Connectivity and the theorems of Menger

Definition 4..1 (Notation for subgraphs)  

If $ G$ graph and $ S \subseteq V(G)$ then $ G-S$ is the induced subgraph with edges in $ S$ deleted. Given $ F \subseteq E(G)$ then $ G-F$ is the spanning subgraph obtained by deleting edges in $ F$.

Definition 4..2 ($ k$ connected)  

A graph is $ k$ connected if $ \vert G\vert > k$ and $ G-S$ is connected for any $ S \subset V(G)$ with $ \vert S\vert < k$.

Observation 4..3  

The only $ k$ connected graph with $ k+1$ vertices is the complete graph $ K_{k+1}$.

Definition 4..4 (Connectivity)  

The connectivity of $ G$ or $ \kappa(G) = \max \{ k : G $is$ k-$connected$ \}$.

Definition 4..5 (Local connectivity)  

If $ a,b \in V(G)$ distinct and $ ab \notin E(G)$ the local connectivity $ \kappa(a,b;G) = \min \{ k : \exists S \subseteq V(G) \setminus \{ a,b
\} : \vert S\vert = k : G - S$ has no $ a-b$ path $ \}$, the minimum number vertices separating $ a$ and $ b$.

Observation 4..6  

If $ G$ is not complete then $ \kappa(G) = \min_{a,b} \{ \kappa(a, b; G) \}$, which is the same as $ \min_{a,b : ab,ba \notin E(G)} \{ \kappa(a, b; G) \}$.

If $ G$ is complete then $ \min_{a,b} \{ \kappa(a, b; G) \}$ does not have a value, but $ \kappa(K_n) = n-1$.

Definition 4..7 ($ k$-edge connected)  

A graph $ G$ is $ k$-edge connected if $ G-F$ is connected for all $ F \subseteq E(G)$ with $ \vert F\vert < k$.

Definition 4..8 (Edge connectivity $ \lambda(G)$)  

The edge connectivity of $ G$ is $ \lambda(G) = \max \{ k : G$ is $ k$ edge connected$ \}$.

Definition 4..9   If $ a,b \in V(G)$ distinct then the local edge connectivity is $ \lambda(a,b; G) = \min \{ k: \exists F \subseteq E(G) : \vert F\vert = k, G -
F$ has no $ a-b$ path $ \}$. Note that $ \lambda(G) =
\min_{a,b} \lambda(a,b;G)$.

Definition 4..10 (Independent paths)   Two $ a-b$ paths are independent if their only common vertices are $ a,b$.

Theorem 4..11 (Menger)  

Vertex form. Let $ a,b \in V(G)$ distinct. $ ab \notin E(G)$ then $ \kappa(a,b;G) = $ maximum number of pairwise independent $ a-b$ paths.

Edge form. $ a,b \in V(G)$ distinct. Then $ \lambda(a,b;G) = $ maximum number of edge disjoint $ a-b$ paths.

Proof. Vertex form. Replace all edges $ uv \in
E(G)$ with two directed edges and give each vertex capacity 1. Apply vertex form of max-flow min-cut to get an integer flow from $ a-b$, $ \kappa(a,b;G)$ since each vertex has capacity $ 1$ or 0.

Edge form. Do the same thing but use the edge form of max-flow min-cut.

$ \qedsymbol$

Definition 4..12 (Line graph)  

The line graph $ L(G)$ is the graph with a vertex $ v_e$ for each $ e
\in E(G)$ and an edge $ v_e v_f$ whenever $ e$ and $ f$ have a common end vertex in $ G$.

Observation 4..13   The edge form can be deduced from the vertex from, using the line graph.

Observation 4..14   Given a set $ P$ of independent $ a-b$ paths then certainly $ \vert P\vert \le
\kappa (a,b;G)$, but it is not necessarily possible to add paths to $ P$ to form $ \kappa(a,b;G)$ independent paths.

Observation 4..15   There is a multiple source - sink version of Menger's theorem.

Lemma 4..16  

Assume $ ab \in E(G)$ and let $ G' = G - ab$. Then $ \kappa(a,b;G') \ge
\kappa(G) - 1$.

Proof. Let $ k = \kappa(a,b;G')$. Choose $ S \subseteq V(G) \setminus \{ a,b \}$ with $ \vert S\vert=k$ and $ G' - S$ disconnected. Let $ x$ and $ y$ be vertices in different components of $ G' - S$, so that $ \vert\{x,y\} \cap \{ a,b \}\vert$ is minimal. Then either
  1. $ a \ne x,y$ so $ G - (S \bigcup {a})$ is disconnected; or
  2. $ b \ne x,y$ so $ G - (S \bigcup {b})$ is disconnected; or

  3. $ \{ x,y \} = \{ a,b \}$ so $ V(G) = S \bigcup \{ a,b \}$, so $ \vert G\vert \le k+2$.

All cases imply that $ \kappa(G) \le k+1$.

$ \qedsymbol$

Corollary 4..17  

A graph is $ k$-connected iff any two vertices are joined by at least $ k$ independent paths.

Proof. If $ G$ is not a complete graph, $ \kappa(G) = \min \kappa(a,b;G)$. Apply Menger's theorem.

$ K_n$ is $ k$-connected iff $ n \ge k+1$, so true for a complete graph.

$ \qedsymbol$

Corollary 4..18  

A graph is $ k$-edge-connected iff any two vertices are joined by at least $ k$ edge disjoint paths.

John Fremlin 2010-02-17