MODULI SPACES AND DEFORMATION THEORY, CLASS 15

RAVI VAKIL

Contents

1. Where we are 1 2. Schlessinger’s criterion for existence of universal deformations and hulls

(miniversal deformations) 2 3. Proof of Schlessinger, take 2 3

In case I forget to tell you on Thursday: I’ll be out of town next Tuesday (Nov. 7), so Jason Starr will kindly be giving a guest lecture. He will likely tell you about applying Schlessinger’s criteria to the Quot functor.

1. Where we are

C, Ĉ. k[V]. A ×_{C} B. k[V ⊕ W] = k[V] ×_{k} k[W].

Lemma. Suppose F is a functor (covariant, on C) such that

F (k[V] ×_{k} k[W]) →^{∼} F (k[W]) × F (k[W])

for finite-dimensional vector spaces V and W over k. Then F (k[V]) and in particular t_{F} = F (k[ϵ]), has a canonical vector space structure, such that F (k[V]) ≅ t_{F} ⊗V.

Patched proof. k[V] is a “vector space object” in Ĉ, via

Hom(A, k[V]) ≅ Der_{k}(A, V).

For the last statement: Note that Hom(k[ϵ]/ϵ^{2}, k[V]) is naturally identified with V. For any element of Hom(k[ϵ]/ϵ^{2}, k[V]) we get a map t_{F} = F (k[ϵ]/ϵ^{2}) → F (k[V]), hence t_{F} × V → F (k[V]). In fact this is ⊗. The desired result is true if V is one-dimensional; then use induction, as V is finite-dimensional, and k[V] = ×_{k}^{dim V} k[ϵ]/(ϵ^{2}). (I said something wrong in class.)

Date: Tuesday, October 31, 2000.

2. Schlessinger’s criterion for existence of universal deformations and hulls (miniversal deformations)

Recall: In C, define a small extension to be a surjection A′′ → A, so A = A′′/I, and m_{A′′}I = 0, and I is one-dimensional.

For the purposes of this course only, define a fairly small extension to be a surjection A′′ → A, so A = A′′/I, and m_{A′′}I = 0, without requiring that I is one-dimensional. Then for any A in C, you can filter A into a series of fairly small extensions (by powers of the maximal ideal). Then you can filter A into a series of small extensions.

Fix our functor F : C → Sets.

Let A′ → A and A′′ → A be morphisms in C, and consider the map

F (A′ ×_{A} A′′) → F (A′) ×_{F (A)} F (A′′). (1)

I explained last time that if F is a prorepresentable functor, by R ∈ C say, then this map is always a bijection — this is because × is a categorical product!

Schlessinger’s Theorem. [Put on one board permanently!]

It has two parts, and I’ll say it slowly, with translations and remarks.

(1) F has a hull iff F has properties H1–H3:

H1. (You can glue.) (1) is a surjection whenever A′′ → A is a small extension.

Equivalently whenever A′′ → A is any surjection. H2. (Uniqueness of gluing on k[ϵ]/ϵ^{2}.) (1) is a bijection when A = k, A′′ = k[ϵ]/ϵ^{2}.

Equivalently, A′′ = k[V]. Then by previous lemma, t_{F} is a k-vector space. H3. (finite-dimensional tangent space) dim_{k}(t_{F}) < ∞.

(2) F is pro-representable if and only if F has the additional property

H4. (bijection for gluing a small extension to itself)

F (A′ ×_{A} A′) → F (A′) ×_{F (A)} F (A′). (2)

is a bijection for any small extension A′ → A.

Recall from last time. Suppose F satisfies H1–H3.

Consider any fairly small extension p : A′ → A, i.e. 0 → I → A′ → A → 0, so m_{A′}I = 0. We have an isomorphism

A′ ×_{A′/I} A′ →^{∼} A′ ×_{k} k[I]

induced by the map (x, y) ↦ (x, x_{0} + y − x) (e.g. A′ = k[u]/u^{3} = k[v]/v^{3}, A = k[u]/u^{2}).

Now given a small extension p : A′ → I, By H2, we get

F (A′ ×_{A} A′) = F (A′ ×_{k} k[I]) →^{∼} F (A′) ×_{F (k)} F (k[I]) = F (A′) × (t_{F} ⊗ I).

Hence we get

F (A′) × (t_{F} ⊗ I) → F (A′) ×_{F (A)} F (A′).

For each η ∈ F (A), this determines a group action of t_{f} ⊗ I on F (p)^{−1}(η), i.e. those F (A′)’s lifting F (A), assuming the set is nonempty. The fact that this is a surjection (H1) means that the action is transitive. H4 is precisely the condition that this set is a principal homogeneous space under t_{F} ⊗ I. (Say more here.)

So explicitly, what this is telling us is explicitly is that if F already has a hull, then its obstruction to be representable is the existence of an automorphism of an object y in some F (A), that cannot be extended to an automorphism of some object y′ ∈ F (A′) for some A′.

3. Proof of Schlessinger, take 2

• Immediate: If F is prorepresentable, then H1–H4 are all satisfied, as (1) is always a bijection, and t_{F} is finite-dimensional.

• Hull and H4 imply prorepresentable (shown last time).

• Next: Then I’ll show that hull implies H1–H3.

• Finally: H1–H3 imply hull.

Hull implies H1–H3.

Recall the definition of a hull or miniversal deformation space (e.g. def space of a node). (R, r ∈ F (R)) → F, formally smooth.

Translation: for each surjection B → A in C, Hom(R, B) → Hom(R, A) ×_{F (A)} F (B) is surjective.

Translation:

(B, b ∈ F (B))
? ↗  ↓
(R, r ∈ F (R)) → (A, a ∈ F (A))

(Note phantom left corner.)

H3: t_{R} ≅ t_{F} proves H3.

H1: Want:

F (A′ ×_{A} A′′) → F (A′) ×_{F (A)} F (A′′).

is a surjection whenever A′′ → A is a small extension (or indeed any surjection).

Suppose we have p′ : (A′, η′ ∈ F (A′)) → (A, η) and p′′ : (A′′, η′′) → (A, η) are morphisms of couples or deformations, and p′′ is a surjection. (Put in diagram, A′ to left, A′′ above).

(A′ ×_{A} A′′,?) → (A′′, a′′)
↓  p′′ ↓
(A′, a′)  →^{p′} (A, a)

Since h_{R} → F is surjective, there is u′ : (R, r) → (A′, η′). By smoothness of h_{R} over F, we get u′′ : (R, r) → (A′′, η′′). We get R maps to A′ ×_{A} A′′; use covariant F to get η there. So H1 is satisfied!

H2 (restate). Suppose now that (A, η) = (k, pt) and A′′ = k[ϵ]/ϵ^{2}. We want there to be only one such (A′ ×_{A} A′′, η). Suppose there are two, η_{1} and η_{2}. Then make that diagram, add in (R, r). Choose u′ as before. Then there is only one u′′ possible, thanks to t_{R} = t_{F}. Then there is only one lift R → A′ ×_{A} A′′, and that’s it.

Diagram (mostly for my own benefit), of maps from (R, r) to various parts of the diagram.

second, lift by smoothness of h_{R} → F → finally, unique by t_{R} = t_{F}
↓  ↓
pick u first  →

Finally, H1–H3 implies hull. This is the trickiest part of the proof. Most of it has the same level of complexity, but there’s a trick at the end.

Assume H1–H3. We will build a hull. By H2, t_{F} is a vector space, and by H3, it is finite-dimensional. Let T_{1}, …, T_{r} be a dual basis for t_{F}. Let S = k[[T_{1}, …, T_{r}]], with maximal ideal 𝔫. Our goal is to get R = S/J, with r ∈ F (R), that is a hull.

R_{2} = S/𝔫^{2}. (Think: S/(J, 𝔫^{2}).) = k[ϵ]/ϵ^{2} ×_{k} ⋯ ×_{k} k[ϵ]/ϵ^{2}.

By H2, there is some r_{2} ∈ F (R_{2}) inducing a bijection between t_{R_{2}} (which is Hom(R_{2}, k[ϵ])) and t_{F}. (Say this differently! Use the lemma; get canonical family over this ring.)

Now we build up R inductively.

Suppose we have (R_{q}, r_{q}), where R_{q} = S/J_{q}. We want J_{q+1} in S, minimal among S = (ideals J in S satisfying (a) 𝔫J_{q} ⊂ J ⊂ J_{q}, and (b) r_{q} lifts to S_{J}). Those satisfying J correspond to subspaces of J_{q}/𝔫J_{q}. (Picture?)

So we need to show that there is a minimal element. So we need to see that S is stable under pairwise intersection. Say J, K ∈ S. Enlarge J so that J + K = J_{q}, without changing their intersection. Then

S/J ×_{S/J_{q}} S/K ≅ S/(J ∩ K)

If you want an example: J_{q} = (x, y)^{2}. J = (x^{2}, xy, y^{3}), K = (x^{3}, xy, y^{2}).

By H1, F (S/J ∩ K) → F (S/J) ×_{F (S/J_{q})} F (S/K) surjects, so that’s okay.

So now let J = ∩_{q>1}J_{q}, and R = S/J. Let r be the inverse limit of the r_{q}. (As J_{q} form a base for the topology, as 𝔫^{q} ⊂ J_{q}, can set r = lim_{←} r_{q} ∈ F (R).) This will be our hull; we’ll check that next day.