[FOM] contra harvey on what number theorists want

Timothy Y. Chow tchow at alum.mit.edu
Tue Apr 4 11:57:25 EDT 2006


Gabriel Stolzenberg <gstolzen at math.bu.edu> wrote:
[Re: obtaining effective versions of theorems of Roth, Faltings, etc.]
>    More generally, I don't believe that any number theorist, leading
> or following, is "very much interested in rectifying this situation."
> Yes, they sometimes talk this way.  But that has more to do with the
> grip of their metaphysics than with number theory.
> 
>    I would change my mind if I was shown, in at least one case, a
> sound mathematical reason for wanting such a constructive proof or
> bound.  A list of number theorists making pronouncements about this
> is surely no substitute for sound mathematical reasons.

What exactly is a "sound mathematical reason for wanting ... a 
constructive proof or bound"?  More generally, what is a "sound 
mathematical reason" for wanting *any* theorem?

For example, lots of mathematicians talk as if they are interested in 
seeing a proof of the Riemann hypothesis.  How are we supposed to tell 
whether they have "sound mathematical reasons" for this, versus simply 
being "metaphysical" and "making pronouncements"?

I can sort of understand what it might mean to have "sound mathematical 
reasons" for, say, *conjecturing* that the Riemann hypothesis is true.  
This would mean partial results, analogous theorems, empirical results, 
etc.  But what on earth does it mean to have "sound mathematical reasons" 
for *wanting* a certain result?

Furthermore, you at least agree that number theorists sometimes do *say* 
they are interested in effectivizing ineffective theorems, and spend time 
proving effective versions of ineffective results.  Why do you think they 
are lying, and that they are *not* in fact interested in these effective 
results, and are spending their careers on things they're not interested 
in?  Note that saying that they're interested for *metaphysical reasons* 
is different from saying that they're *not interested at all*.

Tim


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