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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.
18
votes
Betti numbers of moduli spaces of smooth Riemann surfaces
Calculations of integral homology of $\mathcal{M}_{g, n}$ occur in Abhau, Bodigheimer, Ehrenfried (p. 3) or Godin (p. 4). Of course, in the stable range (degrees $3* \leq 2g$) the rational cohomology …
9
votes
Accepted
Generalizing the Madsen-Weiss Theorem via the scanning map $\mathscr{C}(M,\mathbb{R}^{\infty...
The sequence of papers
S. Galatius, O. Randal-Williams, Stable moduli spaces of high-dimensional manifolds. Acta Math. 212 (2014), no. 2, 257–377. (DOI: 10.1007/s11511-014-0112-7, projecteuclid)
…
5
votes
Is there an analog of compactified moduli spaces(/stacks) for smooth manifolds?
I have a piece of juvenilia on this topic, considering the case of zero-dimensional submanifolds. See Section 9 of
O. Randal-Williams, Embedded cobordism categories and spaces of submanifolds,
IMRN …
6
votes
Is there Harer stability for moduli of curves with level structure?
This is not an answer to your question, but is directly related to your remark so I thought I should mention it.
I have recently proved, though I am afraid that it has not appeared yet, that moduli s …
9
votes
1
answer
1k
views
Picard group of $\mathfrak{M}_g$
Let $\mathfrak{M}_g$ denote the moduli stack of Riemann surfaces of genus $g$, it is a smooth complex analytic stack, and is the analytic stack underlying $\mathsf{M}_g$, the moduli stack of complex a …
2
votes
Accepted
Homology dimension of the mapping class group of a surface with boundary
There is a fibration sequence
$$\mathbb{S}(\Sigma_g) \to \mathcal{M}_{g}^1 \to \mathcal{M}_g$$
where $\mathcal{M}_g$ is the moduli space of Riemann surfaces, $\mathcal{M}_{g}^1$ is the moduli space of …
26
votes
Accepted
Mumford conjecture: Heuristic reasons? Generalizations? ... Algebraic geometry approaches?
All current proofs of Mumford's conjecture in fact prove a far stronger result, the "Strong Mumford conjecture", first formulated by Ib Madsen. This says the following (where by "moduli space" in the …