Let $X$ a complex curve and $x\in X$ a point.

We consider the space of effective divisors $D$ with fixed degree $d$, whic we know is isomorphic to $X^{d}/S_{d}$ where $S_{d}$ is the symmetric group.

Now, we consider the subspace of divisors $D$ with fixed degree $d$ such that:

$\sum\limits_{x untransversal}m_{x}(D)\leq N.$

where the untransversal points means that $m_{x}(D)\geq 2$.

Is this space open in $X^{d}/S_{d}$?

  • 1
    $\begingroup$ I dont really understand your question. First of all, there are perhaps too many question marks, isn't it? Second, space open in which other space? The Picard variety of degree d? $\endgroup$ – diverietti Aug 6 '13 at 9:00

First of all, I assume you're asking about all effective divisors of degree $d$ that satisfy that inequality; if not, then it doesn't make sense to talk about them in $X^d/S_d$.

Assuming this, then you're asking about all the $(x_1,\ldots,x_d)\in X^d/S_d$ such that some "coordinates" are the same, and such that the amount of coordinates that are the same do not add up to more than $N$. This space is open in $X^d/S_d$, for the following reason:

We have the morphism $p_{ij}:X^d/S_d\to X^2/S_2$, where $x_1+\cdots+x_d\mapsto x_i+x_j$. Let $\Delta$ denote the "diagonal" in $X^2/S_2$; that is $\Delta=\{2x:x\in X\}$. This is closed in $X^2/S_2$.

Assume first that we want the set of divisors such that the sum of untransversal points is equal to 2. This is then the closed set $\bigcup_{i,j}p_{ij}^{-1}(\Delta)$.

For similar reasons (which involve more complicated formulas using unions and intersections of the inverse images of these diagonals that I don't feel like writing right now), the set of divisors such that the sum of untransversal points is equal to a given $m\in\mathbb{N}$ (following a user's comment below, let's name these $\mbox{Unt}_m$) is also closed. The set of divisors such that the sum of untransversal points is less than or equal to $N$ is the complement of the union $\bigcup_{N\leq m\leq d}\mbox{Unt}_m$, and so is open.

| cite | improve this answer | |
  • $\begingroup$ it's not closed because it contains all points such that $m_{x}(D)\leq 1$ $\endgroup$ – prochet Aug 15 '13 at 8:22
  • $\begingroup$ This minor misunderstanding should be relatively easy to correct. For each $k$, let $\operatorname{Unt}_k(X)$ be the space of divisors on which the sum of untransversal points is equal to $k$. Then the space in question is the complement of the closed set $\bigcup_{N<k\leq d} \operatorname{Unt}_k(X)$. $\endgroup$ – S. Carnahan Aug 15 '13 at 13:25
  • $\begingroup$ Yes of course! Thanks for the correction, I'll change it now. $\endgroup$ – rfauffar Aug 15 '13 at 14:12

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.