If a graph has exponentially many maximal cliques, what can we say about how many maximal cliques each vertex belongs to?

Question

My question is motivated by the following example: For $n \geq 1$, let $G$ be the complete $n$-partite graph on $2n$ vertices, i.e., every vertex of $G$ is connected to all others except one. Then $G$ has $2^n$ maximal cliques, with each one formed by choosing a vertex from each of the $n$ pairs of disconnected nodes. In this case, it is easy to see from the structure of $G$ that each vertex belongs to $2^n/2 = 2^{n-1}$ maximal cliques.

More generally, what can we say (if anything) about the number of maximal cliques each vertex of $G$ belongs to, if we do not know its structure, but only that $G$ is connected and has exponentially many maximal cliques?

In short, I would like to find out to what extent it is possible to derive local information about maximal clique involvement of vertices in $G$ from the global property of the number of maximal cliques $G$ has.

My hypothesis is the following, though I am unsure about the quantifiers and asymptotics.

Let $G$ be an $n$-vertex connected graph with $\Omega(f(n))$ maximal cliques for some exponential function $f(n)$. Then there always exists a subset $S$ of $\Omega(n)$ vertices such that each vertex of $S$ belongs to $\Omega(f(n))$ maximal cliques.

Proof. Suppose not. That is, suppose that for all subsets $S$ of $G$ containing $\Omega(n)$ vertices, each vertex of $S$ belongs to $O(g(n))$ maximal cliques for some subexponential function $g(n)$. By definition, there exist constants C,D > 0 such that for $n$ sufficiently large we have the following: For all subsets $S$ of size at least $Cn$, each vertex of $S$ belongs to at most $Dg(n)$ maximal cliques. Since $|G|=n$, we can assume that $0 < C \leq 1$. Then letting $S=X$, we have that each vertex of $X$ belongs to at most $Dg(n)$ maximal cliques, so the total number is at most $n\cdot Dg(n)$. However, we have that $n\cdot Dg(n) \neq \Omega(f(n))$, since $f(n)$ is exponential and $g(n)$ is subexponential, a contradiction.

Answer 1

The proof in the question does not work, because it assumes that in a set $S$, either every vertex lies in at least $\Omega(f(n))$ maximal cliques, or at most $O(g(n))$ maximal cliques. This is a false dichotomy for two reasons: not every vertex in $S$ has to do the same thing, and there is a large gap between $\Omega(f(n))$ and $O(g(n))$.

---

However, we can still prove a result along these lines. I have not stated it asymptotically, but we could do that too.

Theorem. For all $c>1$ there is an $\epsilon>0$ with the following property: if an $n$-vertex graph $G$ has $c^n$ maximal cliques, then there is a subset $S \subseteq V(G)$ with $|S| \ge \epsilon n$ such that every vertex in $S$ lies in at least $\epsilon c^n$ maximal cliques.

Proof. For convenience, let the clique degree $\operatorname{cd}(v)$ of a vertex $v$ be the number of maximal cliques of $G$ containing $v$.

Given $c$, we choose $\epsilon$ to satisfy, for all positive integers $n$ at once, the inequality $$\sum_{k=0}^{4\epsilon n} \binom nk < \frac12 c^n.$$ (See the answers to this question for details. It's enough to satisfy the inequality for all sufficiently large $n$, because if we can do that, we can adjust $\epsilon$ to also be sufficiently small to address the finitely many smaller exceptions.)

The result is that if an $n$-vertex graph $G$ has $c^n$ maximal cliques, then at most $\frac12 c^n$ of them can have size less than $4\epsilon n$. Therefore $$\sum_{v \in V(G)} \operatorname{cd}(v) > 2\epsilon n c^n,$$ because a clique of size $k$ is counted $k$ times by this sum, and there are more than $\frac12 c^n$ cliques that are counted at least $4\epsilon n$ times.

Suppose for the sake of contradiction that fewer than $\epsilon n$ vertices have clique degree more than $\epsilon c^n$. Each of those vertices still has clique degree at most $c^n$; that's how many maximal cliques there are! Therefore the vertices with large clique degree contribute at most $\epsilon n \cdot c^n$ to the sum. The remaining vertices have clique degree at most $\epsilon c^n$, and there's at most $n$ of them, so they contribute at most $n \cdot \epsilon c^n$ to the sum.

We get a total of less than $2\epsilon nc^n$, which contradicts our lower bound on the sum! So there must be at least $\epsilon n$ vertices with clique degree at least $\epsilon c^n$.