from affine matroid to measures - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:01:34Z http://mathoverflow.net/feeds/question/95970 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95970/from-affine-matroid-to-measures from affine matroid to measures Dima Pasechnik 2012-05-04T09:26:30Z 2012-05-21T20:06:50Z <p>Let $S$ be an arbitrary finite spanning subset of $\mathbb{R}^d$ of cardinality $N$. Let $W(S)$ be the formal $\mathbb{R}$-vector space generated by all $d$-dimensional simplices (i.e. bases of the affine matroid $\mathcal{M}(S)$ of $S$) with vertices in $S$. In particular</p> <p>$\dim W(S) =$ number of simplices= number of bases of $\mathcal{M}(S)$.</p> <p>For a given simplex $\Delta$ with vertices from $S$ define the volume form $w_\Delta=\chi_\Delta d Vol$, where $\chi_\Delta$ is the characteristic function of $\Delta$. Let $V(S)$ be the $\mathbb{R}$-linear span of $w_\Delta$ where $\Delta$ runs over all simplices with vertices in $S$. We have a natural projection $\Pi: W(S)\to V(S).$ Let $K(S)$ be the kernel of $\Pi.$ Our main goal is to describe this kernel. </p> <p>First we describe some obvious elements from $W(S)$ lying in $K(S)$. Take an arbitrary $(d+2)$-tuple $T$ of points from $S$ spanning $\mathbb{R}^d$. Then there exists a standard element of $K(S)$ associated to $T$. Namely, there exist exactly two triangulations of the convex hull of $T$ by the simplices with vertices in $T$. Considering the difference of these triangulations as an element of $W(S)$ we get the required element of $K(S)$. For example, there are two "different" types of spanning 4-tuples of points in $\mathbb{R}^2.$ Case 1 with the convex hull which is a 4-gon and Case 2 with the convex hull which is a triangle. In Case 1 we have a relation that the sum of 2 triangles = the sum of two other triangles. Thus we have an element of $K(S)$ of the form $\Delta_1+\Delta_2-\Delta_3-\Delta_4$ In Case 2 we have that the biggest triangle $\Delta_1$ equals either the sum of 3 smaller triangles (and so we have an element of $K(S)$ of the form $\Delta_1-\Delta_2-\Delta_3-\Delta_4$) or $\Delta_1$, with one side having 3 points from $S$, equals the sum of 2 smaller triangles (and so we have an element of $K(S)$ of the form $\Delta_1-\Delta_2-\Delta_3$)</p> <p><b>Conjecture</b>. $K(S)$ is spanned by the above standard kernel elements coming from spanning $(d+2)$-tuples of points from $S$.</p> <p>We are able to show that Conjecture holds for $S$ for which any $(d+2)$-subset of points is spanning.</p> <p>This statement sounds to us as an exercise(?) from matroid theory. In particular, we are sure that $\dim K(S)$ is an invariant of $\mathcal{M}(S)$. </p> <p>Can you recognize Conjecture as a known statement from matroid theory (or, even, a version of de Rham's Theorem)?</p> http://mathoverflow.net/questions/95970/from-affine-matroid-to-measures/95976#95976 Answer by Gjergji Zaimi for from affine matroid to measures Gjergji Zaimi 2012-05-04T11:39:04Z 2012-05-21T20:06:50Z <p>This conjecture is proven in theorem 7.4 of <a href="http://www.sciencedirect.com/science/article/pii/S0012365X97001076" rel="nofollow">"Incidence matrices, geometrical bases, combinatorial prebases and matroids"</a> by T.V. Alekseyevskaya and I.M. Gelfand ($n=2$) and theorem 4.5 in <a href="http://arxiv.org/abs/math/9707218" rel="nofollow">"Bases in Systems of Simplices and Chambers"</a> by Alekseyevskaya for higher $n$.</p> <p>A previous version of this answer described the similar situation with scissor congruence groups, but I'm no longer certain that one can get a proof using such theorems anymore. It is worth mentioning here that there is a very similar <a href="http://mathoverflow.net/questions/35061/a-pachner-complex-for-triangulated-manifolds" rel="nofollow">theorem of Pachner</a> which says that two PL-homeomorphic triangulated PL-manifolds are related by a sequence of "bistellar flips", which are essentially the generating relations in your conjecture. I suspect there should be a proof of your conjecture using Pachner's theorem, but I haven't given it much thought.</p>