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It is well known that the set of all polygons with consecutive side lengths $l_1, \dots, l_n$ in $\mathbb{R}^3$, considered up to rigid motions, is a compact complex manifold. Of course, I am assuming, for the sake of simplicity that there are no "straight line" polygons i.e., the length vector $L:= (l_1,\dots, l_n)$ is generic. I will denote this space by $M_L$.

Consider configurations of $n$ not necessarily distinct points $p_i\in \mathbb{P}^1$. Each point $p_i$ is assigned the weight $l_i$. A configuration is called stable if the sum of weights of equal points is less than the total weight $\sum l_j$. The moduli space of stable points on $\mathbb{P}^1$ is defined as the stable configurations modulo projective automorphisms.

It was shown by Kapovich and Millson (and also by Klyachko) that $M_L$ is isomorphic to the moduli space of stable points on $\mathbb{P}^1$.

Before I ask my questions let me recall the definition of one more moduli space. The moduli space of $n$ distinct labeled points on $\mathbb{P}^1$ is obtained from $(\mathbb{P}^1)^n$ by removing the diagonals $z_i = z_j$ for $i\neq j$ and factoring out by the $PSL(2, \mathbb{C})$ action. This space, denoted $M_{0,n}$, has a smooth compactification denoted $M_n$. The way I understand the construction of $M_n$ is as follows: first realize that $M_{0,n}$ is the complement of the braid arrangement in $\mathbb{P}^{n-1}$. Then one reaches $M_n$ by successively blowing along certain intersections.

Now my questions:

  1. For any (generic) choice of length vector one can observe that $M_{0, n} \subset M_L$ as the set of all those polygons for which no to directions are same (i.e., stable configurations where are all points are distinct). In view of this, can one say that $M_L$ is a smooth compactification of $M_{0,n}$?
  2. Is there some nice map from $M_n$ to $M_L$ (or in the other direction) which is an isomorphism on $M_{0,n}$? I have read the following description of $M_n$ in several papers: '' when two points in $M_{0,n}$ come close too each other instead of colliding a new $\mathbb{P}^1$ pops out and these points move on it". I don't understand what this means mathematically but this pictures suggests a map. Given a length vector $L$ the essential information is the collection of all so-called short subsets. I guess this data should help us identify which "popped out" $\mathbb{P}^1$s to contract and obtain a stable configuration (or the other way round?). I know what I have written here is pretty vague but I have a strong feeling that somebody must have thought about this earlier.
  3. The complement $M_L\setminus M_{0,n}$ is certainly not a normal crossing divisor. So, if one were to blow up along the subspaces of colliding points what would one get? Does this even make sense?

Motivation: For the past few months I have been working on some problems related to the moduli space of planar polygons. One day I realized that the planar polygons modulo $O(2)$ contain $M_{0, n}(\mathbb{R})$ as an open subset (which is just a disjoint union of $(n-1)!/2$ open discs). This observation lead me to above questions. I should also confess that I have no expertise in algebraic geometry. I am completely new to the world of moduli space of stable points and labeled points.

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  1. Yes.

  2. I assume that your $M_n$ is what is more usually denoted $\overline{M_{0,n}}$. Then the answer is yes, there is a natural map $\overline{M_{0,n}} \twoheadrightarrow M_L$, for each $L$. Specifically, let $\gamma$ be a tree of $\mathbb P^1$s glued along nodes and with $n$ labeled points (not at the nodes). Then at each node, measure the total weight on either side. By your assumption, on one side it'll be smaller than the other. Contract all those $\mathbb P^1$s on that smaller side to a point, collapsing the labeled points together, and repeat until there is only one $\mathbb P^1$.

The more general statement here is that the polygon spaces are GIT quotients, while $\overline{M_{0,n}}$ is a Chow quotient (in each case, of the $2$-Grassmannian by a torus). The relevant paper is Kapranov's "Chow quotients of Grassmannians", which I'm not finding online (although Keel-Tevelev's followup is available).

  1. You're right that blowing up a divisor does nothing if it's normal crossings, but more generally the condition is that it be "Cartier", i.e. locally given by a single equation. Here the local picture is of $n$ points in $\mathbb C$, and the single equation is the Vandermonde $\prod_{i<j} (z_i-z_j)$. So the answer is that blowing up makes sense but does nothing.
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