# Tuesday March 31st

See notes on Ben's website.
We'll review where we were.

## Deformation Theory

We want to represent certain moduli functors by schemes.
If we know a functor is representable, it's easier to understand the deformation theory of it and still retain a lot of geometric information.
The representability of deformation is much easier to show.
We're considering functors $F: \Art_{/k} \to \Sets$.


:::{.example title="?"}
The Hilbert functor
\[
\hilb_{X_{_{/k}}} (\Sch_{/k})\op \to \sets \\
S \mapsto \theset{ Z  \subset  X \cross S \text{ flat over } S}
.\]

This yields
\[
F: \Art_{/k} \to \sets \\
???
.\]
:::

![Image](figures/2020-03-31-12:44.png)

Recall that we're interested in pro-representability, where $\hat F(R) = \inverselim F(R\mu_R^n)$ is given by a lift of the form

\begin{tikzcd}
\Art_{/k} 
  \ar[r, "F"] 
& \sets 
\\
\hat{\Art_{/k}}
  \ar[u, hook] 
  \ar[ur, "\hat F"']
&
\end{tikzcd}


:::{.question}
Is $\hat F$ representable, i.e. is $F$ pro-representable?
:::

:::{.example title="?"}
The $F$ in the previous example is pro-representable by $\hat F = \hom(\OO_{\hilb, z_0}, \wait)$.
:::


:::{.definition title="Pro-Representable Hull"}
$F$ has a *pro-representable hull* iff there is a formally smooth map $h_R \to F$.
:::

:::{.question}
Does $F$ have a pro-representable hull?
:::

Recall that a map of functors on artinian $k\dash$algebras is **formally smooth** if it can be lifted through nilpotent thickenings.
That is, for $F, G: \Art_{/k} \to \Sets$, $F \to G$ is *formally smooth* if for any thickening $A' \surjects A$, we have

\begin{tikzcd}
& 
& F 
  \ar[d] 
\\
h_{A} 
  \ar[rru] 
  \ar[r] 
& h_{A'} 
  \ar[ru, dotted] 
  \ar[r] 
& G
\\
  \spec A 
  \ar[u, equal] 
  \ar[r] 
& \spec A' 
  \ar[u, equal] 
  \ar[r] 
& G
  \ar[u, equal]
\end{tikzcd}


We proved for $R, A$ finite type over $k$, $\spec R \to \spec A$ smooth is formally smooth.
Given a complete local $k\dash$algebra $R$ and a section $\xi \in \hat F(R)$, we make the following definitions:


:::{.definition title="Versal, Miniversal, Universal"}
The pair $(R, \xi)$ is

- *Versal* for $F$ iff $h_R \mapsvia{\xi} F$ is formally smooth.[^not_unique]

- *Miniversal* for $F$ iff versal and an isomorphism on Zariski tangent spaces.

- *Universal* for $F$ if $h_R \mapsvia{\cong} F$ is an isomorphism, i.e. $h_R$ pro-represents $F$.
  - Pullback by a unique map

[^not_unique]: 
Not a unique map, but still a pullback

:::


:::{.remark}
Note that **versal** means that any formal section $(s, \eta)$ where $\eta \in \hat F(s)$ comes from pullback, i.e there exists a map
\[
R &\to S \\
\hat F(R) &\to \hat F(s) \\
\xi &\mapsto \eta
.\]

**Miniversal** means adds that the derivative is uniquely determined, and universal means that $R\to S$ is unique.
:::


:::{.definition title="Obstruction Theory"}
An **obstruction theory** for $F$ is the data of $\mathrm{def}(F), \mathrm{obs}(F)$ which are finite-dimensional $k\dash$vector spaces, along with a functorial assignment of the following form:
\[
(A' \surjects A) \quad \text{a small thickening } \mapsto \\
\mathrm{def}(F) \selfmap F(A') \to F(A) \mapsvia{\mathrm{obs}} \mathrm{obs}(F)
\]
that is exact[^recall_right_exact] and if $A=k$, it is exact on the left (so the action was faithful on nonempty fibers).

[^recall_right_exact]: 
Recall that right-exactness was a transitive action.

:::

:::{.example title="?"}
We have
\[
\pic_{X_{/k}} : (\sch_{/k})\op &\to \sets \\
S &\mapsto \pic(X\cross X) / \pic(S)
.\]

This yields
\[
F: \Art_{/k} \to \sets \\
A \mapsto L\in \pic(X_A),~ L\tensor k \cong L_0
\]
where $X_{_{/k}}$ is proper and irreducible.
Then $F$ has an obstruction theory with $\mathrm{def}(F) = H^1(\OO_x)$ and $\mathrm{obs}(F) = H^2(\OO_x)$.
The key was to look at the LES of
\[
0 \to \OO_x \to \OO_{X_{A'}}^* \to \OO_{X_A}^* \to 0
.\]

for $0 \to k \to A' \to A \to 0$ small.
:::

:::{.remark title="Summary"}
In both cases, the obstruction theory is exact on the left for any small thickening.
We will prove the following:

- $F$ has an obstruction $\iff$ it has a pro-representable hull, i.e. a versal family

- $F$ has an obstruction theory which is always exact at the left $\iff$ it has a universal family.
:::
  

## Schlessinger's Criterion 

Let $F: \Art_{/k} \to \Set$ be a deformation functor (and it only makes sense to talk about deformation functors when $F(k) = \pt$).
This theorem will tell us when a miniversal and a universal family exists.

:::{.theorem title="Schlessinger"}
$F$ has a miniversal family iff

1. Gluing along common subspaces: ror any small $A' \to A$ and $A'' \to A$ any other thickening, the map
\[
F(A' \cross_A A'') \to F(A') \cross_{F(A)} F(A'')
\]
  is surjective.


2. Unique gluing: if $(A' \to A) = (k[\eps] \to k)$, then the above map is bijective.


3. $t_F = F(k[\eps])$ is a finite dimensional $k\dash$vector space, i.e.
\[
F(k[\eps] \cross_k k[\eps]) \mapsvia{\cong} F(k[\eps]) \cross F(k[\eps])
.\]

4. For $A' \to A$ small,

\begin{tikzcd}
F(A') 
  \ar[r, "f"] 
& F(A) 
\\
t_f\, \selfmap f\inv(\eta) 
  \ar[u, hook, "\subseteq"] 
& \eta 
  \ar[u, "\in"]
\end{tikzcd}
where the action is simply transitive.


$F$ has a miniversal family iff (1)-(3) hold, and universal iff all 4 hold.

:::

:::{.exercise title="?"}
Show that the existence of an obstruction theory which is exact on the left implies (1)-(4).
:::


The following diagram commutes:

\begin{tikzcd}
\mathrm{def} \selfmap F(A' \cross_A A'') \ni \eta 
  \ar[r]
  \ar[d] 
& F(A'') \ni \xi'' 
  \ar[r, "\mathrm{obs}"]
  \ar[d] 
& \mathrm{obs} 
\\
\mathrm{def} \selfmap F(A')\ni \eta'm \xi' 
  \ar[r] 
& F(A')\ni \xi 
  \ar[r, "\mathrm{obs}"] 
& \mathrm{obs} \\
\end{tikzcd}


So we have a map $F(A' \cross_A A'') \to F(A') \cross_{F(A)} F(A'') \ni (\xi',\xi'')$.
Using transitivity of the $\mathrm{def}$ action, we can get $\xi' = \eta' + \theta$ and thus $\eta + \theta$ is the lift.

## Abstract Deformation Theory


:::{.example title="?"}
We start with $\qty{X_0}_{/k}$ and define the functor $F$ sending $A$ to $X/A$ flat families over $A$ with $X_0 \injects^i X$ such that $i \tensor k$ is an isomorphism.
The punchline is that $F$ has an obstruction theory if $X_0$ is smooth with

- $\mathrm{def}(F) = H^1(T_{X_0})$
- $\mathrm{obs}(F) = H^2(T_{X_0})$

:::


:::{.remark}
\envlist

1. If $X$ is a deformation of $X_0$ over $A$ and we have a small extension $k \to A'\to A$ with $X'$ over $A'$ a lift of $X$.
  Then there is an exact sequence
\[
0 \to \text{Der}_R(\OO_{X_0}) \to\aut_{A'}(X') \to \aut_A(X)
.\]

2. If $\qty{X_0}_{/k}$ is smooth and *affine*, then any deformation $X$ over $A$ (a flat family restricting to $X_0$) is trivial, i.e. $X \cong X_0 \cross_k \spec(A)$.

\begin{tikzcd}
&
&
X_0 \cross \spec(A) 
  \ar[d]
\\
X_0 
  \ar[r, hook] 
&
X 
  \ar[r]
  \ar[ru, "f", dotted] 
& \spec(A)
\end{tikzcd}

Thus $X_0 \injects X$ has a section $X\to X_0$, and the claim is that this forces $X$ to be trivial.
:::
  

We have
\begin{tikzcd}
0 
  \ar[r] 
& J \tensor \OO_X 
  \ar[r] 
& \OO_x 
  \ar[r] 
& \OO_{X_0} 
  \ar[r] 
  \ar[l, bend right] 
& 0
\end{tikzcd}


yielding
\[
0 \to K \to \OO_{X_0} \tensor A \to \OO_X \to 0 \\
(\wait \tensor k) \\
1 \to k\tensor k = 0 \to \OO_{X_0} \mapsvia{\cong} \OO_{X_0} \to 0
.\]


:::{.remark}
Why does this involve cohomology of the tangent bundle?
For $X_0$ smooth, $\Der_k(\OO_{X_0}) = \mathcal{H}(T_{X_0})$, but the LHS is equal to $\hom( \Omega_{ \qty{X_0}_{/k}}, \OO_{X_0}) = H^0 (T_{X_0})$.
:::

> Upcoming: proof of Schlessinger so we can use it!