For a curve $C$ over a finite field, I am looking at the map $\phi: \operatorname{Hom}^d(C,\mathbb{P}^1) \to \operatorname{Sym}^d(C)$ where $\operatorname{Hom}^d(C,\mathbb{P}^1)$ are the functions of degree $d$ from $C \to \mathbb{P}^1$ in the $\operatorname{Hom}$-scheme $\operatorname{Hom}(C, \mathbb{P}^1)$, defined by $$\phi: f \mapsto [\sigma]\cdot \Gamma_f$$

i.e. we intersect the graph of $f$ in $C \times \mathbb{P}^1$ with the line $[\sigma]$ in $C \times \mathbb{P}^1$ (which is the inverse image of $\sigma$ under the projection to $\mathbb{P}^1$.

For example the $f$ in the picture would map to $\phi(f) = 2\cdot P_1 + 3\cdot P_2 + 2\cdot P_3 + 2\cdot P_4$ map f: C to P^1

I have looked at Kollár's Rational Curves on Algebraic Varieties and Chapter 9 of Nakajima's Lectures on Hilbert Schemes of Points on Surfaces however I can't find the information I need of this map.

For example, is $\phi$ flat? What is the dimension of the fiber of $\phi$? Etc. Etc.

Does anyone know of a reference for this topic?

EDIT: If we take $d > 2g$ we can view any point in $\operatorname{Sym}^d(C)$ as an effective divisor $D$ of degree $d$ of $C$. Then $D$ is base-point free and so we can construct a map $f: C \to \mathbb P^1$ with fiber $D$ above $0$ (or say, $\sigma$). So $\phi$ is surjective for $d > 2g$. In general, I only care for $d$ much greater than 0 as I want to study it's behavior as $d$ grows.

  • 1
    $\begingroup$ For an example of non-flatness, look at $C$ hyperelliptic of genus $g \geq 2$ with $d = 2$. Then the degree $2$ map $C \to \mathbb P^1$ is unique up to $\operatorname{PGL}_2$. Thus, the image of your map $\phi$ is the locus given by divisors $P+Q$ linearly equivalent to the unique $g^1_2$ (equivalently, $P+Q$ is a fibre of this unique degree $2$ map). Most pairs $(P,Q)$ do not satisfy this, so the image of $\phi$ is a strict closed subvariety. Hence, $\phi$ cannot be flat. $\endgroup$ Jun 14, 2017 at 13:49
  • $\begingroup$ @R.vanDobbendeBruyn As I am mostly interested in large $d$, we may assume that $d > 2g$ so that we can view any point $D \in \operatorname{Sym}^d(C)$ as an effective divisor and so $\phi$ is surjective. I'll edit this in. $\endgroup$
    – Krijn
    Jun 22, 2017 at 22:23

1 Answer 1


Remark. Note that $\operatorname{\mathbf{Hom}}(C,\mathbb P^1)$ is an open subscheme of $\operatorname{\mathbf{Div}}_{C \times \mathbb P^1}$, whereas $\operatorname{Sym}^d(C)$ is a component of $\operatorname{\mathbf{Div}}_C$. Specifically, a homomorphism $\phi \colon C \to \mathbb P^1$ corresponds to its graph, which is a section of a line bundle $\mathscr M$ on $C \times \mathbb P^1$. Since \begin{align*} \operatorname{\mathbf{Pic}}_{C \times \mathbb P^1} &= \operatorname{\mathbf{Pic}}_C \times \operatorname{\mathbf{Pic}}_{\mathbb P^1} \times \operatorname{\mathbf{Hom}}(\operatorname{\mathbf{Alb}}_C,\operatorname{\mathbf{Pic}}^0_{\mathbb P^1})\\ &= \operatorname{\mathbf{Pic}}_C \times \operatorname{\mathbf{Pic}}_{\mathbb P^1}, \end{align*} the line bundle $\mathscr M$ is of the form $\mathscr L \boxtimes \mathcal O(n)$ for some line bundle $\mathscr L$ on $C$ and some $n \in \mathbb Z$. Counting intersections with $\{c\} \times \mathbb P^1$, we see that $n = 1$. Restricting to $C \times \{\sigma\}$, we see that $\mathscr L = \phi^* \mathcal O(1)$. Thus, $\phi$ corresponds to a section $s$ of $\mathscr L \boxtimes \mathcal O(1)$.

Conversely, a section $s$ of $\mathscr L \boxtimes \mathcal O(1)$ comes from a morphism $\phi$ if and only if $\mathscr L$ is effective and $V(s)$ is irreducible. Indeed, since $\mathscr L \boxtimes \mathcal O(1)$ is primitive (not divisible), $V(s)$ is reduced. Hence, the dominant morphism $V(s) \to C$ is flat, since $V(s)$ is torsion-free over $C$ since it is integral. Since it is of degree $1$, it is an isomorphism.

The only other possibility for a section $s$ of $\mathscr L \boxtimes \mathcal O(1)$ is that it contains a vertical component. Indeed, the only decomposition of $\mathscr L \boxtimes \mathcal O(1)$ as a sum of effective divisors must have $\mathscr L' \boxtimes \mathcal O$ as one of its summands, all of whose sections are vertical.

The morphism $\phi \colon C \to \mathbb P^1$ and the pair $(\mathscr L, s)$ determine each other uniquely, where two pairs $(\mathscr L, s)$, $(\mathscr L', s')$ correspond if and only if there exists an isomorphism $\alpha \colon \mathscr L \stackrel \sim \to \mathscr L'$ taking $s$ to $s'$.

Remark. Thus, the map you give can be extended to the rational map \begin{align*} \sigma^* \colon \operatorname{\mathbf{Div}}_{C \times \mathbb P^1} &\dashrightarrow \operatorname{\mathbf{Div}}_C \\ Z &\mapsto Z \cap (C \times \{\sigma\}), \end{align*} which is defined away from the divisors containing $C \times \{\sigma\}$. We will focus on the locus $U$ of $\operatorname{\mathbf{Div}}_{C \times \mathbb P^1}$ of irreducible divisors in $\mathscr L \boxtimes \mathcal O(1)$ for some $\mathscr L$ of degree $d > 2g$. We get a commutative square $$\begin{array}{ccc} \operatorname{\mathbf{Div}}_{C \times \mathbb P^1} & \stackrel{\sigma^*}\dashrightarrow & \operatorname{\mathbf{Div}}_C \\ \downarrow & & \downarrow \\ \operatorname{\mathbf{Pic}}_{C \times \mathbb P^1} & \stackrel{\sigma^*}\rightarrow &\ \operatorname{\mathbf{Pic}}_C. \end{array}$$ Over the locus of $\operatorname{\mathbf{Pic}}_C$ where $\mathscr L$ is of degree $d > 2g$, the right vertical arrow is a $\mathbb P^r$-bundle, where $r = h^0(\mathscr L) - 1 = d - g$ (which does not depend on $\mathscr L$ but only on $d$). Similarly, over the same locus, the left vertical arrow is a $\mathbb P^{r'}$-bundle, where $r' = h^0(\mathscr L \boxtimes \mathcal O(1)) - 1$. By Künneth we have $r' = 2(r+1) - 1 = 2d - 2g + 1$.

Restricting the left hand side to the locus of the form $\mathscr L \boxtimes \mathcal O(1)$, the bottom map becomes an isomorphism. On the locus $U$ of irreducible divisors in $\mathscr L \boxtimes \mathcal O(1)$, the top map on each fibre is a morphism from an open in $\mathbb P^{2d-2g+1}$ to $\mathbb P^{d-g}$.

Remark. To describe the map more explicitly, let $[x:y]$ be coordinates on $\mathbb P^1$. For simplicity, assume $\sigma = [0:1]$. Then $H^0(\mathscr L \boxtimes \mathcal O(1))$ is given by $\{\lambda x + \mu y\ |\ \lambda,\mu \in H^0(\mathscr L)\}$, and a section $\lambda x + \mu y$ is mapped to the section $\mu \in H^0(\mathscr L)$. In particular, this is a linear map, so the fibres are linear spaces.

The locus of divisors $Z$ not containing $C \times \{0\}$ corresponds to the $\lambda x + \mu y$ with $\mu \neq 0$. This is the maximal set where the map on the projective spaces is defined, and there the fibres are $\mathbb A^{d-g+1}$ corresponding to the affine space of values of $\lambda$.

We are further restricting to divisors $Z$ not containing any vertical component; this is probably equivalent to linear independence of $\lambda$ and $\mu$, so this removes an $\mathbb A^1\setminus\{0\}$ from each fibre (corresponding to $\lambda = c \mu$). More canonically, the conditions that $\lambda$ and $\mu$ are linearly independent are cut out in $\mathbb P^{2d-2g+1}$ by the nonvanishing of certain minors.

Any further properties you want to deduce should follow from this description.

  • $\begingroup$ Forgive me my lack of knowledge on these topics, but it is not immediatly clear to me what $V(s)$ should be. $\endgroup$
    – Krijn
    Jun 28, 2017 at 15:05
  • $\begingroup$ I just meant the vanishing locus of $s$, i.e. the divisor cut out by the section $s$ of the line bundle $\mathscr L \boxtimes \mathcal O(1)$, cf. the correspondence between divisors and line bundles. (To make sense of this, you can choose local trivialisations of the line bundle, so that it locally becomes a vanishing locus of an actual function.) $\endgroup$ Jun 29, 2017 at 2:09

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