# References - Course notes [@bakker_8330] - General reference [@hartshorne_2010] - Hilbert schemes/functors of points: [@stromme], [@hartshorne_def]. - Slightly more detailed: [@fantechi_2005] - Curves on surfaces: [@mumford_1985] - Moduli of Curves: [@harris_morrison_1998] (chatty and less rigorous) # Schemes vs Representable Functors (Thursday January 9th) Last time: fix an $S\dash$scheme, i.e. a scheme over $S$. Then there is a map \[ \Sch_{_{/S}} &\to \Fun( \Sch_{_{/S}}\op, \Set ) \\ x &\mapsto h_x(T) = \hom_{\Sch_{_{/S}} }(T, x) .\] where $T' \mapsvia{f} T$ is given by \[ h_x(f): h_x(T) &\to h_x(T') \\ \qty{ T \mapsto x } &\mapsto \text{triangles of the form} .\] \begin{tikzcd} T' \arrow[rr] \arrow[rdd] & & X \\ & & \\ & T \arrow[ruu] & \end{tikzcd} ## Representability :::{.theorem title="?"} $$\hom_{\Fun}(h_x, F) = F(x).$$ ::: :::{.corollary title="?"} $$\hom_{\Sch_{/S}}(x, y) \cong \hom_{\Fun}(h_x, h_y).$$ ::: :::{.definition title="Moduli Functor"} A **moduli functor** is a map \[ F: (\Sch_{/S})\op &\to \Set \\ F(x) &= \text{ "Families of something over $x$" } \\ F(f) &= \text{"Pullback"} .\] ::: :::{.definition title="Moduli Space"} A **moduli space** for that "something" appearing above is an $M \in \mathrm{Obj}(\Sch_{/S})$ such that $F \cong h_M$. ::: :::{.remark} Now fix $S = \spec(k)$, and write $h_m$ for the functor of points over $M$. Then \[ h_m(\spec(k)) = M(\spec(k)) \cong \text{families over } \spec k = F(\spec k) .\] ::: :::{.remark} $h_M(M) \cong F(M)$ are families over $M$, and $\id_M \in \mathrm{Mor}_{\Sch_{/S}}(M, M) = \xi_{Univ}$ is the universal family. Every family is uniquely the pullback of $\xi_{\text{Univ}}$. This makes it much like a classifying space. For $T\in \Sch_{/S}$, \[ h_M &\mapsvia{\cong} F \\ f\in h_M(T) &\mapsvia{\cong} F(T) \ni \xi = F(f)(\xi_{\text{Univ}}) .\] where $T\mapsvia{f} M$ and $f = h_M(f)(\id_M)$. ::: :::{.remark} If $M$ and $M'$ both represent $F$ then $M \cong M'$ up to unique isomorphism. \begin{tikzcd} \xi_M & & \xi_{M'} \\ M \arrow[rr, "f"] & & M' \\ & & \\ M' \arrow[rr, "g"] & & M \\ \xi_{M'} & & \xi_M \end{tikzcd} which shows that $f, g$ must be mutually inverse by using universal properties. ::: :::{.example title="?"} A length 2 subscheme of $\AA^1_k$ (??) then $$ F(S) = \theset{ V(x^2 + bx + c)} \subset \AA^5 $$ where $b, c \in \OO_s(s)$, which is functorially bijective with $\theset{b, c \in \OO_s(s)}$ and $F(f)$ is pullback. Then $F$ is representable by $\AA_k^2(b, c)$ and the universal object is given by $$ V(x^2 + bx + c) \subset \AA^1(?) \cross \AA^2(b, c) $$ where $b, c \in k[b, c]$. Moreover, $F'(S)$ is the set of effective Cartier divisors in $\AA_5'$ which are length 2 for every geometric fiber. $F''(S)$ is the set of subschemes of $\AA_5'$ 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 $\spec k[\varepsilon]$, for which we have the following situation: \begin{tikzpicture}[scale=2.0] \begin{axis}[ hide axis, xmin=-12, xmax=18, ymin=-4, ymax=10, xtick = {0}, ytick = {0}, disabledatascaling] \draw[-][black][opacity=1] (axis cs:-10.0, 9) -- (axis cs:-10, -0); \draw[-][black][opacity=1] (axis cs:-14.0, 0) -- (axis cs:-6, -0); \draw[-][black][opacity=1] (axis cs:0.0, 9) -- (axis cs:-0, -0); \draw[-][black][opacity=1] (axis cs:-4.0, 0) -- (axis cs:4, -0); \draw[-][black][opacity=1] (axis cs:10.0, 9) -- (axis cs: 10, -0); \draw[-][black][opacity=1] (axis cs:6.0, 0) -- (axis cs:14, -0); \node[draw, circle, blue, scale=0.4, fill=blue](left1) at (axis cs:-10, 6) [anchor=center] {}; \node[right=1mm of left1,font=\tiny] {$(\eps+ x - 1)$}; \node[draw, circle, blue, scale=0.4, fill=blue](center1) at (axis cs:0, 6) [anchor=center] {}; \node[right=1mm of center1,font=\tiny] {$(x)(\eps, x-1)$}; \node[draw, circle, blue, scale=0.4, fill=blue](right1) at (axis cs:10, 6) [anchor=center] {}; \node[right=1mm of right1,font=\tiny] {$(x(x-1), \eps)$}; \node[draw, circle, blue, scale=0.4, fill=blue](left2) at (axis cs:-10, 3) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](center2) at (axis cs:0, 3) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](right2) at (axis cs:10, 3) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](left3) at (axis cs:-10, 0) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](center3) at (axis cs:0, 0) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](right3) at (axis cs:10, 0) [anchor=center] {}; \draw[-][blue, very thick][opacity=0.9] (axis cs:-10.0, 6) -- (axis cs:-7, 9); \draw[-][blue, very thick][opacity=0.9] (axis cs:-10.0, 3) -- (axis cs:-7, 3); \draw[-][blue, very thick][opacity=0.9] (axis cs:0, 3) -- (axis cs:3, 3); \draw[-][blue, very thick][opacity=0.9] (axis cs:0, 0) -- (axis cs:3, 0); \draw[-][blue, very thick][opacity=0.9] (axis cs:10, 0) -- (axis cs:13, 0); \end{axis} \end{tikzpicture} \begin{table}[H] \centering \begin{tabular}{l|lll} \hline \\ $F$ & $\checkmark$ & x & x \\ $F'$ & $\checkmark$ & x & x \\ $F''$ & $\checkmark$ & $\checkmark$ & $\checkmark$ \end{tabular} \end{table} \begin{tikzcd} \spec k \arrow[rrr, "i", hook] & & & {\spec k[\varepsilon]} & & =F'(\spec k) \arrow[rd] & \\ {F(\spec k[\varepsilon])} \arrow[rrr, "F(i)"] & & & F(\spec k) \arrow[rru] & & & =F''(\spec k) \\ & & & & & & \\ {T_p F^{', ''}}\arrow[uu, "\subset", hook]& & & P = V(x(x-1)) \arrow[uu, "\in", hook] & && \end{tikzcd} We think of $T_p F^{', ''}$ as the tangent space at $p$. If $F$ is representable, then it is actually the Zariski tangent space. \begin{tikzcd} {M(\spec k[\varepsilon])} \arrow[rr] & & M(\spec k) \\ & & \\ T_p M \arrow[rr] \arrow[uu, "\subset", hook]& & p \arrow[uu, "\subset", hook] \end{tikzcd} \begin{tikzcd} & & \spec k \arrow[rdd, "?"] \arrow[lldd, hook] & \\ & & & \\ {\spec k[\varepsilon]} \arrow[rrr] & & & {\spec \OO_{M, p} \subset M} \\ & & & k \\ & {\OO_{M, p}} \arrow[rru] \arrow[rr] & & {k[\varepsilon]} \arrow[u] \\ & \mfm_p \arrow[u, hook] & & (\varepsilon) \arrow[u, hook] \\ & \mfm_p^2 \arrow[u, hook] & & 0 \arrow[u, hook] \end{tikzcd} Moreover, $T_p M = (\mfm_p / \mfm_p^2)\dual$, and in particular this is a $k\dash$vector space. To see the scaling structure, take $\lambda \in k$. \[ \lambda: k[\varepsilon] & \to k[\varepsilon] \\ \varepsilon &\mapsto \lambda \varepsilon \\ \\ \lambda^*: \spec(k[\varepsilon]) &\to \spec(k[\varepsilon]) \\ \\ \lambda: M(\spec(k[\varepsilon])) &\to M(\spec(k[\varepsilon])) .\] \begin{tikzcd} M(\spec(k[\varepsilon])) \ar[r, "\lambda"] & M(\spec(k[\varepsilon])) \\ T_pM \ar[r] \ar[u, "\subseteq"] & T_pM \ar[u, "\subseteq"] \end{tikzcd} **Conclusion**: If $F$ is representable, for each $p\in F(\spec k)$ there exists a unique point of $T_p F$ that are invariant under scaling. :::{.remark} If $F, F', G \in \Fun( (\Sch_{/S})\op, \Set)$, there exists a fiber product \begin{tikzcd} F \cross_G F' \arrow[rr, dotted] \arrow[dd, dotted] & & F' \arrow[dd] \\ & & \\ F \arrow[rr] & & G \end{tikzcd} where $$ (F \cross_G F')(T) = F(T) \cross_{G(T)} F'(T) .$$ ::: :::{.remark} This works with the functor of points over a fiber product of schemes $X \cross_T Y$ for $X, Y \to T$, where $$ h_{X \cross_T Y}= h_X \cross_{h_t} h_Y .$$ ::: :::{.remark} If $F, F', G$ are representable, then so is the fiber product $F \cross_G F'$. ::: :::{.remark} For any functor $$ F: (\Sch_{/S})\op \to \Set ,$$ for any $T \mapsvia{f} S$ there is an induced functor \[ F_T: (\Sch_{/T}) &\to \Set \\ x &\mapsto F(x) .\] ::: :::{.remark} $F$ is representable by $M_{/S}$ implies that $F_T$ is representable by $M_T = M \cross_S T / T$. ::: ## Projective Space Consider $\PP^n_\ZZ$, i.e. "rank 1 quotient of an $n+1$ dimensional free module". :::{.proposition title="?"} $\PP^n_{/\ZZ}$ represents the following functor \[ F: \Sch\op &\to \Set \\ S &\mapsto \ts{ \OO_S^{n+1} \to L \to 0 } / \sim .\] where $\sim$ identifies diagrams of the following form: \begin{tikzcd} \OO_s^{n+1} \arrow[dd, equal] \arrow[rr] & & L \arrow[dd, "\cong"] \arrow[rr] & & 0 \\ & & & & \\ \OO_s^{n+1} \arrow[rr] & & M \arrow[rr] & & 0 \end{tikzcd} and $F(f)$ is given by pullbacks. ::: :::{.remark} $\PP^n_{/S}$ represents the following functor: \[ F_S: (\Sch_{/S})\op &\to \Sets \\ T &\mapsto F_S(T) = \theset{ \OO_T^{n+1} \to L \to 0} / \sim .\] This gives us a cleaner way of gluing affine data into a scheme. ::: ### Proof of Proposition :::{.remark} Note that $\OO^{n+1} \to L \to 0$ 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 $$ F_i(S) = \theset{\OO_s^{n+1} \to L \to 0}/\sim ,$$ 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. [^what-is-a-subfunction] [^what-is-a-subfunction]: $F\leq G$ is a subfunctor iff $F(s) \injects G(s)$. ::: :::{.claim} It is enough to show that each $F_i$ and each $F_{ij}$ are representable, since we have natural transformations: \begin{tikzcd} F_i \arrow[rr] & & F \\ & & \\ F_{ij} \arrow[rr] \arrow[uu] & & F_j \arrow[uu] \end{tikzcd} and each $F_{ij} \to F_i$ is an open embedding on the level of their representing schemes. ::: :::{.example title="?"} For $n=1$, we can glue along open subschemes \begin{tikzcd} & & F_0 \\ F_{01} \arrow[rru] \arrow[rrd] & & \\ & & F_1 \end{tikzcd} For $n=2$, we get overlaps of the following form: \begin{tikzcd} & & & & F_0 \arrow[rrdd] & & \\ & & & F_{01} \arrow[rd] \arrow[ru] & & & \\ F_{012} \arrow[rr] \arrow[rrru] \arrow[rrrd] & & F_{02} \arrow[ru] \arrow[rd] \arrow[rruu, dotted, bend left=49] \arrow[rrdd, bend right=49] & & F_1 \arrow[rr] & & F \\ & & & F_{12} \arrow[ru] \arrow[rd] & & & \\ & & & & F_2 \arrow[rruu] & & \end{tikzcd} 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 $\OO_s^{n+1} \to L \to 0$, $U_i$ is the locus where $s_i \neq 0$ and by surjectivity, this gives a cover of $S$. The RHS to LHS comes from gluing. ::: :::{.proof title="of claim"} We have $$ F_i(S) = \theset{\OO_S^{n+1} \to L \cong \OO_s \to 0, s_i \neq 0} ,$$ but there are no conditions on the sections other than $s_i$. So specifying $F_i(S)$ is equivalent to specifying $n-1$ functions $f_1 \cdots \hat f_i \cdots f_n \in \OO_S(s)$ with $f_k \neq 0$. We know this is representable by $\AA^n$. 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 $f_0 \cdots \hat f_i \cdots f_j \cdots f_n$ with $f_j \neq 0$. This is representable by $\AA^{n-1} \cross \GG_m$, i.e. $\spec k[x_1, \cdots, \hat x_i, \cdots, x_n, x_j\inv]$. Moreover, $F_{ij} \injects F_i$ is open. What is the compatibility we are using to glue? For any subset $I \subset \theset{0, \cdots, n}$, we can define $$ F_I = \theset{\OO_s^{n+1} \to L \to 0, s_i\neq 0 \text{ for } i\in I} = {\prod{i\in I}}_F F_i ,$$ and $F_I \to F_J$ when $I \supset J$. :::