*After a bit of a hiatus due to difficulties with Wordpress, Mathematical Gemstones is back! Feel free to browse the new layout.*

Today’s post introduces moduli spaces of curves, and one of the many ways combinatorial gemstones arise in this vast area of algebraic geometry.

A brief motivating questionfor studying families of curves: Given 4 general points in the plane, how can you write down all conics that pass through them? (Try it for the four points $(\pm 1, \pm 1)$!)

We will come back to the above question, and how it ties in, in a future post. For now, some background: in previous posts, we’ve discussed *moduli spaces* of $k$-planes in $n$-dimensional space, called Grassmannians (see, for instance, the post on Schubert Calculus). In general, a moduli space is a geometric space whose points correspond to a family of geometric objects.

The moduli spaces $M_g$ (technically stacks, for higher genus) parameterize complex curves with genus $g$, and are a central object in algebraic geometry. In genus $0$, the combinatorial structure is particularly nice. Write $M_{0,n}$ for the moduli space of choices of $n\ge 3$ distinct labeled *marked points* $p_1,\ldots,p_n$ on the complex projective line $\mathbb{P}^1$, up to isomorphism.

Topologically, $\mathbb{P}^1$ over $\mathbb{C}$ is the surface of a sphere, even though we think of it as a complex “curve”, of dimension $1$ over $\mathbb{C}$. For instance, the image below shows a point of $M_{0,5}$.

Since a projective transformation of $\mathbb{P}^1$ is uniquely determined by where it sends three points, we can assume that the first three points have coordinates $0,1,\infty$ (or, in projective notation, $(0:1),(1:1),(1:0)$), and the remaining points can be any distinct points.

One natural compactification of this moduli space, $\overline{M}_{0,n}$, is called the space of **stable curves**, and essentially allows the distinct marked points to collide in the following way. When two points collide (like the two very close on the above picture), they form a new $\mathbb{P}^1$ that attaches to the original by a node at the point of collision, and the two points are now on the new $\mathbb{P}^1$. Since the new $\mathbb{P}^1$ only has three special points - the node and the two collided points - it does not matter, up to isomorphism, where these two points are placed on the new sphere:

In general, if multiple points collide, their coordinates on the new $\mathbb{P}^1$ will reflect the relative speeds at which the points approached the collision point along the chosen path in the moduli space.

Simplifying the pictures somewhat so that we draw each $\mathbb{P}^1$ as a circle rather than a sphere (pretending visually that we are working over $\mathbb{R}$), in general a point of $\overline{M}_{0,n}$ can look like the diagram at left below, with each circle having at least three total *special points* (nodes or marked points).

The genus $g=0$ condition means that the $\mathbb{P}^1$’s are connected in a branching tree structure and do not loop around to form a necklace. This means that its *dual graph* is a tree - we draw one vertex for each $\mathbb{P}^1$ component and one for each marked point, and connect each component to its marked points and also to adjacent components. This forms the **dual tree** of a stable curve, shown at right above. These trees, by the stability condition, have the property that no vertex has degree $2$ (they are “at least trivalent”). The leaves are labeled and the internal vertices are unlabeled.

We now see a combinatorial object - trees - naturally arising in a moduli space of curves. In particular:

Trees play a similar role in $\overline{M}_{0,n}$ as partitions play in the Grassmannian, or permutations in the flag variety.

In particular, $\overline{M}_{0,n}$ has a *boundary stratification* given by the at-least-trivalent trees $T$ on $n$ labeled leaves, where $X_T$ is defined to be the closure of the set of all stable curves with dual tree $T$. These strata have a natural combinatorial condition for compatibility: $X_T$ is contained in $X_S$ if and only if the tree $T$ can be formed by contracting some of its edges to form $S$.

The smallest such strata are the **boundary points** $X_T$ where $T$ is a *trivalent* tree in which each internal vertex has exactly degree $3$. In this case, there is only one curve up to isomorphism with this dual tree, so the entire stratum is literally a single point in the moduli space.

We finally arrive at today’s gemstone: the enumeration of the boundary points in $\overline{M}_{0,n}$!

How many boundary points does $\overline{M}_{0,n}$ have? This is equivalent to enumerating the trivalent trees on $n$ labeled leaves. For $n=3$, there is only one such tree:

For $n=4$, there are three:

and notice that each of these three can be formed by attaching a leaf edge with its leaf vertex labeled $4$ to the midpoint of one of the three edges of the $n=3$ tree. Notice also that by inserting a leaf edge attached to the midpoint of an existing edge, the total number of edges in the tree increases by $2$. Moreover, any trivalent tree on leaves labeled $1,2,\ldots, n+1$ comes from a unique tree on $n$ leaves in this way, by deleting the leaf labeled $n+1$ and the edge it is a vertex of.

By induction, we conclude that there are $2n-3$ edges in any trivalent tree with $n$ labeled leaves, and if $T_n$ is the number of trivalent trees on $n$ labeled leaves, we have \[T_{n+1}=(2n-3) T_{n}\]. Starting from the base case $T_3=1$, we conclude that \[T_{n}=1\cdot 3 \cdot 5 \cdot 7 \cdot \cdots \cdot (2n-5)=(2n-5)!!\]
Here the notation $(2n-5)!!$ denotes the *double factorial*, the product of the odd numbers up to $2n-5$. So for instance, $\overline{M}_{0,5}$ has $1\cdot 3 \cdot 5=15$ boundary points, and $\overline{M}_{0,6}$ has $7!!=105$. A beautiful gemstone of combinatorics within geometry!