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Do sparse graphs contain regular pairs?

Mathematics Asked by alpmu on December 27, 2021

An easy corollary of the Szemerédi Regularity Lemma is that dense graphs contain linear sized $varepsilon$-regular bipartite subgraphs whose density is similar to that of the parent graph. As noted by Tim Gowers in (https://mathoverflow.net/questions/291467/is-there-a-weak-strong-regularity-lemma) there are easier ways of seeing this, with better bounds.

I’m wondering if a meaningful statement with the above flavor holds for sparse graphs, of density $Omega(n^{-1/t})$. That is, graphs which are dense enough to necessarily contain $K_{t,t}$ subgraphs.

What I’m looking for exactly is a subgraph $(A,B)$ that satisfies $|A|=|B|=k$, $e(A,B)=Omega(k^{2-1/t})$, and further, $(A,B)$ is $varepsilon$-regular, in the sense that any subgraph $(A’, B’)$ with $varepsilon k$ vertices on each side satisfies $e(A,B)=Omega(k^{2-1/t})$. Note that this is a lot weaker than the usual notion of $varepsilon$-regularity in that we allow that the density of a subgraph to be off by a constant factor from the density of the the parent graph, all that we insist on is that they are of the same order of magnitude.

I’m okay with aiming for a sub-linear sized regular pair (i.e. take $k=o(n)$) as the graph itself could be almost entirely filled with isolated vertices, except for a small clique, as Misha says. I would expect one could take $k$ polynomial in $n$, but I’m interested in any range where $k$ grows with $n$.

I’m also okay with assuming that the graph is $K_{10t, 10t}$-free (say), although I cannot tell if there is an easy construction that shows the necessity of such an assumption.

From what I can tell, the sparse versions of Regularity Lemma do not say anything immediately meaningful here, as they do not forbid all edges of the sparse graph being between non-regular pairs.

One Answer

Well, without ruling out big complete bipartite graphs, one counterexample is the following sparse graph: a complete graph on about $n^{1-1/(2t)}$ vertices, together with enough isolated vertices to bring the total to $n$.

There are plenty of ways to pick a bipartite subgraph with $epsilon n$ vertices in each part that has the correct density. However, all of these will include a linear number of isolated vertices in each part, so all of them will have linear-sized subgraphs with $0$ edges, and the thing you want cannot hold.

Answered by Misha Lavrov on December 27, 2021

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