Zac Maeder-Wolland: Some Daily AG Problems

Lazarsfeld’s Positivity in Algebraic Geometry: Example 1.2.22

I’m currently reading Lazarsfeld’s book, Positivity in Algebraic Geometry. It is really a joy to read, and I plan to do related exercises or work through examples in his book in the coming months. I still have several Hartshorne problems in the oven, with solutions hopefully soon to come. Here, I go over Example 1.2.22 relating to resolutions of a coherent sheaf. I wanted to work out this exercise as practice for using resolutions of a sheaf. Although Lazarsfeld basically paves the way for this example, I will try here to add all the (maybe unnecessary) details.

$X$ is a projective scheme, $D$ and $E$ are ample Cartier divisors on $X$.

(a) Consider $X\times X$ and let $\delta:X\rightarrow X\times X$ be the closed immersion of $X$ onto the diagonal $\Delta$. This yields an SES

$$ 0\ra \mathcal I_\Delta\ra \OO_{X\times X}\ra \delta_*\OO_\Delta\ra 0.$$

We write $(aD,bE)$ to denote the divisor $\op{pr}_1^*(aD)+\op{pr}_2^*(bE)$ on $X\times X$. Twisting this SES by $(aD,bE)$, and then looking to cohomology, we will show that

$$ H^1\big (X\times X,\mathcal I_\Delta \big ((aD,bE)\big)\big)=0.$$

Suppose we can show this. Then, by the Kunneth formula, we know that

$$ H^0\big (X\times X,\OO_{X\times X}((aD,bE))\big )\cong H^0\big (X\times X,\OO_{X\times X}\big ((aD,0)\big )\big )\otimes H^0\big (X\times X,\OO_{X\times X}\big ((0,bE)\big )\big ).$$

It is a fact that

$$H^i\big (X\times X,\OO_{X\times X}\big ((aD,0)\big )\big )\cong H^i\big (X,\OO_X(aD)\big ),$$

and likewise

$$H^i\big (X\times X,\OO_{X\times X}\big ((0,bE)\big )\big )\cong H^i\big (X,\OO_X(bD)\big ) .$$

for every $i$. You can find this on the stack project. I’ll try and give someintuition for why this works when $i=0$. The global sections of $\OO_{X\times X}\big ((aD,0)\big )$ are precisely pullbacks under $p_1$ of the global sections of $\OO_X(bD)$, because a rational function on $X\times X$ is an element of $\Bbb K(X)\otimes\Bbb K(X)$, hence, it must be homogeneous in both copies of the variables for $\Bbb K(X)$. Its poles are also constrained to $\op{Supp}D$.
So now it follows that we have a surjection,

$$H^0\big (X,\OO_X(aD)\big )\otimes H^0\big (X,\OO_X(bE)\big )\longrightarrow H^0 \big(X,\OO_X(aD+bE)\big ).$$

due to the projection formula and that $\delta^*\big(\op{pr}_1^*(aD)+\op{pr}_2^*(bE)\big )=aD+bE$.

It remains to show the claimed vanishing of the first cohomology group for sufficiently large $a$ and $b$. For this, we produce an appropriate resolution of the coherent sheaf $\mathcal I_\Delta$. The goal will be to obtain a resolution that at some point terminates. At this point, we can split the resulting long exact sequence into many short exact sequences, and since direct sums commute with cohomology, we will obtain a collection of cohomological long exact sequences. Proceeding by induction, we will then be able to show the desired vanishing. To this end, find an integers $p_0,q_0>>0$ such that $p_0D$ and $q_0E$ are very ample. As we showed in Problem 44, it follows that $(p_0D,q_0E)$ is very ample on $X\times X$ (we showed this by using a Segre embedding). At this point it is true that $H^i\big (\mathcal I_\Delta\big ((mp_0D,mq_0E)\big )\big )=0$ for all $m>>0$ and $i>0$, but this only gives the desired result if $a$ and $b$ are the same multiple of the numbers $p_0$ and $q_0$ above. We want something stronger. Notice that it is no loss of generality that $(D,E)$ is ample on $X\times X$, because $\max (p_0,q_0)(D,E)$ is very ample. So, choose $p_0>>0$ such that $\mathcal I_\Delta\Big ((p_0D,p_0E)\Big )$ is globally generated. We then obtain a surjection,

$$\bigoplus _{\text{finite}}\OO_{X\times X}\big ((-p_0D,-p_0E)\big )\longrightarrow \mathcal I_\Delta\ra 0$$

Let $\mathcal K_0$ be the kernel, which is coherent, and find $p_1>>p_0$ such that

$$\bigoplus _{\text{finite}}\OO_{X\times X}\big ((-p_1D,-p_1E)\big )\longrightarrow \mathcal K_0\ra 0$$

is surjective. We continue in this fashion, and obtain a resolution,

$$…\ra\bigoplus _{\text{finite}}\OO_{X\times X}\big ((-p_jD,-p_jE)\big )\ra …\ra \bigoplus _{\text{finite}}\OO_{X\times X}\big ((-p_0D,-p_0E)\big )\longrightarrow \mathcal I_\Delta\ra 0.$$

The resolution may not terminate, but as we will see, that is okay. For any pair of integers $(a,b)\in \Z^2$, we can twist this resolution by $(aD,bE)$. By the Kunneth formula, and because cohomology commutes with direct sums,

$$ H^i\big (X\times X, \bigoplus _{\text{finite}}\OO_{X\times X}\big (((a-p_j)D,(b-p_j)E)\big )\big )$$

$$=\bigoplus _{\text{finite}}\Big (\bigoplus_{s+t=i} H^s\big (X, \OO_X((a-p_j)D)\big )\otimes H^t\big (X, \OO_X((b-p_j)E)\big ) \Big )$$

So, for $i>2\dim X$ and for every $j\ge0$ and every $(a,b)\in \Z^2$, this cohomology vanishes. Now is a good time to make clear the goal of the rest of this argument. We want to find a positive integer $m_0$ such that

$$H^1\big (X\times X,\mathcal_\Delta\big ((aD,bE)\big )\big )=0$$

for all $a,b\ge m_0$.
From the resolution, we obtain an infinite collection of SES’s,

$$ 0\ra \mathcal K_{j} \big ((aD,bE)\big )\longrightarrow \bigoplus _{\text{finite}}\OO_{X\times X}\big (((a-p_j)D,(b-p_j)E)\big )\longrightarrow \mathcal K_{j-1}\big ((aD,bE)\big )\ra 0$$

for $j\ge 0$, where $\mathcal K_{-1}:=\mathcal I_\Delta$. Find positive integers $a_0=a_0(p_0,D),b_0=b_0(p_0,E)$ such that, for $a\ge a_0$ and $b\ge b_0$,

$$H^i\big (X,\OO_X((a-p_0)D)\big )= H^i\big (X,\OO_X((b-p_0)E)\big )=0$$

for all $i>0$. As we can see, it follows that

$$ 0\ra H^1\big (X\times X,\mathcal I_\Delta\big ((aD,bE)\big )\big )\longrightarrow H^2(X\times X,\mathcal K_0)$$

To this end, we find a collection $\{(a_j,b_j)\}_{j=0}^{2\dim X}$ of positive integers, for which each pair satisfies that,

$$ H^i\big (X,\OO_X((a-p_k)D)\big )= H^i\big (X,\OO_X((b-p_k)E)\big )=0$$

for all $i>0$, all $a\ge a_j,b\ge b_j$, and all $k\le j$ (as $a-p_k\ge a_j-p_k\ge a_j-p_j$). Let $m_0:=\max_j{a_j,b_j}$. Then $m_0$ may depend on $D,E$ but not on the $p_j$. For every $a,b\ge m_0$,

$$H^i\big (X\times X,\bigoplus_{\text{finite}}\mathcal O_{X\times X}\big (((a-p_j)D,(b-p_j)E) \big )\big )=0$$

holds for all $i>0$ and all $0\le j\le2\dim X_0$. It also holds for arbitrary $j$ whenever $i>2\dim X$. In particular, it holds for every $i\ge j$. This implies that

$$0\ra H^{j+1}\big (X\times X,\mathcal K_{j-1}\big ((aD,bE)\big )\big )\longrightarrow H^{j+2}\big (X\times X,\mathcal K_{j}\big ((aD,bE)\big )\big )$$

is injective for every $j\ge -1$. But as soon as $j=2\dim X-1$, the rightmost cohomology group becomes trivial! Therefore, it must be that, in fact,

$$H^1\big (X\times X,\mathcal I_\Delta\big ((aD,bE)\big )\big )=0$$

from the start. This is was the last piece to our argument, because now we have the desired result that $(*)$ is surjective.

(b) We now begin by tensoring

$$0\ra \mathcal I_\Delta\ra \OO_{X\times X}\ra \delta_*\OO_\Delta\ra 0.$$

with $\op{pr}_1^*\mathcal F\otimes\op{pr}_2^*\mathcal G$. The sequence will remain exact due to flatness. Then replacing the resolution of $\mathcal I_\Delta$ above with a resolution of $\op{pr}_1^*\mathcal F\otimes\op{pr}_2^*\mathcal G\otimes\mathcal I_\Delta$ will yield the desired result, except now $m_0$ will also depend on $\mathcal F,\mathcal G$.