# Functors as Spaces (Tuesday January 14th) Last time: representability of functors, and specifically projective space $\PP_{/\ZZ}^n$ constructed via a functor of points, i.e. \[ h_{\PP^n_{/\ZZ} }: \Sch\op &\to \Set \\ s &\mapsto \PP^n_{/\ZZ}(s) = \theset{ \OO_s^{n+1} \to L \to 0} .\] for $L$ a line bundle, up to isomorphisms of diagrams: \begin{tikzcd} \OO_{s}^{n+1} \arrow[dd, no head, Rightarrow] \arrow[rr] & & L \arrow[rr] \arrow[dd, "\cong"] & & 0 \\ & & & & \\ \OO_{s}^{n+1} \arrow[rr] & & M \arrow[rr] & & 0 \end{tikzcd} That is, line bundles with $n+1$ sections that globally generate it, up to isomorphism. The point was that for $F_i \subset \PP_{/\ZZ}^n$ where $$ F_i(s) = \theset{\OO_s^{n+1} \to L \to 0 \suchthat s_i \text{ is invertible}} $$ are representable and can be glued together, and projective space represents this functor. :::{.remark} Because projective space represents this functor, there is a universal object: \begin{tikzcd} \OO_{\PP_{\ZZ}^n}^{n+1} \arrow[rr] & & L \arrow[dd, equal] \arrow[rr] & & 0 \\ & & & & \\ & & \OO_{\PP_{\ZZ}^n}(1) & & \end{tikzcd} and other functors are pullbacks of the universal one. (Moduli Space) ::: :::{.exercise title="?"} Show that $\PP_{/\ZZ}^n$ is proper over $\spec \ZZ$. Use the evaluative criterion, i.e. there is a unique lift \begin{tikzcd} \spec k \arrow[dd] \arrow[rrr] & & & \PP^n_{\ZZ} \arrow[dd] \\ & & & \\ \spec R \arrow[rrr] \arrow[rrruu, dashed] & & & \spec \ZZ \end{tikzcd} ::: ## Generalizing Open Covers :::{.definition title="Equalizer"} 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: \begin{tikzcd} X \arrow[rr] & & Y \arrow[rr, shift left=.5ex] \ar[rr, shift right=.5ex] & & Z \\ & & & & \\ & & S \arrow[lluu, dashed, "\exists!"] \arrow[uu] \arrow[rruu] & & \end{tikzcd} where $X$ is the universal object. ::: :::{.example title="?"} For sets, $X = \theset{y \suchthat f(y) = g(y)}$ for $Y \mapsvia{f, g} Z$. ::: :::{.definition title="?"} A **coequalizer** is the dual notion, \begin{tikzcd} & & S & & \\ & & & & \\ Z \arrow[rruu] \arrow[rr, shift left=.5ex] \ar[rr, shift right=.5ex] & & Y \arrow[uu] \arrow[rr] & & X \arrow[lluu, "\exists!"', dashed] \end{tikzcd} ::: :::{.example title="?"} Take $C = \Sch_{/S}$, $X_{/S}$ a scheme, and $X_\alpha \subset X$ an open cover. We can take two fiber products, $X_{\alpha \beta}, X_{\beta, \alpha}$: \begin{tikzcd} X_\alpha \arrow[rr] & & X & & & X_\beta \arrow[rr] & & X \\ & & & & & & & \\ X_{\alpha\beta} \arrow[uu] \arrow[rr] & & X_\beta \arrow[uu] & & & X_{\beta\alpha} \arrow[uu] \arrow[rr] & & X_\alpha \arrow[uu] \end{tikzcd} These are canonically isomorphic. ::: In $\Sch_{/S}$, we have \begin{tikzcd} \disjoint_{\alpha\beta} X_{\alpha\beta} \arrow[rr, shift left=.5ex, "f_{\alpha\beta}"] \arrow[rr, shift right=.5ex,"g_{\alpha\beta}", swap] & & \disjoint_{\alpha} X_\alpha \arrow[rr] & & X \end{tikzcd} where \[ f_{\alpha\beta}: X_{\alpha\beta} &\to X_\alpha \\ g_{\alpha\beta}: X_{\alpha\beta} &\to X_\beta ;\] 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: \begin{tikzpicture}[scale=0.25] \node[draw,circle,minimum size=5cm,inner sep=0pt, label={[xshift=-1.25cm, yshift=-1.0cm] $X_\alpha$ }] at (-3,3) {}; \node[draw,circle,minimum size=5cm,inner sep=0pt, label={[xshift=1.25cm, yshift=-1.0cm] $X_\beta$ }] at (3,3) {}; \node[draw,circle,minimum size=5cm,inner sep=0pt, label={[yshift=-4.25cm] $X_\gamma$ }] at (0,-3) {}; \end{tikzpicture} Then $\varphi_{\alpha\beta}: X_{\alpha\beta}\mapsvia{\cong} X_{\beta\alpha}$ must satisfy the **cocycle condition**: :::{.definition title="Cocycle Condition"} Maps $\varphi_{\alpha\beta}: X_{\alpha\beta}\mapsvia{\cong} X_{\beta\alpha}$ satisfy the **cocycle condition** iff 1. $$\varphi_{\alpha\beta}\inv \qty{ X_{\beta\alpha} \intersect X_{\beta\gamma} } = X_{\alpha\beta} \intersect X_{\alpha \gamma},$$ noting that the intersection is exactly the fiber product $X_{\beta\alpha} \cross_{X_\beta} X_{\beta \gamma}$. 2. The following diagram commutes: \begin{tikzcd} X_{\alpha\beta} \intersect X_{\alpha\gamma} \arrow[rdd, "\varphi_{\alpha\beta}"'] \arrow[rr, "\varphi_{\alpha\gamma}"] && X_{\gamma\alpha} \intersect X_{\gamma\beta} \\ && \\ & X_{\beta\alpha}\intersect X_{\beta\gamma} \arrow[ruu, "\varphi_{\beta\gamma}"'] & \end{tikzcd} ::: Then there exists a scheme $X_{/S}$ such that $\disjoint_{\alpha\beta} X_{\alpha\beta} \rightrightarrows \disjoint X_\alpha \to X$ 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. :::{.definition title="Zariski Sheaf"} A functor $F: (\Sch_{/S})\op \to \Set$ is a **Zariski sheaf** iff for any scheme $T_{/S}$ and any open cover $T_\alpha$, the following is an equalizer: $$ F(T) \to \prod F(T_\alpha) \rightrightarrows \prod_{\alpha\beta} F(T_{\alpha\beta}) $$ where the maps are given by restrictions. ::: :::{.example title="?"} Any representable functor is a Zariski sheaf precisely because the gluing is a coequalizer. Thus if you take the cover $$ \disjoint_{\alpha\beta} T_{\alpha\beta} \to \disjoint_{\alpha}T_\alpha \to T ,$$ 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. ::: :::{.definition title="Subfunctors and Open/Closed Functors"} Suppose we have a morphism $F' \to F$ in the category $\Fun(\Sch_{/S}, \Set)$. - 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 $T' \mapsvia{f} T$, we can take the pullback $f^* \xi$ and $f^*\xi \in F'(T')$ iff $f$ factors through $U$. ::: :::{.remark} 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 $F' \cross_F T$: \begin{tikzcd} F' \ar[r] & F \\ & \\ F' \cross_F T \ar[r, "g"] \ar[uu] & T \ar[uu, "\xi" swap] \end{tikzcd} and $F' \cross_F T \cong U$. Note: this is almost tautological! Thus $F' \to F$ is open/closed/locally closed iff $F' \cross_F T$ is representable and $g$ is open/closed/locally closed. I.e. base change is representable. ::: :::{.exercise title="?"} \envlist 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 ::: :::{.slogan} Treat functors as spaces. ::: We have a definition of open, so now we'll define coverings. :::{.definition title="Open Covers"} 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 $T$: \begin{tikzcd} F_\alpha \ar[r] & F \\ & \\ T_\alpha \ar[uu] \ar[r] & T \ar[uu, "\xi" swap] \end{tikzcd} ::: :::{.example title="?"} Given $$ F(s) = \theset{\OO_s^{n+1} \to L \to 0} $$ 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. ::: ## Results About Zariski Sheaves :::{.proposition title="?"} A Zariski sheaf $F: (\Sch_{/S})\op \to \Set$ with a representable open cover is representable. ::: :::{.proof title="?"} Let $F_\alpha \subset F$ be an open cover, say each $F_\alpha$ is representable by $x_\alpha$. Form the fiber product $F_{\alpha\beta} = F_\alpha \cross_F F_\beta$. 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 \[ F(x) \to \prod F(x_\alpha) \rightrightarrows \prod_{\alpha\beta} F(x_{\alpha\beta}) .\] ::: :::{.lemma title="?"} 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 \begin{tikzcd} E \ar[r] & F \\ & \\ E_\alpha \ar[uu] \ar[r, "\cong"] & F_\alpha \ar[uu] \end{tikzcd} (i.e. these are isomorphisms locally), then the map is an isomorphism. ::: With the following diagram, we're done by the lemma: \begin{tikzcd} X \ar[r] & F \\ & \\ X_\alpha \ar[uu] \ar[r, "\cong"] & F_\alpha \ar[uu] \end{tikzcd} :::{.example title="?"} For $S$ and $E$ a locally free coherent $\OO_s$ module, \[ \PP E(T) = \theset{f^* E \to L \to 0} / \sim \] is a generalization of projectivization, then $S$ admits a cover $U_i$ trivializing $E$. Then the restriction $F_i \to \PP E$ were $F_i(T)$ is the above set if $f$ factors through $U_i$ and empty otherwise. On $U_i$, $E \cong \OO_{U_i}^{n_i}$, so $F_i$ is representable by $\PP_{U_i}^{n_i - 1}$ by the proposition. Note that this is clearly a sheaf. ::: :::{.example title="?"} For $E$ locally free over $S$ of rank $n$, take $r That it's a functor is clear, that it's closed is not. ::: Take $S = \spec k$, then $E$ is a $k\dash$vector space $V$, then sections of the Grassmannian are quotients of $V \tensor \OO$ that are free of rank $n$. Take the subfunctor $G_w \subset \Gr(k, V)$ where $$ G_w(T) = \theset{\OO_T \tensor V \to Q \to 0} \text{ with } Q \cong \OO_t\tensor W \subset \OO_t \tensor V .$$ If we have a splitting $V = W \oplus U$, then $G_W = \AA(\hom(U, W))$. 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 $\PP^n$ of all flat families of subschemes.