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I wondered if it were possible to clarify the basic first lemma in A note on the Intersection of Veronese Surfaces.

Setup: $X_1$ and $X_2$ are two Veronese surfaces in $\mathbb{P}^5$ intersecting along a zero-dimensional scheme $W$.

Statement: Viewing $W$ as a subscheme of $X_1\cong \mathbb{P}^2$, $h^1(I_{W/\mathbb{P}^2}(4))=0$. (So $W$ imposes independent conditions on plane quartics, bounding the length of $W$.)

The proof is just five lines. The argument is to restrict the minimal resolution of $I_{X_2}$ to $X_1$. Using the fact that $I_{X_2}$ is generated by quadrics (with linear relations, linear relations between the relations and so on), we are supposed to conclude something about the resolution of $I_{W/X_1}$.

However, I'm not even sure whether the restriction of the resolution to $X_1$ remains exact or whether that matters.

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The restriction of the resolution of $\mathcal{O}_{X_2}$ to $X_1$ computes $\mathrm{Tor}_i(\mathcal{O}_{X_1},\mathcal{O}_{X_2})$. If the intersection $X_1 \cap X_2$ is dimensionally transverse, higher $\mathrm{Tor}_i$ vanish, so the restriction of the resolution is a resolution of $W$.

EDIT. As @DCT observed, if two Veronese surfaces in $\mathbb{P}^5$ intersect, this is not a dimensionally transverse intersection, so one needs to be more careful.

In fact, if the intersection is "almost dimensionally transverse" (in the sense that the dimension of intersection is by one bigger than the expected dimension) then $$ \mathrm{Tor}_{\ge 2}(\mathcal{O}_{X_1},\mathcal{O}_{X_2}) = 0 $$ and $\mathrm{Tor}_1(\mathcal{O}_{X_1},\mathcal{O}_{X_2})$ is an invertible sheaf on the scheme $X_1 \cap X_2$ (this is a local statement, so it can be proved using the Koszul resolution when the schemes $X_1$ and $X_2$ are locally complete intersections and the ambient scheme is Cohen-Macaulay).

In the case of two Veronese surfaces this can be applied as follows. Recall the resolution $$ 0 \to \to \mathcal{O}(-4)^{\oplus 3} \to \mathcal{O}(-3)^{\oplus 8} \to \mathcal{O}(-2)^{\oplus 6} \to \mathcal{O} \to \mathcal{O}_{X_1} \to 0 $$ of a Veronese surface. Restricting it to $X_2$ we obtain the complex $$ 0 \to \to \mathcal{O}_{X_2}(-8)^{\oplus 3} \to \mathcal{O}_{X_2}(-6)^{\oplus 8} \to \mathcal{O}_{X_2}(-4)^{\oplus 6} \to \mathcal{O}_{X_2} \to 0 $$ whose cohomology is $\mathcal{O}_W$ in degrees $0$ and $-1$ (any invertible sheaf on a zero-dimensional scheme is trivial). Twisting by $\mathcal{O}_{X_2}(4)$ we obtain the complex $$ 0 \to \to \mathcal{O}_{X_2}(-4)^{\oplus 3} \to \mathcal{O}_{X_2}(-2)^{\oplus 8} \to \mathcal{O}_{X_2}^{\oplus 6} \to I_W(4) \to 0 $$ whose cohomology is the sheaf $\mathcal{O}_W$ in the term $\mathcal{O}_{X_2}^{\oplus 6}$. Then the hypercohomology spectral sequence gives you $$ 0 \to H^0(X_2,\mathcal{O}_W) \to \mathbb{C}^6 \to H^0(X_2,I_W(4)) \to \mathbb{C}^{9} \to 0 $$ (the last term is $H^2(X_2,\mathcal{O}_{X_2}(-4)^{\oplus 3})$) and $H^{>0}(X_2,I_W(4)) = 0$. In particular, this bounds the length of $W$ by 6.

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    $\begingroup$ In this case, $X_1$ and $X_2$ are dimension 2 in $\mathbb{P}^5$. If they intersect in finitely many points, does this count as dimensionally transverse? $\endgroup$
    – DCT
    Commented Dec 10, 2019 at 19:15
  • $\begingroup$ @DCT: You are right, one needs a bit more in this case, I will extend the answer. $\endgroup$
    – Sasha
    Commented Dec 10, 2019 at 19:28
  • $\begingroup$ I think I can follow your reasoning to get the last exact sequence, but apparently $W$ could be 10 simple points. Do you know how to resolve the apparent contradiction? $\endgroup$
    – DCT
    Commented Dec 11, 2019 at 3:59
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    $\begingroup$ @DCT: You are right again, and I was a bit hasty. What the spectral sequence gives, it is "roughly" an exact sequence $0 \to H^0(X_2,\mathcal{O}_W) \to \mathbb{C}^6 \oplus \mathbb{C}^9 \to H^0(X_2,I_W(4)) \to 0$, so the actual bound is 15. Of course you can improve it a bit since you know that the second map cannot be identically zero. In fact, it is quite possible that you can establish the bound 10 in this way (for this you have to show that $X_1$ cannot be contained in 2 out of 6 quadrics through $X_2$). $\endgroup$
    – Sasha
    Commented Dec 11, 2019 at 6:12

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