# Hilbert Schemes of Hypersurfaces (Tuesday January 28th) Last time: Twisted cubics, given by $\hilb_{\PP^3_k}^{3t+1}$. \begin{tikzpicture} \node (node_one) at (0,0) { \includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_21-29} }; \node (a) at (0, -4) {$?$}; \node (a) at (-4.5, 3.9) {$A$}; \filldraw[blue](-4.45, 3.4) circle (0.1); \node (a) at (2.44, 2.4) {$B$}; \filldraw[blue](2.04, 2.2) circle (0.1); \node (a) at (-1.8, 1.85) {$C$}; \filldraw[red](-1.38, 1.85) circle (0.1); \node (a) at (-4., 7) {$12$}; \node (a) at (3., 5.5) {$15$}; \end{tikzpicture} > Components of the Scheme of Cubic Curves. We got lower (?) bounds on the dimension by constructing families, but want an exact dimension. The following will be a key fact: :::{.proposition title="?"} Let $Z\subset X$ be a closed $k\dash$dimensional subspace. For $[z] \in \hilb_{X_{_{/k}}}^P(k)$, we have an identification of the Zariski tangent space \[ T_{[z]} \hilb_{X_{_{/k}} }^P = \hom_{\OO_X}(I_z, \OO_Z) \] ::: Say \[ F: (\sch_{_{_{/k}}})\op \to \Set \] is a functor and let $x\in F(k)$. There is an inclusion $i: \spec k \injects \spec k[\eps]$ and an induced map \[ F(\spec k [\eps]) &\mapsvia{i^*} F(\spec k) \\ T_x F \definedas (i^*)\inv(x) &\mapsto x \] So if $F$ is represented by a scheme $H_{/k}$, then \[ T_x h_J = T_x H = (\mfm_x / \mfm_x^2)\dual \,\,\text{over } k \] Will need a criterion for flatness later, esp. for Artinian thickenings. :::{.lemma title="?"} Assume $A'$ is a Noetherian ring and $0 \to J \to A' \to A \to 0$ with $J^2 = 0$. Assume we have $X'_{/ \spec A'}$, and a coherent sheaf $F'$ on $X'$, where $X'$ is Noetherian. Then $F'$ is flat over $A'$ iff 1. $F$ is flat 2. $0 \to F\tensor_A J \to F'$ is exact. \begin{tikzcd} F & F' \\ X \da \spec A' \cross_{\spec A} X \ar[r] \ar[d] & X' \ar[d] \\ \spec A \arrow[ur, phantom, "\scalebox{1.5}{$\ulcorner$}" , very near start, color=black] \ar[r] & \spec A' \end{tikzcd} ::: ### Sketch Proof of Lemma Take the first exact sequence and tensor with $F'$ (which is right-exact), then $J \tensor_{A'} F' = J \tensor_A$ canonically. This follows because $J = J \tensor_{A'} A$, and there is an isomorphism $J \tensor_{A'} A' \to J \tensor_{A'} A$. And $F = F' \tensor_{A'} A$ is a pullback of $F'$. If flat, then tensoring is exact. Note that both conditions in the lemma are necessary since pullbacks of flats are flat by (1), and (2) gives the flatness condition. :::{.definition title="Flat Modules"} Recall that for a module over a Noetherian ring, $M/A$, $M$ is **flat** over $A$ iff \[ \tor_1^A(M, A/p) = 0 && \text{ for all primes } p .\] ::: :::{.remark} Reason: Tor commutes with direct limits, so $M$ is flat iff \[ \Tor_1^A(M, N) = 0 && \text{for all finitely generated } N .\] ::: Since $A$ is Noetherian, $N$ has a finite filtration $N^\cdot$ where $N_i / N_{i+1} \cong A/p_i$. Use the fact that every ideal is contained in a prime ideal. Take $x\in N$, this yields a map $A\to N$ which factors through $A/I$. If we make such a filtration on $A/I$, then we can quotient $N$ by $\im f$ where $f: A/I \to N$. Continuing inductively, the resulting filtration must stabilize. So we can assume $N = A/I$. Then $I$ is contained in a maximal. :::{.exercise title="?"} Finish proof. See Aatiyah Macdonald. ::: ### Proof of Proposition :::{.proof title="of proposition, given lemma"} So it's enough to show that $\tor_1^{A'}(F', A'/p') = 0$ for all primes $p' \subset A'$. :::{.observation} Since $J$ is nilpotent, $J \subset p'$. ::: ## Consequences of Proof Let $p = p'/J$, this is a prime ideal. We have an exact diagram by taking quotients: \begin{tikzcd} & & & & 0 \arrow[dd] & & 0 \arrow[dd] & & \\ & & & & & & & & \\ 0 \arrow[rr] & & J \arrow[rr] & & p' \arrow[rr] \arrow[dd] & & p \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ 0 \arrow[rr] & & J \arrow[rr] & & A' \arrow[rr] \arrow[dd] & & A \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ & & & & A'/p' \arrow[dd] & & A/p \arrow[dd] & & \\ & & & & & & & & \\ & & & & 0 & & 0 & & \end{tikzcd} So we can tensor with $F'$ everywhere, and get a map from kernels to cokernels using the snake lemma: \begin{tikzcd} & & & & 0 \arrow[dd] & & {\tor(A, F) = 0} \arrow[dd] & & \\ & & & & & & & & \\ & & 0 \arrow[rr, "\text{snake}"] \arrow[dd] & & {\tor_1^{A_1}(A'/p', F')} \arrow[dd] & & {\tor_1^{A_1}(A/p, F')} \arrow[dd] & & \\ & & & & & & & & \\ 0 \arrow[rr] & & J \tensor_{A'} F' \arrow[rr, "\text{by commuting square}", hook] & & p' \tensor_{A'} F' \arrow[rr] \arrow[dd] & & p \tensor_{A'} F' \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ 0 \arrow[rr] & & J \tensor_{A'} F' \arrow[rr, "\text{by (2)}"', hook] & & A' \tensor_{A'} F' \arrow[rr] \arrow[dd] & & A \tensor_{A'} F' \arrow[rr] \arrow[dd] & & 0 \\ & & & & & & & & \\ & & 0 \arrow[rr, "\text{snake}"] & & A'/p' \tensor_{A'} F' \arrow[dd] \arrow[rr, "="] & & A/p \tensor_{A'} F' \arrow[dd] & & \\ & & & & & & & & \\ & & & & 0 & & 0 & & \end{tikzcd} Then by (1), we have \[ \tor_1^{A'}(A'/p', F') = \tor_1^{A'}(A/p, F') = 0 .\] ::: We will just need this for $A' = k[\eps]$ and $A=k$. :::{.proposition title="?"} \[ T_z \hilb_{X_{_{/k}}} = \hom_{\OO_x}(I_z, \OO_z) .\] ::: :::{.proof title="?"} Again we have $T_z \hilb_{X_{_{/k}}} \subset \hilb_{X_{_{/k}}}(k[\eps])$, and is given by $$ \theset{Z' \subset X \cross_{\spec k} \spec k[\eps] \st Z' \text{ is flat}_{/k[\eps]},\,\, Z' \cross_{\spec k[\eps]}\spec k = Z} .$$ We have an exact diagram: \begin{tikzcd} & & 0 \arrow[r] & I_{Z'} \arrow[r] & {\OO_{X[\eps]}} \arrow[r] & \OO_{Z'} \arrow[r] & 0 \\ 0 \arrow[d] & & & {} \arrow[d] & {} \arrow[d] & {} \arrow[d] & \\ k \arrow[d] & & {} \arrow[r] & I_Z \arrow[r] \arrow[d] & \OO_x \arrow[r] \arrow[d] & \OO_z \arrow[r] \arrow[d] & {} \\ {k[\eps]} \arrow[d] & & {} \arrow[r] & I_{Z'} \arrow[r] \arrow[d] & {\OO_{x[\eps]}} \arrow[r] \arrow[d] & \OO_{Z'} \arrow[r] \arrow[d] & {} \\ k \arrow[d] & & {} \arrow[r] & I_Z \arrow[r] \arrow[d] & \OO_x \arrow[r] \arrow[d] \arrow[u, dotted, bend right] & \OO_Z \arrow[r] \arrow[d] & {} \\ 0 & & & {} & {} & {} & \end{tikzcd} Note the existence of a splitting above. Given $\phi \in \hom_{\OO_x}(I_Z, \OO_Z)$. We have \[ I_{Z'} = \left\{ f + \eps g \, \middle\vert \, \begin{array}{ll} f,g &\in I_Z, \\ \phi(f) &= g\mod I_Z, \\ \phi(f) &\in \OO_Z, \\ g\mod I_Z &\in \OO_x/I_Z = \OO_Z \end{array} \right\} .\] It's easy to see that $Z' \subset x'$, and 1. $Z'\cross k = Z$ 2. It's flat over $k[\eps]$, looking at $0 \to k\tensor I_{Z'} \to I_{Z'}$. For the converse, take $f\in I_Z$ and lift to $f' = f + \eps g \in I_{Z'}$, then $g\in \OO_x$ is well-defined wrt $I_Z$. Then $g\in \hom_{\OO_x}(I_z, \OO_z)$. ::: The main point here is that these hom sets are extremely computable. :::{.example title="?"} Let $Z$ be a twisted cubic in $\hilb_{\PP^3_{/k}}^{3t+1}(k)$. ::: :::{.observation} \[ \hom_{\OO_x}(I_Z, \OO_Z) = \hom_{\OO_X}(I_Z/I_Z^2, \OO_Z) = \hom_{\OO_Z}(I_Z/I_Z^2, \OO_Z) \] ::: If $I_Z/I_Z^2$ is locally free, these are global sections of the dual, i.e. $H^0((I_Z/I_Z^2)\dual)$. In this case, $Z\injects X$ is regularly embedded, and thus $(I_Z/I_Z^2)\dual$ should be regarded as the normal bundle. Sections of the normal bundle match up with directions to take first-order deformations: \begin{tikzpicture} \definecolor{arrow_color}{HTML}{ba0cff} \node (node_one) at (0,0) { \includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_22-40} }; \node (a) at (-7, 5) {\Huge $\PP^3$}; \node[arrow_color] (a) at (1, -3) {\Huge Deformation}; \end{tikzpicture} For $i:C \injects \PP^3$, there is an exact sequence \[ 0 \to I/I^2 \to &i^* \Omega_{\PP^3} \to \Omega_\eps \to 0 \\ &\Downarrow \quad \text{ taking duals } \\ 0 \to T_C \to &i^* T_{\PP^3} \to N_{C_{/\PP^3} } \to 0 ,\] How do we compute $T_{\PP^3}$? Fit into the exact sequence $$ 0 \to \OO \to i^* \OO(1)^4 \to i^* T_{\PP^3} \to 0 ,$$ which we can restrict to $C$. We have $i^* \OO(1) \cong \OO_{\PP^1}(3)$, so \[ 0 \to H^0 \OO_c \to &H^*(\OO(3)^4) \to H^0(i^* T_{\PP^3}) \to 0 \\ &\Downarrow \\ 0\to k \to &k^{16} \to k^{15} \to 0 .\] This yields \[ 0 \to H^0(T_c) \to &H^0(i^* T_{\PP^3}) \to H^0(N_{C_{ /\PP^3} }) \to H^1 T_c \\ &\Downarrow \\ 0\to k^3 \to &k^{15} \to k^{12} \to 0 \] :::{.example title="?"} $\hilb_{\PP^n_k}^{P_?} \cong \PP H^0(\PP^n, \OO(d))\dual$ which has dimension ${n+1 \choose n} - 1$. Pick $Z$ a $k$ point in this Hilbert scheme, then $T_Z H = \hom(I_Z, \OO_Z)$. Since $I_Z \cong \OO_{\PP}(-d)$ which fits into \[ 0 \to \OO_{\PP^n}(-d) \to \OO_{\PP^n} \to \OO_Z \to 0 .\] We can identify \[ \hom(I_Z,\OO_Z) = H^0( (I_Z/I_Z^2)\dual ) = H^0(\OO_Z(d)) .\] \begin{tikzcd} 0\ar[r] & \OO_{\PP^n}\ar[r] & \OO_{\PP^n}(d)\ar[r] & \OO_Z(d)\ar[r] & 0 \\ & & & & \\ 0\ar[r] & H^0( \OO_{\PP^n} ) \ar[r] & H^0( \OO_{\PP^n}(d) ) \ar[r] & H^0(\OO_Z(d) ) \ar[r] & 0 \\ \text{dim:} & k & k^{n+d \choose n} & k^{{n+d\choose n}-1} & \end{tikzcd} ::: :::{.example title="?"} The tangent space of the following cubic: \begin{tikzpicture} \node (node_one) at (0,0) { \includegraphics{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2020/Spring/Moduli Spaces/sections/figures/2021-01-03_17-27} }; \end{tikzpicture} We can identify \[ \hom_{\OO_k}(I_Z, \OO_Z) = H^0((I_Z/I_Z^2)\dual) = 3 + H^0((I_{Z_0}/I_{Z_0}^2)\dual) ,\] where the latter equals $H^0 \qty{ \OO_1\mid_{z_0} \oplus \OO(\zeta)\mid_{z_0} }$ yielding \[ 3+9 = 12 .\] :::