from affine matroid to measures - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T23:01:34Zhttp://mathoverflow.net/feeds/question/95970http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/95970/from-affine-matroid-to-measuresfrom affine matroid to measuresDima Pasechnik2012-05-04T09:26:30Z2012-05-21T20:06:50Z
<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#95976Answer by Gjergji Zaimi for from affine matroid to measuresGjergji Zaimi2012-05-04T11:39:04Z2012-05-21T20:06:50Z<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>