I am reading Katz' book Enumerative Geometry and String Theory. I have a few questions regarding the moduli space of degree $d$ genus $0$ stable maps into $\mathbb P^n$, denoted $\overline{M}(\mathbb{P}^n, d)$. First let me paraphrase some definitions in the book (given between p.32-37) to the best of my understanding (they're not very precise but neither are they in the book!):

A tree of $\mathbb P^1$s is a union $C=\bigcup_{i=1}^r C_i$ where each $\phi_i:\mathbb P^1\xrightarrow{\cong}C_i$ is a curve, glued along a finite collection of pairs of points (called nodes) $(p_j, q_j)\in C_{k(j)}\times C_{l(j)}$ for distinct indices $1\leq k(j), l(j) \leq r$ and avoiding cycles - finite sequences of distinct nodes $p_{j_1},p_{j_2}, \dots, p_{j_k} = p_{j_1}$ where a single component of $C$ contains both $p_{j_l}$ and $p_{j_{l+1}}$. A morphism $f:C\rightarrow\mathbb P^n$ is a map such that each $f_i :=f\circ\phi_i:\mathbb P^1 \rightarrow\mathbb P^n$ is a morphism of varieties. The degree of $f$ is the sum of the degrees of the $f_i$.

A genus $0$ stable map to $\mathbb P^n$ is a tree morphism $f:C\rightarrow\mathbb P^n$ such that if $f$ is constant on a component $C_i$ of $C$, then $C_i$ contains at least $3$ nodes of the tree. The degree of the stable map is the same thing as its degree as a tree morphism. An isomorphism of stable maps $g:(f:C\rightarrow\mathbb P^n)\rightarrow(f':C'\rightarrow\mathbb P^n)$ is a morphism $g:C\rightarrow C'$ such that

  1. $f'\circ g = f$;
  2. For each $C_i$, $g(C_i) = C'_j$ and for any $j$, there is a unique $i$ with $g(C_i)=C'_j$;
  3. $(\phi_j)^{-1} \circ g\circ \phi_i:\mathbb P^1\rightarrow\mathbb P^1$ is a morphism (hence automorphism?) whenever $g(C_i)\subseteq C'_j$.

The moduli space of genus $0$ stable maps of degree $d$ to $\mathbb P^n$ is then the set $$\overline{M}(\mathbb P^n, d)=\left\{\mbox{isomorphism classes of degree } d \mbox{ genus 0 stable maps into }\mathbb P^n \right\}$$

Here are my first two questions:

  1. Why do we care about trees and cycles? To me, it seems as if they are a convenient (if messy) simplification of something more natural (and apparently related to the concept of genus) but perhaps too complicated to describe in this introductory book.
  2. In the definition of stable map, why three nodes?

In the book, it is claimed that $\overline{M}(\mathbb P^n, d)$ is actually a stack. I don't know what a stack actually is but I know it is a category in some sense. So I'd like to try to make $\overline{M}(\mathbb P^n, d)$ into a category and see how close this is to a "real" stack.

First, let's generalise the notion of an isomorphism between two stable maps to that of a morphism which seems to be well-behaved - a morphism $g:(f:C\rightarrow \mathbb P^n)\rightarrow(f':C'\rightarrow\mathbb P^n)$ of stable maps is a morphism $g:C\rightarrow C'$ such that

  1. $f'\circ g = f$;
  2. For each $C_i$, $g(C_i) = C'_j$ for some $j$;
  3. $(\phi_j)^{-1} \circ g\circ \phi_i:\mathbb P^1\rightarrow\mathbb P^1$ is a morphism (in the sense of algebraic varieties) whenever $g(C_i)\subseteq C'_j$.

Two morphisms $g_1 :f_1\rightarrow f_1'$ and $g_2:f_2\rightarrow f_2'$ between stable maps are isomorphic if there exist isomorphisms $\theta:f_1\rightarrow f_2$ and $\theta':f_1'\rightarrow f_2'$ such that $\theta'\circ g_1 = g_2 \circ\theta$. Isomorphism between stable maps and between their morphisms are equivalence relations. Let's define a category called $\overline{M}(\mathbb P^n, d)$ whose objects are isomorphism classes $[f]=\left[f:C\rightarrow\mathbb P^n\right]$ of degree $d$ genus $0$ stable maps into $\mathbb P^n$ and whose morphisms are isomorphism classes $[g]:[f]\rightarrow [f']$ of morphisms between stable maps, with composition defined by $[h]\circ [g] = [h\circ g]$. This gives us a much richer structure than just the set of all isomorphism classes defined in the book as there are now "asymmetric relations" (described by the morphisms in this category) between the points of the set $\overline{M}(\mathbb P^n, d)$.

My final question is:

  1. In what sense is the category that I have come up with related to the actual stack mentioned in the book?
  • $\begingroup$ My third question is a bit more vague which is why I have accepted Asal Beag Dubh's answer. That said, if anyone can answer my third question I am still interested. Thanks! $\endgroup$ – Alex Saad Apr 30 '15 at 14:27

Here are some brief answers to Questions 1 and 2; you might find them helpful. (I have nothing to say about Question 3.)

  1. Yes, the point is exactly about genus. A tree as defined in your question always has (arithmetic) genus 0, so it can appear as a member of a family of curves whose generic member is a smooth rational curve. That means that compactifying the space of maps from smooth rational curves will naturally involve considering maps from trees. On the other hand, you can check that any curve that contains a cycle must have genus at least 1, so it's not relevant to this moduli problem.

  2. Here the point is about automorphisms. A basic phenomenon in moduli theory is that automorphisms hinder the construction of moduli spaces. (See the reference at the end for details). Now the automorphism group of $\mathbf P^1$ is $PGL(2)$, a 3-dimensional algebraic group; you (and should) can check that an automorphism of $\mathbf P^1$ is the identity iff it fixes 3 points. Now suppose we had a map $f:C \rightarrow \mathbf P^n$ where $C$ is a tree with some contracted component with fewer than 3 nodes. Then $C$ has a positive-dimensional family of automorphisms fixing that component, and precomposing $f$ with any of them gives an automorphism of the point corresponding to $f$ in the moduli space. On the other hand, if every contracted component of $C$ has at least 3 nodes, then automorphisms of this kind are "killed", because automorphisms have to permute the nodes.

The $\epsilon$ I know about this subject, along with much more, can be found in the book Moduli of Curves by Harris and Morrison. You might find it instructive to study that book to get a good handle on the general picture of moduli theory.

  • $\begingroup$ Really clear answers, thank you! $\endgroup$ – Alex Saad Apr 27 '15 at 16:31

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