# Tuesday April 14th Recall that we are looking at $(X_{0})_{/k}$ and $F: \Art_{/k} \to \sets$ where $A$ is sent to $X_{/A}$ flat with $i: X_{0} \injects X$ where $i\tensor k$ is an isomorphism. The second condition is equivalent to a cartesian diagram \begin{tikzcd} X_{0} \ar[r, hook] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{$\ulcorner$}" , very near start, color=black] & X \ar[d] \\ \spec k \ar[r, hook] & \spec A \end{tikzcd} We showed we always have H1 and H2, and H3 if $X_{0}/k$ is projective or $X_{0}$ is affine with isolated singularities. In this situation we have a miniversal family. This occurs iff for $A' \to A$ a small thickening and $(X_{0} \injects X) \in F(A)$, we have a surjection \[ \Aut_{A'}(X_{0} \injects X') \surjects \Aut_{A}(X_{0} \injects X) .\] where the RHS are automorphisms of $X_{/A}$, i.e. those which commute with the identity on $A$ and $X_{0}$. We had a naive functor $F_{n}$ where we don't include the inclusion $X_{0} \injects X$. When $F$ has a hull then the naive functor has a versal family, since there is a forgetful map that is formally smooth. If it's the case that for all $A' \to A$ small and $F_{\text{n}} \to F_{n}(A)$ we have $\Aut_{A'}(X') \surjects \Aut_{A} (X)$, then $F = F_{n}$ and both are pro-representable. The forgetful map is smooth because given $X_{/A}$ in $F_{n}(A)$, we have some inclusion $X_{0} \injects X$, so one gives surjectivity. Using the surjectivity on automorphisms, we get \begin{tikzcd} X_{0}\ar[rd, hook] \ar[rr, hook] & & X\ar[ld, dotted] \\ & X & \end{tikzcd} Deformation theory is better at answering when the following diagrams exist: \begin{tikzcd} X \ar[r, dotted, hook, "\exists?"] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{$\ulcorner$}" , shift right=0.4em, very near start, color=black] & X' \ar[d, dotted, "\exists?"] \\ \spec A \ar[r] & \spec A' \end{tikzcd} i.e., the existence of an extension of $X$ to $A'$. This is different than understanding diagrams of the following type, where we're considering isomorphism classes of the squares, and deformation theory helps understand the blue one: \begin{tikzpicture} [ greenbox/.style={ draw=green, fill=green!3, thick, rounded corners, rectangle }, redbox/.style={ draw=red, fill=red!3, thick, rounded corners, rectangle }, ] \node[ greenbox, minimum height=0.9cm, minimum width=1.2cm ] at (-0.1, 1.3) {}; \node[ redbox, minimum height=0.9cm, minimum width=1.2cm ] at (2.35, 1.3) {}; \node[ greenbox, minimum height=2.4cm, minimum width=8.2cm ] at (0, -0.5) {}; \node[ draw=red, thick, rectangle, minimum height=0.8cm, minimum width=1.2cm ] at (-1.2, -0.6) {}; \node[ draw=blue, thick, rectangle, dotted, minimum height=0.8cm, minimum width=1.2cm ] at (1.2, -0.6) {}; \node at (0, 0) {% \begin{tikzcd} & F(A') \ar[r] & F(A) \\ X_0 \ar[r, hook] \ar[d] & X \ar[r, hook] \ar[d] & X' \ar[d] \\ \spec k \ar[r] & \spec A \ar[r] & \spec A' \end{tikzcd} }; \end{tikzpicture} :::{.example title="Hypersurface Singularities"} Take $S = k[x, y]$ and $B = S/(f)$, then deformations of $\spec B$ to ? Given $k \to k[\eps] \to k$ we can tensor[^tensor_up_to_iso] to obtain \begin{tikzcd} {0} & {B} & {B'} & {B} & {0} \\ {0} & {S} & {S[\eps]} & {S} & {0} \\ {0} & {I} & {I'} & {I} & {0} \\ && {\tiny \gens{f'}} & {\tiny \gens{f}} \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-2, to=2-2] \arrow["{\pi}", from=2-2, to=1-2] \arrow[from=3-3, to=2-3] \arrow["{\pi'}", from=2-3, to=1-3] \arrow[from=3-4, to=2-4] \arrow["{\pi}", from=2-4, to=1-4] \arrow["{\subseteq}" description, from=4-3, to=3-3, no head] \arrow["{\subseteq}" description, from=4-4, to=3-4, no head] \end{tikzcd} > [Link to diagram.](https://q.uiver.app/?q=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) ![Diagram](figures/image_2020-04-14-12-55-58.png){width=350px} We want to understand $F(k[\eps])$. We know $f' = f + \eps g$ for some $g\in S$. :::{.observation} \envlist 1. $g\in B$ and $f'' = f + \eps(g + cf)$ generates the same ideal. 2. We're free to reparameterize, i.e. $x \mapsto x + \eps a$ and $y \mapsto y + \eps b$ and thus \[ g \mapsto g + a f_{x} + b f_{y} \] , i.e. the partial derivatives. ::: Thus isomorphism classes of $B'$ in deformations $B' \to B$ only depend on the isomorphism classes $g\in B/(f_{x}, f_{y}) B$. When the singularities are isolated, this quotient is finite-dimensional as a $k\dash$vector space. [^tensor_up_to_iso]: For flat maps, tensoring up to an isomorphism implies isomorphism. ::: :::{.example title="?"} $F(k[\eps]) = B/(f_{x}, f_{y})B$. Thus H3 holds and there is a miniversal family $h_{R} \to F$. We can describe it explicitly: take $g_{i} \in S$, yielding a $k\dash$basis in $S/(f, f_{x}, f_{y})$. Then \[ V(f + \sum t_{i} g_{i}) \subset \spec k[[t_{1}, \cdots, t_{n}]][x, y] .\] Set $R = k[[t_{1}, \cdots, t_{n}]]$, then this lands in $\AA_{R}^2$. ::: :::{.example title="?"} The nodal curve $y^2 = x^3$, take . \[ S/(y^2-x^3, 2y, -3x^2) = S/(y, x^2) .\] So take $g_{1} = 1, g_{2} = x$, then the miniversal family is . \[ V(y^2 - x^3 + t + t_{2} x) \subset \AA^2_{k[[t_{1}, t_{2}]]} .\] This gives all ways of smoothing the node. ::: :::{.remark} Note that none of these are pro-representable. ::: Given $X$ and $A$, we obtain a miniversal family over the formal spectrum $\mathrm{Spf}(R) = (R, \xi)$ and a unique map: ![Diagram](figures/image_2020-04-14-13-10-21.png){width=350px} We can take two deformations over $A = k[\xi]/ S^n$: - $X_{1} = V(x + y)$?? - $X_{2} = V(x + uy)$?? As deformations over $A$, $X_{1} \cong X_{2}$ where we send , \[ s&\mapsto s, \\ y&\mapsto y, \\ x&\mapsto ux .\] since \[ (xy + us) = (uxy + us) = (u(xy + s)) = (xy + s) .\] But we have two different classifying maps, which do commute up to an automorphism of $A$, but are not equal. Since they pullback to different elements (?), $F$ can not be pro-representable. ![Diagram](figures/image_2020-04-14-13-20-05.png){width=350px} So reparameterization in $A$ yield different objects in $F(A)$. In other words, $\mcx \to \mathrm{Spf}(R)$ has automorphisms inducing reparameterizations of $R$. This indicates why we need maps restricting to the identity. ## The Cotangent Complex For $X \mapsvia{f} Y$, we have $L_{X/Y} \in D \qcoh(X)$, the derived category of quasicoherent sheaves on $X$. This answers the extension question: :::{.answer} For any square-zero thickening $Y \injects Y'$ (a closed immersion) with ideal $I$ yields an $\OO_{Y}\dash$module. 1. An extension exists iff $0 = \mathrm{obs} \in \ext^2(L_{X/Y}, f^* I)$ 2. If so, the set of ways to do so is a torsor over this ext group. 3. The automorphisms of the completion are given by $\hom(L_{X/Y}, f^* I)$. ::: :::{.remark} Some special cases: $X \to Y$ smooth yields $L_{X/Y} = \Omega_{X/Y}[0]$ concentrated in degree zero. ::: :::{.example title="?"} $Y = \spec k$ and $Y' = \spec k[\eps]$ yields \[ \mathrm{obs} \in \Ext_{x}^2(\Omega_{X/Y}, \OO_{x})= H^2(T_{X_{/k}}) .\] For $X\injects Y$ is a regular embedding (closed immersion and locally a regular sequence) $L_{X/Y} = \qty{I/I^2}[1]$, the conormal bundle. ![Diagram](figures/image_2020-04-14-13-32-13.png){width=350px} ::: :::{.example title="?"} For $Y$ smooth, $X \injects Y$ a regular embedding, $L_{X_{/k}} = \Omega_{X_{/k}}$ with $\mathrm{obs}/\mathrm{def} = \Ext^{2/1}(\Omega_{x}, \OO)$ and the infinitesimal automorphisms are the homs. ::: :::{.example title="?"} For $Y = \spec k[x, y] = \AA^2$ and $X = \spec B = V(f) \subset \AA^2$ we get \[ 0 \to I/I^2 \to \Omega_{X_{/k}} \tensor B &\to \Omega_?{X_{/k}} \to 0 \\ \\ & \Downarrow \quad \text{equals} \\ \\ 0 \to B \mapsvia{1 \mapsto (f_{x}, f_{y})} &B^2 \to \Omega_{B_{/k}} = L_{X_{/k}} \to 0 .\] Taking $\hom(\wait, B)$ yields \begin{tikzcd} 0 \ar[r] & \hom(\Omega, B) \ar[r] & B^2 \ar[lld, "{(f_{x}, f_{y})^t}"] \\ \Ext^1(\Omega, B) \ar[r] & 0 \ar[r] & 0 \ar[lld] \\ \Ext^2(\Omega, B) \ar[r] & 0 \ar[r] & 0 \end{tikzcd} So , \[ \mathrm{obs} &= 0 \\ \mathrm{def} &= B/(f_{x}, f_{y})B \\ \aut &\neq 0 .\] and ::: :::{.remark} We have the following obstruction theories: - For abstract deformations, we have \[ X_{0} {}_{/k} \text{ smooth } \implies \aut/\mathrm{def}/\mathrm{obs} = H^{0/1/2}(T_{X_{0}}) .\] - For embedded deformations, $Y_{0}/k$ smooth, $X_{0} \injects Y_{0}$ regular, we have \[ \aut/\mathrm{def}/\mathrm{obs} = 0, H^{0/1}(N_{X_{0}/Y_{0}}) .\] > As an exercise, interpret this in terms of $L_{X_{0}/Y_{0}}$. - For maps $X_{0} \mapsvia{f_{0}} Y_{0}$, i.e. maps \[ X_{0} \cross k[\eps] \mapsvia{f} Y_{0} \cross k[\eps] .\] we consider the graph $\Gamma(f_{0}) \subset X_{0} \cross Y_{0}$. ![Diagram](figures/image_2020-04-14-13-43-40.png){width=350px} Since all of these structures are special cases of the cotangent complex, they place nicely together in the following sense: Given $X \injects_{i} Y$ we have \[ 0 \to T_{X} \to i^* T_{Y} \to N_{X/Y} \to 0 .\] Yielding a LES \[ 0 &\to H^0(T_{X}) \to H^0(i^* T_{Y}) \to H^0(N_{X/Y}) \\ &\to H^1(T_{X}) \to H^1(i^* T_{Y}) \to H^1(N_{X/Y}) \\ &\to H^2(T_{X}) .\] ![Diagram](figures/image_2020-04-14-13-47-05.png){width=350px} ::: :::{.exercise title="?"} Consider $X \subset \PP^3$ a smooth quartic, and show that $\mathrm{def}(X) \cong k^{20}$ but $\mathrm{def}_{\text{embedded}} \cong k^{19}$. This is a quartic K3 surface for which deformations don't lift (non-algebraic, don't sit inside any $\PP^n$). ::: > Next time: > Obstruction theory of sheaves, T1 lifting as a way to show unobstructedness.