Basically, I had a thought about a way to think of the moduli space of hyperelliptic curves. I'm sure it's wrong most likely, but I was hoping someone could maybe point out the flaw in my reasoning.

So one can think of a hyperelliptic curve $C$ as a double-cover $C \to \mathbb{P}^{1}$ of the Riemann sphere, branched over $2n$ points. Thus, there are $2g+2=2n$ dimensions, where $g$ is the genus, and then normally we quotient by $\rm{Aut}(\mathbb{P}^{1})= \rm{PGL}(2, \mathbb{C})$, reducing the dimension by three. However, I want to not quotient by $\rm{Aut}(\mathbb{P}^{1})$. So I guess what I'm interested in is more of a "parameter space" than a genuine moduli space in the algebraic geometry sense.

The above $2n$ bits of data can be thought of as choosing $n$ "double branch points" on $\mathbb{P}^{1}$. The remaining $n$ bits of data correspond to $n$ complex numbers $S_{1}, \ldots, S_{n}$ which I think one can take as periods of the curve. I want to think of the double branch point as splitting off into two branch points, so the $S_{i}$ are simply a single complex parameter giving how this splitting off occurs, (i.e. it happens symmetrically about the double branch point, so maybe angle and modulus or something)

Now, the moduli space of such curves (or perhaps parameter space, more accurately) is some $2n$-dimensional space, call it $\mathcal{M}_{2n}$. I'm hoping I can think of $\mathcal{M}_{2n}$ as a fibration in the following sense. Let $M_{n}$ be the $n$-dimensional "singular" base, the points of which determine a choice of location of the $n$ double branch points. The $n$-dimensional fiber over some point of $M_{n}$ would then essentially be parameterized by the periods $S_{1},\ldots, S_{n}$. Intuitively, sitting at a point in $M_{n}$ all the periods vanish, and you have a "maximally singular" curve. As you move up into the fiber, at least one, possibly all of the periods are non-zero.

Is there anything above that can be made coherent or rigorous in any sense? I'm not too confident, but ultimately I need something of this form so maybe there's a closely related topic I can read up on.

EDIT: I think I've figured this out, with the help of all the comments below. Figured I may as well edit. Sorry I didn't explain very well in OP! So if you take $2n$ points thrown randomly on $\mathbb{P}^{1}$, you can imagine decompactifying to $\mathbb{C}$, and connecting the points pairwise with line segments. The data of the $2n$ points is equivalent to the coordinates of the midpoints of each pair, as well as the coordinates of one of the two of the pair. This is what I was (horribly) trying to explain in this post. I think you can explore the full moduli space by giving the midpoints, as well as the coordinates (what I called $S_{i}$) of one of the two points. Perhaps this isn't quite right.

  • 1
    $\begingroup$ I don't understand your question. For genus $\ge 2$, once you choose your $2g + 2$ points, that's it; there aren't any further choices to make. The resulting parameter space is the configuration space of $2g + 2$ points in $\mathbb{P}^1$. Where are these $S_i$ coming from? $\endgroup$ – Qiaochu Yuan Jun 21 '16 at 3:58
  • 1
    $\begingroup$ ...and if the points are distinct, then it does fiber over the configuration space of n points in P^1, iirc. $\endgroup$ – Mariano Suárez-Álvarez Jun 21 '16 at 4:01
  • $\begingroup$ @QiaochuYuan Right, but I want to have the $n$ double points correspond to when all the $S_{i}$ vanish. This would give a point in what I called $M_{n}$ above. Then, as we turn on the $S_{i}$, these double points split off into two. So the $S_{i}$ are simply a single complex parameter for each of the initial $n$ double points which describes how they split up. So for all $S_{i}$ nonzero, there will be the $2n$ branch points. But if some of the $S_{i}$ vanish, then we have singularities. $\endgroup$ – Benighted Jun 21 '16 at 4:06
  • $\begingroup$ @MarianoSuárez-Alvarez Thank you, that's encouraging. What if some, but not all, of the $2n$ points collide pairwise? Is it still possible to think of this as a fibration? Because if they all collide pairwise, that would correspond to the point in the base. So it seems like if some, but not all collide, this might also be okay. $\endgroup$ – Benighted Jun 21 '16 at 4:19
  • $\begingroup$ @spietro: I don't understand what you mean by "these double points split off into two." What is that supposed to be accomplishing at the level of hyperelliptic curves? $\endgroup$ – Qiaochu Yuan Jun 21 '16 at 4:42

I am not quite sure what you are asking for, at any rate the comment I wanted to make is a bit too long, so I thought will write an `answer'.
Choosing $2n$ (or for that matter any $k$) points in projective line corresponds to a point in $S^{2n}(\mathbb{P}^1)=\mathbb{P}^{2n}$. Thus, we have the incidence variety $\Gamma\subset \mathbb{P}^{2n}\times\mathbb{P}^1$. So, if you want to do what you want in a uniform manner, you should take a double cover of $\mathbb{P}^{2n}\times\mathbb{P}^1$ ramified along $\Gamma$. But, alas, this is not possible, since to do so would require $\Gamma$ to be two times a line bundle. But $\Gamma$ is of type $(1,2n)$ and so not twice a line bundle. So, I do not think your ideas can lead to a satisfactory procedure for taking double covers uniformly. May be I have misunderstood.

  • $\begingroup$ Thanks for your answer. Given that choosing $2n$ points on the projective line determines a hyperelliptic curve, it seems like you're saying that the parameter space of hyperelliptic curves is $\mathbb{P}^{2n}$. If this is indeed the case, what I'm interested in doing is trying to fiber this space over the configuration space on $n$ points on the projective line. In Mariano's comment above, it seems like this might be possible. I think I'm missing how the latter part of your comment applies, admittedly, I'm not yet terribly fluent in algebraic geometry. $\endgroup$ – Benighted Jun 21 '16 at 16:30
  • $\begingroup$ I think your answer might also imply that the configuration space of $n$ points on the projective line is $\mathbb{P}^{n}$. So maybe I'm trying to fiber $\mathbb{P}^{2n}$ over $\mathbb{P}^{n}$? $\endgroup$ – Benighted Jun 21 '16 at 16:36
  • 1
    $\begingroup$ No, the parameter space is an open subset of $\mathbb{P}^{2n}$, namely the open subset where $\Gamma\to\mathbb{P}^{2n}$ is etale. The other points will not give a smooth curve and anyway, it can not be done in a uniform way at those points. $\endgroup$ – Mohan Jun 21 '16 at 16:47
  • $\begingroup$ I see, but for the points in the open subset of $\mathbb{P}^{2n}$ perhaps such a fibering over $\mathbb{P}^{n}$ is possible? $\endgroup$ – Benighted Jun 21 '16 at 16:53
  • $\begingroup$ I am sorry, I haven't understood what fibering you are looking for. What property do you need for it? Your $S_i$s do not make much sense for me. $\endgroup$ – Mohan Jun 21 '16 at 17:25

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.