I am studying the general theory of moduli spaces of stable maps, in particuar of the moduli spaces $\overline{M}_{0,n}(\mathbb{P}^r,d)$ of degree $d$ stable maps from a rational curve with $n$ marked points to $\mathbb{P}^r$.

I see that there are morphisms $f_I:\overline{M}_{0,n}(\mathbb{P}^r,d)\rightarrow \overline{M}_{0,n-|I|}(\mathbb{P}^r,d)$ forgetting the marked points indexed by subsets $I\subset\{1,...,n\}$, and a morphism $g:\overline{M}_{0,n}(\mathbb{P}^r,d)\rightarrow \overline{M}_{0,n}$ forgetting the data of the map.

Let $D_{f_I}$ and $D_{g}$ be the divisors on $\overline{M}_{0,n}(\mathbb{P}^r,d)$ associated to $f_I$ and $g$.

Are the morphisms $f_I$ and $g$ induced by the linear systems $H^0(\overline{M}_{0,n}(\mathbb{P}^r,d),D_{f_I})$ and $H^0(\overline{M}_{0,n}(\mathbb{P}^r,d),D_{g})$ or may it happen that they are induced by proper linear subsystems of $H^0(\overline{M}_{0,n}(\mathbb{P}^r,d),D_{f_I})$ and $H^0(\overline{M}_{0,n}(\mathbb{P}^r,d),D_{g})$ ?

Thank you very much in advance.

  • $\begingroup$ Every contraction between normal, projective varieties comes from a complete linear system. The spaces $\overline{M}_{0,n}(\mathbb{P}^r,d)$ are projective coarse moduli spaces of the algebraic stacks with finite diagonal $\overline{\mathcal{M}}_{0,n}(\mathbb{P}^r,d)$. These algebraic stacks are smooth (cf. Kontsevich's "Enumeration of Rational Curves via Torus Actions" or other sources). Thus the coarse moduli spaces are normal. The forgetful morphisms are contractions: the geometric fibers are integral schemes parameterizing $|I|$-tuples of points on a connected, projective curve. $\endgroup$ Commented Jun 7, 2017 at 10:31
  • $\begingroup$ A similar argument applies for the forgetful morphism $g$. Since the source and target are normal, to prove that $g$ is a contraction it suffices to prove that the geometric generic fiber is connected. The general fiber is birational to the projective space $\mathbb{P} H^0(\mathbb{P}^1,\mathcal{O}(d))^{\oplus (r+1)}$. $\endgroup$ Commented Jun 7, 2017 at 10:49
  • $\begingroup$ One final comment: we do know how to completely describe the nef cones of the spaces $\overline{M}_{0,n}(\mathbb{P}^r,d)$ in terms of the nef cones of the moduli space $\overline{M}_{0,m}$, cf. my papers with Izzet Coskun and Joe Harris. We also know the pullback maps under $f_I$ and $g$ relative to the tautological generators of the Picard groups. So it should be straightforward to find an explicit divisor class on $\overline{M}_{0,n}(\mathbb{P}^r,d)$ that is the pullback under $f_I$, resp., $g$, of an ample divisor class. $\endgroup$ Commented Jun 7, 2017 at 11:23
  • $\begingroup$ Thanks a lot. Just one last question. What do you mean exactly by contratction here? If I got it right you want the fibers to be connected. Do we need to assume that the fibers are of positive dimension? What if we have a birational morphism $f:X\rightarrow Y$ of normal projective varieties? I think that if $f$ is birational we must assume that it is not an isomorphism. $\endgroup$
    – user97096
    Commented Jun 7, 2017 at 15:31
  • $\begingroup$ For normal schemes, a contraction is a proper, finitely presented morphism $f:X\to Y$ such that the induced morphism $f^\#:\mathcal{O}_Y\to f_*\mathcal{O}_X$ is an isomorphism. Using Stein factorization, it suffices to prove that every geometric generic fiber is irreducible and reduced. $\endgroup$ Commented Jun 7, 2017 at 15:37

1 Answer 1


I am posting this as an answer because the comment thread is too long. A proper morphism of integral, Noetherian schemes, $f:X\to Y$, is a contraction if the sheaf homomorphism $f^\#:\mathcal{O}_Y\to f_*\mathcal{O}_X$ is an isomorphism. Typically we only talk about this when $X$, or at least $Y$, is normal, because then we have Zariski's Main Theorem and Stein Factorization. In particular, if $X$ and $Y$ are normal, then $f$ is a contraction if and only if the stalk of $f^\#$ at the generic point $\eta$ of $Y$ is an isomorphism, i.e., if and only if the $\kappa(\eta)$-algebra homomorphism, $$f^\#_\eta:\kappa(\eta) \to H^0(X_\eta,\mathcal{O}_{X_\eta}),$$ is an isomorphism. That holds if the geometric generic fiber of $f$ is integral.

Let $\mathcal{E}$ be any locally free $\mathcal{O}_Y$-module. Then by the Projection Formula, the natural $\mathcal{O}_Y$-module homomorphism, $$\mathcal{E}\otimes_{\mathcal{O}_Y} f_*\mathcal{O}_X \to f_*(f^*\mathcal{E}),$$ is an isomorphism. Thus, if $f$ is a contraction, then the adjunction homomorphism, $$\mathcal{E} \to f_*(f^*\mathcal{E}),$$ is an isomorphism. Thus, the associated map of global sections, $$\Gamma(Y,\mathcal{E})\to \Gamma(Y,f_*(f^*\mathcal{E})),$$ is an isomorphism. By the definition of pushforward, this is the same as the pullback map, $$\Gamma(Y,\mathcal{E})\to \Gamma(X,f^*\mathcal{E}).$$ In particular, for every invertible $\mathcal{O}_Y$-module $\mathcal{L}$, the adjunction homomorphism, $$\Gamma(Y,\mathcal{L})\to \Gamma(X,f^*\mathcal{L}),$$ is an isomorphism. Thus, in the classical language, the pullback of the complete linear system of $\mathcal{L}$ equals the complete linear system of $f^*\mathcal{L}$.

Regarding the specific morphisms $f_I$ and $g$, the moduli spaces involved are all projective, integral, and normal. Normality holds because these are coarse moduli spaces of smooth stacks. For a stack $\mathcal{X}$ that is regular and a dominant $1$-morphism to an integral algebraic space, $p:\mathcal{X}\to X$, by the universal property of the normalization, the morphism $p$ factors uniquely through the normalization $X^{\text{nor}}\to X$. Thus, if $p$ is initial among $1$-morphisms to algebraic spaces, i.e., if $p$ is a coarse moduli space of $\mathcal{X}$, then the normalization $X^{\text{nor}}\to X$ is an isomorphism. Thus the coarse moduli space of $\mathcal{X}$ is normal. Of course projectivity and integrality require other arguments.

Finally, because $f_I$ and $g$ are dominant morphisms of projective, integral, normal schemes, to prove that they are contractions it suffices to prove that the geometric generic fibers are integral. For each type of morphism, this is explained in the comments.


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