# 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 $$S{\hbox{-}}$$scheme, i.e. a scheme over $$S$$. Then there is a map ${\operatorname{Sch}}_{_{/S}} &\to {\operatorname{Fun}}( {\operatorname{Sch}}_{_{/S}}^{\operatorname{op}}, {\operatorname{Set}}) \\ x &\mapsto h_x(T) = \hom_{{\operatorname{Sch}}_{_{/S}} }(T, x) .$ where $$T' \xrightarrow{f} T$$ is given by $h_x(f): h_x(T) &\to h_x(T') \\ \qty{ T \mapsto x } &\mapsto \text{triangles of the form} .$ {=tex} \begin{tikzcd} T' \arrow[rr] \arrow[rdd] & & X \\ & & \\ & T \arrow[ruu] & \end{tikzcd}  ## Representability ::: {.theorem title="?"} $\hom_{{\operatorname{Fun}}}(h_x, F) = F(x).$ ::: ::: {.corollary title="?"} $\hom_{{\operatorname{Sch}}_{/S}}(x, y) \cong \hom_{{\operatorname{Fun}}}(h_x, h_y).$ ::: ::: {.definition title="Moduli Functor"} A **moduli functor** is a map $F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}&\to {\operatorname{Set}}\\ F(x) &= \text{ "Families of something over x" } \\ F(f) &= \text{"Pullback"} .$ ::: ::: {.definition title="Moduli Space"} A **moduli space** for that "something" appearing above is an $$M \in \mathrm{Obj}({\operatorname{Sch}}_{/S})$$ such that $$F \cong h_M$$. ::: ::: {.remark} Now fix $$S = \operatorname{Spec}(k)$$, and write $$h_m$$ for the functor of points over $$M$$. Then $h_m(\operatorname{Spec}(k)) = M(\operatorname{Spec}(k)) \cong \text{families over } \operatorname{Spec}k = F(\operatorname{Spec}k) .$ ::: ::: {.remark} $$h_M(M) \cong F(M)$$ are families over $$M$$, and $$\operatorname{id}_M \in \mathrm{Mor}_{{\operatorname{Sch}}_{/S}}(M, M) = \xi_{Univ}$$ is the universal family. Every family is uniquely the pullback of $$\xi_{\text{Univ}}$$. This makes it much like a classifying space. For $$T\in {\operatorname{Sch}}_{/S}$$, $h_M &\xrightarrow{\cong} F \\ f\in h_M(T) &\xrightarrow{\cong} F(T) \ni \xi = F(f)(\xi_{\text{Univ}}) .$ where $$T\xrightarrow{f} M$$ and $$f = h_M(f)(\operatorname{id}_M)$$. ::: ::: {.remark} If $$M$$ and $$M'$$ both represent $$F$$ then $$M \cong M'$$ up to unique isomorphism. {=tex} \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 $${\mathbb{A}}^1_k$$ (??) then $F(S) = \left\{{ V(x^2 + bx + c)}\right\} \subset {\mathbb{A}}^5$ where $$b, c \in {\mathcal{O}}_s(s)$$, which is functorially bijective with $$\left\{{b, c \in {\mathcal{O}}_s(s)}\right\}$$ and $$F(f)$$ is pullback. Then $$F$$ is representable by $${\mathbb{A}}_k^2(b, c)$$ and the universal object is given by $V(x^2 + bx + c) \subset {\mathbb{A}}^1(?) \times{\mathbb{A}}^2(b, c)$ where $$b, c \in k[b, c]$$. Moreover, $$F'(S)$$ is the set of effective Cartier divisors in $${\mathbb{A}}_5'$$ which are length 2 for every geometric fiber. $$F''(S)$$ is the set of subschemes of $${\mathbb{A}}_5'$$ which are length 2 on all geometric fibers. In both cases, $$F(f)$$ is always given by pullback. ::: Problem: $$F''$$ is not a good moduli functor, as it is not representable. Consider $$\operatorname{Spec}k[\varepsilon]$$, for which we have the following situation: {=tex} \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] {$(\varepsilon+ x - 1)$}; \node[draw, circle, blue, scale=0.4, fill=blue](center1) at (axis cs:0, 6) [anchor=center] {}; \node[right=1mm of center1,font=\tiny] {$(x)(\varepsilon, x-1)$}; \node[draw, circle, blue, scale=0.4, fill=blue](right1) at (axis cs:10, 6) [anchor=center] {}; \node[right=1mm of right1,font=\tiny] {$(x(x-1), \varepsilon)$}; \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}  {=tex} \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}  {=tex} \begin{tikzcd} \operatorname{Spec}k \arrow[rrr, "i", hook] & & & {\operatorname{Spec}k[\varepsilon]} & & =F'(\operatorname{Spec}k) \arrow[rd] & \\ {F(\operatorname{Spec}k[\varepsilon])} \arrow[rrr, "F(i)"] & & & F(\operatorname{Spec}k) \arrow[rru] & & & =F''(\operatorname{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. {=tex} \begin{tikzcd} {M(\operatorname{Spec}k[\varepsilon])} \arrow[rr] & & M(\operatorname{Spec}k) \\ & & \\ T_p M \arrow[rr] \arrow[uu, "\subset", hook]& & p \arrow[uu, "\subset", hook] \end{tikzcd}  {=tex} \begin{tikzcd} & & \operatorname{Spec}k \arrow[rdd, "?"] \arrow[lldd, hook] & \\ & & & \\ {\operatorname{Spec}k[\varepsilon]} \arrow[rrr] & & & {\operatorname{Spec}{\mathcal{O}}_{M, p} \subset M} \\ & & & k \\ & {{\mathcal{O}}_{M, p}} \arrow[rru] \arrow[rr] & & {k[\varepsilon]} \arrow[u] \\ & {\mathfrak{m}}_p \arrow[u, hook] & & (\varepsilon) \arrow[u, hook] \\ & {\mathfrak{m}}_p^2 \arrow[u, hook] & & 0 \arrow[u, hook] \end{tikzcd}  Moreover, $$T_p M = ({\mathfrak{m}}_p / {\mathfrak{m}}_p^2)^\vee$$, and in particular this is a $$k{\hbox{-}}$$vector space. To see the scaling structure, take $$\lambda \in k$$. $\lambda: k[\varepsilon] & \to k[\varepsilon] \\ \varepsilon &\mapsto \lambda \varepsilon \\ \\ \lambda^*: \operatorname{Spec}(k[\varepsilon]) &\to \operatorname{Spec}(k[\varepsilon]) \\ \\ \lambda: M(\operatorname{Spec}(k[\varepsilon])) &\to M(\operatorname{Spec}(k[\varepsilon])) .$ {=tex} \begin{tikzcd} M(\operatorname{Spec}(k[\varepsilon])) \ar[r, "\lambda"] & M(\operatorname{Spec}(k[\varepsilon])) \\ T_pM \ar[r] \ar[u, "\subseteq"] & T_pM \ar[u, "\subseteq"] \end{tikzcd}  **Conclusion**: If $$F$$ is representable, for each $$p\in F(\operatorname{Spec}k)$$ there exists a unique point of $$T_p F$$ that are invariant under scaling. ::: {.remark} If $$F, F', G \in {\operatorname{Fun}}( ({\operatorname{Sch}}_{/S})^{\operatorname{op}}, {\operatorname{Set}})$$, there exists a fiber product {=tex} \begin{tikzcd} F \times_G F' \arrow[rr, dotted] \arrow[dd, dotted] & & F' \arrow[dd] \\ & & \\ F \arrow[rr] & & G \end{tikzcd}  where $(F \times_G F')(T) = F(T) \times_{G(T)} F'(T) .$ ::: ::: {.remark} This works with the functor of points over a fiber product of schemes $$X \times_T Y$$ for $$X, Y \to T$$, where $h_{X \times_T Y}= h_X \times_{h_t} h_Y .$ ::: ::: {.remark} If $$F, F', G$$ are representable, then so is the fiber product $$F \times_G F'$$. ::: ::: {.remark} For any functor $F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}} ,$ for any $$T \xrightarrow{f} S$$ there is an induced functor $F_T: ({\operatorname{Sch}}_{/T}) &\to {\operatorname{Set}}\\ x &\mapsto F(x) .$ ::: ::: {.remark} $$F$$ is representable by $$M_{/S}$$ implies that $$F_T$$ is representable by $$M_T = M \times_S T / T$$. ::: ## Projective Space Consider $${\mathbb{P}}^n_{\mathbb{Z}}$$, i.e. "rank 1 quotient of an $$n+1$$ dimensional free module". ::: {.proposition title="?"} $${\mathbb{P}}^n_{/{\mathbb{Z}}}$$ represents the following functor $F: {\operatorname{Sch}}^{\operatorname{op}}&\to {\operatorname{Set}}\\ S &\mapsto \left\{{ {\mathcal{O}}_S^{n+1} \to L \to 0 }\right\} / \sim .$ where $$\sim$$ identifies diagrams of the following form: {=tex} \begin{tikzcd} {\mathcal{O}}_s^{n+1} \arrow[dd, equal] \arrow[rr] & & L \arrow[dd, "\cong"] \arrow[rr] & & 0 \\ & & & & \\ {\mathcal{O}}_s^{n+1} \arrow[rr] & & M \arrow[rr] & & 0 \end{tikzcd}  and $$F(f)$$ is given by pullbacks. ::: ::: {.remark} $${\mathbb{P}}^n_{/S}$$ represents the following functor: $F_S: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}&\to {\operatorname{Set}}\\ T &\mapsto F_S(T) = \left\{{ {\mathcal{O}}_T^{n+1} \to L \to 0}\right\} / \sim .$ This gives us a cleaner way of gluing affine data into a scheme. ::: ### Proof of Proposition ::: {.remark} Note that $${\mathcal{O}}^{n+1} \to L \to 0$$ is the same as giving $$n+1$$ sections $$s_1, \cdots s_n$$ of $$L$$, where surjectivity ensures that they are not the zero section. So $F_i(S) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0}\right\}/\sim ,$ 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. [^1] ::: ::: {.claim} It is enough to show that each $$F_i$$ and each $$F_{ij}$$ are representable, since we have natural transformations: {=tex} \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 {=tex} \begin{tikzcd} & & F_0 \\ F_{01} \arrow[rru] \arrow[rrd] & & \\ & & F_1 \end{tikzcd}  For $$n=2$$, we get overlaps of the following form: {=tex} \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 $${\mathcal{O}}_s^{n+1} \to L \to 0$$, $$U_i$$ is the locus where $$s_i \neq 0$$ and by surjectivity, this gives a cover of $$S$$. The RHS to LHS comes from gluing. ::: ::: {.proof title="of claim"} We have $F_i(S) = \left\{{{\mathcal{O}}_S^{n+1} \to L \cong {\mathcal{O}}_s \to 0, s_i \neq 0}\right\} ,$ but there are no conditions on the sections other than $$s_i$$.\ So specifying $$F_i(S)$$ is equivalent to specifying $$n-1$$ functions $$f_1 \cdots \widehat{f}_i \cdots f_n \in {\mathcal{O}}_S(s)$$ with $$f_k \neq 0$$. We know this is representable by $${\mathbb{A}}^n$$. We also know $$F_{ij}$$ is obviously the same set of sequences, where now $$s_j \neq 0$$ as well, so we need to specify $$f_0 \cdots \widehat{f}_i \cdots f_j \cdots f_n$$ with $$f_j \neq 0$$. This is representable by $${\mathbb{A}}^{n-1} \times{\mathbb{G}}_m$$, i.e. $$\operatorname{Spec}k[x_1, \cdots, \widehat{x}_i, \cdots, x_n, x_j^{-1}]$$. Moreover, $$F_{ij} \hookrightarrow F_i$$ is open. What is the compatibility we are using to glue? For any subset $$I \subset \left\{{0, \cdots, n}\right\}$$, we can define $F_I = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0, s_i\neq 0 \text{ for } i\in I}\right\} = {\prod{i\in I}}_F F_i ,$ 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 $${\mathbb{P}}_{/{\mathbb{Z}}}^n$$ constructed via a functor of points, i.e. $h_{{\mathbb{P}}^n_{/{\mathbb{Z}}} }: {\operatorname{Sch}}^{\operatorname{op}}&\to {\operatorname{Set}}\\ s &\mapsto {\mathbb{P}}^n_{/{\mathbb{Z}}}(s) = \left\{{ {\mathcal{O}}_s^{n+1} \to L \to 0}\right\} .$ for $$L$$ a line bundle, up to isomorphisms of diagrams: {=tex} \begin{tikzcd} {\mathcal{O}}_{s}^{n+1} \arrow[dd, no head, Rightarrow] \arrow[rr] & & L \arrow[rr] \arrow[dd, "\cong"] & & 0 \\ & & & & \\ {\mathcal{O}}_{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 $$F_i \subset {\mathbb{P}}_{/{\mathbb{Z}}}^n$$ where $F_i(s) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0 {~\mathrel{\Big|}~}s_i \text{ is invertible}}\right\}$ 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: {=tex} \begin{tikzcd} {\mathcal{O}}_{{\mathbb{P}}_{{\mathbb{Z}}}^n}^{n+1} \arrow[rr] & & L \arrow[dd, equal] \arrow[rr] & & 0 \\ & & & & \\ & & {\mathcal{O}}_{{\mathbb{P}}_{{\mathbb{Z}}}^n}(1) & & \end{tikzcd}  and other functors are pullbacks of the universal one. (Moduli Space) ::: ::: {.exercise title="?"} Show that $${\mathbb{P}}_{/{\mathbb{Z}}}^n$$ is proper over $$\operatorname{Spec}{\mathbb{Z}}$$. Use the evaluative criterion, i.e. there is a unique lift {=tex} \begin{tikzcd} \operatorname{Spec}k \arrow[dd] \arrow[rrr] & & & {\mathbb{P}}^n_{{\mathbb{Z}}} \arrow[dd] \\ & & & \\ \operatorname{Spec}R \arrow[rrr] \arrow[rrruu, dashed] & & & \operatorname{Spec}{\mathbb{Z}} \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: {=tex} \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, $$X = \left\{{y {~\mathrel{\Big|}~}f(y) = g(y)}\right\}$$ for $$Y \xrightarrow{f, g} Z$$. ::: ::: {.definition title="?"} A **coequalizer** is the dual notion, {=tex} \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 $$C = {\operatorname{Sch}}_{/S}$$, $$X_{/S}$$ a scheme, and $$X_\alpha \subset X$$ an open cover. We can take two fiber products, $$X_{\alpha \beta}, X_{\beta, \alpha}$$: {=tex} \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 $${\operatorname{Sch}}_{/S}$$, we have {=tex} \begin{tikzcd} {\coprod}_{\alpha\beta} X_{\alpha\beta} \arrow[rr, shift left=.5ex, "f_{\alpha\beta}"] \arrow[rr, shift right=.5ex,"g_{\alpha\beta}", swap] & & {\coprod}_{\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: {=tex} \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 $$\varphi_{\alpha\beta}: X_{\alpha\beta}\xrightarrow{\cong} X_{\beta\alpha}$$ must satisfy the **cocycle condition**: ::: {.definition title="Cocycle Condition"} Maps $$\varphi_{\alpha\beta}: X_{\alpha\beta}\xrightarrow{\cong} X_{\beta\alpha}$$ satisfy the **cocycle condition** iff 1. $\varphi_{\alpha\beta}^{-1}\qty{ X_{\beta\alpha} \cap X_{\beta\gamma} } = X_{\alpha\beta} \cap X_{\alpha \gamma},$ noting that the intersection is exactly the fiber product $$X_{\beta\alpha} \times_{X_\beta} X_{\beta \gamma}$$. 2. The following diagram commutes: {=tex} \begin{tikzcd} X_{\alpha\beta} \cap X_{\alpha\gamma} \arrow[rdd, "\varphi_{\alpha\beta}"'] \arrow[rr, "\varphi_{\alpha\gamma}"] && X_{\gamma\alpha} \cap X_{\gamma\beta} \\ && \\ & X_{\beta\alpha}\cap X_{\beta\gamma} \arrow[ruu, "\varphi_{\beta\gamma}"'] & \end{tikzcd}  ::: Then there exists a scheme $$X_{/S}$$ such that $${\coprod}_{\alpha\beta} X_{\alpha\beta} \rightrightarrows {\coprod}X_\alpha \to X$$ is a coequalizer; this is the gluing. Subfunctors satisfy a patching property because morphisms to schemes are locally determined. Thus representable functors (e.g. functors of points) have to be (Zariski) sheaves. ::: {.definition title="Zariski Sheaf"} A functor $$F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}$$ is a **Zariski sheaf** iff for any scheme $$T_{/S}$$ and any open cover $$T_\alpha$$, the following is an equalizer: $F(T) \to \prod F(T_\alpha) \rightrightarrows \prod_{\alpha\beta} F(T_{\alpha\beta})$ 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 ${\coprod}_{\alpha\beta} T_{\alpha\beta} \to {\coprod}_{\alpha}T_\alpha \to T ,$ 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 $${\operatorname{Fun}}({\operatorname{Sch}}_{/S}, {\operatorname{Set}})$$. - This is a **subfunctor** if $$\iota(T)$$ is injective for all $$T_{/S}$$. - $$\iota$$ is **open/closed/locally closed** iff for any scheme $$T_{/S}$$ and any section $$\xi \in F(T)$$ over $$T$$, then there is an open/closed/locally closed set $$U\subset T$$ such that for all maps of schemes $$T' \xrightarrow{f} T$$, we can take the pullback $$f^* \xi$$ and $$f^*\xi \in F'(T')$$ iff $$f$$ factors through $$U$$. ::: ::: {.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 $$F' \times_F T$$: {=tex} \begin{tikzcd} F' \ar[r] & F \\ & \\ F' \times_F T \ar[r, "g"] \ar[uu] & T \ar[uu, "\xi" swap] \end{tikzcd}  and $$F' \times_F T \cong U$$. Note: this is almost tautological! Thus $$F' \to F$$ is open/closed/locally closed iff $$F' \times_F T$$ is representable and $$g$$ is open/closed/locally closed. I.e. base change is representable. ::: ::: {.exercise title="?"} {=tex} \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 $$T$$: {=tex} \begin{tikzcd} F_\alpha \ar[r] & F \\ & \\ T_\alpha \ar[uu] \ar[r] & T \ar[uu, "\xi" swap] \end{tikzcd}  ::: ::: {.example title="?"} Given $F(s) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0}\right\}$ 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 $$F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}$$ 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 $$F_{\alpha\beta} = F_\alpha \times_F F_\beta$$. Then $$x_\beta$$ yields a section (plus some openness condition?), so $$F_{\alpha\beta} = x_{\alpha\beta}$$ representable. Because $$F_\alpha \subset F$$, the $$F_{\alpha\beta} \to F_\alpha$$ have the correct gluing maps. This follows from Yoneda (schemes embed into functors), and we get maps $$x_{\alpha\beta} \to x_\alpha$$ satisfying the gluing conditions. Call the gluing scheme $$x$$; we'll show that $$x$$ represents $$F$$. First produce a map $$x\to F$$ from the sheaf axioms. We have a map $$\xi \in \prod_\alpha F(x_\alpha)$$, and because we can pullback, we get a unique element $$\xi \in F(X)$$ coming from the diagram $F(x) \to \prod F(x_\alpha) \rightrightarrows \prod_{\alpha\beta} F(x_{\alpha\beta}) .$ ::: ::: {.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 {=tex} \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: {=tex} \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 $${\mathcal{O}}_s$$ module, ${\mathbb{P}}E(T) = \left\{{f^* E \to L \to 0}\right\} / \sim$ is a generalization of projectivization, then $$S$$ admits a cover $$U_i$$ trivializing $$E$$. Then the restriction $$F_i \to {\mathbb{P}}E$$ were $$F_i(T)$$ is the above set if $$f$$ factors through $$U_i$$ and empty otherwise. On $$U_i$$, $$E \cong {\mathcal{O}}_{U_i}^{n_i}$$, so $$F_i$$ is representable by $${\mathbb{P}}_{U_i}^{n_i - 1}$$ by the proposition. Note that this is clearly a sheaf. ::: ::: {.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 = \operatorname{Spec}k$$, then $$E$$ is a $$k{\hbox{-}}$$vector space $$V$$, then sections of the Grassmannian are quotients of $$V \otimes{\mathcal{O}}$$ that are free of rank $$n$$. Take the subfunctor $$G_w \subset {\operatorname{Gr}}(k, V)$$ where $G_w(T) = \left\{{{\mathcal{O}}_T \otimes V \to Q \to 0}\right\} \text{ with } Q \cong {\mathcal{O}}_t\otimes W \subset {\mathcal{O}}_t \otimes V .$ If we have a splitting $$V = W \oplus U$$, then $$G_W = {\mathbb{A}}(\hom(U, W))$$. If you show it's closed, it follows that it's proper by the exercise at the beginning. > Thursday: Define the Hilbert functor, show it's representable. The Hilbert scheme functor gives e.g. for $${\mathbb{P}}^n$$ of all flat families of subschemes. # Thursday January 16th ## Subfunctors ::: {.definition title="Open Functors"} A functor $$F' \subset F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}$$ is **open** iff for all $$T \xrightarrow{\xi} F$$ where $$T = h_T$$ and $$\xi \in F(T)$$. ::: We can take fiber products: {=tex} \begin{tikzcd} F' \ar[r] & F \\ & \\ \parbox{3cm}{\centering $F' \times_F T$ \\ Representable} \ar[r, "\text{Open}"] \ar[uu] & T \ar[uu] \end{tikzcd}  So we can think of "inclusion in $$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$$ in $${\operatorname{Sch}}/T$$, we then have $h_{U/T}(T') = \begin{cases} \emptyset & T' \to T \text{ doesn't factor } \\ {\{\operatorname{pt}\}}& \text{otherwise} \end{cases} .$ which follows because the literal statement is $$h_{U/T}(T') = \hom_T(T', U)$$. By the definition of the fiber product, $(F' \times_F T)(T') = \left\{{ (a,b) \in F'(T) \times T(T) {~\mathrel{\Big|}~}\xi(b) = \iota(a) \text{ in } F(T)}\right\} ,$ where $$F' \xrightarrow{\iota} F$$ and $$T \xrightarrow{\xi} F$$. So note that the RHS diagram here is exactly given by pullbacks, since we identify sections of $$F/T'$$ as sections of $$F$$ over $$T/T'$$ (?). {=tex} \begin{tikzcd} F' \ar[r, "\iota"] & F \\ & & \\ F' \times_F T \ar[uu] \ar[r] & T\ar[uu, "\xi"] \\ & & \\ & & T' \ar[uuuul, "f\circ \xi", bend right] \ar[uul, "f"] \end{tikzcd}  We can thus identify $(F' \times_F T)(T') = h_{U_{/S}}(T') ,$ and so for $$U \subset T$$ in $${\operatorname{Sch}}_{/S}$$ we have $$h_{U_{/S}} \subset h_{T_{/S}}$$ is the functor of maps that factor through $$U$$. We just identify $$h_{U_{/S}}(T') = \hom_S(T', U)$$ and $$h_{T_{/S}}(T') = \hom_S(T', T)$$. ::: {.example title="?"} $${\mathbb{G}}_m, {\mathbb{G}}_a$$. The scheme/functor $${\mathbb{G}}_a$$ represents giving a global function, $${\mathbb{G}}_m$$ represents giving an invertible function. {=tex} \begin{tikzcd} {\mathbb{G}}_m \ar[r] & {\mathbb{G}}_a \\ & \\ T' \arrow[uur, phantom, "\scalebox{1.5}{$\llcorner$}" , very near start, color=black] \ar[uu] \ar[r] & T \ar[uu, "f \in {\mathcal{O}}_T(T)", swap] \end{tikzcd}  where $$T' = \left\{{f\neq 0}\right\}$$ and $${\mathcal{O}}_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$$. ::: {.definition title="The Hilbert Functor"} The **Hilbert functor** is given by $\operatorname{Hilb}_{X_{/S}}: ({\operatorname{Sch}}_{/S})^{op} \to {\operatorname{Set}}$ which sends $$T$$ to closed subschemes $$Z \subset X \times_S T \to T$$ which are flat over $$T$$. ::: Here **flatness** will replace the Cartier condition: ::: {.definition title="Flatness"} For $$X \xrightarrow{f} Y$$ and $${\mathbb{F}}$$ a coherent sheaf on $$X$$, $$f$$ is **flat** over $$Y$$ iff for all $$x\in X$$ the stalk $$F_x$$ is a flat $${\mathcal{O}}_{y, f(x)}{\hbox{-}}$$module. ::: ::: {.remark} Note that $$f$$ is flat if $${\mathcal{O}}_x$$ is. Flatness corresponds to varying continuously. Note that everything works out if we only play with finite covers. ::: ::: {.remark} If $$X_{/k}$$ is projective, so $$X \subset {\mathbb{P}}^n_k$$, we have line bundles $${\mathcal{O}}_x(1) = {\mathcal{O}}(1)$$. For any sheaf $$F$$ over $$X$$, there is a Hilbert polynomial $$P_F(n) = \chi(F(n)) \in {\mathbb{Z}}[n]$$, i.e. we twist by $${\mathcal{O}}(1)$$ $$n$$ times. The cohomology of $$F$$ isn't changed by the pushforward into $${\mathbb{P}}_n$$ since it's a closed embedding, and so $\chi(X, F) = \chi({\mathbb{P}}^n, i_* F) = \sum (-1)^i \dim_k H^i({\mathbb{P}}^n, i_* F(n)) .$ ::: ::: {.fact} For $$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$$, we can define a **Hilbert polynomial**, $P_F(n) = \chi(F\otimes L^n) .$ This is an invariant of a polarized projective variety, and in particular subschemes. Over irreducible bases, flatness corresponds to this invariant being constant. ::: {.proposition title="?"} For $$f:X\to S$$ projective, i.e. there is a factorization: {=tex} \begin{tikzcd} X \arrow[ddr, "f"] \arrow[rr, hook] & & {\mathbb{P}}^n \times S \ni {\mathcal{O}}(1) \ar[ddl] \\ & & \\ & S & \end{tikzcd}  If $$S$$ is reduced, irreducible, locally Noetherian, then $$f$$ is flat $$\iff$$ $$P_{{\mathcal{O}}_{x_s}}$$ is constant for all $$s\in S$$. ::: ::: {.remark} To be more precise, look at the base change to $$X_1$$, and the pullback of the fiber? $${\mathcal{O}}\mathrel{\Big|}_{x_i}$$? Note that we're not using the word "integral" here! $$S$$ is flat $$\iff$$ the Hilbert polynomial over the fibers are constant. ::: ::: {.example title="?"} The zero-dimensional subschemes $$Z \in {\mathbb{P}}^n_k$$, then $$P_Z$$ is the length of $$Z$$, i.e. $$\dim_k({\mathcal{O}}_Z)$$, and $P_Z(n) = \chi({\mathcal{O}}_Z \otimes{\mathcal{O}}(n)) = \chi({\mathcal{O}}_Z) = \dim_k H^0(Z; {\mathcal{O}}_Z) = \dim_k {\mathcal{O}}_Z(Z) .$ For two closed points in $${\mathbb{P}}^2$$, $$P_Z = 2$$. Consider the affine chart $${\mathbb{A}}^2 \subset {\mathbb{P}}^2$$, which is given by $\operatorname{Spec}k[x, y]/(y, x^2) \cong k[x]/(x^2)$ and $$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 = \operatorname{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{\hbox{-}}$$module, and is a free $$A{\hbox{-}}$$module for $$n\gg 0$$. ::: ::: {.proof title="of lemma"} Assumed $$X$$ was projective, so just take $$X = {\mathbb{P}}_A^n$$ and let $$F$$ be the pushforward. There is a correspondence sending $$F$$ to its ring of homogeneous sections constructed by taking the sheaf associated to the graded module\ $\sum_{n\gg0} H^0( {\mathbb{P}}_A^m; F(n) ) = \bigoplus_{n \gg 0} H^0({\mathbb{P}}_A^m; F(n))$ and taking the associated sheaf ($$Y \mapsto \tilde Y$$, as per Hartshorne's notation) which is free, and thus $$F$$ is free. [^2] Conversely, take an affine cover $$U_i$$ of $$X$$. We can compute the cohomology using Čech cohomology, i.e. taking the Čech resolution. We can also assume $$H^i({\mathbb{P}}^m; F(n)) = 0$$ for $$n \gg 0$$, and the Čech complex vanishes in high enough degree. But then there is an exact sequence $0 \to H^0({\mathbb{P}}^m; F(n)) \to \mathcal C^0( \underline{U}; F(n) ) \to \cdots \to C^m( \underline{U}; F(n) ) \to 0 .$ Assuming $$F$$ is flat, and using the fact that flatness is a 2 out of 3 property, the images of these maps are all flat by induction from the right. Finally, local Noetherian and finitely generated flat implies free. ::: By the lemma, we want to show $$H^0({\mathbb{P}}^m; F(n))$$ is free for $$n\gg 0$$ iff the Hilbert polynomials on the fibers $$P_{F_S}$$ are all constant. ::: {.claim title="1"} It suffices to show that for each point $$s\in \operatorname{Spec}A$$, we have $H^0(X_s; F_S(n)) = H^0(X; F(n)) \otimes k(S)$ for $$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, {\,\cdot\,})$$, and for $$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 {\mathbb{Z}}[n]$$ for $$X_{/S}$$ projective, we define a subfunctor by picking only those with Hilbert polynomial $$p$$ fiberwise as $$\operatorname{Hilb}^P_{X_{/S}} \subset \operatorname{Hilb}_{X_{/S}}$$. This is given by $$Z \subset X \times_S T$$ with $$P_{Z} = P$$. ::: ::: {.theorem title="Grothendieck"} If $$S$$ is Noetherian and $$X_{/S}$$ projective, then $$\operatorname{Hilb}_{X_{/S}}^P$$ is representable by a projective $$S{\hbox{-}}$$scheme. > See **cycle spaces** in analytic geometry. ::: # Hilbert Polynomials (Thursday January 23) Some facts about the Hilbert polynomial: 1. For a subscheme $$Z \subset {\mathbb{P}}_k^n$$ with $$\deg P_z = \dim Z = n$$, then $p_z(t) = \deg z \frac{t^n}{n!} + O(t^{n-1}) .$ 2. We have $$p_z(t) = \chi({\mathcal{O}}_z(t))$$, consider the sequence $0 \to I_z(t) \to {\mathcal{O}}_{{\mathbb{P}}^n}^{(t)} \to {\mathcal{O}}_z^{(t)} \to 0 ,$ then $$\chi(I_z(t)) = \dim H^0( {\mathbb{P}}^n, J_z(t) )$$ for $$t \gg 0$$, and $$p_z(0)$$ is the Euler characteristic of $${\mathcal{O}}_Z$$. ::: {.remark} Keywords to look up here: Serre vanishing, Riemann-Roch, ideal sheaf. ::: ::: {.example title="The twisted cubic"} {=tex} \begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1, scale=0.6, every node/.style={scale=0.6}] \node (myfirstpic) at (325,200) {\includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_17-27}}; \node[scale=2.0] at (400, 180) {$C$}; \node[scale=2.0] at (200, 0) {${\mathbb{P}}^3$}; \draw[very thick, blue] (-50,400) -- (-50,100); \draw[thick] (-50-20,400) -- (-50+20,400); \draw[thick] (-50-20,100) -- (-50+20,100); \node[scale=2.0] at (-25, 100-20) {${\mathbb{P}}^1$}; \draw [thick, right hook-latex ] (-50+20, 200) -- (150, 200); \node[scale=2.0] at (50, 180) {$\iota$}; \end{tikzpicture}  Then $p_C(t) = (\deg C)t + \chi({\mathcal{O}}_{{\mathbb{P}}^1}) = 3t + 1 .$ ::: ### Hypersurfaces Recall that length 2 subschemes of $${\mathbb{P}}^1$$ are the same as specifying quadratics that cut them out, each such $$Z \subset {\mathbb{P}}^1$$ satisfies $$Z = V(f)$$ where $$\deg f = d$$ and $$f$$ is homogeneous. So we'll be looking at $${\mathbb{P}}H^0({\mathbb{P}}^n_k, {\mathcal{O}}(d))^\vee$$, and the guess would be that this is $$\operatorname{Hilb}_{{\mathbb{P}}^n_k}$$ Resolve the structure sheaf $0 \to {\mathcal{O}}_{{\mathbb{P}}^n}(-d) \to {\mathcal{O}}_{{\mathbb{P}}^n}(t) \to {\mathcal{O}}_D(t) \to 0 .$ so we can twist to obtain $0 \to {\mathcal{O}}_{{\mathbb{P}}^n}(t-d) \to {\mathcal{O}}_{{\mathbb{P}}^n}(t) \to {\mathcal{O}}_D(t) \to 0 .$ Then $\chi({\mathcal{O}}_D(t)) = \chi({\mathcal{O}}_{{\mathbb{P}}^n}(t)) - \chi({\mathcal{O}}_{{\mathbb{P}}^n}(t-d)) ,$ which is ${n+t \choose n} - {n+t-d \choose n} = \frac{dt^{n-1}}{(n-1)!} + O(t^{n-2}) .$ ::: {.lemma title="?"} Anything with the Hilbert polynomial of a degree $$d$$ hypersurface is in fact a degree $$d$$ hypersurface. ::: We want to write a morphism of functors $\operatorname{Hilb}_{{\mathbb{P}}^n_k}^{P_{n, d}} \to {\mathbb{P}}H^0 ({\mathbb{P}}^n, {\mathcal{O}}(d) )^\vee .$ which sends flat families to families of equations cutting them out. Want $Z \subset {\mathbb{P}}^n \times S \to {\mathcal{O}}_s \otimes H^0( {\mathbb{P}}^n, {\mathcal{O}}(d) )^\vee\to L \to 0 .$ This happens iff $0 \to L^\vee\to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(d))$ with torsion-free quotient. Note that we use $$L^\vee$$ instead of $${\mathcal{O}}_s$$ because of scaling. We have $0 \to I_z &\to {\mathcal{O}}_{{\mathbb{P}}^n \times S} \to {\mathcal{O}}_z \to 0 \\ 0 \to I_z(d) &\to {\mathcal{O}}_{{\mathbb{P}}^n \times S}(d) \to {\mathcal{O}}_z(d) \to 0 \quad\text{by twisting} .$ We then consider $$\pi_s: {\mathbb{P}}^n \times S \to S$$, and apply the pushforward to the above sequence. Notie that it is not right-exact: {=tex} \begin{tikzcd} 0 \ar[r] & \pi_{s*} I_z(d) \ar[r] & \pi_{s*} {\mathcal{O}}_{{\mathbb{P}}^n \times S}(d) \ar[r] & \pi_{s*} {\mathcal{O}}_z(d) \ar[r] & 0 \\ & & & & \\ \ar[uu, equal]0 \ar[r] & {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(d)) \ar[uu, equal]L^\vee= \ar[uu, equal] \ar[r] & \ar[uu, equal]\text{locally free} \ar[r] & 0 \ar[uu, equal] \end{tikzcd}  {=tex} \todo[inline]{Note: above diagram may be off horizontally?}  This equality follows from flatness, cohomology, and base change. In particular, we need the following: ::: {.fact} The scheme-theoretic fibers, given by $$H^0({\mathbb{P}}^n, I_z(d))$$ and $$H^0({\mathbb{P}}^n, {\mathcal{O}}_z(d))$$, are all the same dimension. ::: Using 1. Cohomology and base change, i.e. for $$X \xrightarrow{f} Y$$ a map of Noetherian schemes (or just finite-type) and $$F$$ a sheaf on $$X$$ which is flat over $$Y$$, there is a natural map (not usually an isomorphism) $R^i f_* f \otimes k(y) \to H^i(x_y, {\left.{{F}} \right|_{{x_y}} } ) ,$ but is an isomorphism if $$\dim H^i(x_y, {\left.{{F}} \right|_{{x_y}} } )$$ is constant, in which case $$R^i f_* f$$ is locally free. 2. If $$Z \subset {\mathbb{P}}^n_k$$ is a degree $$d$$ hypersurface, then independently we know $\dim H^0({\mathbb{P}}^n, I_z(d)) = 1 \text{ and } \dim H^0({\mathbb{P}}^n, {\mathcal{O}}_z(d)) = {d+n \choose n} - 1 .$ To get a map going backwards, we take the universal degree 2 polynomial and form $V(a_{00} x_0^2 + a_{11} x_1^2 + a_{12}x_2^2 + a_{01}x_0 x_1 + a_{02} x_0 x_2 + a_{12} x_1 x_2) \subset {\mathbb{P}}^2 \times{\mathbb{P}}^5 .$ ### Example: Twisted Cubics Consider a map $${\mathbb{P}}^1 \to {\mathbb{P}}^3$$ obtained by taking a basis of a homogeneous cubic polynomial. The canonical example is $(x, y) \to (x^3, x^2y, xy^2, y^3) .$ Then $P_C(t) = 3t + 1$ and $$\operatorname{Hilb}_{{\mathbb{P}}_k^3}^{3t+1}$$ has a component with generic point a twisted cubic, and another component with points a curve disjoint union a point, and the overlap are nodal curves with a "fat" 3-dimensional point: ![Components of the Hilbert Scheme](figures/2020-01-23-13:20.png)\ Then $$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''}$$, we take the sequence $0 \to k \to {\mathcal{O}}_{C''} \to {\mathcal{O}}_{C'' \text{reduced}} \to 0 ,$ so $P_{C''} = 1 + P_{C'' \text{red}} = 3t+1 .$ ::: {.remark} Note that flat families *must* have the same (constant) Hilbert polynomial. ::: Note that we can get paths in this space from $$C\to C''$$ and $$C'\to C''$$ by collapsing a twisted cubic onto a plane, and sending a disjoint point crashing into the node on a nodal cubic. We're mapping $${\mathbb{P}}^1 \to {\mathbb{P}}^3$$, and there is a natural action of $$\operatorname{PGL}(4) \curvearrowright{\mathbb{P}}^3$$, so we get a map $\operatorname{PGL}(4) \times{\mathbb{P}}^3 \to {\mathbb{P}}^3 .$ Let $$c\in {\mathbb{P}}^3$$ and let $${\mathcal{C}}$$ be the preimage. This induces (?) a map $\operatorname{PGL}(4) \to \operatorname{Hilb}_{{\mathbb{P}}^3}^{3t+1}$ where the fiber over $$[C]$$ in the latter is $$\operatorname{PGL}(2) = {\operatorname{Aut}}({\mathbb{P}}^1)$$. By dimension counting, we find that the dimension of the twisted cubic component is $$15 - 3 = 12$$. The 15 in the other component comes from 3-dim choices of plane, 3-dim choices of a disjoint point, and ${\mathbb{P}}H^0({\mathbb{P}}^2, {\mathcal{O}}(3))^\vee\cong {\mathbb{P}}^9 ,$ yielding 15 dimensions. To show that these are actually different components, we use Zariski tangent spaces. Let $$T_1$$ be the tangent space of the twisted cubic component, then $\dim T_1 \operatorname{Hilb}_{{\mathbb{P}}_k^3}^{3t+1} = 12 ,$ and similarly the dimension of the tangent space over the $$C'$$ component is 15. ::: {.fact} Let $$A$$ be Noetherian and local, then the dimension of the Zariski tangent space, $$\dim {\mathfrak{m}}/{\mathfrak{m}}^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 \operatorname{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_{{\mathcal{O}}_x}(I_z, {\mathcal{O}}_z)$$. ::: # Hilbert Schemes of Hypersurfaces (Tuesday January 28th) Last time: Twisted cubics, given by $$\operatorname{Hilb}_{{\mathbb{P}}^3_k}^{3t+1}$$. {=tex} \begin{tikzpicture} \node (node_one) at (0,0) { \includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_21-29} }; \node (a) at (0, -4) {$?$}; \node (a) at (-4.5, 3.9) {$A$}; \filldraw[blue](-4.45, 3.4) circle (0.1); \node (a) at (2.44, 2.4) {$B$}; \filldraw[blue](2.04, 2.2) circle (0.1); \node (a) at (-1.8, 1.85) {$C$}; \filldraw[red](-1.38, 1.85) circle (0.1); \node (a) at (-4., 7) {$12$}; \node (a) at (3., 5.5) {$15$}; \end{tikzpicture}  > Components of the Scheme of Cubic Curves. We got lower (?) bounds on the dimension by constructing families, but want an exact dimension. The following will be a key fact: ::: {.proposition title="?"} Let $$Z\subset X$$ be a closed $$k{\hbox{-}}$$dimensional subspace. For $$[z] \in \operatorname{Hilb}_{X_{_{/k}}}^P(k)$$, we have an identification of the Zariski tangent space $T_{[z]} \operatorname{Hilb}_{X_{_{/k}} }^P = \hom_{{\mathcal{O}}_X}(I_z, {\mathcal{O}}_Z)$ ::: Say $F: ({\operatorname{Sch}}_{_{_{/k}}})^{\operatorname{op}}\to {\operatorname{Set}}$ is a functor and let $$x\in F(k)$$. There is an inclusion $$i: \operatorname{Spec}k \hookrightarrow\operatorname{Spec}k[\varepsilon]$$ and an induced map $F(\operatorname{Spec}k [\varepsilon]) &\xrightarrow{i^*} F(\operatorname{Spec}k) \\ T_x F \coloneqq(i^*)^{-1}(x) &\mapsto x$ So if $$F$$ is represented by a scheme $$H_{/k}$$, then\ $T_x h_J = T_x H = ({\mathfrak{m}}_x / {\mathfrak{m}}_x^2)^\vee\,\,\text{over } k$ Will need a criterion for flatness later, esp. for Artinian thickenings. ::: {.lemma title="?"} Assume $$A'$$ is a Noetherian ring and $$0 \to J \to A' \to A \to 0$$ with $$J^2 = 0$$. Assume we have $$X'_{/ \operatorname{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\otimes_A J \to F'$$ is exact. {=tex} \begin{tikzcd} F & F' \\ X \coloneqq\operatorname{Spec}A' \times_{\operatorname{Spec}A} X \ar[r] \ar[d] & X' \ar[d] \\ \operatorname{Spec}A \arrow[ur, phantom, "\scalebox{1.5}{$\ulcorner$}" , very near start, color=black] \ar[r] & \operatorname{Spec}A' \end{tikzcd}  ::: ### Sketch Proof of Lemma Take the first exact sequence and tensor with $$F'$$ (which is right-exact), then $$J \otimes_{A'} F' = J \otimes_A$$ canonically. This follows because $$J = J \otimes_{A'} A$$, and there is an isomorphism $$J \otimes_{A'} A' \to J \otimes_{A'} A$$. And $$F = F' \otimes_{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$$ iff $\operatorname{Tor}_1^A(M, A/p) = 0 && \text{ for all primes } p .$ ::: ::: {.remark} Reason: Tor commutes with direct limits, so $$M$$ is flat iff $\operatorname{Tor}_1^A(M, N) = 0 && \text{for all finitely generated } N .$ ::: Since $$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$$ by $$\operatorname{im}f$$ where $$f: A/I \to N$$. Continuing inductively, the resulting filtration must stabilize. So we can assume $$N = A/I$$. Then $$I$$ is contained in a maximal. ::: {.exercise title="?"} Finish proof. See Aatiyah Macdonald. ::: ### Proof of Proposition ::: {.proof title="of proposition, given lemma"} So it's enough to show that $$\operatorname{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$$, this is a prime ideal. We have an exact diagram by taking quotients: {=tex} \begin{tikzcd} & & & & 0 \arrow[dd] & & 0 \arrow[dd] & & \\ & & & & & & & & \\ 0 \arrow[rr] & & J \arrow[rr] & & p' \arrow[rr] \arrow[dd] & & p \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ 0 \arrow[rr] & & J \arrow[rr] & & A' \arrow[rr] \arrow[dd] & & A \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ & & & & A'/p' \arrow[dd] & & A/p \arrow[dd] & & \\ & & & & & & & & \\ & & & & 0 & & 0 & & \end{tikzcd}  So we can tensor with $$F'$$ everywhere, and get a map from kernels to cokernels using the snake lemma: {=tex} \begin{tikzcd} & & & & 0 \arrow[dd] & & {\operatorname{Tor}(A, F) = 0} \arrow[dd] & & \\ & & & & & & & & \\ & & 0 \arrow[rr, "\text{snake}"] \arrow[dd] & & {\operatorname{Tor}_1^{A_1}(A'/p', F')} \arrow[dd] & & {\operatorname{Tor}_1^{A_1}(A/p, F')} \arrow[dd] & & \\ & & & & & & & & \\ 0 \arrow[rr] & & J \otimes_{A'} F' \arrow[rr, "\text{by commuting square}", hook] & & p' \otimes_{A'} F' \arrow[rr] \arrow[dd] & & p \otimes_{A'} F' \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ 0 \arrow[rr] & & J \otimes_{A'} F' \arrow[rr, "\text{by (2)}"', hook] & & A' \otimes_{A'} F' \arrow[rr] \arrow[dd] & & A \otimes_{A'} F' \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ & & 0 \arrow[rr, "\text{snake}"] & & A'/p' \otimes_{A'} F' \arrow[dd] \arrow[rr, "="] & & A/p \otimes_{A'} F' \arrow[dd] & & \\ & & & & & & & & \\ & & & & 0 & & 0 & & \end{tikzcd}  Then by (1), we have $\operatorname{Tor}_1^{A'}(A'/p', F') = \operatorname{Tor}_1^{A'}(A/p, F') = 0 .$ ::: We will just need this for $$A' = k[\varepsilon]$$ and $$A=k$$. ::: {.proposition title="?"} $T_z \operatorname{Hilb}_{X_{_{/k}}} = \hom_{{\mathcal{O}}_x}(I_z, {\mathcal{O}}_z) .$ ::: ::: {.proof title="?"} Again we have $$T_z \operatorname{Hilb}_{X_{_{/k}}} \subset \operatorname{Hilb}_{X_{_{/k}}}(k[\varepsilon])$$, and is given by $\left\{{Z' \subset X \times_{\operatorname{Spec}k} \operatorname{Spec}k[\varepsilon] {~\mathrel{\Big|}~}Z' \text{ is flat}_{/k[\varepsilon]},\,\, Z' \times_{\operatorname{Spec}k[\varepsilon]}\operatorname{Spec}k = Z}\right\} .$ We have an exact diagram: {=tex} \begin{tikzcd} & & 0 \arrow[r] & I_{Z'} \arrow[r] & {{\mathcal{O}}_{X[\varepsilon]}} \arrow[r] & {\mathcal{O}}_{Z'} \arrow[r] & 0 \\ 0 \arrow[d] & & & {} \arrow[d] & {} \arrow[d] & {} \arrow[d] & \\ k \arrow[d] & & {} \arrow[r] & I_Z \arrow[r] \arrow[d] & {\mathcal{O}}_x \arrow[r] \arrow[d] & {\mathcal{O}}_z \arrow[r] \arrow[d] & {} \\ {k[\varepsilon]} \arrow[d] & & {} \arrow[r] & I_{Z'} \arrow[r] \arrow[d] & {{\mathcal{O}}_{x[\varepsilon]}} \arrow[r] \arrow[d] & {\mathcal{O}}_{Z'} \arrow[r] \arrow[d] & {} \\ k \arrow[d] & & {} \arrow[r] & I_Z \arrow[r] \arrow[d] & {\mathcal{O}}_x \arrow[r] \arrow[d] \arrow[u, dotted, bend right] & {\mathcal{O}}_Z \arrow[r] \arrow[d] & {} \\ 0 & & & {} & {} & {} & \end{tikzcd}  Note the existence of a splitting above. Given $$\phi \in \hom_{{\mathcal{O}}_x}(I_Z, {\mathcal{O}}_Z)$$. We have $I_{Z'} = \left\{ f + \varepsilon g \, \middle\vert \, \begin{array}{ll} f,g &\in I_Z, \\ \phi(f) &= g\pmod I_Z, \\ \phi(f) &\in {\mathcal{O}}_Z, \\ g\pmod I_Z &\in {\mathcal{O}}_x/I_Z = {\mathcal{O}}_Z \end{array} \right\} .$ It's easy to see that $$Z' \subset x'$$, and 1. $$Z'\times k = Z$$ 2. It's flat over $$k[\varepsilon]$$, looking at $$0 \to k\otimes I_{Z'} \to I_{Z'}$$. For the converse, take $$f\in I_Z$$ and lift to $$f' = f + \varepsilon g \in I_{Z'}$$, then $$g\in {\mathcal{O}}_x$$ is well-defined wrt $$I_Z$$. Then $$g\in \hom_{{\mathcal{O}}_x}(I_z, {\mathcal{O}}_z)$$. ::: The main point here is that these hom sets are extremely computable. ::: {.example title="?"} Let $$Z$$ be a twisted cubic in $$\operatorname{Hilb}_{{\mathbb{P}}^3_{/k}}^{3t+1}(k)$$. ::: ::: {.observation} $\hom_{{\mathcal{O}}_x}(I_Z, {\mathcal{O}}_Z) = \hom_{{\mathcal{O}}_X}(I_Z/I_Z^2, {\mathcal{O}}_Z) = \hom_{{\mathcal{O}}_Z}(I_Z/I_Z^2, {\mathcal{O}}_Z)$ ::: If $$I_Z/I_Z^2$$ is locally free, these are global sections of the dual, i.e. $$H^0((I_Z/I_Z^2)^\vee)$$. In this case, $$Z\hookrightarrow X$$ is regularly embedded, and thus $$(I_Z/I_Z^2)^\vee$$ should be regarded as the normal bundle. Sections of the normal bundle match up with directions to take first-order deformations: {=tex} \begin{tikzpicture} \definecolor{arrow_color}{HTML}{ba0cff} \node (node_one) at (0,0) { \includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_22-40} }; \node (a) at (-7, 5) {\Huge ${\mathbb{P}}^3$}; \node[arrow_color] (a) at (1, -3) {\Huge Deformation}; \end{tikzpicture}  For $$i:C \hookrightarrow{\mathbb{P}}^3$$, there is an exact sequence $0 \to I/I^2 \to &i^* \Omega_{{\mathbb{P}}^3} \to \Omega_\varepsilon\to 0 \\ &\Downarrow \quad \text{ taking duals } \\ 0 \to T_C \to &i^* T_{{\mathbb{P}}^3} \to N_{C_{/{\mathbb{P}}^3} } \to 0 ,$ How do we compute $$T_{{\mathbb{P}}^3}$$? Fit into the exact sequence $0 \to {\mathcal{O}}\to i^* {\mathcal{O}}(1)^4 \to i^* T_{{\mathbb{P}}^3} \to 0 ,$ which we can restrict to $$C$$. We have $$i^* {\mathcal{O}}(1) \cong {\mathcal{O}}_{{\mathbb{P}}^1}(3)$$, so $0 \to H^0 {\mathcal{O}}_c \to &H^*({\mathcal{O}}(3)^4) \to H^0(i^* T_{{\mathbb{P}}^3}) \to 0 \\ &\Downarrow \\ 0\to k \to &k^{16} \to k^{15} \to 0 .$ This yields $0 \to H^0(T_c) \to &H^0(i^* T_{{\mathbb{P}}^3}) \to H^0(N_{C_{ /{\mathbb{P}}^3} }) \to H^1 T_c \\ &\Downarrow \\ 0\to k^3 \to &k^{15} \to k^{12} \to 0$ ::: {.example title="?"} $$\operatorname{Hilb}_{{\mathbb{P}}^n_k}^{P_?} \cong {\mathbb{P}}H^0({\mathbb{P}}^n, {\mathcal{O}}(d))^\vee$$ which has dimension $${n+1 \choose n} - 1$$. Pick $$Z$$ a $$k$$ point in this Hilbert scheme, then $$T_Z H = \hom(I_Z, {\mathcal{O}}_Z)$$. Since $$I_Z \cong {\mathcal{O}}_{{\mathbb{P}}}(-d)$$ which fits into $0 \to {\mathcal{O}}_{{\mathbb{P}}^n}(-d) \to {\mathcal{O}}_{{\mathbb{P}}^n} \to {\mathcal{O}}_Z \to 0 .$ We can identify $\hom(I_Z,{\mathcal{O}}_Z) = H^0( (I_Z/I_Z^2)^\vee) = H^0({\mathcal{O}}_Z(d)) .$ {=tex} \begin{tikzcd} 0\ar[r] & {\mathcal{O}}_{{\mathbb{P}}^n}\ar[r] & {\mathcal{O}}_{{\mathbb{P}}^n}(d)\ar[r] & {\mathcal{O}}_Z(d)\ar[r] & 0 \\ & & & & \\ 0\ar[r] & H^0( {\mathcal{O}}_{{\mathbb{P}}^n} ) \ar[r] & H^0( {\mathcal{O}}_{{\mathbb{P}}^n}(d) ) \ar[r] & H^0({\mathcal{O}}_Z(d) ) \ar[r] & 0 \\ \text{dim:} & k & k^{n+d \choose n} & k^{{n+d\choose n}-1} & \end{tikzcd}  ::: ::: {.example title="?"} The tangent space of the following cubic: {=tex} \begin{tikzpicture} \node (node_one) at (0,0) { \includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_17-27} }; \end{tikzpicture}  We can identify $\hom_{{\mathcal{O}}_k}(I_Z, {\mathcal{O}}_Z) = H^0((I_Z/I_Z^2)^\vee) = 3 + H^0((I_{Z_0}/I_{Z_0}^2)^\vee) ,$ where the latter equals $$H^0 \qty{ {\mathcal{O}}_1\mathrel{\Big|}_{z_0} \oplus {\mathcal{O}}(\zeta)\mathrel{\Big|}_{z_0} }$$ yielding $3+9 = 12 .$ ::: # Uniform Vanishing Statements (Thursday January 30th) Recall how we constructed the Hilbert scheme of hypersurfaces $\operatorname{Hilb}_{{\mathbb{P}}_k^n}^{P_{m, d}} = {\mathbb{P}}H^0({\mathbb{P}}^n; {\mathcal{O}}(d))^\vee$ A section $$\operatorname{Hilb}_{{\mathbb{P}}_k^n}^{P}(s)$$ corresponds to $$z\in {\mathbb{P}}^n_s$$. We can look at the exact sequence $0 \to I_Z(m) \to {\mathcal{O}}_{{\mathbb{P}}_S^n} \xrightarrow{\text{restrict}} {\mathcal{O}}_z(m) \to 0 .$ as $${\mathbb{P}}_s^n \xrightarrow{\pi_s} S$$, so we can pushforward along $$\pi$$, which is left-exact, so $0 \to \pi_{s*} I_Z(m) \to \pi_{s*} {\mathcal{O}}_{{\mathbb{P}}_S^n} = {\mathcal{O}}_S \otimes H^0({\mathbb{P}}^n; {\mathcal{O}}(m)) \to {\mathcal{O}}_z(m) \to R^1 \pi_{s*} I_Z(m) \to \cdots .$ *Idea:* $$Z \subset {\mathbb{P}}_k^n$$ will be determined (in families!) by the space of degree $$d$$ polynomials vanishing on $$Z$$ (?), i.e. $H^0({\mathbb{P}}^n, I_z(m)) \subset H^0({\mathbb{P}}^n, {\mathcal{O}}(m))$ for $$m$$ very large. This would give a map of functors $\operatorname{Hilb}_{{\mathbb{P}}_k^n}^{P} \to {\operatorname{Gr}}(N, H^0({\mathbb{P}}^n, {\mathcal{O}}(m) )) .$ If this is a closed subfunctor, a closed subfunctor of a representable functor is representable and we're done . ::: {.remark} We need to get an $$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({\mathbb{P}}^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=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ and $${\mathbb{P}}^n = {\mathbb{P}}_k^n$$. ::: ## $$m{\hbox{-}}$$Regularity {#mhbox-regularity} ::: {.definition title="m-Regularity of Coherent Sheaves"} A coherent sheaf $$F$$ on $${\mathbb{P}}^n$$ is **$$m{\hbox{-}}$$regular** if $$H^i({\mathbb{P}}^n; F(m-i)) = 0$$ for all $$i> 0$$. ::: ::: {.example title="?"} Consider $${\mathcal{O}}_{{\mathbb{P}}^n}$$, this is $$0{\hbox{-}}$$regular. Line bundles on $${\mathbb{P}}_n$$ only have 0 and top cohomology. Just need to check that $$H^n({\mathbb{P}}^n; {\mathcal{O}}(-n)) = 0$$, but by Serre duality this is $H^0({\mathbb{P}}^n; {\mathcal{O}}(n) \otimes\omega_{{\mathbb{P}}^n})^\vee= H^0({\mathbb{P}}^n; {\mathcal{O}}(-1))^\vee= 0 .$ ::: ::: {.proposition title="?"} Assume $$F$$ is $$m{\hbox{-}}$$regular. Then 1. There is a natural multiplication map from linear forms on $${\mathbb{P}}^n$$, $H^0({\mathbb{P}}^n; {\mathcal{O}}(1)) \otimes H^0({\mathbb{P}}^n; F(k)) \to H^0({\mathbb{P}}^n; F(k+1)) ,$ which is surjective for $$k\geq n$$.[^3] 2. $$F$$ is $$m'{\hbox{-}}$$regular for $$m' \geq m$$. 3. $$F(k)$$ is globally generated for $$k\geq m$$, i.e. the restriction $H^0({\mathbb{P}}^n; F(k)) \otimes{\mathcal{O}}_{{\mathbb{P}}^n} \to F(k) \to 0$ is exact (i.e. surjective). ::: ::: {.example title="?"} $${\mathcal{O}}$$ is $$m{\hbox{-}}$$regular for $$m \geq 0$$ implies $${\mathcal{O}}(k)$$ is $$-k{\hbox{-}}$$regular and is also $$m{\hbox{-}}$$regular for$$m\geq -k$$. ::: ### Proof of 2 and 3 Induction on dimension of $$n$$ in $${\mathbb{P}}^n$$. Coherent sheaves on $${\mathbb{P}}^0$$ are vector spaces, so no higher cohomology. ::: {.proof title="Step 1"} Take a generic hyperplane $$H \subset {\mathbb{P}}^n$$, there is an exact sequence $0 \to {\mathcal{O}}(-1) \to {\mathcal{O}}\to {\mathcal{O}}_H \to 0 .$ where $${\mathcal{O}}_H$$ is the structure sheaf. Tensoring with $$H$$ remains exact, so we get $0 \to F(-1) \to F \to F_H \to 0 .$ Why? $${\mathbb{A}}^n \subset {\mathbb{P}}^n$$, let $$A = {\mathcal{O}}_{{\mathbb{P}}^n}({\mathbb{A}}^n)$$ be the polynomial ring over $${\mathbb{A}}^n$$. Then the restriction of the first sequence to $${\mathbb{A}}^n$$ yields $0 \to A \xrightarrow{f} A \to A/f \to 0 ,$ and thus we want $F \xrightarrow{f} F \to F/fF \to 0$ which results after restricting the second sequence to $${\mathbb{A}}^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{\hbox{-}}$$regular. By induction, this proves statements 1 and 2 for all $$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$$ we get statement 2. ::: By multiplication maps, we get a commutative diagram: {=tex} \begin{tikzcd} & & H^0({\mathcal{O}}(1)) \otimes H^0(F(k)) \arrow[dd, "\beta"] \arrow[rrr] \arrow[rrrdd] & & & H^0({\mathcal{O}}(1))\otimes H^0(F_H(k)) \arrow[dd] \\ & & & & & \\ H^0(F(k)) \arrow[rr, "H"] \arrow[rruu, "H \otimes\operatorname{id}"] & & H^0(F(k+1)) \arrow[rrr, "\alpha", dashed] & & & H^0(F_H(k+1)) \end{tikzcd}  We'd like to show the diagonal map is surjective. ::: {.observation} {=tex} \envlist  1. The top map is a surjection, since $H^0(F(k)) \to H^0(F_H(k)) \to H^1(F(k-1)) = 0$ for $$k\geq m$$ by (2). 2. The right-hand map is surjective for $$k\geq m$$. 3. $$\ker(\alpha) \subset \operatorname{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 {\mathbb{Q}}[t]$$ be a Hilbert polynomial. There exists an $$m_0$$ only depending on $$P$$ such that for all subschemes $$Z \subset {\mathbb{P}}^n_k$$ with Hilbert polynomial $$P_Z = P$$, the ideal sheaf $$I_z$$ is $$m_0{\hbox{-}}$$regular. ::: ### Proof of Theorem Induct on $$n$$. For $$n=0$$, again clear because higher cohomology vanishes and there are no nontrivial subschemes. For a fixed $$Z$$, pick $$H$$ in $${\mathbb{P}}^n$$ (and setting $$I \coloneqq 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_{{\mathcal{O}}_{{\mathbb{P}}^n}} - P_Z$$. By induction, there exists some $$m_1$$ depending only on $$P$$ such that $$I_H$$ is $$m_1{\hbox{-}}$$regular. We get $H^{i-1}(I_H(k)) \to H^i(I(k-1)) \to H^i(I(k)) \to H^i(I_H(k)) ,$ and for $$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}$$ is *strictly* decreasing.[^4] ::: ::: {.remark} Given the lemma, it's enough to take $$m_0 \geq m_1 + h^1(I(m_1 - 1))$$. Consider the LES we have a surjection $H^0({\mathcal{O}}_Z(m_1 - 1)) \to H^1(I(m_1 - 1)) \to 0 .$ So the dimension of the LHS is equal to $$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)$$. ::: ::: {.proof title="of Lemma"} Considering the LES $H^0(I(k+1)) \xrightarrow{\alpha_{k+1}} H^0(I_H(k+1)) \to H^1(I(k)) \to H^1(I(k+1)) \to 0 ,$ where the last term is zero because $$I_H$$ is $$m_1{\hbox{-}}$$regular. So the sequence $$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}$$ is surjective, since $H^0({\mathcal{O}}(1)) \otimes H^0(I_H(k+1)) \twoheadrightarrow H^0(I_H(k+2)) .$ ::: So if one is surjective, everything above it is surjective, but by Serre vanishing we eventually get zeros. So $$\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 {\mathbb{P}}^n_k$$ with $$P_Z = P$$, we uniformly know that $$I_Z$$ is $$m_0{\hbox{-}}$$regular for some $$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({\mathcal{O}}(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)$$ is constant. ::: It is determined by these polynomials because we have a sequence $0 \to I_Z(m_0) \to {\mathcal{O}}(m_0) \to {\mathcal{O}}_Z(m_0) \to 0 .$ We can get a commuting diagram over it $0 \to H^0(I_Z(m_0)) \otimes{\mathcal{O}}_{{\mathbb{P}}^n} \to H^0({\mathcal{O}}(m_0)) \otimes{\mathcal{O}}_{{\mathbb{P}}^n} \to \cdots$ where the middle map down is just evaluation and.the first map down is a surjection. Hence $$I_Z(m_0)$$, hence $${\mathcal{O}}_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=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$, and $$C_{/k}$$ a smooth projective curve, then $$\operatorname{Hilb}_{C_{/k}}^n = \operatorname{Sym}^n C$$. ::: {.definition title="The Hilbert-Chow Map"} For $$X_{_{/k}}$$ a smooth projective *surface*, $$\operatorname{Hilb}_{X_{_{/k}}}^n \neq \operatorname{Sym}^n X$$, there is a map (the Hilbert-Chow map) $\operatorname{Hilb}_{X_{_{/k}}}^n &\to \operatorname{Sym}^n X \\ Z &\mapsto {\operatorname{supp}}(Z) \\ U = \text{reduced subschemes} &\mapsto U' = \text{ reduced multisets } \\ {\mathbb{P}}^1 &\mapsto (x, x) .$ ::: ::: {.example title="?"} Consider $${\mathbb{A}}^2 \times{\mathbb{A}}^2$$ under the $${\mathbb{Z}}/2{\mathbb{Z}}$$ action $( (x_1, y_1), (x_2, y_2)) \mapsto ((x_2, y_2), (x_1, y_1)) .$ Then $({\mathbb{A}}^2)^2 / {\mathbb{Z}}/2{\mathbb{Z}} &= \operatorname{Spec}k[x_1, y_1, x_2, y_2]^{{\mathbb{Z}}/2{\mathbb{Z}}} \\ &= \operatorname{Spec}k[x_1 x_2, y_1 y_2, x_1 + x_2, y_1 + y_2, x_1 y_2 + x_2 y_1, \cdots]$ with a bunch of symmetric polynomials adjoined. ::: ::: {.example title="?"} Take $${\mathbb{A}}^2$$ and consider $$\operatorname{Hilb}_{{\mathbb{P}}^2}^3$$. If $$I$$ is a monomial ideal in $${\mathbb{A}}^2$$, there is a nice picture. We can identify the tangent space $T_Z \operatorname{Hilb}_{{\mathbb{P}}^2}^n = \hom_{{\mathcal{O}}_{{\mathbb{P}}^2}} ( I_2, {\mathcal{O}}_Z) = \bigoplus \hom(I_{Z_i}, {\mathcal{O}}_{Z_i}) .$ if $$Z = {\coprod}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$$. Write $${\mathcal{O}}= \left\{{1, x}\right\}$$, then note that this relation is trivial in $${\mathcal{O}}$$ since $$y=x^2=0$$. Thus $$\hom(I, {\mathcal{O}}) = \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 \times C^n$$, then quotient by the symmetric group $$S^n$$ (need to show this can be done), then $$Z/S^n \subset C \times\operatorname{Sym}^n C$$ and composing with the functor $$\operatorname{Hilb}$$ represents yields a map $$\operatorname{Sym}^n C \to \operatorname{Hilb}_{C_{/k}}^n$$. This is bijective on points, and a tangent space computation shows it's an isomorphism. ::: ::: {.example title="?"} Consider the nodal cubic in $${\mathbb{P}}^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 \operatorname{Hilb}_{C_{/k}}^2$$ of points $$z \subset U$$ for $$U \subset C$$ open. We can normalize: ![Normalized cubic](figures/2020-02-06-13:03.png){width="350px"} This yields a map fro $${\mathbb{P}}^1 \setminus\text{2 points}$$. This gives us a stratification, i.e. a locally closed embedding $(\text{z supported on U}) {\coprod}(\text{1 point at p}) {\coprod}(\text{both points at p}) \to \operatorname{Hilb}_{C_{/k}}^2 .$ The first locus is given by the complement of two lines: ![Locus 1](figures/2020-02-06-13:08.png){width="350px"} The third locus is given by arrows at $$p$$ pointing in any direction, which gives a copy of $${\mathbb{P}}^1$$. The second is $${\mathbb{P}}^1$$ minus two points. Above each point is a nodal cubic with two marked points, and moving the base point towards a line correspond to moving one of the points toward the node: ![Moving base toward the point](figures/2020-02-06-13:11.png){width="350px"} More precisely, we're considering the cover $${\mathbb{P}}^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$$ on $${\mathbb{P}}_k^n$$ for $$k$$ a field is $$m{\hbox{-}}$$regular iff $$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 {\mathbb{P}}^n_k$$ with $$P_Z = P$$ satisfies $$I_Z$$ is $$m{\hbox{-}}$$regular. ::: ::: {.remark title="1"} $$F$$ is $$m{\hbox{-}}$$ regular iff $$\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu = F \times_{\operatorname{Spec}k} \operatorname{Spec}\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ is $$m{\hbox{-}}$$regular. ::: ::: {.remark title="2"} The $$m_0$$ produced does not depend on $$k$$. ::: ::: {.lemma title="?"} For $$m_0 = m_0(P)$$ and $$N = N(P)$$, we have an embedding as a subfunctor $\operatorname{Hilb}_{{\mathbb{P}}^m_{\mathbb{Z}}}^P \to {\operatorname{Gr}}(N, H^0( {\mathbb{P}}^n_{\mathbb{Z}}, {\mathcal{O}}(m_0) )^\vee) .$ ::: For any $$Z \subset {\mathbb{P}}^n_S$$ flat over $$S$$ with $$P_{Z_s} = P$$ for all $$s\in S$$ points, we want to send this to $0\to R^\vee\to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n_{\mathbb{Z}}, {\mathcal{O}}(m_0))^\vee\to Q \to 0$ or equivalently $0 \to Q^\vee\to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n_{\mathbb{Z}}, {\mathcal{O}}(m_0)) \to R \to 0$ with $$R$$ locally free. So instead of the quotient $$Q$$ being locally free, we can ask for the sub $$Q^\vee$$ to be locally free instead, which is a weaker condition. We thus send $$Z$$ to $0 \to \pi_{s*} I_Z(m_0) \to \pi_{s*} {\mathcal{O}}_{{\mathbb{P}}^n_s}(m_0) = {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(m_0))$ which we obtain by taking the pushforward from this square: {=tex} \begin{tikzcd} {\mathbb{P}}^n_s \arrow[dd, "\pi_s"] \arrow[rr] & & {\mathbb{P}}^n_Z \arrow[dd] \\ & & \\ S \arrow[rr] & & \operatorname{Spec}{\mathbb{Z}} \end{tikzcd}  We have a sequence $$0 \to I_Z(m_0) \to {\mathcal{O}}(m_0) \to {\mathcal{O}}_Z(m_0) \to 0$$. Thus we get a sequence $0 \to \pi_{s*}I_Z(m_0) \to \pi_{s*}{\mathcal{O}}(m_o) \to \pi_{s*} {\mathcal{O}}_Z(m_0) \to R^1 \pi_{s*}I_Z(m_0) \to \cdots .$ ### Step 1 $R^1\pi_* I_Z(m_0) = 0 .$ By base change, it's enough to show that $$H^1(Z_s, I_{Z_s}(m_0)) = 0$$. This follows by $$m_0{\hbox{-}}$$regularity. ### Step 2 $$\pi_{s*}I_Z(m_0)$$ and $$\pi_{s*} {\mathcal{O}}_Z(m_0)$$ are locally free. For all $$i>0$$, we have - $$R^i \pi_{s*} I_Z(m_0) = 0$$ by $$m_0{\hbox{-}}$$regularity, - $$R^i \pi_{s*} {\mathcal{O}}(m_0) = 0$$ by base change, - and thus $$R^i \pi_{s*} {\mathcal{O}}_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) \otimes 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{\hbox{-}}$$regularity and $h^0(I_{Z_S}(m_0)) = P_{\mathcal{O}}(m_0) - P_{{\mathcal{O}}_{Z_s}}(m_0) = P_{\mathcal{O}}(m_0) - P(m_0) .$ This yields a well-defined functor $\operatorname{Hilb}_{{\mathbb{P}}^n_{\mathbb{Z}}}^P \to {\operatorname{Gr}}(N, H^0({\mathbb{P}}^n, {\mathcal{O}}(m_0))^\vee) .$ ::: {.remark} Note that we've just said what happens to objects; strictly speaking we should define what happens for morphisms, but they're always give by pullback. ::: We want to show injectivity, i.e. that we can recover $$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 {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(m_0))$$. Given $0 \to Q^\vee\to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(m_0)) = \pi_{s*} {\mathcal{O}}_{{\mathbb{P}}^n_S}(m_0)$ we get a diagram {=tex} \begin{tikzcd} \pi_{s}^* Q^\vee\arrow[rrdd] \arrow[rrr] & & & {\mathcal{O}}_{{\mathbb{P}}^n_s}(m_0) \\ & & & \\ & & I(m_0) \arrow[ruu, hook] & \end{tikzcd}  where $$Q^\vee= \pi_{s*} I_Z(m_0)$$, so we're looking at {=tex} \begin{tikzcd} Q^\vee= \pi_{s*}^* \pi_{s*} I_Z(m_0) \arrow[rrdd, twoheadrightarrow] \arrow[rrr] & & & {\mathcal{O}}_{{\mathbb{P}}^n_s}(m_0) \\ & & & \\ & & I(m_0) \arrow[ruu, hook] & \end{tikzcd}  The surjectivity here follows from $${\mathcal{O}}_{Z_s} \otimes H^0(I_{Z_s}(m_0)) \to I_{Z_s}(m_0)$$ (?). Given a universal family $$G = {\operatorname{Gr}}( N, H^0({\mathcal{O}}(m_0))^\vee)$$ and $$Q^\vee\subset {\mathcal{O}}_G \otimes H^0({\mathcal{O}}(m_0))^\vee$$, we obtain $$I_W \subset {\mathcal{O}}_G$$ and $$W \subset {\mathbb{P}}^n_G$$. # Tuesday February 18th ::: {.theorem title="?"} Let $$X/S$$ be a projective subscheme (i.e. $$X\subset {\mathbb{P}}^n$$ for some $$n$$). The Hilbert functor of flat families $$\operatorname{Hilb}_{X/S}^p$$ is representable by a projective $$S{\hbox{-}}$$scheme. ::: ::: {.remark} Note that without a fixed $$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$$, there is a smooth family $${\mathcal{C}}\xrightarrow{\pi} S$$ with $$S$$ finite-type over $${\mathbb{Z}}$$ where every genus $$g$$ curve appears as a fiber. I.e., genus $$g$$ curves form a *bounded family* (here there are only finitely many algebraic parameters to specify a curve). How did we construct? Take the third power of the canonical bundle and show it's very ample, so it embeds into some projective space and has a Hilbert polynomial. ::: In fact, there is a finite type *moduli stack* $${\mathcal{M}}_g / {\mathbb{Z}}$$ of genus $$g$$ curves. There will be a map $$S \twoheadrightarrow{\mathcal{M}}_g$$, noting that $${\mathcal{C}}$$ is not a moduli space since it may have redundancy. We'll use the fact that a finite-type scheme surjects onto $${\mathcal{M}}_g$$ to show it is finite type. ::: {.remark} If $$X/S$$ is proper, we can't talk about the Hilbert polynomial, but the functor $$\operatorname{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\hookrightarrow{\mathbb{P}}^n_S$$), then $$\operatorname{Hilb}_{X/S}^P(T) \coloneqq\left\{{z\in X_T \text{ projective, flat over S with fiberwise Hilbert polynomial P }}\right\}$$ is still representable, but now by a quasiprojective scheme. ::: ::: {.example title="?"} Length $$Z$$ subschemes of $${\mathbb{A}}^1$$: representable by $${\mathbb{A}}^2$$. ![Image](figures/2020-02-18-12:46.png)\ Upstairs: parametrizing length 1 subschemes, i.e. points. ::: ::: {.remark} If $$X\subset {\mathbb{P}}_S^n$$ and $$E$$ is a coherent sheaf on $$X$$, then $\operatorname{Quot}_{E, X/S}^{P}(T) = \left\{{ j^*E \to F \to 0, \text{ over } X_T \to T,~F \text{ flat with fiberwise Hilbert polynomial } P }\right\}$ where $$T \xrightarrow{g} S$$ is representable by an $$S{\hbox{-}}$$projective scheme. ::: ::: {.example title="?"} Take $$E = {\mathcal{O}}_x$$, $$X$$ and $$S$$ a point, and $$E$$ is a vector space, then $$\operatorname{Quot}_{E/S}^P = {\operatorname{Gr}}(\operatorname{rank}, E)$$. ::: ::: {.warnings} The Hilbert scheme of 2 points on a surface is more complicated than just the symmetric product. ::: ::: {.example title="?"} $\qty{{\mathbb{A}}^2}^3 &\to \qty{{\mathbb{A}}^2}^2 \\ \supseteq \Delta\coloneqq\Delta_{01} \times\Delta_{02} &\to \qty{{\mathbb{A}}^2}^2$ where $$\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{{\mathbb{A}}^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$$ of a $$\operatorname{Hilb}_{{\mathbb{P}}^3_C}^P$$ whose generic point corresponds to a smooth irreducible $$C\subset {\mathbb{P}}^3$$ which is generically non-reduced. ::: ## Cubic Surfaces > See Hartshorne Chapter 5. Let $$X\subset {\mathbb{P}}^3$$ be a smooth cubic surface, then $${\mathcal{O}}(1)$$ on $${\mathbb{P}}^3$$ restricts to a divisor class $$H$$ of a hyperplane section, i.e. the associated line bundle $${\mathcal{O}}_x(H) = {\mathcal{O}}_x(1)$$. ::: {.fact title="Important fact 1"} $$X$$ is the blowup of $${\mathbb{P}}^2$$ minus 6 points (replace each point with a curve). There is thus a blowdown map $$X \xrightarrow{\pi} {\mathbb{P}}^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$$ into $${\mathbb{P}}^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$$ in the linear system $${\left\lvert {4H + 2L} \right\rvert}$$. Fact: $${\mathcal{O}}(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({\mathcal{O}}_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)\mathrel{\Big|}_C$$. Since $$X\subset {\mathbb{P}}^3$$, we have $\omega_X &= \omega_{{\mathbb{P}}^3}(X) \mathrel{\Big|}_X \\ &= {\mathcal{O}}(-4) \oplus {\mathcal{O}}(3)\mathrel{\Big|}_X \\ &= {\mathcal{O}}_X(-1) \\ &= {\mathcal{O}}_X(-H) .$ We also have $\omega_C &= \omega_X(C)\mathrel{\Big|}_X \\ &= { \left.{{ \qty{ {\mathcal{O}}_X(-H) \oplus {\mathcal{O}}_X(4H + 2L)}}} \right|_{{C}} } \\ \\ &\Downarrow \qquad \text{taking degrees} \\ \\ \deg \omega_C &= C\cdot(3H + 2L) \\ &= (4H+2L)(3H+2L) \\ &= 12H^2 + 14HL + 4L^2 \\ &= 36 + 14 + (-4) \\ &= 46 .$ Since this equals $$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 {\mathbb{P}}^3_S$$ with fiberwise Hilbert polynomial $$P$$ of cures of this form such that the image of the map $$S \to \operatorname{Hilb}_{{\mathbb{P}}^3}^P$$ has dimension 56. ::: ::: {.proof title="of claim"} We can compute the dimension of the space of smooth cubic surfaces, since these live in $${\mathbb{P}}H^0({\mathbb{P}}^3, {\mathcal{O}}(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$$ in the linear system $${\left\lvert {4H + 2L} \right\rvert}$$ will add $$\dim {\left\lvert {4H + 2L} \right\rvert} = \dim {\mathbb{P}}H^0(x, {\mathcal{O}}_x(C))$$. We have an exact sequence $0 \to {\mathcal{O}}_X \to {\mathcal{O}}_X(C) \to {\mathcal{O}}_C(C) \to 0 \\ H^0\qty{ 0 \to {\mathcal{O}}_X \to {\mathcal{O}}_X(C) \to {\mathcal{O}}_C(C) \to 0 } \\ .$ Since the first $$H^0$$ vanishes (?) we get an isomorphism. By Riemann-Roch, we have $\deg {\mathcal{O}}_C(C) = C^2 = (4H+2L)^2 = 16H^2 + 16 HL + 4L^2 = 64 - 4 = 60 .$ We can also compute $$\chi({\mathcal{O}}_C(C)) = 60 - 23 = 37$$. We have $h^0({\mathcal{O}}_C(C)) - h^1({\mathcal{O}}_C(C)) = h^0({\mathcal{O}}_C(C)) - h^0(\omega_C(-C))) = 2(23) - 60 < 0 ,$ so there are no sections. So $$\dim {\left\lvert {4H + 2L} \right\rvert} = 37$$. Thus letting $$S$$ be the space of cubic surfaces $$X$$, a line $$L$$, and a general $$C \in {\left\lvert {4H + 2L} \right\rvert}$$, $$\dim S = 56$$. We get a map $$S \to \operatorname{Hilb}_{{\mathbb{P}}^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\cap X'$$ is degree 14 curve sitting inside a degree 6 curve, which can't happen. Thus if $$H$$ is a component of $$\operatorname{Hilb}_{{\mathbb{P}}^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/{\mathbb{P}}^3})$$. There's an exact sequence $0 \to N_{C/X} = {\mathcal{O}}_C(C) \to N_{C/{\mathbb{P}}^3} \to N_{X/{\mathbb{P}}^3}\mathrel{\Big|}_C = {\mathcal{O}}_C(x)\mathrel{\Big|}_C = {\mathcal{O}}_C(3H)\mathrel{\Big|}_C \to 0 .$ > Note $$\omega_C = {\mathcal{O}}_C(3H + 2L)$$. As we computed, $H^0({\mathcal{O}}_C(C)) &= 37 \\ H^1({\mathcal{O}}_C(C)) &= 0 .$ So we need to understand the right-hand term $$H^0({\mathcal{O}}_C(3H))$$. By Serre duality, this equals $$h^1(\omega_C(-3H)) = h^1({\mathcal{O}}_C(3L))$$. We get an exact sequence $0 \to {\mathcal{O}}_X(2L-C) \to {\mathcal{O}}_X(2L) \to {\mathcal{O}}_C(2L) \to 0 .$ Taking homology, we have $$0\to 0 \to 1 \to 1 \to 0$$ since $$2L-C = -4H$$. Computing degrees yields $$h^0 ({\mathcal{O}}_C(3H)) = 20$$. Thus the original exact sequence yields $0 \to 37 \to ? \to 20 \to 0 ,$ so $$? = 57$$ and thus $$\dim N_{C/{\mathbb{P}}^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'}$$ be the ideal of this curve in $${\mathbb{P}}^3$$, there is a SES $0\to I(4) \to {\mathcal{O}}(4) \to {\mathcal{O}}_C(4) \to 0 .$ It can be shown that $$\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$$. It can be shown that the picard rank of such a quartic is 1, generated by $${\mathcal{O}}(1)$$, so this is a *strict* inequality, which is a contradiction since $$\dim \operatorname{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{\hbox{-}}$$points $$\operatorname{Hilb}_{X_{_{/k}}}^P(k)$$ corresponds to closed subschemes $$Z\subset X$$ with hilbert polynomial $$P_z = P$$. Given a $$P$$, we want to understand the local structure of $$\operatorname{Hilb}_{X_{_{/k}}}^p$$, i.e. diagrams of the form {=tex} \begin{tikzcd} & & & & \operatorname{Hilb}_{X_{_{/k}}}^P \arrow[dd] \\ & & & & \\ \operatorname{Spec}(k) \arrow[rrrruu, "p"] \arrow[rr] & & \operatorname{Spec}(A) \arrow[rruu, "?", dashed] \arrow[rr] & & \operatorname{Spec}(k) \\ & & & & \\ & & A_{/k} \text{ Artinian local} \arrow[uu] & & \end{tikzcd}  ::: {.example title="?"} For $$A = k[\varepsilon]$$, the set of extensions is the Zariski tangent space. ::: ::: {.definition title="Category of Artinian Algebras"} Let $$(\operatorname{Art}_{/k})$$ be the category of local Artinian $$k{\hbox{-}}$$algebras with local residue field $$k$$. ::: Note that these will be the types of algebras appearing in the above diagrams. ::: {.remark} This category has fiber coproducts, i.e. there are pushouts: {=tex} \begin{tikzcd} C \arrow[dd] \arrow[rr] & & A \arrow[dd, dashed] \\ & & \\ B \arrow[rr, dashed] & & A \otimes_C B \end{tikzcd}  There are also fibered products, {=tex} \begin{tikzcd} A \times_C B \arrow[rr, dashed] \arrow[dd, dashed] & & B \arrow[dd] \\ & & \\ A \arrow[rr] & & C \end{tikzcd}  Here, $$A \times_C B \coloneqq\left\{{(a, b) {~\mathrel{\Big|}~}f(a) = g(b)}\right\} \subset A\times B$$. ::: ::: {.example title="?"} If $$A = B = k[\varepsilon]/(\varepsilon^2)$$ and $$C = k$$, then $$A\times_C B = k[\varepsilon_1, \varepsilon_2]/(\varepsilon_1, \varepsilon_2)^2$$ Note that on the $$\operatorname{Spec}$$ side, these should be viewed as $\operatorname{Spec}(A) {\coprod}_{\operatorname{Spec}(C)} \operatorname{Spec}(B) = \operatorname{Spec}(A\times_C B) .$ ::: ::: {.definition title="Deformation Functor (Preliminary Definition)"} A *deformation functor* is a functor $$F: (\operatorname{Art}_{/k}) \to {\operatorname{Set}}$$ such that $$F(k) = {\{\operatorname{pt}\}}$$ is a singleton. ::: ::: {.example title="?"} Let $$X_{_{/k}}$$ be any scheme and let $$x\in X(k)$$ be a $$k{\hbox{-}}$$point. We can consider the deformation functor $$F$$ such that $$F(A)$$ is the set of extensions $$f$$ of the following form: {=tex} \begin{tikzcd} & & & & X \arrow[dd] \\ & & & & \\ \operatorname{Spec}(k) \arrow[rrrruu, "x"] \arrow[rr, hook] & & \operatorname{Spec}(A) \arrow[rruu, "f", dashed] \arrow[rr] & & \operatorname{Spec}(k) \end{tikzcd}  If $$A' \to A$$ is a morphism, then we define $$F(A') \to F(A)$$ is defined because we can precompose to fill in the following diagram {=tex} \begin{tikzcd} & & & & & & & & X \arrow[ddd] \\ & & & & & & & & \\ & & & & & & & & \\ \operatorname{Spec}(k) \arrow[rrd] \arrow[rrrrrrrruuu] & & & & & & & & \operatorname{Spec}(k) \\ & & \operatorname{Spec}(A) \arrow[rr] \arrow[rrrrrruuuu, "\exists \tilde f"] & & \operatorname{Spec}(A') \arrow[rrrru] \arrow[rrrruuuu, "f", dashed] & & & & \end{tikzcd}  So this is indeed a deformation functor. ::: ::: {.example title="a motivating example"} The Zariski tangent space on the nodal cubic doesn't "see" the two branches, so we allow "second order" tangent vectors. ::: We can consider parametrizing the functors above as $$F_{X, x}(A)$$, which is isomorphic to $$F_{\operatorname{Spec}({\mathcal{O}}_x)_{X, x}}$$ and further isomorphic to $$F_{\operatorname{Spec}\widehat{{\mathcal{O}}_x}_{x, X} }$$. This is because for Artinian algebras, we have maps $\operatorname{Spec}({\mathcal{O}}_{x, X})/{\mathfrak{m}}^N \to \operatorname{Spec}{\mathcal{O}}_{X, x} \to X .$ ::: {.remark} $$\widehat{ {\mathcal{O}}}_{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) \pmod t$$. ![Image](figures/2020-02-25-13:20.png)\ Note that the 'second order' tangent vector comes from $$\operatorname{Spec}k[t]/t^3$$. We can write $$F_{X, x}(A) = \pi^{-1}(x)$$ where $\hom_{{\operatorname{Sch}}_{/k}}(\operatorname{Spec}k, X) \xrightarrow{\pi} \hom_{{\operatorname{Sch}}_{/k}}(\operatorname{Spec}k, x) \ni x .$ Thus $F_{X, x}(A) = \hom_{{\operatorname{Sch}}_{/k}}(\operatorname{Spec}A, \operatorname{Spec}{\mathcal{O}}_{x, X}) = \hom_{k{\hbox{-}}\mathrm{Alg}}(\widehat{{\mathcal{O}}}_{X, x}, A) .$ ::: ::: {.example title="?"} Given any local $$k{\hbox{-}}$$algebra $$R$$, we can consider $h_R: (\operatorname{Art}_{/k}) &\to {\operatorname{Set}}\\ A &\mapsto \hom(R, A) .$ and $h_{\operatorname{Spec}R}: (\operatorname{Art}{\operatorname{Sch}}_{/k})^{\operatorname{op}}\to {\operatorname{Set}}\\ \operatorname{Spec}(A) &\mapsto \hom(\operatorname{Spec}A, \operatorname{Spec}R) .$ ::: ::: {.definition title="Representable Deformation"} A deformation $$F$$ is **representable** if it is of the form $$h_R$$ as above for some $$R \in \operatorname{Art}_{/k}$$. ::: ::: {.remark} There is a Yoneda Lemma for $$A\in \operatorname{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{\hbox{-}}$$algebra (i.e. a limit of Artinian local $$k{\hbox{-}}$$algebras). ::: ::: {.remark} We will see that there are simple criteria for a deformation functor to be pro-representable. This will eventually give us the complete local ring, which will give us the scheme representing the functor we want. ::: ::: {.remark} It is difficult to understand even $$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{\hbox{-}}$$algebras of the form $0 \to J \to A' \to A \to 0 .$ such that $$J$$ is annihilated by the maximal ideal fo $$A'$$. ::: ::: {.lemma title="?"} Given any quotient $$B\to A \to 0$$ of Artinian $$k{\hbox{-}}$$algebras, there is a sequence of small extensions (quotients): {=tex} \begin{tikzcd} 0 & & & & & & \\ & & & & & & \\ B_0 \arrow[uu] & & B_1 \arrow[lluu] & & \cdots & & B_n = A \arrow[lllllluu] \\ & & & & & & \\ B \arrow[uu] \arrow[rruu] \arrow[rrrrrruu] & & & & & & \end{tikzcd}  This yields {=tex} \begin{tikzcd} \operatorname{Spec}A \arrow[rrrr, hook] \arrow[rrrrdddddd, Rightarrow] & & & & \operatorname{Spec}B \\ & & & & \\ & & & & \operatorname{Spec}B_0 \arrow[uu, hook] \\ & & & & \\ & & & & \vdots \arrow[uu, hook] \\ & & & & \\ & & & & \operatorname{Spec}B_n \arrow[uu, hook] \end{tikzcd}  where the $$\operatorname{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=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ and let $$X_{_{/k}}$$ be connected. We have a picard functor ${\operatorname{Pic}}_{X_{_{/k}}}: ({\operatorname{Sch}}_{/k})^{\operatorname{op}}&\to {\operatorname{Set}}\\ S &\mapsto {\operatorname{Pic}}(X_S) / {\operatorname{Pic}}(S) .$ If we take a point $$x\in {\operatorname{Pic}}_{X_{_{/k}}}(k)$$, which is equivalent to line bundles on $$X$$ up to equivalence, we obtain a deformation functor $F \coloneqq F_{{\operatorname{Pic}}_{ X_{_{/k}}, x }} &\to {\operatorname{Set}}\\ A \mapsto \pi^{-1}(x)$ where $\pi: {\operatorname{Pic}}_{X_{_{/k}}}(\operatorname{Spec}A) &\to {\operatorname{Pic}}_{X_{_{/k}}} (\operatorname{Spec}k) \\ \pi^{-1}(x) &\mapsto x .$ This is given by taking a line bundle on the thickening and restricting to a closed point. Thus the functor is given by sending $$A$$ to the set of line bundles on $$X_A$$ which restrict to $$X_x$$. That is, $$F(A) \subset {\operatorname{Pic}}_{X_{_{/k}}}(\operatorname{Spec}A)$$ which restrict to $$x$$. So just pick the subspace $${\operatorname{Pic}}(X_A)$$ (base changing to $$A$$) which restrict. There is a natural identification of $${\operatorname{Pic}}(X_A) = H^1(X_A, {\mathcal{O}}_{X_A}^*)$$. If $0\to J \to A' \to A \to 0 .$ is a thickening of Artinian $$k{\hbox{-}}$$algebras, there is a restriction map of invertible functions ${\mathcal{O}}_{X_A}^* \to {\mathcal{O}}_{X_A'}^* \to 0 .$ which is surjective since the map on structure sheaves is surjective and its a nilpotent extension. The kernel is then just $${\mathcal{O}}_{X_{A'}} \otimes J$$. If this is a small extension, we get a SES $0 \to {\mathcal{O}}_X \otimes J \to {\mathcal{O}}_{X_{A'}}^* \to {\mathcal{O}}_{x_A}^* \to 0 .$ Taking the LES in cohomology, we obtain $H^1 {\mathcal{O}}_X \otimes J \to H^1 {\mathcal{O}}_{X_{A'}}^* \to H^1{\mathcal{O}}_{x_A}^* \to H^0 {\mathcal{O}}_X \otimes J .$ Thus there is an obstruction class in $$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$$ is the **obstruction space**. # Deformation Theory (Thursday February 27th) Big picture idea: We have moduli functors, such as $F_{S'}: ({\operatorname{Sch}}_{/k})^{\operatorname{op}}&\to {\operatorname{Set}}\\ \operatorname{Hilb}: S &\to \text{flat subschemes of } X_S \\ {\operatorname{Pic}}: S &\to {\operatorname{Pic}}(X_S)/{\operatorname{Pic}}(S) \\ \mathrm{Def}: S &\to \text{flat families } / S,~ \text{smooth, finite, of genus } g .$ ::: {.definition title="Deformation Theory"} Choose a point $$f$$ the scheme representing $$F_{S'}$$ with $$\xi_0 \in F_{gl}(\operatorname{Spec}K)$$. Define $F_{\text{loc}}: (\text{Artinian local schemes} / K)^{\operatorname{op}}\to {\operatorname{Set}} .$ {=tex} \begin{tikzcd} \operatorname{Spec}(K) \arrow[rr, "i", hook] & & \operatorname{Spec}(A) \arrow[rr] & & F(i)^{-1}(\xi_0) \arrow[rr] & & F_{gr}(\operatorname{Spec}K) \arrow[dd, "F(i)"] \\ & & & & & & \\ & & & & & & F_{gl}(\operatorname{Spec}K) \end{tikzcd}  ::: ::: {.definition title="Deformation Functors"} Let $$F: (\operatorname{Art}_{/k}) \to {\operatorname{Set}}$$ where $$F(k)$$ is a point. Denote $$\widehat{\operatorname{Art}}_{/k}$$ the set of complete local $$k{\hbox{-}}$$algebras. Since $$\operatorname{Art}_{/k} \subset \widehat{\operatorname{Art}} / k$$, we can make extensions $$\widehat{F}$$ by just taking limits: {=tex} \begin{tikzcd} & \operatorname{Art}_{/k} \arrow[rrr, "F"] & & & {\operatorname{Set}}\\ & & & & \\ \lim_{\leftarrow} R/{\mathfrak{m}}_R^n = R \in & \widehat{\operatorname{Art}}_{/k} \arrow[uu] \arrow[rrruu, "\widehat{F}"] & & & \end{tikzcd}  where we define $\widehat{F}(R) \coloneqq\varprojlim F(R/{\mathfrak{m}}_R^n) .$ ::: ::: {.question} When is $$F$$ pro-representable, which happens iff $$\widehat{F}$$ is representable? In particular, we want $$h_R \xrightarrow{\cong} \widehat{F}$$ for $$R\in \widehat{\operatorname{Art}}_{/k}$$, so $h_R = \hom_{\widehat{\operatorname{Art}}_{/k}}(R, {\,\cdot\,}) = \hom_{?}({\,\cdot\,}, \operatorname{Spec}k) .$ ::: ::: {.example title="?"} Let $$F_{\text{gl}} = \operatorname{Hilb}_{X_{_{/k}}}^p$$, which is represented by $$H_{/k}$$. Then . $\xi_0 = F_{\text{gl}}(k) = H(k) = \left\{{Z\subset X {~\mathrel{\Big|}~}P_z = f}\right\} .$ Then $$F_{\text{loc} }$$ is representable by $$\widehat{{\mathcal{O}}}_{H/\xi_0}$$. ::: ::: {.definition title="Thickening"} Given an Artinian $$k{\hbox{-}}$$algebra $$A \in \operatorname{Art}_{/k}$$, a *thickening* is an $$A' \in \operatorname{Art}_{/k}$$ such that $$0 \to J \to A' \to A \to 0$$, so $$\operatorname{Spec}A \hookrightarrow\operatorname{Spec}A'$$. ::: ::: {.definition title="Small Thickening"} A **small thickening** is a thickening such that $$0 = {\mathfrak{m}}_{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$$, fits into a diagram: {=tex} \begin{tikzcd} & & & & 0 & & & & \\ & & & & & & & & \\ & & J \arrow[rr] & & A' \arrow[uu] \arrow[rr] & & A \arrow[dd, Rightarrow] \arrow[rr] & & 0 \\ & & & & & & & & \\ 0 \arrow[rr] & & I \arrow[rr] \arrow[uu] & & B \arrow[uu] \arrow[rr] & & A \arrow[rr] & & 0 \\ & & & & & & & & \\ & & I' \arrow[rr, Rightarrow] \arrow[uu] & & I' \arrow[uu] & & & & \\ & & & & & & & & \\ & & 0 \arrow[uu] & & 0 \arrow[uu] & & & & \end{tikzcd}  ::: ::: {.proof title="of lemma"} We just need $$I' \subset I$$ with $${\mathfrak{m}}_S I \subset J' \subset I \iff J {\mathfrak{m}}_B = 0$$. Choose $$J'$$ to be a preimage of a codimension 1 vector space in $$I/{\mathfrak{m}}_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$$ and thus $$\widehat{F}$$ are determined by their behavior under small extensions. ### Example Consider $${\operatorname{Pic}}$$, fix $$X_{_{/k}}$$, start with a line bundle $$L_0 \in {\operatorname{Pic}}(x) /{\operatorname{Pic}}(k) = {\operatorname{Pic}}(x)$$ and the deformation functor $$F(A)$$ being the set of line bundles $$L$$ on $$X_A$$with $${\left.{{L}} \right|_{{x}} } \cong L_0$$, modulo isomorphism. Note that this yields a diagram {=tex} \begin{tikzcd} x \arrow[rr] \arrow[dd, hook] & & k \arrow[dd, "\text{unique closed point}"] \\ & & \\ X_A \arrow[rr] & & \operatorname{Spec}A \end{tikzcd}  This is equal to $$(I_x)^{-1}(L_0)$$, where $${\operatorname{Pic}}(X_a) \xrightarrow{I_x} {\operatorname{Pic}}(x)$$. If $0 \to J \to A' \to A \to 0 .$ is a small thickening, we can identify {=tex} \begin{tikzcd} 0 \arrow[rr] & & J \otimes_x {\mathcal{O}}_{x} \cong {\mathcal{O}}_x \arrow[rr] & & {\mathcal{O}}_{X_{A'}} \arrow[rr] & & {\mathcal{O}}_{X_{A}} \arrow[rr] & & 0 & & \\ & & & & & & & & & &\in\text{AbSheaves} \\ 0 \arrow[rr] & & {\mathcal{O}}_x \arrow[rr, "f\mapsto 1+f"] & & {\mathcal{O}}_{X_{A'}}^* \arrow[rr] \arrow[uu, hook] & & {\mathcal{O}}_{X_{A}}^* \arrow[rr] & & 0 & & \end{tikzcd}  This yields a LES {=tex} \begin{tikzcd}[column sep=tiny] 0 \arrow[rr] & & {H^0(X, {\mathcal{O}}_x) = k} \arrow[rr] & & {H^0(X_{A'}, {\mathcal{O}}_{x_{A'}}^*) = {A'}^*} \arrow[rr] & & {H^0(X_{A'}, {\mathcal{O}}_{x_{A}}^*) = A^*} \arrow[lllldd] \arrow[rr] & & \therefore 0 \\ & & & & & & & & \\ \therefore 0 \arrow[rr] & & {H^1(X, {\mathcal{O}}_{x})} \arrow[rr] & & {H^1(X_{A'}, {\mathcal{O}}_{x_{A'}}^*) = {\operatorname{Pic}}(X_{A'})} \arrow[rr, "\scriptsize\text{restriction to } X_A", outer sep=1em] & & {H^1(X_{A}, {\mathcal{O}}_{x_{A}}^*) = {\operatorname{Pic}}(X_A)} \arrow[lllldd, "\text{obs}"] & & \\ & & & & & & & & \\ & & {H^2(X, {\mathcal{O}}_x)} \ar[rr] & &\cdots & & & & \end{tikzcd}  ::: {.remark} To understand $$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 {\operatorname{Pic}}(X_A)$$, when does it extend? ::: By exactness, $$L'$$ exists iff $$\text{obs}(L) = 0\in H^2(X, {\mathcal{O}}_x)$$, which answers 1. To answer 2, $$(I_x)^{-1}(L)$$ is the set of extensions of $$L$$, which is a torsor under $$H^1(x, {\mathcal{O}}_x)$$. Note that these are fixed $$k{\hbox{-}}$$vector spaces. ::: {.remark} $$H^1(X, {\mathcal{O}}_x)$$ is interpreted as the **tangent space** of the functor $$F$$, i.e. $$F_{\text{loc}}(K[\varepsilon])$$. Note that if $$X$$ is projective, line bundles can be unobstructed without the group itself being zero. ::: For (3), just play with $$A = k[\varepsilon]$$, which yields $$0 \to k \xrightarrow{\varepsilon} k[\varepsilon] \to k \to 0$$, then {=tex} \begin{tikzcd} 0 \arrow[rr] & & {H^1(X, {\mathcal{O}}_x)} \arrow[rr] & & {H^1(X_{k[\varepsilon]}, {\mathcal{O}}_{k[\varepsilon]}^*)} \arrow[rr, "I_x"] & & {H^1(X, {\mathcal{O}}_x^*)} \arrow[ll, bend right=49] \\ & & & & {(I_x)^{-1}(L_0) \in {\operatorname{Pic}}(X_{k[\varepsilon]})} & & L_0 \in {\operatorname{Pic}}(x) \end{tikzcd}  i.e., there is a canonical trivial extension $$L_0[\varepsilon]$$. ::: {.example title="?"} Let $$X \supset Z_0 \in \operatorname{Hilb}_{X_{_{/k}}}(k)$$, we computed $T_{Z_0} \operatorname{Hilb}_{X_{_{/k}}} = \hom_{{\mathcal{O}}_x}(I_{Z_0}, {\mathcal{O}}_z) .$ We took $$Z_0 \subset X$$ and extended to $$Z' \subset X_{k[\varepsilon]}$$ by base change. In this case, $$F_{\text{loc}}(A)$$ was the set of $$Z'\subset X_A$$ which are flat over $$A$$, such that base-changing $$Z' \times_{\operatorname{Spec}A} \operatorname{Spec}k \cong Z$$. This was the same as looking at the preimage restricted to the closed point, $\operatorname{Hilb}_{X_{_{/k}}}(A) \xrightarrow{i^*} \operatorname{Hilb}_{X_{_{/k}}}(k) \\ (i^*)^{-1}(z_0) \mapsfrom z_0 .$ Recall how we did the thickening: we had $$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\otimes_{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')$$. By tensoring the two exact sequences here, we get an exact grid: {=tex} \begin{tikzcd} 0 \arrow[rr] \arrow[dd] & & I_{Z'} \arrow[rr] & & {\mathcal{O}}_{X_{A'}} \arrow[rr] & & {\mathcal{O}}_{Z'} \arrow[rr] & & 0 \\ & & 0 \arrow[d] & & 0 \arrow[d] & & 0 \arrow[d] & & \\ J \arrow[dd] & 0 \arrow[r] & I_{Z_0} \arrow[dd] \arrow[rr] & & {\mathcal{O}}_X \arrow[dd] \arrow[rr] & & {\mathcal{O}}_{Z_0} \arrow[dd] \arrow[r] & 0 & \\ & & & & & & & & \\ A' \arrow[dd] & 0 \arrow[r] & I_{Z'} \arrow[rr] \arrow[dd] & & {\mathcal{O}}_{X_{A'}} \arrow[rr] \arrow[dd] & & {\mathcal{O}}_{Z'} \arrow[dd] \arrow[r] & 0 & \\ & & & & & & & & \\ A \arrow[dd] & 0 \arrow[r] & I_Z \arrow[d] \arrow[rr] & & {\mathcal{O}}_{X_A} \arrow[rr] \arrow[d] & & {\mathcal{O}}_Z \arrow[d] \arrow[r] & 0 & \\ & & 0 & & 0 & & 0 & & \\ 0 & & & & & & & & \end{tikzcd}  The space of extension should be a torsor under $$\hom_{{\mathcal{O}}_X}(I_{Z_0}, {\mathcal{O}}_{Z_0})$$, which we want to think of as $$\hom_{{\mathcal{O}}_X}(I_{Z_0}, {\mathcal{O}}_{Z_0})$$. Picking a $$\phi$$ in this hom space, we want to take an extension $$I_{Z'} \xrightarrow{\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: \operatorname{Art}_{/k} \to {\operatorname{Set}}$$. ::: {.example title="?"} The Hilbert functor $\operatorname{Hilb}_{X_{_{/k}}} ({\operatorname{Sch}}_{/k})^{\operatorname{op}}\to {\operatorname{Set}}\\ S \mapsto \left\{{ Z \subset X \times S \text{ flat over } S}\right\} .$ This yields $F: \operatorname{Art}_{/k} \to {\operatorname{Set}}\\ ??? .$ ::: ![Image](figures/2020-03-31-12:44.png) Recall that we're interested in pro-representability, where $$\widehat{F}(R) = \varprojlim F(R\mu_R^n)$$ is given by a lift of the form {=tex} \begin{tikzcd} \operatorname{Art}_{/k} \ar[r, "F"] & {\operatorname{Set}} \\ \widehat{\operatorname{Art}_{/k}} \ar[u, hook] \ar[ur, "\widehat{F}"'] & \end{tikzcd}  ::: {.question} Is $$\widehat{F}$$ representable, i.e. is $$F$$ pro-representable? ::: ::: {.example title="?"} The $$F$$ in the previous example is pro-representable by $$\widehat{F} = \hom({\mathcal{O}}_{\operatorname{Hilb}, z_0}, {\,\cdot\,})$$. ::: ::: {.definition title="Pro-Representable Hull"} $$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{\hbox{-}}$$algebras is **formally smooth** if it can be lifted through nilpotent thickenings. That is, for $$F, G: \operatorname{Art}_{/k} \to {\operatorname{Set}}$$, $$F \to G$$ is *formally smooth* if for any thickening $$A' \twoheadrightarrow A$$, we have {=tex} \begin{tikzcd} & & F \ar[d] \\ h_{A} \ar[rru] \ar[r] & h_{A'} \ar[ru, dotted] \ar[r] & G \\ \operatorname{Spec}A \ar[u, equal] \ar[r] & \operatorname{Spec}A' \ar[u, equal] \ar[r] & G \ar[u, equal] \end{tikzcd}  We proved for $$R, A$$ finite type over $$k$$, $$\operatorname{Spec}R \to \operatorname{Spec}A$$ smooth is formally smooth. Given a complete local $$k{\hbox{-}}$$algebra $$R$$ and a section $$\xi \in \widehat{F}(R)$$, we make the following definitions: ::: {.definition title="Versal, Miniversal, Universal"} The pair $$(R, \xi)$$ is - *Versal* for $$F$$ iff $$h_R \xrightarrow{\xi} F$$ is formally smooth.[^5] - *Miniversal* for $$F$$ iff versal and an isomorphism on Zariski tangent spaces. - *Universal* for $$F$$ if $$h_R \xrightarrow{\cong} F$$ is an isomorphism, i.e. $$h_R$$ pro-represents $$F$$. - Pullback by a unique map ::: ::: {.remark} Note that **versal** means that any formal section $$(s, \eta)$$ where $$\eta \in \widehat{F}(s)$$ comes from pullback, i.e there exists a map $R &\to S \\ \widehat{F}(R) &\to \widehat{F}(s) \\ \xi &\mapsto \eta .$ **Miniversal** means adds that the derivative is uniquely determined, and universal means that $$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{\hbox{-}}$$vector spaces, along with a functorial assignment of the following form: $(A' \twoheadrightarrow A) \quad \text{a small thickening } \mapsto \\ \mathrm{def}(F) {\circlearrowleft}F(A') \to F(A) \xrightarrow{\mathrm{obs}} \mathrm{obs}(F)$ that is exact[^6] and if $$A=k$$, it is exact on the left (so the action was faithful on nonempty fibers). ::: ::: {.example title="?"} We have ${\operatorname{Pic}}_{X_{/k}} : ({\operatorname{Sch}}_{/k})^{\operatorname{op}}&\to {\operatorname{Set}}\\ S &\mapsto {\operatorname{Pic}}(X\times X) / {\operatorname{Pic}}(S) .$ This yields $F: \operatorname{Art}_{/k} \to {\operatorname{Set}}\\ A \mapsto L\in {\operatorname{Pic}}(X_A),~ L\otimes k \cong L_0$ where $$X_{_{/k}}$$ is proper and irreducible. Then $$F$$ has an obstruction theory with $$\mathrm{def}(F) = H^1({\mathcal{O}}_x)$$ and $$\mathrm{obs}(F) = H^2({\mathcal{O}}_x)$$. The key was to look at the LES of $0 \to {\mathcal{O}}_x \to {\mathcal{O}}_{X_{A'}}^* \to {\mathcal{O}}_{X_A}^* \to 0 .$ for $$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: \operatorname{Art}_{/k} \to {\operatorname{Set}}$$ be a deformation functor (and it only makes sense to talk about deformation functors when $$F(k) = {\{\operatorname{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$$ any other thickening, the map $F(A' \times_A A'') \to F(A') \times_{F(A)} F(A'')$ is surjective. 2. Unique gluing: if $$(A' \to A) = (k[\varepsilon] \to k)$$, then the above map is bijective. 3. $$t_F = F(k[\varepsilon])$$ is a finite dimensional $$k{\hbox{-}}$$vector space, i.e. $F(k[\varepsilon] \times_k k[\varepsilon]) \xrightarrow{\cong} F(k[\varepsilon]) \times F(k[\varepsilon]) .$ 4. For $$A' \to A$$ small, {=tex} \begin{tikzcd} F(A') \ar[r, "f"] & F(A) \\ t_f\, {\circlearrowleft}f^{-1}(\eta) \ar[u, hook, "\subseteq"] & \eta \ar[u, "\in"] \end{tikzcd}  where the action is simply transitive. $$F$$ has a miniversal family iff (1)-(3) hold, and universal iff all 4 hold. ::: ::: {.exercise title="?"} Show that the existence of an obstruction theory which is exact on the left implies (1)-(4). ::: The following diagram commutes: {=tex} \begin{tikzcd} \mathrm{def} {\circlearrowleft}F(A' \times_A A'') \ni \eta \ar[r] \ar[d] & F(A'') \ni \xi'' \ar[r, "\mathrm{obs}"] \ar[d] & \mathrm{obs} \\ \mathrm{def} {\circlearrowleft}F(A')\ni \eta'm \xi' \ar[r] & F(A')\ni \xi \ar[r, "\mathrm{obs}"] & \mathrm{obs} \\ \end{tikzcd}  So we have a map $$F(A' \times_A A'') \to F(A') \times_{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 \hookrightarrow^i X$$ such that $$i \otimes 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})$$ ::: ::: {.remark} {=tex} \envlist  1. If $$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$$. Then there is an exact sequence $0 \to \text{Der}_R({\mathcal{O}}_{X_0}) \to\operatorname{Aut}_{A'}(X') \to \operatorname{Aut}_A(X) .$ 2. If $$\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 \times_k \operatorname{Spec}(A)$$. {=tex} \begin{tikzcd} & & X_0 \times\operatorname{Spec}(A) \ar[d] \\ X_0 \ar[r, hook] & X \ar[r] \ar[ru, "f", dotted] & \operatorname{Spec}(A) \end{tikzcd}  Thus $$X_0 \hookrightarrow X$$ has a section $$X\to X_0$$, and the claim is that this forces $$X$$ to be trivial. ::: We have {=tex} \begin{tikzcd} 0 \ar[r] & J \otimes{\mathcal{O}}_X \ar[r] & {\mathcal{O}}_x \ar[r] & {\mathcal{O}}_{X_0} \ar[r] \ar[l, bend right] & 0 \end{tikzcd}  yielding $0 \to K \to {\mathcal{O}}_{X_0} \otimes A \to {\mathcal{O}}_X \to 0 \\ ({\,\cdot\,}\otimes k) \\ 1 \to k\otimes k = 0 \to {\mathcal{O}}_{X_0} \xrightarrow{\cong} {\mathcal{O}}_{X_0} \to 0 .$ ::: {.remark} Why does this involve cohomology of the tangent bundle? For $$X_0$$ smooth, $$\operatorname{Der}_k({\mathcal{O}}_{X_0}) = \mathcal{H}(T_{X_0})$$, but the LHS is equal to $$\hom( \Omega_{ \qty{X_0}_{/k}}, {\mathcal{O}}_{X_0}) = H^0 (T_{X_0})$$. ::: > Upcoming: proof of Schlessinger so we can use it! # Thursday April 2nd ## Abstract Deformations Let $$X_0$$ be smooth and consider the deformation functor $F : \operatorname{Art}_{_{/k}} &\to {\operatorname{Set}}\\ A &\mapsto (X_{/A} , \iota)$ where $$X$$ is flat (and thus smooth) and $$i$$ is a closed embedding $$i: X_0 \hookrightarrow X$$ with $$i\otimes 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$$. ::: {.remark title="Key points"} {=tex} \envlist  - All deformations of smooth affine schemes are trivial - Automorphisms of a deformation $$X/A$$ which are the identity on $$X_0$$ are $$\operatorname{id}+ \delta$$ for $$\delta$$ a derivation in $$\operatorname{Der}_k({\mathcal{O}}_{X_0}) = \hom_{{\mathcal{O}}_{X_0}}(\Omega_{\qty{X_0}_{_{/k}}}, {\mathcal{O}}_{X_0})$$. > See screenshot. ::: Suppose we have a small thickening $$k \to {\mathbb{A}}^1 \to {\mathbb{A}}$$ and $$X/{\mathbb{A}}$$ with an affine cover $$X_\alpha$$ of $$X$$. This comes with gluing information $$\phi_{\alpha\beta}: X_{\alpha\beta} \to X_{\beta\alpha} = X_\alpha \cap X_\beta$$. These maps satisfy a cocycle condition: {=tex} \begin{tikzcd} X_{\alpha\beta} \cap X_{\alpha\gamma} \ar[rr] \ar[rd] & & X_{\gamma\alpha} \cap X_{\gamma\beta} \ar[ld] \\ & X_{\beta\alpha} \cap X_{\beta\gamma} & \end{tikzcd}  ::: {.question} Can we extend this to $$X'/{\mathbb{A}}$$? ::: We have $$X_\alpha \cong X_\alpha^\mathrm{red} \times{\mathbb{A}}$$? Choose $$\phi'_{\alpha\beta}$$ such that {=tex} \begin{tikzcd} X'_{\alpha\beta} \ar[r, "\phi'_{\alpha\beta}"] & X'_{\beta\alpha} = X_{\beta\alpha}^\mathrm{red} \times{\mathbb{A}} \\ X_{\alpha\beta}\ar[u, hook] \ar[r, "\phi_{\alpha\beta}"] & X_{\beta\alpha} \ar[u, hook] \end{tikzcd}  We need $$\phi'_{\alpha\beta}$$ to satisfy the cocycle condition in order to glue. We want the following map to be the identity: $$(\phi'_{\alpha\gamma})^{-1}\phi'_{\beta\gamma} \phi'_{\alpha\beta}$$. This is an automorphism of $$X'_{\alpha\beta} \cap X'_{\alpha\beta}$$ and is thus the identity in $$\operatorname{Aut}(X_{\alpha\beta} \cap X_{\alpha\gamma})$$. So it makes sense to talk about $\delta_{\alpha\beta\gamma} \coloneqq (\phi'_{\alpha\gamma})^{-1} \phi'_{\beta\gamma} \phi'_{\alpha\beta} - \operatorname{id}\in M^0(T_{X^\mathrm{red}_{\alpha\beta\gamma}}) .$ ::: {.exercise title="?"} In parts, 1. $$\delta_{\alpha\beta\gamma}$$ is a $$2{\hbox{-}}$$cocycle for $$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} = {{\partial}}\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}' \xrightarrow{\phi_{\alpha\beta}'} X_{\beta\alpha}'$$ where the LHS is isomorphic to $$(X_{\alpha\beta}')^\mathrm{red} \times{\mathbb{A}}^r$$? Given $$\eta_{\alpha\beta} \in H^0(T_{X_{\alpha\beta}})$$, then $$\phi'_{\alpha\beta} + \eta_{\alpha\beta} = \phi_{\alpha\beta}''$$ is another such identification. ::: ::: {.exercise title="?"} In parts 1. $${{\partial}}\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']$$ then this isomorphism can be modified by by $$\varepsilon$$ defined by $$\eta-\eta' = {{\partial}}\varepsilon$$, and it patches. ::: ::: {.remark} This kind of patching is ubiquitous -- essentially patching together local obstructions to get a global one. In general, there is a local-to-global spectral sequence that computes the obstruction space ::: ## Proving Schlessinger ### The Schlessinger Axioms #### H1 For any two small thickenings $A' &\to A \\ A'' &\to A$ we have a natural map $F(A' \times_A A'') \to F(A') \times_{F(A)} F(A'')$ and we require that this map is surjective. So deformations agreeing on the sub glue together. #### H2 When $$(A' \to A) = (k[\varepsilon] \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[\varepsilon])$$, and we require that $$\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$$ small, $t_F {\circlearrowleft}F(A') \to F(A)$ is exact in the middle and left. ::: {.remark} Note that the existence of this action uses H2. ::: ::: {.warnings} For $$(R, \xi)$$ a complete local ring and $$\xi \in \widehat{F}(R)$$ a formal family, this is a hull $$\iff$$ miniversal, i.e. for $$h_R \xrightarrow{\xi} F$$, this is smooth an isomorphism on tangent spaces. ::: ::: {.theorem title="1, Schlessinger"} {=tex} \envlist  a. $$F$$ has a miniversal family $$(R, \xi)$$ with $$\dim t_R < \infty$$, noting that $$t_R = {\mathfrak{m}}_R / {\mathfrak{m}}_R^2$$, iff H1-H3 hold. b. $$F$$ has a universal family $$(R, \xi)$$ with $$\dim t_R < \infty$$ iff h1-H4 hold. ::: ::: {.theorem title="2"} {=tex} \envlist  a. $$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{\hbox{-}}$$algebra with $$t_R$$ finite dimensional has a strong obstruction theory. ::: Can always find a surjection from a power series ring: $S \coloneqq k[[t_R^\vee]] \twoheadrightarrow R$ which yields an obstruction theory - $$\mathrm{def} = t_R$$ - $$\mathrm{obs} = I/{\mathfrak{m}}_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' \times_A A'') \twoheadrightarrow F(A') \times_{F(A)} F(A'')$$ For versal, if we have $$h_R \xrightarrow{\xi} F$$ smooth, we have {=tex} \begin{tikzcd} & & & h_r \ar[d] \\ h_k \ar[r] \ar[rrru] & h_A \ar[r] \ar[rru, dotted] \ar[rr, bend right, "\eta"] & h_{A'} \ar[ur, dotted] \ar[r] & F \end{tikzcd}  and we can find a lift from $$h_{A''}$$ as well, so we get a diagram {=tex} \begin{tikzcd} & & F \\ h_{A''} \ar[r] & h_R \ar[ru] \\ h_A \ar[u] \ar[r] & h_{A'} \ar[u] \end{tikzcd}  and thus {=tex} \begin{tikzcd} {A''} \ar[r] & R \\ A \ar[u] \ar[r] & {A'} \ar[u] \end{tikzcd}  So we get the left $$\tilde \eta$$ of $$(\eta', \eta'')$$ we want from {=tex} \begin{tikzcd} h_{A' \times_A A''} \ar[r, "f"] \ar[rr, "\tilde\eta", bend right] & h_R \ar[r] & F \end{tikzcd}  If $$(R, \xi)$$ is miniversal, then H2 holds. We want to show that the map $F(A'' \times_K k[\varepsilon]) \xrightarrow{\sim} ??$ is a bijection. Suppose we have two maps {=tex} \begin{tikzcd} & & h_R \\ h_{A''} \ar[rru, bend left] \ar[r] & h_{A'' \times k[\varepsilon]} \ar[ru, shift left=0.75ex] \ar[ru, shift right=0.75ex] \ar[r, shift left=0.75ex] \ar[r, shift right=0.75ex] & F \\ & h_{k[\varepsilon]} \ar[ur, bend right] & \end{tikzcd}  Then the two lifts are in fact equal, and {=tex} \begin{tikzcd} R \ar[r, shift left=0.75ex] \ar[r, shift right=0.75ex] & A'' \times k[\varepsilon] \ar[r] \ar[d] & k[\varepsilon] \\ & A'' & \end{tikzcd}  If $$(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$$ be small, then $A' \times_A A' &= A' \times_k k[\varepsilon] \\ (x, y) &\mapsto ?? .$ Using H2, we can identify $$F(A; \times_A A') \cong t_F \times 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$$ such that $${\mathfrak{m}}_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 \cap I'$$ does as well. So $$I, I'$$ are determined by their images $$v, v'$$ in the vector space $$I_q \otimes k$$. ::: So enlarge either $$v$$ or $$v'$$ such that $$v + v' = I_q \otimes k$$ but $$v \cap v'$$ is the same. We can thus assume that $$I + I' = I_q$$, and so $S / I \cap I' = S/I \times_{S/I_q} S/I'$ which by H1 yields a map $F(S/I\cap I') &\to F(S/I) \times_{F(S/I_q)} F(S/I')$ So $$I\cap 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 = \varinjlim R_q$$ and $$\xi = \varinjlim\xi_q$$, the claim is that $$(R, \xi)$$ is miniversal. ::: We already have $$h_R \xrightarrow{\xi} F$$ and thus $$t_R \xrightarrow{\cong}t_F$$ is fulfilled. We need to show formal smoothness, i.e. for $$A' \to A$$ a small thickening, suppose we have a lift {=tex} \begin{tikzcd} & & h_R \ar[d, "\xi"] \\ h_a \ar[rru, "n"] \ar[rr, bend right] \ar[r] & h_{A'} \ar[r] & F \end{tikzcd}  If we have a $$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$$ acts transitively on fibers, yielding {=tex} \begin{tikzcd} t_R {\circlearrowleft}\ar[d, "\cong"] & u'\in h_R(A') \ar[r] \ar[d] \ar[r] & u\in h_R(A) \ar[d]\\ t_F {\circlearrowleft}& \eta' \in F(A') \ar[r] \ar[r] & \eta \in F(A) \\ \end{tikzcd}  Then $$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$$. So we can modify the lift to make these agree. Thus it suffices to show {=tex} \begin{tikzcd} A' \ar[r] & A & R_q \ar[l] \\ S \ar[u, "v"] \ar[r] & \ar[lu, dotted, "\exists_? u'"] \ar[u, "u"] \ar[ur] & \end{tikzcd}  We get a diagram of the form {=tex} \begin{tikzcd} S \ar[d] \ar[r, "w"] & A' \times_A R_1 \ar[r] \ar[d, "{ \pi_2, \text{small} }"] & A' \ar[d, "{ \text{small} }"] \\ R \ar[r] & R_q \ar[r] & A \end{tikzcd}  ::: {.observation} {=tex} \envlist  - $$S \to R_q$$ is surjective. - $$\operatorname{im}(w) \subset A' \times_A R_1$$ is a subring, so either - $$\operatorname{im}(w) \xrightarrow{\cong} R_q$$ if it doesn't meet the kernel, or - $$\operatorname{im}(w) = A' \times_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$$ is surjective and consider {=tex} \begin{tikzcd} 0 \ar[r] & I \ar[r] & S \ar[r] & A' \times_A R_q \ar[r]\ar[d, "\text{small}"] & 0 \\ & & & R_q & \end{tikzcd}  and we have $${\mathfrak{m}}_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$$, and we have $\xi \in F(S/I) \twoheadrightarrow\xi = \xi' \times\xi_q ? .$ Therefore $$I_{q+1} \subset I$$ and we have a factorization {=tex} \begin{tikzcd} S \ar[rr] \ar[dr, dotted] & & S/I \\ & R_{q+1}\ar[ur, dotted] & \end{tikzcd}  Recall that we had ![](figures/image_2020-04-07-13-17-11.png) {=tex} \todo[inline]{Image to diagram}  where the diagonal map $$u'$$ gives us the desired lift, and thus {=tex} \begin{tikzcd} R \ar[r] \ar[rr, bend left] & R_{q+1} \ar[r] & A' \end{tikzcd}  exists. This concludes showing miniversality. ## Part of Proof To finish, we want to show that H4 implies that the map on sections $$h_R \xrightarrow{\xi} F$$ is bijective. {=tex} \begin{tikzcd} & & & h_R \ar[d, "\xi"] \\ & h_A \ar[rr, bend right, "\eta"] \ar[rru, bend left, "u"] & h_{A'} \ar[r, "\eta'"] \ar[ru, "\exists ! u'"] & F \end{tikzcd}  where the map $$\xi$$ is "formal etale", which will necessarily imply that it's a bijection over all artinian rings. So we just need to show formal étaleness. We have a diagram {=tex} \begin{tikzcd} t_R {\circlearrowleft}u'\in h_R(A') \ar[r] \ar[d] & u\in h_R(A) \ar[d] \\ t_F {\circlearrowleft}\eta' \in h_R(A') \ar[r] & \eta \in h_R(A) \end{tikzcd}  where $$u'$$ exists by smoothness. Assume that are two $$u', u''$$, then $$u' = u'' + \theta$$ and $$\operatorname{im}(u') = \operatorname{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 \xrightarrow{\xi }F$$. The obstruction theory of $$F$$ *gives* an obstruction theory of $$h_R$$: given $$A' \to A$$ a small thickening, we need a functorial assignment $t_R = \mathrm{def} {\circlearrowleft}h_R(A') \to h_R(A) \xrightarrow{\mathrm{obs}} \mathrm{obs} \\ \mathrm{def} {\circlearrowleft}F(A') \to F(A) \xrightarrow{\mathrm{obs}} \mathrm{obs}$ where there are vertical maps with equality on the edges. ![Vertical maps](figures/image_2020-04-07-13-35-00.png) By formal smoothness, $$\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$$, since we can always find a quotient $I \to S = k[[t_R^\vee]] \twoheadrightarrow R$ and $$h_K$$ has an obstruction theory - $$\mathrm{def} = t_R = \qty{{\mathfrak{m}}_R/{\mathfrak{m}}_R^2}^\vee$$ - $$\mathrm{obs} = \qty{I/{\mathfrak{m}}_S I}^\vee$$ ::: ::: {.fact title="proof can be found in FGA"} Any other obstruction theory $$(\mathrm{def}', \mathrm{obs}')$$ of $$h_R$$ admits an injection $$\qty{I/{\mathfrak{m}}_S I}^\vee\hookrightarrow\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 \xrightarrow{\xi }F$$ with $$R = S/ I$$ a quotient of the formal power series ring over some ideal, where $$S = k[[t_F^\vee]]$$. It follows that $$\dim(I/{\mathfrak{m}}_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 = \widehat{{\mathcal{O}}}_{{\mathfrak{m}}, 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: \operatorname{Art}_{/k} \to {\operatorname{Set}}$$ be a deformation functor with an obstruction theory. Then H1-H3 imply the existence of a miniversal family, and gives us some control on the hull $$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)^\vee]] = k[[ t_{F}^\vee]]$$. ::: {.example title="?"} Let $$M = \operatorname{Hilb}_{{\mathbb{P}}^n_{/k}}^{dt + (1-g)}$$ where $$k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$, and suppose $$[Z] \in M$$ is a smooth point. Then $\mathrm{def} = \hom_{ {{ {\mathcal{O}}_{x} }{\hbox{-}}\operatorname{mod}} }(I_{Z}, {\mathcal{O}}_{Z}) = \hom_{Z}(I_{Z}/I_{Z}^2, {\mathcal{O}}_{Z}) = H^0(N_{Z/X}) .$ the normal bundle $$N_{Z/X} = (I/I^2)^\vee$$ of the regular embedding, and $$\mathrm{obs} = H^1(N_{Z/X})$$. ::: {.claim} If $$H^1({\mathcal{O}}_{Z}(1)) = 0$$ (e.g. if $$d > 2g-2)$$ then $$M$$ is smooth. ::: ::: {.proof title="of claim"} The tangent bundle of $${\mathbb{P}}^n$$ sits in the Euler sequence $0 \to {\mathcal{O}}\to {\mathcal{O}}(1)^{n+1} \to T_{{\mathbb{P}}^n} \to 0 .$ And the normal bundles satisfies $0 \to T_{Z} &\to T_{{\mathbb{P}}^n}\mathrel{\Big|}_{Z} \to N_{Z/{\mathbb{P}}^n} \to 0 \\ \\ &\Downarrow \text{ is the dual of }\\ \\ 0 \to I/I^2 &\to \Omega \mathrel{\Big|}_{Z} \to \Omega \to 0 .$ There is another SES: $????? .$ Taking the LES in cohomology yields $H^1({\mathcal{O}}_{Z}(1)^{n+1})=0 \to H^1(N_{Z/{\mathbb{P}}^n}) =0 \to 0$ and thus $$M$$ is smooth at $$[Z]$$. We can compute the dimension using Riemann-Roch: $\dim_{[Z]} M &= \dim H^0(N_{Z/{\mathbb{P}}^n}) \\ &= \chi(N_{Z/{\mathbb{P}}^n}) \\ &= \deg N + {\operatorname{rank}}N(1-g) \\ &= \deg T_{{\mathbb{P}}^n} \mathrel{\Big|}_Z - \deg T_{Z} + (n-1)(1-g) \\ &= d(n+1) + (2-2g) + (n-1)(1-g) .$ ::: ::: ::: {.remark} This is one of the key outputs of obstruction theory: being able to compute these dimensions. ::: ::: {.example title="?"} Let $$X \subset {\mathbb{P}}^5$$ be a smooth cubic hypersurface and let $$H = \operatorname{Hilb}_{X_{/k}}^{\text{lines} = t+1} \subset \operatorname{Hilb}_{{\mathbb{P}}^5/k}^{t+1} = {\operatorname{Gr}}(1, {\mathbb{P}}^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})$$ We have an exact sequence $0 \to N_{\ell/X} \to N_{\ell/{\mathbb{P}}} \to N_{X/{\mathbb{P}}}\mathrel{\Big|}_\ell \to 0 \\ .$ There are surjections from $${\mathcal{O}}_\ell(1)^6$$ onto the last two terms. ::: {.claim title="Subclaim"} For $$N = N_{\ell/{\mathbb{P}}}$$ or $$N_{X/{\mathbb{P}}}\mathrel{\Big|}_\ell$$, we have $$H^1(N) = 0$$ and $${\mathcal{O}}(1)^6 \twoheadrightarrow N$$ is surjective on global sections. ::: ::: {.proof title="of subclaim"} Because $$\ell$$ is a line, $${\mathcal{O}}_\ell(1) = {\mathcal{O}}(1)$$ and $$H^1({\mathcal{O}}_\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 = {\mathcal{O}}(2)$$, and the LES for $$0 \to {\mathcal{O}}\to K \to T_\ell$$ shows $$H^1(K) = 0$$. Looking at the horizontal SES $$0 \to K \to {\mathcal{O}}_\ell(1)^6 \twoheadrightarrow N_{\ell/{\mathbb{P}}}$$ yields the surjection claim. We have ![Diagram](figures/a.png) and taking the LES in cohomology yields ![Diagram](figures/image_2020-04-09-12-55-01.png)\ Therefore $$H$$ is smooth at $$\ell$$ and $\dim_\ell H &= \chi(N_{\ell/X}) \\ &= \deg T_{X} - \deg T_\ell + 3 \\ &= \deg T_{\mathbb{P}}- \deg N_{X/{\mathbb{P}}} - \deg T_\ell + 3 \\ &= 6 - 3 - 2 + 3 = 4 .$ ::: ::: ::: {.remark} It turns out that the Hilbert scheme of lines on a cubic has some geometry: the Hilbert scheme of two points on a K3 surface. ::: ## Abstract Deformations Revisited Take $$X_{0} / k$$ some scheme and consider the deformation functor $$F(A)$$ taking $$A$$ to $$X/A$$ flat with an embedding $$\iota: X_{0} \hookrightarrow X$$ with $$\iota \otimes k$$ an isomorphism. Start with H1, the gluing axiom (regarding small thickenings $$A' \to A$$ and a thickening $$A'' \to A$$). Suppose $X_{0} \hookrightarrow X' \in F(A') \to F(A) .$ which restricts to $$X_{0} \hookrightarrow X$$. Then in $$F(A)$$, we have $$X_{0} \hookrightarrow X' \otimes_{A'} A$$, and we obtain a commutative diagram where $$X' \otimes A \hookrightarrow X'$$ is a closed immersion: ![???](figures/abcdefg.png){width="350px"} The restriction $$X' \to X$$ means that there exists a diagram {=tex} \begin{tikzcd} X' & & X \ar[ll, dotted, "\exists"] \\ & X \ar[ur, hook] \ar[ul, hook] \end{tikzcd}  Note that this is not necessarily unique. We have ![Diagram?](figures/image_2020-04-09-13-06-40.png){width="350px"} This means that we can find embeddings such that {=tex} \begin{tikzcd} X'' & \ar[l, "\exists", hook] X \ar[r, "\exists", hook] & X' \\ & X_{0} \ar[ul, hook] \ar[u, hook] \ar[ur, hook] \end{tikzcd}  ![Diagram](figures/image_2020-04-09-13-08-19.png){width="350px"} And thus if we have ![Diagram](figures/image_2020-04-09-13-08-42.png){width="350px"} then $$X_{0} \hookrightarrow Z$$ is **a** required lift (again not unique). ::: {.question} When is such a lift unique? ::: Suppose $$X_{0} \hookrightarrow 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$$, which is thus an isomorphism.[^7] ::: {.remark} Thus the lift is unique if - $$X = X_{0}$$, then the following diagrams commute by taking the identity and the embedding you have. Note that in particular, this implies H2. ![Diagram](figures/image_2020-04-09-13-15-09.png){width="350px"} - Generally, these diagrams can be completed (and thus the gluing maps are bijective) if the map $\operatorname{Aut}(X_{0}\hookrightarrow X') \to \operatorname{Aut}(X_{0} \hookrightarrow X) .$ of automorphisms of $$X'$$ commuting with $$X_{0} \hookrightarrow 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_{{\mathcal{O}}_{k}}({\mathcal{O}}_{X_{0}}, {\mathcal{O}}_{X_{0}})$$, then the gluing map is bijective. ::: ::: {.proof title="?"} The claim is that $$\operatorname{Aut}(X_{0} \hookrightarrow X) = 1$$ are always trivial. This would imply that all random choices lead to triangles that commute. Proceeding by induction, for the base case $$\operatorname{Aut}(X_{0} \hookrightarrow X_{0}) = 1$$ trivially. Assume $$X_{0} \hookrightarrow X_{i}$$ lifts $$X_{0} \hookrightarrow X$$, then there's an exact sequence $0 \to \operatorname{Der}_{k}({\mathcal{O}}_{X_{0}}, {\mathcal{O}}_{X_{0}}) \to {\operatorname{Aut}}(X_{0} \hookrightarrow X_0') \to {\operatorname{Aut}}(X_{0} \hookrightarrow X) .$ ::: Thus $$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[\varepsilon]$$ is finite, i.e. $$\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 {\mathbb{A}}^n$$, and for simplicity, $$(f=0) \subset {\mathbb{A}}^2$$, and let - $$S = {\mathbb{C}}[x, y]$$. - $$B = {\mathbb{C}}[x, y] / (f)$$ ::: {.question} What are the deformations over $$A \coloneqq k[\varepsilon]$$? ::: This means we have a ring $$B'$$ flat over $$k$$ and tensors to an isomorphism, so tensoring $$k\to A\to k$$ yields the following: {=tex} \begin{tikzcd} 0 \ar[r] & B \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & S \ar[u] \ar[r] & S[\varepsilon] \ar[u, "\exists", twoheadrightarrow] \ar[r] & S \ar[r] \ar[u] & 0 \\ 0 \ar[r] & S \cong I \ar[u] \ar[r] & I'= \left\langle{f'}\right\rangle \ar[u] \ar[r] & I = \left\langle{f}\right\rangle \ar[u, "\cong"] \ar[r] & S \end{tikzcd}  Thus any such $$B'$$ is the quotient of $$S[\varepsilon]$$ by an ideal, and we have $$f' = f + \varepsilon g$$. ::: {.question} When do two $$f'$$s give the same $$B'$$? ::: We have $$\varepsilon f' = \varepsilon f$$, so $$\varepsilon f \in (f')$$ and we can modify $$g$$ by any $$cf$$ where $$c\in S$$, where only the equivalence class $$g\in S/(f)$$ matters. Now consider $$\operatorname{Aut}(B \hookrightarrow B')$$, i.e. maps of the form $x &\mapsto x + \varepsilon a \\ y &\mapsto y + cb$ for $$a, b\in S$$. Under this map, $f_0' = f + \varepsilon g \mapsto & f(x + \varepsilon a, y + \varepsilon b) + \varepsilon g(x ,y) \\ \\ &\Downarrow \quad\text{implies} \\ \\ f(x, y) &= \varepsilon a {\frac{\partial }{\partial x}\,} f + \varepsilon b {\frac{\partial }{\partial y}\,} f + \varepsilon g(x ,y) ,$ so in fact only the class of $$g\in S/(f, {\partial}_{x} f, {\partial}_{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 = {\mathbb{C}}[[t_{1}, \cdots, t_{m} ]]$$ and taking the locus $$V(f + \sum t_{i} g_{i}) \subset {\mathbb{A}}_{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 {\mathbb{C}}[[t_{1}]][x, y]$$. The greater generality is needed because there are deformation functors with a hull but no universal families. ::: # Tuesday April 14th Recall that we are looking at $$(X_{0})_{/k}$$ and $$F: \operatorname{Art}_{/k} \to {\operatorname{Set}}$$ where $$A$$ is sent to $$X_{/A}$$ flat with $$i: X_{0} \hookrightarrow X$$ where $$i\otimes k$$ is an isomorphism. The second condition is equivalent to a cartesian diagram {=tex} \begin{tikzcd} X_{0} \ar[r, hook] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{$\ulcorner$}" , very near start, color=black] & X \ar[d] \\ \operatorname{Spec}k \ar[r, hook] & \operatorname{Spec}A \end{tikzcd}  We showed we always have H1 and H2, and H3 if $$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} \hookrightarrow X) \in F(A)$$, we have a surjection ${\operatorname{Aut}}_{A'}(X_{0} \hookrightarrow X') \twoheadrightarrow{\operatorname{Aut}}_{A}(X_{0} \hookrightarrow X) .$ where the RHS are automorphisms of $$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} \hookrightarrow 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)$$ we have $${\operatorname{Aut}}_{A'}(X') \twoheadrightarrow{\operatorname{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} \hookrightarrow X$$, so one gives surjectivity. Using the surjectivity on automorphisms, we get {=tex} \begin{tikzcd} X_{0}\ar[rd, hook] \ar[rr, hook] & & X\ar[ld, dotted] \\ & X & \end{tikzcd}  Deformation theory is better at answering when the following diagrams exist: {=tex} \begin{tikzcd} X \ar[r, dotted, hook, "\exists?"] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{$\ulcorner$}" , shift right=0.4em, very near start, color=black] & X' \ar[d, dotted, "\exists?"] \\ \operatorname{Spec}A \ar[r] & \operatorname{Spec}A' \end{tikzcd}  i.e., the existence of an extension of $$X$$ to $$A'$$. This is different than understanding diagrams of the following type, where we're considering isomorphism classes of the squares, and deformation theory helps understand the blue one: {=tex} \begin{tikzpicture} [ greenbox/.style={ draw=green, fill=green!3, thick, rounded corners, rectangle }, redbox/.style={ draw=red, fill=red!3, thick, rounded corners, rectangle }, ] \node[ greenbox, minimum height=0.9cm, minimum width=1.2cm ] at (-0.1, 1.3) {}; \node[ redbox, minimum height=0.9cm, minimum width=1.2cm ] at (2.35, 1.3) {}; \node[ greenbox, minimum height=2.4cm, minimum width=8.2cm ] at (0, -0.5) {}; \node[ draw=red, thick, rectangle, minimum height=0.8cm, minimum width=1.2cm ] at (-1.2, -0.6) {}; \node[ draw=blue, thick, rectangle, dotted, minimum height=0.8cm, minimum width=1.2cm ] at (1.2, -0.6) {}; \node at (0, 0) {% \begin{tikzcd} & F(A') \ar[r] & F(A) \\ X_0 \ar[r, hook] \ar[d] & X \ar[r, hook] \ar[d] & X' \ar[d] \\ \operatorname{Spec}k \ar[r] & \operatorname{Spec}A \ar[r] & \operatorname{Spec}A' \end{tikzcd} }; \end{tikzpicture}  ::: {.example title="Hypersurface Singularities"} Take $$S = k[x, y]$$ and $$B = S/(f)$$, then deformations of $$\operatorname{Spec}B$$ to ? Given $$k \to k[\varepsilon] \to k$$ we can tensor[^8] to obtain {=tex} \begin{tikzcd} {0} & {B} & {B'} & {B} & {0} \\ {0} & {S} & {S[\varepsilon]} & {S} & {0} \\ {0} & {I} & {I'} & {I} & {0} \\ && {\tiny \left\langle{f'}\right\rangle} & {\tiny \left\langle{f}\right\rangle} \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-2, to=2-2] \arrow["{\pi}", from=2-2, to=1-2] \arrow[from=3-3, to=2-3] \arrow["{\pi'}", from=2-3, to=1-3] \arrow[from=3-4, to=2-4] \arrow["{\pi}", from=2-4, to=1-4] \arrow["{\subseteq}" description, from=4-3, to=3-3, no head] \arrow["{\subseteq}" description, from=4-4, to=3-4, no head] \end{tikzcd}  > [Link to diagram.](https://q.uiver.app/?q=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) ![Diagram](figures/image_2020-04-14-12-55-58.png){width="350px"} We want to understand $$F(k[\varepsilon])$$. We know $$f' = f + \varepsilon g$$ for some $$g\in S$$. ::: {.observation} {=tex} \envlist  1. $$g\in B$$ and $$f'' = f + \varepsilon(g + cf)$$ generates the same ideal. 2. We're free to reparameterize, i.e. $$x \mapsto x + \varepsilon a$$ and $$y \mapsto y + \varepsilon 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{\hbox{-}}$$vector space. ::: ::: {.example title="?"} $$F(k[\varepsilon]) = 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{\hbox{-}}$$basis in $$S/(f, f_{x}, f_{y})$$. Then $V(f + \sum t_{i} g_{i}) \subset \operatorname{Spec}k[[t_{1}, \cdots, t_{n}]][x, y] .$ Set $$R = k[[t_{1}, \cdots, t_{n}]]$$, then this lands in $${\mathbb{A}}_{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$$, then the miniversal family is . $V(y^2 - x^3 + t + t_{2} x) \subset {\mathbb{A}}^2_{k[[t_{1}, t_{2}]]} .$ This gives all ways of smoothing the node. ::: ::: {.remark} Note that none of these are pro-representable. ::: Given $$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)$$. In other words, $${\mathcal{X}}\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 \xrightarrow{f} Y$$, we have $$L_{X/Y} \in D {\mathrm{QCoh}}(X)$$, the derived category of quasicoherent sheaves on $$X$$. This answers the extension question: ::: {.answer} For any square-zero thickening $$Y \hookrightarrow Y'$$ (a closed immersion) with ideal $$I$$ yields an $${\mathcal{O}}_{Y}{\hbox{-}}$$module. 1. An extension exists iff $$0 = \mathrm{obs} \in \operatorname{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 = \operatorname{Spec}k$$ and $$Y' = \operatorname{Spec}k[\varepsilon]$$ yields $\mathrm{obs} \in \operatorname{Ext}_{x}^2(\Omega_{X/Y}, {\mathcal{O}}_{x})= H^2(T_{X_{/k}}) .$ For $$X\hookrightarrow 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 \hookrightarrow Y$$ a regular embedding, $$L_{X_{/k}} = \Omega_{X_{/k}}$$ with $$\mathrm{obs}/\mathrm{def} = \operatorname{Ext}^{2/1}(\Omega_{x}, {\mathcal{O}})$$ and the infinitesimal automorphisms are the homs. ::: ::: {.example title="?"} For $$Y = \operatorname{Spec}k[x, y] = {\mathbb{A}}^2$$ and $$X = \operatorname{Spec}B = V(f) \subset {\mathbb{A}}^2$$ we get $0 \to I/I^2 \to \Omega_{X_{/k}} \otimes B &\to \Omega_?{X_{/k}} \to 0 \\ \\ & \Downarrow \quad \text{equals} \\ \\ 0 \to B \xrightarrow{1 \mapsto (f_{x}, f_{y})} &B^2 \to \Omega_{B_{/k}} = L_{X_{/k}} \to 0 .$ Taking $$\hom({\,\cdot\,}, B)$$ yields {=tex} \begin{tikzcd} 0 \ar[r] & \hom(\Omega, B) \ar[r] & B^2 \ar[lld, "{(f_{x}, f_{y})^t}"] \\ \operatorname{Ext}^1(\Omega, B) \ar[r] & 0 \ar[r] & 0 \ar[lld] \\ \operatorname{Ext}^2(\Omega, B) \ar[r] & 0 \ar[r] & 0 \end{tikzcd}  So , $\mathrm{obs} &= 0 \\ \mathrm{def} &= B/(f_{x}, f_{y})B \\ \operatorname{Aut}&\neq 0 .$ and ::: ::: {.remark} We have the following obstruction theories: - For abstract deformations, we have $X_{0} {}_{/k} \text{ smooth } \implies \operatorname{Aut}/\mathrm{def}/\mathrm{obs} = H^{0/1/2}(T_{X_{0}}) .$ - For embedded deformations, $$Y_{0}/k$$ smooth, $$X_{0} \hookrightarrow Y_{0}$$ regular, we have $\operatorname{Aut}/\mathrm{def}/\mathrm{obs} = 0, H^{0/1}(N_{X_{0}/Y_{0}}) .$ > As an exercise, interpret this in terms of $$L_{X_{0}/Y_{0}}$$. - For maps $$X_{0} \xrightarrow{f_{0}} Y_{0}$$, i.e. maps $X_{0} \times k[\varepsilon] \xrightarrow{f} Y_{0} \times k[\varepsilon] .$ we consider the graph $$\Gamma(f_{0}) \subset X_{0} \times 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 \hookrightarrow_{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 {\mathbb{P}}^3$$ a smooth quartic, and show that $$\mathrm{def}(X) \cong k^{20}$$ but $$\mathrm{def}_{\text{embedded}} \cong k^{19}$$. This is a quartic K3 surface for which deformations don't lift (non-algebraic, don't sit inside any $${\mathbb{P}}^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 \xrightarrow{f} Y$$ and $$Y \hookrightarrow Y'$$ a square zero thickening. When can the pullback diagram be filled in? {=tex} \begin{tikzcd} X \ar[r, dotted] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{\color{black}$\lrcorner$}" , very near start, color=black] & X' \ar[d, dotted] \\ Y \ar[r] & Y' \end{tikzcd}  - The existence is governed by $$\mathrm{obs} \in \operatorname{Ext}^2( L_{X/Y}, f^* I)$$ - The number of extensions by $$\operatorname{Ext}^1( L_{X/Y}, f^* I)$$ - The automorphisms by $$\operatorname{Ext}^0( L_{X/Y}, f^* I)$$ Suppose we're considering $$k[\varepsilon] \to k$$, where $$L_{X_{/k}} = \Omega_{X_{/k}}$$, and $$H^*(T_{X_{/k}})$$ houses the obstruction theory. For an embedded deformation $$X \hookrightarrow Y$$, we have {=tex} \begin{tikzcd} X \ar[r, dotted] & X' \ar[d, dotted] \\ Y \ar[r] & Y \times_{\operatorname{Spec}k} \operatorname{Spec}k[\varepsilon] \end{tikzcd}  then $$L_{X/Y} = I/I^2 [1] = N_{X/Y}^\vee[1]$$ and $\mathrm{obs} \in \operatorname{Ext}^2(N^\vee[1], {\mathcal{O}}) = \operatorname{Ext}^1(N^\vee, {\mathcal{O}}) = H^1(N) .$ and similarly $$\mathrm{def} = H^0(N)$$ and $$\operatorname{Aut}= 0$$. For $$X \xrightarrow{f} Y$$, we can think of this as an embedded deformation of $$\Gamma \subset X \times Y$$, in which case $$N^\vee= F^* \Omega_{Y_{/k}}$$. Then $$\mathrm{obs}, \mathrm{def} \in H^{1, 0}(f^* T_{X_{/k}})$$ respectively and $$\operatorname{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}$$ smooth projective, the picard functor $${\operatorname{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({\mathcal{O}}_{X})$$, which may be nonzero. We could also have abstract deformations given by $$H^2(T_{X})$$ Given $$A \in \operatorname{Art}_{/k}$$ and $$M$$ a finite length $$A{\hbox{-}}$$module, we can form the ring $$A \oplus M$$ where $$M$$ is square zero and $$A\curvearrowright 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')$$. ::: ::: {.proposition title="Characterization of Smoothness"} Assume $$\operatorname{ch}k =0$$ and $$F$$ is a pro-representable deformation functor, so $$F = \hom(R, \cdot)$$ where $$R$$ is a complete local $$k{\hbox{-}}$$algebra with $$\dim t_{R} < \infty$$. Then $$R$$ is smooth[^9] over $$k$$ $$\iff$$ for all $$A\in \operatorname{Art}_{/k}$$ and all $$M, M' \in A{\hbox{-}}\text{mod}$$ finite dimensional with $$M \twoheadrightarrow M'$$, we have $F(A\oplus M) \twoheadrightarrow F(A\oplus M') .$ ::: ### Proof of Proposition ::: {.observation} First observe that $$\ker(F(A\oplus M) \to F(A)) = \ker(\hom(R, A\oplus M) \to \hom(R, A))$$, note that if we have two morphisms {=tex} \begin{tikzcd} R \ar[r] & R \ar[r, shift left=0.75ex, "g \oplus g"] \ar[r, shift right=0.75ex, "f \oplus g'"'] & A \oplus M \end{tikzcd}  denoting these maps $$h, h'$$ we have 1. $$g-g' \in \operatorname{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 \operatorname{Der}_{k}(R, M)$$, then $$g + \theta: R \to A\oplus M$$. ::: We conclude that the fibers are naturally torsors for $$\operatorname{Der}_{k}(R, M)$$ if nonempty. It is in fact a canonically trivial torsor, since there is a distinguished element in each fiber. Thus to show the following, it is enough to show surjection on fibers and trivial extensions go to trivial ones, then $$\operatorname{Der}_{k}(R, M) \to \operatorname{Der}_{k}(R, M')$$ with $$0\mapsto 0$$. {=tex} \begin{tikzcd} F(A\oplus M) \ar[rr] \ar[rd] & & F(A\oplus M') \ar[ld] \\ & F(A) & \end{tikzcd}  The criterion for $$F$$ being surjective is equivalent to $\operatorname{Der}_{k}(R, M) &\twoheadrightarrow\operatorname{Der}_{k}(R, M') \\ \\ &\Downarrow \qquad \text{identified as }\\ \\ \hom_{R}(\Omega_{R_{/k}}, M) &\twoheadrightarrow\hom(\Omega_{R_{/k}'}, M') .$ ::: {.warnings} $$\Omega_{R_{/k}}$$ is complicated. An example is $\Omega_{k[[x]]/k} \otimes k((x)) = \Omega_{k((x))/k} .$ which is an infinite dimensional $$k((x))$$ vector space. ::: Here we only need to consider the completions $$\hom_{R}(\widehat{\Omega}_{R_{/k}}, M) \twoheadrightarrow\hom(\widehat{\Omega}_{R_{/k}'}, M') = k[[x]]~dx$$. ::: {.fact} In characteristic zero, $$R?k$$ is smooth iff $$\widehat{\Omega}_{R_{/k}}$$ is free. ::: Thus the surjectivity condition is equivalent to checking that $$\hom(\widehat{\Omega}_{R_{/k}}, {\,\cdot\,})$$ is right-exact on finite length modules. This happens iff $$\widehat{\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}$$ is **Calabi-Yau** iff $\omega_{x} \cong {\mathcal{O}}_{x} ,$ i.e. the canonical bundle is trivial. ::: ::: {.proposition title="?"} $$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 {\mathbb{P}}^4$$ a smooth quintic threefold. - By adjunction, this is Calabi-Yau since $\omega_{x} = \omega_{{\mathbb{P}}^4}(5) \mathrel{\Big|}_{X} = {\mathcal{O}}_{x} .$ - By Lefschetz, $H^i_\mathrm{sing} ({\mathbb{P}}^4, {\mathbb{C}}) &\xrightarrow{\cong} H^i_{\mathrm{sing}}(X, {\mathbb{C}}) && \text{except in middle dimension} \\ \\ &\Downarrow \quad \text{ implies} \\ \\ H^{3, 1} &= H^{1, 3} = 0 .$ - By Serre duality, $H^0(T_{x}) &= 0 \cong H^4(\Omega_{x} \otimes\omega_{x}) \\ \\ &\Downarrow \quad\text{implies} \\ \\ H^3(\Omega_{x}) &= H^{3, 1} = 0 .$ ::: ::: {.exercise title="?"} There are nontrivial embedded deformations that yield the same abstract deformations, write them down for the quintic threefold. ::: ::: {.claim} The abstract moduli space here is given by $$\operatorname{PGL}(5) \setminus\operatorname{Hilb}$$ where $$\operatorname{Hilb}$$ is smooth. ::: ### Proof that obstructions to deformations of Calabi-Yaus are unobstructed We need to show that for any $$M \twoheadrightarrow M'$$ that $F(A\oplus M) \twoheadrightarrow F(A\oplus M') .$ The fibers of the LHS are extensions from $$A$$ to $$A\oplus M$$, and the RHS are extensions of $$X/A$$? By dualizing, we need to show $$H^1(T_{X/A}\otimes M ) \twoheadrightarrow H^1(T_{X/A} \otimes M')$$ since the LHS is $$\operatorname{Ext}^1(\Omega_{X/A}, M)$$. We want the bottom map here to be surjective: {=tex} \begin{tikzcd} X \ar[d] & X' \ar[d] \\ \operatorname{Spec}A \ar[r, hook] & \operatorname{Spec}A \oplus M \end{tikzcd}  ::: {.fact title="Important"} For $$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}) \otimes M \twoheadrightarrow H^1(T_{X/A}) \otimes M'$$ by exactness. This uses the fact that there's a spectral sequence $\operatorname{Tor}_{q}(H^p(T_{X/A}), M) \implies H^{p+q} (T_{X/A} \otimes M)$ which follows from base change and uses the fact that $$T_{X/A}$$ is flat. ::: We'll be looking at $$\operatorname{Tor}_{1}(H^0(T_{X/A}), M)$$ which is zero by freeness. Hodge theory is now used: by Deligne-Illusie, for $$X\xrightarrow{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} = {\mathcal{O}}_{X}$$ is trivial. Using Deligne-Illusie, since $$\omega$$ is trivial on the special fiber, $$H^0(\omega_{X/A}) = A$$ is free of rank 1. We thus have a section $${\mathcal{O}}_{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} \otimes\omega_{X/A}) ^\vee= H^{n-1}(\Omega_{X/A})^\vee$$, which is free by Deligne-Illusie. This also holds for $$H^0(T_{X/A}) = H^n(\Omega_{X/A})^\vee$$ is free. ::: Thus deformations of Calabi-Yaus are unobstructed. ### Remarks ::: {.remark} In fact we need much less. Take $$A_{n} = k[t] / t^n$$, then consider {=tex} \begin{tikzcd} 0 \ar[r] & A_n \ar[r] & A_n[\varepsilon] \ar[r] & A_n \\ 0 \ar[r] \ar[u, equal] & A_n \ar[r] \ar[u, equal] & A_n \oplus \varepsilon A_n \ar[r] \ar[u, equal] & A_n \ar[u, equal] \end{tikzcd}  For a deformation $$X/A_{n}$$, let $$T^1(X/A_{n}) = \ker(F(A_{n}[\varepsilon]) \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}$$, there is a map $T^1(X/A_{n+1}) \to T^1(X\otimes A_{n} / A_{n}) .$ and surjectivity is equivalent to the lifting condition. In the CY situation, the extension group $$T^1(X/A_{n+1}) = H^1(T_{X/A_{n+1}})$$ and the RHS is $$H^1(T_{X\otimes 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}}) \otimes 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$$. We'll prove that there's a correspondence $\left\{{\substack{ \ell_{/k}, \text{ twisted forms} \\ Y \text{ of } X_{/k} }}\right\} &\rightleftharpoons H^1(\ell_{/k}, \operatorname{Aut}(X_{/\ell})) .$ Recall that $$\operatorname{PGL}(n ,\ell) \coloneqq\operatorname{GL}(n ,\ell) / \ell^{\times}$$. ::: {.example title="?"} Let $$X = {\mathbb{P}}^{n-1}/k$$, then $$H^1(\ell_{/k}, \operatorname{PGL}(n, \ell)$$ parameterizes twisted forms of $${\mathbb{P}}^{n-1}$$, e.g. for $$n=2$$ twisted forms of $${\mathbb{P}}^1$$ and plane curves. ::: ::: {.example title="?"} Take $$X = M_{n}(k)$$ the algebra of $$n\times n$$ matrices. Then by a theorem (Skolern-Noether) $$\operatorname{Aut}(M_{n}(k)) = \operatorname{PGL}(n, k)$$. Thus $$H^1(\ell_{/k}, \operatorname{PGL}(n, k))$$ also parameterizes twisted forms of $$M_{n}(k)$$ in the category of unital (not necessarily commutative) $$k{\hbox{-}}$$algebras. These are exactly central simple algebras $$A_{/k}$$ where $$\dim_{k} A = n^2$$ with center $$Z(A) = k$$ with no nontrivial two-sided ideals. By taking $$\ell = k^{s}$$, we get a correspondence $\left\{{\substack{\text{CSAs} A_{/k} \text{ of degree } n}}\right\} &\rightleftharpoons \left\{{\substack{\text{ Severi-Brauer varieties of dimension n-1} }}\right\} .$ Taking $$n=2$$ we obtain $\left\{{\substack{\text{Quaternion algebras } A_{/k}}}\right\} &\rightleftharpoons \left\{{\substack{\text{Genus 0 curves } \ell_{/k}}}\right\} .$ ::: ## The Weil Descent Criterion Fix $$\ell_{/k}$$ finite Galois with $$g \coloneqq\operatorname{Aut}(\ell_{/k})$$. 1. $$X_{/k} \to X_{/\ell}$$ with a $$g{\hbox{-}}$$action. 2. What additional data on an $$\ell{\hbox{-}}$$variety $$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{\hbox{-}}$$algebra via $$\sigma: \ell \to \ell^\sigma$$. If $$X_{/\ell}$$ is a variety, so is $$X^\sigma_{/\ell}$$? {=tex} \begin{tikzcd} X^\sigma\ar[dr, dotted] \ar[r]\ar[d] & X \ar[d] \\ \operatorname{Spec}\ell^\sigma \ar[r, "f"] & \operatorname{Spec}\ell \end{tikzcd}  where $$f$$ is the map induced on $$\operatorname{Spec}$$ by $$\sigma$$. We can also think of these on defining equations: $X &= \operatorname{Spec}\ell[t_{1}, \cdots, t_{n}] / \left\langle{p_{1}, \cdots, p^n}\right\rangle \\ X^\sigma &= \operatorname{Spec}\ell[t_{1}, \cdots, t_{n}] / \left\langle{\sigma_{p_{1}}, \cdots, \sigma{p^n}}\right\rangle \\ .$ For $$X_{/k}, X_{/\ell}$$, we canonically identify $$X$$ with $$X^\sigma$$ by the map $$f_\sigma: X \xrightarrow{\cong} X^\sigma$$, a canonical isomorphism of $$\ell{\hbox{-}}$$varieties. We thus have {=tex} \begin{tikzcd} X \ar[r, "f_\sigma"] \ar[rr, bend left, "f_{\sigma \tau}"] & X^\sigma \ar[r, "f_\sigma"] & X^{\sigma \tau} \end{tikzcd}  under a "cocycle condition" $$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\xrightarrow{\cong} Y^\sigma$$ satisfying the above cocycle condition, and there exists a unique $$X_{/k}$$ such that $$X_{/\ell} \xrightarrow{\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}, \operatorname{Aut}X)$$. Conversely, we have the following: ::: {.definition title="Twisted Descent Data"} Let $$a_{0}$$ be such a cocycle and $$\left\{{s_\sigma: X\to X^\sigma}\right\}$$ be descent datum attached to $$X$$. Define twisted descent datum $$g_\sigma \coloneqq f_\sigma \circ a_\sigma$$ from $X /\ell\xrightarrow{a_\sigma} X_{/\ell} \xrightarrow{f_\sigma} X^\sigma / \ell .$ ::: ::: {.exercise title="?"} Check that $$g_\sigma$$ satisfies the cocycle condition, so by Weil uniquely determines a ($$k{\hbox{-}}$$model) $$Y_{/k}$$ of $$X_{/\ell}$$. ::: ::: {.example title="?"} Let $$G_{/k}$$ be a smooth algebraic group and $$X_{/k}$$ a torsor under $$G$$. Then $${\operatorname{Aut}}(G) \supset \operatorname{Aut}_{G{\hbox{-}}\text{torsor}} (G) = G$$, since in general the translations will only be a subgroup of the full group of automorphisms. Then $H^1(\ell_{/k}, G) \to H^1(\ell_{/k}, \operatorname{Aut}G)$ defines a twisted form $$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}$$ be a smooth connected commutative algebraic group where $$\operatorname{ch}k$$ does not divide $$n$$, so the map $$[n]: G \to G$$ is an isogeny. Then $0 \to G[n] (k^{s} ) \to G(k^{s} ) \xrightarrow{[n]} G(k^{s} ) \to 0$ is a SES of $$g = \operatorname{Aut}(k^{s}_{/k}){\hbox{-}}$$modules. ::: {.claim} Taking the associated cohomology sequence yields the Kummer sequence: $0 \to G(k) / nG(k) \to H^1(k, G[n]) \to H^1(k, G)[n] \to 0$ where the RHS is the **Weil--Châtelet** group and the LHS is the **Mordell-Weil** group. ::: For $$g$$ a profinite group, a commutative discrete $$g{\hbox{-}}$$group is by definition a $$g{\hbox{-}}$$module. These form an abelian category with enough injectives, so we can take right-derived functors of left-exact functors. We will consider the functor $A \mapsto A^g \coloneqq\left\{{x\in A {~\mathrel{\Big|}~}\sigma x = x ~\forall \sigma\in g}\right\} ,$ then define $$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) = {\operatorname{Map}}(g^n, A)$$ with $d: C^n(g, A) &\to C^{n+1}(g, A) \\ (df)(\sigma_{1}, \cdots, \sigma_{n+1} &\coloneqq \sigma_{1} f(\sigma_{2}, \cdots, \sigma_{n+1}) \\ &\qquad + \sum_{i=1}^n (-1) f(\sigma _1, \cdots, \sigma_{i-1}, \sigma_{i}, \sigma_{i+1}, \cdots, \sigma_{n+1}) \\ &\qquad + (-1)^{n+1} f(\sigma_{1}, \cdots, \sigma_{n}) .$ Then $$d^2 = 0$$, $$H^n$$ is kernels mod images, and this agrees with $$H^1$$ as defined before with $$H^0 = A^g$$. We'll see that that $H^i(g, A) = \varinjlim_{U} G^i(g/U, A^U) .$ If $$g$$ is finite, $$A$$ is a $$g{\hbox{-}}$$module $$\iff$$ $$A$$ is a $${\mathbb{Z}}[g]{\hbox{-}}$$module, and thus $A^g = \hom_{{\mathbb{Z}}[g]{\hbox{-}}\text{mod}}({\mathbb{Z}}, A) .$ where $${\mathbb{Z}}$$ is equipped with a trivial $$g{\hbox{-}}$$action. We can thus think of $H^i(g, A) = \operatorname{Ext}^i_{{\mathbb{Z}}[g]}({\mathbb{Z}}, A) .$ > The end! [^1]: $$F\leq G$$ is a subfunctor iff $$F(s) \hookrightarrow G(s)$$. [^2]: See tilde construction in Hartshorne, essentially amounts to localizing free tings. [^3]: Think of this as a graded module, this tells you the lowest number of small grade pieces needed to determine the entire thing. [^4]: Note: $$h^1 = \dim H^1$$. [^5]: Not a unique map, but still a pullback [^6]: Recall that right-exactness was a transitive action. [^7]: Recall that by Nakayama, a nonzero module tensor $$k$$ can not be zero. [^8]: For flat maps, tensoring up to an isomorphism implies isomorphism. [^9]: I.e. $$R \cong k[[t_{R}^\vee]]$$.