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Let there be two curves of degree $d=2 m^2$ in $\mathbb{A}^3$ having $\geq c d^2$ points of intersection, where $c>0$ is a constant. Then their union $V$ is a curve of degree $2 d = 4 m^2$. Can $V$ be the intersection of two surfaces of degree $\leq C m$, where $C$ is a constant depending on $c$? (That is, can that happen for $m$ arbitrarily large and $c$, $C$ fixed?)

[The original question had $c=1$. In that case, as pointed out in an answer below, there are no such two curves, unless they lie on a hyperplane.)

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  • $\begingroup$ Do you have explicit examples of such curves (not contained in a hyperplane)? $\endgroup$ Feb 10, 2021 at 9:21
  • $\begingroup$ @FrancescoPolizzi No. $\endgroup$ Feb 10, 2021 at 9:38
  • $\begingroup$ I suspect they do not exist $\endgroup$ Feb 10, 2021 at 9:40
  • $\begingroup$ And I would like to see a proof :). More generally, can there be two such curves having $\gg d^2$ points of intersection, not lying on a hypersurface of bounded degree? $\endgroup$ Feb 10, 2021 at 9:41
  • $\begingroup$ It does seem to me that there is a constant $c>0$ such that $c d^2$ points in general position in space do not determine a single curve (but rather many curves pass through them). That would give us two (in fact many) curves not on a hyperplane. $\endgroup$ Feb 10, 2021 at 9:45

2 Answers 2

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Edit. This is an answer to the original question (in which $c=1$.)

I think these examples do not exist. In fact, there is the following result, that can be found in

S. Diaz: Space curves that intersect often, Pac. J. Math. 123, 263-267 (1986). ZBL0562.14008.

Theorem. Let $X$, $Y$ be two irreducible, reduced curves in $\mathbb{P}^3$, neither of which is contained in a hyperplane. Then the number $s$ of intersection points of $X$ and $Y$, counted without multiplicity, satisfies $$s \leq (\deg X -1)(\deg Y -1)+1.$$ The bound is sharp and, if equality holds, there is a quadric hypersurface containing $X \cup Y$.

In your case, passing to the projective completion, we have $\deg X = \deg Y =d$, and so $$s \leq (d-1)^2-1=d^2-2d < d^2.$$

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  • $\begingroup$ Thanks. I'll amend the question to make it softer. $\endgroup$ Feb 10, 2021 at 9:48
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$\newcommand{\P}{\mathbb{P}}\newcommand{\C}{\mathbb{C}}$ Consider irreducible surfaces $P,Q\subset \P^3$ of degrees $p$, $q$. Let $C=P\cap Q$ be the scheme-theoretic intersection, reduced or not. We have a complex of line bundles on $\P^3$. $$ X=O(-P-Q)\to O(-P)\oplus O(-Q) \to O. $$ It is quasi-isomorphic to $O_P(-Q)\to O_P$ and therefore has only cohomology in degree $2$, and the cohomology is isomorphic to $O_C$, the structure sheaf of the curve $C$. Hence we have $$ H^{*+2}(\P^3, X) = H^*(\P^3, O_C) = H^*(C, O_C). $$ Let us compute the cohomology of $X$. Note that $H_0(\P^3, O)=\C$ and other cohomologies vanish. Note that $H^i(\P^3,O(-P))=0$ for $i=0,1,2$ and the same for $O(-Q)$ and $O(-P-Q)$. Drawing the spectral sequence we see that $H^0(C, O_C) = \C$. The Euler characteristic of $X$ is given by $$ \chi(C,O_C) = \chi(\P^3, X) = 1 - \binom{1-p}{3} - \binom{1-q}{3} + \binom{1-p-q}{3} = -\frac{p q (p+q-4)}2 $$ Now the structure sheaf of $C$ fits into exact sequence $$ 0\to J \to O_C \to O_C^{red}\to 0, $$ where $J$ contains all the nilpotent elements. So we obtain a long exact sequence $$ 0\to H^0(C, J) \to H^0(C, O_C) \to H^0(C, O_C^{red}) \to H^1(C, J) \to H^1(C, O_C) \to H^1(C, O_C^{red}) \to 0. $$ So we see that $H^0(C,J)=0$ and therefore $\chi(C, J)\leq 0$ and $\chi(C, O_C^{red})\geq -\frac{p q (p+q-4)}2$. Let $\pi:\tilde C\to C$ be the normalization of $C$. Suppose $\tilde C$ has components of genera $g_1,\dots,g_k$. Consider the short exact sequence $$ 0\to O_C\to \pi_* O_{\tilde C} \to T \to 0 $$ where $T$ is a torsion sheaf supported at the singularities of $C$. We obtain $$ \sum_{i=1}^k (1-g_i) - \dim H^0(C, T) \geq -\frac{p q (p+q-4)}2 $$ Each self-intersection of $C$ contributes at least one to $T$, so the number of self-intersections is bounded by $$ \sum_{i=1}^k (1-g_i) + \frac{p q (p+q-4)}2 \leq k + \frac{p q (p+q-4)}2. $$ This is basically elaboration of Will Sawin's comment, so I am making this a community wiki.

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  • $\begingroup$ And what would be an easy bound on $k$ in terms of $p$ and $q$? $\endgroup$ Feb 12, 2021 at 20:43
  • $\begingroup$ @HAHelfgott in your original question it would be $k=2$, in general I don't know, the method does not seem to produce that, good question. On the other hand, we do seem to know that $C$ is connected ($H^0(C, O_c)=\C$), so at least $k-1$ self-intersections will be spent to make it connected. So if your want to estimate intersections between any fixed pair of components you can set $k=2$ in the formula. $\endgroup$ Feb 12, 2021 at 20:52
  • $\begingroup$ @HAHelfgott $k$ can be as high as $pq$: take $P$, $Q$ cones over some plane curves. The intersection is $pq$ lines, all meeting at a point. $\endgroup$ Feb 13, 2021 at 9:43

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