User pasten - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:01:19Z http://mathoverflow.net/feeds/user/26218 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/106649#106649 Answer by Pasten for Philosophy behind Mochizuki's work on the ABC conjecture Pasten 2012-09-08T06:22:45Z 2012-09-10T13:43:34Z <p>I want to point out a bibliographical information that perhaps is not very well-known and can be taken as "evidence" for the possibility of applying anabelian geometry to the ABC conjecture successfully. However, I am not claiming that this is related in any sort of way to Mochizuki's work. </p> <p>Here is the fact: There is a $\pi_1$ proof of the function field Szpiro conjecture (over the complex numbers, as far as I know). The proof is indeed easy and conceptually clear, you can find a nice exposition of it in some (expository) paper of Zhang, whose title is lost somewhere in my memories. (EDIT: the paper is "Geometry of algebraic points").</p> <p>Anyway, I can tell you what is the key point of the argument. Let E be an elliptic fibration over the projective line L over the complex numbers. Assume that E has only multiplicative bad reduction. You can read the order of the discriminant at a point of L from the Kodaira type of the fibre, which in turn can be recovered in terms of monodromy representations of the fundamental group of L minus the points with bad fibres: smooth fibres have trivial monodromy, and the monodromy of singular fibres is determined by Dehn twists (assuming multiplicative reduction). You can look at all these local representations at once, after choosing loops to link bad points to some generic point p of L and then study the image of the global monodromy representation on the homology of the fibre above p. Choosing loops appropriately gives the usual commutator relation which in the image of the global representation gives a relation R=1 among generators of the local reps (and they "know" what the discriminant is). Everything here is inside $SL_2(Z)=Aut(Z^2)=Aut(H_1(E_p,Z))$ which acts on the real plane, and up to scalars it acts on the projective real line whose universal covering you already know (yes, the real line). One can lift the relation R=1 to a relation among automorphisms of the real line to get a relation R'=1' where now 1' knows the number of terms appearing on R, that is to say the number of singular fibres, which is the conductor of E in this setting. Then the Szpiro bound can be recovered from the relation R'=1'. </p> <p>And there you have, a derivative-free proof of the Szpiro conjecture for function fields (a bit shocking at least for me the first time I saw it). All the diophantine information being supplied by fundamental groups.</p> http://mathoverflow.net/questions/106321/mochizukis-proof-and-siegel-zeros/106399#106399 Answer by Pasten for Mochizuki's proof and Siegel zeros Pasten 2012-09-05T04:41:13Z 2012-09-09T21:52:46Z <p>I don't think so. Mochizuki claims to have proved a diophantine result for points of bounded degree, while you need a uniform form of the ABC conjecture for the application that you mention.</p> <p>EDIT: About your question below, on the version of the ABC conjecture claimed in Mochizuki's work, it is clearly stated in Theorem A of the 4th paper. Anyways, for the benefit of the people that might read this question, I will state in very elementary terms a corollary of Theorem A in the following context: X is the projective line with the usual projective coordinates [x:y], and D is the divisor $[0:1] + [1:0] + [1:1]$ which makes the curve U=X\D hyperbolic (the degree of the canonical divisor $\omega$ of X in this case is -2 and the degree of D is 3, hence the degree of $\omega(D)$ is 1>0). Ok, here is the corollary (the notation is explained below):</p> <p><em>Statement:</em> Let $d$ be a positive integer and let $\epsilon>0$. There is a constant $C>0$ depending only on $d$ and $\epsilon$ such that the following is true: If $A,B$ are non-zero algebraic numbers with $A+B=1$, and if the degree over Q of the number field $K=Q(A)$ is at most d, then we have $H(A,B,1) &lt; C(\Delta_K N_K(A,B,1))^{1+\epsilon}.$</p> <p><em>Notation:</em> Here I am using the same definition of $\Delta_K$, H(a,b,c) and $N_K(a,b,c)$ as in the paper on Siegel zeros of the question (this notation is explained in the first page of the paper). Well, if you check the reference you'll see that actually there is one difference: the paper uses N(a,b,c), not $N_K(a,b,c)$. However, in the above statement it is crucial that we must compute N(A,B,1) using the number filed K=Q(A), that's why I added this subscript.</p> <p>I hope that the readers can see the difference between this version and the uniform ABC conjecture for the paper on Siegel zeros: the fact that here the constant C also depends on d, not only $\epsilon$.</p> <p>A last trivial remark. To get the classical ABC conjecture with coprime integers a+b=c you take A=a/c, B=b/c and hence K=Q which makes $\Delta_K=1$, and N(A,B,1)=rad(abc).</p> http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/107386#107386 Comment by Pasten Pasten 2012-09-17T21:42:20Z 2012-09-17T21:42:20Z (cont.) The table in page 27 of IUTT-I gives an idea of what are the roles played by some of the main objects introduced by Mochizuki (and as VD pointed out, the hyperbolic curve &quot;is&quot; the number field, not the elliptic curve). Anybody can read this directly from the paper, but the only reason why I am mentioning it is the following: I was very curious about the papers (as everybody else), but the first couple pages seemed very intimidating. However, after spending some time with the papers on Frobenioids then the introduction of IUTT-I became readable after all. I hope this suggestion helps! http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/107386#107386 Comment by Pasten Pasten 2012-09-17T21:28:11Z 2012-09-17T21:28:11Z For people wanting to known more details without having to read all the 500 pages: the answer provided by VD is a nice survey of the first 16 pages of IUTT-I (avoiding technicalities of how the several types of Hodge theaters are actually constructed, or what are the prime-strips). For the interested reader, the first 27 pages of IUTT-I indeed give a very good introduction. However, it is better to get used with the language of Frobenioids FIRST, otherwise the exposition can be intimidating. Unfortunately, it does not hint on the actual &quot;source of inequality&quot; (I mean, not beyond analogies). http://mathoverflow.net/questions/44620/undergraduate-logic-textbook/107322#107322 Comment by Pasten Pasten 2012-09-16T23:58:35Z 2012-09-16T23:58:35Z Very nice suggestion! http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/107279#107279 Comment by Pasten Pasten 2012-09-16T13:58:31Z 2012-09-16T13:58:31Z @VD: I think one should read more carefully the hypothesis. Also, I would not be surprised if the final Diophatine statement is not 100% correct as stated and needs to be refined - it is such a long and complicated work!. However, I think that the whole point is the technique: if it is correct$-\epsilon$ then people will make it work at some point. I do not remember the solution of a BIG problem that was 100% correct the first time it was released (perhaps I am exaggerating a little bit). http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/107279#107279 Comment by Pasten Pasten 2012-09-16T03:58:19Z 2012-09-16T03:58:19Z If you want to apply the theorem 1.10 with initial theta data having F=Q then you have a problem: F must contain i (square root of -1). If E was already semi-stable over Q then I guess that nothing happens, but otherwise the height of E gets smaller. Also, you need semi-stable reduction of E over F as part of the conditions. For example, for the Frey curve associated to an ABC triple to be semi-stable over Q, you need that the ABC triple (a,b,a+b) must be primitive (no common factor) and 16 must divide ab(a+b) (perhaps not 16...). This slightly reduces the list &quot;too-good-to-be-true&quot; examples. http://mathoverflow.net/questions/106859/beautiful-theorems-with-short-proof Comment by Pasten Pasten 2012-09-11T00:08:22Z 2012-09-11T00:08:22Z Zagier's paper &quot;values of zeta functions and their applications&quot; has a nice short proof of $\zeta(2)=\pi^2/6$ due to Calabi. http://mathoverflow.net/questions/106848/at-what-times-were-people-interested-in-prime-numbers Comment by Pasten Pasten 2012-09-10T23:44:42Z 2012-09-10T23:44:42Z The Ishango bone is pretty old and curiously has some suspicious prime numbers on it. I'm adding this as a comment because of lack of reasons for considering it as relevant, but I could not resist. P. http://mathoverflow.net/questions/106321/mochizukis-proof-and-siegel-zeros/106399#106399 Comment by Pasten Pasten 2012-09-10T02:58:21Z 2012-09-10T02:58:21Z @GH: Perhaps the word &quot;clearly&quot; should be omitted in my post. Once I heard that the word &quot;clearly&quot; should be omitted in all the mathematical literature: if something is &quot;so clear&quot; then it is pointless to say that it is clear, while on the other hand if we use the word &quot;clearly&quot; to hide an argument that we don't want to write then we should perhaps be honest and at least give some hint. In this case, what I meant is that Theorem A alone (modulo notation) gives the main result without having to prove further propositions before using it. In any case, sorry about using the word &quot;clearly&quot;. http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/106649#106649 Comment by Pasten Pasten 2012-09-08T14:26:21Z 2012-09-08T14:26:21Z yes, you found it! http://mathoverflow.net/questions/106321/mochizukis-proof-and-siegel-zeros/106399#106399 Comment by Pasten Pasten 2012-09-06T01:47:28Z 2012-09-06T01:47:28Z I don't know if the claimed result is effective or not, I started to read the papers just about a week ago and they are certainly hard. However, the first main Diophantine consequence claimed in the 4th paper is for a somewhat restricted class of curves which is nonetheless &quot;sufficiently general&quot;. Then the author reduces the general case to this sufficiently general case (sec. 2), and it is remarkable that this reduction step is performed keeping track of explicit constants. Does this indicate that the ultimate goal is an effective result? no idea, I guess that we have to read, not speculate.