$\require{mathtools}
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% .. over and under
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% .. one-part
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Adapted from lecture 3 of Yufei Zhao’s Fall 2019 class on “Graph theory and additive combinatorics” at MIT.
Say I give you some constant-size graph $H$ (e.g. a 4-clique, a 5-cycle) and tell you make a very dense graph $G$ that doesn’t contain $H$ as a subgraph.
If $H$ is not bipartite (i.e. it has an odd cycle), then this is super easy: all you need to do is to make $G$ bipartite. For example, you can make $G$ the complete bipartite graph $K_{\frac{n}{2},\frac{n}{2}}$, which has $\frac{n^2}{4}$ edges, so you get edge density $\approx 1/2$. Sometimes you can do even better than $1/2$, and the right answer is always known:
- When $H$ is an $(r+1)$-clique, Turán’s graph theorem pinpoints the exact number of edges you can get: about $(1-\frac{1}{r})\frac{n^2}{2}$.
- In general, the Erdős–Stone–Simonovits theorem tells you that the best edge density tends to $1-\frac{1}{\chi(H)-1}$ as $n$ grows, where $\chi(H)$ is the coloring number of $H$.
On the other hand, if $H$ is bipartite, then there is no “cheap trick” that allows you to avoid $H$. Indeed, even when $H$ is the complete bipartite graph $K_{s,t}$, we’ll see that the edge density will tend to $0$ as $n$ grows. More precisely, the Kővári–Sós–Turán theorem says that
\[
m \leq O\p{t^{1/s} n^{2-1/s}} = O\p{n^{2-1/s}},
\]
which is strongly subquadratic! As we’ll see in the lower bounds section, this is mostly tight.
The idea is that forbidding $K_{s,t}$ puts a cap on the number of $s$-stars in $G$, and that you cannot have high density without creating many $s$-stars.
Proof
The proof works by upper and lower bounding the number of $s$-stars in $G$.
For an upper bound, we can count $s$-stars by enumerating through all choices $S \in \binom{[n]}{s}$ of the $s$ “ends” of the star, then enumerating the choices of the center. Since $G$ is $K_{s,t}$-free, for any fixed $S$, there must be $<t$ possible choices for the center, so
\[
\#\text{$s$-stars} < t\binom{n}{s}.
\]
(This is roughly $O(n^s)$; without restrictions, $G$ could have had as many as $(n-s) \binom{n}{s} \ge \Omega(n^{s+1})$ $s$-stars, so this upper bound is actually quite stringent! The factor $n$ that separates these two expressions is exactly what will give the $n^{1/s}$ loss in the number of edges.)
For the lower bound, we start out by computing the number of $s$-stars as a function of the degrees $d(u)$ of each node:
\[
\#\text{$s$-stars} = \sum_u \binom{d(u)}{s}.
\]
Now, it seems like for some fixed average degree $D$, the best way to minimize the number of $s$-stars is to have all degrees equal: indeed, if we let some degrees get bigger, then those nodes would produce proportionally many more $s$-stars. Formally, since $\binom{d(u)}{s}$ is convex in $d(u)$, we have
\[
\#\text{$s$-stars} = \sum_u \binom{d(u)}{s} \ge n \binom{D}{s}.
\]
Putting the bounds together, we get
\[
\al{
&&n\binom{D}{s} &\leq t\binom{n}{s}\\
&\Rightarrow& n \p{\frac{D}{s}}^s &\leq t \cdot O\p{\frac{n}{s}}^s\\
&\Rightarrow& D &\le O\p{t^{1/s} n^{1-1/s}}.
}
\]
Lower bounds
The $m \leq O(n^{2-1/s})$ bound is believed to be tight, but surprisingly, matching constructions are only known in a few cases.
Probabilistic constructions
Let $v(H)$, $e(H)$ be the number of vertices and edges in the graph $H$ you want to forbid. Take an Erdős-Rényi random graph $G$ with edge probability $p$. There are $\leq n^{v(H)}$ possible “sites” where $H$ could appear, and for each of them, the probability that it appears is $p^{e(H)}$. So if we make sure that
\[
n^{v(H)}p^{e(H)} \le 1/2 \stackrel{\sim}{\Leftrightarrow} p \le n^{-\frac{v(H)}{e(H)}},
\]
then with constant probability we’ll get an $H$-free graph with $\Omega(pn^2) = \Omega\p{n^{2-\frac{v(H)}{e(H)}}}$ edges.
For $K_{s,t}$ (with $s \le t$), this gives $\Omega\p{n^{2-\frac{s+t}{st}}} \geq \Omega(n^{2-2/s})$, so $n^{2-\Theta(1/s)}$ is the right answer. In fact, when $t$ tends to infinity, we get $n^{2-\frac{1+o(1)}{s}}$, so the constant in the numerator is tight. Unfortunately though, the exponent is not exactly right for any $s,t \geq 2$.
Algebraic constructions
There are algebraic constructions that give the right exponent for some specific values of $s,t$.
Points and lines: $s=t=2$
Let $G$ be the (bipartite) graph of incidences between points and lines in $\Z_p^2$. This graph has $n=\Theta(p^2)$ vertices and $\Theta(p^3) = \Theta(n^{3/2})$ edges. Since two distinct lines cannot meet at two distinct points, there is no $K_{2,2}$.
Spheres and points: $s=t=3$
This time, consider incidences between points and “spheres” of a fixed radius in $\Z_p^3$. It’s not super clear what a “sphere” is in $\Z_p^3$, but at least in Euclidean space, three distinct spheres of the same radius cannot meet at three distinct points, so there is no $K_{3,3}$. When the radius is chosen well, this graph has $n=\Theta(p^3)$ vertices and $\Theta(p^5) = \Theta(n^{5/3})$ edges.
General: Alon–Kollár–Rónyai–Szabó
In general, for $t \geq (s-1)!+1$, there are $K_{s,t}$-free graphs with $\Omega(n^{2-1/s})$ edges. But the answer remains unknown (everywhere?) when $s \leq t \leq (s-1)!$.