1
$\begingroup$

This question is inspired by a Andrew D. King's comment in Linear intersection number and coloring (not chromatic) number

A linear hypergraph is a pair $\pi=(X, L)$ where $X\neq \emptyset$ is a set and $L\subseteq {\cal P}(X)$ has the following properties:

  1. for $e\in L$ we have $|e|\geq 2$;
  2. if $e_1\neq e_2 \in L$ then $|e_1\cap e_2|\leq 1$.

We set $X(\pi)=X$ and $L(\pi)=L$. The graph $G_\pi$ associated to a linear hypergraph $\pi$ is given by $G=(V,E)$ where $V = L$ and $E = \{\{e_1, e_2\} \subseteq L: e_1\neq e_2\text{ and } e_1\cap e_2\neq \emptyset\}$. It turns out that for any graph $G$ there is a linear hypergraph $\pi$ such that $G\cong G_\pi$. For any graph $G$ the we set $$\ell(G) := \text{min}\{|X(\pi)|:\pi \text{ is a linear hypergraph such that } G_{\pi} \cong G\}$$ and call this the linear intersection number of $G$. (For infinite graphs, this concept is boring: $\ell(G) = |V(G)|$ for infinite graphs.)

For any finite graph $G$ let $\Delta(G)$ be the maximum degree.

Is there a finite graph $G$ such that $\Delta(G) > \ell(G)$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

Yes there is such a graph. Consider the following hypergraph $\pi = (X, L)$ where

  • $X =\{1,2,\ldots 6\}$;
  • $L = \big\{\{1,2,3\}\big\} \cup \big\{\{a,z\}: a\in \{1,2,3\}, z\in\{4,5,6\}\big\}$.

Let $v_0 = \{1,2,3\}$ and we set $G:=G_\pi$. Then clearly $\ell(G) \leq 6$, but $\Delta(G) \geq \text{deg}(v_0) = 9$. So $\Delta(G) > \ell(G)$.

Improve this answer
$\endgroup$

You must log in to answer this question.