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Does someone knows of any good reference for a proof of the Riemann-Hurwitz Formula (of Riemann Surfaces) that uses Spectral-Sequences and Homology ?

Thanks in advance !

[ I think I know how to start this kind of proof, but have no idea on how to finish it...If no one has a good reference, I might ask for the specialists' help in finishing my idea]

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Seems like overkill for such an elementary fact. –  Felipe Voloch Jun 11 '12 at 16:20
    
Sure, the proof in Hartshorne refers to Riemann-Roch for the fact that the canonical sheaf has degree $2g-2$, and you can prove Riemann-Roch using arbitrarily fancy tools. What sort of enlightenment are you expecting from such a proof? –  S. Carnahan Jun 11 '12 at 16:42
    
Though it's obviously an overkill, I'm currently studying homolgical algebra from a categorical viewpoint, and I found a remark in one of the online notes that says that such structures have applications to algebraic geometry (including proving the genus formula I stated)... So I thought it might be interesting to find this kind of proof in order to see some concrete applications... Where can I find Hartshorne's proof? Can you help me? Thanks a lot ! –  jason mfash Jun 11 '12 at 17:06

2 Answers 2

up vote 6 down vote accepted

I'll try to sketch a proof of Riemann-Hurwitz using the Leray spectral sequence. It has the feel of a fun exercise.

To fix notation, let $X$ and $Y$ be compact Riemann surfaces and let $f : X \to Y$ be a finite surjective morphism of degree $d$. We want to compare the topological Euler numbers of these surfaces; these are the numbers defined as $$ \chi(X) = h^0(X,\mathbb C) - h^1(X,\mathbb C) + h^2(X,\mathbb C) $$ and similarly for $Y$. We'll use arbitrarily fancy facts of sheaf cohomology to do this.

If $\mathcal F$ is a sheaf on $X$, then the first terms of the Leray spectral sequence read $$ E_2^{p,q} = H^q(Y, \mathcal R^p f_* \mathcal F) \Rightarrow H^{p+q}(X,\mathcal F). $$ As the fibers of $f$ are 0-dimensional, we have $\mathcal R^p f_* \mathcal F = 0$ for any $p \geq 1$. Combined with the annihilation of cohomology on $Y$ for dimension reasons, we find that $E^{p,q}_2 = 0$ for any $p \geq 1$ and $q \geq 3$. The second page of the Leray spectral sequence is thus just $$ E_2^{0,0} \qquad E_2^{0,1} \qquad E_2^{0,2} $$ and all other entries are zero, so the sequence degenerates at the $E_2$-level. It follows that $H^k(Y,f_*\mathcal F) = H^k(X,\mathcal F)$ for any $k$.

Consider now a point $y$ on $Y$ that is not in the image of the ramification locus of $f$, in other words the preimage $f^{-1}(y)$ consists of $d$ distinct points. Then we see that $f_{\ast} {\mathbb C} = {\mathbb C}^{\oplus d}$. This line of though yields a short exact sequence $$ 0 \longrightarrow f_* \mathbb C \longrightarrow \mathbb C^{\oplus d} \longrightarrow \mathcal G \longrightarrow 0 $$ where $\mathcal G$ is a skyskraper sheaf supported on the image of the ramification divisor of the morphism $f$. Taking Euler characteristics we get $$ d\,\chi(Y) = \chi(f_*\mathbb C) + \chi(\mathcal G) = \chi(X) + h^0(Y,\mathcal G).$$ Expressing $h^0(Y,\mathcal G)$ in terms of the degrees of $f$ at its ramification points, and thus showing that it has the expected form, should not be a source of great trouble.

The thing that makes this proof relatively painless is that the Leray spectral sequence degenerates straight away (at least without recourse to heavy machinery) and that calculating the cohomology of a sheaf supported on a finite number of points is easy. The spectral sequence will again degenerate at the $E_2$-level in the case of a morphism between surfaces, but there a finer analysis is needed to calculate the cohomology of the corresponding sheaf $\mathcal G$. In any case the proof points the way to a similar statement for finite surjective morphisms between higher dimensional varieties, though it also seems to indicate that this is not a path one wants to take unless one really needs to.

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Hi Magnusson, Thanks a lot for your detailed answer, but there are few things I couldn't understand right at the beginning: First, when you say that the morphism $f$ is of degree $d$, what do you mean by that? Do you mean the number $ #f^{-1}(z)=[m_X:\mathbb{C}(z)] $ ? (where $m_X$ is the field of all meromorphic functions from X to the riemann sphere)? Second, what do you mean by the notation $h^i (X,\mathbb{C} ) $ ? The last thing I'll be glad to know is what do you mean by "the fibers of f" ? I'll try to read your answer again after I'll understand these notions... Thanks a lot again ! –  jason mfash Jun 11 '12 at 18:54
    
OK... SO as far as I can understand, the degree is excatly what I meant, the $h^i$ are the dimension of the sheaf cohomology. The only thing I miss is what do you mean by the fibers of f? Thanks again, I'll re-read the answer and try to complete the details myself...Hope I'll manage –  jason mfash Jun 11 '12 at 19:03
    
Dear Jason, I usually think of the degree as being the number of points in the preimage of a "generic" point on $Y$ (but I'm not really an algebraic geometer). It is the number of sheets that $f$ would have if it were an honest covering map. The $h^i$ are indeed what you think they are. Finally the fibers of $f$ are the subvarieties $f^{-1}(y)$ of $X$, where $y$ are points on $Y$. In general these consist of $d$ different points, but there are subtleties at some special points (basically one has to count points in the inverse image with multiplicity, but this is not important for the answer). –  Gunnar Magnusson Jun 11 '12 at 19:54
    
Hi again Gunner, it seems that I have a lot of "holes" in my algebraic geometry knowledge. Cn you please explain what do you mean by "the skyskraper sheaf supported on the image of the ramification divisor of the morphism f "? In addition, what does the notation $d_\chi (Y)$ stands for? I really hope you'll be able to explain me these notions. Thanks a lot again (I'm pretty sure I'll have a few more questions after I'll reread the answer on the weekend. Hope you'll have enough patience to help me :) ) BTW-do you have a good book that contains this kind of sheaf cohomology? Thanks again! –  jason mfash Jun 12 '12 at 14:45
    
In this case "skyscraper sheaf" means that it is zero except at the images of points where $f$ is ramified (i.e. is not a d-fold covering). The notation is $d$ times the Euler characteristic $\chi(Y)$ (since $\chi(\mathbb C^{\oplus d}) = d \cdot \chi(Y)$). For books you can try Huybrecht's "Complex geometry", or Hartshorne's "Algebraic geometry", or go to the classic Godement's "Topologie algebrique et theorie des faisceaux". –  Gunnar Magnusson Jun 12 '12 at 15:15

Just in case you don't know how overkill this machinery really is (especially in the case of Riemann surfaces which is what you asked for) I have posted a sketch of the basic argument. Since it doesn't really answer your question, I have made the answer CW.

Let $f: X \to Y$ be a holomorphic map of Riemann surfaces of degree $d$, i.e. the preimage of all but finitely many points in $Y$ has $d$ points in it. Let $x_i$ be the branch points of $f$ and $y_j$ be the images of these branch points.

Triangulate $Y$ so that each $y_i$ is a vertex. Lift this triangulation of $Y$ to a triangulation of $X$ via $f$. Let $F,E,V$ be the number of faces, edges, and vertexes in the triangulation of $Y$. Then there are $dF$ faces and $dE$ edges in $X$. There are almost $dV$ vertexes, but I get less vertexes because at a ramification point in $Y$ there are less than the full $d$ preimages. How many less? Exactly the sum of one less than the degrees of the $x_i$!

Feel free to edit this post if it is unclear.

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OK, using spectral sequences to prove Riemann-Hurwitz may seem like an overkill (although it is a fun exercise in itself; +1 to the question and to the accepted answer), but proving that every surface can be triangulated is not a piece of cake either (pardon the pun). In my opinion, different approaches just use different definitions of Euler characteristics. –  Margaret Friedland Jun 12 '12 at 15:51
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@Margaret I agree. It is just that, in this crazy mixed up world of ours, it is entirely possible that one work's through all of Hartshorne and sees a proof of Riemann-Hurwitz via sheaf cohomology before you see the very simple geometry at work in my answer (this happened to me). I just wanted to make sure the OP knew this very intuitive way to think about it. I don't even remember RH anymore - it is much easier to take a second to think through the above line of reasoning. –  Steven Gubkin Jun 12 '12 at 15:57
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Huh. I'm almost in the same position as you Steven, in that I'd only seen cohomological or low-tech residue proofs of RH, and can never remember the statement of the damn thing. That argument however is very nice, very nice indeed. Bravo. –  Gunnar Magnusson Jun 12 '12 at 19:58
    
Thank you Gunnar. I liked your answer a lot too. I have been reading Joseph Taylor's SCV book, and I liked seeing your approach - uses a lot of ideas present in that book. –  Steven Gubkin Jun 12 '12 at 20:32

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