# Smooth bases of matroids

Motivated by algebraic geometry, I've come up with a purely combinatorial definition within the theory of matroids. The question is: is this concept known?

If you like matroids but not algebraic geometry, skip to the definition below.

Let $n\choose k$ denote the collection of all $k$-element subsets of $[1,n]$ (rather than the number thereof). We can and will identify this collection both with the set of $T$-fixed points on the $k$-Grassmannian $Gr(k,n)$, where $T$ is the $n$-torus that acts (unfaithfully), and also with the set of Plücker coordinates.

Let $C \subseteq {n\choose k}$ be a subcollection. Then, following Neil White, we can define a subscheme $\Pi_C$ of $Gr(k,n)$ by killing all Plücker coordinates $p_S, S \notin C$. This subscheme is $T$-invariant, and its $T$-fixed points are exactly $C$.

Easy fact: if $\Pi_C$ is irreducible, then $C$ is a matroid. The non-Pappus matroid shows the converse is false. (This is my own motivation for matroids -- they serve as combinatorial stand-ins for subvarieties of Grassmannians.)

I'm interested in the smooth points of $\Pi_M$, where $M$ is a matroid. Perhaps the most efficient way to describe $M$ is by listing its connected flats $F$, and for each, giving the rank. (Saying that $rank(F) \leq r$ means that for each $S$ that intersects $F$ too much, $p_F = 0$. I'm pretty sure that the connected flats gives the shortest list of $F$s to give all the $S$.)

If $M \subseteq {n\choose k}$ is a matroid, call $\lambda \in M$ a smooth base if for any connected flat $F$, $rank(F) = |\lambda \cap F|$.

Note that $\geq$ is required for $\lambda$ to be a base at all. It's pretty easy to prove that $\lambda$ is a smooth point of $\Pi_M$ iff $\lambda$ is a smooth base of $M$ in the sense above.

Is this concept known to matroid theorists? Is this characterization of smooth points known to anybody?

Example: let $M$ be the Schubert matroid for a $\lambda \in {n\choose k}$, i.e. for each $i \notin \lambda, i+1 \in \lambda$, we have a connected flat $[1,i]$ with rank $|[1,i] \cap \lambda|$. Then $\lambda$ is a smooth base of $M$. And indeed, $\lambda$ is the point in the Bruhat cell whose closure is the Schubert variety $\Pi_M$.

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I hope this question gets a good answer. In the mean time I'll mention a concept I've seen which seems somewhat related to your condition of smoothness.

When you have a matroid $\mathcal M$ with a base $B$ with the property that all cyclic flats $F$ are spanned by $F\cap B$, this is called a fundamental transversal matroid. The base $B$ is called a fundamental base.

Schubert matroids are a special case of fundamental transversal matroids, and the base $\lambda$ is a fundamental base in the sense above. At some point I was convinced that your definition implies at least that the matroid is transversal, but now I'm not so sure anymore.

Some further characterizations in terms of some rank inequalities, or affine representations on simplices are proved in "Characterizations of transversal and fundamental transversal matroids" by J.E. Bonin, J.P.S. Kung and A. de Mier. See also these slides for pictures. I should also point out that I haven't seen the geometric connection which you explain being mentioned in the literature on fundamental transversal matroids.

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Allen's condition for cyclic flats is equivalent to the same condition for connected flats. First, note that any connected flat is cyclic (possibly excepting flats consisting of a single element, depending on how you handle that boundary case, but Allen must not have meant to include that or the Schubert example wouldn't work.) Then notice that any cyclic flat is a disjoint rank-additive union of connected flats. – David Speyer Jun 21 '12 at 11:31
So then a smooth base is the same as a fundamental base in the sense above. To see why matroids with a smooth base have to be transversal, place the elements of the fundamental base at the vertices of a simplex and the rest of the elements on the faces in such a way that cyclic flats of rank $r$ are the collections of elements in an $r-1$ dimensional subsimplex. It is a theorem of Brylwaski that matroids admitting such a representation are transversal. (This is all in the paper/slides linked above.) – Gjergji Zaimi Jun 21 '12 at 11:44
"I hope this question gets a good answer" -- it did! – Allen Knutson Jun 25 '12 at 17:48
BTW I asked Bonin, who had not made the geometric connection. – Allen Knutson Jun 25 '12 at 17:49