and checkthat each such cycle (if such exists) is not knotted in the embedding of Figure 8. Acknowledgements
We thank Ramin Naimi for encouragement and for sharing an unknotted em-bedding of the graph E and Joel Foisy for helpful conversations. This study wasinspired by the Master’s thesis of Chris Morris. References [BBFFHL] P. Blain, G. Bowlin, T. Fleming, J. Foisy, J. Hendricks, and J. LaCombe, ‘Some Resultson Intrinsically Knotted Graphs,’
J. Knot Theory Ramifications , (2007), 749–760.[B] T. B¨ohme, ‘On spatial representations of graphs,’ Contemporary methods in graph theory ,Bibliographisches Inst., Mannheim, (1990), 151–167.[CMOPRW] J. Campbell, T.W. Mattman, R. Ottman, J. Pyzer, M. Rodrigues, and S. Williams,‘Intrinsic knotting and linking of almost complete graphs,’
Kobe J. Math , (2008), 39–58.math.GT/0701422[JKM] B. Johnson, M.E. Kidwell, and T.S. Michael, ‘Intrinsically knotted graphs have at least21 edges,’ (to appear in J. Knot Theory Ramifications ).[KS] T. Kohara and S. Suzuki, ‘Some remarks on knots and links in spatial graphs’, in
Knots90, Osaka, 1990 , de Gruyter (1992) 435–445.[MRS] R. Motwani, A. Raghunathan, and H. Saran, ‘Constructive Results from Graph Minors:Linkless Embeddings,’ (1998), 298–409.[N] R. Naimi, private communication.[OT] M. Ozawa and Y. Tsutsumi, ‘Primitive Spatial Graphs and Graph Minors,’
Rev. Mat.Complut. (2007), 391–406.[S] H. Sachs, ‘On Spatial Representation of Finite Graphs’, in: A. Hajnal, L. Lovasz, V.T. S`os(Eds.), Colloq. Math. Soc. J´anos Bolyai , North-Holland, Amsterdam, (1984), 649–662.
Department of Mathematics and Statistics, California State University, Chico, Chico,CA 95929-0525
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