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Japanese guy may have proved the \(abc\) conjecture

Up to a few exceptions to be proved separately, a strengthening of Fermat's Last Theorem

Four days ago, Nature described a potentially exciting development in mathematics, namely number theory:

Proof claimed for deep connection between primes
Shinichi Mochizuki of Kyoto divided the steps needed to prove the 1985 conjecture by Oesterlé and Masser into four papers listed at the bottom of the Nature article above.



The newly revealed proof works with mathematical structures such as the Hodge theaters (a theater with Patricia Hodge is above, I hope it's close enough) and with canonical splittings of the log-theta lattice (yes, the word "splitting" is appropriate above, too).

What is the conjecture about and why it's important, perhaps more important than Fermat's Last Theorem itself?




First, before I tell you what it is about, let me say that, as shown by Goldfeld in 1996, it is "almost stronger" than Fermat's Last Theorem (FLT) i.e. it "almost implies" FLT. What does "almost" mean? It means that it only implies a weakened FLT in which the exponent has to be larger than a certain large finite number.

I am not sure whether all the exponents for which the \(abc\) theorem doesn't imply FLT have been proved before Wiles or whether the required Goldfeld bound is much higher. Please tell me if you know the answer: what's the minimum Goldfeld exponent?



Off-topic: When Ben Bernanke was a kid... Via AA

Recall that Wiles proved Fermat's Last Theorem in 1996 and his complicated proof is based on elliptic curves. That's also true for Mochizuki's new (hopefully correct) proof. However, Mochizuki also uses Teichmüller theory, Hodge-Arejekekiv theory, log-volume computations and log-theta lattices, and various sophisticated algebraic structures generalizing simple sets, permutations, topologies and matrices. To give an example, one of these object is the "Hodge theater" which sounds pretty complicated and cultural. ;-)

I am not gonna verify the proof although I hope that some readers will try to do it. But let me just tell everybody what the FLT theorem and the \(abc\) theorem are.

Fermat's Last Theorem says that if positive integers \(a,b,c,n\) obey\[

a^n+b^n = c^n,

\] then it must be that \(n\leq 2\). Indeed, you can find solutions with powers \(1,2\) such as \(2+3=5\) and \(3^2+4^2=5^2\) but you will fail for all higher powers. Famous mathematicians have been trying to prove the theorem for centuries but at most, they were able to prove it for individual exponents \(n\), not the no-go theorem for all values of \(n\).

The \(abc\) conjecture says the following.

For any (arbitrarily small) \(\epsilon\gt 0\), there exists a (large enough but fixed) constant \(C_\epsilon\) such that each triplet of relatively prime (i.e. having no common divisors) integers \(a,b,c\) that satisfies \[

a+b=c

\] the following inequality still holds:\[

\Large \max (\abs a, \abs b, \abs c) \leq C_\epsilon \prod_{p|(abc)} p^{1+\epsilon}.

\] That's it. I used larger fonts because it's a key inequality of this blog entry.

In other words, we're trying to compare the maximum of the three numbers \(\abs a,\abs b,\abs c\) with the "square-free part" of their product (product in which we eliminate all copies of primes that appear more than once in the decomposition). The comparison is such that if the "square-free part" is increased by exponentiating it to a power \(1+\epsilon\), slightly greater than one but arbitrarily close, the "square-free part" will be typically be smaller than \(abc\) and even \(a,b,c\) themselves but the ratio how much it is smaller than either \(a\) or \(b\) or \(c\) will never exceed a bound, \(C_\epsilon\), that may be chosen to depend on \(\epsilon\) but it isn't allowed to depend on \(a,b,c\).

See e.g. Wolfram Mathworld for a longer introduction.

Now, you may have been motivated to jump to the hardcore maths and verify all the implications and proofs that have been mentioned above or construct your own.

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snail feedback (15) :


reader Jason said...

Hi Luboš, regarding the minimum size of the exponent implicit in the statement of Goldfeld's theorem. I don't have an answer for you but pages 16 through 19 of this 1995 PDF by Barry Mazur of your old alma mater are closely related to your question. It says that under certain circumstances that seem very lax (but unfortunately nonetheless are unproven), ABC implies Fermat for exponents greater than 5.

http://www.math.harvard.edu/~mazur/papers/scanQuest.pdf


reader tlss said...

This proof is the most arcane and forbidding I've ever seen. Only a handful of mathematicians could begin to understand it for now (and most of them no doubt are trying to). There is no way any non-specialist could verify the proof, at least without so many years of study that it would have been verified (or, perhaps, debunked) long before.


reader Mitchell Porter said...

"Square free" can mean fermions: http://en.wikipedia.org/wiki/M%C3%B6bius_function#Physics

How long until we have "Mochizuki for physicists"? :-)


reader MikeN said...

If A+B=C, what's the point of the MAX function?


reader Luboš Motl said...

They use a more symmetric convention in which each a,b,c may be either positive or negative.


reader David Nataf said...

I wonder if Hodge theatres will be a standard string theory calculation tool in 20 years.


reader James Gallagher said...

Glorying in the "sophistication" of the proof, is pretty shit modern thing (supports my academic field etc etc,) I hope this proof proves wrong, and I hope some-one comes up with a "proof" (convincing to humans) not involving the axiom of choice ala Wiles for these numerical oddities


reader Jason said...

Oh, please, The axiom of choice is a perfectly cromulent axiom.


reader James Gallagher said...

well, maybe, but I'm with Gauss, although the famous bad judgement he made was that roots of quintics weren't worth pursuing (led to group theory)). But he was right otherwise, ie Mathematics has become a ridiculously complicated game that no one but the players or unfortunate conscripts (because they're semi-autistic) care about


reader AntiWigner said...

Is there a part of maths that can be proved to be useless for physics?


reader Tobias Sander said...

Yes, so the product should run over the prime factors of the magitude of abc as well. (Sorry for the unimportant nitpick.)
Very interesting developments, thanks for writing about them.


reader Luboš Motl said...

No, Tobias, the absolute value you mention is not necessary because the divisors of abc and -abc are exactly the same.


reader Tobias Sander said...

Yes, Lubos, you are basically right, of course. Except that the textbooks are usually careful to define prime factors only for positive integers, see also here:
http://en.wikipedia.org/wiki/Prime_factor
I already regret my admittedly useless comment. I just got excited about the beautiful equation so I looked at every detail of it... ;-)


reader Luboš Motl said...

Again, Tobias, I just want to make you sure that when one is ready to be *completely* accurate, and I am arguably more strict than you are right now ;-), what you considered a mistake isn't a mistake.


The equation doesn't use the term "prime factors". It involves a product over all primes - which are positive numbers 2,3,5... - that are divisors of abc. And they're the same primes regardless of the sign of abc by the most common definition of "divisors".


Whether "prime factors" are only defined for positive integers or all integers is irrelevant because the equation doesn't use the phrase "prime factors", neither verbally nor in notation.


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