Suppose that we have a compact Riemann surface (complete complex curve) and we remove one point from it. Then we can easily recover the pair from . As a result, different communities of mathematicians have different opinions as to whether or is the more fundamental object. People from mathematical physics like to talk about curves with punctures, adopting the first perspective, while algebraic geometers like to talk about curves with marked points, adopting the latter. What I want to do in this post is explain why we algebraic geometers are correct — if you allow us to choose all the definitions :). (Perhaps AJ will write a post explaining why the physicists are correct if you let them choose the definitions.) Specifically, I want to explain the following fact which, when I was first informed of, I could not believe:

Let , which should be considered as an infinitesimally small neighborhood of the origin in the complex plane. Let be the trivial family of projective lines over and let and be two closed subsets of : namely, we take and . (More precisely, and .) Geometrically, and should each be viewed as four infinitesimal discs, each projecting down isomorphically to . There are three discs which and share in common. The fourth disc of is different from the fourth disc of , but tangent to it over the origin .

Now, it turns out that there is no automorphism of taking to . However, and are isomorphic! Thus, the operation of “filling in punctures”, which is so common when working with individual Riemmann surfaces, can not be done in a well defined manner in families over schemes like . If you are presented with the family , you have no way of knowing whether I obtained this family as or as . So this is why algebraic geometers talk about marked points, rather than forgetting information by passing to the family of punctured curves.

Below the fold, I will work out this example carefully. I will then use it to advertise many of the further subtleties which arise in deformation theory. I should say that this post was entirely inspired by Prof. Hartshorne’s course on deformation theory in spring 2005; the notes from which are still online. I came up with this example in the course of trying to understand his far more subtle ones.

First, let me sketch why there is no automorphism of taking to . It will be convenient to introduce the notation for the ring . One can check that the Picard group of is still . Therefore, the line bundle on can be identified from the fact that its space of global sections is so any automorphsim of must pull back to a bundle isomorphic to . From here, one obtains in the usual way that the automorphism group of is , where is the group of matrices with entries in whose determinant is a unit of . The in comes from the choice of isomorphism between the pulled-back copy of and itself. So, all the autmorphisms of look like Mobius transformations , with , , and in and a unit of . A little work will check that only the identity Mobius transformation fixes three points, even over this strange ring .

The argument in the previous paragraph is just the standard argument that crossratio is a well defined invariant of four points on , done carefully enough to work over the ring . So you shouldn’t find it surprising. What you should find surprising is that is isomorphic to . The easiest way to see this is so slick that you’ll think I’m cheating. is simply . But, in this ring, is already invertible; it’s inverse is . So . Similarly, .

This feels like cheating to me, so let’s redo the argument in the coordinate . In this coordinate, and are and respectively. But, in the former ring, is already a unit, with inverse . So

. Yup, the argument is still right!

Of course, we can mix punctures and marked points together. For example, suppose that we delete from , leaving behind the affine line . Let’s mark and . Is there an automorphism of taking to ? Find out in the next installment!

This is indeed a pathology (one which turns some people off from the algebraic geometry approach to deformation theory). But, of course, you know there is nothing special about the projective line in this “pathology”. Every smooth algebra is isomorphic to , i.e., every smooth, affine scheme over is isomorphic to a “trivial deformation”.

Hi David,

Fun post!

FWIW, I’m used to mathematical physicists talking about Riemann surfaces with boundary circles. Families of marked curves (especially nodal marked curves) only emerge after some work.

The rough story goes as follows: A 2d QFT gives you a vector space for every circle and a morphism for every Riemann surface (with metric ) having incoming circles and outgoing circles. If you choose vectors and , you can define the amplitude

This will be some function of the metric . In a CFT, it should only depend on the conformal equivalence class of .

If your QFT is nice, then each factor in is the vector associated by the QFT functor to a disc, which we think of as a bordism from the empty set to . Likewise for . In this case, we can get the amplitude as the number associated by the QFT functor to the closed Riemann surface that we get by capping off the boundaries of . If our QFT is conformal, then this function should only depend on the complex structure on , and on marked points which remember where we inserted the vectors and . In other words, the amplitude should descend to a function on the stack of smooth marked curves, obtained as metrics modulo diffeomorphism and conformal equivalence.

Then we have to explain why this function extends to the boundary of . I might try to write that up in a separate post. Or maybe we can harass Scott C. into doing it.

Boundary circles of positive length also play a role in Mirzakhani’s proof of Witten’s conjecture. As I understand it, Mirzakhani studies the variation of the volume of the moduli spaces (w.r.t the Weil-Peterson metric) as the lengths of the boundary circles vary.

Nice post! But I’m not sure I would call it a pathology.. One way to look at the issue is that when you consider objects with large automorphism groups their moduli is tricky. An affine variety has a huge

automorphism FORMAL group. More specifically, if you take an affine curve then it has an infinite-dimensional k-Lie algebra of infinitesimal

automorphisms — all sections of the tangent bundle (with arbitrary

poles “at infinity”). We can exponentiate all of these formally to

give a (“Virasoro”) formal group of automorphisms — meaning

if we look over rings with nilpotents our curve has a huge number

of automorphisms, in particular enough to identify all

of the punctured curves under consideration.

So this tells us that to get a nicely

behaved moduli problem we might want to rigidify somewhat –

e.g. by looking at those that extend to the complete curve.

Hi AJ!

One can also look at the marked points vs boundary circles as the

state-field correspondence in the CFT — marked points is where you insert local operators in any QFT,

but in a CFT (or a TFT in any dimension) the space of local operators is the same as the space of states on the link of the insertion point, ie a circle in the 2d story. (Likewise e.g. loop operators in a 4d TFT are labeled by states on the link of the loop, ie S^2 times S^1..)

David,

Here’s my favorite example of a related phenomenon: a family

of varieties with a flat connection, so infinitesimally nearby fibers are

CANONICALLY isomorphic, but no two fibers over closed points are

isomorphic. (As opposed to the example of any family of affine

curves, where the nearby fibers are not canonically isomorphic).

Take your favorite family of projective curves (eg M_g)

and consider over it the moduli space of curves with rank n

bundles with flat connection — ie the fiber over a curve X

is the space of rank n flat connections on X. Complex analytically these fibers are just the character varieties — homomorphisms

from pi_1(X) to GL_n up to conjugacy — i.e. connections are given

by their monodromy. Algebraically no two fibers are the same!

(I believe this is due to Indranil Biswas in general – “Torelli

theorem” for the de Rham space of a curve).

However there is a flat connection on the family — ie

you can canonically lift any infinitesimal deformation of the curve X to a deformation of the moduli space. However this flat

connection does not integrate at all – you can’t compare

any two distinct fibers! Kind of cool but baffling.

(It also means that these spaces of connections, or

the categories of algebraic coherent sheaves on them,

don’t seem to form a topological field theory —

meaning that the Kapustin-Witten story, if it is algebraizable,

is really about the character varieties not the moduli of flat connections..)

Hi David!

In fact, I was trying to describe the state-operator correspondence. But now that I think about it, the state-operator correspondence can’t quite be shoehorned into this language: For TFT and CFT, the incoming disc can’t possibly give you all vectors! So my story is a bit useless, except as vague motivation.

For the state-operator correspondence, we need something more like Freed’s language, where we associate 1-dimensional linear gadgets to pairs consisting of a Riemann surface and a field . In particular, we can ask that the vector be created by some field on the disc.

Hi David B-Z,

It’s a good point that infinite-dimensional automorphisms usually create bad moduli. But I don’t think that is the end of this phenomenon. For instance, following is an example of a smooth projective scheme over and a closed subscheme of which is also flat over such that the open complement is not a constant family over , i.e., not isomorphic to the basechange of a family over , but for every integer the restriction of to is a constant family. Of course David S. just gave us such an example where is the projective line. And you rightly point out this is explained by the fact that is HUGE! However in the following example, is zero.

So take to be your favorite projective surface which has discrete automorphism group, e.g., the blowing up of the projective plane in 4 general points. Let be the product of with . Let be the union of two sections

and of such that equals but does not equal . So is a family of quasi-projective schemes which are basically the complements of two points in which are coalescing. The family is not constant, because the generic fiber of has two points, whereas the closed fiber of has only one point. But for every integer , the restriction of over is simply the restriction of minus one closed point. On the other hand, by Hartog’s phenomenon (or the fact that has Serre’s property in the algebraic category), equals which is zero.

Perhaps you will object that this is too easy, e.g., is not smooth over , it is only flat. But it is easy to come up with examples where is smooth over , e.g., let be over and let be a family of balanced quartic scrolls degenerating to an unbalanced quartic scroll. All of this adds up to the same moral, “For many reasons, deformations of quasi-projective schemes are less well-behaved than deformations of projective schemes.”

Hi guys,

On further reflection, I think the quartic scrolls example is bogus. So is there a “pathological” pair over where is smooth over but is not an infinitely-generated module over ?

Hi guys,

On further, further reflection, the quartic scroll example isn’t bogus :)

In fact there are much easier examples than either of the ones in Post 8. The real point of all these examples is that for any closed subscheme of a scheme , all infinitesimal deformations of as a closed subscheme of give the (same) trivial infinitesimal deformation of as an abstract scheme. The reason is that the deformation of over is simply the trivial deformation of minus the closed set of the infinitesimal deformation of . This set doesn’t see the infinitesimal deformation of — it literally is just as a set.

Stated more succinctly: the two schemes and in the post by David S. are not just abstractly isomorphic, they are literally equal as open subschemes of !