# Hilbert Polynomials (Thursday January 23)

Some facts about the Hilbert polynomial:

1. For a subscheme $Z \subset \PP_k^n$ with $\deg P_z = \dim Z = n$, then
  $$
  p_z(t) = \deg z \frac{t^n}{n!} + O(t^{n-1})
  .$$

2. We have $p_z(t) = \chi(\OO_z(t))$, consider the sequence
  $$
  0 \to I_z(t) \to \OO_{\PP^n}^{(t)} \to \OO_z^{(t)} \to 0
  ,$$
  then $\chi(I_z(t)) = \dim H^0( \PP^n, J_z(t)  )$ for $t \gg 0$, and $p_z(0)$ is the Euler characteristic of $\OO_Z$.

:::{.remark}
Keywords to look up here: Serre vanishing, Riemann-Roch, ideal sheaf.
:::

:::{.example title="The twisted cubic"}

\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1, scale=0.6, every node/.style={scale=0.6}]

\node (myfirstpic) at (325,200) {\includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_17-27}};
\node[scale=2.0] at (400, 180) {$C$};
\node[scale=2.0] at (200, 0) {$\PP^3$};


\draw[very thick, blue] (-50,400) -- (-50,100);
\draw[thick] (-50-20,400) -- (-50+20,400);
\draw[thick] (-50-20,100) -- (-50+20,100);
\node[scale=2.0] at (-25, 100-20) {$\PP^1$};

\draw [thick, right hook-latex ] (-50+20, 200) -- (150, 200);
\node[scale=2.0] at (50, 180) {$\iota$};

\end{tikzpicture}

Then
\[
p_C(t) = (\deg C)t + \chi(\OO_{\PP^1}) = 3t + 1
.\]

:::

### Hypersurfaces

Recall that length 2 subschemes of $\PP^1$ are the same as specifying quadratics that cut them out, each such $Z \subset \PP^1$ satisfies $Z = V(f)$ where $\deg f = d$ and $f$ is homogeneous.
So we'll be looking at $\PP H^0(\PP^n_k, \OO(d))\dual$, and the guess would be that this is $\hilb_{\PP^n_k}$
Resolve the structure sheaf

\[
0 \to \OO_{\PP^n}(-d) \to \OO_{\PP^n}(t) \to \OO_D(t) \to 0
.\]
so we can twist to obtain
\[
0 \to \OO_{\PP^n}(t-d) \to \OO_{\PP^n}(t) \to \OO_D(t) \to 0
.\]
Then
\[
\chi(\OO_D(t)) = \chi(\OO_{\PP^n}(t)) - \chi(\OO_{\PP^n}(t-d))
,\]
which is
\[
{n+t \choose n} - {n+t-d \choose n} = \frac{dt^{n-1}}{(n-1)!} + O(t^{n-2})
.\]

:::{.lemma title="?"}
Anything with the Hilbert polynomial of a degree $d$ hypersurface is in fact a degree $d$ hypersurface.
:::

We want to write a morphism of functors
\[
\hilb_{\PP^n_k}^{P_{n, d}} \to \PP H^0 (\PP^n, \OO(d) )\dual
.\]
which sends flat families to families of equations cutting them out.
Want
$$
Z \subset \PP^n \cross S \to \OO_s \tensor H^0( \PP^n, \OO(d) )\dual \to L \to 0
.$$
This happens iff
$$
0 \to L\dual \to \OO_s \tensor H^0(\PP^n, \OO(d))
$$
with torsion-free quotient.
Note that we use $L\dual$ instead of $\OO_s$ because of scaling.
We have
\[
0 \to I_z &\to \OO_{\PP^n \cross S} \to \OO_z \to 0 \\
0 \to I_z(d) &\to \OO_{\PP^n \cross S}(d) \to \OO_z(d) \to 0 \quad\text{by twisting}
.\]
We then consider $\pi_s: \PP^n \cross S \to S$, and apply the pushforward to the above sequence.
Notie that it is not right-exact:

\begin{tikzcd}
  0 
  \ar[r]  
& \pi_{s*} I_z(d) 
  \ar[r] 
& \pi_{s*} \OO_{\PP^n \cross S}(d) 
  \ar[r] 
& \pi_{s*} \OO_z(d) 
  \ar[r] 
&  0 
\\
& & & & 
\\
  \ar[uu, equal]0 
  \ar[r] 
& \OO_s \tensor H^0(\PP^n, \OO(d)) 
  \ar[uu, equal]L\dual = 
  \ar[uu, equal]
  \ar[r] 
& 
  \ar[uu, equal]\text{locally free} 
  \ar[r] 
& 0
\ar[uu, equal]
\end{tikzcd}

\todo[inline]{Note: above diagram may be off horizontally?}

This equality follows from flatness, cohomology, and base change.
In particular, we need the following:

:::{.fact}
The scheme-theoretic fibers, given by $H^0(\PP^n, I_z(d))$ and $H^0(\PP^n, \OO_z(d))$, are all the same dimension.
:::
     

Using

1. Cohomology and base change, i.e. for $X \mapsvia{f} Y$ a map of Noetherian schemes (or just finite-type) and $F$ a sheaf on $X$ which is flat over $Y$, there is a natural map (not usually an isomorphism)
$$
R^i f_* f \tensor k(y) \to H^i(x_y, \restrictionof{F}{x_y})
,$$
but is an isomorphism if $\dim H^i(x_y, \restrictionof{F}{x_y})$ is constant, in which case $R^i f_* f$ is locally free.

2. If $Z \subset \PP^n_k$ is a degree $d$ hypersurface, then independently we know
  $$
  \dim H^0(\PP^n, I_z(d)) = 1 \text{ and } \dim H^0(\PP^n, \OO_z(d)) = {d+n \choose n} - 1
  .$$


To get a map going backwards, we take the universal degree 2 polynomial and form
$$
V(a_{00} x_0^2 + a_{11} x_1^2 + a_{12}x_2^2 + a_{01}x_0 x_1 + a_{02} x_0 x_2 + a_{12} x_1 x_2) \subset \PP^2 \cross \PP^5
.$$

### Example: Twisted Cubics

Consider a map $\PP^1 \to \PP^3$ obtained by taking a basis of a homogeneous cubic polynomial.
The canonical example is
\[
(x, y) \to (x^3, x^2y, xy^2, y^3)
.\]
Then
\[
P_C(t) = 3t + 1
\]
and $\hilb_{\PP_k^3}^{3t+1}$ has a component with generic point a twisted cubic, and another component with points a curve disjoint union a point, and the overlap are nodal curves with a "fat" 3-dimensional point:

![Components of the Hilbert Scheme](figures/2020-01-23-13:20.png)\

Then $P_{C'} = 1 + \tilde P$, the Hilbert polynomial of just the base without the disjoint point, so this equals $1 + P_{2, 3} = 1 + (3t + 0) = 3t +1$.
For $P_{C''}$, we take the sequence
$$
0 \to k \to \OO_{C''} \to \OO_{C'' \text{reduced}} \to 0
,$$
so
$$
P_{C''} = 1 + P_{C'' \text{red}} = 3t+1
.$$

:::{.remark}
Note that flat families *must* have the same (constant) Hilbert polynomial.
:::

Note that we can get paths in this space from $C\to C''$ and $C'\to C''$ by collapsing a twisted cubic onto a plane, and sending a disjoint point crashing into the node on a nodal cubic.
We're mapping $\PP^1 \to \PP^3$, and there is a natural action of $\PGL(4) \actson \PP^3$, so we get a map
\[
\PGL(4) \cross \PP^3 \to \PP^3
.\]

Let $c\in \PP^3$ and let $\mcc$ be the preimage.
This induces (?) a map
\[
\PGL(4) \to \hilb_{\PP^3}^{3t+1}
\]
where the fiber over $[C]$ in the latter is $\PGL(2) = \Aut(\PP^1)$.
By dimension counting, we find that the dimension of the twisted cubic component is $15 - 3 = 12$.
The 15 in the other component comes from 3-dim choices of plane, 3-dim choices of a disjoint point, and
\[
\PP H^0(\PP^2, \OO(3))\dual \cong \PP^9
,\]
yielding 15 dimensions.
To show that these are actually different components, we use Zariski tangent spaces.
Let $T_1$ be the tangent space of the twisted cubic component, then
$$
\dim T_1 \hilb_{\PP_k^3}^{3t+1} = 12
,$$
and similarly the dimension of the tangent space over the $C'$ component is 15.

:::{.fact}
Let $A$ be Noetherian and local, then the dimension of the Zariski tangent space, $\dim \mfm /\mfm^2 \geq \dim A$, the Krull dimension.
If this is an equality, then $A$ is regular.
:::

:::{.slogan}
Dimensions of tangent spaces give an upper bound.
:::

:::{.proposition title="?"}
If $X_{/k}$ is projective and $P$ is a Hilbert polynomial, then $[Z] \in \hilb_{X_{/k}}^P$, i.e. a closed subscheme of $X$ with Hilbert polynomial $p$ (note there's an ample bundle floating around) then the tangent space is $\hom_{\OO_x}(I_z, \OO_z)$.
:::