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I have read two proofs of Riemann-Roch : one very quick in Forster, Lecture on Riemann Surfaces which use cohomology of sheaf, and results from functional analysis.

Another one is in the book of Miranda about Riemann surfaces, which is more elementary, but use lot of intermediate results and especially snake lemma.

Each time I'm reading one of these proofs, I just can't convince myself that is true because I have to believe these results in functionnal analysis, or the snake lemma. To me it looks really like powerful and a bit mysterious results (even it's probably a basic result for most of mathematician). So my question is

Can we find a reasonably short proof of Riemann-Roch which not use homological algebra or functional analysis and which is "almost" elementary ?

I'm aware that this theorem is quite powerful so we probably need a bit of work or powerful theorem. But, what is the most "effective" proof which use not too much big results ?

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    This isn't an answer to your question, but the snake lemma shouldn't be so mysterious - it's possible to give a quite concrete proof. – Alex Kruckman Mar 16 '15 at 17:48
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    If you have one, I'm interested by this proof. –  Mar 16 '15 at 17:49
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    https://www.youtube.com/watch?v=etbcKWEKnvg – Alex Kruckman Mar 16 '15 at 17:50
  • The are lots of details to check, of course, but the thing that looks mysterious about the snake is where the connecting map from the last cokernel to the first kernel comes from. Here's a good attempt to motivate the existence of this map. – Alex Kruckman Mar 16 '15 at 17:59
  • There is another proof of RR for curves due to Weil (and presented tersely in Serre's Algebraic Groups and Class fields). It is conceptually simpler but does use Adeles which might need some getting used to (if you have never seen them before). The benefit of this approach is that it gives a uniform proof for smooth curve over any field without mentioning sheaves. – DBS Mar 16 '15 at 20:23

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WARNING: Miranda does not prove the Riemann-Roch for a Riemann surface $X$ but only for a smooth projective curve $C$.
It is impossible to prove Riemann-Roch for $X$ in the holomorphic case without some highly non-trivial analysis, which essentially proves that there are exist non constant meromorphic functions on $X$.
This is acknowledged by Miranda at the beginning of chapter VI of his (excellent) book.

Answer to your question
The snake lemma is small beer but there is no easy way of proving Riemann-Roch for a compact Riemann surface not known a priori to be algebraic: any proof involves difficult (functional) analysis.
Riemann's remarkable result is that the Riemann surface $X$ actually is an algebraic curve $C$, but to prove this is the core difficulty.

Georges Elencwajg
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  • You say it's impossible to prove analytic Riemann-Roch without highly non-trivial analysis --- is there possibly a route via Chow's theorem? For instance, I found this proof of Chow's theorem https://darknmt.github.io/html/chow-theorem.html which doesn't seem to use much analysis (but I'm not an expert so there many be analysis hidden in some fact being used a black box in this proof). – D.R. Jan 13 '25 at 05:02
  • Chow's theorem assumes that the variety is already embedded in projective space, which is by no means evident for an arbitrary Riemann surface. – Georges Elencwajg Jan 13 '25 at 08:47
  • Thanks. I would like to use your comment (+ your comments https://math.stackexchange.com/questions/2332562/prove-that-every-compact-riemann-surface-is-an-algebraic-curve#comment4802094_2332562) to correct the Wikipedia page https://en.wikipedia.org/wiki/Riemann%E2%80%93Roch_theorem#Proof_for_compact_Riemann_surfaces, but I am not an expert and I feel like I can't write it well. Would you mind editing it so that it is finally correct? If writing it on Wikipedia is not possible, would it be possible for you to write a correct version of the Wikipedia entry here on MSE, and I can push it to Wiki? – D.R. Jan 13 '25 at 21:21