# VE and PE

A lot of things in algebraic geometry (if not all) can be thought of taking some commutative algebra construction and ‘schemifying’ it; that is making some construction about schemes that reduces to the original commutative algebra construction in the affine case.  The functors $\mathbb{V}E$ and $\mathbb{P}E$ are no exceptions.

### Commutative Algebra $\mathbb{V}E$

Let be R a ring and E an R module.  Denote the category of R algebras as $\mathcal{A}_R$.  Then define the functor

$\mathbb{V}E \colon \mathcal{A}_R \to$ SETS

$A \mapsto \hom_{mod}(E,A)$

Then $\mathbb{V}E$ is represented by $S^\bullet E$, the symmetric algebra on E.  That is, for any map $\phi \colon E \to A$, there is a canonical inclusion $E \to S^\bullet E$ and a unique map $S^\bullet E \to A$ making the following diagram commute.

$\begin{array}{ccc} E & \to & A \\ \downarrow & \nearrow & \mbox{} \\ S^\bullet E & \mbox{} & \mbox{} \end{array}$

### Scheme $\mathbb{V}E$

Now E is qcoh on a scheme X. For $T \in \mbox{Sch}_X$, let $p_T \colon T \to X$ be the map to the base.

Define the functor

$\begin{array}{ccc} \mathbb{V}E \colon \mbox{Sch}_X & \longrightarrow & \mbox{SETS} \\ T & \mapsto & \hom_{\mathcal{O}_X}(E, p_{T*}\mathcal{O}_T) \\ \mbox{} & \mbox{} & =\hom_{\mathcal{O}_T}(p^*_TE,\mathcal{O}_T) \end{array}$

This functor is represented by $\mbox{Spec}_X S^\bullet E$ (said relative spec; the construction of relative spec was one of the applications Ogus did of the all the open subfunctor business).  That is, $S^\bullet E$ is a qcho sheaf of $\mathcal{O}_X$ algebras, there is a sheaf map $E \to S^\bullet E$ over X, and for any sheaf map $E \to p_{T*}\mathcal{O}_T$ there is a unique map etc. making the following diagram commute

$\begin{array}{ccc} E & \to & p_{T*}\mathcal{O}_T \\ \downarrow & \nearrow & \mbox{} \\ S^\bullet E & \mbox{} & \mbox{} \end{array}$

And (by universal property of relative spec) such a diagram gives a unique scheme map $T \to \mbox{Spec}_X S^\bullet E$.

Again, by adjunction I can look at sheaf maps into $\mathcal{O}_T$ instead, then any $a \in \Gamma(T, \mathcal{O}_T) =: \mathbb{A}^1(T)$, there is a sheaf map $\mathcal{O}_T \to \mathcal{O}_T$ given by multiplication by a, and as $p_T^* E$ is an $\mathcal{O}_T$ module, multiplication by a also gives a sheaf map to itself; denote these sheaf maps as $a_T, a_E$ respectively.  Then there is a commutative diagram

$\begin{array}{ccc} p_T^*E & \xrightarrow{a_E} & p_T^*E \\ \downarrow & \mbox{} & \downarrow \\ \mathcal{O}_T & \xrightarrow{a_T} & \mathcal{O}_T \end{array}$

Or, in other words, $\mathbb{V}E$ is an $\mathbb{A}^1$ module. Note, $\mathbb{A}^1$ is represented by $\mbox{Spec}_X \mathcal{O}_X[t]$.

Exercise: As $\mathbb{V}E$ has a module structure, the scheme which represents it $\mbox{Spec}_X S^\bullet E$ has a group action and a multiplication action of $\mathbb{A}^1$, this is encoded as saying there are morphisms

$\mbox{Spec}_X S^\bullet E \times \mbox{Spec}_X S^\bullet E \to \mbox{Spec}_X S^\bullet E$

$\mbox{Spec}_X S^\bullet E \times \mbox{Spec}_X \mathcal{O}_X[t] \to \mbox{Spec}_X S^\bullet E$

Describe the backwards map on structure sheaves.

### Commutative Algebra $\mathbb{P}E$

The motivation is that $\mathbb{P}E$ should be ‘lines’ through $\mathbb{V}E$.  So consider

$\begin{array}{cc} \mathbb{P}E(k) & = \mbox{lines through 0 in }\hom(E,k) \\ \mbox{} & = \frac{\hom(E,k) \backslash 0}{k^*} \\ \mbox{} & = \mbox{hyperplanes in E} \end{array}$

So to get scheme $\mathbb{P}E$ the notion of removing 0 and modding out by units need to be defined.

### Scheme $\mathbb{P}E$

There is always the 0 map $E \to \mathcal{O}_X$, and by universal property of relative spec, this gives a section $X \to \mbox{Spec}_X S^\bullet E$, this is the ‘zero section.’  This zero section characterizes the set of points of X where all sections of E vanish ( or some generating set for E vanish).  In particular, for a given map $m \colon E \to p_{T*} \mathcal{O}_T$ (recall this gives a map $T \to \mbox{Spec}_X S^\bullet E$), if there is some point of X such that all sections of E map to 0, then the map $T \to \mbox{Spec}_X S^\bullet E$ must hit the zero section.

Now consider the complement of the zero section; it represents a functor Ogus calls $\mathbb{U}E$.  Now $\mathbb{U}E$ is a subfunctor of $\mathbb{V}E$ and in particular, by the previous paragraph, $\mathbb{U}E(T)$ consists of sheaf maps $E \to p_{T*}\mathcal{O}_T$ where for every point of X, some section of E doesn’t vanish.  Now by adjunction, this becomes a sheaf map $p_T^*E \to \mathcal{O}_T$, and this has the property that for every point of T, there is some section of E that doesn’t vanish there (i.e. not in the maximal ideal of the local ring).  In other words, every point has a small enough nbd s.t. some section of E maps to a unit in $\mathcal{O}_T$, i.e. the sheaf map is locally surjective, hence surjective.  So

$\mathbb{U}E(T) = \{m\colon p_T^*E \to \mathcal{O}_T | m \mbox{ is surjective } \}$

so in particular, $\mathbb{U}E(T)$ consists of sheaf maps $E \to p_{T*}\mathcal{O}_T$, or by adjunction $p_T^*E \to \mathcal{O}_T$.

The analogue to modding out by units requires the scheme $\mathbb{G}_m := \mbox{Spec}k[x,1/x]$.  Then $\mathbb{P}E$ can be defined as taking the quotient of $\mathbb{U}E$ by some action of $\mathbb{G}_m$, but a more explicit definition comes from defining

a hyperplane $H$ of a qcoh sheaf of modules $G$ is a submodule such that $G/H$ is invertible, and an invertible quotient is a surjection $G \to L$ where $L$ is invertible.  Then $G \to L$ is isomorphic to $G \to L'$ if $L \cong L'$.  Then

$\begin{array}{rl} \mathbb{P}E(T) & =\{\mbox{hyperplanes in } p_T^*E \} \\ \mbox{} & =\{\mbox{isom. classes of inv. quot.}\} \end{array}$

Enough to show its representable when applied to affines.  Two things need to be shown (using Thm1 from last post)

### PE is representable

1. Show $\mathbb{P}E$ is a sheaf (this is true essentially because qcho sheafs are determined locally)
2. $\mathbb{P}E$ has an open cover by representable open subfunctors.

For the second step, introduce $\mathbb{P}_e E(T) := \{H \subset p_T^* E =: E_T| \overline{e} \in E_T/H \mbox{ is nowhere zero} \}$, where $e \in \Gamma(X,E)$, note $\bar e$ defines an isomorphism $E_T/H \cong \mathcal{O}_T$.  Now show $\mathbb{P}_e E$ is an open subfunctor, this amounts to looking at some fiber product diagrams an showing $\mathbb{P}_eE(T)$ is represented by $\{ t \in T | 0 \ne \bar e (t) \in (E_T/H)_t \}$ where $\bar e(t)$ is the restriction of $\bar e$ to the stalk at $t \in T$.

Now check that $\mathbb{P}_e E$ is represented by $\mbox{Spec}_X S^\bullet E/(e - 1)$, and if $\{e_i\}$ generate E, then $\{\mathbb{P}_{e_i} E\}$ cover $\mathbb{P}E$.

It seems like you also have $\mathbb{P}E \cong \mbox{Proj} S^\bullet E$ for a ring R and E and R module.