My friend, who is currently taking an algebraic geometry course from an unnamed prolific poster on MO, told me about the following bonus question on one of his problem sets a few weeks ago.

Preliminary discussion. By Bézout's Theorem (which we will prove later in the course), two plane cubics in general position with regard to one another, and hence any two memebrs of the pencil that they span, intersect in $9$ points. The previous question shows that in fact this pencil is already determined by $8$ of the $9$ intersection points.

From this you can deduce two things:

(a) $9$ points in general position cannot arise as the intersection of two cubic projective curves.

(b) $8$ points in general position determine a $9$th point (namely the $9$th common point of intersection of the pencil of cubics determined by the $9$ give points).

We can regard this process as defining a kind of $8$-ary operation$$(\mathbb{P}^2)^8\text{ }-\to \mathbb{P}^2.$$(We use a broken arrow because this operation is not really defined on all $8$-tuples of points in $\mathbb{P}^2$, but only on those in general position. It is an example of what is called a rational map in algebraic geometry.)

Actual bonus question. Can you say anything about what this formula would look like (e.g. its degree)?

This begs the natural question, does there exist a natural moduli space $X$ of $9$ points so that our map $X \to$ $9$th point is actually a map and not just a rational thing? We could start with $\text{Hilb}^8 \,\mathbb{P}^2$ and figure out how we need to modify it.

At this point, we could remark the following. Presumably, it would be an open subscheme of $\text{Hilb}^8\,\mathbb{P}^2$? We can not get general position already as an open subscheme of $(\mathbb{P}^2)^8$ because we would have to account for infinitely many closed things to throw out. but perhaps we could get this to work on passing to the Hilbert scheme. Actually intuitively, probably not – we would probably have to add extra parameters or something to parameterize all the possible things we would need to throw out? But it definitely seems possible.

Or wait, linear dependence is definitely finitely encodable, so we can get finitely many things to throw out already in in the $(\mathbb{P}^2)^8$ case. Obviously, since it is a rational map, it is defined on an open subset. Basic degree questions like "if we fix $7$ points, and a target point, what is the locus of the possible $8$th point" can be addressed in that context geometrically too. So does the Hilbert scheme buy us anything? Obviously, the symmetry of the point arrangements and stuff – so we guess that is a goal in itself. And it would be useful, for example, if we wanted to consider "families" of points over something, the Hilbert scheme would allow us to see the possible behaviors more cleanly. We do not know if we could convince someone else that the Hilbert scheme adds something here, though.

Also, but wait, the weird blowup behavior of the Hilbert scheme of points only happens at diagonals, so in this case, the open subscheme of the Hilbert scheme of points would literally just be the symmetric quotient of $(\mathbb{P}^2)^8$, so in particular, any behavior of it is recoverable in an extremely simple way from the latter.

With regards to those remarks, the entire point of using the Hilbert scheme is we want to understand what happens at nongeneric points to do a lot of things (e.g. intersection theory) because we need properness, which we are going to get from modifying the Hilbert scheme in some, hopefully small, way.

At this point, we could remark that we do not see an obvious way to extend the rational map to all of $\text{Hilb}$. Conceivably, $\text{Hilb}$ might help, since the blowups give us extra information about how points approach the degenerate positions, but we do not know how to actually do it.

With regard to that remark, right, the rational map does not extend to $\text{Hilb}$, and that is why we need to modify $\text{Hilb}$ first.

At this point, we could remark yeah true, but we do not see why a priori modifying $\text{Hilb}$ to have this behavior is easier than just modifying $(\mathbb{P}^2)^8$.

With regard to that remark, we think $\text{Hilb}$ is the correct starting point because the intersection of $2$ different cubic curves is a length $9$ subscheme.

Question. Hopefully building on the blath I have above, is there anything more we can say about what this formula looks like?

  • $\begingroup$ Just to make sure I actually understand the question: you have a rational map $(\mathbb P^2)^8 \dashrightarrow \mathbb P^2$, and you want to know how to resolve the map by some blow-up $X \to (\mathbb P^2)^8$, and also want a geometric interpretation of the points of $X$? $\endgroup$ – user47305 Feb 17 '16 at 3:20
  • 10
    $\begingroup$ As far as actually finding the formula, see this nice paper: arxiv.org/abs/1405.6438 (some version of this was known to Cayley) $\endgroup$ – user47305 Feb 17 '16 at 3:24
  • 1
    $\begingroup$ I've long wondered if there's an elegant and simple ruler-only construction of this map. $\endgroup$ – Gro-Tsen Feb 24 '16 at 5:16

This is very much a question of recent research, solved with varying degrees of generality in the papers listed below. Here is a summary.

The rational map $\mathbb{P}^{2[8]}\dashrightarrow \mathbb{P}^2$ from the Hilbert scheme of $8$ points in $\mathbb{P}^2$ is the map corresponding to an extremal effective divisor on the Hilbert scheme. This suggests that you should run the minimal model program for the Hilbert scheme, studying the birational modifications that arise on the way from the standard model of the Hilbert scheme until you reach the model where the map to $\mathbb{P}^2$ becomes a morphism. (Note that the Hilbert scheme is a Mori dream space, so this goal is actually realistic.)

Along the way, you'll find that the models you obtain can be realized as moduli spaces of certain Bridgeland semistable objects. In the general case of $n$ points, the Hilbert scheme will be replaced by a Grassmannian bundle over a moduli space of representations of a Kronecker quiver; in your $n=8$ point case this simplifies to a $G(2,9)$-bundle over $\mathbb{P}^2$, as follows.

Consider the space consisting of pairs $(p,\Lambda)$ where $p\in\mathbb{P}^2$ is a point and $$\Lambda\in G(2, H^0(\mathcal{O}_{\mathbb{P}^2}(3) \otimes I_p))$$ is a two-plane in the space of cubics passing through $p$. Thus, this space is a $G(2,9)$-bundle over $\mathbb{P}^2$. It is evidently birational to the Hilbert scheme, since a general $(p,\Lambda)$ determines a length $8$ scheme residual to $p$ in the intersection of the cubics in $\Lambda$. And, the forgetful map to $\mathbb{P}^2$ is your Cayley-Bacharach map.

The references for more general questions of this type are:

Daniele Arcara, Aaron Bertram, Izzet Coskun, and Jack Huizenga, The minimal model program for the Hilbert scheme of points on $\Bbb{P}^2$ and Bridgeland stability, Adv. Math. 235 (2013), 580--626. (Explicitly runs the MMP for $n\leq 9$ points, so your question is a special case.)

Coskun, H., Woolf, "The effective cone of the moduli space of sheaves on the plane."

H., "Effective divisors on the Hilbert scheme of points in the plane and interpolation for stable bundles."

Also there is a survey which might help get you started:

Izzet Coskun and Jack Huizenga, The birational geometry of the moduli spaces of sheaves on $\Bbb P^2$, Proceedings of the Gökova Geometry-Topology Conference 2014 (2015), 114--155.

  • 5
    $\begingroup$ Beautiful answer; I would say the only thing not covered that one could dream of would be a space $X$ with birational maps $f: X\to \mathbb{P}^{2[8]}$ and $g: X\to Y$ (where $Y$ is your moduli space) such that $X, f, g$ have modular interpretations. But maybe this is too much to ask for. $\endgroup$ – Daniel Litt Feb 17 '16 at 4:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy