# Uniform Vanishing Statements (Thursday January 30th) Recall how we constructed the Hilbert scheme of hypersurfaces $$ \hilb_{\PP_k^n}^{P_{m, d}} = \PP H^0(\PP^n; \OO(d))\dual $$ A section $\hilb_{\PP_k^n}^{P}(s)$ corresponds to $z\in \PP^n_s$. We can look at the exact sequence \[ 0 \to I_Z(m) \to \OO_{\PP_S^n} \mapsvia{\text{restrict}} \OO_z(m) \to 0 .\] as $\PP_s^n \mapsvia{\pi_s} S$, so we can pushforward along $\pi$, which is left-exact, so \[ 0 \to \pi_{s*} I_Z(m) \to \pi_{s*} \OO_{\PP_S^n} = \OO_S \tensor H^0(\PP^n; \OO(m)) \to \OO_z(m) \to R^1 \pi_{s*} I_Z(m) \to \cdots .\] *Idea:* $Z \subset \PP_k^n$ will be determined (in families!) by the space of degree $d$ polynomials vanishing on $Z$ (?), i.e. $$ H^0(\PP^n, I_z(m)) \subset H^0(\PP^n, \OO(m)) $$ for $m$ very large. This would give a map of functors $$ \hilb_{\PP_k^n}^{P} \to \Gr(N, H^0(\PP^n, \OO(m) )) .$$ If this is a closed subfunctor, a closed subfunctor of a representable functor is representable and we're done . :::{.remark} We need to get an $m$ uniform in $Z$, and more concretely: 1. First need to make sense of what it means for $Z$ to be determined by $H^0(\PP^n, I_Z(m))$ for $m$ only depending on $P$. 2. This works point by point, but we need to do this in families. I.e. we'll use the previous exact sequence, and want the $R^1$ to vanish. ::: :::{.slogan} We need *uniform* vanishing statements. There is a convenient way to package the vanishing requirements needed here. From now on, take $k=\bar k$ and $\PP^n = \PP_k^n$. ::: ## $m\dash$Regularity :::{.definition title="m-Regularity of Coherent Sheaves"} A coherent sheaf $F$ on $\PP^n$ is **$m\dash$regular** if $H^i(\PP^n; F(m-i)) = 0$ for all $i> 0$. ::: :::{.example title="?"} Consider $\OO_{\PP^n}$, this is $0\dash$regular. Line bundles on $\PP_n$ only have 0 and top cohomology. Just need to check that $H^n(\PP^n; \OO(-n)) = 0$, but by Serre duality this is \[ H^0(\PP^n; \OO(n) \tensor \omega_{\PP^n})\dual = H^0(\PP^n; \OO(-1))\dual = 0 .\] ::: :::{.proposition title="?"} Assume $F$ is $m\dash$regular. Then 1. There is a natural multiplication map from linear forms on $\PP^n$, \[ H^0(\PP^n; \OO(1)) \tensor H^0(\PP^n; F(k)) \to H^0(\PP^n; F(k+1)) ,\] which is surjective for $k\geq n$.[^graded_module_note] 2. $F$ is $m'\dash$regular for $m' \geq m$. 3. $F(k)$ is globally generated for $k\geq m$, i.e. the restriction \[ H^0(\PP^n; F(k)) \tensor \OO_{\PP^n} \to F(k) \to 0 \] is exact (i.e. surjective). [^graded_module_note]: Think of this as a graded module, this tells you the lowest number of small grade pieces needed to determine the entire thing. ::: :::{.example title="?"} $\OO$ is $m\dash$regular for $m \geq 0$ implies $\OO(k)$ is $-k\dash$regular and is also $m\dash$regular for$m\geq -k$. ::: ### Proof of 2 and 3 Induction on dimension of $n$ in $\PP^n$. Coherent sheaves on $\PP^0$ are vector spaces, so no higher cohomology. :::{.proof title="Step 1"} Take a generic hyperplane $H \subset \PP^n$, there is an exact sequence \[ 0 \to \OO(-1) \to \OO \to \OO_H \to 0 .\] where $\OO_H$ is the structure sheaf. Tensoring with $H$ remains exact, so we get \[ 0 \to F(-1) \to F \to F_H \to 0 .\] Why? $\AA^n \subset \PP^n$, let $A = \OO_{\PP^n}(\AA^n)$ be the polynomial ring over $\AA^n$. Then the restriction of the first sequence to $\AA^n$ yields $$ 0 \to A \mapsvia{f} A \to A/f \to 0 ,$$ and thus we want $$ F \mapsvia{f} F \to F/fF \to 0 $$ which results after restricting the second sequence to $\AA^n$. Thus we just want $f$ to not be a zero divisor. If we take $f$ not vanishing on any associated point of $F$, then this will be exact. Associated points: generic points arising by supports of sections of $F$. $F$ is coherent, so it has finitely many associated points. If $H$ does not contain any of the associated points of $F$, then the second sequence is indeed exact. ::: :::{.proof title="Step 2"} Twist up by $k$ to obtain \[ 0 \to F(k-1) \to F(k) \to F_H(k) \to 0 .\] Look at the LES in cohomology to get \[ H^i(F(m-i)) \to H^i(F_H(m-i)) \to H^{i+1}(F(m - (i+1))) .\] So $F_H$ is $m\dash$regular. By induction, this proves statements 1 and 2 for all $F_H$. So take $k = m+1-i$ and consider \[ H^i(F(m-i)) \to H^i(F(m+1-i)) \to H^i(F_H(m+1-i)) .\] We know 2 is satisfied, so the RHS is zero, and we know the LHS is zero, so the middle term is zero. Thus $F$ itself is $m+1$ regular, and by inducting on $m$ we get statement 2. ::: By multiplication maps, we get a commutative diagram: \begin{tikzcd} & & H^0(\OO(1)) \tensor H^0(F(k)) \arrow[dd, "\beta"] \arrow[rrr] \arrow[rrrdd] & & & H^0(\OO(1))\tensor H^0(F_H(k)) \arrow[dd] \\ & & & & & \\ H^0(F(k)) \arrow[rr, "H"] \arrow[rruu, "H \tensor \id"] & & H^0(F(k+1)) \arrow[rrr, "\alpha", dashed] & & & H^0(F_H(k+1)) \end{tikzcd} We'd like to show the diagonal map is surjective. :::{.observation} \envlist 1. The top map is a surjection, since $$ H^0(F(k)) \to H^0(F_H(k)) \to H^1(F(k-1)) = 0 $$ for $k\geq m$ by (2). 2. The right-hand map is surjective for $k\geq m$. 3. $\ker(\alpha) \subset \im(\beta)$ by a small diagram chase, so $\beta$ is surjective. This shows (1) and (2) completely. ::: :::{.proof title="of 3"} We know $F(k)$ is globally generated for $k\gg 0$. Thus for all $k\geq m$, $F(k)$ is globally generated by (1). ::: :::{.theorem title="?"} Let $P \in \QQ[t]$ be a Hilbert polynomial. There exists an $m_0$ only depending on $P$ such that for all subschemes $Z \subset \PP^n_k$ with Hilbert polynomial $P_Z = P$, the ideal sheaf $I_z$ is $m_0\dash$regular. ::: ### Proof of Theorem Induct on $n$. For $n=0$, again clear because higher cohomology vanishes and there are no nontrivial subschemes. For a fixed $Z$, pick $H$ in $\PP^n$ (and setting $I \definedas I_z$ for notation) such that \[ 0 \to I(-1) \to I \to I_H \to 0 .\] is exact. Note that the Hilbert polynomial $P_{I_H}(t) = P_I(t) - P_I(t-1)$ and $P_I = P_{\OO_{\PP^n}} - P_Z$. By induction, there exists some $m_1$ depending only on $P$ such that $I_H$ is $m_1\dash$regular. We get \[ H^{i-1}(I_H(k)) \to H^i(I(k-1)) \to H^i(I(k)) \to H^i(I_H(k)) ,\] and for $k\geq m_1 - i$ the LHS and RHS vanish so we get an isomorphism in the middle. By Serre vanishing, for $k \gg 0$ we have $H^i(I(k)) = 0$ and thus $H^i(I(k)) = 0$ for $k\geq m_i - i$. This works for all $i > 1$, we have $H^i(I(m_i - i)) = 0$. We now need to find $m_0 \geq m_1$ such that $H^1(I(m_0 - 1)) = 0$ (trickiest part of the proof). :::{.lemma title="?"} The sequence $\qty{\dim H^1(I(k))}_{k\geq m_i - 1}$ is *strictly* decreasing.[^explain_little_h] [^explain_little_h]: Note: $h^1 = \dim H^1$. ::: :::{.remark} Given the lemma, it's enough to take $m_0 \geq m_1 + h^1(I(m_1 - 1))$. Consider the LES we have a surjection $$ H^0(\OO_Z(m_1 - 1)) \to H^1(I(m_1 - 1)) \to 0 .$$ So the dimension of the LHS is equal to $P_Z(m_1 - 1)$, using the fact that terms vanish and make the Euler characteristic equal to $P_Z$. Thus we can take $m_0 = m_1 + P(m_1 - 1)$. ::: :::{.proof title="of Lemma"} Considering the LES \[ H^0(I(k+1)) \mapsvia{\alpha_{k+1}} H^0(I_H(k+1)) \to H^1(I(k)) \to H^1(I(k+1)) \to 0 ,\] where the last term is zero because $I_H$ is $m_1\dash$regular. So the sequence $h^1(I(k))$ is non-increasing. :::{.observation} If it does *not* strictly decrease for some $k$, then there is an equality on the RHS, which makes $\alpha_{k+1}$ surjective. This means that $\alpha_{k+2}$ is surjective, since \[ H^0(\OO(1)) \tensor H^0(I_H(k+1)) \surjects H^0(I_H(k+2)) .\] ::: So if one is surjective, everything above it is surjective, but by Serre vanishing we eventually get zeros. So $\alpha_{k+i}$ is surjective for all $i\geq 1$, contradicting Serre vanishing, since the RHS are isomorphisms for all $k$. ::: Thus for any $Z\subset \PP^n_k$ with $P_Z = P$, we uniformly know that $I_Z$ is $m_0\dash$regular for some $m_0$ depending only on $P$. :::{.claim} $Z$ is determined by the degree $m_0$ polynomials vanishing on $Z$, i.e. $H^0(I_z(m_0))$ as a subspace of all degree $m_0$ polynomials $H^0(\OO(m_0))$ and has fixed dimension. We have $H^i(I_Z(m_0)) = 0$ for all $i> 0$, and in particular $h^0(I_Z(m_0)) = P(m_0)$ is constant. ::: It is determined by these polynomials because we have a sequence \[ 0 \to I_Z(m_0) \to \OO(m_0) \to \OO_Z(m_0) \to 0 .\] We can get a commuting diagram over it $$ 0 \to H^0(I_Z(m_0)) \tensor \OO_{\PP^n} \to H^0(\OO(m_0)) \tensor \OO_{\PP^n} \to \cdots $$ where the middle map down is just evaluation and.the first map down is a surjection. Hence $I_Z(m_0)$, hence $\OO_Z$, hence $Z$ is determined by $H^0(I_Z(m_0))$. > Next time: we'll show that this is a subfunctor that is locally closed.