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This fact was apparently known to Riemann. How did Riemann think about this?

Bruno Joyal
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1 Answers1

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At that time, it was not known that there existed a variety of moduli of smooth curves, so his count was really a moduli count, in the latin sense. I think the strategy in Riemann's count was to count covers of the projective line by all smooth curves at a time, in two ways.

First way. Fix a smooth curve $C$. Let us study degree $d$ covers $C\to\mathbb P^1$. To give one of these is the same as to give two linearly independent sections of some $L\in \textrm{Pic}^d(C)$. Now, fix one such $L$, and assume $d$ is big enough (say $d>2g-2$). Then $h^0(L)=d-g+1$. This means that, having fixed $C$ and $L$, there is a $2(d-g+1)-1=2d-2g+1$dimensional family of degree covers $C\to\mathbb P^1$ attached to $L$. (The $-1$ is just to identify $(s,t)$ with $(as,at)$, for $a\in\mathbb C^\times$). Now, $\textrm{Pic}^d(C)\cong \textrm{Pic}^0(C)$ has dimension $g$. So, let us vary the line bundle $L$ and the curve to get the count we want: we get $$(2d-2g+1)+g+\dim M_g=2d-g+1+\dim M_g.$$

Second way. Using (the currently named) Riemann-Hurwitz formula. Any cover has then $2g+2d-2$ branch points and thanks to Riemann's Existence Theorem, which classifies covers of the projective line, we know that no correction term is needed, so that $$2d-g+1+\dim M_g=2g+2d-2,$$ whence $\dim M_g=3g-3$.

You may want to look up to E. Looijenga lecture notes, where he explains this in another, yet similar fashion in the section "Riemann moduli count".

By the way, since the title of your question starts with "what is the best way", I cannot help doing this remark: from deformation theory, the Zariski tangent space of $M_g$ at a point $[C]$ is isomorphic to $$H^1(C,\mathcal T_C),$$ which in turn - by Serre duality - is isomorphic to $$H^0(C,\omega_C^{\otimes 2})^\prime,$$ and by Riemann-Roch the latter has dimension $$h^0(C,\omega_C^{\otimes 2})=2(2g-2)+1-g=3g-3,$$ having observed that $\deg \omega_C^{\otimes 2}>2g-2$.

Brenin
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  • Lovely answer!$\ $ –  Oct 04 '13 at 10:48
  • Fantastic answer Brenin, thanks for taking the time to write it up! – Bruno Joyal Oct 04 '13 at 14:27
  • It has been a pleasure, I love this topic! – Brenin Oct 04 '13 at 15:12
  • Dear Brenin, do you know a reference for where Riemann gave this argument? I am giving a short talk on this subject, and I would like to include a bit of history. –  Mar 04 '14 at 12:28
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    I think a reference could be: Theorie der Abel’schen Funktionen, which I only found in German (http://www.maths.tcd.ie/pub/HistMath/People/Riemann/AbelFn/AbelFn.pdf). The genus is denoted $p$, there. In English, I found it quoted here. Good luck! :-) – Brenin Mar 04 '14 at 16:12
  • Dear @Brenin: thanks for your response (which I only saw now). I can read German, so I will have a look at Riemann's paper. Thanks a lot! –  Mar 17 '14 at 13:27
  • @AsalBeagDubh: You are welcome! hope that helps. – Brenin Mar 17 '14 at 15:09
  • @Brenin. I like the answe but there are two things I don't understand: the first one is that I think you should write First/second "step" and not "way" because both parts are needed to get the conclusion as far as I get it – Heitor Fontana Jun 12 '18 at 22:24
  • @Brenin. The second thing: to my understanding (sorry if I am wrong) there are two things that you count in the wrong way (or at least I don't get it) and the result add up to the good answer (3g-3). In fact choosing a pencil means to chose a 2-dimensional subspace of $H^0(L)$, i.e. a point in $Gr(2, H^0(L))$ which has dimension $2h^0(L) -4 = 2d-2g-2$. The second thing is that the cover depends on the number of branched point minus the projective transformations of $\Bbb{P}^1$ (you can always choose three branch points to be $0,1.\infty$) hence on $2g+2d-5$ points of $\Bbb{P}^1$. – Heitor Fontana Jun 12 '18 at 22:32
  • By first and second "way" I mean ways to count covers of $\mathbb P^1$, see first paragraph. I do not see what is counted wrong. But if you wish to replace "second way" paragraph with your $\dim Gr(2,H^0(L))$ I think it's fine. – Brenin Jun 16 '18 at 11:39