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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 $X^{d}/S_{d}$?

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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? –  diverietti Aug 6 '13 at 9:00
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1 Answer 1

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.

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it's not closed because it contains all points such that $m_{x}(D)\leq 1$ –  prochet Aug 15 '13 at 8:22
    
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 $\bigcup_{N<k\leq d} \operatorname{Unt}_k(X)$. –  S. Carnahan Aug 15 '13 at 13:25
    
Yes of course! Thanks for the correction, I'll change it now. –  Robert Auffarth Aug 15 '13 at 14:12
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