Elementary Results on Forbidden Minors
Arnold Tan Junhan

TL;DR
This paper explores the theory of forbidden minors in graphs, proves Wagner's Theorem via Kuratowski's Theorem, and discusses how forbidding certain minors constrains the chromatic number, relating to Hadwiger's Conjecture.
Contribution
It provides elementary proofs of key theorems and establishes connections between forbidden minors and graph coloring, advancing understanding of graph minor theory.
Findings
Proved Wagner's Theorem using Kuratowski's Theorem
Linked forbidden minors to bounds on chromatic number
Discussed implications for Hadwiger's Conjecture
Abstract
We start by building up some theory to state Wagner's Theorem, and then prove it using Kuratowski's Theorem, a proof of which is found in Diester (2000). Following this, we establish some connections between the chromatic number of a graph and some of its forbidden minors. The idea is that if we forbid to have certain graphs as a minor, then the chromatic number of cannot be too large. Intuitively, this makes sense: if we disallow from having too many edges, then this makes it easier to colour the graph with fewer colours; we will of course make this precise. We close by explaining how this all relates to Hadwiger's Conjecture.
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Taxonomy
TopicsLimits and Structures in Graph Theory · Advanced Graph Theory Research · Advanced Topology and Set Theory
Elementary Results on Forbidden Minors
Arnold Tan Junhan
**Michaelmas 2018 Mini Projects: Graph Theory
University of Oxford**
Contents
1 Abstract
Throughout this report , , , etc. will denote graphs. Our graphs are finite, always have at least one vertex, and do not have loops or multi-edges.
We first build up some theory to state Wagner’s Theorem, and then prove it using a well-known result called Kuratowski’s Theorem. Actually, the two are equivalent, but we will not prove the latter. (For a proof of Kuratowski’s Theorem, see Diester (2000).)
Following this, we establish some connections between the chromatic number of a graph and some of its forbidden minors. The idea is that if we forbid to have certain graphs as a minor, then the chromatic number of cannot be too large. Intuitively, this makes sense: if we disallow from having too many edges, then this makes it easier to colour the graph with fewer colours; we will of course make this precise.
We end by stating a well-known conjecture that generalises our work.
2 Wagner’s Theorem
Definition 2.1**.**
A graph is a minor of a graph (or * has an -minor**) if there is a sequence such that, for , is obtained from by contracting an edge, or deleting an edge or a vertex.*
(We allow to be a minor of itself. Also, in practice, we only care what the minor is up to isomorphism: if has a -minor, and , then we may simply say has an -minor.)
There is an equivalent formulation of minors:
Proposition 2.2**.**
Graph is a minor of graph if and only if, writing , there are nonempty disjoint subsets of such that is connected for each and whenever .
Definition 2.3**.**
Subdividing an edge of a graph deletes the edge and adds a new vertex with neighbours and . is a subdivision of if it can be obtained from by a sequence of subdivisions of edges – that is, there is a sequence ,, such that for each , such that . is a topological minor of if contains a subdivision of as a subgraph.
Note that being a subdivision of means precisely that we can replace each edge in by some path from to (adding some vertices to ), such that these paths are internally vertex disjoint, and is the resulting graph.
Notation 2.4**.**
Write to mean contains , to mean has an -minor, and to mean contains a subdivision of .
We will use either equivalent formulation of minors, depending on which is convenient to us. There are some obvious properties of minors and subdivisions.
Lemma 2.5**.**
If and , then .
Proof.
Suppose that the sequence witnesses , and the sequence witnesses . Then the sequence witnesses . ∎
It is also clear that if has as a subgraph, then has as a minor. (Remove the vertices in one by one, in any order, to get a sequence of subgraphs that witnesses .) In fact:
Corollary 2.6**.**
.
Proof.
Certainly if is a subgraph of , then contains a subdivision of , namely itself. This proves the first implication. For the second, suppose contains a subdivision, say , of . That is, , and there is a sequence witnessing that is a subdivision of . It is enough to show that is a minor of , for then we would have , and we can finish by Lemma 2.5. Define a sequence by setting for each . This sequence witnesses that is a minor of , since each graph in the sequence is obtained from the previous graph (where ) by an edge contraction. To be pedantic, recalling that , we have
[TABLE]
This completes the proof that is a minor of , hence the result holds. ∎
Corollary 2.7**.**
If is a subdivision of , then is a minor of .
Proof.
We have already shown this in the proof of Corollary 2.6, but it also follows from the corollary, since if is a subdivision of , then , so , so is a minor of . ∎
On the other hand, we might ask, if is a minor of , must contain a subdivision of as a subgraph? The answer, in general, is no. Famously, is a minor of the Petersen graph, but not a topological minor.
Proposition 2.8**.**
Let be the Petersen graph, defined by
[TABLE]
*where subscripts are read modulo .
is a minor of . but does not contain a subdivision of as a subgraph.*
Proof.
Note that each is neighbours precisely with , and , and each is neighbours precisely with , and . In either case, there are exactly three distinct neighbours, since we are reading subscripts modulo . Therefore is -regular.
For the first claim, let us use our alternative formulation of minors. To see that is a minor of , consider the disjoint nonempty subsets . Each is obviously connected, so it remains to show that whenever (). If differ modulo by , then without loss so shows that . If differ modulo by , then without loss so shows that . There are no other possibilities, so we are done.
Let us prove the second claim. It follows by definition that subdividing an edge of a graph creates a new vertex of degree , but does not change the degree of any vertex of . Therefore, if is a subdivision of and , then , and so if contains a subdivision of , then . In particular, when is the Petersen graph , and is , then this says that if the Petersen graph contained a subdivision of , then we would have : an obvious contradiction. Therefore . ∎
We might wonder if the result becomes true under some simple additional assumption. This is indeed the case, but before stating the result we give a useful lemma.
Lemma 2.9**.**
Let and be graphs, where . Let be a minor of , and let this be witnessed by the nonempty disjoint subsets from Proposition 2.2. Suppose that no proper subgraph of contains as a minor. Then,
Each is minimally connected, i.e., a tree. 2. 2.
Whenever , we have . 3. 3.
Whenever then . 4. 4.
For every leaf of a tree of size larger than , we can find some such that . 5. 5.
* has at most leaves.* 6. 6.
* cover .*
Proof.
If the first claim were false, then we could remove an edge from to get a proper subgraph of , where is still connected, so the same subsets would witness that the proper subgraph has an -minor.
If the second claim were false, then there would be at least two edges from some to some , and if we remove one of these we get a proper subgraph of where for each i, and there is still an edge in from to whenever . Hence the same subsets would witness that the proper subgraph has an -minor.
If the third claim were false, then we could find with an edge between and . Removing this edge gives a proper subgraph of where for each i. Hence the same subsets would witness that the proper subgraph has an -minor.
If the fourth claim were false, then we could remove to get a tree . This is still nonempty and connected, and there is still an edge in the proper subgraph of from to whenever . Hence the subsets would witness that the proper subgraph has an -minor.)
We will use the claims above to prove the fifth claim. Without loss has more than one vertex (otherwise has a single vertex, of degree zero – this is not a leaf). There is an injective function from the set of leaves of to the set , given by sending a leaf to any fixed choice of given by the fourth claim: since , by the third claim is a neighbor of , so this function is well-defined. It is an injection since if two leaves are sent to the same , then there are edges from into and from into . By the second claim, .
If the sixth claim were false, then would witness that the proper subgraph of has an -minor. ∎
Proposition 2.10**.**
Suppose is a minor of and . Then contains a subdivision of .
Proof.
We prove by induction on that for any with and , we have . If , then so the result is trivial: if and only if if and only if .
Let , and suppose the result holds for all smaller values of – that is, whenever we have graphs and , where , , and , then . The condition holds whenever is a proper subgraph of . Therefore, it suffices to prove the following claim:
[TABLE]
(Once we have the claim, then for general , if a proper subgraph contains an -minor, then we may apply the inductive hypothesis to to deduce that . Hence contains , which contains a subdivision of , so certainly contains a subdivision of .)
Let us prove the claim. Writing , there are disjoint nonempty subsets of such that each is connected, and whenever . By the lemma above, each is minimally connected, and whenever .
If a tree has a single vertex , then we can find some such that if (and only if, by the lemma above) is not isolated in . Moreover, for every leaf of a tree of size larger than , we can find some such that .
Denote by the unique leaf with (if it exists). In this case, fixing and running over all neighbours of in gives us a collection of (not necessarily distinct) leaves (where varies) in . By the lemma above, this collection has size at most , since . This means the tree has at most leaves. In particular there is always a vertex of , and paths from to each that are pairwise disjoint except at . Indeed, if there is only one leaf in , take , and the trivial path from to itself. If there are only two leaves, let be one of these, and take our two paths to be the trivial path from to itself, and a path from to the other leaf. (Actually, it is unique, since is a tree.) If there are three leaves , , , first consider a path from to , and a path from to . The last edge in both of these paths has to be the unique edge that meets . Let us take to be the first vertex in the path from to that is also in the path from to , and take our three paths to be the three unique paths starting at and ending at each respective leaf. By construction, none of these paths intersect other than at .
In the special case where has a single vertex and for each , then is isolated in ; we can still define .
This exhibits, as desired, a subdivision of as a subgraph of . Explicitly, take to be the union of subgraphs of of the form:
- •
paths , where , is the unique path in from to , is an edge in , and is the unique path in from to , and we have concatenated them together to get a path in ;
- •
trees on a single vertex , such that for each .
Note that the collection of paths are internally vertex disjoint. It remains to see that is a subdivision of . This is clear, if you let the vertices in correspond to the vertices of . Then, whenever , the path from to corresponds to a repeated subdivision of the edge . ∎
Lemma 2.11**.**
If is a minor of , then contains a subdivision of either or .
Proof.
If is a minor of , let be a minimal subgraph of containing a -minor – that is, contains a -minor, but no proper subgraph of contains a -minor. By Proposition 2.2, there exist disjoint nonempty subsets such that each is connected, and for distinct . By Lemma 2.9, each is a tree, and for distinct . That is, fixing , there is exactly one edge from to each of the other four . Consider the tree obtained from the tree by adding these four edges. ( is indeed a tree since, for instance, adding these edges does not create cycles or destroy connectivity.) Note that has exactly leaves, one in each of the other than itself. (Adding the four edges to created these four leaves, and also increased the degree of any leaf of .) We consider two possibilities for the form of .
First, note that a tree has at least leaves. (Let have degree ; for each edge consider a maximal path starting with ; by maximality this path ends at a leaf. We get paths starting at and ending at different leaves, because these paths are necessarily disjoint except at – otherwise would have a cycle.) Therefore . The handshaking lemma now says , where denotes the number of vertices of degree . This simplifies to , so we have the two cases below for .
- •
has one vertex of degree , four leaves, possibly some vertices of degree , and no other vertices. (That is, is a subdivision of .)
- •
has two vertices of degree , four leaves, possibly some vertices of degree , and no other vertices.
If each of the five fall under the first case, then we have that is a subdivision of . (Consider the vertices . Between each two of these there is a path in ; these paths are internally vertex disjoint.)
If some falls under the second case, we claim has a -minor. Once we have this, then we are done: since , by Proposition 2.10 (and hence ) contains a subdivision of . It remains to show has a -minor. Well, if we contract onto the two vertices of having degree , and contract the other onto singletons, then we get six-vertex graph that contains .
∎
Theorem 2.12** (Kuratowski’s Theorem).**
A graph is planar if and only if does not contain a subdivision of or of .
For a proof of Kuratowski’s Theorem, see Diestel (2000).
We are now in position to state and prove Wagner’s Theorem from Kuratowski’s Theorem.
Theorem 2.13** (Wagner’s Theorem).**
A graph is planar if and only if contains neither nor as a minor.
Proof.
For the forward direction, suppose is planar. By Kuratowski’s Theorem, does not contain a subdivision of or of . By Lemma 2.11, cannot contain as a minor. Moreover, by Proposition 2.10, cannot contain as a minor, otherwise would contain a subdivision of , because .
For the backward direction, suppose contains neither nor as a minor. Then, whenever is a subgraph of , we must have and . (If , then would imply that , but this is forbidden. Similarly, we cannot have .) By Corollary 2.7, is neither a subdivision of nor of . We have shown that no subgraph of is a subdivision of or of . Therefore by Kuratowski’s Theorem, is planar. ∎
3 The Chromatic Number and Forbidden Minors
In this section we prove several results of the form: if the chromatic number of a graph is large enough, then this forces to contain a complete graph as its minor.
Definition 3.1**.**
Let , and . An -fan is a set of paths from to that are pairwise disjoint except at . The size of the fan is the number of paths in the collection.
Lemma 3.2** (Fan Lemma).**
If is -connected, and the size of is at least . Then there exists an -fan of size .
Proof.
Add a new vertex to , along with edges from to each vertex of . This new graph is again -connected, since and no set of size at most separates . Indeed, suppose there was such a set .
- •
If , then as is disconnected, hence so is . Indeed, has a component containing , and at least one other component not containing , so has at least two components, namely D and some component of , since is nonempty: it contains some element of , since the size of was less than .
- •
If , then is disconnected.
Now, is -connected, so by Menger’s Theorem, there exist independent paths from to in . (Menger’s Theorem says there exist independent paths from to its non-neighbour , but , where the first inequality is because no set of size less than separates from .) Each such path is necessarily of the form for some , where if . Therefore the paths in form our desired -fan of size . ∎
Lemma 3.3**.**
If a graph has a cycle, then contains a -minor.
Proof.
Suppose has a cycle as a subgraph. By Corollary 2.6, has a -minor. Now note that in turn has a -minor, as witnessed by the sequence , each obtained from the previous by contracting an edge. Hence, Lemma 2.5 implies that is a minor of . ∎
Proposition 3.4**.**
If a graph has then contains a -minor.
Proof.
This is straightforward, by noting some equivalent characterisations of bipartite graphs: a graph is bipartite if and only if it is -colourable, if and only if it contains no odd cycles. Hence, means is not -colourable, therefore is not bipartite, therefore has an odd cycle, therefore contains a -minor by Lemma 3.3. ∎
Proposition 3.5**.**
If a graph has then contains a -minor.
Proof.
Let us proceed by induction on . If , there is nothing to show; if and then , so must be . Now suppose , and that the result holds for all graphs on fewer vertices. Without loss, we may assume is connected. (Indeed, is the maximum of the chromatic numbers of the components of , so has a component with . If is not connected then is a proper subgraph of , so by induction – and hence – must have a -minor.) Now we split into four cases.
- •
Suppose . is not -connected, so as , we have that is disconnected. This contradicts our assumption.
- •
Suppose . is not -connected, so as , we have that some set separates . will have a partition into two disjoint subgraphs and (these will be unions of the components of ). Let , for . Then , and are complete (they are both equal to , so . Without loss, . is a proper subgraph of , as ), so by induction (and hence ) must have a -minor.
- •
Suppose , and is a separating set: there are subgraphs and such that , , and for both . Suppose for a contradiction that both and can be -coloured. Both colourings must colour and with different colours, so after possibly permuting one of the colourings, we can take their union to get a -colouring on ; this is a contradiction. Hence, assume without loss that . has fewer vertices than (as ), so by induction it has a -minor. We are not quite done yet, as need not be an edge of . If it is, we are done, so suppose not. Note that there is a path from to in – without loss has a neighbour in , but cannot be a cutvertex, so removing it shows that there is a path from to that does not use ; concatenating with a minimal such path produces a path from to in . This path ’substitutes’ for the missing edge ; we can just contract it and pretend as though we have the edge in our graph. In other words, shows that has a -minor.
- •
Finally, suppose . Let . Then , so is -connected, and in particular contains a cycle. (Otherwise is connected and acyclic, i.e., a tree. It has a leaf ; if we delete the unique neighbour of from this tree we get a disconnected graph, because is not another leaf, since has at least vertices. This contradicts that is -connected.) Take to be the set of vertices of this cycle, so its size is at least . Since is -connected, by Lemma 3.2 there exist three paths from to this cycle, that are disjoint except at . Let be the three distinct end-vertices of the . Then the sets witness that has a -minor, since we can contract the cycle onto the three vertices , and contract the paths to edges – in the resulting graph, is neighbours with each of the three vertices in a triangle. This is a copy of .
∎
In light of the previous two propositions, we might conjecture:
Conjecture 3.6**.**
For every , if then contains a -minor.
This is precisely the famous Hadwiger’s Conjecture.
Let us prove a weaker statement:
Theorem 3.7**.**
For every , there exists a constant such that if then contains a -minor.
Proof.
See proof of the restated version, Theorem 3.12. ∎
Notation 3.8**.**
For a graph , write for the length of a shortest path between vertices and (if it exists). This is the distance from to . Let be a connected graph (so the distance between any two vertices is well-defined), and be a vertex of . Define V(G) for each . For instance, .
The idea is that if we fix a vertex , then the sets partition into ’layers’ around , where each vertex in an outer layer has some edge entering the layer below it, and there are no edges between any two layers that are not adjacent. Let us state this formally.
Lemma 3.9**.**
Let be a connected graph, and a vertex of . If and , then such that . Furthermore, whenever , we have .
Proof.
For the first claim, note that implies there is a path of length from to . Let this path be with , . Consider the path in from to , of length . This must be a shortest path from to , for if there were a path of length less than length from to , then concatenating it with the edge would result in a path of length less than from to , then we would have , contradicting . Therefore . Since , this proves the first claim.
For the second claim, suppose for a contradiction that there exists some and some , such that . Let be a shortest path from to , where . Then, concatenating with the edge , we have a path from to of length . This shows that , contradicting . ∎
Lemma 3.10**.**
Let be a connected graph, and a vertex of . Then there is some such that is empty whenever , and nonempty whenever .
Proof.
First observe that is always empty for large enough . For instance, take to be the length of a longest path in , then is empty whenever . Then can only be nonempty for a finite number of values . Let be the largest among these such that is nonempty. Then, for all , will also be nonempty. Indeed, suppose not, then there is such that . Let us assume is maximal as such, so . The lemma above implies, in particular, that there is some . This contradicts . Therefore we must indeed conclude that for all , is nonempty. ∎
Proposition 3.11**.**
Let be a connected graph, and a vertex of . Then there is some such that .
Proof.
Let be as in the lemma above. Choose which maximizes . Let . It is enough to show that , or equivalently, that is -colourable. Let us -colour starting with the outermost layer and working inwards. By the maximality assumption, we can certainly (properly) -colour each layer (where ). Therefore we can colour with colours from , with colours from , with colours from , and alternating so on and so forth, until we reach (which gets a single colour, of course). This procedure colours all of , and indeed gives us a proper colouring on . (Edges within a single layer do not cause problems, since this colouring on arises from properly colouring each layer; on the other hand an edge between layers can only exist between adjacent layers by Lemma 3.9, and such an edge can never receive the same colour on both its edges, since we are using entirely different colour schemes on alternating layers!) ∎
Theorem 3.12**.**
For every , if then contains a -minor.
Proof.
We will proceed by induction on . The base case is . In this case, if then in particular, since , has an edge, so certainly contains a -minor.
Now assume , and suppose the result holds for the case . That is, suppose that whenever is any graph with , then contains a -minor. We must show that if then contains a -minor. Without loss of generality, we may assume is connected. (For general , is the maximum of the chromatic numbers of its components. Therefore if , then has a component with . If the result holds for connected graphs, then contains a -minor, hence so does .)
Consider the induced subgraph , where , is any fixed vertex of and is as in Proposition 3.11. Then , so by the inductive hypothesis applied to , contains a -minor.
Now it is not hard to show that contains a -minor – already the layer has a -minor, but by Lemma 3.9 there is a path from each vertex of to . We can contract all but one edge in each such path, and finish by noting that adding a new vertex to adjacent to all the other vertices produces .
Let us argue in more detail, using our alternative formulation of minors. First note that . (Indeed, if , then shows that , a contradiction.) Let . Since , there are disjoint nonempty subsets , …, of such that each is connected, and whenever .
Consider the graph . This is a copy of obtained by adding a new vertex to , adjacent to every other vertex. I show this a minor of . Well, consider the nonempty subsets of , where . These sets witness that has a -minor:
- •
They are pairwise disjoint, because , …, were already pairwise disjoint, and .
- •
Each is connected: for each , , which is connected by assumption. (Both these graphs have vertex set and edge set .) is connected since for all , there are paths in from to and from to , which concantenated produce a path in from to . To be clear, say , where . If then there is a trivial path from to , of length [math]. Otherwise, apply Lemma 3.9 repeatedly to see that such that , such that , , such that . This yields a path from to . Similarly, there is a path from to .
- •
whenever : this holds if is an edge in the copy of we started with, so suppose we have an edge , where . We must show that . First, note that , so . Then Lemma 3.9 says such that . Since and , this shows as required.
This completes the proof that has a -minor. ∎
4 Conclusion
We have shown that for every , there is a constant such that if then contains as a minor. Our choice of was exponential in , but Kostochka (1984) achieved .
Hadwiger’s Conjecture says we can do better still:
Conjecture 4.1** (Hadwiger’s Conjecture).**
For every , if then contains as a minor.
This is essentially the nicest result we can hope to get. For instance if and then has no edges, so it cannot contain a -minor.
Hadwiger’s Conjecture is known to hold for the cases :
k = 2:
trivial; a graph with chromatic number larger than must have an edge, so it contains (as a subgraph, hence as a minor).
k = 3, 4:
we proved these cases above.
k = 5, 6:
these cases are implied by the Four Colour Theorem. See Robertson, Seymour, Thomas (1993).
k ¿ 6 :
higher cases remain unresolved.
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