Did Grothendieck have a plan for proving Riemann Existence algebraically? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:41:49Z http://mathoverflow.net/feeds/question/80786 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80786/did-grothendieck-have-a-plan-for-proving-riemann-existence-algebraically Did Grothendieck have a plan for proving Riemann Existence algebraically? James D. Taylor 2011-11-13T00:27:41Z 2011-11-13T17:35:19Z <p>A <a href="http://mathoverflow.net/questions/80770/reference-request-riemanns-existence-theorem" rel="nofollow">recent question</a> reminded me of a question I've had in the back of my mind for a long time. It is said that Grothendieck wanted the center-piece of SGA1 to be a completely algebraic proof (without topology) of the following theorem: $\pi_1^{et}(\mathbb{P}^1_{\mathbb{C}}\smallsetminus a_1,...,a_r)\cong$ the profinite completion of $\langle \alpha_1,...,\alpha_r|\alpha_1...\alpha_r=1\rangle$.</p> <p>As you may know, he did not succeed.</p> <p>From my experience with Grothendieck's ideas, he often has a proposed proof in mind that would take years (if it all) to be realized. Did Grothendieck have an idea of how to prove this fact algebraically? If so, what was the missing element in his proposed proof?</p> http://mathoverflow.net/questions/80786/did-grothendieck-have-a-plan-for-proving-riemann-existence-algebraically/80827#80827 Answer by Joël for Did Grothendieck have a plan for proving Riemann Existence algebraically? Joël 2011-11-13T16:39:08Z 2011-11-13T17:35:19Z <p>Nice question. I hope that someone more knowledgeable than me will step in and bring more information, but let me share what I think.</p> <p>It doesn't seem to me that Grothendieck had an idea of how to prove this fact algebraically. It is true that Grothendieck had many ideas, and that some of them was so ambitiously genial that they would take year even to him to be realized, the most notable example being the idea of the proof of the Weil's conjecture. But in this case, he would say that he had the idea. For example he announced his strategy for the Weil conjectures very early, at the ICM 1958, and in the introduction of the EGA. From my readings of Récoltes et Semailles, I believe he would have considered unethical to raise a question without saying everything he had in mind on the subject. </p> <p>This being said, back on the subject of determining algebraically the étale fundamental group of the projective line minus n points, here is what he says in SGA 1 (remarks 2.7 page 267):</p> <p>"A cet égard, la situation de la droite rationnelle privée de $n$ points, et l’étude des revêtements d’icelle modérément ramifiés en ces points, est plus sympathique, puisque la considération des groupes de ramification en ces $n$ points fournit $n$ éléments du groupe fondamental à étudier, dont on montre en effet qu’ils engendrent topologiquement ce groupe fondamental, comme nous verrons ultérieurement. Mais même dans ce cas particulièrement concret, il ne semble pas exister de démonstration purement algébrique. Une telle démonstration serait évidemment extrêmement intéressante."</p> <p>So he says that one can algebraically construct $n$ elements in the $\pi^1$, and using transcendental methods prove that those elements topologically generates the $\pi^1$. But he doesn't seem to imply that he has the slightest idea about a purely algebraic proof.</p> <p>Edit: An other argument came to my mind. The subject of the étale fundamental group of the projective line minus n points is one of the very few question that interested Grothendieck a lot in the two part of his mathematical life (before his abrupt depart from IHES in 1970, and after that). So if he had an idea about how to prove this fact around 1960, and not the time to follow it, would he not have come back to it later when in the nineties if he was interested in the "Dessins d'enfants" and all those anabelian things. To my knowledge he didn't. (Again this is just a guess, that would happily yield to facts or even more educated guesses) </p>