# Thursday March 5th :::{.remark} Which functors have a good deformation theory? - $\pic$ - $\Hilb$ - Maps $X \to Y, X \to Y_A, X_A \to Y_A$ - Sheaves $E_0$ on $X_0$, $E$ on $X_A$, and $E$ on $X$. Note the having an obstruction theory is equivalent to left-exactness in $$ \text{Def}\selfmap F(A') \to F(A) \to \text{Obs} $$ for $F$ and $A' \to A$ A small thickening. ::: :::{.theorem title="Schlessinger's Criterion"} \envlist - $F$ has a **pro-representable hull** iff $F$ has an obstruction theory. - $F$ is **pro-representable** iff $F$ has a *left-exact* deformation theory. ::: ## Abstract Deformation Theory :::{.remark} There is a naïve deformation functor: given ${X_0}_{/k}$ a finite-type scheme, take $F_{\text{naïve}}(A)$ to be the set of $X_{/A}$, flat over $A$ such that $X\tensor k \cong x_0$. Isomorphisms are given by diagrams over $\spec A$: \begin{tikzcd} X && {X'} \\ \\ & {\spec A} \arrow[from=1-1, to=3-2] \arrow[from=1-3, to=3-2] \arrow[from=1-1, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJYIl0sWzIsMCwiWCciXSxbMSwyLCJcXHNwZWMgQSJdLFswLDJdLFsxLDJdLFswLDFdXQ==) Note that this isn't quite a set, and is instead a point with non-trivial automorphisms, i.e. a connected groupoid. To fix this, take $F(A)$ to be $X_{/A}$ flat over $A$, and take maps $X_0 \mapsvia{i} X$ such that $i\tensor k$ is an isomorphism. Here isomorphisms are given by diagrams \begin{tikzcd} X_0 \arrow[d, "i"] \arrow[rr, "="] & & X_0\arrow[d, "i"] \\ X \arrow[rd] \arrow[rr] & & X' \arrow[dl] \\ & \spec A & \end{tikzcd} ::: :::{.remark} Upshot: If ${X_0}_{/k}$ is smooth, then this $F$ has an obstruction theory given by \[ \text{Def} &= H^1(x_0, T_{x_0}) \\ \text{Obs} &= H^2(x_0, T_{x_0}) .\] ::: :::{.remark} Recall that for $X/k$, we can always define the sheaf of Kähler differentials $\Omega_{x_0/k}$, and $T_{x_0} = \hom_{x_0}(\Omega_{x_o/k}, \OO_{x_0})$. In general, if $x_0/k$ is *not* smooth, then $F$ (if it's a locally complete intersection?) has an obstruction theory given by \[ \text{Def} &= \ext_{x_0}^1(\Omega^1_{x_0}, \OO_{x_0}) \\ \text{Obs} &= \ext_{x_0}^2(\Omega^1_{x_0}, \OO_{x_0}) .\] There is a local-to-global spectral sequence $$ H^i \ext_{x_0}^j(\Omega_{x_0}^1, \OO_{x_0}) \abuts \Ext^{i+j}(\Omega^1_{x_0}, \OO_{x_0}) $$ In complete generality, there's a good answer using the cotangent complex. In general $H^1(x_0, T_{x_0})$ and $H^2(x_0, T_{x_0})$ are Def and Obs of *locally trivial* deformations. In this case $F(A)$ is given by $X_{/A}$ flat etc such that there exists an open cover $U_i$ of $X$ for which $U_i \cong U_i^{\text{red}} \tensor A$. ::: :::{.proposition title="Infinitesimal lifting property"} Let $A/k$ be a finite type algebra and $R/k$ finite type and smooth (so $\spec R$ is smooth over $\spec A$). Suppose we have a nilpotent thickening $0 \to J \to B' \to B$ of $k\dash$algebras with $J^n = 0$ for some $n$. Then ??? \begin{tikzcd} &&&& {\spec R} \\ \\ {\spec B} && {\spec B'} && {\spec A} \arrow[from=3-3, to=3-5] \arrow[dashed, from=3-3, to=1-5] \arrow[from=1-5, to=3-5] \arrow[from=3-1, to=3-3] \arrow[from=3-1, to=1-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbNCwwLCJcXHNwZWMgUiJdLFs0LDIsIlxcc3BlYyBBIl0sWzIsMiwiXFxzcGVjIEInIl0sWzAsMiwiXFxzcGVjIEIiXSxbMiwxXSxbMiwwLCIiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMCwxXSxbMywyXSxbMywwXV0=) ::: :::{.remark} This lifting property is called *formal smoothness*, and is equivalent to smoothness for finite-type maps. ::: ## Proof ### Step 1 Given $f: \spec(B) \to \spec(R)$, for simplicity call induced map $f: R\to B$ given by pullback $f$ as well. Any nilpotent thickening is a sequence of square zero thickenings, so we can assume $J^2 = 0$. - Claim 1: Any two lifts $g-g' \in \text{Der}_A(R, J)$, so there is a torsor action - Claim 2: $\OO\in \Der_A(R, J) \implies g + \theta$ is lift. Any two lifts $g, g'$ of $A\dash$algebras yield a diagram \begin{tikzcd} J && {B'} && B \\ \\ &&&& R \arrow["f"', from=3-5, to=1-5] \arrow["g", from=3-5, to=1-3] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJKIl0sWzIsMCwiQiciXSxbNCwwLCJCIl0sWzQsMiwiUiJdLFszLDIsImYiLDJdLFszLDEsImciXSxbMCwxXSxbMSwyXV0=) Recall that $\delta \in \text{Der}_A (R, M)$ with $\delta(ab) = a\delta(b) + b\delta(a)$ with $\delta(A\cdot 1) = 0$. We have $$ (g-g')(ab) &= g(ab) - g'(ab) \\ &= g(a) g(b) - g'(a) g'(b) \\ &= g(a) g(b) - g(a) g'(b) + g(a) g'(b) - g'(a) g'(b) \\ &= g(a) (g(b) - g'(b)) + g'(b) (g(a) - g'(a)) \\ &= g(a)(J) + g'(b)(J) \\ &= a(g-g')(b) + b(g-g')(a) ,$$ so $\delta \definedas g-g'$ is a derivation. That $\delta(A\cdot 1) = 0$ follows from $\delta(1) = 0$ and $\delta(a\cdot 1) = a\cdot \delta(1)$. For claim 2, we just need to show that $g + \theta$ is an $A\dash$algebra map. $$ (g+\theta)(ab) &= g(ab) + \theta(ab) \\ &= g(a)g(b) + a\theta(b) + b\theta(a) \\ &= g(a) g(b) + g(a)\theta(b) + g(b)\theta(a) (+ \theta(a)\theta(b)) \quad\text{since } J^2 =0 \\ &= ((g+\theta)(a)) ((g+\theta)(b)) .$$ So this gives us step 1. ### Step 2 Write a presentation for $R$ as $R = P / I$ where $P = A[x_1, \cdots, x_n]$. We get some map $h$ in the following diagram where $h(I) \to J$ \begin{tikzcd} J \arrow[r] & B' \arrow[r] & B \\ I \arrow[r]\arrow[u] & P \arrow[r]\arrow[u, "h"] & R\arrow[u, "f"] \end{tikzcd} Note that $\bar h: I/I^2 \to J$ is a map of $A\dash$modules. ### Step 3 By Hartshorne, any smooth thing regularly embeds into a smooth thing, yielding a SES $$ 0 \to I/I^2 \to \Omega_{P/A}\tensor R \to \Omega_{R/A} \to 0 $$ So take $\hom_{\rmod}(\wait, J)$, which potentially yields an $\ext_R^1(\Omega_{R/A}, J)$. Again by the local-to-global spectral sequence, this is zero because the $H^1$ is zero by local-freeness and another $H^*$ is zero by smoothness. We can then take $\bar h \in \Hom_{\rmod}(I/I^2, J)$, and by surjectivity we obtain a \[ \theta \in \Hom(\Omega_{P/A} \tensor R, J) = \Hom_P(\Omega_{P/A}, J) = \Der_A(P, J) .\] ### Step 4 $h - \theta$ is an algebra morphism that kills $I$ and thus yields the desired lift. Since $J^2 = 0$, $\restrictionof{h-\theta}{I}$ factors through $I/I^2$, and by construction is zero. :::{.definition title="Formally smooth"} Given functor $F, G: \Art_{/k} \to \Set$, we say $G\to F$ is **formally smooth** if it satisfies the infinitesimal lifting property for $A' \to A$ a thickening in $\Art_{/k}$. See diagram. ::: :::{.remark} By Yoneda, we have $\hom_{\text{Fun}}(h_A, F) = F(A)$, i.e. $G(A') \surjects F(A') \cross_{F(A)} G(A)$. ::: :::{.definition title="Pro-representability, pro-representable hull"} For $F: \Art_{/k} \to \Set$ a deformation functor, then - $F$ is **pro-representable** if there exists a complete $k\dash$algebra $R$ such that the functor of points $h_R = F$, i.e. $h_R = \hat F$. - A map $h_R \mapsvia{\xi} F$ is a **pro-representable hull** if $R$ is a complete $k\dash$algebra and $\xi$ is a formally smooth map. ::: :::{.remark} Upgraded Yoneda: If $R$ is a complete $k\dash$algebra, then \[ \Hom_{\Fun(\Art_{/k}, \Set)} (h_R, F) = \hat F(R) = \inverselim_{n} F(R/\mfm^n) ,\] where $\hat F(R)$ is interpreted as *formal families* over $R$. Thus $h_R \mapsvia{\xi} F$ is the same data as a section $\xi \in \hat F(R)$. ::: :::{.definition title="Versal, miniversal, universal"} A pair $(R, xi)$ with $\xi \in \hat F(R)$ is - **Versal** for $F$ $h_r\mapsvia{\xi} F$ is formally smooth. - I.e. any section of $F$ is obtained by pulling this one back. For $\eta \in F(A)$, this means $\spec(A) \mapsvia{f} \spec R/\mfm^n$ for $n\gg 0$, in which case $\eta = f^* \xi$. - **Miniversal** or **semi-universal** if it is versal, formally smooth, and is an isomorphism on tangent spaces. - The map need not be unique. - $(R, \xi)$ **pro-represents** (or is **universal** for) $F$ if $h_R \mapsvia{\cong} f$. - I.e. any section of $F$ is obtained by a *unique* pullback. ::: Back to abstract deformations. :::{.lemma title="?"} For ${X_0}_{/k}$ and $X/k[\eps]$ flat with $X\tensor k \cong X_0$, the following sequence is exact: $$ 0\to \Der_k(\OO_{X_0}, \OO_{X_0}) \to \Aut_{k[\eps]} (X) \to \Aut_{k}(X_0) $$ ::: :::{.proof title="?"} As in the previous argument, take $f-1 \in \Der_k(\OO_{X_0}, \OO_{X_0})$, then $1 + \theta$ is an automorphism for any such derivation $\theta$. :::