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\newcommand{\containing}[0]{\supseteq} \newcommand{\cat}[1]{\mathcal{#1}} \newcommand{\thecat}[1]{\mathbf{#1}} \newcommand{\sheaf}[1]{\operatorname{\mathcal{#1}}} \newcommand\rrarrows{\rightrightarrows} \newcommand\rrrarrows{ \mathrel{\substack{\textstyle\rightarrow\\[-0.6ex] \textstyle\rightarrow \\[-0.6ex] \textstyle\rightarrow}} } \newcommand{\colim}{\operatornamewithlimits{\underset{\longrightarrow}{colim}}} \newcommand\fp[1]{\underset{\scriptscriptstyle {#1} }{\times}} \newcommand\ul[1]{\underline{#1}} \newcommand\Hc[0]{{\check{H}}} \newcommand\Cc[0]{{\check{C}}} # 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 $I \supset J$. ::: # 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. # Thursday January 16th ## Subfunctors :::{.definition title="Open Functors"} A functor $F' \subset F: (\Sch_{/S})\op \to \Set$ is **open** iff for all $T \mapsvia{\xi} F$ where $T = h_T$ and $\xi \in F(T)$. ::: We can take fiber products: \begin{tikzcd} F' \ar[r] & F \\ & \\ \parbox{3cm}{\centering $F' \cross_F T$ \\ Representable} \ar[r, "\text{Open}"] \ar[uu] & T \ar[uu] \end{tikzcd} 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 $$ F(f)(\xi) = f^*(\xi) \in F'(T') $$ iff $f$ factors through $U$. Suppose $U \subset T$ in $\Sch/T$, we then have \[ h_{U/T}(T') = \begin{cases} \emptyset & T' \to T \text{ doesn't factor } \\ \pt & \text{otherwise} \end{cases} .\] which follows because the literal statement is $h_{U/T}(T') = \hom_T(T', U)$. By the definition of the fiber product, \[ (F' \cross_F T)(T') = \theset{ (a,b) \in F'(T) \cross T(T) \suchthat \xi(b) = \iota(a) \text{ in } F(T)} ,\] where $F' \mapsvia{\iota} F$ and $T \mapsvia{\xi} F$. So note that the RHS diagram here is exactly given by pullbacks, since we identify sections of $F/T'$ as sections of $F$ over $T/T'$ (?). \begin{tikzcd} F' \ar[r, "\iota"] & F \\ & & \\ F' \cross_F T \ar[uu] \ar[r] & T\ar[uu, "\xi"] \\ & & \\ & & T' \ar[uuuul, "f\circ \xi", bend right] \ar[uul, "f"] \end{tikzcd} We can thus identify \[ (F' \cross_F T)(T') = h_{U_{/S}}(T') ,\] and so for $U \subset T$ in $\Sch_{/S}$ 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 $h_{T_{/S}}(T') = \hom_S(T', T)$. :::{.example title="?"} $\GG_m, \GG_a$. The scheme/functor $\GG_a$ represents giving a global function, $\GG_m$ represents giving an invertible function. \begin{tikzcd} \GG_m \ar[r] & \GG_a \\ & \\ T' \arrow[uur, phantom, "\scalebox{1.5}{$\llcorner$}" , very near start, color=black] \ar[uu] \ar[r] & T \ar[uu, "f \in \OO_T(T)", swap] \end{tikzcd} where $T' = \theset{f\neq 0}$ and $\OO_T(T)$ are global functions. ::: ## Actual Geometry: Hilbert Schemes > The best moduli space! :::{.warnings} 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 $X$. :::{.definition title="The Hilbert Functor"} The **Hilbert functor** is given by \[ \Hilb_{X_{/S}}: (\Sch_{/S})^{op} \to \Set \] which sends $T$ to closed subschemes $Z \subset X \cross_S T \to T$ which are flat over $T$. ::: Here **flatness** will replace the Cartier condition: :::{.definition title="Flatness"} For $X \mapsvia{f} Y$ and $\FF$ a coherent sheaf on $X$, $f$ is **flat** over $Y$ iff for all $x\in X$ the stalk $F_x$ is a flat $\OO_{y, f(x)}\dash$module. ::: :::{.remark} Note that $f$ is flat if $\OO_x$ is. Flatness corresponds to varying continuously. Note that everything works out if we only play with finite covers. ::: :::{.remark} If $X_{/k}$ is projective, so $X \subset \PP^n_k$, we have line bundles $\OO_x(1) = \OO(1)$. For any sheaf $F$ over $X$, there is a Hilbert polynomial $P_F(n) = \chi(F(n)) \in \ZZ[n]$, i.e. we twist by $\OO(1)$ $n$ times. The cohomology of $F$ isn't changed by the pushforward into $\PP_n$ since it's a closed embedding, and so \[ \chi(X, F) = \chi(\PP^n, i_* F) = \sum (-1)^i \dim_k H^i(\PP^n, i_* F(n)) .\] ::: :::{.fact} 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**, \[ P_F(n) = \chi(F\tensor L^n) .\] This is an invariant of a polarized projective variety, and in particular subschemes. Over irreducible bases, flatness corresponds to this invariant being constant. :::{.proposition title="?"} For $f:X\to S$ projective, i.e. there is a factorization: \begin{tikzcd} X \arrow[ddr, "f"] \arrow[rr, hook] & & \PP^n \cross S \ni \OO(1) \ar[ddl] \\ & & \\ & S & \end{tikzcd} If $S$ is reduced, irreducible, locally Noetherian, then $f$ is flat $\iff$ $P_{\OO_{x_s}}$ is constant for all $s\in S$. ::: :::{.remark} To be more precise, look at the base change to $X_1$, and the pullback of the fiber? $\OO\mid_{x_i}$? Note that we're not using the word "integral" here! $S$ is flat $\iff$ the Hilbert polynomial over the fibers are constant. ::: :::{.example title="?"} The zero-dimensional subschemes $Z \in \PP^n_k$, then $P_Z$ is the length of $Z$, i.e. $\dim_k(\OO_Z)$, and $$ P_Z(n) = \chi(\OO_Z \tensor \OO(n)) = \chi(\OO_Z) = \dim_k H^0(Z; \OO_Z) = \dim_k \OO_Z(Z) .$$ For two closed points in $\PP^2$, $P_Z = 2$. Consider the affine chart $\AA^2 \subset \PP^2$, which is given by $$ \spec k[x, y]/(y, x^2) \cong k[x]/(x^2) $$ and $P_Z = 2$. I.e. in flat families, it has to record how the tangent directions come together. ::: :::{.example title="?"} Consider the flat family $xy = 1$ (flat because it's an open embedding) over $k[x]$, here we have points running off to infinity. ::: :::{.proposition title="Modified Characterization of Flatness for Sheaves"} A sheaf $F$ is flat iff $P_{F_S}$ is constant. ::: ## Proof That Flat Sheaves Have Constant Hilbert Polynomials Assume $S = \spec A$ for $A$ a local Noetherian domain. :::{.lemma title="?"} 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 $A\dash$module, and is a free $A\dash$module for $n\gg 0$. ::: :::{.proof title="of lemma"} Assumed $X$ was projective, so just take $X = \PP_A^n$ 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 \ \[ \sum_{n\gg0} H^0( \PP_A^m; F(n) ) = \bigoplus_{n \gg 0} H^0(\PP_A^m; F(n)) \] and taking the associated sheaf ($Y \mapsto \tilde Y$, as per Hartshorne's notation) which is free, and thus $F$ is free. [^tilde-construction] [^tilde-construction]: See tilde construction in Hartshorne, essentially amounts to localizing free tings. 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 $H^i(\PP^m; F(n)) = 0$ for $n \gg 0$, and the Čech complex vanishes in high enough degree. But then there is an exact sequence \[ 0 \to H^0(\PP^m; F(n)) \to \mathcal C^0( \underline{U}; F(n) ) \to \cdots \to C^m( \underline{U}; F(n) ) \to 0 .\] 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 $H^0(\PP^m; F(n))$ is free for $n\gg 0$ iff the Hilbert polynomials on the fibers $P_{F_S}$ are all constant. :::{.claim title="1"} It suffices to show that for each point $s\in \spec A$, we have $$ H^0(X_s; F_S(n)) = H^0(X; F(n)) \tensor k(S) $$ for $k(S)$ the residue field, for $n\gg 0$. ::: :::{.claim title="2"} $P_{F_S}$ measures the rank of the LHS. ::: :::{.proof title="of claim 2"} $\implies$: 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 $$ 0 \to A^n \to M \to Q $$ is surjective after tensoring with $\mathrm{Frac}(A)$, and tensoring with $k(S)$ for a closed point, if $\dim A^n = \dim M$ then $Q = 0$. :::{.proof title="of claim 1"} By localizing, we can assume $s$ is a closed point. Since $A$ is Noetherian, its ideal is f.g. and we have $$ A^m \to A \to k(S) \to 0 .$$ We can tensor with $F$ (viewed as restricting to fiber) to obtain $$ F(n)^m \to F(n) \to F_S(n) \to 0 .$$ Because $F$ is flat, this is still exact. We can take $H^*(x, \wait)$, and for $n\gg 0$ only $H^0$ survives. This is the same as tensoring with $H^0(x, F(n))$. ::: :::{.definition title="Hilbert Polynomial Subfunctor"} Given a polynomial $P \in \ZZ[n]$ for $X_{/S}$ projective, we define a subfunctor by picking only those with Hilbert polynomial $p$ fiberwise as $\Hilb^P_{X_{/S}} \subset \Hilb_{X_{/S}}$. This is given by $Z \subset X \cross_S T$ with $P_{Z} = P$. ::: :::{.theorem title="Grothendieck"} If $S$ is Noetherian and $X_{/S}$ projective, then $\Hilb_{X_{/S}}^P$ is representable by a projective $S\dash$scheme. > See **cycle spaces** in analytic geometry. ::: # 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 $\hom_{\OO_x}(I_z, \OO_z)$. ::: # Hilbert Schemes of Hypersurfaces (Tuesday January 28th) Last time: Twisted cubics, given by $\hilb_{\PP^3_k}^{3t+1}$. \begin{tikzpicture} \node (node_one) at (0,0) { \includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_21-29} }; \node (a) at (0, -4) {$?$}; \node (a) at (-4.5, 3.9) {$A$}; \filldraw[blue](-4.45, 3.4) circle (0.1); \node (a) at (2.44, 2.4) {$B$}; \filldraw[blue](2.04, 2.2) circle (0.1); \node (a) at (-1.8, 1.85) {$C$}; \filldraw[red](-1.38, 1.85) circle (0.1); \node (a) at (-4., 7) {$12$}; \node (a) at (3., 5.5) {$15$}; \end{tikzpicture} > Components of the Scheme of Cubic Curves. We got lower (?) bounds on the dimension by constructing families, but want an exact dimension. The following will be a key fact: :::{.proposition title="?"} Let $Z\subset X$ be a closed $k\dash$dimensional subspace. For $[z] \in \hilb_{X_{_{/k}}}^P(k)$, we have an identification of the Zariski tangent space \[ T_{[z]} \hilb_{X_{_{/k}} }^P = \hom_{\OO_X}(I_z, \OO_Z) \] ::: Say \[ F: (\sch_{_{_{/k}}})\op \to \Set \] is a functor and let $x\in F(k)$. There is an inclusion $i: \spec k \injects \spec k[\eps]$ and an induced map \[ F(\spec k [\eps]) &\mapsvia{i^*} F(\spec k) \\ T_x F \definedas (i^*)\inv(x) &\mapsto x \] So if $F$ is represented by a scheme $H_{/k}$, then \[ T_x h_J = T_x H = (\mfm_x / \mfm_x^2)\dual \,\,\text{over } k \] Will need a criterion for flatness later, esp. for Artinian thickenings. :::{.lemma title="?"} Assume $A'$ is a Noetherian ring and $0 \to J \to A' \to A \to 0$ with $J^2 = 0$. Assume we have $X'_{/ \spec A'}$, and a coherent sheaf $F'$ on $X'$, where $X'$ is Noetherian. Then $F'$ is flat over $A'$ iff 1. $F$ is flat 2. $0 \to F\tensor_A J \to F'$ is exact. \begin{tikzcd} F & F' \\ X \da \spec A' \cross_{\spec A} X \ar[r] \ar[d] & X' \ar[d] \\ \spec A \arrow[ur, phantom, "\scalebox{1.5}{$\ulcorner$}" , very near start, color=black] \ar[r] & \spec A' \end{tikzcd} ::: ### Sketch Proof of Lemma Take the first exact sequence and tensor with $F'$ (which is right-exact), then $J \tensor_{A'} F' = J \tensor_A$ canonically. This follows because $J = J \tensor_{A'} A$, and there is an isomorphism $J \tensor_{A'} A' \to J \tensor_{A'} A$. And $F = F' \tensor_{A'} A$ is a pullback of $F'$. If flat, then tensoring is exact. Note that both conditions in the lemma are necessary since pullbacks of flats are flat by (1), and (2) gives the flatness condition. :::{.definition title="Flat Modules"} Recall that for a module over a Noetherian ring, $M/A$, $M$ is **flat** over $A$ iff \[ \tor_1^A(M, A/p) = 0 && \text{ for all primes } p .\] ::: :::{.remark} Reason: Tor commutes with direct limits, so $M$ is flat iff \[ \Tor_1^A(M, N) = 0 && \text{for all finitely generated } N .\] ::: Since $A$ is Noetherian, $N$ has a finite filtration $N^\cdot$ where $N_i / N_{i+1} \cong A/p_i$. Use the fact that every ideal is contained in a prime ideal. Take $x\in N$, this yields a map $A\to N$ which factors through $A/I$. If we make such a filtration on $A/I$, then we can quotient $N$ by $\im f$ where $f: A/I \to N$. Continuing inductively, the resulting filtration must stabilize. So we can assume $N = A/I$. Then $I$ is contained in a maximal. :::{.exercise title="?"} Finish proof. See Aatiyah Macdonald. ::: ### Proof of Proposition :::{.proof title="of proposition, given lemma"} So it's enough to show that $\tor_1^{A'}(F', A'/p') = 0$ for all primes $p' \subset A'$. :::{.observation} Since $J$ is nilpotent, $J \subset p'$. ::: ## Consequences of Proof Let $p = p'/J$, this is a prime ideal. We have an exact diagram by taking quotients: \begin{tikzcd} & & & & 0 \arrow[dd] & & 0 \arrow[dd] & & \\ & & & & & & & & \\ 0 \arrow[rr] & & J \arrow[rr] & & p' \arrow[rr] \arrow[dd] & & p \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ 0 \arrow[rr] & & J \arrow[rr] & & A' \arrow[rr] \arrow[dd] & & A \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ & & & & A'/p' \arrow[dd] & & A/p \arrow[dd] & & \\ & & & & & & & & \\ & & & & 0 & & 0 & & \end{tikzcd} So we can tensor with $F'$ everywhere, and get a map from kernels to cokernels using the snake lemma: \begin{tikzcd} & & & & 0 \arrow[dd] & & {\tor(A, F) = 0} \arrow[dd] & & \\ & & & & & & & & \\ & & 0 \arrow[rr, "\text{snake}"] \arrow[dd] & & {\tor_1^{A_1}(A'/p', F')} \arrow[dd] & & {\tor_1^{A_1}(A/p, F')} \arrow[dd] & & \\ & & & & & & & & \\ 0 \arrow[rr] & & J \tensor_{A'} F' \arrow[rr, "\text{by commuting square}", hook] & & p' \tensor_{A'} F' \arrow[rr] \arrow[dd] & & p \tensor_{A'} F' \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ 0 \arrow[rr] & & J \tensor_{A'} F' \arrow[rr, "\text{by (2)}"', hook] & & A' \tensor_{A'} F' \arrow[rr] \arrow[dd] & & A \tensor_{A'} F' \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ & & 0 \arrow[rr, "\text{snake}"] & & A'/p' \tensor_{A'} F' \arrow[dd] \arrow[rr, "="] & & A/p \tensor_{A'} F' \arrow[dd] & & \\ & & & & & & & & \\ & & & & 0 & & 0 & & \end{tikzcd} Then by (1), we have \[ \tor_1^{A'}(A'/p', F') = \tor_1^{A'}(A/p, F') = 0 .\] ::: We will just need this for $A' = k[\eps]$ and $A=k$. :::{.proposition title="?"} \[ T_z \hilb_{X_{_{/k}}} = \hom_{\OO_x}(I_z, \OO_z) .\] ::: :::{.proof title="?"} Again we have $T_z \hilb_{X_{_{/k}}} \subset \hilb_{X_{_{/k}}}(k[\eps])$, and is given by $$ \theset{Z' \subset X \cross_{\spec k} \spec k[\eps] \st Z' \text{ is flat}_{/k[\eps]},\,\, Z' \cross_{\spec k[\eps]}\spec k = Z} .$$ We have an exact diagram: \begin{tikzcd} & & 0 \arrow[r] & I_{Z'} \arrow[r] & {\OO_{X[\eps]}} \arrow[r] & \OO_{Z'} \arrow[r] & 0 \\ 0 \arrow[d] & & & {} \arrow[d] & {} \arrow[d] & {} \arrow[d] & \\ k \arrow[d] & & {} \arrow[r] & I_Z \arrow[r] \arrow[d] & \OO_x \arrow[r] \arrow[d] & \OO_z \arrow[r] \arrow[d] & {} \\ {k[\eps]} \arrow[d] & & {} \arrow[r] & I_{Z'} \arrow[r] \arrow[d] & {\OO_{x[\eps]}} \arrow[r] \arrow[d] & \OO_{Z'} \arrow[r] \arrow[d] & {} \\ k \arrow[d] & & {} \arrow[r] & I_Z \arrow[r] \arrow[d] & \OO_x \arrow[r] \arrow[d] \arrow[u, dotted, bend right] & \OO_Z \arrow[r] \arrow[d] & {} \\ 0 & & & {} & {} & {} & \end{tikzcd} Note the existence of a splitting above. Given $\phi \in \hom_{\OO_x}(I_Z, \OO_Z)$. We have \[ I_{Z'} = \left\{ f + \eps g \, \middle\vert \, \begin{array}{ll} f,g &\in I_Z, \\ \phi(f) &= g\mod I_Z, \\ \phi(f) &\in \OO_Z, \\ g\mod I_Z &\in \OO_x/I_Z = \OO_Z \end{array} \right\} .\] It's easy to see that $Z' \subset x'$, and 1. $Z'\cross k = Z$ 2. It's flat over $k[\eps]$, looking at $0 \to k\tensor I_{Z'} \to I_{Z'}$. For the converse, take $f\in I_Z$ and lift to $f' = f + \eps g \in I_{Z'}$, then $g\in \OO_x$ is well-defined wrt $I_Z$. Then $g\in \hom_{\OO_x}(I_z, \OO_z)$. ::: The main point here is that these hom sets are extremely computable. :::{.example title="?"} Let $Z$ be a twisted cubic in $\hilb_{\PP^3_{/k}}^{3t+1}(k)$. ::: :::{.observation} \[ \hom_{\OO_x}(I_Z, \OO_Z) = \hom_{\OO_X}(I_Z/I_Z^2, \OO_Z) = \hom_{\OO_Z}(I_Z/I_Z^2, \OO_Z) \] ::: If $I_Z/I_Z^2$ is locally free, these are global sections of the dual, i.e. $H^0((I_Z/I_Z^2)\dual)$. In this case, $Z\injects X$ is regularly embedded, and thus $(I_Z/I_Z^2)\dual$ should be regarded as the normal bundle. Sections of the normal bundle match up with directions to take first-order deformations: \begin{tikzpicture} \definecolor{arrow_color}{HTML}{ba0cff} \node (node_one) at (0,0) { \includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_22-40} }; \node (a) at (-7, 5) {\Huge $\PP^3$}; \node[arrow_color] (a) at (1, -3) {\Huge Deformation}; \end{tikzpicture} For $i:C \injects \PP^3$, there is an exact sequence \[ 0 \to I/I^2 \to &i^* \Omega_{\PP^3} \to \Omega_\eps \to 0 \\ &\Downarrow \quad \text{ taking duals } \\ 0 \to T_C \to &i^* T_{\PP^3} \to N_{C_{/\PP^3} } \to 0 ,\] How do we compute $T_{\PP^3}$? Fit into the exact sequence $$ 0 \to \OO \to i^* \OO(1)^4 \to i^* T_{\PP^3} \to 0 ,$$ which we can restrict to $C$. We have $i^* \OO(1) \cong \OO_{\PP^1}(3)$, so \[ 0 \to H^0 \OO_c \to &H^*(\OO(3)^4) \to H^0(i^* T_{\PP^3}) \to 0 \\ &\Downarrow \\ 0\to k \to &k^{16} \to k^{15} \to 0 .\] This yields \[ 0 \to H^0(T_c) \to &H^0(i^* T_{\PP^3}) \to H^0(N_{C_{ /\PP^3} }) \to H^1 T_c \\ &\Downarrow \\ 0\to k^3 \to &k^{15} \to k^{12} \to 0 \] :::{.example title="?"} $\hilb_{\PP^n_k}^{P_?} \cong \PP H^0(\PP^n, \OO(d))\dual$ which has dimension ${n+1 \choose n} - 1$. Pick $Z$ a $k$ point in this Hilbert scheme, then $T_Z H = \hom(I_Z, \OO_Z)$. Since $I_Z \cong \OO_{\PP}(-d)$ which fits into \[ 0 \to \OO_{\PP^n}(-d) \to \OO_{\PP^n} \to \OO_Z \to 0 .\] We can identify \[ \hom(I_Z,\OO_Z) = H^0( (I_Z/I_Z^2)\dual ) = H^0(\OO_Z(d)) .\] \begin{tikzcd} 0\ar[r] & \OO_{\PP^n}\ar[r] & \OO_{\PP^n}(d)\ar[r] & \OO_Z(d)\ar[r] & 0 \\ & & & & \\ 0\ar[r] & H^0( \OO_{\PP^n} ) \ar[r] & H^0( \OO_{\PP^n}(d) ) \ar[r] & H^0(\OO_Z(d) ) \ar[r] & 0 \\ \text{dim:} & k & k^{n+d \choose n} & k^{{n+d\choose n}-1} & \end{tikzcd} ::: :::{.example title="?"} The tangent space of the following cubic: \begin{tikzpicture} \node (node_one) at (0,0) { \includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_17-27} }; \end{tikzpicture} We can identify \[ \hom_{\OO_k}(I_Z, \OO_Z) = H^0((I_Z/I_Z^2)\dual) = 3 + H^0((I_{Z_0}/I_{Z_0}^2)\dual) ,\] where the latter equals $H^0 \qty{ \OO_1\mid_{z_0} \oplus \OO(\zeta)\mid_{z_0} }$ yielding \[ 3+9 = 12 .\] ::: # Uniform Vanishing Statements (Thursday January 30th) Recall how we constructed the Hilbert scheme of hypersurfaces $$ \hilb_{\PP_k^n}^{P_{m, d}} = \PP H^0(\PP^n; \OO(d))\dual $$ A section $\hilb_{\PP_k^n}^{P}(s)$ corresponds to $z\in \PP^n_s$. We can look at the exact sequence \[ 0 \to I_Z(m) \to \OO_{\PP_S^n} \mapsvia{\text{restrict}} \OO_z(m) \to 0 .\] as $\PP_s^n \mapsvia{\pi_s} S$, so we can pushforward along $\pi$, which is left-exact, so \[ 0 \to \pi_{s*} I_Z(m) \to \pi_{s*} \OO_{\PP_S^n} = \OO_S \tensor H^0(\PP^n; \OO(m)) \to \OO_z(m) \to R^1 \pi_{s*} I_Z(m) \to \cdots .\] *Idea:* $Z \subset \PP_k^n$ will be determined (in families!) by the space of degree $d$ polynomials vanishing on $Z$ (?), i.e. $$ H^0(\PP^n, I_z(m)) \subset H^0(\PP^n, \OO(m)) $$ for $m$ very large. This would give a map of functors $$ \hilb_{\PP_k^n}^{P} \to \Gr(N, H^0(\PP^n, \OO(m) )) .$$ If this is a closed subfunctor, a closed subfunctor of a representable functor is representable and we're done . :::{.remark} We need to get an $m$ uniform in $Z$, and more concretely: 1. First need to make sense of what it means for $Z$ to be determined by $H^0(\PP^n, I_Z(m))$ for $m$ only depending on $P$. 2. This works point by point, but we need to do this in families. I.e. we'll use the previous exact sequence, and want the $R^1$ to vanish. ::: :::{.slogan} We need *uniform* vanishing statements. There is a convenient way to package the vanishing requirements needed here. From now on, take $k=\bar k$ and $\PP^n = \PP_k^n$. ::: ## $m\dash$Regularity :::{.definition title="m-Regularity of Coherent Sheaves"} A coherent sheaf $F$ on $\PP^n$ is **$m\dash$regular** if $H^i(\PP^n; F(m-i)) = 0$ for all $i> 0$. ::: :::{.example title="?"} Consider $\OO_{\PP^n}$, this is $0\dash$regular. Line bundles on $\PP_n$ only have 0 and top cohomology. Just need to check that $H^n(\PP^n; \OO(-n)) = 0$, but by Serre duality this is \[ H^0(\PP^n; \OO(n) \tensor \omega_{\PP^n})\dual = H^0(\PP^n; \OO(-1))\dual = 0 .\] ::: :::{.proposition title="?"} Assume $F$ is $m\dash$regular. Then 1. There is a natural multiplication map from linear forms on $\PP^n$, \[ H^0(\PP^n; \OO(1)) \tensor H^0(\PP^n; F(k)) \to H^0(\PP^n; F(k+1)) ,\] which is surjective for $k\geq n$.[^graded_module_note] 2. $F$ is $m'\dash$regular for $m' \geq m$. 3. $F(k)$ is globally generated for $k\geq m$, i.e. the restriction \[ H^0(\PP^n; F(k)) \tensor \OO_{\PP^n} \to F(k) \to 0 \] is exact (i.e. surjective). [^graded_module_note]: Think of this as a graded module, this tells you the lowest number of small grade pieces needed to determine the entire thing. ::: :::{.example title="?"} $\OO$ is $m\dash$regular for $m \geq 0$ implies $\OO(k)$ is $-k\dash$regular and is also $m\dash$regular for$m\geq -k$. ::: ### Proof of 2 and 3 Induction on dimension of $n$ in $\PP^n$. Coherent sheaves on $\PP^0$ are vector spaces, so no higher cohomology. :::{.proof title="Step 1"} Take a generic hyperplane $H \subset \PP^n$, there is an exact sequence \[ 0 \to \OO(-1) \to \OO \to \OO_H \to 0 .\] where $\OO_H$ is the structure sheaf. Tensoring with $H$ remains exact, so we get \[ 0 \to F(-1) \to F \to F_H \to 0 .\] Why? $\AA^n \subset \PP^n$, let $A = \OO_{\PP^n}(\AA^n)$ be the polynomial ring over $\AA^n$. Then the restriction of the first sequence to $\AA^n$ yields $$ 0 \to A \mapsvia{f} A \to A/f \to 0 ,$$ and thus we want $$ F \mapsvia{f} F \to F/fF \to 0 $$ which results after restricting the second sequence to $\AA^n$. Thus we just want $f$ to not be a zero divisor. If we take $f$ not vanishing on any associated point of $F$, then this will be exact. Associated points: generic points arising by supports of sections of $F$. $F$ is coherent, so it has finitely many associated points. If $H$ does not contain any of the associated points of $F$, then the second sequence is indeed exact. ::: :::{.proof title="Step 2"} Twist up by $k$ to obtain \[ 0 \to F(k-1) \to F(k) \to F_H(k) \to 0 .\] Look at the LES in cohomology to get \[ H^i(F(m-i)) \to H^i(F_H(m-i)) \to H^{i+1}(F(m - (i+1))) .\] So $F_H$ is $m\dash$regular. By induction, this proves statements 1 and 2 for all $F_H$. So take $k = m+1-i$ and consider \[ H^i(F(m-i)) \to H^i(F(m+1-i)) \to H^i(F_H(m+1-i)) .\] We know 2 is satisfied, so the RHS is zero, and we know the LHS is zero, so the middle term is zero. Thus $F$ itself is $m+1$ regular, and by inducting on $m$ we get statement 2. ::: By multiplication maps, we get a commutative diagram: \begin{tikzcd} & & H^0(\OO(1)) \tensor H^0(F(k)) \arrow[dd, "\beta"] \arrow[rrr] \arrow[rrrdd] & & & H^0(\OO(1))\tensor H^0(F_H(k)) \arrow[dd] \\ & & & & & \\ H^0(F(k)) \arrow[rr, "H"] \arrow[rruu, "H \tensor \id"] & & H^0(F(k+1)) \arrow[rrr, "\alpha", dashed] & & & H^0(F_H(k+1)) \end{tikzcd} We'd like to show the diagonal map is surjective. :::{.observation} \envlist 1. The top map is a surjection, since $$ H^0(F(k)) \to H^0(F_H(k)) \to H^1(F(k-1)) = 0 $$ for $k\geq m$ by (2). 2. The right-hand map is surjective for $k\geq m$. 3. $\ker(\alpha) \subset \im(\beta)$ by a small diagram chase, so $\beta$ is surjective. This shows (1) and (2) completely. ::: :::{.proof title="of 3"} We know $F(k)$ is globally generated for $k\gg 0$. Thus for all $k\geq m$, $F(k)$ is globally generated by (1). ::: :::{.theorem title="?"} Let $P \in \QQ[t]$ be a Hilbert polynomial. There exists an $m_0$ only depending on $P$ such that for all subschemes $Z \subset \PP^n_k$ with Hilbert polynomial $P_Z = P$, the ideal sheaf $I_z$ is $m_0\dash$regular. ::: ### Proof of Theorem Induct on $n$. For $n=0$, again clear because higher cohomology vanishes and there are no nontrivial subschemes. For a fixed $Z$, pick $H$ in $\PP^n$ (and setting $I \definedas I_z$ for notation) such that \[ 0 \to I(-1) \to I \to I_H \to 0 .\] is exact. Note that the Hilbert polynomial $P_{I_H}(t) = P_I(t) - P_I(t-1)$ and $P_I = P_{\OO_{\PP^n}} - P_Z$. By induction, there exists some $m_1$ depending only on $P$ such that $I_H$ is $m_1\dash$regular. We get \[ H^{i-1}(I_H(k)) \to H^i(I(k-1)) \to H^i(I(k)) \to H^i(I_H(k)) ,\] and for $k\geq m_1 - i$ the LHS and RHS vanish so we get an isomorphism in the middle. By Serre vanishing, for $k \gg 0$ we have $H^i(I(k)) = 0$ and thus $H^i(I(k)) = 0$ for $k\geq m_i - i$. This works for all $i > 1$, we have $H^i(I(m_i - i)) = 0$. We now need to find $m_0 \geq m_1$ such that $H^1(I(m_0 - 1)) = 0$ (trickiest part of the proof). :::{.lemma title="?"} The sequence $\qty{\dim H^1(I(k))}_{k\geq m_i - 1}$ is *strictly* decreasing.[^explain_little_h] [^explain_little_h]: Note: $h^1 = \dim H^1$. ::: :::{.remark} Given the lemma, it's enough to take $m_0 \geq m_1 + h^1(I(m_1 - 1))$. Consider the LES we have a surjection $$ H^0(\OO_Z(m_1 - 1)) \to H^1(I(m_1 - 1)) \to 0 .$$ So the dimension of the LHS is equal to $P_Z(m_1 - 1)$, using the fact that terms vanish and make the Euler characteristic equal to $P_Z$. Thus we can take $m_0 = m_1 + P(m_1 - 1)$. ::: :::{.proof title="of Lemma"} Considering the LES \[ H^0(I(k+1)) \mapsvia{\alpha_{k+1}} H^0(I_H(k+1)) \to H^1(I(k)) \to H^1(I(k+1)) \to 0 ,\] where the last term is zero because $I_H$ is $m_1\dash$regular. So the sequence $h^1(I(k))$ is non-increasing. :::{.observation} If it does *not* strictly decrease for some $k$, then there is an equality on the RHS, which makes $\alpha_{k+1}$ surjective. This means that $\alpha_{k+2}$ is surjective, since \[ H^0(\OO(1)) \tensor H^0(I_H(k+1)) \surjects H^0(I_H(k+2)) .\] ::: So if one is surjective, everything above it is surjective, but by Serre vanishing we eventually get zeros. So $\alpha_{k+i}$ is surjective for all $i\geq 1$, contradicting Serre vanishing, since the RHS are isomorphisms for all $k$. ::: Thus for any $Z\subset \PP^n_k$ with $P_Z = P$, we uniformly know that $I_Z$ is $m_0\dash$regular for some $m_0$ depending only on $P$. :::{.claim} $Z$ is determined by the degree $m_0$ polynomials vanishing on $Z$, i.e. $H^0(I_z(m_0))$ as a subspace of all degree $m_0$ polynomials $H^0(\OO(m_0))$ and has fixed dimension. We have $H^i(I_Z(m_0)) = 0$ for all $i> 0$, and in particular $h^0(I_Z(m_0)) = P(m_0)$ is constant. ::: It is determined by these polynomials because we have a sequence \[ 0 \to I_Z(m_0) \to \OO(m_0) \to \OO_Z(m_0) \to 0 .\] We can get a commuting diagram over it $$ 0 \to H^0(I_Z(m_0)) \tensor \OO_{\PP^n} \to H^0(\OO(m_0)) \tensor \OO_{\PP^n} \to \cdots $$ where the middle map down is just evaluation and.the first map down is a surjection. Hence $I_Z(m_0)$, hence $\OO_Z$, hence $Z$ is determined by $H^0(I_Z(m_0))$. > Next time: we'll show that this is a subfunctor that is locally closed. # Thursday February 6th > Review base-change! For $k=\bar k$, and $C_{/k}$ a smooth projective curve, then $\hilb_{C_{/k}}^n = \sym^n C$. :::{.definition title="The Hilbert-Chow Map"} For $X_{_{/k}}$ a smooth projective *surface*, $\hilb_{X_{_{/k}}}^n \neq \sym^n X$, there is a map (the Hilbert-Chow map) \[ \hilb_{X_{_{/k}}}^n &\to \sym^n X \\ Z &\mapsto \supp(Z) \\ U = \text{reduced subschemes} &\mapsto U' = \text{ reduced multisets } \\ \PP^1 &\mapsto (x, x) .\] ::: :::{.example title="?"} Consider $\AA^2 \cross \AA^2$ under the $\ZZ/2\ZZ$ action \[ ( (x_1, y_1), (x_2, y_2)) \mapsto ((x_2, y_2), (x_1, y_1)) .\] Then \[ (\AA^2)^2 / \ZZ/2\ZZ &= \spec k[x_1, y_1, x_2, y_2]^{\ZZ/2\ZZ} \\ &= \spec k[x_1 x_2, y_1 y_2, x_1 + x_2, y_1 + y_2, x_1 y_2 + x_2 y_1, \cdots] \] with a bunch of symmetric polynomials adjoined. ::: :::{.example title="?"} Take $\AA^2$ and consider $\hilb_{\PP^2}^3$. If $I$ is a monomial ideal in $\AA^2$, there is a nice picture. We can identify the tangent space \[ T_Z \hilb_{\PP^2}^n = \hom_{\OO_{\PP^2}} ( I_2, \OO_Z) = \bigoplus \hom(I_{Z_i}, \OO_{Z_i}) .\] if $Z = \disjoint Z_i$. If $I$ is supported at 0, then we can identify the ideal with the generators it leaves out. ::: :::{.example title="?"} $I = (x^2, xy, y^2)$: ![Image](figures/2020-02-06-12:48.png){width=350px} ::: :::{.example title="?"} $I = (x^6, x^2y^2, xy^4, y^5)$: ![Image](figures/2020-02-06-12:49.png){width=350px} ::: :::{.example title="?"} $I = (x^2, y)$. Let $e=x^2, f = y$. ![Image](figures/2020-02-06-12:54.png){width=350px} By comparing rows to columns, we obtain a relation $ye = x^2 f$. Write $\OO = \theset{1, x}$, then note that this relation is trivial in $\OO$ since $y=x^2=0$. Thus $\hom(I, \OO) = \hom(k^2, k^2)$ is 4-dimensional. ::: :::{.remark} Note that $C_{_{/k}}$ for curves is an important case to know. Take $Z \subset C \cross C^n$, then quotient by the symmetric group $S^n$ (need to show this can be done), then $Z/S^n \subset C \cross \sym^n C$ and composing with the functor $\hilb$ represents yields a map $\sym^n C \to \hilb_{C_{/k}}^n$. This is bijective on points, and a tangent space computation shows it's an isomorphism. ::: :::{.example title="?"} Consider the nodal cubic in $\PP^2$: ![Nodal cubic](figures/2020-02-06-13:01.png){width=350px} > The nodal cubic $zy^2 = x^2(x+z)$. Consider the open subscheme $V \subset \hilb_{C_{/k}}^2$ of points $z \subset U$ for $U \subset C$ open. We can normalize: ![Normalized cubic](figures/2020-02-06-13:03.png){width=350px} This yields a map fro $\PP^1 \setminus\text{2 points}$. This gives us a stratification, i.e. a locally closed embedding \[ (\text{z supported on U}) \disjoint (\text{1 point at p}) \disjoint (\text{both points at p}) \to \hilb_{C_{/k}}^2 .\] The first locus is given by the complement of two lines: ![Locus 1](figures/2020-02-06-13:08.png){width=350px} The third locus is given by arrows at $p$ pointing in any direction, which gives a copy of $\PP^1$. The second is $\PP^1$ minus two points. Above each point is a nodal cubic with two marked points, and moving the base point towards a line correspond to moving one of the points toward the node: ![Moving base toward the point](figures/2020-02-06-13:11.png){width=350px} More precisely, we're considering the cover $\PP^1 \setminus\text{2 points} \to C$ and thinking about ways in which two points and approach the missing points. These give specific tangent directions at the node on the cubic, depending on how this approach happens -- either both points approach missing point #1, both approach missing point #2, or each approach a separate missing point. ::: :::{.remark} Useful example to think about. Not normal, reduced, but glued in a weird way. Possibly easier to think about: cuspidal cubic. ::: ## Representability Recall the following definition: :::{.definition title="$m\dash$Regularity"} A coherent sheaf $F$ on $\PP_k^n$ for $k$ a field is $m\dash$regular iff $H^i(F(m-i)) = 0$ for all $i> 0$. ::: :::{.proposition title="?"} For every Hilbert polynomial $P$, there exists some $m_0$ depending on $P$ such that any $Z \subset \PP^n_k$ with $P_Z = P$ satisfies $I_Z$ is $m\dash$regular. ::: :::{.remark title="1"} $F$ is $m\dash$ regular iff $\bar F = F \cross_{\spec k} \spec \bar k$ is $m\dash$regular. ::: :::{.remark title="2"} The $m_0$ produced does not depend on $k$. ::: :::{.lemma title="?"} For $m_0 = m_0(P)$ and $N = N(P)$, we have an embedding as a subfunctor \[ \hilb_{\PP^m_\ZZ}^P \to \Gr(N, H^0( \PP^n_\ZZ, \OO(m_0) )\dual ) .\] ::: For any $Z \subset \PP^n_S$ flat over $S$ with $P_{Z_s} = P$ for all $s\in S$ points, we want to send this to $$ 0\to R\dual \to \OO_s \tensor H^0(\PP^n_\ZZ, \OO(m_0))\dual \to Q \to 0 $$ or equivalently $$ 0 \to Q\dual \to \OO_s \tensor H^0(\PP^n_\ZZ, \OO(m_0)) \to R \to 0 $$ with $R$ locally free. So instead of the quotient $Q$ being locally free, we can ask for the sub $Q\dual$ to be locally free instead, which is a weaker condition. We thus send $Z$ to $$ 0 \to \pi_{s*} I_Z(m_0) \to \pi_{s*} \OO_{\PP^n_s}(m_0) = \OO_s \tensor H^0(\PP^n, \OO(m_0)) $$ which we obtain by taking the pushforward from this square: \begin{tikzcd} \PP^n_s \arrow[dd, "\pi_s"] \arrow[rr] & & \PP^n_Z \arrow[dd] \\ & & \\ S \arrow[rr] & & \spec \ZZ \end{tikzcd} We have a sequence $0 \to I_Z(m_0) \to \OO(m_0) \to \OO_Z(m_0) \to 0$. Thus we get a sequence \[ 0 \to \pi_{s*}I_Z(m_0) \to \pi_{s*}\OO(m_o) \to \pi_{s*} \OO_Z(m_0) \to R^1 \pi_{s*}I_Z(m_0) \to \cdots .\] ### Step 1 \[ R^1\pi_* I_Z(m_0) = 0 .\] By base change, it's enough to show that $H^1(Z_s, I_{Z_s}(m_0)) = 0$. This follows by $m_0\dash$regularity. ### Step 2 $\pi_{s*}I_Z(m_0)$ and $\pi_{s*} \OO_Z(m_0)$ are locally free. For all $i>0$, we have - $R^i \pi_{s*} I_Z(m_0) = 0$ by $m_0\dash$regularity, - $R^i \pi_{s*} \OO(m_0) = 0$ by base change, - and thus $R^i \pi_{s*} \OO_Z(m_0) = 0$. ### Step 3 $\pi_{s*}I_Z(m_0)$ has rank $N = N(P)$. Again by base change, there is a map $\pi_* I_Z(m_0) \tensor k(s) \to H^0(Z_S, I_{Z_s}(m_0))$ which we know is an isomorphism. Because $h^i ( I_{Z_S}(m_0) ) = 0$ for $i>0$ by $m\dash$regularity and \[ h^0(I_{Z_S}(m_0)) = P_\OO(m_0) - P_{\OO_{Z_s}}(m_0) = P_\OO (m_0) - P(m_0) .\] This yields a well-defined functor \[ \hilb_{\PP^n_\ZZ}^P \to \Gr(N, H^0(\PP^n, \OO(m_0))\dual ) .\] :::{.remark} Note that we've just said what happens to objects; strictly speaking we should define what happens for morphisms, but they're always give by pullback. ::: We want to show injectivity, i.e. that we can recover $Z$ from the data of a number f polynomials vanishing on it, which is the data $0 \to \pi_{s*} I_Z(m_0) \to \OO_s \tensor H^0(\PP^n, \OO(m_0))$. Given $$ 0 \to Q\dual \to \OO_s \tensor H^0(\PP^n, \OO(m_0)) = \pi_{s*} \OO_{\PP^n_S}(m_0) $$ we get a diagram \begin{tikzcd} \pi_{s}^* Q\dual \arrow[rrdd] \arrow[rrr] & & & \OO_{\PP^n_s}(m_0) \\ & & & \\ & & I(m_0) \arrow[ruu, hook] & \end{tikzcd} where $Q\dual = \pi_{s*} I_Z(m_0)$, so we're looking at \begin{tikzcd} Q\dual = \pi_{s*}^* \pi_{s*} I_Z(m_0) \arrow[rrdd, twoheadrightarrow] \arrow[rrr] & & & \OO_{\PP^n_s}(m_0) \\ & & & \\ & & I(m_0) \arrow[ruu, hook] & \end{tikzcd} The surjectivity here follows from $\OO_{Z_s} \tensor H^0(I_{Z_s}(m_0)) \to I_{Z_s}(m_0)$ (?). Given a universal family $G = \Gr( N, H^0(\OO(m_0))\dual )$ and $Q\dual \subset \OO_G \tensor H^0(\OO(m_0))\dual$, we obtain $I_W \subset \OO_G$ and $W \subset \PP^n_G$. # Tuesday February 18th :::{.theorem title="?"} Let $X/S$ be a projective subscheme (i.e. $X\subset \PP^n$ for some $n$). The Hilbert functor of flat families $\hilb_{X/S}^p$ is representable by a projective $S\dash$scheme. ::: :::{.remark} Note that without a fixed $P$, this is *locally* of finite type but not finite type. After fixing $P$, it becomes finite type. ::: :::{.example title="?"} For a curve of genus $g$, there is a smooth family $\mcc \mapsvia{\pi} S$ with $S$ finite-type over $\ZZ$ where every genus $g$ curve appears as a fiber. I.e., genus $g$ curves form a *bounded family* (here there are only finitely many algebraic parameters to specify a curve). How did we construct? Take the third power of the canonical bundle and show it's very ample, so it embeds into some projective space and has a Hilbert polynomial. ::: In fact, there is a finite type *moduli stack* $\mcm_g / \ZZ$ of genus $g$ curves. There will be a map $S \surjects \mcm_g$, noting that $\mcc$ is not a moduli space since it may have redundancy. We'll use the fact that a finite-type scheme surjects onto $\mcm_g$ to show it is finite type. :::{.remark} If $X/S$ is proper, we can't talk about the Hilbert polynomial, but the functor $\hilb_{X/S}$ is still representable by a locally finite-type scheme with connected components which are proper over $S$. ::: :::{.remark} If $X/S$ is *quasiprojective* (so locally closed, i.e. $X\injects \PP^n_S$), then $\hilb_{X/S}^P(T) \definedas \theset{z\in X_T \text{ projective, flat over S with fiberwise Hilbert polynomial P }}$ is still representable, but now by a quasiprojective scheme. ::: :::{.example title="?"} Length $Z$ subschemes of $\AA^1$: representable by $\AA^2$. ![Image](figures/2020-02-18-12:46.png)\ Upstairs: parametrizing length 1 subschemes, i.e. points. ::: :::{.remark} If $X\subset \PP_S^n$ and $E$ is a coherent sheaf on $X$, then \[ \Quot_{E, X/S}^{P}(T) = \theset{ j^*E \to F \to 0, \text{ over } X_T \to T,~F \text{ flat with fiberwise Hilbert polynomial } P } \] where $T \mapsvia{g} S$ is representable by an $S\dash$projective scheme. ::: :::{.example title="?"} Take $E = \OO_x$, $X$ and $S$ a point, and $E$ is a vector space, then $\Quot_{E/S}^P = \Gr(\rank, E)$. ::: :::{.warnings} The Hilbert scheme of 2 points on a surface is more complicated than just the symmetric product. ::: :::{.example title="?"} \[ \qty{\AA^2}^3 &\to \qty{\AA^2}^2 \\ \supseteq \Delta\definedas \Delta_{01} \cross \Delta_{02} &\to \qty{\AA^2}^2 \] where $\Delta_{ij}$ denote the diagonals on the $i, j$ factors. Here all associate points of $\Delta$ dominate the image, but it is not flat. Note that if we take the complement of the diagonal in the image, then the restriction $\Delta' \to \qty{\AA^2}^2\setminus D$ is in fact flat. ::: :::{.example title="Mumford"} The Hilbert scheme may have nontrivial scheme structure, i.e. this will be a "nice" Hilbert scheme with is generally not reduced. We will find a component $H$ of a $\hilb_{\PP^3_C}^P$ whose generic point corresponds to a smooth irreducible $C\subset \PP^3$ which is generically non-reduced. ::: ## Cubic Surfaces > See Hartshorne Chapter 5. Let $X\subset \PP^3$ be a smooth cubic surface, then $\OO(1)$ on $\PP^3$ restricts to a divisor class $H$ of a hyperplane section, i.e. the associated line bundle $\OO_x(H) = \OO_x(1)$. :::{.fact title="Important fact 1"} $X$ is the blowup of $\PP^2$ minus 6 points (replace each point with a curve). There is thus a blowdown map $X \mapsvia{\pi} \PP^2$. ![Image](figures/2020-02-18-13:07.png) Let $\ell = \pi^*(\text{line})$, then a fact is that $3\ell - E_1 -\cdots - E_6$ (where $E_i$ are the curves about the $p_i$) is very ample and embeds $X$ into $\PP^3$ as a cubic. ::: :::{.fact title="Important fact 2"} Every smooth cubic surface $X$ has *precisely* 27 lines. Any 6 pairwise skew lines arise as $E_1, \cdots, E_6$ as in the previous construction. ::: Take an $X$ and a line $L\subset X$. Consider any $C$ in the linear system $\abs{4H + 2L}$. Fact: $\OO(4H + 2L)$ is very ample, so embeds into a big projective space, and thus $C$ is smooth and irreducible by Bertini. Then the Hilbert polynomial of $C$ is of the form $at + b$ where $b = \chi(\OO_c)$, the Euler characteristic of the structure sheaf of $C$, and $a = \deg C$. So we'll compute these. We have $\deg C = H \cdot C$ (intersection) $= H \cdot(4H + 2L) = 4H^2 + 2H\cdot L = 4\cdot 3 + 2 = 14$. The intersections here correspond to taking hyperplane sections, intersecting with $X$ to get a curve, and counting intersection points: ![Image](figures/2020-02-18-13:14.png)\ In general, for $X$ a surface and $C\subset X$ a smooth curve, then $\omega_C = \omega_X(C)\mid_C$. Since $X\subset \PP^3$, we have \[ \omega_X &= \omega_{\PP^3}(X) \mid_X \\ &= \OO(-4) \oplus \OO(3)\mid_X \\ &= \OO_X(-1) \\ &= \OO_X(-H) .\] We also have \[ \omega_C &= \omega_X(C)\mid_X \\ &= \ro{ \qty{ \OO_X(-H) \oplus \OO_X(4H + 2L)}}{C} \\ \\ &\Downarrow \qquad \text{taking degrees} \\ \\ \deg \omega_C &= C\cdot(3H + 2L) \\ &= (4H+2L)(3H+2L) \\ &= 12H^2 + 14HL + 4L^2 \\ &= 36 + 14 + (-4) \\ &= 46 .\] Since this equals $2g(C) - 2$, we can conclude that the genus is given by $g(C) = 24$. Thus $P$ is given by $14t + (1-g) = 14t - 23$. :::{.remark} Good to know: moving a cubic surface moves the lines, you get a monodromy action, and the Weyl group of $E_6$ acts transitively so lines look the same. ::: :::{.claim title="1"} There is a flat family $Z\subset \PP^3_S$ with fiberwise Hilbert polynomial $P$ of cures of this form such that the image of the map $S \to \hilb_{\PP^3}^P$ has dimension 56. ::: :::{.proof title="of claim"} We can compute the dimension of the space of smooth cubic surfaces, since these live in $\PP H^0(\PP^3, \OO(3))$, which has dimension ${3+3\choose 3} -1 = 19$. Since there are 27 lines, the dimension of the space of such cubics with a choices of a line is also 19. Choose a general $C$ in the linear system $\abs{4H + 2L}$ will add $\dim \abs{4H + 2L} = \dim \PP H^0(x, \OO_x(C))$. We have an exact sequence \[ 0 \to \OO_X \to \OO_X(C) \to \OO_C(C) \to 0 \\ H^0\qty{ 0 \to \OO_X \to \OO_X(C) \to \OO_C(C) \to 0 } \\ .\] Since the first $H^0$ vanishes (?) we get an isomorphism. By Riemann-Roch, we have $$ \deg \OO_C(C) = C^2 = (4H+2L)^2 = 16H^2 + 16 HL + 4L^2 = 64 - 4 = 60 .$$ We can also compute $\chi(\OO_C(C)) = 60 - 23 = 37$. We have $$ h^0(\OO_C(C)) - h^1(\OO_C(C)) = h^0(\OO_C(C)) - h^0(\omega_C(-C))) = 2(23) - 60 < 0 ,$$ so there are no sections. So $\dim \abs{4H + 2L} = 37$. Thus letting $S$ be the space of cubic surfaces $X$, a line $L$, and a general $C \in \abs{4H + 2L}$, $\dim S = 56$. We get a map $S \to \hilb_{\PP^3}^P$, and we need to check that the fibers are 0-dimensional (so there are no redundancies). We then just need that every such $C$ lies on a unique cubic. Why does this have to be the case? If $C \subset X, X'$ then $C \subset X\intersect X'$ is degree 14 curve sitting inside a degree 6 curve, which can't happen. Thus if $H$ is a component of $\hilb_{\PP^3}^P$ containing the image of $S$, the $\dim H \geq 56$. ::: :::{.claim title="2"} For any $C$ above, we have $\dim T_C H = 57$. ::: When the subscheme is smooth, we have an identification with sections of the normal bundle $T_C H = H^0(C, N_{C/\PP^3})$. There's an exact sequence \[ 0 \to N_{C/X} = \OO_C(C) \to N_{C/\PP^3} \to N_{X/\PP^3}\mid_C = \OO_C(x)\mid_C = \OO_C(3H)\mid_C \to 0 .\] > Note $\omega_C = \OO_C(3H + 2L)$. As we computed, \[ H^0(\OO_C(C)) &= 37 \\ H^1(\OO_C(C)) &= 0 .\] So we need to understand the right-hand term $H^0(\OO_C(3H))$. By Serre duality, this equals $h^1(\omega_C(-3H)) = h^1(\OO_C(3L))$. We get an exact sequence \[ 0 \to \OO_X(2L-C) \to \OO_X(2L) \to \OO_C(2L) \to 0 .\] Taking homology, we have $0\to 0 \to 1 \to 1 \to 0$ since $2L-C = -4H$. Computing degrees yields $h^0 (\OO_C(3H)) = 20$. Thus the original exact sequence yields \[ 0 \to 37 \to ? \to 20 \to 0 ,\] so $? = 57$ and thus $\dim N_{C/\PP^3} = 57$. :::{.claim title="3"} \[ \dim H = 56 .\] ::: ### Proof That the Dimension is 56 Suppose otherwise. Then we have a family over $H^\mathrm{red}$ of *smooth* curves, where $f(S) \subset H^\mathrm{red}$, where the generic element is not on a cubic or any lower degree surface. Let $C'$ be a generic fiber. Then $C'$ lies on a pencil of quartics, i.e. 2 linearly independent quartics. Let $I = I_{C'}$ be the ideal of this curve in $\PP^3$, there is a SES $$ 0\to I(4) \to \OO(4) \to \OO_C(4) \to 0 .$$ It can be shown that $\dim H^0(I(4)) \geq 2$. :::{.fact} A generic quartic in this pencil is *smooth* (can be argued because of low degree and smoothness). ::: We can compute the dimension of quartics, which is ${4+3 \choose 3} - 1 = 35 - 1 = 34$. The dimension of $C'$s lying on a fixed quartic is $24$. But then the dimension of the image in the Hilbert scheme is at most $24 + 34 - 1 = 57$. It can be shown that the picard rank of such a quartic is 1, generated by $\OO(1)$, so this is a *strict* inequality, which is a contradiction since $\dim \hilb = 56$. This proves the theorem. :::{.remark} Use the fact that these curves are $K3$ surfaces? Get the fact about the generator of the Picard group from Hodge theory. So we can deform curves a bit, but not construct an algebraic family that escapes a particular cubic. ::: # Obstruction and Deformation (Tuesday February 25th) Let $k$ be a field, $X_{_{/k}}$ projective, then the $k\dash$points $\hilb_{X_{_{/k}}}^P(k)$ corresponds to closed subschemes $Z\subset X$ with hilbert polynomial $P_z = P$. Given a $P$, we want to understand the local structure of $\hilb_{X_{_{/k}}}^p$, i.e. diagrams of the form \begin{tikzcd} & & & & \hilb_{X_{_{/k}}}^P \arrow[dd] \\ & & & & \\ \spec(k) \arrow[rrrruu, "p"] \arrow[rr] & & \spec(A) \arrow[rruu, "?", dashed] \arrow[rr] & & \spec(k) \\ & & & & \\ & & A_{/k} \text{ Artinian local} \arrow[uu] & & \end{tikzcd} :::{.example title="?"} For $A = k[\eps]$, the set of extensions is the Zariski tangent space. ::: :::{.definition title="Category of Artinian Algebras"} Let $(\Art_{/k})$ be the category of local Artinian $k\dash$algebras with local residue field $k$. ::: Note that these will be the types of algebras appearing in the above diagrams. :::{.remark} This category has fiber coproducts, i.e. there are pushouts: \begin{tikzcd} C \arrow[dd] \arrow[rr] & & A \arrow[dd, dashed] \\ & & \\ B \arrow[rr, dashed] & & A \tensor_C B \end{tikzcd} There are also fibered products, \begin{tikzcd} A \cross_C B \arrow[rr, dashed] \arrow[dd, dashed] & & B \arrow[dd] \\ & & \\ A \arrow[rr] & & C \end{tikzcd} Here, $A \cross_C B \definedas \theset{(a, b) \suchthat f(a) = g(b)} \subset A\cross B$. ::: :::{.example title="?"} If $A = B = k[\eps]/(\eps^2)$ and $C = k$, then $A\cross_C B = k[\eps_1, \eps_2]/(\eps_1, \eps_2)^2$ Note that on the $\spec$ side, these should be viewed as $$ \spec(A) \disjoint_{\spec(C)} \spec(B) = \spec(A\cross_C B) .$$ ::: :::{.definition title="Deformation Functor (Preliminary Definition)"} A *deformation functor* is a functor $F: (\Art_{/k}) \to \Set$ such that $F(k) = \pt$ is a singleton. ::: :::{.example title="?"} Let $X_{_{/k}}$ be any scheme and let $x\in X(k)$ be a $k\dash$point. We can consider the deformation functor $F$ such that $F(A)$ is the set of extensions $f$ of the following form: \begin{tikzcd} & & & & X \arrow[dd] \\ & & & & \\ \spec(k) \arrow[rrrruu, "x"] \arrow[rr, hook] & & \spec(A) \arrow[rruu, "f", dashed] \arrow[rr] & & \spec(k) \end{tikzcd} If $A' \to A$ is a morphism, then we define $F(A') \to F(A)$ is defined because we can precompose to fill in the following diagram \begin{tikzcd} & & & & & & & & X \arrow[ddd] \\ & & & & & & & & \\ & & & & & & & & \\ \spec(k) \arrow[rrd] \arrow[rrrrrrrruuu] & & & & & & & & \spec(k) \\ & & \spec(A) \arrow[rr] \arrow[rrrrrruuuu, "\exists \tilde f"] & & \spec(A') \arrow[rrrru] \arrow[rrrruuuu, "f", dashed] & & & & \end{tikzcd} So this is indeed a deformation functor. ::: :::{.example title="a motivating example"} The Zariski tangent space on the nodal cubic doesn't "see" the two branches, so we allow "second order" tangent vectors. ::: We can consider parametrizing the functors above as $F_{X, x}(A)$, which is isomorphic to $F_{\spec (\OO_x)_{X, x}}$ and further isomorphic to $F_{\spec \hat{\OO_x}_{x, X} }$. This is because for Artinian algebras, we have maps $$ \spec (\OO_{x, X})/\mfm^N \to \spec \OO_{X, x} \to X .$$ :::{.remark} $\hat{ \OO }_{X, x}$ will be determined by $F_{X, x}$. ::: :::{.example title="?"} Consider $y^2 = x^2(x+1)$, and think about solving this over $k[t]/t^n$ with solutions equivalent to $(0, 0) \mod t$. ![Image](figures/2020-02-25-13:20.png)\ Note that the 'second order' tangent vector comes from $\spec k[t]/t^3$. We can write $F_{X, x}(A) = \pi\inv(x)$ where $$ \hom_{\Sch_{/k}}(\spec k, X) \mapsvia{\pi} \hom_{\Sch_{/k}}(\spec k, x) \ni x .$$ Thus $$ F_{X, x}(A) = \hom_{\Sch_{/k}}(\spec A, \spec \OO_{x, X}) = \hom_{k\dash\alg}(\hat \OO_{X, x}, A) .$$ ::: :::{.example title="?"} Given any local $k\dash$algebra $R$, we can consider \[ h_R: (\Art_{/k}) &\to \Set \\ A &\mapsto \hom(R, A) .\] and \[ h_{\spec R}: (\Art\Sch_{/k})\op \to \Set \\ \spec(A) &\mapsto \hom(\spec A, \spec R) .\] ::: :::{.definition title="Representable Deformation"} A deformation $F$ is **representable** if it is of the form $h_R$ as above for some $R \in \Art_{/k}$. ::: :::{.remark} There is a Yoneda Lemma for $A\in \Art_{/k}$, \[ \hom_{\mathrm{Fun}}(h_A, F) = F(A) .\] We are thus looking for things that are representable in a larger category, which restrict. ::: :::{.definition title="Pro-Representability"} A deformation functor is *pro-representable* if it is of the form $h_R$ for $R$ a complete local $k\dash$algebra (i.e. a limit of Artinian local $k\dash$algebras). ::: :::{.remark} We will see that there are simple criteria for a deformation functor to be pro-representable. This will eventually give us the complete local ring, which will give us the scheme representing the functor we want. ::: :::{.remark} It is difficult to understand even $F_{X, x}(A)$ directly, but it's easier to understand small extensions. ::: :::{.definition title="Small Extensions"} A *small extension* is a SES of Artinian $k\dash$algebras of the form \[ 0 \to J \to A' \to A \to 0 .\] such that $J$ is annihilated by the maximal ideal fo $A'$. ::: :::{.lemma title="?"} Given any quotient $B\to A \to 0$ of Artinian $k\dash$algebras, there is a sequence of small extensions (quotients): \begin{tikzcd} 0 & & & & & & \\ & & & & & & \\ B_0 \arrow[uu] & & B_1 \arrow[lluu] & & \cdots & & B_n = A \arrow[lllllluu] \\ & & & & & & \\ B \arrow[uu] \arrow[rruu] \arrow[rrrrrruu] & & & & & & \end{tikzcd} This yields \begin{tikzcd} \spec A \arrow[rrrr, hook] \arrow[rrrrdddddd, Rightarrow] & & & & \spec B \\ & & & & \\ & & & & \spec B_0 \arrow[uu, hook] \\ & & & & \\ & & & & \vdots \arrow[uu, hook] \\ & & & & \\ & & & & \spec B_n \arrow[uu, hook] \end{tikzcd} where the $\spec B_i$ are all small. ::: :::{.remark} In most cases, extending deformations over small extensions is easy. ::: ## First Example of Deformation and Obstruction Spaces Suppose $k=\bar k$ and let $X_{_{/k}}$ be connected. We have a picard functor \[ \pic_{X_{_{/k}}}: (\Sch_{/k})\op &\to \Set \\ S &\mapsto \pic(X_S) / \pic(S) .\] If we take a point $x\in \pic_{X_{_{/k}}}(k)$, which is equivalent to line bundles on $X$ up to equivalence, we obtain a deformation functor \[ F \definedas F_{\pic_{ X_{_{/k}}, x }} &\to \Set\\ A \mapsto \pi\inv(x) \] where \[ \pi: \pic_{X_{_{/k}}}(\spec A) &\to \pic_{X_{_{/k}}} (\spec k) \\ \pi\inv(x) &\mapsto x .\] This is given by taking a line bundle on the thickening and restricting to a closed point. Thus the functor is given by sending $A$ to the set of line bundles on $X_A$ which restrict to $X_x$. That is, $F(A) \subset \pic_{X_{_{/k}}}(\spec A)$ which restrict to $x$. So just pick the subspace $\pic(X_A)$ (base changing to $A$) which restrict. There is a natural identification of $\pic(X_A) = H^1(X_A, \OO_{X_A}^*)$. If \[ 0\to J \to A' \to A \to 0 .\] is a thickening of Artinian $k\dash$algebras, there is a restriction map of invertible functions \[ \OO_{X_A}^* \to \OO_{X_A'}^* \to 0 .\] which is surjective since the map on structure sheaves is surjective and its a nilpotent extension. The kernel is then just $\OO_{X_{A'}} \tensor J$. If this is a small extension, we get a SES \[ 0 \to \OO_X \tensor J \to \OO_{X_{A'}}^* \to \OO_{x_A}^* \to 0 .\] Taking the LES in cohomology, we obtain \[ H^1 \OO_X \tensor J \to H^1 \OO_{X_{A'}}^* \to H^1\OO_{x_A}^* \to H^0 \OO_X \tensor J .\] Thus there is an obstruction class in $H^2$, and the ambiguity is detected by $H^1$. Thus $H^1$ is referred to as the **deformation space**, since it counts the extensions, and $H^2$ is the **obstruction space**. # Deformation Theory (Thursday February 27th) Big picture idea: We have moduli functors, such as \[ F_{S'}: (\Sch_{/k})\op &\to \Set \\ \hilb: S &\to \text{flat subschemes of } X_S \\ \pic: S &\to \pic(X_S)/\pic(S) \\ \mathrm{Def}: S &\to \text{flat families } / S,~ \text{smooth, finite, of genus } g .\] :::{.definition title="Deformation Theory"} Choose a point $f$ the scheme representing $F_{S'}$ with $\xi_0 \in F_{gl}(\spec K)$. Define \[ F_{\text{loc}}: (\text{Artinian local schemes} / K)\op \to \set .\] \begin{tikzcd} \spec(K) \arrow[rr, "i", hook] & & \spec(A) \arrow[rr] & & F(i)\inv(\xi_0) \arrow[rr] & & F_{gr}(\spec K) \arrow[dd, "F(i)"] \\ & & & & & & \\ & & & & & & F_{gl}(\spec K) \end{tikzcd} ::: :::{.definition title="Deformation Functors"} Let $F: (\Art_{/k}) \to \Set$ where $F(k)$ is a point. Denote $\hat{\Art}_{/k}$ the set of complete local $k\dash$algebras. Since $\Art_{/k} \subset \hat{\Art} / k$, we can make extensions $\hat F$ by just taking limits: \begin{tikzcd} & \Art_{/k} \arrow[rrr, "F"] & & & \Set \\ & & & & \\ \lim_{\from} R/\mfm_R^n = R \in & \hat{\Art}_{/k} \arrow[uu] \arrow[rrruu, "\hat F"] & & & \end{tikzcd} where we define \[ \hat{F}(R) \da \inverselim F(R/\mfm_R^n) .\] ::: :::{.question} When is $F$ pro-representable, which happens iff $\hat F$ is representable? In particular, we want $h_R \mapsvia{\cong} \hat F$ for $R\in \hat{\Art}_{/k}$, so \[ h_R = \hom_{\hat{\Art}_{/k}}(R, \wait) = \hom_{?}(\wait, \spec k) .\] ::: :::{.example title="?"} Let $F_{\text{gl}} = \hilb_{X_{_{/k}}}^p$, which is represented by $H_{/k}$. Then . \[ \xi_0 = F_{\text{gl}}(k) = H(k) = \theset{Z\subset X \suchthat P_z = f} .\] Then $F_{\text{loc} }$ is representable by $\hat \OO_{H/\xi_0}$. ::: :::{.definition title="Thickening"} Given an Artinian $k\dash$algebra $A \in \Art_{/k}$, a *thickening* is an $A' \in \Art_{/k}$ such that $0 \to J \to A' \to A \to 0$, so $\spec A \injects \spec A'$. ::: :::{.definition title="Small Thickening"} A **small thickening** is a thickening such that $0 = \mfm_{A'} J$, so $J$ becomes a module for the residue field, and $\dim_k J = 1$. ::: :::{.lemma title="?"} Any thickening of $A$, say $B\to A$, fits into a diagram: \begin{tikzcd} & & & & 0 & & & & \\ & & & & & & & & \\ & & J \arrow[rr] & & A' \arrow[uu] \arrow[rr] & & A \arrow[dd, Rightarrow] \arrow[rr] & & 0 \\ & & & & & & & & \\ 0 \arrow[rr] & & I \arrow[rr] \arrow[uu] & & B \arrow[uu] \arrow[rr] & & A \arrow[rr] & & 0 \\ & & & & & & & & \\ & & I' \arrow[rr, Rightarrow] \arrow[uu] & & I' \arrow[uu] & & & & \\ & & & & & & & & \\ & & 0 \arrow[uu] & & 0 \arrow[uu] & & & & \end{tikzcd} ::: :::{.proof title="of lemma"} We just need $I' \subset I$ with $\mfm_S I \subset J' \subset I \iff J \mfm_B = 0$. Choose $J'$ to be a preimage of a codimension 1 vector space in $I/\mfm_B I$. Thus $J = I/I'$ is 1-dimensional. ::: Thus any thickening $A$ can be obtained by a sequence of small thickenings. By the lemma, in principle $F$ and thus $\hat{F}$ are determined by their behavior under small extensions. ### Example Consider $\pic$, fix $X_{_{/k}}$, start with a line bundle $L_0 \in \pic(x) /\pic(k) = \pic(x)$ and the deformation functor $F(A)$ being the set of line bundles $L$ on $X_A$with $\restrictionof{L}{x} \cong L_0$, modulo isomorphism. Note that this yields a diagram \begin{tikzcd} x \arrow[rr] \arrow[dd, hook] & & k \arrow[dd, "\text{unique closed point}"] \\ & & \\ X_A \arrow[rr] & & \spec A \end{tikzcd} This is equal to $(I_x)\inv (L_0)$, where $\pic(X_a) \mapsvia{I_x} \pic(x)$. If \[ 0 \to J \to A' \to A \to 0 .\] is a small thickening, we can identify \begin{tikzcd} 0 \arrow[rr] & & J \tensor_x \OO_{x} \cong \OO_x \arrow[rr] & & \OO_{X_{A'}} \arrow[rr] & & \OO_{X_{A}} \arrow[rr] & & 0 & & \\ & & & & & & & & & &\in\text{AbSheaves} \\ 0 \arrow[rr] & & \OO_x \arrow[rr, "f\mapsto 1+f"] & & \OO_{X_{A'}}^* \arrow[rr] \arrow[uu, hook] & & \OO_{X_{A}}^* \arrow[rr] & & 0 & & \end{tikzcd} This yields a LES \begin{tikzcd}[column sep=tiny] 0 \arrow[rr] & & {H^0(X, \OO_x) = k} \arrow[rr] & & {H^0(X_{A'}, \OO_{x_{A'}}^*) = {A'}^*} \arrow[rr] & & {H^0(X_{A'}, \OO_{x_{A}}^*) = A^*} \arrow[lllldd] \arrow[rr] & & \therefore 0 \\ & & & & & & & & \\ \therefore 0 \arrow[rr] & & {H^1(X, \OO_{x})} \arrow[rr] & & {H^1(X_{A'}, \OO_{x_{A'}}^*) = \pic(X_{A'})} \arrow[rr, "\scriptsize\text{restriction to } X_A", outer sep=1em] & & {H^1(X_{A}, \OO_{x_{A}}^*) = \pic(X_A)} \arrow[lllldd, "\text{obs}"] & & \\ & & & & & & & & \\ & & {H^2(X, \OO_x)} \ar[rr] & &\cdots & & & & \end{tikzcd} :::{.remark} To understand $F$ on small extensions, we're interested in 1. Given $L \in F_{\text{loc}}(A)$, i.e. $L$ on $X_A$ restricting to $L_0$, when does it extend to $L' \in F_{\text{loc}}(A')$? I.e., does there exist an $L'$ on $X_{A'}$ restricting to $L$? 2. Provided such an extension $L'$ exists, how many are there, and what is the structure of the space of extensions? ::: :::{.question} We have an $L\in \pic(X_A)$, when does it extend? ::: By exactness, $L'$ exists iff $\text{obs}(L) = 0\in H^2(X, \OO_x)$, which answers 1. To answer 2, $(I_x)\inv(L)$ is the set of extensions of $L$, which is a torsor under $H^1(x, \OO_x)$. Note that these are fixed $k\dash$vector spaces. :::{.remark} $H^1(X, \OO_x)$ is interpreted as the **tangent space** of the functor $F$, i.e. $F_{\text{loc}}(K[\eps])$. Note that if $X$ is projective, line bundles can be unobstructed without the group itself being zero. ::: For (3), just play with $A = k[\eps]$, which yields $0 \to k \mapsvia{\eps} k[\eps] \to k \to 0$, then \begin{tikzcd} 0 \arrow[rr] & & {H^1(X, \OO_x)} \arrow[rr] & & {H^1(X_{k[\eps]}, \OO_{k[\eps]}^*)} \arrow[rr, "I_x"] & & {H^1(X, \OO_x^*)} \arrow[ll, bend right=49] \\ & & & & {(I_x)\inv(L_0) \in \pic(X_{k[\eps]})} & & L_0 \in \pic(x) \end{tikzcd} i.e., there is a canonical trivial extension $L_0[\eps]$. :::{.example title="?"} Let $X \supset Z_0 \in \hilb_{X_{_{/k}}}(k)$, we computed \[ T_{Z_0} \hilb_{X_{_{/k}}} = \hom_{\OO_x}(I_{Z_0}, \OO_z) .\] We took $Z_0 \subset X$ and extended to $Z' \subset X_{k[\eps]}$ by base change. In this case, $F_{\text{loc}}(A)$ was the set of $Z'\subset X_A$ which are flat over $A$, such that base-changing $Z' \cross_{\spec A} \spec k \cong Z$. This was the same as looking at the preimage restricted to the closed point, \[ \hilb_{X_{_{/k}}}(A) \mapsvia{i^*} \hilb_{X_{_{/k}}}(k) \\ (i^*)\inv(z_0) \mapsfrom z_0 .\] Recall how we did the thickening: we had $0 \to J \to A' \to A \to 0$ with $J^2 = 0$, along with $F$ on $X_A$ which is flat over $A$ with $X_{_{/k}}$ projective, and finally an $F'$ on $X_{A'}$ restricting to $F$. The criterion we had was $F'$ was flat over $A'$ iff $0 \to J\tensor_{A'} F' \to F'$, i.e. this is injective. Suppose $z\in F_{\text{loc}}(A)$ and an extension $z' \in F_{\text{loc}}(A')$. By tensoring the two exact sequences here, we get an exact grid: \begin{tikzcd} 0 \arrow[rr] \arrow[dd] & & I_{Z'} \arrow[rr] & & \OO_{X_{A'}} \arrow[rr] & & \OO_{Z'} \arrow[rr] & & 0 \\ & & 0 \arrow[d] & & 0 \arrow[d] & & 0 \arrow[d] & & \\ J \arrow[dd] & 0 \arrow[r] & I_{Z_0} \arrow[dd] \arrow[rr] & & \OO_X \arrow[dd] \arrow[rr] & & \OO_{Z_0} \arrow[dd] \arrow[r] & 0 & \\ & & & & & & & & \\ A' \arrow[dd] & 0 \arrow[r] & I_{Z'} \arrow[rr] \arrow[dd] & & \OO_{X_{A'}} \arrow[rr] \arrow[dd] & & \OO_{Z'} \arrow[dd] \arrow[r] & 0 & \\ & & & & & & & & \\ A \arrow[dd] & 0 \arrow[r] & I_Z \arrow[d] \arrow[rr] & & \OO_{X_A} \arrow[rr] \arrow[d] & & \OO_Z \arrow[d] \arrow[r] & 0 & \\ & & 0 & & 0 & & 0 & & \\ 0 & & & & & & & & \end{tikzcd} The space of extension should be a torsor under $\hom_{\OO_X}(I_{Z_0}, \OO_{Z_0})$, which we want to think of as $\hom_{\OO_X}(I_{Z_0}, \OO_{Z_0})$. Picking a $\phi$ in this hom space, we want to take an extension $I_{Z'} \mapsvia{\phi} I_{Z''}$. ::: > We'll cover how to make this extension next time. # Tuesday March 31st See notes on Ben's website. We'll review where we were. ## Deformation Theory We want to represent certain moduli functors by schemes. If we know a functor is representable, it's easier to understand the deformation theory of it and still retain a lot of geometric information. The representability of deformation is much easier to show. We're considering functors $F: \Art_{/k} \to \Sets$. :::{.example title="?"} The Hilbert functor \[ \hilb_{X_{_{/k}}} (\Sch_{/k})\op \to \sets \\ S \mapsto \theset{ Z \subset X \cross S \text{ flat over } S} .\] This yields \[ F: \Art_{/k} \to \sets \\ ??? .\] ::: ![Image](figures/2020-03-31-12:44.png) Recall that we're interested in pro-representability, where $\hat F(R) = \inverselim F(R\mu_R^n)$ is given by a lift of the form \begin{tikzcd} \Art_{/k} \ar[r, "F"] & \sets \\ \hat{\Art_{/k}} \ar[u, hook] \ar[ur, "\hat F"'] & \end{tikzcd} :::{.question} Is $\hat F$ representable, i.e. is $F$ pro-representable? ::: :::{.example title="?"} The $F$ in the previous example is pro-representable by $\hat F = \hom(\OO_{\hilb, z_0}, \wait)$. ::: :::{.definition title="Pro-Representable Hull"} $F$ has a *pro-representable hull* iff there is a formally smooth map $h_R \to F$. ::: :::{.question} Does $F$ have a pro-representable hull? ::: Recall that a map of functors on artinian $k\dash$algebras is **formally smooth** if it can be lifted through nilpotent thickenings. That is, for $F, G: \Art_{/k} \to \Sets$, $F \to G$ is *formally smooth* if for any thickening $A' \surjects A$, we have \begin{tikzcd} & & F \ar[d] \\ h_{A} \ar[rru] \ar[r] & h_{A'} \ar[ru, dotted] \ar[r] & G \\ \spec A \ar[u, equal] \ar[r] & \spec A' \ar[u, equal] \ar[r] & G \ar[u, equal] \end{tikzcd} We proved for $R, A$ finite type over $k$, $\spec R \to \spec A$ smooth is formally smooth. Given a complete local $k\dash$algebra $R$ and a section $\xi \in \hat F(R)$, we make the following definitions: :::{.definition title="Versal, Miniversal, Universal"} The pair $(R, \xi)$ is - *Versal* for $F$ iff $h_R \mapsvia{\xi} F$ is formally smooth.[^not_unique] - *Miniversal* for $F$ iff versal and an isomorphism on Zariski tangent spaces. - *Universal* for $F$ if $h_R \mapsvia{\cong} F$ is an isomorphism, i.e. $h_R$ pro-represents $F$. - Pullback by a unique map [^not_unique]: Not a unique map, but still a pullback ::: :::{.remark} Note that **versal** means that any formal section $(s, \eta)$ where $\eta \in \hat F(s)$ comes from pullback, i.e there exists a map \[ R &\to S \\ \hat F(R) &\to \hat F(s) \\ \xi &\mapsto \eta .\] **Miniversal** means adds that the derivative is uniquely determined, and universal means that $R\to S$ is unique. ::: :::{.definition title="Obstruction Theory"} An **obstruction theory** for $F$ is the data of $\mathrm{def}(F), \mathrm{obs}(F)$ which are finite-dimensional $k\dash$vector spaces, along with a functorial assignment of the following form: \[ (A' \surjects A) \quad \text{a small thickening } \mapsto \\ \mathrm{def}(F) \selfmap F(A') \to F(A) \mapsvia{\mathrm{obs}} \mathrm{obs}(F) \] that is exact[^recall_right_exact] and if $A=k$, it is exact on the left (so the action was faithful on nonempty fibers). [^recall_right_exact]: Recall that right-exactness was a transitive action. ::: :::{.example title="?"} We have \[ \pic_{X_{/k}} : (\sch_{/k})\op &\to \sets \\ S &\mapsto \pic(X\cross X) / \pic(S) .\] This yields \[ F: \Art_{/k} \to \sets \\ A \mapsto L\in \pic(X_A),~ L\tensor k \cong L_0 \] where $X_{_{/k}}$ is proper and irreducible. Then $F$ has an obstruction theory with $\mathrm{def}(F) = H^1(\OO_x)$ and $\mathrm{obs}(F) = H^2(\OO_x)$. The key was to look at the LES of \[ 0 \to \OO_x \to \OO_{X_{A'}}^* \to \OO_{X_A}^* \to 0 .\] for $0 \to k \to A' \to A \to 0$ small. ::: :::{.remark title="Summary"} In both cases, the obstruction theory is exact on the left for any small thickening. We will prove the following: - $F$ has an obstruction $\iff$ it has a pro-representable hull, i.e. a versal family - $F$ has an obstruction theory which is always exact at the left $\iff$ it has a universal family. ::: ## Schlessinger's Criterion Let $F: \Art_{/k} \to \Set$ be a deformation functor (and it only makes sense to talk about deformation functors when $F(k) = \pt$). This theorem will tell us when a miniversal and a universal family exists. :::{.theorem title="Schlessinger"} $F$ has a miniversal family iff 1. Gluing along common subspaces: ror any small $A' \to A$ and $A'' \to A$ any other thickening, the map \[ F(A' \cross_A A'') \to F(A') \cross_{F(A)} F(A'') \] is surjective. 2. Unique gluing: if $(A' \to A) = (k[\eps] \to k)$, then the above map is bijective. 3. $t_F = F(k[\eps])$ is a finite dimensional $k\dash$vector space, i.e. \[ F(k[\eps] \cross_k k[\eps]) \mapsvia{\cong} F(k[\eps]) \cross F(k[\eps]) .\] 4. For $A' \to A$ small, \begin{tikzcd} F(A') \ar[r, "f"] & F(A) \\ t_f\, \selfmap f\inv(\eta) \ar[u, hook, "\subseteq"] & \eta \ar[u, "\in"] \end{tikzcd} where the action is simply transitive. $F$ has a miniversal family iff (1)-(3) hold, and universal iff all 4 hold. ::: :::{.exercise title="?"} Show that the existence of an obstruction theory which is exact on the left implies (1)-(4). ::: The following diagram commutes: \begin{tikzcd} \mathrm{def} \selfmap F(A' \cross_A A'') \ni \eta \ar[r] \ar[d] & F(A'') \ni \xi'' \ar[r, "\mathrm{obs}"] \ar[d] & \mathrm{obs} \\ \mathrm{def} \selfmap F(A')\ni \eta'm \xi' \ar[r] & F(A')\ni \xi \ar[r, "\mathrm{obs}"] & \mathrm{obs} \\ \end{tikzcd} So we have a map $F(A' \cross_A A'') \to F(A') \cross_{F(A)} F(A'') \ni (\xi',\xi'')$. Using transitivity of the $\mathrm{def}$ action, we can get $\xi' = \eta' + \theta$ and thus $\eta + \theta$ is the lift. ## Abstract Deformation Theory :::{.example title="?"} We start with $\qty{X_0}_{/k}$ and define the functor $F$ sending $A$ to $X/A$ flat families over $A$ with $X_0 \injects^i X$ such that $i \tensor k$ is an isomorphism. The punchline is that $F$ has an obstruction theory if $X_0$ is smooth with - $\mathrm{def}(F) = H^1(T_{X_0})$ - $\mathrm{obs}(F) = H^2(T_{X_0})$ ::: :::{.remark} \envlist 1. If $X$ is a deformation of $X_0$ over $A$ and we have a small extension $k \to A'\to A$ with $X'$ over $A'$ a lift of $X$. Then there is an exact sequence \[ 0 \to \text{Der}_R(\OO_{X_0}) \to\aut_{A'}(X') \to \aut_A(X) .\] 2. If $\qty{X_0}_{/k}$ is smooth and *affine*, then any deformation $X$ over $A$ (a flat family restricting to $X_0$) is trivial, i.e. $X \cong X_0 \cross_k \spec(A)$. \begin{tikzcd} & & X_0 \cross \spec(A) \ar[d] \\ X_0 \ar[r, hook] & X \ar[r] \ar[ru, "f", dotted] & \spec(A) \end{tikzcd} Thus $X_0 \injects X$ has a section $X\to X_0$, and the claim is that this forces $X$ to be trivial. ::: We have \begin{tikzcd} 0 \ar[r] & J \tensor \OO_X \ar[r] & \OO_x \ar[r] & \OO_{X_0} \ar[r] \ar[l, bend right] & 0 \end{tikzcd} yielding \[ 0 \to K \to \OO_{X_0} \tensor A \to \OO_X \to 0 \\ (\wait \tensor k) \\ 1 \to k\tensor k = 0 \to \OO_{X_0} \mapsvia{\cong} \OO_{X_0} \to 0 .\] :::{.remark} Why does this involve cohomology of the tangent bundle? For $X_0$ smooth, $\Der_k(\OO_{X_0}) = \mathcal{H}(T_{X_0})$, but the LHS is equal to $\hom( \Omega_{ \qty{X_0}_{/k}}, \OO_{X_0}) = H^0 (T_{X_0})$. ::: > Upcoming: proof of Schlessinger so we can use it! # Thursday April 2nd ## Abstract Deformations Let $X_0$ be smooth and consider the deformation functor \[ F : \Art_{_{/k}} &\to \sets \\ A &\mapsto (X_{/A} , \iota) \] where $X$ is flat (and thus smooth) and $i$ is a closed embedding $i: X_0 \injects X$ with $i\tensor k$ an isomorphism. Then $F$ has an obstruction theory with - $\mathrm{def}(F) = H^1(X_0, T_0)$ of the tangent bundle - $\mathrm{obs}(F) = H^2(X_0, T_0)$. Additionally assume $X_0$ is smooth and projective, which will force the above cohomology groups to be finite-dimensional over $k$. :::{.remark title="Key points"} \envlist - All deformations of smooth affine schemes are trivial - Automorphisms of a deformation $X/A$ which are the identity on $X_0$ are $\id + \delta$ for $\delta$ a derivation in $\Der_k(\OO_{X_0}) = \hom_{\OO_{X_0}}(\Omega_{\qty{X_0}_{_{/k}}}, \OO_{X_0})$. > See screenshot. ::: Suppose we have a small thickening $k \to \AA^1 \to \AA$ and $X/\AA$ with an affine cover $X_\alpha$ of $X$. This comes with gluing information $\phi_{\alpha\beta}: X_{\alpha\beta} \to X_{\beta\alpha} = X_\alpha \intersect X_\beta$. These maps satisfy a cocycle condition: \begin{tikzcd} X_{\alpha\beta} \intersect X_{\alpha\gamma} \ar[rr] \ar[rd] & & X_{\gamma\alpha} \intersect X_{\gamma\beta} \ar[ld] \\ & X_{\beta\alpha} \intersect X_{\beta\gamma} & \end{tikzcd} :::{.question} Can we extend this to $X'/\AA$? ::: We have $X_\alpha \cong X_\alpha^\mathrm{red} \cross \AA$? Choose $\phi'_{\alpha\beta}$ such that \begin{tikzcd} X'_{\alpha\beta} \ar[r, "\phi'_{\alpha\beta}"] & X'_{\beta\alpha} = X_{\beta\alpha}^\mathrm{red} \cross \AA \\ X_{\alpha\beta}\ar[u, hook] \ar[r, "\phi_{\alpha\beta}"] & X_{\beta\alpha} \ar[u, hook] \end{tikzcd} We need $\phi'_{\alpha\beta}$ to satisfy the cocycle condition in order to glue. We want the following map to be the identity: $(\phi'_{\alpha\gamma})\inv \phi'_{\beta\gamma} \phi'_{\alpha\beta}$. This is an automorphism of $X'_{\alpha\beta} \intersect X'_{\alpha\beta}$ and is thus the identity in $\aut(X_{\alpha\beta} \intersect X_{\alpha\gamma})$. So it makes sense to talk about \[ \delta_{\alpha\beta\gamma} \da (\phi'_{\alpha\gamma})\inv \phi'_{\beta\gamma} \phi'_{\alpha\beta} - \id \in M^0(T_{X^\mathrm{red}_{\alpha\beta\gamma}}) .\] :::{.exercise title="?"} In parts, 1. $\delta_{\alpha\beta\gamma}$ is a $2\dash$cocycle for $T_{X_0}$, so it has trivial boundary in terms of Cech cocycles. Thus $[\delta_{\alpha\beta\gamma}] \in H^2(T_{X_0})$. 2. The class $[\delta_{\alpha\beta\gamma}]$ is independent of choice of $\phi'_{\alpha\beta}$, i.e. $\phi'_{\alpha\beta} - \phi_{\alpha\beta}'' \in H^0((T_X)_{\alpha\beta})$ gives a coboundary $\eta$ and thus $\delta = \delta' + \eta$. This yields $\mathrm{obs}(X) \in H^2(T_{X_0})$. 3. $\mathrm{obs}(X) = 0 \iff X$ lifts to some $X'$ (i.e. a lift exists) ::: :::{.remark} For the sufficiency, we have $\delta_{\alpha\beta\gamma} = \bd \eta_{\alpha\beta} \in H^0(T_{X_{\alpha\beta}})$. Let $\phi_{\alpha\beta}'' = \phi_{\alpha\beta}' - \eta_{\alpha\beta}$, the claim is that $\phi_{\alpha\beta}''$ satisfies the gluing condition. This covers the obstruction, so now we need to show that the set of lifts is a torsor for the action of the deformation space $\mathrm{def}(F) = H^1(T_{X_0})$. From an $X'$, we obtain $X_{\alpha\beta}' \mapsvia{\phi_{\alpha\beta}'} X_{\beta\alpha}'$ where the LHS is isomorphic to $(X_{\alpha\beta}')^\mathrm{red} \cross \AA^r$? Given $\eta_{\alpha\beta} \in H^0(T_{X_{\alpha\beta}})$, then $\phi'_{\alpha\beta} + \eta_{\alpha\beta} = \phi_{\alpha\beta}''$ is another such identification. ::: :::{.exercise title="?"} In parts 1. $\bd \eta_{\alpha\beta} = 0$. 2. Given an $X'$ and 1-coboundary $\eta$, we get a new lift $X'' = X' + \eta$. If $[\eta] = [\eta'] \in H^1(T_{X_0})$, then $X' + \eta \cong X' + \eta'$. By construction, $(X' + \eta)_\alpha \cong (X' + \eta')_\alpha$, but these may not patch together. However, if $[\eta] = [\eta']$ then this isomorphism can be modified by by $\eps$ defined by $\eta-\eta' = \bd \eps$, and it patches. ::: :::{.remark} This kind of patching is ubiquitous -- essentially patching together local obstructions to get a global one. In general, there is a local-to-global spectral sequence that computes the obstruction space ::: ## Proving Schlessinger ### The Schlessinger Axioms #### H1 For any two small thickenings \[ A' &\to A \\ A'' &\to A \] we have a natural map \[ F(A' \cross_A A'') \to F(A') \cross_{F(A)} F(A'') \] and we require that this map is surjective. So deformations agreeing on the sub glue together. #### H2 When $(A' \to A) = (k[\eps] \to k)$ is the trivial extension, the map in H1 is an isomorphism. > Doing things to first order is especially simple. #### H3 The tangent space of $F$ is given by $t_F = F(k[\eps])$, and we require that $\dim_k t_F < \infty$, which makes sense due to H2. #### H4 If we have two equal small thickenings $(A' \to A) = (A'' \to A)$, then the map in H1 is an isomorphism. #### H4' For $A' \to A$ small, \[ t_F \selfmap F(A') \to F(A) \] is exact in the middle and left. :::{.remark} Note that the existence of this action uses H2. ::: :::{.warnings} For $(R, \xi)$ a complete local ring and $\xi \in \hat F(R)$ a formal family, this is a hull $\iff$ miniversal, i.e. for $h_R \mapsvia{\xi} F$, this is smooth an isomorphism on tangent spaces. ::: :::{.theorem title="1, Schlessinger"} \envlist a. $F$ has a miniversal family $(R, \xi)$ with $\dim t_R < \infty$, noting that $t_R = \mfm_R / \mfm_R^2$, iff H1-H3 hold. b. $F$ has a universal family $(R, \xi)$ with $\dim t_R < \infty$ iff h1-H4 hold. ::: :::{.theorem title="2"} \envlist a. $F$ having an obstruction theory implies H1-H3. b. $F$ having a strong obstruction theory (exact on the left) is equivalent to H1-H4. ::: Some preliminary observations: :::{.exercise title="Easy, fun, diagram chase"} If $F$ has an obstruction theory, then H1-H3 hold. ::: :::{.exercise title="?"} An obstruction theory being exact on the left implies H4. ::: ### Example :::{.exercise title="?"} For $R$ a complete local $k\dash$algebra with $t_R$ finite dimensional has a strong obstruction theory. ::: Can always find a surjection from a power series ring: \[ S \da k[[t_R\dual]] \surjects R \] which yields an obstruction theory - $\mathrm{def} = t_R$ - $\mathrm{obs} = I/\mfm_S I$ i.e., if $F$ is pro-representable, then it has a strong obstruction theory. Suppose that $(R, \xi)$ is versal for $F$, this implies H1. We get $F(A' \cross_A A'') \surjects F(A') \cross_{F(A)} F(A'')$ For versal, if we have $h_R \mapsvia{\xi} F$ smooth, we have \begin{tikzcd} & & & h_r \ar[d] \\ h_k \ar[r] \ar[rrru] & h_A \ar[r] \ar[rru, dotted] \ar[rr, bend right, "\eta"] & h_{A'} \ar[ur, dotted] \ar[r] & F \end{tikzcd} and we can find a lift from $h_{A''}$ as well, so we get a diagram \begin{tikzcd} & & F \\ h_{A''} \ar[r] & h_R \ar[ru] \\ h_A \ar[u] \ar[r] & h_{A'} \ar[u] \end{tikzcd} and thus \begin{tikzcd} {A''} \ar[r] & R \\ A \ar[u] \ar[r] & {A'} \ar[u] \end{tikzcd} So we get the left $\tilde \eta$ of $(\eta', \eta'')$ we want from \begin{tikzcd} h_{A' \cross_A A''} \ar[r, "f"] \ar[rr, "\tilde\eta", bend right] & h_R \ar[r] & F \end{tikzcd} If $(R, \xi)$ is miniversal, then H2 holds. We want to show that the map \[ F(A'' \cross_K k[\eps]) \mapsvia{\sim} ?? \] is a bijection. Suppose we have two maps \begin{tikzcd} & & h_R \\ h_{A''} \ar[rru, bend left] \ar[r] & h_{A'' \cross k[\eps]} \ar[ru, shift left=0.75ex] \ar[ru, shift right=0.75ex] \ar[r, shift left=0.75ex] \ar[r, shift right=0.75ex] & F \\ & h_{k[\eps]} \ar[ur, bend right] & \end{tikzcd} Then the two lifts are in fact equal, and \begin{tikzcd} R \ar[r, shift left=0.75ex] \ar[r, shift right=0.75ex] & A'' \cross k[\eps] \ar[r] \ar[d] & k[\eps] \\ & A'' & \end{tikzcd} If $(R, \xi)$ is miniversal with $t_R$ finite dimensional, then H3 holds immediately. If $(R, \xi)$ is universal, then H4 holds. :::{.question} Why are H4 and H4' connected? ::: :::{.answer} Let $A' \to A$ be small, then \[ A' \cross_A A' &= A' \cross_k k[\eps] \\ (x, y) &\mapsto ?? .\] Using H2, we can identify $F(A; \cross_A A') \cong t_F \cross F(A')$. We can thus define an action \[ (\theta, \xi) &\mapsto (\theta + \xi, \xi) .\] If this is an isomorphism, then this action is simply transitive. The map $\theta \mapsto \theta + \xi$ gives an isomorphism on the fiber of $F(A') \to F(A)$. ::: > Next time we'll show the interesting part of the sufficiency proof. # Tuesday April 7th > (Missing first few minutes.) Take $I_{q+1}$ to be the minimal $I$ such that $\mfm_q I_q \subset I \subset I_1$ and $\xi_q$ lifts to $S/I$. :::{.claim} Such a minimal $I$ exists, i.e. if $I, I'$ satisfy the two conditions then $I \intersect I'$ does as well. So $I, I'$ are determined by their images $v, v'$ in the vector space $I_q \tensor k$. ::: So enlarge either $v$ or $v'$ such that $v + v' = I_q \tensor k$ but $v \intersect v'$ is the same. We can thus assume that $I + I' = I_q$, and so \[ S / I \intersect I' = S/I \cross_{S/I_q} S/I' \] which by H1 yields a map \[ F(S/I\intersect I') &\to F(S/I) \cross_{F(S/I_q)} F(S/I') \] So $I\intersect I'$ satisfies both conditions and thus a minimal $I_{q+1}$ exists. Let $\xi_{q+1}$ be a lift of $\xi_q$ over $S/I_{q+1}$ (noting that there may be many lifts). ## Showing Miniversality :::{.claim} Define $R = \directlim R_q$ and $\xi = \directlim \xi_q$, the claim is that $(R, \xi)$ is miniversal. ::: We already have $h_R \mapsvia{\xi} F$ and thus $t_R \mapsvia{\cong}t_F$ is fulfilled. We need to show formal smoothness, i.e. for $A' \to A$ a small thickening, suppose we have a lift \begin{tikzcd} & & h_R \ar[d, "\xi"] \\ h_a \ar[rru, "n"] \ar[rr, bend right] \ar[r] & h_{A'} \ar[r] & F \end{tikzcd} If we have a $u'$ such that commutativity in square 1 holds (?) then we can form a lift $u'$ satisfying commutativity in both squares 1 and 2. We can restrict sections to get a map $F(A') \to F(A)$ and using representability obtain $h_R(A') \to h_R(A)$. Combining H1 and H2, we know $t_F$ acts transitively on fibers, yielding \begin{tikzcd} t_R \selfmap \ar[d, "\cong"] & u'\in h_R(A') \ar[r] \ar[d] \ar[r] & u\in h_R(A) \ar[d]\\ t_F \selfmap & \eta' \in F(A') \ar[r] \ar[r] & \eta \in F(A) \\ \end{tikzcd} Then $u' \mapsto u$ is equivalent to (1), and $u' \mapsto \eta'$ is equivalent to (2). Let $\eta_0$ be the image of $u'$ and define $\eta' = \eta_0 + \theta, \theta \in t_F$ then $u' = u' + \theta, \theta \in t_R$. So we can modify the lift to make these agree. Thus it suffices to show \begin{tikzcd} A' \ar[r] & A & R_q \ar[l] \\ S \ar[u, "v"] \ar[r] & \ar[lu, dotted, "\exists_? u'"] \ar[u, "u"] \ar[ur] & \end{tikzcd} We get a diagram of the form \begin{tikzcd} S \ar[d] \ar[r, "w"] & A' \cross_A R_1 \ar[r] \ar[d, "{ \pi_2, \text{small} }"] & A' \ar[d, "{ \text{small} }"] \\ R \ar[r] & R_q \ar[r] & A \end{tikzcd} :::{.observation} \envlist - $S \to R_q$ is surjective. - $\im(w) \subset A' \cross_A R_1$ is a subring, so either - $\im(w) \mapsvia{\cong} R_q$ if it doesn't meet the kernel, or - $\im(w) = A' \cross_A R_q$ In case (a), this yields a section of the middle map and we'd get a map $R_q \to A'$ and thus the original map we were after $R \to A$. ::: So assume $w$ is surjective and consider \begin{tikzcd} 0 \ar[r] & I \ar[r] & S \ar[r] & A' \cross_A R_q \ar[r]\ar[d, "\text{small}"] & 0 \\ & & & R_q & \end{tikzcd} and we have $\mfm_S I_1 \subset I \subset I_q$ where the second containment is because $I$ a quotient of $R_q$ factors through $S/I$ and the first is because $S/I$ is a small thickening of $R_q$. But $\xi_q$ lifts of $S/I$, and we have \[ \xi \in F(S/I) \surjects \xi = \xi' \cross \xi_q ? .\] Therefore $I_{q+1} \subset I$ and we have a factorization \begin{tikzcd} S \ar[rr] \ar[dr, dotted] & & S/I \\ & R_{q+1}\ar[ur, dotted] & \end{tikzcd} Recall that we had ![](figures/image_2020-04-07-13-17-11.png) \todo[inline]{Image to diagram} where the diagonal map $u'$ gives us the desired lift, and thus \begin{tikzcd} R \ar[r] \ar[rr, bend left] & R_{q+1} \ar[r] & A' \end{tikzcd} exists. This concludes showing miniversality. ## Part of Proof To finish, we want to show that H4 implies that the map on sections $h_R \mapsvia{\xi} F$ is bijective. \begin{tikzcd} & & & h_R \ar[d, "\xi"] \\ & h_A \ar[rr, bend right, "\eta"] \ar[rru, bend left, "u"] & h_{A'} \ar[r, "\eta'"] \ar[ru, "\exists ! u'"] & F \end{tikzcd} where the map $\xi$ is "formal etale", which will necessarily imply that it's a bijection over all artinian rings. So we just need to show formal étaleness. We have a diagram \begin{tikzcd} t_R \selfmap u'\in h_R(A') \ar[r] \ar[d] & u\in h_R(A) \ar[d] \\ t_F \selfmap \eta' \in h_R(A') \ar[r] & \eta \in h_R(A) \end{tikzcd} where $u'$ exists by smoothness. Assume that are two $u', u''$, then $u' = u'' + \theta$ and $\im(u') = \im(u'') + \theta \implies \theta = 0$ and thus $u' = u''$. ## Revisiting Goals We originally had two goals: 1. Given a representable moduli functor (such as the Hilbert functor), we wanted to understand the local structure by analyzing the deformation functor at a given point. 2. We want to use representability of the deformation functors to get global representability of the original functor. :::{.question} What can we now deduce about the local structure of functors using their deformation theory? ::: :::{.fact title="1"} Any two hulls $h_R \to F$ are isomorphic but not canonically. We can lift maps at every finite level and induct up, which is an isomorphism on tangent spaces and thus an isomorphism. The sketch: use smoothness to get the map, and the tangent space condition will imply the full isomorphism. ::: :::{.fact title="3"} Suppose that $F$ has an obstruction theory (not necessarily strong). This implies there exists a hull $h_R \mapsvia \xi F$. The obstruction theory of $F$ *gives* an obstruction theory of $h_R$: given $A' \to A$ a small thickening, we need a functorial assignment \[ t_R = \mathrm{def} \selfmap h_R(A') \to h_R(A) \mapsvia{\mathrm{obs}} \mathrm{obs} \\ \mathrm{def} \selfmap F(A') \to F(A) \mapsvia{\mathrm{obs}} \mathrm{obs} \] where there are vertical maps with equality on the edges. ![Vertical maps](figures/image_2020-04-07-13-35-00.png) By formal smoothness, $\eta'$ lifts to some $\xi'$, but using the transitivity of the action of the tangent space can fix this. We already had an obstruction theory of $R$, since we can always find a quotient \[ I \to S = k[[t_R\dual]] \surjects R \] and $h_K$ has an obstruction theory - $\mathrm{def} = t_R = \qty{\mfm_R/\mfm_R^2}\dual$ - $\mathrm{obs} = \qty{I/\mfm_S I}\dual$ ::: :::{.fact title="proof can be found in FGA"} Any other obstruction theory $(\mathrm{def}', \mathrm{obs}')$ of $h_R$ admits an injection $\qty{I/\mfm_S I}\dual \injects \mathrm{obs}'$. ::: Combining these three facts, we conclude the following: If $F$ has an obstruction theory $\mathrm{def}(F), \mathrm{obs}(F)$, then $F$ has a miniversal family $h_R \mapsvia \xi F$ with $R = S/ I$ a quotient of the formal power series ring over some ideal, where $S = k[[t_F\dual]]$. It follows that $\dim(I/\mfm_S I) \leq \dim \mathrm{obs}(F)$, and thus the minimal number of generators of $I$ (equal to the LHS by Nakayama) is bounded by the RHS. Thus \[ \dim_k \mathrm{def}(F) \geq \dim R \geq \dim \mathrm{def}(F) - \dim \mathrm{obs}(F) .\] In particular, if $\dim(R) = \dim \mathrm{def}(F) - \dim \mathrm{obs}(F)$, then $R$ is a complete intersection. If $\dim(R) = \dim \mathrm{def}(R)$, the ideal doesn't have any generators, and $R \cong S$. In particular, if $\mathrm{obs}(F) = 0$, then $R \cong S$ is isomorphic to this power series ring. Finally, if $F$ is the deformation functor for a global representable functor, then $R = \hat{\OO}_{\mfm, p}$ is the completion of this local ring and the same things hold for this completion. Thus regularity can be checked on the completion. So if you have a representable functor with an obstruction theory (e.g. the Hilbert Scheme) with zero obstruction, then we have smoothness at that point. If we know something about the dimension at a point relative to the obstruction, we can deduce information about being a local intersection. So the deformation tells you the dimension of a minimal smooth embedding, and the obstruction is the maximal number of equations needed to cut it out locally. :::{.remark} The content here: see Hartshorne's *Deformation Theory*. The section in FGA is in less generality but has many good examples. See "Fundamental Algebraic Geometry". See also representability of the Picard scheme. ::: # Thursday April 9th Let $F: \Art_{/k} \to \sets$ be a deformation functor with an obstruction theory. Then H1-H3 imply the existence of a miniversal family, and gives us some control on the hull $h_{R} \to F$, namely \[ \dim \mathrm{def}(F) \geq \dim R \geq \dim \mathrm{def}(F) - \dim \mathrm{obs}(F) .\] In particular, if $\mathrm{obs}(F) = 0$, then $R \cong k[[\mathrm{def}(F)\dual]] = k[[ t_{F}\dual ]]$. :::{.example title="?"} Let $M = \hilb_{\PP^n_{/k}}^{dt + (1-g)}$ where $k=\bar k$, and suppose $[Z] \in M$ is a smooth point. Then \[ \mathrm{def} = \hom_{ \mods{ \OO_{x} } }(I_{Z}, \OO_{Z}) = \hom_{Z}(I_{Z}/I_{Z}^2, \OO_{Z}) = H^0(N_{Z/X}) .\] the normal bundle $N_{Z/X} = (I/I^2)\dual$ of the regular embedding, and $\mathrm{obs} = H^1(N_{Z/X})$. :::{.claim} If $H^1(\OO_{Z}(1)) = 0$ (e.g. if $d > 2g-2)$ then $M$ is smooth. ::: :::{.proof title="of claim"} The tangent bundle of $\PP^n$ sits in the Euler sequence \[ 0 \to \OO \to \OO(1)^{n+1} \to T_{\PP^n} \to 0 .\] And the normal bundles satisfies \[ 0 \to T_{Z} &\to T_{\PP^n}\mid_{Z} \to N_{Z/\PP^n} \to 0 \\ \\ &\Downarrow \text{ is the dual of }\\ \\ 0 \to I/I^2 &\to \Omega \mid_{Z} \to \Omega \to 0 .\] There is another SES: \[ ????? .\] Taking the LES in cohomology yields \[ H^1(\OO_{Z}(1)^{n+1})=0 \to H^1(N_{Z/\PP^n}) =0 \to 0 \] and thus $M$ is smooth at $[Z]$. We can compute the dimension using Riemann-Roch: \[ \dim_{[Z]} M &= \dim H^0(N_{Z/\PP^n}) \\ &= \chi(N_{Z/\PP^n}) \\ &= \deg N + \rk N(1-g) \\ &= \deg T_{\PP^n} \mid _Z - \deg T_{Z} + (n-1)(1-g) \\ &= d(n+1) + (2-2g) + (n-1)(1-g) .\] ::: ::: :::{.remark} This is one of the key outputs of obstruction theory: being able to compute these dimensions. ::: :::{.example title="?"} Let $X \subset \PP^5$ be a smooth cubic hypersurface and let $H = \hilb_{X_{/k}}^{\text{lines} = t+1} \subset \hilb_{\PP^5/k}^{t+1} = \Gr(1, \PP^5)$, the usual Grassmannian. :::{.claim} Let $[\ell] \in H$, then the claim is that $H$ is smooth at $[\ell]$ of dimension 4. ::: :::{.proof title="of claim"} We have - $\mathrm{def} = H^0(N_{\ell/X})$ - $\mathrm{obs} = H^1(N_{\ell/X})$ We have an exact sequence \[ 0 \to N_{\ell/X} \to N_{\ell/\PP} \to N_{X/\PP}\mid_\ell \to 0 \\ .\] There are surjections from $\OO_\ell(1)^6$ onto the last two terms. :::{.claim title="Subclaim"} For $N = N_{\ell/\PP}$ or $N_{X/\PP}\mid_\ell$, we have $H^1(N) = 0$ and $\OO(1)^6 \surjects N$ is surjective on global sections. ::: :::{.proof title="of subclaim"} Because $\ell$ is a line, $\OO_\ell(1) = \OO(1)$ and $H^1(\OO_\ell(1)) = 0$ and the previous proof applies, so $H^1(N) = 0$. ::: We thus have a diagram: ![Image](figures/image_2020-04-09-12-51-51.png) In particular, $T_\ell = \OO(2)$, and the LES for $0 \to \OO \to K \to T_\ell$ shows $H^1(K) = 0$. Looking at the horizontal SES $0 \to K \to \OO_\ell(1)^6 \surjects N_{\ell/\PP}$ yields the surjection claim. We have ![Diagram](figures/a.png) and taking the LES in cohomology yields ![Diagram](figures/image_2020-04-09-12-55-01.png)\ Therefore $H$ is smooth at $\ell$ and \[ \dim_\ell H &= \chi(N_{\ell/X}) \\ &= \deg T_{X} - \deg T_\ell + 3 \\ &= \deg T_\PP - \deg N_{X/\PP} - \deg T_\ell + 3 \\ &= 6 - 3 - 2 + 3 = 4 .\] ::: ::: :::{.remark} It turns out that the Hilbert scheme of lines on a cubic has some geometry: the Hilbert scheme of two points on a K3 surface. ::: ## Abstract Deformations Revisited Take $X_{0} / k$ some scheme and consider the deformation functor $F(A)$ taking $A$ to $X/A$ flat with an embedding $\iota: X_{0} \injects X$ with $\iota \tensor k$ an isomorphism. Start with H1, the gluing axiom (regarding small thickenings $A' \to A$ and a thickening $A'' \to A$). Suppose \[ X_{0} \injects X' \in F(A') \to F(A) .\] which restricts to $X_{0} \injects X$. Then in $F(A)$, we have $X_{0} \injects X' \tensor_{A'} A$, and we obtain a commutative diagram where $X' \tensor A \injects X'$ is a closed immersion: ![???](figures/abcdefg.png){width=350px} The restriction $X' \to X$ means that there exists a diagram \begin{tikzcd} X' & & X \ar[ll, dotted, "\exists"] \\ & X \ar[ur, hook] \ar[ul, hook] \end{tikzcd} Note that this is not necessarily unique. We have ![Diagram?](figures/image_2020-04-09-13-06-40.png){width=350px} This means that we can find embeddings such that \begin{tikzcd} X'' & \ar[l, "\exists", hook] X \ar[r, "\exists", hook] & X' \\ & X_{0} \ar[ul, hook] \ar[u, hook] \ar[ur, hook] \end{tikzcd} ![Diagram](figures/image_2020-04-09-13-08-19.png){width=350px} And thus if we have ![Diagram](figures/image_2020-04-09-13-08-42.png){width=350px} then $X_{0} \injects Z$ is **a** required lift (again not unique). :::{.question} When is such a lift unique? ::: Suppose $X_{0} \injects W$ is another lift, then it restricts to both $X, X'$ and we can fill in the following diagrams: ![Diagram](figures/image_2020-04-09-13-10-44.png){width=350px} Using the universal property of $Z$, which is the coproduct of this diagram: ![Diagram](figures/image_2020-04-09-13-11-13.png){width=350px} However, there may be no such way to fill in the following diagram: ![Diagram](figures/image_2020-04-09-13-11-58.png){width=350px} But if there exists a map making this diagram commute: ![Diagram](figures/image_2020-04-09-13-12-25.png){width=350px} Then there is a map $Z\to W$ which is flat after tensoring with $k$, which is thus an isomorphism.[^nakayama_rmk] :::{.remark} Thus the lift is unique if - $X = X_{0}$, then the following diagrams commute by taking the identity and the embedding you have. Note that in particular, this implies H2. ![Diagram](figures/image_2020-04-09-13-15-09.png){width=350px} - Generally, these diagrams can be completed (and thus the gluing maps are bijective) if the map \[ \aut(X_{0}\injects X') \to \aut(X_{0} \injects X) .\] of automorphisms of $X'$ commuting with $X_{0} \injects X$ is surjective. ::: So in this situation, there is only *one* way to fill in this diagram up to isomorphism: ![Diagram](figures/image_2020-04-09-13-18-59.png){width=350px} If we had two ways of filling it in, we obtain bridging maps: ![Diagram](figures/image_2020-04-09-13-20-07.png){width=350px} :::{.lemma title="?"} If $H^0(X_{0}, T_{X_{0}}) = 0$ (where the tangent bundle always makes sense as the dual of the sheaf of Kahler differentials) which we can identify as derivations $D_{\OO_{k}}(\OO_{X_{0}}, \OO_{X_{0}})$, then the gluing map is bijective. ::: :::{.proof title="?"} The claim is that $\aut(X_{0} \injects X) = 1$ are always trivial. This would imply that all random choices lead to triangles that commute. Proceeding by induction, for the base case $\aut(X_{0} \injects X_{0}) = 1$ trivially. Assume $X_{0} \injects X_{i}$ lifts $X_{0} \injects X$, then there's an exact sequence \[ 0 \to \Der_{k}(\OO_{X_{0}}, \OO_{X_{0}}) \to \Aut(X_{0} \injects X_0') \to \Aut(X_{0} \injects X) .\] ::: Thus $F$ always satisfies H1 and H2, and $H^0(T_{X_{0}}) = 0$ (so no "infinitesimal automorphism") implies H4. Recall that the dimension of deformations of $F$ over $k[\eps]$ is finite, i.e. $\dim t_{F} < \infty$ This is where some assumptions are needed. If $X_{/K}$ is either - Projective, or - Affine with isolated singularities, this is enough to imply H3. Thus by Schlessinger, under these conditions $F$ has a miniversal family. Moreover, if $H^0(T_{X_{0}}) = 0$ then $F$ is pro-representable. :::{.example title="?"} If $X_{0}$ is a smooth projective genus $g\geq 2$ curve, then - Obstruction theory gives the existence of a miniversal family - We have $\mathrm{obs} = H^2(T_{X_{0}}) = 0$, and thus the base of the miniversal family is smooth of dimension $\mathrm{def}(F) \dim H^1(T_{X_{0}})$, - $H^0(T_{X_{0}}) = 0$ and $\deg T_{X_{0}} = 2-2g < 0$, which implies that the miniversal family is universal. We can conclude \[ \dim H^1(T_{X_{0}}) = -\chi(T_{X_{0}}) = -\deg T_{X_{0}} + g-1 = 3(g-1) .\] ::: :::{.remark} Note that the global deformation functor is not representable by a scheme, and instead requires a stack. However, the same fact shows smoothness in that setting. ::: ## Hypersurface Singularities Consider $X(f) \subset \AA^n$, and for simplicity, $(f=0) \subset \AA^2$, and let - $S = \CC[x, y]$. - $B = \CC[x, y] / (f)$ :::{.question} What are the deformations over $A \da k[\eps]$? ::: This means we have a ring $B'$ flat over $k$ and tensors to an isomorphism, so tensoring $k\to A\to k$ yields the following: \begin{tikzcd} 0 \ar[r] & B \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & S \ar[u] \ar[r] & S[\eps] \ar[u, "\exists", twoheadrightarrow] \ar[r] & S \ar[r] \ar[u] & 0 \\ 0 \ar[r] & S \cong I \ar[u] \ar[r] & I'= \gens{f'} \ar[u] \ar[r] & I = \gens{f} \ar[u, "\cong"] \ar[r] & S \end{tikzcd} Thus any such $B'$ is the quotient of $S[\eps]$ by an ideal, and we have $f' = f + \eps g$. :::{.question} When do two $f'$s give the same $B'$? ::: We have $\eps f' = \eps f$, so $\eps f \in (f')$ and we can modify $g$ by any $cf$ where $c\in S$, where only the equivalence class $g\in S/(f)$ matters. Now consider $\aut(B \injects B')$, i.e. maps of the form \[ x &\mapsto x + \eps a \\ y &\mapsto y + cb \] for $a, b\in S$. Under this map, \[ f_0' = f + \eps g \mapsto & f(x + \eps a, y + \eps b) + \eps g(x ,y) \\ \\ &\Downarrow \quad\text{implies} \\ \\ f(x, y) &= \eps a \dd{}{x} f + \eps b \dd{}{y} f + \eps g(x ,y) ,\] so in fact only the class of $g\in S/(f, \del_{x} f, \del_{y} f)$. This is the ideal of the singular locus, and will be Artinian (and thus finite-dimensional) if the singularities are isolated, which implies H3. We can in fact exhibit the miniversal family explicitly by taking $g_{i} \in S$, yielding a basis of the above quotient. The hull will be given by setting $R = \CC[[t_{1}, \cdots, t_{m} ]]$ and taking the locus $V(f + \sum t_{i} g_{i}) \subset \AA_{R}^2$. :::{.example title="simple"} For $f = xy$, then the ideal is $I = (xy, y, x) = (x, y)$ and $C/I$ is 1-dimensional, so the miniversal family is given by $V(xy + t) \subset \CC[[t_{1}]][x, y]$. The greater generality is needed because there are deformation functors with a hull but no universal families. ::: [^nakayama_rmk]: Recall that by Nakayama, a nonzero module tensor $k$ can not be zero. # Tuesday April 14th Recall that we are looking at $(X_{0})_{/k}$ and $F: \Art_{/k} \to \sets$ where $A$ is sent to $X_{/A}$ flat with $i: X_{0} \injects X$ where $i\tensor k$ is an isomorphism. The second condition is equivalent to a cartesian diagram \begin{tikzcd} X_{0} \ar[r, hook] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{$\ulcorner$}" , very near start, color=black] & X \ar[d] \\ \spec k \ar[r, hook] & \spec A \end{tikzcd} We showed we always have H1 and H2, and H3 if $X_{0}/k$ is projective or $X_{0}$ is affine with isolated singularities. In this situation we have a miniversal family. This occurs iff for $A' \to A$ a small thickening and $(X_{0} \injects X) \in F(A)$, we have a surjection \[ \Aut_{A'}(X_{0} \injects X') \surjects \Aut_{A}(X_{0} \injects X) .\] where the RHS are automorphisms of $X_{/A}$, i.e. those which commute with the identity on $A$ and $X_{0}$. We had a naive functor $F_{n}$ where we don't include the inclusion $X_{0} \injects X$. When $F$ has a hull then the naive functor has a versal family, since there is a forgetful map that is formally smooth. If it's the case that for all $A' \to A$ small and $F_{\text{n}} \to F_{n}(A)$ we have $\Aut_{A'}(X') \surjects \Aut_{A} (X)$, then $F = F_{n}$ and both are pro-representable. The forgetful map is smooth because given $X_{/A}$ in $F_{n}(A)$, we have some inclusion $X_{0} \injects X$, so one gives surjectivity. Using the surjectivity on automorphisms, we get \begin{tikzcd} X_{0}\ar[rd, hook] \ar[rr, hook] & & X\ar[ld, dotted] \\ & X & \end{tikzcd} Deformation theory is better at answering when the following diagrams exist: \begin{tikzcd} X \ar[r, dotted, hook, "\exists?"] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{$\ulcorner$}" , shift right=0.4em, very near start, color=black] & X' \ar[d, dotted, "\exists?"] \\ \spec A \ar[r] & \spec A' \end{tikzcd} i.e., the existence of an extension of $X$ to $A'$. This is different than understanding diagrams of the following type, where we're considering isomorphism classes of the squares, and deformation theory helps understand the blue one: \begin{tikzpicture} [ greenbox/.style={ draw=green, fill=green!3, thick, rounded corners, rectangle }, redbox/.style={ draw=red, fill=red!3, thick, rounded corners, rectangle }, ] \node[ greenbox, minimum height=0.9cm, minimum width=1.2cm ] at (-0.1, 1.3) {}; \node[ redbox, minimum height=0.9cm, minimum width=1.2cm ] at (2.35, 1.3) {}; \node[ greenbox, minimum height=2.4cm, minimum width=8.2cm ] at (0, -0.5) {}; \node[ draw=red, thick, rectangle, minimum height=0.8cm, minimum width=1.2cm ] at (-1.2, -0.6) {}; \node[ draw=blue, thick, rectangle, dotted, minimum height=0.8cm, minimum width=1.2cm ] at (1.2, -0.6) {}; \node at (0, 0) {% \begin{tikzcd} & F(A') \ar[r] & F(A) \\ X_0 \ar[r, hook] \ar[d] & X \ar[r, hook] \ar[d] & X' \ar[d] \\ \spec k \ar[r] & \spec A \ar[r] & \spec A' \end{tikzcd} }; \end{tikzpicture} :::{.example title="Hypersurface Singularities"} Take $S = k[x, y]$ and $B = S/(f)$, then deformations of $\spec B$ to ? Given $k \to k[\eps] \to k$ we can tensor[^tensor_up_to_iso] to obtain \begin{tikzcd} {0} & {B} & {B'} & {B} & {0} \\ {0} & {S} & {S[\eps]} & {S} & {0} \\ {0} & {I} & {I'} & {I} & {0} \\ && {\tiny \gens{f'}} & {\tiny \gens{f}} \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-2, to=2-2] \arrow["{\pi}", from=2-2, to=1-2] \arrow[from=3-3, to=2-3] \arrow["{\pi'}", from=2-3, to=1-3] \arrow[from=3-4, to=2-4] \arrow["{\pi}", from=2-4, to=1-4] \arrow["{\subseteq}" description, from=4-3, to=3-3, no head] \arrow["{\subseteq}" description, from=4-4, to=3-4, no head] \end{tikzcd} > [Link to diagram.](https://q.uiver.app/?q=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) ![Diagram](figures/image_2020-04-14-12-55-58.png){width=350px} We want to understand $F(k[\eps])$. We know $f' = f + \eps g$ for some $g\in S$. :::{.observation} \envlist 1. $g\in B$ and $f'' = f + \eps(g + cf)$ generates the same ideal. 2. We're free to reparameterize, i.e. $x \mapsto x + \eps a$ and $y \mapsto y + \eps b$ and thus \[ g \mapsto g + a f_{x} + b f_{y} \] , i.e. the partial derivatives. ::: Thus isomorphism classes of $B'$ in deformations $B' \to B$ only depend on the isomorphism classes $g\in B/(f_{x}, f_{y}) B$. When the singularities are isolated, this quotient is finite-dimensional as a $k\dash$vector space. [^tensor_up_to_iso]: For flat maps, tensoring up to an isomorphism implies isomorphism. ::: :::{.example title="?"} $F(k[\eps]) = B/(f_{x}, f_{y})B$. Thus H3 holds and there is a miniversal family $h_{R} \to F$. We can describe it explicitly: take $g_{i} \in S$, yielding a $k\dash$basis in $S/(f, f_{x}, f_{y})$. Then \[ V(f + \sum t_{i} g_{i}) \subset \spec k[[t_{1}, \cdots, t_{n}]][x, y] .\] Set $R = k[[t_{1}, \cdots, t_{n}]]$, then this lands in $\AA_{R}^2$. ::: :::{.example title="?"} The nodal curve $y^2 = x^3$, take . \[ S/(y^2-x^3, 2y, -3x^2) = S/(y, x^2) .\] So take $g_{1} = 1, g_{2} = x$, then the miniversal family is . \[ V(y^2 - x^3 + t + t_{2} x) \subset \AA^2_{k[[t_{1}, t_{2}]]} .\] This gives all ways of smoothing the node. ::: :::{.remark} Note that none of these are pro-representable. ::: Given $X$ and $A$, we obtain a miniversal family over the formal spectrum $\mathrm{Spf}(R) = (R, \xi)$ and a unique map: ![Diagram](figures/image_2020-04-14-13-10-21.png){width=350px} We can take two deformations over $A = k[\xi]/ S^n$: - $X_{1} = V(x + y)$?? - $X_{2} = V(x + uy)$?? As deformations over $A$, $X_{1} \cong X_{2}$ where we send , \[ s&\mapsto s, \\ y&\mapsto y, \\ x&\mapsto ux .\] since \[ (xy + us) = (uxy + us) = (u(xy + s)) = (xy + s) .\] But we have two different classifying maps, which do commute up to an automorphism of $A$, but are not equal. Since they pullback to different elements (?), $F$ can not be pro-representable. ![Diagram](figures/image_2020-04-14-13-20-05.png){width=350px} So reparameterization in $A$ yield different objects in $F(A)$. In other words, $\mcx \to \mathrm{Spf}(R)$ has automorphisms inducing reparameterizations of $R$. This indicates why we need maps restricting to the identity. ## The Cotangent Complex For $X \mapsvia{f} Y$, we have $L_{X/Y} \in D \qcoh(X)$, the derived category of quasicoherent sheaves on $X$. This answers the extension question: :::{.answer} For any square-zero thickening $Y \injects Y'$ (a closed immersion) with ideal $I$ yields an $\OO_{Y}\dash$module. 1. An extension exists iff $0 = \mathrm{obs} \in \ext^2(L_{X/Y}, f^* I)$ 2. If so, the set of ways to do so is a torsor over this ext group. 3. The automorphisms of the completion are given by $\hom(L_{X/Y}, f^* I)$. ::: :::{.remark} Some special cases: $X \to Y$ smooth yields $L_{X/Y} = \Omega_{X/Y}[0]$ concentrated in degree zero. ::: :::{.example title="?"} $Y = \spec k$ and $Y' = \spec k[\eps]$ yields \[ \mathrm{obs} \in \Ext_{x}^2(\Omega_{X/Y}, \OO_{x})= H^2(T_{X_{/k}}) .\] For $X\injects Y$ is a regular embedding (closed immersion and locally a regular sequence) $L_{X/Y} = \qty{I/I^2}[1]$, the conormal bundle. ![Diagram](figures/image_2020-04-14-13-32-13.png){width=350px} ::: :::{.example title="?"} For $Y$ smooth, $X \injects Y$ a regular embedding, $L_{X_{/k}} = \Omega_{X_{/k}}$ with $\mathrm{obs}/\mathrm{def} = \Ext^{2/1}(\Omega_{x}, \OO)$ and the infinitesimal automorphisms are the homs. ::: :::{.example title="?"} For $Y = \spec k[x, y] = \AA^2$ and $X = \spec B = V(f) \subset \AA^2$ we get \[ 0 \to I/I^2 \to \Omega_{X_{/k}} \tensor B &\to \Omega_?{X_{/k}} \to 0 \\ \\ & \Downarrow \quad \text{equals} \\ \\ 0 \to B \mapsvia{1 \mapsto (f_{x}, f_{y})} &B^2 \to \Omega_{B_{/k}} = L_{X_{/k}} \to 0 .\] Taking $\hom(\wait, B)$ yields \begin{tikzcd} 0 \ar[r] & \hom(\Omega, B) \ar[r] & B^2 \ar[lld, "{(f_{x}, f_{y})^t}"] \\ \Ext^1(\Omega, B) \ar[r] & 0 \ar[r] & 0 \ar[lld] \\ \Ext^2(\Omega, B) \ar[r] & 0 \ar[r] & 0 \end{tikzcd} So , \[ \mathrm{obs} &= 0 \\ \mathrm{def} &= B/(f_{x}, f_{y})B \\ \aut &\neq 0 .\] and ::: :::{.remark} We have the following obstruction theories: - For abstract deformations, we have \[ X_{0} {}_{/k} \text{ smooth } \implies \aut/\mathrm{def}/\mathrm{obs} = H^{0/1/2}(T_{X_{0}}) .\] - For embedded deformations, $Y_{0}/k$ smooth, $X_{0} \injects Y_{0}$ regular, we have \[ \aut/\mathrm{def}/\mathrm{obs} = 0, H^{0/1}(N_{X_{0}/Y_{0}}) .\] > As an exercise, interpret this in terms of $L_{X_{0}/Y_{0}}$. - For maps $X_{0} \mapsvia{f_{0}} Y_{0}$, i.e. maps \[ X_{0} \cross k[\eps] \mapsvia{f} Y_{0} \cross k[\eps] .\] we consider the graph $\Gamma(f_{0}) \subset X_{0} \cross Y_{0}$. ![Diagram](figures/image_2020-04-14-13-43-40.png){width=350px} Since all of these structures are special cases of the cotangent complex, they place nicely together in the following sense: Given $X \injects_{i} Y$ we have \[ 0 \to T_{X} \to i^* T_{Y} \to N_{X/Y} \to 0 .\] Yielding a LES \[ 0 &\to H^0(T_{X}) \to H^0(i^* T_{Y}) \to H^0(N_{X/Y}) \\ &\to H^1(T_{X}) \to H^1(i^* T_{Y}) \to H^1(N_{X/Y}) \\ &\to H^2(T_{X}) .\] ![Diagram](figures/image_2020-04-14-13-47-05.png){width=350px} ::: :::{.exercise title="?"} Consider $X \subset \PP^3$ a smooth quartic, and show that $\mathrm{def}(X) \cong k^{20}$ but $\mathrm{def}_{\text{embedded}} \cong k^{19}$. This is a quartic K3 surface for which deformations don't lift (non-algebraic, don't sit inside any $\PP^n$). ::: > Next time: > Obstruction theory of sheaves, T1 lifting as a way to show unobstructedness. # Characterization of Smoothness (Thursday April 16th) > Recap from last time: the cotangent complex answers an extension problem. Given $X \mapsvia{f} Y$ and $Y \injects Y'$ a square zero thickening. When can the pullback diagram be filled in? \begin{tikzcd} X \ar[r, dotted] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{\color{black}$\lrcorner$}" , very near start, color=black] & X' \ar[d, dotted] \\ Y \ar[r] & Y' \end{tikzcd} - The existence is governed by $\mathrm{obs} \in \Ext^2( L_{X/Y}, f^* I)$ - The number of extensions by $\Ext^1( L_{X/Y}, f^* I)$ - The automorphisms by $\Ext^0( L_{X/Y}, f^* I)$ Suppose we're considering $k[\eps] \to k$, where $L_{X_{/k}} = \Omega_{X_{/k}}$, and $H^*(T_{X_{/k}})$ houses the obstruction theory. For an embedded deformation $X \injects Y$, we have \begin{tikzcd} X \ar[r, dotted] & X' \ar[d, dotted] \\ Y \ar[r] & Y \cross_{\spec k} \spec k[\eps] \end{tikzcd} then $L_{X/Y} = I/I^2 [1] = N_{X/Y}\dual[1]$ and \[ \mathrm{obs} \in \ext^2(N\dual[1], \OO) = \Ext^1(N\dual, \OO) = H^1(N) .\] and similarly $\mathrm{def} = H^0(N)$ and $\aut = 0$. For $X \mapsvia{f} Y$, we can think of this as an embedded deformation of $\Gamma \subset X \cross Y$, in which case $N\dual = F^* \Omega_{Y_{/k}}$. Then $\mathrm{obs}, \mathrm{def} \in H^{1, 0}(f^* T_{X_{/k}})$ respectively and $\aut = 0$. There is an exact triangle \[ f^* L_{Y_{/k}} \to L_{X_{/k}} \to L_{X/Y} \to f^* L_{Y_{/k}}[1] .\] ## T1 Lifting This will give a criterion for a pro-representable functor to be smooth. We've seen a condition on $F$ with obstruction theory for the hull to be smooth, namely $\mathrm{obs}(F) = 0$. However, often $F = h_{R}$ will have $R$ smooth with a natural obstruction theory for which $\mathrm{obs}(F) \neq 0$. :::{.example title="?"} For $X_{/k}$ smooth projective, the picard functor $\pic_{X_{/k}}$ is smooth because we know it's an abelian variety. We also know that the natural obstruction space is $\mathrm{obs} = H^2(\OO_{X})$, which may be nonzero. We could also have abstract deformations given by $H^2(T_{X})$ Given $A \in \Art_{/k}$ and $M$ a finite length $A\dash$module, we can form the ring $A \oplus M$ where $M$ is square zero and $A\actson M$ by the module structure. This yields \[ 0 \to M \to A \oplus M \to A \to 0 \] The explicit ring structure is given by $(x, y) \cdot (x, y') = (xx', x'y + xy')$. ::: :::{.proposition title="Characterization of Smoothness"} Assume $\ch k =0$ and $F$ is a pro-representable deformation functor, so $F = \hom(R, \cdot)$ where $R$ is a complete local $k\dash$algebra with $\dim t_{R} < \infty$. Then $R$ is smooth[^smoothness_reminder] over $k$ $\iff$ for all $A\in \Art_{/k}$ and all $M, M' \in A\dash\text{mod}$ finite dimensional with $M \surjects M'$, we have \[ F(A\oplus M) \surjects F(A\oplus M') .\] [^smoothness_reminder]: I.e. $R \cong k[[t_{R}\dual]]$. ::: ### Proof of Proposition :::{.observation} First observe that $\ker(F(A\oplus M) \to F(A)) = \ker(\hom(R, A\oplus M) \to \hom(R, A))$, note that if we have two morphisms \begin{tikzcd} R \ar[r] & R \ar[r, shift left=0.75ex, "g \oplus g"] \ar[r, shift right=0.75ex, "f \oplus g'"'] & A \oplus M \end{tikzcd} denoting these maps $h, h'$ we have 1. $g-g' \in \Der_{k}(R, M)$, since \[ (h-h')(x, y) &= h(x)h(y) - h'(x) h'(y) \\ &= (f(x)f(y), f(x)g(y) + f(y)g(x) ) - (f(x)f(y), f(x) g'(y) + f(y) g'(x)) \\ &= f(x)(g-g')(y) + f(y)(g-g')(x) .\] 2. Given $g: R\to A\oplus M$ and $\theta \in \Der_{k}(R, M)$, then $g + \theta: R \to A\oplus M$. ::: We conclude that the fibers are naturally torsors for $\Der_{k}(R, M)$ if nonempty. It is in fact a canonically trivial torsor, since there is a distinguished element in each fiber. Thus to show the following, it is enough to show surjection on fibers and trivial extensions go to trivial ones, then $\Der_{k}(R, M) \to \Der_{k}(R, M')$ with $0\mapsto 0$. \begin{tikzcd} F(A\oplus M) \ar[rr] \ar[rd] & & F(A\oplus M') \ar[ld] \\ & F(A) & \end{tikzcd} The criterion for $F$ being surjective is equivalent to \[ \Der_{k}(R, M) &\surjects \Der_{k}(R, M') \\ \\ &\Downarrow \qquad \text{identified as }\\ \\ \hom_{R}(\Omega_{R_{/k}}, M) &\surjects \hom(\Omega_{R_{/k}'}, M') .\] :::{.warnings} $\Omega_{R_{/k}}$ is complicated. An example is \[ \Omega_{k[[x]]/k} \tensor k((x)) = \Omega_{k((x))/k} .\] which is an infinite dimensional $k((x))$ vector space. ::: Here we only need to consider the completions $\hom_{R}(\hat \Omega_{R_{/k}}, M) \surjects \hom(\hat \Omega_{R_{/k}'}, M') = k[[x]]~dx$. :::{.fact} In characteristic zero, $R?k$ is smooth iff $\hat \Omega_{R_{/k}}$ is free. ::: Thus the surjectivity condition is equivalent to checking that $\hom(\hat \Omega_{R_{/k}}, \wait)$ is right-exact on finite length modules. This happens iff $\hat \Omega$ are projective iff they are free. :::{.fact title="from algebra"} Uses an algebra fact: for a complete finitely-generated module $M$ over a complete ring, then $M$ is free if $M$ projective with respect to sequences of finite-length modules. Over a local ring, finitely-generated and projective implies free. ::: :::{.remark} This is powerful -- allows showing deformations of Calabi-Yaus are unobstructed! ::: :::{.definition title="Calabi-Yau"} A smooth projective $X_{/k}$ is **Calabi-Yau** iff \[ \omega_{x} \cong \OO_{x} ,\] i.e. the canonical bundle is trivial. ::: :::{.proposition title="?"} $X_{/k}$ CY with $H^0(T_{X}) = 0$ (implying that the deformation functor $F$ of $X$ is pro-representable, say by $R$, and has no infinitesimal automorphisms) has unobstructed deformations, i.e. $R$ is smooth of dimension $H^1(T_{X})$. ::: Note that $H^2(T_{X}) \neq 0$ in general, so this is a finer criterion. :::{.example title="?"} Take $X \subset \PP^4$ a smooth quintic threefold. - By adjunction, this is Calabi-Yau since \[ \omega_{x} = \omega_{\PP^4}(5) \mid_{X} = \OO_{x} .\] - By Lefschetz, \[ H^i_\mathrm{sing} (\PP^4, \CC) &\mapsvia{\cong} H^i_{\mathrm{sing}}(X, \CC) && \text{except in middle dimension} \\ \\ &\Downarrow \quad \text{ implies} \\ \\ H^{3, 1} &= H^{1, 3} = 0 .\] - By Serre duality, \[ H^0(T_{x}) &= 0 \cong H^4(\Omega_{x} \tensor \omega_{x}) \\ \\ &\Downarrow \quad\text{implies} \\ \\ H^3(\Omega_{x}) &= H^{3, 1} = 0 .\] ::: :::{.exercise title="?"} There are nontrivial embedded deformations that yield the same abstract deformations, write them down for the quintic threefold. ::: :::{.claim} The abstract moduli space here is given by $\PGL(5) \sm \hilb$ where $\hilb$ is smooth. ::: ### Proof that obstructions to deformations of Calabi-Yaus are unobstructed We need to show that for any $M \surjects M'$ that \[ F(A\oplus M) \surjects F(A\oplus M') .\] The fibers of the LHS are extensions from $A$ to $A\oplus M$, and the RHS are extensions of $X/A$? By dualizing, we need to show $H^1(T_{X/A}\tensor M ) \surjects H^1(T_{X/A} \tensor M')$ since the LHS is $\ext^1(\Omega_{X/A}, M)$. We want the bottom map here to be surjective: \begin{tikzcd} X \ar[d] & X' \ar[d] \\ \spec A \ar[r, hook] & \spec A \oplus M \end{tikzcd} :::{.fact title="Important"} For $X/A$ a deformation of a CY, $H^*(T_{X/A})$ is free. This will finish the proof, since the map is given by $H^1(T_{X/A}) \tensor M \surjects H^1(T_{X/A}) \tensor M'$ by exactness. This uses the fact that there's a spectral sequence \[ \Tor_{q}(H^p(T_{X/A}), M) \implies H^{p+q} (T_{X/A} \tensor M) \] which follows from base change and uses the fact that $T_{X/A}$ is flat. ::: We'll be looking at $\Tor_{1}(H^0(T_{X/A}), M)$ which is zero by freeness. Hodge theory is now used: by Deligne-Illusie, for $X\mapsvia{f} S$ smooth projective, taking pushforwards $R^p f_* \Omega^q_{X_{/S}}$ are free (coming from degeneration of Hodge to de Rham) and commutes with base change. :::{.remark} This implies that $\omega_{X/A} = \OO_{X}$ is trivial. Using Deligne-Illusie, since $\omega$ is trivial on the special fiber, $H^0(\omega_{X/A}) = A$ is free of rank 1. We thus have a section $\OO_{X} \to \omega_{X/A}$ which is an isomorphism by flatness, since it's an isomorphism on the special fiber. ::: :::{.remark} By Serre duality, $H^1(T_{X/A}) = H^{n-1}(\Omega_{X/A} \tensor \omega_{X/A}) \dual = H^{n-1}(\Omega_{X/A})\dual$, which is free by Deligne-Illusie. This also holds for $H^0(T_{X/A}) = H^n(\Omega_{X/A})\dual$ is free. ::: Thus deformations of Calabi-Yaus are unobstructed. ### Remarks :::{.remark} In fact we need much less. Take $A_{n} = k[t] / t^n$, then consider \begin{tikzcd} 0 \ar[r] & A_n \ar[r] & A_n[\eps] \ar[r] & A_n \\ 0 \ar[r] \ar[u, equal] & A_n \ar[r] \ar[u, equal] & A_n \oplus \eps A_n \ar[r] \ar[u, equal] & A_n \ar[u, equal] \end{tikzcd} For a deformation $X/A_{n}$, let $T^1(X/A_{n}) = \ker(F(A_{n}[\eps]) \to F(A_{n}) )$, the fiber above $X/A_{n}$. Then Kuramata shows that one only needs to show surjectivity for these kinds of extensions, which is quite a bit less. ::: In the T1 lifting theorem, the condition is equivalent to the following: For any deformation $X/A_{n+1}$, there is a map \[ T^1(X/A_{n+1}) \to T^1(X\tensor A_{n} / A_{n}) .\] and surjectivity is equivalent to the lifting condition. In the CY situation, the extension group $T^1(X/A_{n+1}) = H^1(T_{X/A_{n+1}})$ and the RHS is $H^1(T_{X\tensor A_{n} / A_{n}})$. So the slogan for the T1 lifting property is the following: :::{.slogan} If the deformation space is free and commutes with base change, then deformations are unobstructed. ::: Commuting with base change means the RHS is $H^1(T_{X/A_{n}}) \tensor A_{n}$, so we just need to show it's free? # Monday April 27th ## Principle of Galois Cohomology Let $\ell_{/k}$ a galois extension and $X_{/k}$ some "object" for which it makes sense to associate another object over $\ell$. We'll prove that there's a correspondence \[ \correspond{ \ell_{/k}, \text{ twisted forms} \\ Y \text{ of } X_{/k} } &\mapstofrom H^1(\ell_{/k}, \aut(X_{/\ell})) .\] Recall that $\PGL(n ,\ell) \da \GL(n ,\ell) / \ell\units$. :::{.example title="?"} Let $X = \PP^{n-1}/k$, then $H^1(\ell_{/k}, \PGL(n, \ell)$ parameterizes twisted forms of $\PP^{n-1}$, e.g. for $n=2$ twisted forms of $\PP^1$ and plane curves. ::: :::{.example title="?"} Take $X = M_{n}(k)$ the algebra of $n\times n$ matrices. Then by a theorem (Skolern-Noether) $\aut(M_{n}(k)) = \PGL(n, k)$. Thus $H^1(\ell_{/k}, \PGL(n, k))$ also parameterizes twisted forms of $M_{n}(k)$ in the category of unital (not necessarily commutative) $k\dash$algebras. These are exactly central simple algebras $A_{/k}$ where $\dim_{k} A = n^2$ with center $Z(A) = k$ with no nontrivial two-sided ideals. By taking \(\ell = k^{s} \), we get a correspondence \[ \correspond{\text{CSAs} A_{/k} \text{ of degree } n} &\mapstofrom \correspond{\text{ Severi-Brauer varieties of dimension n-1} } .\] Taking $n=2$ we obtain \[ \correspond{\text{Quaternion algebras } A_{/k}} &\mapstofrom \correspond{\text{Genus 0 curves } \ell_{/k}} .\] ::: ## The Weil Descent Criterion Fix $\ell_{/k}$ finite Galois with $g \da \aut(\ell_{/k})$. 1. $X_{/k} \to X_{/\ell}$ with a $g\dash$action. 2. What additional data on an $\ell\dash$variety $Y_{/\ell}$ do we need in order to "descend the base" from $\ell$ to $k$? For $\sigma \in g$, write $\ell^\sigma$ to denote $\ell$ given the structure of an $\ell\dash$algebra via $\sigma: \ell \to \ell^\sigma$. If $X_{/\ell}$ is a variety, so is $X^\sigma_{/\ell}$? \begin{tikzcd} X^\sigma\ar[dr, dotted] \ar[r]\ar[d] & X \ar[d] \\ \spec \ell^\sigma \ar[r, "f"] & \spec \ell \end{tikzcd} where $f$ is the map induced on $\spec$ by $\sigma$. We can also think of these on defining equations: \[ X &= \spec \ell[t_{1}, \cdots, t_{n}] / \gens{p_{1}, \cdots, p^n} \\ X^\sigma &= \spec \ell[t_{1}, \cdots, t_{n}] / \gens{\sigma_{p_{1}}, \cdots, \sigma{p^n}} \\ .\] For $X_{/k}, X_{/\ell}$, we canonically identify $X$ with $X^\sigma$ by the map $f_\sigma: X \mapsvia{\cong} X^\sigma$, a canonical isomorphism of $\ell\dash$varieties. We thus have \begin{tikzcd} X \ar[r, "f_\sigma"] \ar[rr, bend left, "f_{\sigma \tau}"] & X^\sigma \ar[r, "f_\sigma"] & X^{\sigma \tau} \end{tikzcd} under a "cocycle condition" $f_{\sigma \tau} = {}^\sigma f_\tau \circ f_\sigma$. :::{.theorem title="Weil"} Given $Y_{/\ell}$ quasi-projective and $\forall \sigma \in g$ we have descent datum $f_\sigma: Y\mapsvia{\cong} Y^\sigma$ satisfying the above cocycle condition, and there exists a unique $X_{/k}$ such that $X_{/\ell} \mapsvia{\cong} Y_{/\ell}$ and the descent data coincide. ::: ### An Application Let $X_{/k}$ be a quasiprojective variety and $Y_{/k}$ and $\ell_{/k}$ twisted forms. Then $a_{0} \in Z' (\ell_{/k}, \aut X)$. Conversely, we have the following: :::{.definition title="Twisted Descent Data"} Let $a_{0}$ be such a cocycle and $\theset{s_\sigma: X\to X^\sigma}$ be descent datum attached to $X$. Define twisted descent datum $g_\sigma \da f_\sigma \circ a_\sigma$ from \[ X /\ell\mapsvia{a_\sigma} X_{/\ell} \mapsvia{f_\sigma} X^\sigma / \ell .\] ::: :::{.exercise title="?"} Check that $g_\sigma$ satisfies the cocycle condition, so by Weil uniquely determines a ($k\dash$model) $Y_{/k}$ of $X_{/\ell}$. ::: :::{.example title="?"} Let $G_{/k}$ be a smooth algebraic group and $X_{/k}$ a torsor under $G$. Then $\Aut(G) \supset \aut_{G\dash\text{torsor}} (G) = G$, since in general the translations will only be a subgroup of the full group of automorphisms. Then \[ H^1(\ell_{/k}, G) \to H^1(\ell_{/k}, \aut G) \] defines a twisted form $X$ of $G$. How do you descend the torsor structure? This is possible, but not covered in Bjoern's book! This requires expressing the descent data more functorially -- see the book on Neron models. ::: ## The Cohomology Theory ### Motivation Let $G_{/k}$ be a smooth connected commutative algebraic group where $\ch k$ does not divide $n$, so the map $[n]: G \to G$ is an isogeny. Then \[ 0 \to G[n] (k^{s} ) \to G(k^{s} ) \mapsvia{[n]} G(k^{s} ) \to 0 \] is a SES of $g = \aut(k^{s}_{/k})\dash$modules. :::{.claim} Taking the associated cohomology sequence yields the Kummer sequence: \[ 0 \to G(k) / nG(k) \to H^1(k, G[n]) \to H^1(k, G)[n] \to 0 \] where the RHS is the **Weil-Châtelet** group and the LHS is the **Mordell-Weil** group. ::: For $g$ a profinite group, a commutative discrete $g\dash$group is by definition a $g\dash$module. These form an abelian category with enough injectives, so we can take right-derived functors of left-exact functors. We will consider the functor $$ A \mapsto A^g \da \theset{x\in A \suchthat \sigma x = x ~\forall \sigma\in g} ,$$ then define $H^i(g, A)$ to be the $i$th right-derived functor of $A \mapsto A^\sigma$. This is abstractly defined by taking an injective resolution, applying the functor, then taking cohomology. A concrete description is given by $C^n(g, A) = \Map(g^n, A)$ with \[ d: C^n(g, A) &\to C^{n+1}(g, A) \\ (df)(\sigma_{1}, \cdots, \sigma_{n+1} &\da \sigma_{1} f(\sigma_{2}, \cdots, \sigma_{n+1}) \\ &\qquad + \sum_{i=1}^n (-1) f(\sigma _1, \cdots, \sigma_{i-1}, \sigma_{i}, \sigma_{i+1}, \cdots, \sigma_{n+1}) \\ &\qquad + (-1)^{n+1} f(\sigma_{1}, \cdots, \sigma_{n}) .\] Then $d^2 = 0$, $H^n$ is kernels mod images, and this agrees with $H^1$ as defined before with $H^0 = A^g$. We'll see that that \[ H^i(g, A) = \directlim_{U} G^i(g/U, A^U) .\] If $g$ is finite, $A$ is a $g\dash$module $\iff$ $A$ is a $\ZZ[g]\dash$module, and thus \[ A^g = \hom_{\ZZ[g]\dash\text{mod}}(\ZZ, A) .\] where $\ZZ$ is equipped with a trivial $g\dash$action. We can thus think of \[ H^i(g, A) = \ext^i_{\ZZ[g]}(\ZZ, A) .\] > The end!