Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

From the my limited experience it seems that often a parameter space (miraculously) has some of the properties of the elements it parameterizes. For example:

  • The parameter space of all plane conics is an algebraic variety.
  • The moduli space of genus $g$ curves $\mathcal{M}_{g}$ is itself an algebraic variety.
  • The Mandelbrot set is connected and can be thought of as the parameter space: $$\{c\in \mathbb{C}\; : \; J_{f_c}\; \text{is connected}\}$$ where $J_{f_c}$ is the Julia set of the polynomial $f_c=x^2+c$.
  • The Teichmuller space parameterizes complex structures on a surface $X$ and itself is a complex manifold.

I am wondering if there is a reason for this occurring. State a bit more formally my question is:

Question: Why is it the case that if parameter space $M$ parameterizes spaces of with property $P$ then $M$ (often) also has property $P$?

Note I say often as there are examples where this does not happen. For example, there exists quadratic polynomials whose Julia set is not locally connect and there exists quadratic polynomials whose Julia set is locally connected, but of course the Mandelbrot set is either locally connected or not.

share|improve this question
3  
Three of these examples are structures, not properties. And $M_g$ is a stack, not just a variety. –  Qiaochu Yuan Feb 25 at 20:40
    
(And one has to say more about the relevance of these structures to the parameterization problem, which is what the point of view of moduli functors is all about. Of course for any set parameterizing any sort of thing I can decide to arbitrarily give it any structure I want, but the compatibility of this structure with the moduli problem makes it an interesting question to find such a structure, which is now uniquely determined, at least morally, by the Yoneda lemma.) –  Qiaochu Yuan Feb 26 at 17:56
add comment

2 Answers

I disagree with the premise: I think the correct statement is "moduli spaces in algebraic/complex geometry are unusually nice, in that they are also spaces with algebraic/complex structure."

By way of contrast:

  • The "moduli space" of topological manifolds of dimension $d$ is a discrete set and has no interesting topological structure.

  • The moduli space of smooth structures on a topological manifold is usually discrete (eg the 24 smooth structures on $S^7$). I have the impression that the reverse is true for $\mathbb{R}^4$ -- that the space of exotic $\mathbb{R}^4$'s is infinite dimensional -- but this is very far from my expertise and I couldn't find a paper stating it.

  • The moduli space of Riemannian metrics on a topological manifold is infinite dimensional.

I think it is quite amazing that, in complex and algebraic geometry, moduli spaces are so often finite dimensional but not discrete.

share|improve this answer
add comment

Moduli spaces should represent moduli functors. Moduli functors should describe what it means to have an "$X$-parameterized family" of objects in terms of a suitable kind of map $E \to X$, the idea being that the fibers of this map are the family. To say that such a functor is represented by a moduli space $M$ is to say that giving such a map is equivalent to giving a map $X \to M$; in particular, it's natural to ask for $M$ to be an object in the same category that $X$ and $E$ (and the fibers of the map $E \to X$) are.

And even if a moduli functor is not, strictly speaking, representable by an object in the category you care about, morally speaking it describes a "generalized object" in the category anyway. Formally, the category of presheaves on a category $C$ is the free cocompletion of $C$, so any presheaf on $X$ (including a moduli functor) can be thought of as a formal colimit of objects in $C$. (There's an additional complications in the case of $M_g$, which is a stack: this reflects the fact that the moduli functor in this case should be thought of as landing in groupoids, not sets.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.