TITLE: Have you seen my matroid?
QUESTION [28 upvotes]: Let $M(n,k)$ be the matroid on the ground set $\{\pm 1,\ldots,\pm n\}$ for which a set is independent if and only if it contains at most $k$ pairs $\pm i$.  Note that the signed permutation group (the Coxeter group of type $B_n$) acts on this matroid. Questions: 

Does this matroid have a name?
Has it been studied before?
Is there a nice formula for its characteristic polynomial?

Here are some boring special cases:

$M(n,n)$ is the Boolean matroid on $2n$ elements.
$M(n,n-1)$ is the uniform matroid of rank $2n-1$ on $2n$ elements.
$M(n,0)$ is the direct sum of $n$ copies of the uniform matroid of rank 1 on 2 elements.

The first interesting case is $M(3,1)$, which has rank 4 and characteristic polynomial
$$q^4 - 6q^3 + 15q^2 - 17q + 7$$
I am also interested in truncations of this matroid.  That is, let $M(n,k,d)$ be the matroid on the ground set $\{\pm 1,\ldots,\pm n\}$ for which a set is independent if and only if it contains at most $k$ pairs $\pm i$ and has size at most $d$.  All of the same questions apply!
Remark:  I would like to regard these matroids as type B analogues of uniform matroids.  Uniform matroids are the permutation-invariant matroids on the ground set $\{1,\ldots,n\}$, while these are the signed-permutation-invariant matroids on the ground set $\{\pm 1,\ldots,\pm n\}$.

REPLY [14 votes]: Let $U$ be the uniform matroid of rank $k$ on $n$. Since $U$ is orientable one can consider the Lawrence oriented matroid $\Lambda(U)$ associated with any orientation of $U$ (the Lawrence construction doesn't care about which orientation you take). Then $M(n,k)$ is precisely the underlying unoriented matroid $\underline{\Lambda(U)}$ of $\Lambda(U)$.
Also, the dual matroid $M^*(n,k)$ is a symplectic matroid, which explains why the group $B_n$ acts on the primal.