I want to find a condition on $\delta(G)$ (ex. $\delta(G) \geq an$) that guarantees $\kappa(G)=\delta(G)$ where $\kappa(G)$ is the vertex-connectivity of a bipartite graph $G$, and $\delta(G)$ is the minimum degree of $G$.  



In other words, I want to prove that the statement



> 
> If $\delta(G) \geq an$, then $\kappa(G)=\delta(G)$
> 
> 
> 


holds for bipartite graphs.


I know that $\delta(G) \geq \frac{n+1}{4}$ implies $\kappa'(G)=\delta(G)$ where $\kappa'(G)$ is the edge-connectivity of $G$, and it is sharp.  

Also, I proved that bipartite $G$ is connected if $\delta(G) \geq \frac{n+1}{4}$.  

But I have no idea with the **vertex-connectivity**.  

Here are my ideas to find such boundary $an$:


1. (Contrapositive) Prove the statement "If $\kappa(G)<\delta(G)$, then $\delta(G)<an$."
2. Use the statement $\delta(G) \geq \frac{n+1}{4}\Rightarrow \kappa'(G)=\delta(G)$. Find the boundary of $\delta(G)$ that implies $\kappa(G)=\kappa'(G)$, and then show this boundary also satisfies $\delta(G) \geq \frac{n+1}{4}$.


Would you help me?



Connectivity and the minimum degree of bipartite graph
This becomes true at $a = \frac{1}{3}$.


**Claim.** If $G$ is an $n$-vertex bipartite graph such that $\delta(G) \geq \frac{1}{3}n$, then $\delta(G)=\kappa(G)$.


*Proof.* Let $(A,B)$ be the bipartition of $G$. Suppose $\kappa(G)<\delta(G)$ and let $X \subseteq V(G)$ be such that $G-X$ is disconnected and $|X|=\kappa(G)$. Let $A=A\_1 \sqcup A\_2 \sqcup A\_3$ and $B=B\_1 \sqcup B\_2 \sqcup B\_3$ be such that $A\_1=X \cap A$, $B\_1=X \cap B$, and $G[A\_2 \cup B\_2]$ and $G[A\_3 \cup B\_3]$ are both a union of connected components of $G-X$. Since $\kappa(G) < \delta(G)$, none of $A\_2, A\_3, B\_2, B\_3$ are empty. Observe that the degree in $G$ of each vertex in $A\_2, A\_3, B\_2, B\_3$ are at most $|B\_2|+|B\_1|, |B\_3|+|B\_1|, |A\_2|+|A\_1|, |A\_3|+|A\_1|$, respectively. By taking the average of these four numbers, it follows that for some $Y \in \{A\_2, A\_3,B\_2,B\_3\}$ every vertex in $Y$ has degree at most $(n+\kappa(G))/4$. Thus $\delta(G) \leq (n+\kappa(G))/4 < (n+\delta(G))/4$, which implies $\delta(G) < \frac{n}{3}$.


On the other hand, the following claim shows that $\frac{1}{3}$ is best possible.


**Claim.** For all $\ell \in \mathbb{N}$ exists a bipartite graph $G$ on $3\ell+1$ vertices such that $\delta(G)=\ell$, but $\kappa(G) < \delta(G)$.


*Proof.* Let $G\_1$ and $G\_2$ be copies of $K\_{\ell, \ell}$ with bipartitions $(A\_1,B\_1)$ and $(A\_2,B\_2)$, respectively. Let $X\_1 \subseteq A\_1$ and $X\_2 \subseteq A\_2$ both be of size $k$, and let $G$ be the graph obtained from $G\_1$ and $G\_2$ by identifying $X\_1$ and $X\_2$. Note that $G$ is a bipartite graph with $4\ell-k$ vertices, minimum degree $\ell$, and vertex-connectivity $k$. Setting $k=\ell-1$ proves the claim.


