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Let $d$ be a positive integer and $0 \leq k \leq 3d-2$ another integer.

Question 1. If $p_1,...,p_k$ are $k$ general points in ${\mathbb P}^2$, is it known in general whether the space parametrizing degree $d$ plane rational curves passing through $p_1,...,p_k$ is irreducible?

I think it's clear that the answer is yes, but I don't know if this is supposed to be easy to prove or not. Of course, if $k=0$, this is trivial, but I'm more curios about $k$ at the opposite end of the spectrum.

I was imprecise about the specific type of compactification we're using (e.g. it could be a closed subvariety of the Severi variety, or a geometric fiber of the evaluation map $ev:\overline{\mathcal M}_{0,k}({\mathbb P}^2,d) \to ({\mathbb P}^2)^k$), since this seems rather irrelevant.

Some thoughts. If we think in terms of the evaluation map -- since the source is smooth (at least as a stack) -- it looks like probably it should be enough to prove that the fibers are connected. If, on the other hand, we think in terms of Severi varieties, then imposing "incidence with points" conditions amounts to taking hyperplane sections, but I'm not sure if that's useful, since (closures of) Severi varieties are quite singular and these hyperplane sections are not general.

Question 2 (Generalization). What if instead we take $k \leq 3d+g-2$ and ask the same question about curves of geometric genus $g$? Does this follow somehow from the irreducibility of Severi varieties?

EDIT: Marked as solved, thanks to Jason Starr's answer below. However, comments on the second question are still appreciated!

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  • $\begingroup$ For $k$ quite close to the upper bound, you should have just a finite number of rational curves, so a disconnected space $\endgroup$
    – Giulio
    Commented Sep 30, 2016 at 8:16
  • $\begingroup$ For $k=3d-1$, you do get a finite number of rational curves. But for smaller $k$, the dimensions of all irreducible components should be precisely $3d-1-k \geq 1$ if the points are general and at least that if they're arbitrary. $\endgroup$
    – azaha89
    Commented Sep 30, 2016 at 9:01
  • $\begingroup$ @azah89. If you e-mail me, I can give you more details. For small values of $k$, this also follows from Damiano Testa's thesis: consider the blowing up of $\mathbb{P}^2$ at $k$ general points as a del Pezzo surface, and use Testa's theorem. There might also be some work on this type of thing by Shevchishin. $\endgroup$ Commented Sep 30, 2016 at 10:31

1 Answer 1

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Yes, for every integer $k$ with $0\leq k \leq 3d-2$, the following evaluation morphism is surjective and the generic fiber is geometrically irreducible, $$\text{ev}_{1,2,\dots,k}:\overline{\mathcal{M}}_{0,k}(\mathbb{P}^2,d)\to (\mathbb{P}^2)^k.$$ One proof uses the version of Bertini's connectedness theorem as formulated by Cristian Minoccheri in Proposition 3.1 of "On the Arithmetic of Weighted Complete Intersections of Low Degree".

Bertini's Connectedness Theorem [Cristian Minoccheri] Let $h:M\to X$ be a projective morphism of $k$-schemes with $X$ a smooth variety that is algebraically simply connected, and with $M$ a normal, quasi-projective variety. If the closed subscheme of $M$ where $h$ is not smooth has codimension at least $2$, then $h$ is surjective and the generic fiber is geometrically irreducible.

In this theorem, it suffices for $M$ to be normal and "pure" in the sense of SGA2. In particular, this is true for the coarse moduli space of a smooth Deligne-Mumford stack, as with $\overline{\mathcal{M}}_{0,k}(\mathbb{P}^2,d)$. Also, we may as well assume that $k$ equals $3d-2$, since the case of $k\leq 3d-2$ follows from the case of $k=3d-2$.

It is convenient to begin with a result about the branching behavor of the evaluation morphism when $k$ equals $3d-1$. In this case, the evaluation morphism is surjective and generically etale. By the Purity Theorem from SGA2, the non-smooth locus of the evaluation morphism in $\overline{\mathcal{M}}_{0,3d-1}(\mathbb{P}^2,d)$ is a Cartier divisor. We know the divisor class group of this moduli space. In particular, the restriction map to the boundary locus $\Delta_{(3,\{1,\dots,8\})(d-3,\{9,\dots,3d-1\}}$ is injective. Thus, by analyzing the case that $d=3$ and $3d-1=8$, the non-smooth locus has multiplicity $1$ (i.e., simple branching) and every generic point parameterizes a stable map with one cusp.

By Minoccheri's version of Bertini's theorem, to prove that the generic fiber of $\text{ev}_{1,2,\dots,k}$ is geometrically irreducible, it suffices to prove that over codimension one points of $(\mathbb{P}^2)^k$, the fiber is generically nonreduced. Because $k<3d-1$, every fiber of the evaluation morphism has dimension $\geq 1$. Because $k$ equals $3d-2$, which is $\geq 2$, by the famous Bend-and-Break theorem, every irreducible component of every fiber of the evaluation morphism has nonempty intersection with the boundary divisor $\Delta$ of $\overline{\mathcal{M}}_{0,k}(\mathbb{P}^2_k,d)$. Thus, it suffices to prove that over codimension $1$ points of $(\mathbb{P}^2)^{3d-2}$, the intersection of the fiber with the boundary $\Delta$ is generically nonreduced.

The boundary divisors are indexed by partitions $d=d'+d''$ and a corresponding partitions of $P'\sqcup P'' = \{1,\dots,3d-2\}$. In order for that boundary divisor to dominate a divisor in $(\mathbb{P}^2)^{3d-2}$, the sizes of the partition sets must be either $(3d',3d''-2)$, $(3d'-1,3d''-1)$, or $(3d'-2,3d'')$. The first and last are symmetric: one component has degree $e$ and contains $3e$ of the marked points. For degree $e$, genus $0$ stable maps, for $k=3e$, the evaluation morphism $ev_{3e}$ is generically an embedding to a Cartier divisor $D$ in $(\mathbb{P}^2)^{3e}$. For a point of $(\mathbb{P}^2)^{3d-2}$, it is already a codimension one condition for the image point in $(\mathbb{P}^2)^{3e}$ to be contained in $D$. Thus, the remaining $3d-3e-2$ points are unconstrained. So the fiber is a product of the reduced fiber of $\text{ev}_{3e}$ (just one reduced point) and the reduced fiber of $ev_{1,\dots,3(d-e)-2}$ for the component of degree $d-e$.

The main case to consider is when the two partition sets have sizes $3d'-1$ and $3d''-1$. In this case, in order for a stable map in the boundary to lie over a codimension one point of $(\mathbb{P}^2)^{3d-2}$ and also to be a nonsmooth point of $\text{ev}_{1,\dots,3d-2}$, the restriction of the stable map to precisely one of the two components of the domain, say the component with degree $d'$, must be a ramification point for the evaluation morphism $\text{ev}_{1,\dots,3d'-1}$. But we know what are these ramification points. In particular, so long as the other irreducible component is generic, then the union is a smooth point of $\text{ev}_{1,\dots,3d-2}$.

This is the same deformation theory analysis as in my paper with Graber and Harris about section of rationally connected fibrations. So long as the stable map is unramified at the node where the two components of the domain intersect, then the normal sheaf of the stable map restricted to each component equals the normal sheaf of that component "twisted up" by the degree $1$ invertible sheaf of the node as a reduced Cartier divisor on that component. Since the normal sheaf of the component was just shy of being $(3d'-1)$-globally generated (by the analysis of ramification points), twisting up by a degree $1$ invertible sheaf is enough to kill the $H^1$.

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  • $\begingroup$ Thank you very much -- this is exactly what I was looking for! Minoccheri's version of Bertini is clearly the ingredient on which I couldn't put my finger. It's interesting how the simple connectivity of $({\mathbb P}^2)^k$ is, in some sense, "the" crucial ingredient. [This is probably too obvious to point out, but: in the fifth paragraph "generically nonreduced" should be read "generically reduced", right?] $\endgroup$
    – azaha89
    Commented Sep 30, 2016 at 16:49

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