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Can anyone help me prove this theorem?

Let $G$ be a simple graph with $2n$ vertices and $n^2$ edges. If $G$ has no triangles, then $G$ is the complete bipartite graph $K_{n,n}$.

What I know: Assuming the $G$ has no triangles it follows that for every 3 vertices there are only 2 edges connecting them. For instance, having two sets of vertices i.e a's and b's. If I have vertices $a_1,b_1,b_2$. Then the edges are $a_1b_1,a_1b_2$. If I add another vertex i.e $a_2$, the edges added would be $a_2b_1,a_2b_2$. I added the edges such that there would be no triangles formed.

I get the idea of the complete bipartite graph through sketching it but I am having difficulty how can I come up to the definition of complete bipartite graph in order to prove the theorem.

Any insight is appreciated.

RobPratt
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JiSeung
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1 Answers1

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In fact, a triangle-free (simple) graph with $2n$ vertices and at least $n^2$ edges can only be a complete bipartite graph $K_{n,n}$. (This is Mantel's theorem for graphs of even order.)

Let $G=(V,E)$ be such a graph. Since the degree-sum is at least $2n^2$, the average degree is at least $n$, so the maximum degree is $\Delta(G)=n+t\ge n$. Choose a vertex $v$ of degree $\deg v=\Delta(G)=n+t$ where $t\ge0$. So $|N(v)|=\deg v=n+t$, and $|V\setminus N(v)|=n-t$.

Since $G$ is triangle-free, each vertex $w$ in $N(v)$ has neighbors only in $V\setminus N(v)$, so $\deg w\le n-t$. Thus $G$ has $n+t$ vertices of degree $\le n-t$, and the remaining $n-t$ vertices have degree $\le\Delta(G)=n+t$, so the degree-sum is $\le2(n-t)(n+t)=2(n^2-t^2)$, and $|E|\le n^2-t^2$. Since $G$ has at least $n^2$ edges, we must have $t=0$, that is, $\Delta(G)=n$. Since the maximum degree and the average4 degree are both equal to $n$, the graph is $n$-regular.

Let $V_1=V\setminus N(v)$ and $V_2=N(v)$, so that $|V_1|=|V_2|=n$. Since $V_2$ is independent, and each vertex has degree $n$, it follows that every vertex in $V_2$ is joined to every vertex in $V_1$. There can be no other edges since $G$ is triangle-free. Hence $G$ is a complete bipartite graph $K_{n,n}$.

P.S. A similar argument proves Mantel's theorem for graphs of odd order, i.e., that a triangle-free graph of order $2n+1$ with at least $n(n+1)$ edges is a complete bipartite graph $K_{n,n+1}$.

Let $G=(V,E)$ be such a graph. Choose a vertex $v$ of maximum degree, $\deg v=\Delta(G)$, and let $V_1=V\setminus N(v)$ and $V_2=N(v)$. By similar reasoning as before, we can see that $|V_1|=n$ and $|V_2|=n+1=\Delta(G)$. Let $E_1$ be the set of edges with both endpoints in $V_1$. Since $V_2$ is independent, we have $$|E_1|+n(n+1)\le|E_1|+|E|=\sum_{w\in V_1}\deg(w)\le n(n+1),$$ whence $|E_1|=0$, i.e., $V_1$ is also independent, and $|E|=n(n+1)$, so that $G$ is a the complete bipartite graph with partite sets $V_1$ and $V_2$.

bof
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