• Course notes 

• General reference 

• Hilbert schemes/functors of points: , .

• Slightly more detailed: 
• Curves on surfaces: 

• Moduli of Curves:  (chatty and less rigorous)

# 2 Schemes vs Representable Functors (Thursday January 9th)

Last time: fix an $$S{\hbox{-}}$$scheme, i.e. a scheme over $$S$$. Then there is a map \begin{align*} {\operatorname{Sch}}_{_{/S}} &\to {\operatorname{Fun}}( {\operatorname{Sch}}_{_{/S}}^{\operatorname{op}}, {\operatorname{Set}}) \\ x &\mapsto h_x(T) = \hom_{{\operatorname{Sch}}_{_{/S}} }(T, x) .\end{align*}

where $$T' \xrightarrow{f} T$$ is given by

\begin{align*} h_x(f): h_x(T) &\to h_x(T') \\ \qty{ T \mapsto x } &\mapsto \text{triangles of the form} .\end{align*} ## 2.1 Representability

\begin{align*}\hom_{{\operatorname{Fun}}}(h_x, F) = F(x).\end{align*}

\begin{align*}\hom_{{\operatorname{Sch}}_{/S}}(x, y) \cong \hom_{{\operatorname{Fun}}}(h_x, h_y).\end{align*}

A moduli functor is a map

\begin{align*} F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}&\to {\operatorname{Set}}\\ F(x) &= \text{ "Families of something over $x$" } \\ F(f) &= \text{"Pullback"} .\end{align*}

A moduli space for that “something” appearing above is an $$M \in \mathrm{Obj}({\operatorname{Sch}}_{/S})$$ such that $$F \cong h_M$$.

Now fix $$S = \operatorname{Spec}(k)$$, and write $$h_m$$ for the functor of points over $$M$$. Then \begin{align*} h_m(\operatorname{Spec}(k)) = M(\operatorname{Spec}(k)) \cong \text{families over } \operatorname{Spec}k = F(\operatorname{Spec}k) .\end{align*}

$$h_M(M) \cong F(M)$$ are families over $$M$$, and $$\operatorname{id}_M \in \mathrm{Mor}_{{\operatorname{Sch}}_{/S}}(M, M) = \xi_{Univ}$$ is the universal family.

Every family is uniquely the pullback of $$\xi_{\text{Univ}}$$. This makes it much like a classifying space. For $$T\in {\operatorname{Sch}}_{/S}$$, \begin{align*} h_M &\xrightarrow{\cong} F \\ f\in h_M(T) &\xrightarrow{\cong} F(T) \ni \xi = F(f)(\xi_{\text{Univ}}) .\end{align*}

where $$T\xrightarrow{f} M$$ and $$f = h_M(f)(\operatorname{id}_M)$$.

If $$M$$ and $$M'$$ both represent $$F$$ then $$M \cong M'$$ up to unique isomorphism. which shows that $$f, g$$ must be mutually inverse by using universal properties.

A length 2 subscheme of $${\mathbb{A}}^1_k$$ (??) then \begin{align*} F(S) = \left\{{ V(x^2 + bx + c)}\right\} \subset {\mathbb{A}}^5 \end{align*} where $$b, c \in {\mathcal{O}}_s(s)$$, which is functorially bijective with $$\left\{{b, c \in {\mathcal{O}}_s(s)}\right\}$$ and $$F(f)$$ is pullback. Then $$F$$ is representable by $${\mathbb{A}}_k^2(b, c)$$ and the universal object is given by \begin{align*} V(x^2 + bx + c) \subset {\mathbb{A}}^1(?) \times{\mathbb{A}}^2(b, c) \end{align*} where $$b, c \in k[b, c]$$. Moreover, $$F'(S)$$ is the set of effective Cartier divisors in $${\mathbb{A}}_5'$$ which are length 2 for every geometric fiber. $$F''(S)$$ is the set of subschemes of $${\mathbb{A}}_5'$$ which are length 2 on all geometric fibers. In both cases, $$F(f)$$ is always given by pullback.

Problem: $$F''$$ is not a good moduli functor, as it is not representable. Consider $$\operatorname{Spec}k[\varepsilon]$$, for which we have the following situation:  We think of $$T_p F^{', ''}$$ as the tangent space at $$p$$. If $$F$$ is representable, then it is actually the Zariski tangent space.  Moreover, $$T_p M = ({\mathfrak{m}}_p / {\mathfrak{m}}_p^2)^\vee$$, and in particular this is a $$k{\hbox{-}}$$vector space. To see the scaling structure, take $$\lambda \in k$$.

\begin{align*} \lambda: k[\varepsilon] & \to k[\varepsilon] \\ \varepsilon &\mapsto \lambda \varepsilon \\ \\ \lambda^*: \operatorname{Spec}(k[\varepsilon]) &\to \operatorname{Spec}(k[\varepsilon]) \\ \\ \lambda: M(\operatorname{Spec}(k[\varepsilon])) &\to M(\operatorname{Spec}(k[\varepsilon])) .\end{align*} Conclusion: If $$F$$ is representable, for each $$p\in F(\operatorname{Spec}k)$$ there exists a unique point of $$T_p F$$ that are invariant under scaling.

If $$F, F', G \in {\operatorname{Fun}}( ({\operatorname{Sch}}_{/S})^{\operatorname{op}}, {\operatorname{Set}})$$, there exists a fiber product where \begin{align*} (F \times_G F')(T) = F(T) \times_{G(T)} F'(T) .\end{align*}

This works with the functor of points over a fiber product of schemes $$X \times_T Y$$ for $$X, Y \to T$$, where \begin{align*} h_{X \times_T Y}= h_X \times_{h_t} h_Y .\end{align*}

If $$F, F', G$$ are representable, then so is the fiber product $$F \times_G F'$$.

For any functor \begin{align*} F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}} ,\end{align*} for any $$T \xrightarrow{f} S$$ there is an induced functor \begin{align*} F_T: ({\operatorname{Sch}}_{/T}) &\to {\operatorname{Set}}\\ x &\mapsto F(x) .\end{align*}

$$F$$ is representable by $$M_{/S}$$ implies that $$F_T$$ is representable by $$M_T = M \times_S T / T$$.

## 2.2 Projective Space

Consider $${\mathbb{P}}^n_{\mathbb{Z}}$$, i.e. “rank 1 quotient of an $$n+1$$ dimensional free module.”

$${\mathbb{P}}^n_{/{\mathbb{Z}}}$$ represents the following functor \begin{align*} F: {\operatorname{Sch}}^{\operatorname{op}}&\to {\operatorname{Set}}\\ S &\mapsto \left\{{ {\mathcal{O}}_S^{n+1} \to L \to 0 }\right\} / \sim .\end{align*}

where $$\sim$$ identifies diagrams of the following form: and $$F(f)$$ is given by pullbacks.

$${\mathbb{P}}^n_{/S}$$ represents the following functor: \begin{align*} F_S: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}&\to {\operatorname{Set}}\\ T &\mapsto F_S(T) = \left\{{ {\mathcal{O}}_T^{n+1} \to L \to 0}\right\} / \sim .\end{align*}

This gives us a cleaner way of gluing affine data into a scheme.

### 2.2.1 Proof of Proposition

Note that $${\mathcal{O}}^{n+1} \to L \to 0$$ is the same as giving $$n+1$$ sections $$s_1, \cdots s_n$$ of $$L$$, where surjectivity ensures that they are not the zero section. So \begin{align*} F_i(S) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0}\right\}/\sim ,\end{align*} with the additional condition that $$s_i \neq 0$$ at any point. There is a natural transformation $$F_i \to F$$ by forgetting the latter condition, and is in fact a subfunctor.1

It is enough to show that each $$F_i$$ and each $$F_{ij}$$ are representable, since we have natural transformations: and each $$F_{ij} \to F_i$$ is an open embedding on the level of their representing schemes.

For $$n=1$$, we can glue along open subschemes For $$n=2$$, we get overlaps of the following form: This claim implies that we can glue together $$F_i$$ to get a scheme $$M$$. We want to show that $$M$$ represents $$F$$. $$F(s)$$ (LHS) is equivalent to an open cover $$U_i$$ of $$S$$ and sections of $$F_i(U_i)$$ satisfying the gluing (RHS). Going from LHS to RHS isn’t difficult, since for $${\mathcal{O}}_s^{n+1} \to L \to 0$$, $$U_i$$ is the locus where $$s_i \neq 0$$ and by surjectivity, this gives a cover of $$S$$. The RHS to LHS comes from gluing.

We have \begin{align*} F_i(S) = \left\{{{\mathcal{O}}_S^{n+1} \to L \cong {\mathcal{O}}_s \to 0, s_i \neq 0}\right\} ,\end{align*} but there are no conditions on the sections other than $$s_i$$.
So specifying $$F_i(S)$$ is equivalent to specifying $$n-1$$ functions $$f_1 \cdots \widehat{f}_i \cdots f_n \in {\mathcal{O}}_S(s)$$ with $$f_k \neq 0$$. We know this is representable by $${\mathbb{A}}^n$$. We also know $$F_{ij}$$ is obviously the same set of sequences, where now $$s_j \neq 0$$ as well, so we need to specify $$f_0 \cdots \widehat{f}_i \cdots f_j \cdots f_n$$ with $$f_j \neq 0$$. This is representable by $${\mathbb{A}}^{n-1} \times{\mathbb{G}}_m$$, i.e. $$\operatorname{Spec}k[x_1, \cdots, \widehat{x}_i, \cdots, x_n, x_j^{-1}]$$. Moreover, $$F_{ij} \hookrightarrow F_i$$ is open.

What is the compatibility we are using to glue? For any subset $$I \subset \left\{{0, \cdots, n}\right\}$$, we can define \begin{align*} F_I = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0, s_i\neq 0 \text{ for } i\in I}\right\} = {\prod{i\in I}}_F F_i ,\end{align*} and $$F_I \to F_J$$ when $$I \supset J$$.

# 3 Functors as Spaces (Tuesday January 14th)

Last time: representability of functors, and specifically projective space $${\mathbb{P}}_{/{\mathbb{Z}}}^n$$ constructed via a functor of points, i.e.

\begin{align*} h_{{\mathbb{P}}^n_{/{\mathbb{Z}}} }: {\operatorname{Sch}}^{\operatorname{op}}&\to {\operatorname{Set}}\\ s &\mapsto {\mathbb{P}}^n_{/{\mathbb{Z}}}(s) = \left\{{ {\mathcal{O}}_s^{n+1} \to L \to 0}\right\} .\end{align*}

for $$L$$ a line bundle, up to isomorphisms of diagrams: That is, line bundles with $$n+1$$ sections that globally generate it, up to isomorphism. The point was that for $$F_i \subset {\mathbb{P}}_{/{\mathbb{Z}}}^n$$ where \begin{align*} F_i(s) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0 {~\mathrel{\Big|}~}s_i \text{ is invertible}}\right\} \end{align*} are representable and can be glued together, and projective space represents this functor.

Because projective space represents this functor, there is a universal object: and other functors are pullbacks of the universal one. (Moduli Space)

Show that $${\mathbb{P}}_{/{\mathbb{Z}}}^n$$ is proper over $$\operatorname{Spec}{\mathbb{Z}}$$. Use the evaluative criterion, i.e. there is a unique lift ## 3.1 Generalizing Open Covers

For a category $$C$$, we say a diagram $$X \to Y \rightrightarrows Z$$ is an equalizer iff it is universal with respect to the following property: where $$X$$ is the universal object.

For sets, $$X = \left\{{y {~\mathrel{\Big|}~}f(y) = g(y)}\right\}$$ for $$Y \xrightarrow{f, g} Z$$.

A coequalizer is the dual notion, Take $$C = {\operatorname{Sch}}_{/S}$$, $$X_{/S}$$ a scheme, and $$X_\alpha \subset X$$ an open cover. We can take two fiber products, $$X_{\alpha \beta}, X_{\beta, \alpha}$$: These are canonically isomorphic.

In $${\operatorname{Sch}}_{/S}$$, we have where \begin{align*} f_{\alpha\beta}: X_{\alpha\beta} &\to X_\alpha \\ g_{\alpha\beta}: X_{\alpha\beta} &\to X_\beta ;\end{align*}

form a coequalizer. Conversely, we can glue schemes. Given $$X_\alpha \to X_{\alpha\beta}$$ (schemes over open subschemes), we need to check triple intersections: Then $$\varphi_{\alpha\beta}: X_{\alpha\beta}\xrightarrow{\cong} X_{\beta\alpha}$$ must satisfy the cocycle condition:

Maps $$\varphi_{\alpha\beta}: X_{\alpha\beta}\xrightarrow{\cong} X_{\beta\alpha}$$ satisfy the cocycle condition iff

1. \begin{align*}\varphi_{\alpha\beta}^{-1}\qty{ X_{\beta\alpha} \cap X_{\beta\gamma} } = X_{\alpha\beta} \cap X_{\alpha \gamma},\end{align*} noting that the intersection is exactly the fiber product $$X_{\beta\alpha} \times_{X_\beta} X_{\beta \gamma}$$.

2. The following diagram commutes: Then there exists a scheme $$X_{/S}$$ such that $${\coprod}_{\alpha\beta} X_{\alpha\beta} \rightrightarrows {\coprod}X_\alpha \to X$$ is a coequalizer; this is the gluing.

Subfunctors satisfy a patching property because morphisms to schemes are locally determined. Thus representable functors (e.g. functors of points) have to be (Zariski) sheaves.

A functor $$F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}$$ is a Zariski sheaf iff for any scheme $$T_{/S}$$ and any open cover $$T_\alpha$$, the following is an equalizer: \begin{align*} F(T) \to \prod F(T_\alpha) \rightrightarrows \prod_{\alpha\beta} F(T_{\alpha\beta}) \end{align*} where the maps are given by restrictions.

Any representable functor is a Zariski sheaf precisely because the gluing is a coequalizer. Thus if you take the cover \begin{align*} {\coprod}_{\alpha\beta} T_{\alpha\beta} \to {\coprod}_{\alpha}T_\alpha \to T ,\end{align*} since giving a local map to $$X$$ that agrees on intersections if enough to specify a map from $$T\to X$$.

Thus any functor represented by a scheme automatically satisfies the sheaf axioms.

Suppose we have a morphism $$F' \to F$$ in the category $${\operatorname{Fun}}({\operatorname{Sch}}_{/S}, {\operatorname{Set}})$$.

• This is a subfunctor if $$\iota(T)$$ is injective for all $$T_{/S}$$.

• $$\iota$$ is open/closed/locally closed iff for any scheme $$T_{/S}$$ and any section $$\xi \in F(T)$$ over $$T$$, then there is an open/closed/locally closed set $$U\subset T$$ such that for all maps of schemes $$T' \xrightarrow{f} T$$, we can take the pullback $$f^* \xi$$ and $$f^*\xi \in F'(T')$$ iff $$f$$ factors through $$U$$.

This says that we can test if pullbacks are contained in a subfunctors by checking factorization. This is the same as asking if the subfunctor $$F'$$, which maps to $$F$$ (noting a section is the same as a map to the functor of points), and since $$T\to F$$ and $$F' \to F$$, we can form the fiber product $$F' \times_F T$$: and $$F' \times_F T \cong U$$. Note: this is almost tautological! Thus $$F' \to F$$ is open/closed/locally closed iff $$F' \times_F T$$ is representable and $$g$$ is open/closed/locally closed. I.e. base change is representable.

1. If $$F' \to F$$ is open/closed/locally closed and $$F$$ is representable, then $$F'$$ is representable as an open/closed/locally closed subscheme

2. If $$F$$ is representable, then open/etc subschemes yield open/etc subfunctors

Treat functors as spaces.

We have a definition of open, so now we’ll define coverings.

A collection of open subfunctors $$F_\alpha \subset F$$ is an open cover iff for any $$T_{/S}$$ and any section $$\xi \in F(T)$$, i.e. $$\xi: T\to F$$, the $$T_\alpha$$ in the following diagram are an open cover of $$T$$: Given \begin{align*} F(s) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0}\right\} \end{align*} and $$F_i(s)$$ given by those where $$s_i \neq 0$$ everywhere, the $$F_i \to F$$ are an open cover. Because the sections generate everything, taking the $$T_i$$ yields an open cover.

## 3.2 Results About Zariski Sheaves

A Zariski sheaf $$F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}$$ with a representable open cover is representable.

Let $$F_\alpha \subset F$$ be an open cover, say each $$F_\alpha$$ is representable by $$x_\alpha$$. Form the fiber product $$F_{\alpha\beta} = F_\alpha \times_F F_\beta$$. Then $$x_\beta$$ yields a section (plus some openness condition?), so $$F_{\alpha\beta} = x_{\alpha\beta}$$ representable. Because $$F_\alpha \subset F$$, the $$F_{\alpha\beta} \to F_\alpha$$ have the correct gluing maps.

This follows from Yoneda (schemes embed into functors), and we get maps $$x_{\alpha\beta} \to x_\alpha$$ satisfying the gluing conditions. Call the gluing scheme $$x$$; we’ll show that $$x$$ represents $$F$$. First produce a map $$x\to F$$ from the sheaf axioms. We have a map $$\xi \in \prod_\alpha F(x_\alpha)$$, and because we can pullback, we get a unique element $$\xi \in F(X)$$ coming from the diagram \begin{align*} F(x) \to \prod F(x_\alpha) \rightrightarrows \prod_{\alpha\beta} F(x_{\alpha\beta}) .\end{align*}

If $$E \to F$$ is a map of functors and $$E, F$$ are Zariski sheaves, where there are open covers $$E_\alpha \to E, F_\alpha \to F$$ with commutative diagrams (i.e. these are isomorphisms locally), then the map is an isomorphism.

With the following diagram, we’re done by the lemma: For $$S$$ and $$E$$ a locally free coherent $${\mathcal{O}}_s$$ module, \begin{align*} {\mathbb{P}}E(T) = \left\{{f^* E \to L \to 0}\right\} / \sim \end{align*} is a generalization of projectivization, then $$S$$ admits a cover $$U_i$$ trivializing $$E$$. Then the restriction $$F_i \to {\mathbb{P}}E$$ were $$F_i(T)$$ is the above set if $$f$$ factors through $$U_i$$ and empty otherwise. On $$U_i$$, $$E \cong {\mathcal{O}}_{U_i}^{n_i}$$, so $$F_i$$ is representable by $${\mathbb{P}}_{U_i}^{n_i - 1}$$ by the proposition. Note that this is clearly a sheaf.

For $$E$$ locally free over $$S$$ of rank $$n$$, take $$r<n$$ and consider the functor \begin{align*} {\operatorname{Gr}}(k, E)(T) = \left\{{f^*E \to Q \to 0}\right\} /\sim \end{align*} (a Grassmannian) where $$Q$$ is locally free of rank $$k$$.

1. Show that this is representable

2. For the Plucker embedding \begin{align*} {\operatorname{Gr}}(k, E) \to {\mathbb{P}}\wedge^k E ,\end{align*} a section over $$T$$ is given by $$f^*E \to Q \to 0$$ corresponding to \begin{align*} \wedge^k f^*E \to \wedge^k Q \to 0 ,\end{align*} noting that the left-most term is $$f^* \wedge^k E$$. Show that this is a closed subfunctor.

That it’s a functor is clear, that it’s closed is not.

Take $$S = \operatorname{Spec}k$$, then $$E$$ is a $$k{\hbox{-}}$$vector space $$V$$, then sections of the Grassmannian are quotients of $$V \otimes{\mathcal{O}}$$ that are free of rank $$n$$. Take the subfunctor $$G_w \subset {\operatorname{Gr}}(k, V)$$ where \begin{align*} G_w(T) = \left\{{{\mathcal{O}}_T \otimes V \to Q \to 0}\right\} \text{ with } Q \cong {\mathcal{O}}_t\otimes W \subset {\mathcal{O}}_t \otimes V .\end{align*} If we have a splitting $$V = W \oplus U$$, then $$G_W = {\mathbb{A}}(\hom(U, W))$$. If you show it’s closed, it follows that it’s proper by the exercise at the beginning.

Thursday: Define the Hilbert functor, show it’s representable. The Hilbert scheme functor gives e.g. for $${\mathbb{P}}^n$$ of all flat families of subschemes.

# 4 Thursday January 16th

## 4.1 Subfunctors

A functor $$F' \subset F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}$$ is open iff for all $$T \xrightarrow{\xi} F$$ where $$T = h_T$$ and $$\xi \in F(T)$$.

We can take fiber products: So we can think of “inclusion in $$F$$” as being an open condition: for all $$T_{/S}$$ and $$\xi \in F(T)$$, there exists an open $$U \subset T$$ such that for all covers $$f: T' \to T$$, we have \begin{align*} F(f)(\xi) = f^*(\xi) \in F'(T') \end{align*} iff $$f$$ factors through $$U$$.

Suppose $$U \subset T$$ in $${\operatorname{Sch}}/T$$, we then have \begin{align*} h_{U/T}(T') = \begin{cases} \emptyset & T' \to T \text{ doesn't factor } \\ {\{\operatorname{pt}\}}& \text{otherwise} \end{cases} .\end{align*}

which follows because the literal statement is $$h_{U/T}(T') = \hom_T(T', U)$$. By the definition of the fiber product, \begin{align*} (F' \times_F T)(T') = \left\{{ (a,b) \in F'(T) \times T(T) {~\mathrel{\Big|}~}\xi(b) = \iota(a) \text{ in } F(T)}\right\} ,\end{align*} where $$F' \xrightarrow{\iota} F$$ and $$T \xrightarrow{\xi} F$$. So note that the RHS diagram here is exactly given by pullbacks, since we identify sections of $$F/T'$$ as sections of $$F$$ over $$T/T'$$ (?). We can thus identify \begin{align*} (F' \times_F T)(T') = h_{U_{/S}}(T') ,\end{align*} and so for $$U \subset T$$ in $${\operatorname{Sch}}_{/S}$$ we have $$h_{U_{/S}} \subset h_{T_{/S}}$$ is the functor of maps that factor through $$U$$. We just identify $$h_{U_{/S}}(T') = \hom_S(T', U)$$ and $$h_{T_{/S}}(T') = \hom_S(T', T)$$.

$${\mathbb{G}}_m, {\mathbb{G}}_a$$. The scheme/functor $${\mathbb{G}}_a$$ represents giving a global function, $${\mathbb{G}}_m$$ represents giving an invertible function. where $$T' = \left\{{f\neq 0}\right\}$$ and $${\mathcal{O}}_T(T)$$ are global functions.

## 4.2 Actual Geometry: Hilbert Schemes

The best moduli space!

Unless otherwise stated, assume all schemes are Noetherian.

We want to parameterize families of subschemes over a fixed object. Fix $$k$$ a field, $$X_{/k}$$ a scheme; we’ll parameterize subschemes of $$X$$.

The Hilbert functor is given by \begin{align*} \operatorname{Hilb}_{X_{/S}}: ({\operatorname{Sch}}_{/S})^{op} \to {\operatorname{Set}} \end{align*} which sends $$T$$ to closed subschemes $$Z \subset X \times_S T \to T$$ which are flat over $$T$$.

Here flatness will replace the Cartier condition:

For $$X \xrightarrow{f} Y$$ and $${\mathbb{F}}$$ a coherent sheaf on $$X$$, $$f$$ is flat over $$Y$$ iff for all $$x\in X$$ the stalk $$F_x$$ is a flat $${\mathcal{O}}_{y, f(x)}{\hbox{-}}$$module.

Note that $$f$$ is flat if $${\mathcal{O}}_x$$ is. Flatness corresponds to varying continuously. Note that everything works out if we only play with finite covers.

If $$X_{/k}$$ is projective, so $$X \subset {\mathbb{P}}^n_k$$, we have line bundles $${\mathcal{O}}_x(1) = {\mathcal{O}}(1)$$. For any sheaf $$F$$ over $$X$$, there is a Hilbert polynomial $$P_F(n) = \chi(F(n)) \in {\mathbb{Z}}[n]$$, i.e. we twist by $${\mathcal{O}}(1)$$ $$n$$ times. The cohomology of $$F$$ isn’t changed by the pushforward into $${\mathbb{P}}_n$$ since it’s a closed embedding, and so \begin{align*} \chi(X, F) = \chi({\mathbb{P}}^n, i_* F) = \sum (-1)^i \dim_k H^i({\mathbb{P}}^n, i_* F(n)) .\end{align*}

For $$n \gg 0$$, $$\dim_k H^0 = \dim M_n$$, the $$n$$th graded piece of $$M$$, which is a graded module over the homogeneous coordinate ring whose $$i_*F = \tilde M$$.

In general, for $$L$$ ample of $$X$$ and $$F$$ coherent on $$X$$, we can define a Hilbert polynomial, \begin{align*} P_F(n) = \chi(F\otimes L^n) .\end{align*}

This is an invariant of a polarized projective variety, and in particular subschemes. Over irreducible bases, flatness corresponds to this invariant being constant.

For $$f:X\to S$$ projective, i.e. there is a factorization: If $$S$$ is reduced, irreducible, locally Noetherian, then $$f$$ is flat $$\iff$$ $$P_{{\mathcal{O}}_{x_s}}$$ is constant for all $$s\in S$$.

To be more precise, look at the base change to $$X_1$$, and the pullback of the fiber? $${\mathcal{O}}\mathrel{\Big|}_{x_i}$$? Note that we’re not using the word “integral” here! $$S$$ is flat $$\iff$$ the Hilbert polynomial over the fibers are constant.

The zero-dimensional subschemes $$Z \in {\mathbb{P}}^n_k$$, then $$P_Z$$ is the length of $$Z$$, i.e. $$\dim_k({\mathcal{O}}_Z)$$, and \begin{align*} P_Z(n) = \chi({\mathcal{O}}_Z \otimes{\mathcal{O}}(n)) = \chi({\mathcal{O}}_Z) = \dim_k H^0(Z; {\mathcal{O}}_Z) = \dim_k {\mathcal{O}}_Z(Z) .\end{align*}

For two closed points in $${\mathbb{P}}^2$$, $$P_Z = 2$$. Consider the affine chart $${\mathbb{A}}^2 \subset {\mathbb{P}}^2$$, which is given by \begin{align*} \operatorname{Spec}k[x, y]/(y, x^2) \cong k[x]/(x^2) \end{align*} and $$P_Z = 2$$. I.e. in flat families, it has to record how the tangent directions come together.

Consider the flat family $$xy = 1$$ (flat because it’s an open embedding) over $$k[x]$$, here we have points running off to infinity.

A sheaf $$F$$ is flat iff $$P_{F_S}$$ is constant.

## 4.3 Proof That Flat Sheaves Have Constant Hilbert Polynomials

Assume $$S = \operatorname{Spec}A$$ for $$A$$ a local Noetherian domain.

For $$F$$ a coherent sheaf on $$X_{/A}$$ is flat, we can take the cohomology via global sections $$H^0(X; F(n))$$. This is an $$A{\hbox{-}}$$module, and is a free $$A{\hbox{-}}$$module for $$n\gg 0$$.

Assumed $$X$$ was projective, so just take $$X = {\mathbb{P}}_A^n$$ and let $$F$$ be the pushforward. There is a correspondence sending $$F$$ to its ring of homogeneous sections constructed by taking the sheaf associated to the graded module

\begin{align*} \sum_{n\gg0} H^0( {\mathbb{P}}_A^m; F(n) ) = \bigoplus_{n \gg 0} H^0({\mathbb{P}}_A^m; F(n)) \end{align*} and taking the associated sheaf ($$Y \mapsto \tilde Y$$, as per Hartshorne’s notation) which is free, and thus $$F$$ is free.2

Conversely, take an affine cover $$U_i$$ of $$X$$. We can compute the cohomology using Čech cohomology, i.e. taking the Čech resolution. We can also assume $$H^i({\mathbb{P}}^m; F(n)) = 0$$ for $$n \gg 0$$, and the Čech complex vanishes in high enough degree. But then there is an exact sequence \begin{align*} 0 \to H^0({\mathbb{P}}^m; F(n)) \to \mathcal C^0( \underline{U}; F(n) ) \to \cdots \to C^m( \underline{U}; F(n) ) \to 0 .\end{align*} Assuming $$F$$ is flat, and using the fact that flatness is a 2 out of 3 property, the images of these maps are all flat by induction from the right. Finally, local Noetherian and finitely generated flat implies free.

By the lemma, we want to show $$H^0({\mathbb{P}}^m; F(n))$$ is free for $$n\gg 0$$ iff the Hilbert polynomials on the fibers $$P_{F_S}$$ are all constant.

It suffices to show that for each point $$s\in \operatorname{Spec}A$$, we have \begin{align*} H^0(X_s; F_S(n)) = H^0(X; F(n)) \otimes k(S) \end{align*} for $$k(S)$$ the residue field, for $$n\gg 0$$.

$$P_{F_S}$$ measures the rank of the LHS.

$$\implies$$: The dimension of RHS is constant, whereas the LHS equals $$P_{F_S}(n)$$.

$$\impliedby$$: If the dimension of the RHS is constant, so the LHS is free.

For a f.g. module over a local ring, testing if localization at closed point and generic point have the same rank. For $$M$$ a finitely generated module over $$A$$, we find that \begin{align*} 0 \to A^n \to M \to Q \end{align*} is surjective after tensoring with $$\mathrm{Frac}(A)$$, and tensoring with $$k(S)$$ for a closed point, if $$\dim A^n = \dim M$$ then $$Q = 0$$.

By localizing, we can assume $$s$$ is a closed point. Since $$A$$ is Noetherian, its ideal is f.g. and we have \begin{align*} A^m \to A \to k(S) \to 0 .\end{align*} We can tensor with $$F$$ (viewed as restricting to fiber) to obtain \begin{align*} F(n)^m \to F(n) \to F_S(n) \to 0 .\end{align*} Because $$F$$ is flat, this is still exact. We can take $$H^*(x, {\,\cdot\,})$$, and for $$n\gg 0$$ only $$H^0$$ survives. This is the same as tensoring with $$H^0(x, F(n))$$.

Given a polynomial $$P \in {\mathbb{Z}}[n]$$ for $$X_{/S}$$ projective, we define a subfunctor by picking only those with Hilbert polynomial $$p$$ fiberwise as $$\operatorname{Hilb}^P_{X_{/S}} \subset \operatorname{Hilb}_{X_{/S}}$$. This is given by $$Z \subset X \times_S T$$ with $$P_{Z} = P$$.

If $$S$$ is Noetherian and $$X_{/S}$$ projective, then $$\operatorname{Hilb}_{X_{/S}}^P$$ is representable by a projective $$S{\hbox{-}}$$scheme.

See cycle spaces in analytic geometry.

# 5 Hilbert Polynomials (Thursday January 23)

Some facts about the Hilbert polynomial:

1. For a subscheme $$Z \subset {\mathbb{P}}_k^n$$ with $$\deg P_z = \dim Z = n$$, then \begin{align*} p_z(t) = \deg z \frac{t^n}{n!} + O(t^{n-1}) .\end{align*}

2. We have $$p_z(t) = \chi({\mathcal{O}}_z(t))$$, consider the sequence \begin{align*} 0 \to I_z(t) \to {\mathcal{O}}_{{\mathbb{P}}^n}^{(t)} \to {\mathcal{O}}_z^{(t)} \to 0 ,\end{align*} then $$\chi(I_z(t)) = \dim H^0( {\mathbb{P}}^n, J_z(t) )$$ for $$t \gg 0$$, and $$p_z(0)$$ is the Euler characteristic of $${\mathcal{O}}_Z$$.

Keywords to look up here: Serre vanishing, Riemann-Roch, ideal sheaf. Then \begin{align*} p_C(t) = (\deg C)t + \chi({\mathcal{O}}_{{\mathbb{P}}^1}) = 3t + 1 .\end{align*}

### 5.0.1 Hypersurfaces

Recall that length 2 subschemes of $${\mathbb{P}}^1$$ are the same as specifying quadratics that cut them out, each such $$Z \subset {\mathbb{P}}^1$$ satisfies $$Z = V(f)$$ where $$\deg f = d$$ and $$f$$ is homogeneous. So we’ll be looking at $${\mathbb{P}}H^0({\mathbb{P}}^n_k, {\mathcal{O}}(d))^\vee$$, and the guess would be that this is $$\operatorname{Hilb}_{{\mathbb{P}}^n_k}$$ Resolve the structure sheaf

\begin{align*} 0 \to {\mathcal{O}}_{{\mathbb{P}}^n}(-d) \to {\mathcal{O}}_{{\mathbb{P}}^n}(t) \to {\mathcal{O}}_D(t) \to 0 .\end{align*} so we can twist to obtain \begin{align*} 0 \to {\mathcal{O}}_{{\mathbb{P}}^n}(t-d) \to {\mathcal{O}}_{{\mathbb{P}}^n}(t) \to {\mathcal{O}}_D(t) \to 0 .\end{align*} Then \begin{align*} \chi({\mathcal{O}}_D(t)) = \chi({\mathcal{O}}_{{\mathbb{P}}^n}(t)) - \chi({\mathcal{O}}_{{\mathbb{P}}^n}(t-d)) ,\end{align*} which is \begin{align*} {n+t \choose n} - {n+t-d \choose n} = \frac{dt^{n-1}}{(n-1)!} + O(t^{n-2}) .\end{align*}

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 \begin{align*} \operatorname{Hilb}_{{\mathbb{P}}^n_k}^{P_{n, d}} \to {\mathbb{P}}H^0 ({\mathbb{P}}^n, {\mathcal{O}}(d) )^\vee .\end{align*} which sends flat families to families of equations cutting them out. Want \begin{align*} Z \subset {\mathbb{P}}^n \times S \to {\mathcal{O}}_s \otimes H^0( {\mathbb{P}}^n, {\mathcal{O}}(d) )^\vee\to L \to 0 .\end{align*} This happens iff \begin{align*} 0 \to L^\vee\to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(d)) \end{align*} with torsion-free quotient. Note that we use $$L^\vee$$ instead of $${\mathcal{O}}_s$$ because of scaling. We have \begin{align*} 0 \to I_z &\to {\mathcal{O}}_{{\mathbb{P}}^n \times S} \to {\mathcal{O}}_z \to 0 \\ 0 \to I_z(d) &\to {\mathcal{O}}_{{\mathbb{P}}^n \times S}(d) \to {\mathcal{O}}_z(d) \to 0 \quad\text{by twisting} .\end{align*} We then consider $$\pi_s: {\mathbb{P}}^n \times S \to S$$, and apply the pushforward to the above sequence. Notie that it is not right-exact: This equality follows from flatness, cohomology, and base change. In particular, we need the following:

The scheme-theoretic fibers, given by $$H^0({\mathbb{P}}^n, I_z(d))$$ and $$H^0({\mathbb{P}}^n, {\mathcal{O}}_z(d))$$, are all the same dimension.

Using

1. Cohomology and base change, i.e. for $$X \xrightarrow{f} Y$$ 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) \begin{align*} R^i f_* f \otimes k(y) \to H^i(x_y, {\left.{{F}} \right|_{{x_y}} } ) ,\end{align*} but is an isomorphism if $$\dim H^i(x_y, {\left.{{F}} \right|_{{x_y}} } )$$ is constant, in which case $$R^i f_* f$$ is locally free.

2. If $$Z \subset {\mathbb{P}}^n_k$$ is a degree $$d$$ hypersurface, then independently we know \begin{align*} \dim H^0({\mathbb{P}}^n, I_z(d)) = 1 \text{ and } \dim H^0({\mathbb{P}}^n, {\mathcal{O}}_z(d)) = {d+n \choose n} - 1 .\end{align*}

To get a map going backwards, we take the universal degree 2 polynomial and form \begin{align*} 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 {\mathbb{P}}^2 \times{\mathbb{P}}^5 .\end{align*}

### 5.0.2 Example: Twisted Cubics

Consider a map $${\mathbb{P}}^1 \to {\mathbb{P}}^3$$ obtained by taking a basis of a homogeneous cubic polynomial. The canonical example is \begin{align*} (x, y) \to (x^3, x^2y, xy^2, y^3) .\end{align*} Then \begin{align*} P_C(t) = 3t + 1 \end{align*} and $$\operatorname{Hilb}_{{\mathbb{P}}_k^3}^{3t+1}$$ 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: