Smooth bases of matroids - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T01:33:55Z http://mathoverflow.net/feeds/question/100015 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100015/smooth-bases-of-matroids Smooth bases of matroids Allen Knutson 2012-06-19T15:29:32Z 2012-06-21T09:16:15Z <p>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?</p> <p>If you like matroids but not algebraic geometry, skip to the definition below.</p> <p>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.</p> <p>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 $C$.</p> <p>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.)</p> <p>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 $S$.)</p> <blockquote> <p>If $M \subseteq {n\choose k}$ is a matroid, call $\lambda \in M$ a <em>smooth base</em> if for any connected flat $F$, $rank(F) = |\lambda \cap F|$.</p> </blockquote> <p>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.</p> <blockquote> <p>Is this concept known to matroid theorists? Is this characterization of smooth points known to anybody?</p> </blockquote> <p>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 $\Pi_M$.</p> http://mathoverflow.net/questions/100015/smooth-bases-of-matroids/100219#100219 Answer by Gjergji Zaimi for Smooth bases of matroids Gjergji Zaimi 2012-06-21T09:16:15Z 2012-06-21T09:16:15Z <p>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.</p> <p>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. </p> <p>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.</p> <p>Some further characterizations in terms of some rank inequalities, or affine representations on simplices are proved in <a href="http://arxiv.org/abs/1009.3435" rel="nofollow">"Characterizations of transversal and fundamental transversal matroids"</a> by J.E. Bonin, J.P.S. Kung and A. de Mier. See also <a href="http://home.gwu.edu/~jbonin/Fundamental.pdf" rel="nofollow">these slides</a> 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.</p>