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What is a good example of a fact about the moduli space of some object telling us something useful about a specific one of the objects?

I am currently learning about moduli spaces (in the context of the moduli space of elliptic curves). While moduli spaces do seem to be fascinating objects in themselves, I am after examples in which facts about a moduli space tell us something interesting about the specific objects that they parametrise. For example, does the study of $\mathbb RP^n$ tell us anything we don't already know about some given line through the origin (say, the $x_1$-axis) in $\mathbb R^{n+1}$?

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Mazur's deep study of modular curves over $\mathbf{Q}$, especially understanding the arithmetic of their Jacobians (something impossible to express without the moduli space itself), was the key to his determination of the possibilities for torsion groups in Mordell-Weil groups of elliptic curves over $\mathbf{Q}$. Refinement of these ideas was used by Merel for number fields. –  BCnrd May 28 '10 at 1:11
Similar in nature to Brian's example, there is the arithmetic moduli of principally polarized abelian varieties (and compactifications). Analyzing the height function on this space was essential to the deep results concerning Galois actions on the Tate module of an abelian variety, such as its semi-simplicity or the Tate conjecture. I was going to mention the Shafarevich conjecture as well (inextricably tied to the other statements), but that's not about a single abelian variety. –  Minhyong Kim May 28 '10 at 9:52
It's also amusing to note that the proof of the Mordell conjecture (using $M_g$) or the modularity conjecture for elliptic curves (using modular curves) are not examples of the sort we want, strictly speaking. –  Minhyong Kim May 28 '10 at 10:02
It's fun, however, to speculate that the above two are examples. –  Minhyong Kim May 28 '10 at 10:19

6 Answers 6

up vote 17 down vote accepted

The easiest example I can think of is the natural incidence correspondence between $\mathbb{P}^3$ and the parameter space of cubic surfaces. This can be used to show that every cubic surface contains a line; from this it follows easily that every smooth cubic surface contains exactly $27$ lines.

Another example is the moduli space of stable maps constructed by Kontsevich; this parametrizes certain maps from curves to (to stick with a simple case) $\mathbb{P}^2$. It can be used to answer the following question: given $3d-1$ points in $\mathbb{P}^2$ in general position, compute the number $N_d$ of rational curves of degree $d$ passing through these points. It turns out that the values $N_d$ satisfy a certain recursive relation which allows you to compute all these numbers starting from the obvious $N_1 = 1$ (through $2$ points passes exactly one line). You can find the formula here; it yields for instance $N_2 = 1$ and $N_3 = 12$.

Yet another example, again more elementary is the following. The Grassmannian $G = \mathop{Gr}(1, \mathbb{P}^3)$ parametrizes lines in $\mathbb{P}^3$. The computation of the cohomology of $G$ allows you to compute the number of lines which are incident to $4$ fixed lines in general position (it turns out this number is $2$).

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This is a fantastic example: really using geometry of moduli space, not just language about it, to do something about concrete objects. –  BCnrd May 28 '10 at 1:07

Here is one example.

Let $M^3$ be a hyperbolic manifold. Consider the moduli space curvature $-1$ metrics on $M^3$ modulo $Diff_0(M^3)$. This is a point. Conclusion: every diffeomorphism of $M^3\to M^3$ is homotopic to an isometry.

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I don't know much about hyperbolic geometry, so maybe this is a naive question: does the proof that it is a point involve something other than proving directly the conclusion that you mention? –  BCnrd May 28 '10 at 1:08
BCnrd, this question gives a summary of the proof: mathoverflow.net/questions/21986/… . –  HJRW May 28 '10 at 2:27
@Henry: OK, so it seems one does give a direct proof. But then I don't understand why rephrasing this as saying "moduli space is a point" is anything deeper than linguistics. That is, I don't see how this is an example of a moduli space being used in an essential way, since what we are learning is actually proved directly, without recourse to any concept of moduli space. I am not saying that this reformulation is a bad thing, but just that I don't see how introducing the moduli space viewpoint provides insight in this case. –  BCnrd May 28 '10 at 3:58
BCnrd, this is just lingustics indeed. I was just afraid that in algebraic geometry there will not be a non-trivial understandable statement, that motivates introduction of moduli spaces from the point of view of the preson who asked the question. By the way, can you formulate someting in algebraic geometry the seems to be interesting (while it is only linguistics), and uses the same trick - that the moduli space is a point? –  Dmitri May 28 '10 at 6:30
@Dmitri: Some not-so-linguistic examples come to mind when moduli space is etale. Deformation theory (vanishing of tangent space) shows moduli spaces of polarized CM ab. var. in char. 0 and Hom-schemes between abelian varieties are etale, so CM ab. var. are defined over number field and "geometric" Homs between ab. var. are defined over sep'ble ext'n of ground field. (Pfs can be more direct.) Likewise, finiteness of set of finite maps between two curves of genus $> 1$ over alg. closed field of any char.: can prove suitable Hilbert scheme has tang. spaces = 0, hence etale, hence finite. –  BCnrd May 28 '10 at 17:31

You have chosen an example where the moduli space (a projective space) is a homogeneous space. So, geometrically, all the objects it parametrises are "the same", and the nature of the moduli space merely confirms that.

Perhaps it should be said first, therefore, that moduli spaces are not always homogeneous spaces. Not all points on the moduli space look the same, and therefore questions arise. This is seen classically for elliptic curves, where the typical automorphism group (preserving the identity) of an elliptic curve is of order 2, but in a few cases it may be of order 4 or order 6. Does this show up in the moduli space? Yes, when you construct it in the classical way from a fundamental domain in the upper half-plane. Moral: if there are "special" points in the moduli space, there is a geometrical reason they are special.

There are actually three levels to look at: the structure of the moduli space qua space (manifold-like, let's say, for complex geometry); for sophisticates using scheme theory the so-called infinitesimal structure in the sheaf given on the space; and the "moduli" themselves, such as the classical j-invariant, namely the parameters used to describe the space. It depends from what direction you are coming, but certainly for arithmetic special values of the moduli read back in an interesting way.

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The real test is moduli space telling us something couldn't see without it. It is easy to express transitivity of ${\rm{PGL}}_n$-action on linear subspaces of a fixed dimension in projective space without knowing anything about moduli spaces. Likewise, the quirky Aut-groups of CM elliptic curves are well-understood without mention of moduli. Nifty examples: (1) expectation of density of some locus in a moduli space motivating use of deformation theory in proofs (even if logically doesn't require a global moduli space), (2) cohomology of moduli space used for arithmetic of modular forms... –  BCnrd May 28 '10 at 1:05

Here is an example. The generic Riemann surface of genus $g>2$ has no automorphisms.
The proof uses the following ingredients:

  • Dimension of the moduli space of curves is $3g-3$;

  • If a automorphism of a non-hyperelliptic Riemann surface $C$ fixes the Weierstrass points, then it is the identity;

  • Riemann-Hurwitz formula.

Now, keep in mind that every Riemann surface can be embedded into $\mathbb P^3$, however the theorem says that in a generic case, not even a automorphism of $\mathbb P^3$ (which is linear change of variables) induces a automorphism of the curve! (genus >2).

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In fact, even the precise meaning of the concept of "generic [compact connected] Riemann surface" rests crucially on the existence of the moduli space, and in particular its irreducibility. –  BCnrd May 29 '10 at 0:29

Another example also in the flavor enumerative geometry: by considering the (Deligne- Mumford compactification of) moduli space of elliptic curves, we see that the point at infinity represents the genus one curve with one node, while each other point represents curve with fix j-invariant. The moduli space now look somewhat like $\mathbb{P}^1$ (except that it is not, for some stacky reasons, but for current purpose, we can pretend that it is ) and we can view the nodal curve as having $j = \infty$. Since any two points on $\mathbb{P}^1$ are equivalent as divisors , the nodal curve is "equivalent" to any other curve with fix j-invariant. ( except when j is 0,1,1728)

This property translates well into the Kontsevich moduli space of stable maps of elliptic curves to say, $\mathbb{P}^2$. The locus of maps of nodal curves are equivalent to the locus of maps of any curves with fix j-invariant as divisors, hence they should have the same enumerative invariants, because after all, enumerative invariants are just intersection numbers on the moduli space of maps. Thus, the problem of counting elliptic curves with fix j-invariant is the same as counting nodal curves, and in $\mathbb{P}^2$, nodal curves are the same as rational curves ( as rational plane curves necessarily have nodes if the degree is more than 2 ). This argument was used by Pandharipande to compute the characteristic numbers of plane elliptic curves with fix j-invariant.

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Another example: If $X$ is a smooth projective variety over a field $k$, then the relative Picard functor is representable by a smooth group scheme $Pic_{X/k}$. This is the moduli space of line bundles on $X$, in particular $Pic_{X/k}(k)$ (Edit: If $X(k)\neq \emptyset$) is the usual Picard group. The geometry of the Picard scheme tells us a lot about the line bundles; for example one can look at the connected component $Pic_{X/k}^0$ of the identity of $Pic_{X/k}$, and the points $L\in Pic_{X/k}^0(k)$ are precisely the line bundles belonging to divisors algebraically equivalent to $0$.

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If $X(k)$ is empty then the natural map ${\rm{Pic}}(X) \rightarrow {\rm{Pic}}_ {X/k}(k)$ may not be an isomorphism (unless $k$ has trivial Brauer group). This already arises for $k = \mathbf{R}$ and $k = \mathbf{Q}$. Logically, it is tied up with the actual definition of the relative Picard functor. A nice application is that if $k$ has characteristic 0 then by Cartier's theorem the Picard scheme is then smooth, so there is no obstruction to infinitesimal deformation of line bundles if $X(k)$ is non-empty (false in char. > 0). Could otherwise be hard to see purely algebraically. –  BCnrd May 28 '10 at 1:00
Thanks for the correction! –  Lars May 28 '10 at 5:57
@BCnrd: It is easy to see directly using the cohomological description of the Picard group that infinitesimal lifting is true in characteristic $0$ (with or without rational points) so this is (unfortunately) not an example of the kind the poser wants. –  Torsten Ekedahl May 28 '10 at 6:18
@Torsten: I put emphasis on a "purely algebraic" argument (to rule out complex-analytic methods). What is it about char. 0 that you use in a purely algebraic argument if not the smoothness of lft groups? –  BCnrd May 28 '10 at 16:21
@BCnrd: Characteristic $0$ comes in when one uses the fact that $exp(r)$ and $log(1+r)$ are defined when $r$ is nilpotent (this is actually the basis for the fact that group schemes are smooth so they are equally algebraic). This implies that $mR$ as an additive group and $1+mR$ as multiplicative group are isomorphic when $m$ is a nilpotent ideal in $R$. This is false when we are not in characteristic $0$, see for instance the case of $1+tF[[t]]$ and $tF[[t]]$ when $F$ is a finite field. Mumford in his book "Curves on surfaces" has a nice description in these terms of what happens for $p>0$. –  Torsten Ekedahl May 30 '10 at 9:42

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