# Some oddities of Deformation Theory

Suppose that we have a compact Riemann surface (complete complex curve) $X$ and we remove one point $x$ from it. Then we can easily recover the pair $(X, x)$ from $X \setminus x$. As a result, different communities of mathematicians have different opinions as to whether $X \setminus x$ or $(X,x)$ 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 $S=\mathrm{Spec} \ \mathbb{C}[t]/t^N$, which should be considered as an infinitesimally small neighborhood of the origin in the complex plane. Let $\mathbb{P}^1_S$ be the trivial family of projective lines over $S$ and let $Z_1$ and $Z_2$ be two closed subsets of $\mathbb{P}^1_S$: namely, we take $Z_1=\{0, 1, -1, \infty \}$ and $Z_2=\{ 0, 1, -1, t^{-1} \}$. (More precisely, $Z_1=\{ (0:1), (1:1), (-1:1), (1:0) \}$ and $Z_2=\{(0:1), (1:1), (-1:1), (1:t) \}$.) Geometrically, $Z_1$ and $Z_2$ should each be viewed as four infinitesimal discs, each projecting down isomorphically to $S$. There are three discs which $Z_1$ and $Z_2$ share in common. The fourth disc of $Z_2$ is different from the fourth disc of $Z_1$, but tangent to it over the origin $t=0$.

Now, it turns out that there is no automorphism of $\mathbb{P}^1_S$ taking $Z_1$ to $Z_2$. However, $P^1_S \setminus Z_1$ and $P^1_S \setminus Z_2$ 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 $S$. If you are presented with the family $P^1_S \setminus Z_1$, you have no way of knowing whether I obtained this family as $P^1_S \setminus Z_1$ or as $P^1_S \setminus Z_2$. 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 $\mathbb{P}^1_S$ taking $Z_1$ to $Z_2$. It will be convenient to introduce the notation $R$ for the ring $\mathbb{C}[t]/t^N$. One can check that the Picard group of $\mathbb{P}^1_S$ is still $\mathbb{Z}$. Therefore, the line bundle $O(1)$ on $P^1(S)$ can be identified from the fact that its space of global sections is $R^2$ so any automorphsim of $\mathbb{P}^1_S$ must pull back $O(1)$ to a bundle isomorphic to $O(1)$. From here, one obtains in the usual way that the automorphism group of $\mathbb{P}^1_S$ is $PGL_2(R)$, where $GL_2(R)$ is the group of $2 \times 2$ matrices with entries in $R$ whose determinant is a unit of $R$. The $P$ in $PGL$ comes from the choice of isomorphism between the pulled-back copy of $O(1)$ and $O(1)$ itself. So, all the autmorphisms of $\mathbb{P}^1_S$ look like Mobius transformations $z \mapsto (az+b)/(cz+d)$, with $a$, $b$, $c$ and $d$ in $R$ and $ad-bc$ a unit of $R$. A little work will check that only the identity Mobius transformation fixes three points, even over this strange ring $R$.

The argument in the previous paragraph is just the standard argument that crossratio is a well defined invariant of four points on $\mathbb{P}^1$, done carefully enough to work over the ring $R$. So you shouldn’t find it surprising. What you should find surprising is that $P^1_S \setminus Z_1$ is isomorphic to $P^1_S \setminus Z_2$. The easiest way to see this is so slick that you’ll think I’m cheating. $P^1_S \setminus Z_1$ is simply $\mathrm{Spec} R[x][\left( x(x-1)(x+1) \right)^{-1}]$. But, in this ring, $1-xt$ is already invertible; it’s inverse is $1+xt+x^2t^2+\cdots x^{N-1} T^{N-1}$. So $P^1_S \setminus Z_1 \cong P^1_S \setminus (Z_1 \cap Z_2)$. Similarly, $P^1_S \setminus Z_2 \cong P^1_S \setminus (Z_1 \cap Z_2)$.

This feels like cheating to me, so let’s redo the argument in the coordinate $y=1/x$. In this coordinate, $P^1_S \setminus Z_1$ and $P^1_S \setminus Z_2$ are $\mathrm{Spec} R[y][\left( y(y-1)(y+1) \right)^{-1}]$ and $\mathrm{Spec} R[y][\left( (y-t)(y-1)(y+1) \right)^{-1}]$ respectively. But, in the former ring, $y-t$ is already a unit, with inverse $y^{-1}+y^{-2} t + y^{-3} t^2 + \cdots + y^{-N} t^{N-1}$. So $R[y][\left( y(y-1)(y+1) \right)^{-1}]=R[y][\left( y(y-t)(y-1)(y+1) \right)^{-1}]=$
$R[y][\left( (y-t)(y-1)(y+1) \right)^{-1}]$. Yup, the argument is still right!

Of course, we can mix punctures and marked points together. For example, suppose that we delete $\infty$ from $\mathbb{P}^1_S$, leaving behind the affine line $\mathbb{A}^1_S$. Let’s mark $W_1=\{-1,0,1 \}$ and $W_2=\{-1, t, 1\}$. Is there an automorphism of $\mathbb{A}^1_S$ taking $W_1$ to $W_2$? Find out in the next installment!

## 10 thoughts on “Some oddities of Deformation Theory”

1. Jason Starr says:

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 $k[t]/t^n$ algebra $A$ is isomorphic to $k[t]/t^n \otimes_k (A/tA)$, i.e., every smooth, affine scheme over $Spec k[t]/t^n$ is isomorphic to a “trivial deformation”.

2. A.J. Tolland says:

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 $H$ for every circle and a morphism $m_\eta: H^{\otimes I} \to H^{\otimes J}$ for every Riemann surface (with metric $\eta$) having $I$ incoming circles and $J$ outgoing circles. If you choose vectors $v \in H^{\otimes I}$ and $w \in (H^{\otimes J})^{\vee}$, you can define the amplitude

$\langle w | v \rangle = \langle w, m_{\eta}(v)\rangle$

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

If your QFT is nice, then each factor $v_i$ in $v = \otimes_i v_i$ is the vector associated by the QFT functor to a disc, which we think of as a bordism from the empty set to $S^1$. Likewise for $w$. In this case, we can get the amplitude $\langle w|v \rangle$ as the number associated by the QFT functor to the closed Riemann surface $\Sigma'$ that we get by capping off the boundaries of $\Sigma$. If our QFT is conformal, then this function should only depend on the complex structure on $\Sigma'$, and on marked points which remember where we inserted the vectors $v$ and $w$. 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 $\overline{\mathcal{M}}_{g,I \sqcup J}$. I might try to write that up in a separate post. Or maybe we can harass Scott C. into doing it.

3. 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.

4. David Ben-Zvi says:

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.

5. David Ben-Zvi says:

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..)

6. David Ben-Zvi says:

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..)

7. A.J. Tolland says:

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 $(\Sigma,\phi)$ consisting of a Riemann surface and a field $\phi$. In particular, we can ask that the vector $v_i$ be created by some field on the disc.

8. 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 $P$ over $k[[t]]$ and a closed subscheme $Z$ of $P$ which is also flat over $k[[t]]$ such that the open complement $U$ is not a constant family over $k[[t]]$, i.e., not isomorphic to the basechange of a family over $k$, but for every integer $n$ the restriction of $U$ to $k[[t]]/t^n$ is a constant family. Of course David S. just gave us such an example where $P$ is the projective line. And you rightly point out this is explained by the fact that $H^0(U,T_U)$ is HUGE! However in the following example, $H^0(U,T_U)$ is zero.

So take $P_0$ 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 $P$ be the product of $P_0$ with $\text{Spec} k[[t]]$. Let $Z$ be the union of two sections
$a$ and $b$ of $P\rightarrow \text{Spec} k[[t]]$ such that $a(0)$ equals $b(0)$ but $a$ does not equal $b$. So $U$ is a family of quasi-projective schemes which are basically the complements of two points in $P_0$ which are coalescing. The family is not constant, because the generic fiber of $Z$ has two points, whereas the closed fiber of $Z$ has only one point. But for every integer $n$, the restriction of $U$ over $\text{Spec} k[[t]]/t^n$ is simply the restriction of $P$ minus one closed point. On the other hand, by Hartog’s phenomenon (or the fact that $P$ has Serre’s property $S_2$ in the algebraic category), $H^0(U,T_U)$ equals $H^0(P,T_P)$ which is zero.

Perhaps you will object that this is too easy, e.g., $Z$ is not smooth over $k[[t]]$, it is only flat. But it is easy to come up with examples where $Z$ is smooth over $k[[t]]$, e.g., let $P$ be $P^5$ over $k[[t]]$ and let $Z$ 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.”

9. Hi guys,

On further reflection, I think the quartic scrolls example is bogus. So is there a “pathological” pair $(P,Z)$ over $k[[t]]$ where $Z$ is smooth over $k[[t]]$ but $H^0(U,T_U)$ is not an infinitely-generated module over $k[[t]]$?

10. 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 $Z_0$ of a scheme $P_0$, all infinitesimal deformations of $Z_0$ as a closed subscheme of $P_0$ give the (same) trivial infinitesimal deformation of $U_0$ as an abstract scheme. The reason is that the deformation of $U_0$ over $\text{Spec} k[[t]]/t^n$ is simply the trivial deformation of $P_0$ minus the closed set of the infinitesimal deformation of $Z_0$. This set doesn’t see the infinitesimal deformation of $Z_0$ — it literally is just $Z_0$ as a set.

Stated more succinctly: the two schemes $P^1_S-Z_1$ and $P^1_S-Z_2$ in the post by David S. are not just abstractly isomorphic, they are literally equal as open subschemes of $P^1_S$!