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Theorem. Fix $\epsilon > 0$; for sufficiently large n, any graph with n vertices and $\epsilon \binom{n}{2}$ edges contains many (nondegenerate) cycles of length 4.

The proof is simple; put an indicator variable $\delta_{x, y}$ for each pair of vertices corresponding to whether or not there is an edge there; then start with

$n^8 \epsilon^4 = (\sum \delta_{x, y})^4$

and apply Cauchy-Schwarz twice; finally, note that there are $O(n^3)$ "degenerate 4-cycles".

A basic corollary of this is the following fact:

Corollary. Any graph with girth at least 5 and n vertices has $o(n^2)$ edges.

This seems like it should be possible to prove without resorting to "analytic" machinery like Cauchy-Schwarz; indeed, it seems like it should be weak enough to prove almost by arguing "locally." But none of the obvious lines of reasoning seem to provide a proof.

Is it possible to get a good bound on the density of large-girth graphs without using Cauchy-Schwarz or equivalent?

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up vote 18 down vote accepted

It's known more specifically that any graph with girth ≥ 5 has $O(n^{3/2})$ edges — see e.g. Wikipedia on the Zarankiewicz problem.

Here's a combinatorial proof. Suppose that graph $G$ has $\ge kn^{3/2}$ edges for a sufficiently large constant $k$. As long as there are vertices with degree smaller than some appropriate constant times $\sqrt n$ one can remove them and get a smaller graph with the same property of having at least $kn^{3/2}$ edges, so eventually one can reach a state where every vertex has degree at least $\Omega(\sqrt n)$. Once this happens, there are $O(n^2)$ possible pairs of neighbors that a vertex might have, and each vertex has $\Omega(n)$ pairs of neighbors, so some pair of neighbors appears twice causing the graph to have a 4-cycle.

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