Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Given a concrete category C, with objects denoted Obj(C), and an equivalence relation ~ on Obj(C) given by morphisms in C. The moduli set for Obj(C) is the set of equivalence classes with respect to ~; denoted Iso(C). When Iso(C) is an object in the category Top, then the moduli set is called a moduli space.
10
votes
Tangent space of the stack $\overline{\mathcal{M}}_{g,n}(X,\beta)$.
The classical reference is Illusie's PhD thesis "complexe cotangent et deformations".There the result without the marked points is proven in a very, very general context using the cotangent complex; t …
6
votes
Proving that a map is a morphism
Emerton has already answered, but let me summarize all the steps:
1) to a perfect torsion complex in the derived category associate a Cartier divisor;
2) to every sequence of morphisms $C\to X\to S$ …
5
votes
There are many varieties with ample canonical bundle
If $X$ is a surface, then the Hilbert polynomial of the canonical divisor determines $d:=1-\chi(T_X)$ which is a lower bound for the dimension of any component of the moduli at the point $[X]$. Hence …