# Thursday February 6th

> Review base-change!

For $k=\bar k$, and $C_{/k}$ a smooth projective curve, then $\hilb_{C_{/k}}^n = \sym^n C$.

:::{.definition title="The Hilbert-Chow Map"}
For $X_{_{/k}}$ a smooth projective  *surface*, $\hilb_{X_{_{/k}}}^n \neq \sym^n X$, there is a map (the Hilbert-Chow map)
\[
\hilb_{X_{_{/k}}}^n &\to \sym^n X \\
Z &\mapsto \supp(Z) \\
U  = \text{reduced subschemes} &\mapsto U' = \text{ reduced multisets } \\
\PP^1 &\mapsto (x, x)
.\]
:::

:::{.example title="?"}
Consider $\AA^2 \cross \AA^2$ under the $\ZZ/2\ZZ$ action
\[
( (x_1, y_1), (x_2, y_2)) \mapsto ((x_2, y_2), (x_1, y_1))
.\]


Then
\[
(\AA^2)^2 / \ZZ/2\ZZ
&= \spec k[x_1, y_1, x_2, y_2]^{\ZZ/2\ZZ} \\
&= \spec k[x_1 x_2, y_1 y_2, x_1 + x_2, y_1 + y_2, x_1 y_2 + x_2 y_1, \cdots]
\]
with a bunch of symmetric polynomials adjoined.
:::

:::{.example title="?"}
Take $\AA^2$ and consider $\hilb_{\PP^2}^3$.
If $I$ is a monomial ideal in $\AA^2$, there is a nice picture.
We can identify the tangent space
\[
T_Z \hilb_{\PP^2}^n = \hom_{\OO_{\PP^2}} ( I_2, \OO_Z) = \bigoplus \hom(I_{Z_i}, \OO_{Z_i})
.\]
if $Z = \disjoint Z_i$.
If $I$ is supported at 0, then we can identify the ideal with the generators it leaves out.
:::


:::{.example title="?"}
$I = (x^2, xy, y^2)$:

![Image](figures/2020-02-06-12:48.png){width=350px}
:::


:::{.example title="?"}
$I = (x^6, x^2y^2, xy^4, y^5)$:

![Image](figures/2020-02-06-12:49.png){width=350px}
:::



:::{.example title="?"}
$I = (x^2, y)$.
Let $e=x^2, f = y$.

![Image](figures/2020-02-06-12:54.png){width=350px}

By comparing rows to columns, we obtain a relation $ye = x^2 f$.
Write $\OO = \theset{1, x}$, then note that this relation is trivial in $\OO$ since $y=x^2=0$.
Thus $\hom(I, \OO) = \hom(k^2, k^2)$ is 4-dimensional.
:::



:::{.remark}
Note that $C_{_{/k}}$ for curves is an important case to know.
Take $Z \subset C \cross C^n$, then quotient by the symmetric group $S^n$ (need to show this can be done), then $Z/S^n \subset C \cross \sym^n C$ and composing with the functor $\hilb$ represents yields a map $\sym^n C \to \hilb_{C_{/k}}^n$.
This is bijective on points, and a tangent space computation shows it's an isomorphism.
:::


:::{.example title="?"}
Consider the nodal cubic in $\PP^2$:

![Nodal cubic](figures/2020-02-06-13:01.png){width=350px}

> The nodal cubic $zy^2 = x^2(x+z)$.

Consider the open subscheme $V \subset \hilb_{C_{/k}}^2$ of points $z \subset U$ for $U \subset C$ open.
We can normalize:

![Normalized cubic](figures/2020-02-06-13:03.png){width=350px}

This yields a map fro $\PP^1 \setminus\text{2 points}$.
This gives us a stratification, i.e. a locally closed embedding
\[
(\text{z supported on U}) \disjoint (\text{1 point at p}) \disjoint (\text{both points at p}) \to \hilb_{C_{/k}}^2
.\]

The first locus is given by the complement of two lines:

![Locus 1](figures/2020-02-06-13:08.png){width=350px}

The third locus is given by arrows at $p$ pointing in any direction, which gives a copy of $\PP^1$.
The second is $\PP^1$ minus two points.
Above each point is a nodal cubic with two marked points, and moving the base point towards a line correspond to moving one of the points toward the node:

![Moving base toward the point](figures/2020-02-06-13:11.png){width=350px}

More precisely, we're considering the cover $\PP^1 \setminus\text{2 points} \to C$ and thinking about ways in which two points and approach the missing points.
These give specific tangent directions at the node on the cubic, depending on how this approach happens -- either both points approach missing point #1, both approach missing point #2, or each approach a separate missing point.
:::

:::{.remark}
Useful example to think about. Not normal, reduced, but glued in a weird way.
Possibly easier to think about: cuspidal cubic.
:::
  

## Representability


Recall the following definition:

:::{.definition title="$m\dash$Regularity"}
A coherent sheaf $F$ on $\PP_k^n$ for $k$ a field is $m\dash$regular iff $H^i(F(m-i)) = 0$ for all $i> 0$.
:::


:::{.proposition title="?"}
For every Hilbert polynomial $P$, there exists some $m_0$ depending on $P$  such that any $Z \subset \PP^n_k$ with $P_Z = P$ satisfies $I_Z$ is $m\dash$regular.
:::
  


:::{.remark title="1"}
$F$ is $m\dash$ regular iff $\bar F = F \cross_{\spec k} \spec \bar k$ is $m\dash$regular.
:::


:::{.remark title="2"}
The $m_0$ produced does not depend on $k$.
:::



:::{.lemma title="?"}
For $m_0 = m_0(P)$ and $N = N(P)$, we have an embedding as a subfunctor
\[
\hilb_{\PP^m_\ZZ}^P \to \Gr(N, H^0( \PP^n_\ZZ, \OO(m_0)  )\dual )
.\]
:::

For any $Z \subset \PP^n_S$ flat over $S$ with $P_{Z_s} = P$ for all $s\in S$ points, we want to send this to
$$
0\to R\dual \to \OO_s \tensor H^0(\PP^n_\ZZ, \OO(m_0))\dual \to Q \to 0
$$
or equivalently
$$
0 \to Q\dual \to \OO_s \tensor H^0(\PP^n_\ZZ, \OO(m_0)) \to R \to 0
$$
with $R$ locally free.


So instead of the quotient $Q$ being locally free, we can ask for the sub $Q\dual$ to be locally free instead, which is a weaker condition.

We thus send $Z$ to
$$
0 \to \pi_{s*} I_Z(m_0) \to \pi_{s*} \OO_{\PP^n_s}(m_0) = \OO_s \tensor H^0(\PP^n, \OO(m_0))
$$
which we obtain by taking the pushforward from this square:


\begin{tikzcd}
\PP^n_s \arrow[dd, "\pi_s"] \arrow[rr] &  & \PP^n_Z \arrow[dd] \\
                                &  &                    \\
S \arrow[rr]                           &  & \spec \ZZ
\end{tikzcd}

We have a sequence $0 \to I_Z(m_0) \to \OO(m_0) \to \OO_Z(m_0) \to 0$.
Thus we get a sequence

\[
0 \to \pi_{s*}I_Z(m_0) \to \pi_{s*}\OO(m_o) \to \pi_{s*} \OO_Z(m_0) \to R^1 \pi_{s*}I_Z(m_0) \to \cdots
.\]

### Step 1

\[
R^1\pi_* I_Z(m_0) = 0
.\]

By base change, it's enough to show that $H^1(Z_s, I_{Z_s}(m_0)) = 0$.
This follows by $m_0\dash$regularity.

### Step 2

$\pi_{s*}I_Z(m_0)$ and $\pi_{s*} \OO_Z(m_0)$ are locally free.
For all $i>0$, we have

- $R^i \pi_{s*} I_Z(m_0) = 0$ by $m_0\dash$regularity,
- $R^i \pi_{s*} \OO(m_0) = 0$ by base change,
- and thus $R^i \pi_{s*} \OO_Z(m_0) = 0$.

### Step 3

$\pi_{s*}I_Z(m_0)$ has rank $N = N(P)$.

Again by base change, there is a map $\pi_* I_Z(m_0) \tensor k(s) \to H^0(Z_S, I_{Z_s}(m_0))$ which we know is an isomorphism.
Because $h^i ( I_{Z_S}(m_0) ) = 0$ for $i>0$ by $m\dash$regularity and
\[
h^0(I_{Z_S}(m_0)) = P_\OO(m_0) - P_{\OO_{Z_s}}(m_0) = P_\OO (m_0) - P(m_0)
.\]

This yields a well-defined functor 
\[
\hilb_{\PP^n_\ZZ}^P \to \Gr(N, H^0(\PP^n, \OO(m_0))\dual )
.\]

:::{.remark}
Note that we've just said what happens to objects; strictly speaking we should define what happens for morphisms, but they're always give by pullback.
:::

We want to show injectivity, i.e. that we can recover $Z$ from the data of a number f polynomials vanishing on it, which is the data $0 \to \pi_{s*} I_Z(m_0) \to \OO_s \tensor H^0(\PP^n, \OO(m_0))$.

Given
$$
0 \to Q\dual \to \OO_s \tensor H^0(\PP^n, \OO(m_0)) = \pi_{s*} \OO_{\PP^n_S}(m_0)
$$
we get a diagram


\begin{tikzcd}
\pi_{s}^* Q\dual \arrow[rrdd] \arrow[rrr] &  &                          & \OO_{\PP^n_s}(m_0) \\
                                  &  &                          &                    \\
                                  &  & I(m_0) \arrow[ruu, hook] &
\end{tikzcd}


where $Q\dual = \pi_{s*} I_Z(m_0)$, so we're looking at

\begin{tikzcd}
Q\dual = \pi_{s*}^* \pi_{s*} I_Z(m_0) 
  \arrow[rrdd, twoheadrightarrow] 
  \arrow[rrr] 
&  
&                          
& \OO_{\PP^n_s}(m_0) 
\\
&  
&                          
&                    
\\
&  
& I(m_0) 
  \arrow[ruu, hook] 
&
\end{tikzcd}

The surjectivity here follows from $\OO_{Z_s} \tensor H^0(I_{Z_s}(m_0)) \to I_{Z_s}(m_0)$ (?).
Given a universal family $G = \Gr( N, H^0(\OO(m_0))\dual )$ and $Q\dual \subset \OO_G \tensor H^0(\OO(m_0))\dual$, we obtain $I_W \subset \OO_G$ and $W \subset \PP^n_G$.