## from affine matroid to measures

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

$\dim W(S) =$ number of simplices= number of bases of $\mathcal{M}(S)$.

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.

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 $\Delta_1-\Delta_2-\Delta_3$)

Conjecture. $K(S)$ is spanned by the above standard kernel elements coming from spanning $(d+2)$-tuples of points from $S$.

We are able to show that Conjecture holds for $S$ for which any $(d+2)$-subset of points is spanning.

This statement sounds to us as an exercise(?) from matroid theory. In particular, we are sure that $\dim K(S)$ is an invariant of $\mathcal{M}(S)$.

Can you recognize Conjecture as a known statement from matroid theory (or, even, a version of de Rham's Theorem)?

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Are you aware of the corresponding papers "Fiber polytopes" by Billera and Sturmfels, and "The polytope of all triangulations of a point configuration" by de Loera, Hosten, Santos and Sturmfels? Perhaps also "Exterior algebras and projections of polytopes" by Filliman might be useful. – Camilo Sarmiento May 4 2012 at 16:09
We certainly are aware of the construction fiber polytopes... – Dima Pasechnik May 4 2012 at 17:02

This conjecture is proven in theorem 7.4 of "Incidence matrices, geometrical bases, combinatorial prebases and matroids" by T.V. Alekseyevskaya and I.M. Gelfand ($n=2$) and theorem 4.5 in "Bases in Systems of Simplices and Chambers" by Alekseyevskaya for higher $n$.

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 theorem of Pachner 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.

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Thanks, this is a very useful connection that we were not aware about! – Dima Pasechnik May 5 2012 at 5:01
Near the bottom, you write: "Now, because all the polytopes involved have vertices in $S$, the only triangulations involved are the ones with vertices in $S$." This seems to be unclear for $d\geq 3$, as then there are polytopes which cannot be triangulated without extra vertices. e.g. en.wikipedia.org/wiki/Sch%C3%B6nhardt_polyhedron Could you clarify, perhaps? – Dima Pasechnik May 5 2012 at 6:14
I finally found the result in the literature, so I updated my answer accordingly. I guess my previous idea was a bit of wishful thinking. I haven't been able to turn it into a rigorous proof. – Gjergji Zaimi May 8 2012 at 6:02
Thanks for digging up these references, they are very useful! Regarding a possible connection with Pachner's Theorem, it does not seem likely. Indeed, in this setting, of point configurations, two triangulations of the same set need not be connected in this way, see e.g. ams.org/journals/jams/2000-13-03/… – Dima Pasechnik May 21 2012 at 17:08
That was a typo. The theorems are 7.4 in the first paper, and it deals with dimension 2; and theorem 4.5 in the second paper which deals with the higher values. – Gjergji Zaimi May 21 2012 at 20:18