1
$\begingroup$

Let us denote by $Mor_3(\mathbb{P}^1,\mathbb{P}^3)$ the spaces of degree three morphisms $f:\mathbb{P}^1\rightarrow\mathbb{P}^3$, $$f(x_0,x_1)=[f_0(x_0,x_1):f_1(x_0,x_1):f_2(x_0,x_1):f_3(x_0,x_1)]$$ where $f_0,f_1,f_2,f_3$ do not have common factors. Then $Mor_3(\mathbb{P}^1,\mathbb{P}^3)$ sits inside $\mathbb{P}^{15}$ as a dense open subset.

Do there exist "good" compactifications (different from $\mathbb{P}^{15}$) of $Mor_3(\mathbb{P}^1,\mathbb{P}^3)$? Here by "good" I mean for instance that the boundary divisor is simple normal crossing.

$\endgroup$
7
  • 2
    $\begingroup$ Yes: Kontsevich's moduli space of stable maps. A standard reference is Notes on stable maps and quantum cohomology, by Fulton-Pandharipande, in the Proceedings of the SantaCruz conference. $\endgroup$
    – abx
    Jan 31, 2018 at 15:37
  • $\begingroup$ The specific moduli space of stable maps for your case is usually denoted $\overline{\mathcal{M}}_{0,0}(\mathbb{P}^1\times \mathbb{P}^3,\beta)$, where $\beta$ is the curve class on $\mathbb{P}^1\times \mathbb{P}^3$ whose degree with respect to $\text{pr}_1^*\mathcal{O}_{\mathbb{P}^1}(1)$, resp. $\text{pr}_2^*\mathcal{O}_{\mathbb{P}^3}(1)$, equals $1$, resp. $3$. As a Deligne-Mumford stack this is smooth, and theboundary is a simple normal crossings divisor. However, the coarse moduli space is singular. $\endgroup$ Jan 31, 2018 at 16:28
  • $\begingroup$ Thanks a lot. I think this is exactly what I am looking for. Is it know if it is $\mathbb{Q}$-factorial and its Picard number? I think since the stack is DM and smooth the coarse moduli space should be normal and $\mathbb{Q}$-factorial. $\endgroup$
    – user117617
    Jan 31, 2018 at 16:41
  • 1
    $\begingroup$ The coarse moduli space is normal and $\mathbb{Q}$-factorial (away from characteristics $2$ and $3$). The Picard groups of stable map spaces, such as this one, are worked out in an article by Pandharipande. The list of generators is extended to generators for the entire Chow ring in the Ph.D. thesis of Dragos Oprea. In this case, if memory serves, the $\mathbb{Q}$-Picard group is generated by the images of the three generators of $\text{CH}^2(\mathbb{P}^1\times \mathbb{P}^3)\otimes \mathbb{Q}$ together with the divisor classes of the three irreducible components of the boundary. $\endgroup$ Jan 31, 2018 at 17:25
  • $\begingroup$ Actually $\text{CH}^2(\mathbb{P}^1\times \mathbb{P}^3)\otimes \mathbb{Q}$ has dimension $2$ as a $\mathbb{Q}$-vector space, so it appears that the Picard rank equals $2+3=5$. $\endgroup$ Feb 1, 2018 at 0:28

1 Answer 1

1
$\begingroup$

I started writing an answer about the generators and relations for the $\mathbb{Q}$-Picard group of the stack of stable maps of genus $0$ curves to an arbitrary projective homogeneous variety, but it quickly got too long. So here is an explanation of the extra divisor class relation that I was missing in my comments. This uses Lemma 5.2 of my article with A. J. de Jong.

MR3644251
de Jong, A. J.(1-CLMB); Starr, Jason(1-SUNYS)
Divisor classes and the virtual canonical bundle for genus 0 maps.
Geometry over nonclosed fields, 97–126,
Simons Symp., Springer, Cham, 2017.
https://arxiv.org/abs/math/0602642

The $\mathbb{Q}$-vector space $\text{CH}^2(\mathbb{P}^1\times \mathbb{P}^3)_{\mathbb{Q}}$ is generated by two generators $g_1=\text{pr}_1^*c_1(\mathcal{O}(1))\cdot \text{pr}_2^*c_1(\mathcal{O}(1))$ and $g_2=\text{pr}_2^* c_1(\mathcal{O}(1)^2$. Those give two divisor classes on the moduli space $\overline{\mathcal{M}}_{0,0}(\mathbb{P}^1\times \mathbb{P}^3,\beta)$. However the divisor class relation from Lemma 5.2 proves that, modulo linear combinations of boundary divisors, these two divisors are in the span of the two divisor classes that are pushforwards via $\pi$ from the universal family of stable maps, $$\pi:\mathcal{C}\to \overline{\mathcal{M}}_{0,0}(\mathbb{P}^1\times \mathbb{P}^3,\beta),\ \ f:\mathcal{C}\to \mathbb{P}^1\times \mathbb{P}^3,$$ of the codimension $2$ classes $f^*\text{pr}_1^*c_1(\mathcal{O}(1))\cdot c_1(\omega_\pi)$ and $f^*\text{pr}_2^*c_1(\mathcal{O}(1))\cdot c_1(\omega_\pi)$. Finally, when we restrict on the complement of the boundary divisors, the morphism $$(\text{pr}_1\circ f,\pi):\mathcal{C}\to \mathbb{P}^1\times \overline{\mathcal{M}}_{0,0}(\mathbb{P}^1\times \mathbb{P}^3,\beta),$$ is an isomorphism. In particular, on this open subset, the pullback $f^*\text{pr}_1^*\mathcal{O}(-2)$ is isomorphic to $\omega_\pi$. Thus, modulo the boundary, the $\pi_*$-pushforward of $f^*\text{pr}_1^*c_1(\mathcal{O}(1))\cdot c_1(\omega_\pi)$ is a $\mathbb{Q}$-multiple of the $\pi^*$ pushforward of $f^*\text{pr}_1^* c_1(\mathcal{O}(1))^2$. However, on $\mathbb{P}^1$, the square of $c_1(\mathcal{O}(1))$ equals $0$. Thus, modulo the boundary, the two cycle classes $g_1$ and $g_2$ map to divisor classes that are $\mathbb{Q}$-linearly dependent. So, together with the $3$ boundary divisor classes, this gives $4$ divisor classes, not $5$ as I wrote in the comment above.

$\endgroup$

Your Answer

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