Does every geometric lattice of rank $r$ contain the Boolean $B_r$ as a sublattice?

Question

A finite lattice is geometric if it is semimodular and atomistic. Geometric lattices can have arbitrarily high rank $r$, as evidenced by the Boolean lattice $B_r$ (power set of $r$ elements with the subset relation).

Question: Does every geometric lattice of rank $r$ contain $B_r$ as a sublattice?

I could not find this in the standard references I tried (Grätzer's Lattice Theory: Foundation, Stanley's Enumerative Combinatorics Vol 1, Stern's Semimodular Lattices). For all I know it might be stated somewhere else, or be a direct consequence of something I don't know (perhaps something with matroids or with the Möbius function).

Motivation

I would like to better understand the structure of long (high-rank) geometric lattices. The vast majority of geometric lattices are short, indeed rank-three. For example, with 32 elements, the numbers of nonisomorphic geometric lattices of ranks 2, 3, 4, 5 are 1, 215, 3, 1. (The last one is Boolean.)

Thus long geometric lattices are "rare", but they might be interesting to study, for example, with respect to the unimodality conjecture (apparently due to Rota; see also Dowling and Wilson below), which states that a geometric lattice cannot have a "narrow level" $k$ such that $W_{k-1} > W_k < W_{k+1}$, where $W_k$ means the number of elements of rank $k$. Obviously a rank-3 lattice cannot be non-unimodal, so you need to look at higher ranks. A possible follow-up question could be: Given a high rank $r$, how does one create many examples of rank-$r$ geometric lattices other than the Boolean one? It might help a little if we at least knew that they always contain the Boolean.

Assorted observations

1. The answer is "yes" for geometric lattices of at most 37 elements, by inspecting all of them; there are not very many (a few thousand).

2. The answer is "yes" for the lattice of partitions of $[r+1] = \{1,2,\ldots,r+1\}$, because $B_r$ can be embedded there by mapping each subset $S \subseteq [r]$ to the partition that has a block $S \cup \{r+1\}$ and all elements of $[r] \setminus S$ as singleton blocks.

3. At least the levels of any rank-$r$ geometric lattice $L$ are wide enough to contain the corresponding levels of $B_r$. First, $L$ has at least $r$ atoms. This is because if its rank function is $\rho$, we have $\rho(a \wedge b) + \rho(a \vee b) \le \rho(a) + \rho(b)$ for any two elements $a,b \in L$. Suppose the atoms are $a_1,\ldots,a_n$ in some order, and consider the join of first $i$ atoms; by induction one sees that it has rank at most $i$. So the top (join of all atoms) has rank at most $n$. Thus $n \ge r$. For levels other than the atoms, we use a nice lower bound by Dowling and Wilson (1974). Let $W_k$ be the number of elements of rank $k$ in $L$, and $n=W_1$ the number of atoms. Then $$ W_k \ge \binom{r-2}{k-1}(n-r) + \binom{r}{k}, \qquad\text{for $0 \le k \le r$}. $$ Since $n\ge r$, we have $W_k \ge \binom{r}{k}$. This also implies that $|L| \ge 2^r$.

4. Incidentally, if a geometric rank-$r$ lattice $L$ has exactly $r$ atoms, then $|L|=2^r$. It cannot be more, because from $r$ atoms one can form at most $2^r$ different joins. In this case $L$ has the binomial rank sequence $W_k = \binom{r}{k}$. I don't know whether this is enough to show that it actually is $B_r$; there is a somehow related MO question "Enumerative characterisation of boolean lattices II" (whether a graded lattice of rank $r$, with $\binom{r}{k}$ elements of rank $k$, and $r2^{r-1}$ edges, is necessarily $B_r$).

References

• Thomas A. Dowling and Richard M. Wilson (1974): The slimmest geometric lattices. Transactions of the American Mathematical Society 196: 203-215.

Answer 1

According to Wikipedia, "geometric lattice" is equivalent to "lattice of flats of a finite matroid". If this is true, the answer is yes.

Let $M$ be a matroid of rank $r$, and choose a basis $(e_1, e_2, \ldots, e_r)$. Then the flats spanned by the subsets of this basis form a boolean lattice.