I am looking for a specific [matroid](https://en.wikipedia.org/wiki/Matroid). I found a source that claimed to discuss these matroids, but then, only discusses [geometric lattice](https://en.wikipedia.org/wiki/Geometric_lattice). Even more, in that paper, the geometric lattice that seems to be the right one was described as



> 
> ... the lattice associated with the [Steiner system](https://en.wikipedia.org/wiki/Steiner_system) $S(3,6,22)$.
> 
> 
> 


It might be clear to some, how to translate between all these different constructs, but I have a hard time finding any source explaining to me (in short) how these concepts are linked.


I suppose, that the matroid of the geometric lattice $\mathcal L$ is defined on the set of atoms of $\mathcal L$, and independence of atoms $a\_1,...,a\_n\in\mathcal L$ means that the supremum $a\_1\vee \cdots \vee a\_n$ has rank $n$. 
But this is just a guess. 
Furthermore, the lattice that comes from $S(3,6,22)$ is said to be of rank at least 3, but there is not much more said about this.


Can someone tell me how to obtain the matroid from $S(3,6,22$)?




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There are actually two papers I am talking about:


* [Basis Transitive Matroids](https://link.springer.com/content/pdf/10.1023%2FA%3A1022411901450.pdf), by A. Delandtsheer, H. Li., which discusses matroids purely in terms of geometric lattices and dimensional linear spaces (DLS).
* [Homogeneous Designs and Geometric Lattices](https://reader.elsevier.com/reader/sd/pii/0097316585900226?token=97090CC92B3E729C65E22D6EACE74C5E72BD01CFD609BB19D7B88AE952A540A3FA10D898DD584A899A074FB911CE653D), by W. M. Kantor (Theorem 2), which lists $S(3,6,22)$ as a Steiner system associated with a lattice.



From Steiner systems to geometric lattices to matroids
Let $B$ be the subsets of $\{1,2,...,22\}$ of size $4$ which are not contained in any block of $S(3,6,22)$. $B$ satisfies the basis exchange property. 


Proof: 


Let $X,Y\in B$, $a\in X$. $X \setminus a$ is a 3-element set, so it is contained in exactly one block of $S(3,6,22)$.


If $∀y\in Y :(X \setminus a) \cup y \notin B$, it follows that all elements of $Y$ are in some block containing $X \setminus a$, and there's only one such block, call it $A$. It follows that $Y\subset A$, contradiction.


Corollary. $B$ is the basis of a matroid with rank $4$.


