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I am trying to understand some single steps in this recap of Bernhard Riemann's original proof by Alexandre Eremenko that the moduli space $M_g$ of compact Riemann surfaces with genus $g \ge 2$ has dimension $3g-3$ (or in original context the $3g-3$ 'degrees of freedom' are what Riemann called 'moduli')

Riemann combines what is called Riemann-Roch and Riemann-Hurwitz nowadays. He considers the dimension of the space of holomorphic maps of degree $d$ from the Riemann surface of genus $g$ to the sphere. He computes this dimension in two ways. By Riemann-Roch this dimension is $2d-g+1$, for a fixed Riemann surface. (Indeed, Riemann-Roch says that the dimension of the space of such functions with $d$ poles fixed is $d-g+1$ (when $d\geq 2g-1$ which we may assume), but these poles can be moved, so one has to add $d$ parameters).

On the other hand, such a function has $2(d+g-1)$ critical points by Riemann-Hurwitz. Generically, the critical values are distinct, and can be arbitrarily assigned, and this gives the dimension of the set of all such maps on all Riemann surfaces of genus $g$.

So the space of all Riemann surfaces of genus $g$ must be of dimension $$2(d+g-1)-(2d-g+1)=3g-3.$$

There are essentially two things I struggle with.

Point 1: Why the dimension of the space of holomorphic maps of degree $d$ from the Riemann surface $S$ of genus $g$ to the sphere equals $2d-g+1$?

As far as can follow we are going to apply RR. Let $h$ any holomorphic map $S \to \mathbb{P}^1$ of degree $d$ which we regard as meromorphic function $h$ which behaves compatible with degree $d$ in it's poles and zeros, set $(h) :=D$ and apply RR to it. RR tells $l(D)= d -g+1$ where $l(D)$ is the $C$-dimension of meromorphic functions $f$ on $S$ such that all the coefficients of the divisors $(f) + D$ are non-negative.

Eremenko argues then that the additional $d$ dimensions we obtain because 'the poles of can be moved, so one has to add $d$ parameters.'

I not understand why moving of the $d$ poles gives additionally exaxt $d$ parameters.

Point 2: Why the difference

$$ 2(d+g-1)-(2d-g+1) $$

between the number of critical points of these moromorphic functions (that's Riemann-Hurwitz-Thm) and the counted dimension of the space of holomorphic maps of degree $d$ from point 1 gives exactly the dimension of the moduli?

user267839
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1 Answers1

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  1. To specify a degree $d$ map $f: X \to \mathbb{P}^1$, one first chooses an effective divisor $D$ of degree $d$ to specify the poles of $f$. Since $D$ has support consisting of $d$ points, this amounts to $d$ choices. So this by this "the poles can be moved" comment, he means that choosing a different divisor $D'$ will result in a different map with poles given by the support of $D'$ instead. Then one computes $\ell(D)$ as you have described above to get the dimension of the space of degree $d$ maps $f: X \to \mathbb{P}^1$ with $\renewcommand{\div}{\operatorname{div}} \div_\infty(f) = D$.

  2. We want to compute the dimension of $\mathcal{M}_g$, the moduli space of curves of genus $g$, but what we've really done is computed the dimension of the moduli space of pairs $(X,f)$ where $X$ is a curve of genus $g$ and $f$ is a degree $d$ map $X \to \mathbb{P}^1$. This related moduli space is called a Hurwitz space, denoted $\DeclareMathOperator{\Hur}{Hur} \Hur_{d,g}$. In order to remove this choice of a map $f: X \to \mathbb{P}^1$, we subtract off the dimension of the space of degree $d$ maps $X \to \mathbb{P}^1$ which was computed above.

    You can interpret this using the projection \begin{align*} \pi: \Hur_{d,g} &\to \mathcal{M}_g\\ (X,f) &\mapsto X \, . \end{align*} We've computed the dimension of $\Hur_{d,g}$ and, for a given $X$, the dimension of the fiber $\pi^{-1}(X)$. Assuming $\pi$ is sufficiently "nice", we should have \begin{align*} \dim(\mathcal{M}_g) = \dim(\Hur_{d,g}) - \dim(\pi^{-1}(X)) = 2d + 2g - 2 - (2d - g + 1) = 3g-3 \, , \end{align*} as desired.

This is covered in greater detail in chapter 2, section 3 (p. 255) of Principles Of Algebraic Geometry by Griffiths and Harris. Jarod Alper also recently covered this in a lecture for a course he's teaching, which you can watch here. (See here for the dimension calculation you've asked about.)

Viktor Vaughn
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  • See also this post. – Viktor Vaughn Feb 22 '21 at 05:39
  • The point $2$ meanwhile presumably I understand. In summary there exist an dense open subset $U \subset \DeclareMathOperator{\Hur}{Hur} \Hur_{d,g}$ over which the restriction of the projection of $\pi$ to $\pi \vert U$ behaves 'like a fibre bundle' from dimensional conting point of view, thus the the formula $\dim(\mathcal{M}_g) = \dim(\Hur{d,g}) - \dim(\pi^{-1}(X))$, right? – user267839 Feb 23 '21 at 01:28
  • About the point $1$ I still not sure if I understand it completely. I would to explain the proble i see. We fix a Riemann surface $X$ and want to deduce that the dimension of the holom maps of degree $d$ $X \to P^1$ equals exactly $\ell(D)+ d$. As you said every such map has an associated divisor $D$ of deg $d$ which is supported in $d$ points (counted with multiplimities). And RR says $\ell(D)=d-g +1$. Then we vary these $d$ points slowly say from $D:= {p_1,...,p_d}$ to $D':={p'_1,...,p'_d}$. – user267839 Feb 23 '21 at 01:29
  • Since all these points live on $X$, which is a $1$-dim manifold, so the variation of every point $p \in X$ separately contributes exactly one 'degree of freedom' =$\mathbb{C}$-dimension, as far correct? But now I see a problem: – user267839 Feb 23 '21 at 01:30
  • essentially, up to now counted the dimensions like in case of a product vector spaces $V and W$: $\operatorname{dim}(V \times W)= \operatorname{dim}(V) + \operatorname{dim}(W)$ where the role of $L$ plays for a fixed divisor $D$ the space $L(D)$ and of $W$ the 'naked' variation of $d$ points (which corresponds to a $d$-dimensional space. But here I see a problem that it could be not so easy just to add these dimensions: In case of $V \times W$ that works since for $w \neq w' \in W$ the spaces $V \times {w} $ and $V \times {w'} $ are disjoint. – user267839 Feb 23 '21 at 01:33
  • In our case it could happen that for different divisors $D \neq D'$ the intersection $L(D) \cap L(D')$ could have a positive dimension: $\operatorname{dim}(L(D) \cap L(D')) > 0$(see below why), especially these might be not disjoint, so I'm not sure why we can here just add the dimensions of $L(D)$ (for arbitrary $D$) together with the 'space of variation of $d$ points' like in case of product space $V \times W$. – user267839 Feb 23 '21 at 01:33
  • (on why) recall that for a fixed divisor $D$ the space of holom maps of degree $d$ with resp $D$ is defined as $L(D):=\left{f: X\rightarrow {\mathbb {P}}^{1} {\mbox{ holomomorphic}},|,\left(f\right) -D \geq 0 \right}$ and therefore by this definition for different $D \neq D'$ it could happen that $\operatorname{dim}(L(D) \cap L(D')) > 0$. – user267839 Feb 23 '21 at 01:34
  • Do you see how to resolve this problem? – user267839 Feb 23 '21 at 01:39
  • I'm not sure I fully understand your objection, but I think the answer is that the sets you're concerned about are of strictly smaller dimension, so they don't affect the dimension. Note that not every function in $L(D)$ has pole divisor $D$. Indeed, there are many that don't, the constant functions, for instance. However the elements $L(D)$ with pole divisor $\neq D$ are in a Zariski-closed subset of strictly smaller dimension. So really we don't want to compute the dimension of $L(D)$, but of the dense open subset $U_D$ of $L(D)$ consisting of functions $f$ with pole divisor $D$. – Viktor Vaughn Feb 23 '21 at 03:02
  • These $U_D$ are disjoint for different choices of $D$ by definition. And indeed, any function in $L(D) \cap L(D’)$ can’t have pole divisor $D$ or $D’$: it must have some pole divisor $D’’$ made up of some combination of the supports of $D$ and $D’$. Let me know if that resolves your concern. – Viktor Vaughn Feb 23 '21 at 03:03
  • My objection was that in point 1 the calulation of the dimension of the holom maps of degree $d$ for fixed Riemann surface $X$ based on addition of the dimensions of the space $L(D)$ ($D$ could be choosen arbitrary in spirit of RR) and the dimension of $d$-dimensional "space" $Div(X, d)$ of degree $d$ divisors as if these both $C$-vectorspaces were regarded (at least locally) as parts of product space $L(D) \times D(X,d)$ but in order to assure that the dimension of the space of the holom maps $X \to P^1$ of degree $d$ coincides – user267839 Feb 24 '21 at 01:19
  • with $dim(L(D))+dim(Div(X, d))= d-g+1+g$ we have to check that (at least locally) the space of the holom maps of degree $d$ looks like the product $L(D) \times D(X,d)$. But in contast to 'classical' product space the problem here was that for two $D \neq D' \in D(X,d)$ the $L(D)$ and $L(D')$ may be not disjoint where in order to add dimensions like for products we have to be sure that $L(D) $ (which we identify with $L(D) \times {D} $) and $L(D') $ (which we identify with $L(D) \times {D'} $) has to be disjoint. That was my concern in previous comments. – user267839 Feb 24 '21 at 01:19
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    But I think now you precisely hit the essential point that answers my concern. We want to count the dimension generically, that's it's suffice to pass somehow to open dense sets $V \subset D(X,d)$ and family of open dense subsets $U_D \subset L(D)$ for $D \in V$, with property $U_D \cap U_{D'} = \emptyset$ for $D \neq D' \in V$, that was your argument, right? – user267839 Feb 24 '21 at 01:20
  • A nitpick: for fixed divisor $D$ with respect which topology $U_D$ is open dense in $L(D)$ in your considerations? You remaked that the elements of $L(D)$ with pole divisor $\neq D$ live in a a Zariski-closed subset of strictly smaller dimension. I'm confused here, do you mean that in the sense that you endow here each $L(D)$ with affine Zariski topology using the identification $L(D)= \mathbb{C}^{d-g+1}$ and by alg closeness identify elements in $\mathbb{C}^{d-g+1}$ with primes? – user267839 Feb 24 '21 at 01:46
  • Okay, glad that cleared up your doubt. Yes, that was my argument. Right, $L(D)$ is a vector space, so we can identify it with affine space $\mathbb{A}^{d-g+1}$ and equip it with the Zariski topology. I think the set of elements with pole divisor $\neq D$ are a linear subspace: by choosing a basis $h_1, \ldots, h_{d-g+1}$ of $L(D)$ with distinct pole orders, the set of such elements should correspond to setting some of the coefficients $a_i=0$ in a linear combination $a_1 h_1 + \cdots + a_{d-g+1} h_{d-g+1}$. – Viktor Vaughn Feb 24 '21 at 06:03