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It is well known that imposing vanishing at general points of $\mathbb P^2$ gives independent conditions on curves of degree $d$. Also, it is known that a small number ($\le d+1$) points always impose independent conditions (without genericity assumptions). But can we mix these?

More precisely, pick numbers $n$ and $m$ with $\chi(\mathcal O(d))>n+m$. Can one pick $m$ points $Z$ so generally that so that for ANY set $W$ of $n$ points that $W$ and $Z$ together impose independent conditions on curves of degree $d$?

Said differently and more generally: for a general enough $Z$, is it true that for ALL 0-dimensional schemes $W$ of length $n$, $H^1(I_Z \otimes I_W(d))=0$?

(Here $I_W$ is the ideal sheaf of the subscheme.)

Of course if $n$ is too big (e.g. greater than or equal to $d$) you are doomed. Also if $m$ is too big $H^0(I_Z(d))$ will have base points so even $n=1$ will fail. But with $n$ and $m$ both small it is trivially 'yes'. For some fixed $n$ and $d$, how large can you make $m$ so that the answer is 'yes'? I think already $n=1$ and $n=2$ are not obvious.

Conjecture (Inspired by Will Sawin's comment)

Just by doing some dimension counts, one can see that a for a general dimension $3n$ sub-linear series, every $n$ points impose independent conditions. So one might guess that one can take $m=\chi(\mathcal O(d))-3n$ points to make a linear series that is general in this sense. This agrees with the $n=1$ case which we can check rigorously.

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For $n=1$ and $d\geq 5$, we may take $m=\chi(\mathcal O(d))-3$. Take the complete intersection of two general degree $d$ curves. These points define a two-dimensional linear system. We may find among these points $\chi(\mathcal O(d))-3$ points that apply linearly independent conditions, so that they define a three-dimensional linear system.

Let's check that these $m$ points are linearly independent from any other point.

Any point that is not among the original $d^2$ points fails to lie on one of the two curves and hence provides a linearly independent condition. So we need to check that none of the other points is a base point.

We can do this using a monodromy argument. Observe that the set of base points among the original $d^2$ is determined by the choice of $m$ points. We have a canonical way to, from a choice of $m$ points in the intersection of two generic degree $d$ curves, obtain a subset of the complement. Because it is canonical, it must be invariant by the monodromy of the family of $d^2$ points. As long as the monodromy group is $S_{d^2}$, the only invariant such functions are the function that sends a subset to the empty set or to the whole complement.

Because the linear system is $3$-dimensional, there is one function in it that is not in the original linear system, which must fail to vanish at one of the remaining points. So they are not all base points. Because we know not every point in the complement is a base point, the set of base points must be empty, and we are done (conditional on the monodromy group being $S_{d^2}$).

Let's show the monodromy group is $S_{d^2}$. By the classification of $k$-transitive permutation groups for $k\geq 4$, it is sufficient to show that it is $6$-transitive. This is equivalent to saying that the expectation of, for a random element of the monodromy group, the number ordered $6$-tuples of fixed points, is $1$.

We can compute this expectation using Deligne's equidistribution theorem. One consequence of this is that for a variety $X$ over a finite field $\mathbb F_q$ and a map from $\pi_1(X)$ to a finite group that is surjective on the geometric fundamental group, the distribution of Frobenius elements of a random point in $X(\mathbb F_q)$ is close to the distribution of random elements of $G$, and the distribution gets closer as $q$ grows.

Here we take $X$ to be the space of pairs of degree $d$ plane curves that intersect transversely and the map to be the map to the monodromy group as a subgroup of $S_{d^2}$. We need to know that the expectation of the number of ordered $6$-tuples of $\mathbb F_p$-points of a complete intersection of two random degree $d$ curves over $\mathbb F_p$ is $1+o(1)$. This is true by counting: The number of $6$-tuples of points of $\mathbb P^2$ is $\frac{(p^2+p+1)!}{(p^2+p-5)!}$. Because $d\geq 5$, the points apply linearly independent conditions, so the probability of such a tuple lying on the complete intersection of two random degree $d$ curves is $1/p^{12}$. Hence the total expectation is:

$$\frac{(p^2+p+1)!}{p^{12}(p^2+p-5)!}=1+o(1)$$

This monodromy computation is due to Alexei Entin.

There is an alternative interpretation of the same arugment that does not use finite fields. The complete intersections form a $d^2$-fold cover of the space of pairs of degree $d$ curves with transverse intersection. The set of $6$-tuples of points in this cover forms a $\frac{ (d^2)!}{(d^2-6)!}$-fold cover. We want to show that the monodromy group acts transitively on this cover. Equivalently, we want to show that it is irreducible. It is sufficient to show that the space of pairs of two degree $d$ curves and $6$ distinct points on both curves is irreducible, without the transverse intersection condition. We can do this by viewing it as a bundle over the space of $6$-uples of points on $\mathbb P^2$, which is irreducible. The fibers of this bundle are the spaces of pairs of degree $d$ curves through $6$ distinct points, which are a product of projective spaces. Because the conditions imposed by the points are linearly independent, the projective spaces form a fiber bundle. Because we have an irreducible bundle on an irreducible base, the scheme is irreducible.

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  • $\begingroup$ Thanks, this looks great. I think I followed everything except the one sentence starting with "Because Frob_p in large characteristic..." Why are we suddenly looking at random degree d curves? Maybe I need to see the precise statement of this theorem. $\endgroup$ – Drew Apr 27 '15 at 18:23
  • $\begingroup$ Is there any hope that something like this could work for larger n? $\endgroup$ – Drew Apr 27 '15 at 18:23
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    $\begingroup$ @Drew I added more details, and I've been thinking about this same question. I think you want to take $m= \chi(O(d))-3n$ so that the $m$ points define a rank $3n$ linear system. I think for a generic rank $3n$ linear system, every $n$ distinct points impose independent conditions, but I don't know how to prove the linear system is generic. Perhaps somehow trying a similar argument on $(\mathbb P^2)^n$ would work, but I don't see how. $\endgroup$ – Will Sawin Apr 27 '15 at 18:50
  • $\begingroup$ Thanks for your contribution! I was hoping someone would have an answer for more $n$, but I will award you the bounty. $\endgroup$ – Drew Apr 30 '15 at 19:52
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Here is another argument for $n=1$ (and distinct points) by Thomas Goller. A special case of the Cayley-Bacharach theorem CB5 in Eisenbug, Green, and Harris says

If $C$ and $C'$ (of degree $d$) intersect in finitely many points $\Gamma$, then for any partition $\Gamma=\Gamma' \coprod \Gamma''$ then the dimension of curves of degree $d-3$ though $\Gamma'$ (modulo those through $\Gamma$) is equal to the failure of $\Gamma''$ to impose independent conditions on curves of degree $d$.

Now choose $\chi(\mathcal O(d-3))+1$ points generally, and pick $C$ and $C'$ to be two degree $d$ curves through the chosen points, and let $\Gamma'$ be the chosen points. Then $|\Gamma''| = \chi(\mathcal O(d))-3$ (this will be our $m$). According to the theorem, $\Gamma''$ imposes independent conditions on curves of degree $d$ since there is no curve of degree $d-3$ through $\Gamma'$. We now want to show that the linear system defined by $\Gamma''$ has no other base points. The only other possibilities for base points are the points $\Gamma'$, but if you move a point from $\Gamma'$ to $\Gamma''$ you see that the theorem still says that $\Gamma'$ still imposes independent conditions.

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