I’ll basically be stealing from my other blog at http://categorymath.wordpress.com for the next three posts, since I really want to post but I am extremely busy. This post can be found here.

I had already learnt about this formula a few years ago, and again a few weeks ago while perusing Hartshorne’s *Algebraic Geometry*, and came across it *yet *again while reading (or should I say glancing through) Diamond and Shurman’s *A First Course in Modular Forms *(which I call [DS] from now)*.* I *love* category theory, and am not an expert at number theory – this book seemed like an interesting book, and so I borrowed it from my local library. (The only parts I fully understood were the sections on topology, charts, and Riemann surfaces.)

Now to actual math. In [DS], section 3.1, one has the Riemann-Hurwitz formula. I’m going to try to explain this from a number-theorist point of view, so please feel free to point out any mistakes.

Let and be two compact Riemann surfaces, with a nonconstant holomorphic map . The first theorem is:

**Theorem: *** is surjective.*

**Proof:** The image must be closed and open, thus is open, since compact sets are closed in Hausdorff spaces. Thus, is disconnected, hence a contradiction. Q.E.D.

has a number associated to it, known as the *degree*, a number such that for all but finitely many , which is defined by the following construction. Let denote the “multiplicity with which takes to as a map in local coordinates, making an e_x-to- map about ” ([DS]). Then, the degree is the positive integer such that for all . (For a proof of the existence of , see [DS, section 3.1].)

Let and denote the genera of X and respectively (“genera” sounds weird, at least to me – but so does “genuses” – I guess “genii” would make sense, but that sounds a lot like the plural of “genie”…). The Riemann-Hurwitz formula is as follows:

**Theorem (Riemann-Hurwitz Formula): **

**Proof Sketch: **There’s a very interesting proof (sketch) of this theorem, as [DS] gives. First, define . Triangulate using vertices including all points of , faces and edges. Under , we may lift this to a triangulation of with edges, faces, and by ramification, vertices (recall is as above). Since and , the Riemann-Hurwitz formula follows.

A very detailed proof, which is in essence what I did when I expanded out the proof sketch in [DS], is here.

Best,

S.D.

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