Elementary Results on Forbidden Minors

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 $G$ to have certain graphs as a minor, then the chromatic number of $G$ cannot be too large. Intuitively, this makes sense: if we disallow $G$ 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.

Equations

$$H_i/(v_i{w}_i)$$

$$=((V(H_{i-1})\backslash \{w_i\})\cup \{w_i^{\prime }\},(E(H_{i-1})\backslash \{\text{edges containing }{w}_i\})\cup \{w_i^{\prime }x:x\text{ adjacent in }{H}_i\text{ to }{v}_i\text{ or }{w}_i;x\ne v_i,w_i\})$$

$$=((V(H_{i-1})\backslash \{w_i\})\cup \{w_i^{\prime }\},(E(H_{i-1})\backslash \{\text{edges containing }{w}_i\})\cup \{w_i^{\prime }x:x\text{ adjacent in }{H}_{i-1}\text{ to }{w}_i\})$$

$$\cong (V(H_{i-1}),E(H_{i-1}))$$

$$=H_{i-1}.$$

$$(V(P),E(P))=(\{u_0,\dots ,u_4,v_0,\dots ,v_4\},\{u_i{u}_{i+1},u_i{v}_i,v_i{v}_{i+2}:i=0,\dots ,4\})$$

$$\text{If }G\text{ is such that }H{\le }_{m}G\text{, but no proper subgraph of }G\text{ has an }H\text{-minor, where }\Delta (H)\le 3\text{, then }H{\le }_{t}G.$$

Full text

1. 1 Abstract
2. 2 Wagner’s Theorem
3. 3 The Chromatic Number and Forbidden Minors
4. 4 Conclusion

1 Abstract

Throughout this report $G$, $H$, $K$, 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 $G$ to have certain graphs as a minor, then the chromatic number of $G$ cannot be too large. Intuitively, this makes sense: if we disallow $G$ 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 $H$ is a minor of a graph $G$ (or $G$* has an $H$-minor**) if there is a sequence $G=G_{0},G_{1},\ldots,G_{k}=H$ such that, for $i=1,\ldots,k$, $G_{i}$ is obtained from $G_{i-1}$ by contracting an edge, or deleting an edge or a vertex.*

(We allow $G$ to be a minor of itself. Also, in practice, we only care what the minor $H$ is up to isomorphism: if $G$ has a $K$-minor, and $H\cong K$, then we may simply say $G$ has an $H$-minor.)

There is an equivalent formulation of minors:

Proposition 2.2**.**

Graph $H$ is a minor of graph $G$ if and only if, writing $V(H)={v_{1},\ldots,v_{h}}$, there are nonempty disjoint subsets $V_{1},\ldots,V_{h}$ of $V(G)$ such that $G[V_{i}]$ is connected for each $i$ and $e(V_{i},V_{j})>0$ whenever $v_{i}v_{j}\in E(H)$.

Definition 2.3**.**

Subdividing an edge $xy$ of a graph $H$ deletes the edge $xy$ and adds a new vertex $v$ with neighbours $x$ and $y$. $G$ is a subdivision of $H$ if it can be obtained from $H$ by a sequence of subdivisions of edges – that is, there is a sequence $H=H_{0}$,$\ldots$,$H_{k}=G$ such that for each $1\leq i\leq k$, $\exists v_{i}\in V(H_{i})\backslash V(H_{i-1}),w_{i}u_{i}\in E(H_{i-1})$ such that $H_{i}=(V(H_{i-1})\cup\{v_{i}\},(E(H_{i-1})\backslash\{w_{i}u_{i}\})\cup\{v_{i}w_{i},v_{i}u_{i}\})$. $H$ is a topological minor of $G$ if $G$ contains a subdivision of $H$ as a subgraph.

Note that $G$ being a subdivision of $H$ means precisely that we can replace each edge $uv$ in $H$ by some path from $u$ to $v$ (adding some vertices to $H$), such that these paths are internally vertex disjoint, and $G$ is the resulting graph.

Notation 2.4**.**

Write $H\leq G$ to mean $G$ contains $H$, $H\leq_{m}G$ to mean $G$ has an $H$-minor, and $H\leq_{t}G$ to mean $G$ contains a subdivision of $H$.

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 $H\leq_{m}G$ and $K\leq_{m}H$, then $K\leq_{m}G$.

Proof.

Suppose that the sequence $G=G_{0},G_{1},\ldots,G_{k}=H$ witnesses $H\leq_{m}G$, and the sequence $H=H_{0},H_{1},\ldots,H_{l}=K$ witnesses $K\leq_{m}H$. Then the sequence $G=G_{0},G_{1},\ldots,G_{k}=H=H_{0},H_{1},\ldots,H_{l}=K$ witnesses $K\leq_{m}G$. ∎

It is also clear that if $G$ has $H$ as a subgraph, then $G$ has $H$ as a minor. (Remove the vertices in $V(G)\backslash V(H)$ one by one, in any order, to get a sequence of subgraphs that witnesses $H\leq_{m}G$.) In fact:

Corollary 2.6**.**

$H\leq G\Rightarrow H\leq_{t}G\Rightarrow H\leq_{m}G$.

Proof.

Certainly if $H$ is a subgraph of $G$, then $G$ contains a subdivision of $H$, namely $H$ itself. This proves the first implication. For the second, suppose $G$ contains a subdivision, say $H^{\prime}$, of $H$. That is, $H^{\prime}\leq G$, and there is a sequence $H=H_{0},\ldots,H_{k}=H^{\prime}$ witnessing that $H^{\prime}$ is a subdivision of $H$. It is enough to show that $H$ is a minor of $H^{\prime}$, for then we would have $H\leq_{m}H^{\prime}\leq_{m}G$, and we can finish by Lemma 2.5. Define a sequence $H^{\prime}=H^{\prime}_{0},\ldots,H^{\prime}_{k}=H$ by setting $H^{\prime}_{j}=H_{k-j}$ for each $0\leq j\leq k$. This sequence witnesses that $H$ is a minor of $H^{\prime}$, since each graph $H^{\prime}_{k-i+1}=H_{i-1}$ in the sequence is obtained from the previous graph $H^{\prime}_{k-i}=H_{i}$ (where $1\leq i\leq k$) by an edge contraction. To be pedantic, recalling that $H_{i}=(V(H_{i-1})\cup\{v_{i}\},(E(H_{i-1})\backslash\{w_{i}u_{i}\})\cup\{v_{i}w_{i},v_{i}u_{i}\})$, we have

[TABLE]

This completes the proof that $H$ is a minor of $H^{\prime}$, hence the result holds. ∎

Corollary 2.7**.**

If $G$ is a subdivision of $H$, then $H$ is a minor of $G$.

Proof.

We have already shown this in the proof of Corollary 2.6, but it also follows from the corollary, since if $G$ is a subdivision of $H$, then $H\leq_{t}G$, so $H\leq_{m}G$, so $H$ is a minor of $G$. ∎

On the other hand, we might ask, if $H$ is a minor of $G$, must $G$ contain a subdivision of $H$ as a subgraph? The answer, in general, is no. Famously, $K_{5}$ is a minor of the Petersen graph, but not a topological minor.

Proposition 2.8**.**

Let $P$ be the Petersen graph, defined by

[TABLE]

*where subscripts are read modulo $5$.

$K_{5}$ is a minor of $P$. but $P$ does not contain a subdivision of $K_{5}$ as a subgraph.*

Proof.

Note that each $u_{i}$ is neighbours precisely with $u_{i-1}$, $u_{i+1}$ and $v_{i}$, and each $v_{i}$ is neighbours precisely with $v_{i-2}$, $v_{i+2}$ and $u_{i}$. In either case, there are exactly three distinct neighbours, since we are reading subscripts modulo $5$. Therefore $P$ is $3$-regular.

For the first claim, let us use our alternative formulation of minors. To see that $K_{5}$ is a minor of $P$, consider the disjoint nonempty subsets $V_{0}=\{u_{0},v_{0}\},\ldots,V_{4}=\{u_{4},v_{4}\}$. Each $G[V_{i}]=(\{u_{i},v_{i}\},\{u_{i}v_{i}\})$ is obviously connected, so it remains to show that $e(V_{i},V_{j})>0$ whenever $i\neq j$ ($i,j\in\{0,\ldots,4\}$). If $i,j$ differ modulo $5$ by $1$, then without loss $j=i+1$ so $u_{i}u_{j}\in E(P)$ shows that $e(V_{i},V_{j})>0$. If $i,j$ differ modulo $5$ by $2$, then without loss $j=i+2$ so $v_{i}v_{j}\in E(P)$ shows that $e(V_{i},V_{j})>0$. 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 $H$ creates a new vertex of degree $2$, but does not change the degree of any vertex of $H$. Therefore, if $H^{\prime}$ is a subdivision of $H$ and $\Delta(H)\geq 2$, then $\Delta(H^{\prime})=\Delta(H)$, and so if $G$ contains a subdivision of $H$, then $\Delta(G)\geq\Delta(H)$. In particular, when $G$ is the Petersen graph $P$, and $H$ is $K_{5}$, then this says that if the Petersen graph contained a subdivision of $K_{5}$, then we would have $3=\Delta(P)\geq\Delta(K_{5})=4$: an obvious contradiction. Therefore $K_{5}\nleq_{t}P$. ∎

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 $G$ and $H$ be graphs, where $V(H)=\{v_{1},\ldots,v_{h}\}$. Let $H$ be a minor of $G$, and let this be witnessed by the nonempty disjoint subsets $V_{1},\ldots,V_{h}$ from Proposition 2.2. Suppose that no proper subgraph of $G$ contains $H$ as a minor. Then,

Each $G[V_{i}]$ is minimally connected, i.e., a tree. 2. 2.

Whenever $v_{i}v_{j}\in E(H)$, we have $e(V_{i},V_{j})=1$. 3. 3.

Whenever $v_{i}v_{j}\notin E(H)$ then $e(V_{i},V_{j})=0$. 4. 4.

For every leaf $w$ of a tree $G[V_{i}]$ of size larger than $1$, we can find some $j\neq i$ such that $e(\{w\},V_{j})>0$. 5. 5.

$G[V_{i}]$* has at most $d(v_{i})$ leaves.* 6. 6.

$V_{1},\ldots,V_{h}$* cover $V(G)$.*

Proof.

If the first claim were false, then we could remove an edge from $V_{i}$ to get a proper subgraph $G_{1}$ of $G$, where $G_{1}[V_{i}]$ is still connected, so the same subsets $V_{1},\ldots,V_{h}$ would witness that the proper subgraph $G_{1}$ has an $H$-minor.

If the second claim were false, then there would be at least two edges from some $V_{i}$ to some $V_{j}$, and if we remove one of these we get a proper subgraph $G_{2}$ of $G$ where $G[V_{i}]=G_{2}[V_{i}]$ for each i, and there is still an edge in $G_{2}$ from $V_{i}$ to $V_{j}$ whenever $v_{i}v_{j}\in E(H)$. Hence the same subsets $V_{1},\ldots,V_{h}$ would witness that the proper subgraph $G_{2}$ has an $H$-minor.

If the third claim were false, then we could find $v_{i}v_{j}\notin E(H)$ with an edge between $V_{i}$ and $V_{j}$. Removing this edge gives a proper subgraph $G_{3}$ of $G$ where $G[V_{i}]=G_{3}[V_{i}]$ for each i. Hence the same subsets $V_{1},\ldots,V_{h}$ would witness that the proper subgraph $G_{2}$ has an $H$-minor.

If the fourth claim were false, then we could remove $w$ to get a tree $G[V_{i}]-w=G[V_{i}\backslash\{w\}]$. This is still nonempty and connected, and there is still an edge in the proper subgraph $G_{4}=G-w$ of $G$ from $V_{i}\backslash\{w\}$ to $V_{j}$ whenever $v_{i}v_{j}\in E(H)$. Hence the subsets $V_{1},\ldots,V_{i}\backslash\{w\},\ldots,V_{h}$ would witness that the proper subgraph $G_{4}=G-w$ has an $H$-minor.)

We will use the claims above to prove the fifth claim. Without loss $G[V_{i}]$ has more than one vertex (otherwise $G[V_{i}]$ has a single vertex, of degree zero – this is not a leaf). There is an injective function from the set of leaves of $G[V_{i}]$ to the set $\{j\in\{1,\ldots,h\}:v_{j}\text{ is a neighbour of }v_{i}\}$, given by sending a leaf $w$ to any fixed choice of $j$ given by the fourth claim: since $e(\{w\},V_{j})>0$, by the third claim $v_{j}$ is a neighbor of $v_{i}$, so this function is well-defined. It is an injection since if two leaves $w,w^{\prime}$ are sent to the same $j$, then there are edges from $w$ into $V_{j}$ and from $w^{\prime}$ into $V_{j}$. By the second claim, $w=w^{\prime}$.

If the sixth claim were false, then $V_{1},\ldots,V_{h}$ would witness that the proper subgraph $G[V_{1}\cup\ldots\cup V_{h}]$ of $G$ has an $H$-minor. ∎

Proposition 2.10**.**

Suppose $H$ is a minor of $G$ and $\Delta(H)\leq 3$. Then $G$ contains a subdivision of $H$.

Proof.

We prove by induction on $n=e(G)+v(G)$ that for any $H$ with $H\leq_{m}G$ and $\Delta(H)\leq 3$, we have $H\leq_{t}G$. If $n=1$, then $V(G)=1$ so the result is trivial: $H\leq_{m}G$ if and only if $H\leq_{t}G$ if and only if $H=G$.

Let $n=e(G)+v(G)>1$, and suppose the result holds for all smaller values of $n$ – that is, whenever we have graphs $H$ and $G^{\prime}$, where $\Delta(H)\leq 3$, $H\leq_{m}G^{\prime}$, and $e(G^{\prime})+v(G^{\prime})<n$, then $H\leq_{t}G^{\prime}$. The condition $e(G^{\prime})+v(G^{\prime})<n$ holds whenever $G^{\prime}$ is a proper subgraph of $G$. Therefore, it suffices to prove the following claim:

[TABLE]

(Once we have the claim, then for general $G$, if a proper subgraph $G^{\prime}$ contains an $H$-minor, then we may apply the inductive hypothesis to $G^{\prime}$ to deduce that $H\leq_{t}G^{\prime}$. Hence $G$ contains $G^{\prime}$, which contains a subdivision of $H$, so $G$ certainly contains a subdivision of $H$.)

Let us prove the claim. Writing $V(H)=\{v_{1},\ldots,v_{h}\}$, there are disjoint nonempty subsets $V_{1},\ldots,V_{h}$ of $V(G)$ such that each $G[V_{i}]$ is connected, and $e(V_{i},V_{j})>0$ whenever $v_{i}v_{j}\in E(H)$. By the lemma above, each $G[V_{i}]$ is minimally connected, and $e(V_{i},V_{j})=1$ whenever $v_{i}v_{j}\in E(H)$.

If a tree $G[V_{k}]$ has a single vertex $w_{k}$, then we can find some $j\neq k$ such that $e(\{w_{k}\},V_{j})>0$ if (and only if, by the lemma above) $v_{k}$ is not isolated in $H$. Moreover, for every leaf $w$ of a tree $G[V_{i}]$ of size larger than $1$, we can find some $j\neq i$ such that $e(\{w\},V_{j})>0$.

Denote by $w_{ij}$ the unique leaf $w\in G[V_{i}]$ with $e(\{w\},V_{j})>0$ (if it exists). In this case, fixing $i$ and running over all neighbours $v_{j}$ of $v_{i}$ in $H$ gives us a collection of (not necessarily distinct) leaves $w_{ij}$ (where $j$ varies) in $G[V_{i}]$. By the lemma above, this collection has size at most $3$, since $\Delta(H)\leq 3$. This means the tree $G[V_{i}]$ has at most $3$ leaves. In particular there is always a vertex $x_{i}$ of $G[V_{i}]$, and paths from $x_{i}$ to each $w_{ij}$ that are pairwise disjoint except at $x_{i}$. Indeed, if there is only one leaf $w$ in $G[V_{i}]$, take $x_{i}=w$, and the trivial path from $x_{i}$ to itself. If there are only two leaves, let $x_{i}$ be one of these, and take our two paths to be the trivial path from $x_{i}$ to itself, and a path from $x_{i}$ to the other leaf. (Actually, it is unique, since $G[V_{i}]$ is a tree.) If there are three leaves $w_{ij_{1}}$, $w_{ij_{2}}$, $w_{ij_{3}}$, first consider a path from $w_{ij_{1}}$ to $w_{ij_{3}}$, and a path from $w_{ij_{2}}$ to $w_{ij_{3}}$. The last edge in both of these paths has to be the unique edge $xw_{ij_{3}}$ that meets $w_{ij_{3}}$. Let us take $x_{i}$ to be the first vertex in the path from $w_{ij_{1}}$ to $w_{ij_{3}}$ that is also in the path from $w_{ij_{2}}$ to $w_{ij_{3}}$, and take our three paths to be the three unique paths starting at $x_{i}$ and ending at each respective leaf. By construction, none of these paths intersect other than at $x_{i}$.

In the special case where $G[V_{k}]$ has a single vertex $w_{k}$ and $e(\{w_{k}\},V_{j})=0$ for each $j\neq k$, then $v_{k}$ is isolated in $H$; we can still define $x_{k}=w_{k}$.

This exhibits, as desired, a subdivision of $H$ as a subgraph $H^{\prime}$ of $G$. Explicitly, take $H^{\prime}$ to be the union of subgraphs of $G$ of the form:

• •

paths $Q_{ij}=P_{x_{i}w_{ij}}(w_{ij}w_{ji})P_{w_{ji}x_{j}}$, where $v_{i}v_{j}\in E(H)$, $P_{x_{i}w_{ij}}$ is the unique path in $G[V_{i}]$ from $x_{i}$ to $w_{ij}$, $w_{ij}w_{ji}$ is an edge in $G$, and $P_{w_{ji}x_{j}}$ is the unique path in $G[V_{j}]$ from $x_{j}$ to $w_{ji}$, and we have concatenated them together to get a path $Q_{ij}$ in $G$;

• •

trees $G[V_{k}]$ on a single vertex $w_{k}$, such that $e(\{w_{k}\},V_{j})=0$ for each $j$.

Note that the collection of paths $Q_{ij}$ are internally vertex disjoint. It remains to see that $H^{\prime}$ is a subdivision of $H$. This is clear, if you let the vertices $x_{1},\ldots,x_{h}$ in $H^{\prime}$ correspond to the vertices $v_{1},\ldots,v_{h}$ of $H$. Then, whenever $v_{i}v_{j}\in E(H)$, the path $Q_{ij}$ from $x_{i}$ to $x_{j}$ corresponds to a repeated subdivision of the edge $v_{i}v_{j}$. ∎

Lemma 2.11**.**

If $K_{5}$ is a minor of $G$, then $G$ contains a subdivision of either $K_{3,3}$ or $K_{5}$.

Proof.

If $K_{5}$ is a minor of $G$, let $H$ be a minimal subgraph of $G$ containing a $K_{5}$-minor – that is, $H$ contains a $K_{5}$-minor, but no proper subgraph of $H$ contains a $K_{5}$-minor. By Proposition 2.2, there exist disjoint nonempty subsets $V_{1},\ldots,V_{5}$ such that each $H[V_{i}]$ is connected, and $e(V_{i},V_{j})>0$ for distinct $i,j\in\{1,\ldots,5\}$. By Lemma 2.9, each $H[V_{i}]$ is a tree, and $e(V_{i},V_{j})$ for distinct $i,j$. That is, fixing $i$, there is exactly one edge from $V_{i}$ to each of the other four $V_{j}$. Consider the tree $T_{i}$ obtained from the tree $H[V_{i}]$ by adding these four edges. ($T_{i}$ is indeed a tree since, for instance, adding these edges does not create cycles or destroy connectivity.) Note that $T_{i}$ has exactly $4$ leaves, one in each of the $V_{j}$ other than $V_{i}$ itself. (Adding the four edges to $H[V_{i}]$ created these four leaves, and also increased the degree of any leaf of $H[V_{i}]$.) We consider two possibilities for the form of $T_{i}$.

First, note that a tree $T$ has at least $\Delta(T)$ leaves. (Let $v\in V(T)$ have degree $\Delta(T)$; for each edge $vu$ consider a maximal path starting with $vu$; by maximality this path ends at a leaf. We get $\Delta(T)$ paths starting at $v$ and ending at different leaves, because these paths are necessarily disjoint except at $v$ – otherwise $T$ would have a cycle.) Therefore $\Delta(T_{i})\leq 4$. The handshaking lemma now says $2(3+r_{2}+r_{3}+r_{4})=4+2r_{2}+3r_{3}+4r_{4}$, where $r_{i}$ denotes the number of vertices of degree $i$. This simplifies to $2=r_{3}+2r_{4}$, so we have the two cases below for $T_{i}$.

• •

$T_{i}$ has one vertex $x_{i}$ of degree $4$, four leaves, possibly some vertices of degree $2$, and no other vertices. (That is, $T_{i}$ is a subdivision of $K_{1,4}$.)

• •

$T_{i}$ has two vertices $y^{1}_{i},y^{2}_{i}$ of degree $3$, four leaves, possibly some vertices of degree $2$, and no other vertices.

If each of the five $T_{i}$ fall under the first case, then we have that $H$ is a subdivision of $K_{5}$. (Consider the vertices $x_{1},\ldots,x_{5}$. Between each two of these there is a path in $H$; these paths are internally vertex disjoint.)

If some $T_{i}$ falls under the second case, we claim $H$ has a $K_{3,3}$-minor. Once we have this, then we are done: since $\Delta(K_{3,3})=3$, by Proposition 2.10 $H$ (and hence $G$) contains a subdivision of $K_{3,3}$. It remains to show $H$ has a $K_{3,3}$-minor. Well, if we contract $G[V_{i}]$ onto the two vertices $y^{1}_{i},y^{2}_{i}$ of $T_{i}$ having degree $3$, and contract the other $G[V_{j}]$ onto singletons, then we get six-vertex graph that contains $K_{3,3}$.

∎

Theorem 2.12** (Kuratowski’s Theorem).**

A graph $G$ is planar if and only if $G$ does not contain a subdivision of $K_{5}$ or of $K_{3,3}$.

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 $G$ is planar if and only if $G$ contains neither $K_{5}$ nor $K_{3,3}$ as a minor.

Proof.

For the forward direction, suppose $G$ is planar. By Kuratowski’s Theorem, $G$ does not contain a subdivision of $K_{5}$ or of $K_{3,3}$. By Lemma 2.11, $G$ cannot contain $K_{5}$ as a minor. Moreover, by Proposition 2.10, $G$ cannot contain $K_{3,3}$ as a minor, otherwise $G$ would contain a subdivision of $K_{3,3}$, because $\Delta(K_{3,3})=3$.

For the backward direction, suppose $G$ contains neither $K_{5}$ nor $K_{3,3}$ as a minor. Then, whenever $H$ is a subgraph of $G$, we must have $K_{5}\nleq_{m}H$ and $K_{3,3}\nleq_{m}H$. (If $K_{5}\leq_{m}H$, then $K_{5}\leq_{m}H\leq G$ would imply that $K_{5}\leq_{m}G$, but this is forbidden. Similarly, we cannot have $K_{3,3}\leq_{m}H$.) By Corollary 2.7, $H$ is neither a subdivision of $K_{5}$ nor of $K_{3,3}$. We have shown that no subgraph $H$ of $G$ is a subdivision of $K_{5}$ or of $K_{3,3}$. Therefore by Kuratowski’s Theorem, $G$ 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 $G$ is large enough, then this forces $G$ to contain a complete graph as its minor.

Definition 3.1**.**

Let $x\in V(G)$, and $U\subseteq V(G)\backslash\{x\}$. An $x,U$-fan is a set of paths from $x$ to $U$ that are pairwise disjoint except at $x$. The size of the fan is the number of paths in the collection.

Lemma 3.2** (Fan Lemma).**

If $G$ is $k$-connected, and the size of $U$ is at least $k$. Then there exists an $x,U$-fan of size $k$.

Proof.

Add a new vertex $u$ to $G$, along with edges from $u$ to each vertex of $U$. This new graph $G^{\prime}$ is again $k$-connected, since $v(G^{\prime})\geq k+1$ and no set of size at most $k-1$ separates $G^{\prime}$. Indeed, suppose there was such a set $S$.

• •

If $S\subseteq V(G)$, then as $G^{\prime}-S$ is disconnected, hence so is $G-S=G^{\prime}-u-S=G^{\prime}-S-u$. Indeed, $G^{\prime}-S$ has a component $C$ containing $u$, and at least one other component $D$ not containing $u$, so $G^{\prime}-S-u$ has at least two components, namely D and some component of $C-u$, since $C-u$ is nonempty: it contains some element of $U$, since the size of $S$ was less than $k$.

• •

If $u\in S$, then $G-S=G^{\prime}-u-S=G^{\prime}-S$ is disconnected.

Now, $G^{\prime}$ is $k$-connected, so by Menger’s Theorem, there exist $k$ independent paths $P_{1},\ldots,P_{k}$ from $x$ to $u$ in $G^{\prime}$. (Menger’s Theorem says there exist $K_{G^{\prime}}(x,u)$ independent paths from $x$ to its non-neighbour $u$, but $K_{G^{\prime}}(x,u)\geq\kappa(G^{\prime})\geq k$, where the first inequality is because no set of size less than $\kappa(G^{\prime})$ separates $x$ from $u$.) Each such path is necessarily of the form $x\ldots u_{i}u$ for some $u_{i}\in U$, where $u_{i}\neq u_{j}$ if $i\neq j$. Therefore the paths $x\ldots u_{i}$ in $G$ form our desired $x,U$-fan of size $k$. ∎

Lemma 3.3**.**

If a graph $G$ has a cycle, then $G$ contains a $K_{3}$-minor.

Proof.

Suppose $G$ has a cycle $C_{n}$ $(n\geq 3)$ as a subgraph. By Corollary 2.6, $G$ has a $C_{n}$-minor. Now note that $C_{n}$ in turn has a $C_{3}$-minor, as witnessed by the sequence $C_{n},C_{n-1},\ldots,C_{3}$, each obtained from the previous by contracting an edge. Hence, Lemma 2.5 implies that $K_{3}\cong C_{3}$ is a minor of $G$. ∎

Proposition 3.4**.**

If a graph $G$ has $\chi(G)\geq 3$ then $G$ contains a $K_{3}$-minor.

Proof.

This is straightforward, by noting some equivalent characterisations of bipartite graphs: a graph $G$ is bipartite if and only if it is $2$-colourable, if and only if it contains no odd cycles. Hence, $\chi(G)\geq 3$ means $G$ is not $2$-colourable, therefore $G$ is not bipartite, therefore $G$ has an odd cycle, therefore $G$ contains a $K_{3}$-minor by Lemma 3.3. ∎

Proposition 3.5**.**

If a graph $G$ has $\chi(G)\geq 4$ then $G$ contains a $K_{4}$-minor.

Proof.

Let us proceed by induction on $v(G)$. If $v(G)\leq 3$, there is nothing to show; if $v(G)=4$ and $\chi(G)\geq 4$ then $\chi(G)=4$, so $G$ must be $K_{4}$. Now suppose $v(G)>4$, and that the result holds for all graphs on fewer vertices. Without loss, we may assume $G$ is connected. (Indeed, $\chi(G)$ is the maximum of the chromatic numbers of the components of $G$, so $G$ has a component $C$ with $\chi(C)\geq 4$. If $G$ is not connected then $C$ is a proper subgraph of $G$, so by induction $C$ – and hence $G$ – must have a $K_{4}$-minor.) Now we split into four cases.

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Suppose $\kappa(G)=0$. $G$ is not $1$-connected, so as $v(G)\geq\chi(G)\geq 4\geq 2$, we have that $G$ is disconnected. This contradicts our assumption.

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Suppose $\kappa(G)=1$. $G$ is not $2$-connected, so as $v(G)\geq\chi(G)\geq 4\geq 3$, we have that some set $\{c\}\subseteq V(G)$ separates $G$. $G-c$ will have a partition into two disjoint subgraphs $G^{\prime}_{1}$ and $G^{\prime}_{2}$ (these will be unions of the components of $G-c$). Let $G_{i}=G[V(G^{\prime}_{i})\cup\{c\}]$, for $i=1,2$. Then $V(G_{1})\cap V(G_{2})=\{c\}$, and $G_{1}[\{c\}]=G_{2}[\{c\}]$ are complete (they are both equal to $(\{c\},\emptyset)\cong K_{1})$, so $\chi(G)=\chi(G_{1}\cup G_{2})=\text{max}\{\chi(G_{1}),\chi(G_{2})\}$. Without loss, $\chi(G_{1})=\chi(G)\geq 4$. $G_{1}$ is a proper subgraph of $G$, as $\emptyset\neq V(G^{\prime}_{2})\subseteq V(G)\backslash V(G_{1})$), so by induction $G_{1}$ (and hence $G$) must have a $K_{4}$-minor.

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Suppose $\kappa(G)=2$, and $\{x,y\}$ is a separating set: there are subgraphs $G_{1}$ and $G_{2}$ such that $G=G_{1}\cup G_{2}$, $V(G_{1})\cap V(G_{2})=\{x,y\}$, $e(V(G_{1})\backslash\{x,y\},V(G_{2})\backslash\{x,y\})=0$ and $V(G_{i})\backslash\{x,y\}\neq\emptyset$ for both $i$. Suppose for a contradiction that both $G_{1}+xy$ and $G_{2}+xy$ can be $3$-coloured. Both colourings must colour $x$ and $y$ with different colours, so after possibly permuting one of the colourings, we can take their union to get a $3$-colouring on $G$; this is a contradiction. Hence, assume without loss that $\chi(G_{1}+xy)\geq 4$. $G_{1}+xy$ has fewer vertices than $G$ (as $V(G_{2})\backslash\{x,y\}\neq\emptyset$), so by induction it has a $K_{4}$-minor. We are not quite done yet, as $xy$ need not be an edge of $G$. If it is, we are done, so suppose not. Note that there is a path from $x$ to $y$ in $G_{2}$ – without loss $x$ has a neighbour $z$ in $G_{2}$, but $x$ cannot be a cutvertex, so removing it shows that there is a path from $z$ to $y$ that does not use $x$; concatenating $xz$ with a minimal such path produces a path $P_{xy}$ from $x$ to $y$ in $G_{2}$. This path ’substitutes’ for the missing edge $xy$; we can just contract it and pretend as though we have the edge $xy$ in our graph. In other words, $K_{4}\leq_{m}G_{1}+xy\leq_{m}G_{1}\cup P_{xy}\leq G$ shows that $G$ has a $K_{4}$-minor.

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Finally, suppose $\kappa(G)\geq 3$. Let $v\in V(G)$. Then $\kappa(G-v)\geq\kappa(G)-1\geq 2$, so $G-v$ is $2$-connected, and in particular contains a cycle. (Otherwise $G-v$ is connected and acyclic, i.e., a tree. It has a leaf $l$; if we delete the unique neighbour $n$ of $l$ from this tree we get a disconnected graph, because $n$ is not another leaf, since $G-v$ has at least $3$ vertices. This contradicts that $G-v$ is $2$-connected.) Take $U$ to be the set of vertices of this cycle, so its size is at least $3$. Since $G$ is $3$-connected, by Lemma 3.2 there exist three paths $P_{1},P_{2},P_{3}$ from $v$ to this cycle, that are disjoint except at $v$. Let $v_{1},v_{2},v_{3}$ be the three distinct end-vertices of the $P_{i}$. Then the sets $\{v\},\{v_{1}\},\{v_{2}\},\{v_{3}\}$ witness that $G$ has a $K_{4}$-minor, since we can contract the cycle onto the three vertices $v_{1},v_{2},v_{3}$, and contract the paths $P_{i}$ to edges – in the resulting graph, $x$ is neighbours with each of the three vertices in a triangle. This is a copy of $K_{4}$.

∎

In light of the previous two propositions, we might conjecture:

Conjecture 3.6**.**

For every $k\geq 2$, if $\chi(G)\geq k$ then $G$ contains a $K_{k}$-minor.

This is precisely the famous Hadwiger’s Conjecture.

Let us prove a weaker statement:

Theorem 3.7**.**

For every $k\geq 2$, there exists a constant $c(k)$ such that if $\chi(G)\geq c(k)$ then $G$ contains a $K_{k}$-minor.

Proof.

See proof of the restated version, Theorem 3.12. ∎

Notation 3.8**.**

For a graph $G$, write $d_{G}(x,y)$ for the length of a shortest path between vertices $x$ and $y$ (if it exists). This is the distance from $x$ to $y$. Let $G$ be a connected graph (so the distance between any two vertices is well-defined), and $x$ be a vertex of $G$. Define $S_{d}(x)=\{y\in$V(G)$:d_{G}(x,y)=d\}$ for each $d\geq 0$. For instance, $S_{0}(x)=\{x\}$.

The idea is that if we fix a vertex $x$, then the sets $S_{d}(x)$ partition $V(G)$ into ’layers’ around $x$, where each vertex $y$ 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 $G$ be a connected graph, and $x$ a vertex of $G$. If $d>0$ and $y\in S_{d}(x)$, then $\exists z\in S_{d-1}(x)$ such that $zy\in E(G)$. Furthermore, whenever $0\leq c<d-1$, we have $e(S_{c}(x),S_{d}(x))=0$.

Proof.

For the first claim, note that $y\in S_{d}(x)$ implies there is a path of length $d$ from $x$ to $y$. Let this path be $v_{0}\ldots v_{d}$ with $v_{0}=x$, $v_{d}=y$. Consider the path $v_{0}\ldots v_{d-1}$ in $G$ from $x$ to $v_{d-1}$, of length $d-1$. This must be a shortest path from $x$ to $v_{d-1}$, for if there were a path of length less than length $d-1$ from $x$ to $v_{d-1}$, then concatenating it with the edge $v_{d-1}v_{d}$ would result in a path of length less than $d$ from $x$ to $y$, then we would have $d_{G}(x,y)<d$, contradicting $y\in S_{d}(x)$. Therefore $v_{d-1}\in S_{d-1}(x)$. Since $v_{d-1}v_{d}\in E(G)$, this proves the first claim.

For the second claim, suppose for a contradiction that there exists some $0\leq c<d-1$ and some $w\in S_{c}(x)$, $y\in S_{d}(x)$ such that $wy\in E(G)$. Let $w_{0}\ldots w_{c}$ be a shortest path from $x$ to $w$, where $w_{0}=x,w_{c}=w$. Then, concatenating with the edge $w_{c}y$, we have a path $w_{0}\ldots w_{c}y$ from $x$ to $y$ of length $c+1<d$. This shows that $d_{G}(x,y)<d$, contradicting $y\in S_{d}(x)$. ∎

Lemma 3.10**.**

Let $G$ be a connected graph, and $x$ a vertex of $G$. Then there is some $D\geq 0$ such that $S_{d}(x)$ is empty whenever $d>D$, and nonempty whenever $0\leq d\leq D$.

Proof.

First observe that $S_{d}(x)$ is always empty for large enough $d$. For instance, take $D^{\prime}$ to be the length of a longest path in $G$, then $S_{d}(x)$ is empty whenever $d>D^{\prime}$. Then $S_{d}(x)$ can only be nonempty for a finite number of values $d=0,1,\ldots,D^{\prime}$. Let $D$ be the largest among these such that $S_{D}(x)$ is nonempty. Then, for all $0\leq d\leq D$, $S_{d}(x)$ will also be nonempty. Indeed, suppose not, then there is $d\in\{0,\ldots,D-1\}$ such that $S_{d}(x)=\emptyset$. Let us assume $d$ is maximal as such, so $\exists y\in S_{d+1}(x)$. The lemma above implies, in particular, that there is some $z\in S_{d}(x)$. This contradicts $S_{d}(x)=\emptyset$. Therefore we must indeed conclude that for all $0\leq d\leq D$, $S_{d}(x)$ is nonempty. ∎

Proposition 3.11**.**

Let $G$ be a connected graph, and $x$ a vertex of $G$. Then there is some $d\geq 0$ such that $\chi(G[S_{d}(x)])\geq\chi(G)/2$.

Proof.

Let $D$ be as in the lemma above. Choose $d\in\{0,\ldots,D\}$ which maximizes $\chi(G[S_{d}(x)])$. Let $\chi=\chi(G[S_{d}(x)])$. It is enough to show that $\chi\geq\chi(G)/2$, or equivalently, that $G$ is $2\chi$-colourable. Let us $2\chi$-colour $G$ starting with the outermost layer $G[S_{D}(x)]$ and working inwards. By the maximality assumption, we can certainly (properly) $\chi$-colour each layer $G[S_{b}(x)]$ (where $0\leq b\leq D$). Therefore we can colour $G[S_{D}(x)]$ with colours from $\{1,\ldots,\chi\}$, $G[S_{D-1}(x)]$ with colours from $\{\chi+1,\ldots,2\chi\}$, $G[S_{D-2}(x)]$ with colours from $\{1,\ldots,\chi\}$, and alternating so on and so forth, until we reach $G[S_{0}(x)]={x}$ (which gets a single colour, of course). This procedure colours all of $G$, and indeed gives us a proper colouring on $G$. (Edges within a single layer $G[S_{b}(x)]$ do not cause problems, since this colouring on $G$ 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 $k\geq 2$, if $\chi(G)\geq 2^{k}$ then $G$ contains a $K_{k}$-minor.

Proof.

We will proceed by induction on $k$. The base case is $k=2$. In this case, if $\chi(G)\geq 4$ then in particular, since $\chi(G)>1$, $G$ has an edge, so $G$ certainly contains a $K_{2}$-minor.

Now assume $k>2$, and suppose the result holds for the case $k-1$. That is, suppose that whenever $H$ is any graph with $\chi(H)\geq 2^{k-1}$, then $H$ contains a $K_{k-1}$-minor. We must show that if $\chi(G)\geq 2^{k}$ then $G$ contains a $K_{k}$-minor. Without loss of generality, we may assume $G$ is connected. (For general $G$, $\chi(G)$ is the maximum of the chromatic numbers of its components. Therefore if $\chi(G)\geq 2^{k}$, then $G$ has a component $C$ with $\chi(C)\geq 2^{k}$. If the result holds for connected graphs, then $C$ contains a $K_{k}$-minor, hence so does $G$.)

Consider the induced subgraph $H=G[S_{d}(x)]$, where $\chi(G)\geq 2^{k}$, $x$ is any fixed vertex of $G$ and $d\geq 0$ is as in Proposition 3.11. Then $\chi(H)=\chi(G[S_{d}(x)])\geq\chi(G)/2\geq 2^{k-1}$, so by the inductive hypothesis applied to $H$, $H$ contains a $K_{k-1}$-minor.

Now it is not hard to show that $G$ contains a $K_{k}$-minor – already the layer $G[S_{d}(x)]$ has a $K_{k-1}$-minor, but by Lemma 3.9 there is a path from each vertex of $G[S_{d}(x)]$ to $x$. We can contract all but one edge in each such path, and finish by noting that adding a new vertex to $K_{k-1}$ adjacent to all the other vertices produces $K_{k}$.

Let us argue in more detail, using our alternative formulation of minors. First note that $d>0$. (Indeed, if $d=0$, then $1=\chi(G[{x}])=\chi(G[S_{0}(x)])\geq\chi(G)/2$ shows that $2\geq\chi(G)\geq 2^{k}>2$, a contradiction.) Let $V(K_{k-1})=\{v_{1},\ldots,v_{k-1}\}$. Since $K_{k-1}\leq_{m}H$, there are disjoint nonempty subsets $V_{1}$, …, $V_{k-1}$ of $V(H)=S_{d}(x)$ such that each $H[V_{i}]$ is connected, and $e(V_{i},V_{j})>0$ whenever $v_{i}v_{j}\in E(K_{k-1})$.

Consider the graph $K_{k}\cong(V(K_{k-1})\cup\{x_{0}\},E(K_{k-1})\cup\{v_{1}x_{0},\ldots,v_{k-1}x_{0}\})$. This is a copy of $K_{k}$ obtained by adding a new vertex $x_{0}$ to $K_{k-1}$, adjacent to every other vertex. I show this a minor of $G$. Well, consider the nonempty subsets $V_{1},\ldots,V_{k-1},V_{k}$ of $V(G)$, where $V_{k}:=\bigcup_{b=0}^{d-1}S_{b}(x)\ni x$. These sets witness that $G$ has a $K_{k}$-minor:

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They are pairwise disjoint, because $V_{1}$, …, $V_{k-1}$ were already pairwise disjoint, and $\bigcup_{i=1}^{k-1}V_{i}\subseteq V(H)=S_{d}(x)\subseteq V(G)\backslash\bigcup_{b=0}^{d-1}S_{b}(x)=V(G)\backslash V_{k}$.

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Each $G[V_{i}]$ is connected: for each $1\leq i\leq k-1$, $G[V_{i}]=H[V_{i}]$, which is connected by assumption. (Both these graphs have vertex set $V_{i}$ and edge set $E(G)\cap V_{i}^{(2)}=E(G)\cap V_{i}^{(2)}\cap V(H)^{(2)}=E(H)\cap V_{i}^{(2)}$.) $G[V_{k}]$ is connected since for all $u,w\in V(G[V_{k}])=V_{k}$, there are paths in $G$ from $u$ to $x$ and from $x$ to $w$, which concantenated produce a path in $G$ from $u$ to $w$. To be clear, say $u\in S_{b}(x)$, where $0\leq b\leq d-1$. If $b=0$ then there is a trivial path from $u$ to $x$, of length [math]. Otherwise, apply Lemma 3.9 repeatedly to see that $\exists u_{1}\in S_{b-1}(x)$ such that $u_{1}u\in E(G)$, $\exists u_{2}\in S_{b-2}(x)$ such that $u_{2}u_{1}\in E(G)$, $\ldots$, $\exists u_{b}\in S_{0}(x)$ such that $u_{b}u_{b-1}\in E(G)$. This yields a path $uu_{1}\ldots u_{b}$ from $u$ to $x$. Similarly, there is a path from $w$ to $x$.

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$e(V_{i},V_{j})>0$ whenever $v_{i}v_{j}\in E(K_{k})$: this holds if $v_{i}v_{j}$ is an edge in the copy of $K_{k-1}$ we started with, so suppose we have an edge $v_{i}x_{0}\in E(K_{k})$, where $1\leq i<k$. We must show that $e(V_{i},V_{k})>0$. First, note that $V_{i}\neq\emptyset$, so $\exists y\in V_{i}$. Then Lemma 3.9 says $\exists z\in S_{d-1}(x)$ such that $zy\in E(G)$. Since $y\in V_{i}$ and $z\in V_{k}$, this shows $e(V_{i},V_{k})>0$ as required.

This completes the proof that $G$ has a $K_{k}$-minor. ∎

4 Conclusion

We have shown that for every $k\geq 2$, there is a constant $c(k)$ such that if $\chi(G)\geq c(k)$ then $G$ contains $K_{k}$ as a minor. Our choice of $c(k)$ was exponential in $k$, but Kostochka (1984) achieved $c(k)=\mathcal{O}(k\sqrt{log(k)})$.

Hadwiger’s Conjecture says we can do better still:

Conjecture 4.1** (Hadwiger’s Conjecture).**

For every $k\geq 2$, if $\chi(G)\geq k$ then $G$ contains $K_{k}$ as a minor.

This is essentially the nicest result we can hope to get. For instance if $k=2$ and $\chi(G)=1$ then $G$ has no edges, so it cannot contain a $K_{2}$-minor.

Hadwiger’s Conjecture is known to hold for the cases $k\leq 6$:

k = 2:

trivial; a graph with chromatic number larger than $1$ must have an edge, so it contains $K_{2}$ (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.