Timeline for Finite unramified analytic coverings vs finite etale coverings
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| Feb 23, 2021 at 3:15 | history | edited | Sándor Kovács | CC BY-SA 4.0 |
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| Dec 30, 2010 at 2:30 | vote | accept | Hugo Chapdelaine | ||
| Dec 30, 2010 at 2:30 | history | bounty ended | Hugo Chapdelaine | ||
| Dec 28, 2010 at 3:09 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
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| Dec 26, 2010 at 14:28 | comment | added | Hugo Chapdelaine | You wrote <<Once you learn finite analytic maps better, you'll see removing branch locus is unnecessary to make various arguments (such as algebraicity constructions) and it simplifies things to keep branch pts in the picture.>> I'm willing to buy that and I would really like to learn more about finite analytic maps. It is still a bit annoying (at least to myself) not being able to see the heuristic about how this key notion of COHERENCE of this sheaf of algebras $\mathcal{F}$ can be used to construct enough meromorphic functions on $Y$. There must be some yoga behind that! | |
| Dec 26, 2010 at 14:21 | comment | added | Hugo Chapdelaine | Hi BCnrd, <<I meant just that coherence is as ubiquitous in C-analytic geom. as in alg. geom.>>. I certainly agree with this statement of yours. Also you wrote <<fn on compact Riemann surface requires serious analysis with any approach.>> I completely agree with that! In fact I never looked at all the fine details of this proof. For sure it only works in complex dimension $1$. | |
| Dec 26, 2010 at 12:43 | comment | added | BCnrd | Dear Hugo: when I said "everything uses coherence", I meant just that coherence is as ubiquitous in C-analytic geom. as in alg. geom. Producing a non-constant mero. fn on compact Riemann surface requires serious analysis with any approach. Once you learn finite analytic maps better, you'll see removing branch locus is unnecessary to make various arguments (such as algebraicity constructions) and it simplifies things to keep branch pts in the picture. The tool you seem to be missing is more effective use of finite analytic maps. Also think more about why finite over projective is projective. | |
| Dec 26, 2010 at 4:35 | comment | added | Hugo Chapdelaine | Second approach: Prove the existence of a single meromorphic non-constant function $f:X\rightarrow P^1(C)$ where $X$ is your Riemann surface. Let $X'$ be $X$ minus the ramification locus. Then $f$ gives you a finite unramified analytic cover of $P^1(C)$ minus the branch points. Then you conclude that $X'$ is an an algebraic curve which implies that $X$ is an algebraic curve over $C$ I find the second approach fascinating since no harmonic analysis is involved but we still need the analytic map $f$ as an input which makes it weaker than approach 1. | |
| Dec 26, 2010 at 4:31 | comment | added | Hugo Chapdelaine | So coming back to what I have saying see here are two ways of showing that a compact Riemann surface is algebraic. First approach: Using harmonic analysis, show that there ()is enough meromorphic functions which separate points and tangents. Now using () and some basic cohomology prove Riemann-Roch. Now using Riemann-Roch you may fin a very ample line bundle to embedd your Riemann surface in a projective space. | |
| Dec 26, 2010 at 4:24 | comment | added | Hugo Chapdelaine | You wrote <<Everything uses coh>> Well I'm not sure that I would agree that coherent sheaves are ubiquitous to mathematics but for sure this is a key finiteness notion which appear at many places. Nevertheless, I still think you can come up with a better answer and point me out the key places in the argument where the finiteness condition of coherence is used :) In any case, thanks a lot for your comments, I appreciate. | |
| Dec 26, 2010 at 4:21 | comment | added | Hugo Chapdelaine | As you know in complex dimension > 1 this is precisely the lack of meromorphic functions (or differentials) which prevent an compact complex manifold from being algebraic. So here from a single analytic map $f:Y\rightarrow X$ we get the existence of meromorphic functions on $Y$ which separate points and tangents. You see, there is no harmonic analysis involved in this argument. It seems to be purely some "homological algebras" and some probably clever commutative algebra. | |
| Dec 26, 2010 at 4:16 | comment | added | Hugo Chapdelaine | Hi BCnrd, you wrote <<Your amazement about many rat'l fns must already occur for the proof of projectivity of compact Riemann surfaces (crux is to make one non-constant mero. fn), >> That is exactly it! You see in general I know how to prove that a compact Riemann is projective. First using some harmonic analysis you prove that you have () enough functions which separate points and tangent and from there you prove Riemann-Roch using only () as the input. See for example Miranda's book on Comapct Riemann surfaces and algebraic. | |
| Dec 26, 2010 at 3:04 | comment | added | BCnrd | Dear Hugo: Stein stuff is irrelevant. Your (2) expresses coherence, so if you mean "how do we use coh.?" then I ask: how can you do anything without it? Everything uses coh., like in algebraic geometry, so asking where (2) gets used is a puzzling point of emphasis. Your amazement about many rat'l fns must already occur for the proof of projectivity of compact Riemann surfaces (crux is to make one non-constant mero. fn), so focus on that, not higher-dim'l stuff. Probably you should understand finite analytic maps better; I recommend the awesome book "Coh. Analytic Sheaves" by Grauert-Remmert. | |
| Dec 26, 2010 at 2:26 | comment | added | Hugo Chapdelaine | And by many rational functions I mean enough functions so that you can separate points and tangents. This is kind of fascinating since a priori I don't quite see how to use the mere existence of $f$ to construct a meromorphic function $g:Y\rightarrow\mathbf{C}$ such that $g(P)=0$ and $g(Q)=1$ where $P$ and $Q$ are 2 points in the same fiber of the map $f:Y\rightarrow X$. | |
| Dec 26, 2010 at 2:06 | comment | added | Hugo Chapdelaine | So you see the thing which I find fascinating about the algebraicity of $Y$ is the following. So we use the same notation as in the question. So from the existence of this analytic map $f:Y\rightarrow X$ we get that $Y$ is quasi-projective which implies that $Y$ has a lot of rational functions (so meromorphic)! So from the existence of only one analytic map we get the existence of many rational functions (even many regular functions if $X$ is affine for example)! So I'm trying to understand the heuristic behind that! | |
| Dec 26, 2010 at 2:05 | comment | added | Hugo Chapdelaine | Are we somehow also using in the course of the argument the fact that on a Stein manifold $W$ a coherent sheaf of $O_W$-module is (1) generated by global sections and (2) has trivial cohomology groups in degrees larger than $0$. | |
| Dec 26, 2010 at 2:00 | comment | added | Hugo Chapdelaine | Hi BCnrd, thanks a lot for Houzel's references. So in total Houzel's 4 papers "Geometrie analytique locale" consists of $12+22+25+15=74$ pages which is quite a lot of pages to read. So you made a very good remark by emphasizing the fact that $\mathcal{F}$ is not only a coherent sheaf of modules but of algebras. This remark is quite important. Nevertheless I would like to see at what places do we use the exact sequence which appears in (2) of my question. | |
| Dec 26, 2010 at 1:12 | comment | added | BCnrd | For $\mathbf{C}$-scheme $S$ loc. of f.type & coh. $O_S$-algebra $A$, $q:S' = {\rm{Spec}}_S(A) \rightarrow S$ as loc. ringed spaces over $\mathbf{C}$ and canonical $A \simeq q_{\ast}(O_{S'})$ is final among pairs $(h:T \rightarrow S, \phi:A \rightarrow h_{\ast}(O_T))$ with $\mathbf{C}$-morphism $h$ & $O_S$-alg. map $\phi$. Use $T = {\rm{Spec}}_{S^{\rm{an}}}(A^{\rm{an}})$ & canonical $h$ and $\phi$ to get $T \rightarrow S'$ as loc. ringed spaces over $S$. Univ. property of analytification gives $f:T \rightarrow {S'}^{\rm{an}}$ over $S^{\rm{an}}$. This $f$ is isom, via completed stalk argument. | |
| Dec 26, 2010 at 0:59 | history | edited | BCnrd | CC BY-SA 2.5 |
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| Dec 26, 2010 at 0:54 | history | answered | Sándor Kovács | CC BY-SA 2.5 |