D*****r
发帖数: 6791 | 1 http://golem.ph.utexas.edu/category/2013/05/bounded_gaps_betwee
Guest post by Emily Riehl
Whether we grow up to become category theorists or applied mathematicians,
one thing that I suspect unites us all is that we were once enchanted by
prime numbers. It comes as no surprise then that a seminar given yesterday
afternoon at Harvard by Yitang Zhang of the University of New Hampshire
reporting on his new paper “Bounded gaps between primes” attracted a
diverse audience. I don’t believe the paper is publicly available yet, but
word on the street is that the referees at the Annals say it all checks out.
What follows is a summary of his presentation. Any errors should be ascribed
to the ignorance of the transcriber (a category theorist, not an analytic
number theorist) rather than to the author or his talk, which was lovely.
Prime gaps
Let us write p 1,p 2,… for the primes in increasing cardinal order. We know
of course that this list is countably infinite. A prime gap is an integer
p n+1?p n. The Prime Number Theorem tells us that p n+1?p n is approximately
log(p n) as n approaches infinity.
The twin primes conjecture, on the other hand asserts that
liminf n→∞(p n+1?p n)=2
i.e., that there are infinitely many pairs of twin primes for which the
prime gap is just two. A generalization, attributed to Alphonse de Polignac,
states that for any positive even integer, there are infinitely many prime
gaps of that size. This conjecture has been neither proven nor disproven in
any case. These conjectures are related to the Hardy-Littlewood conjecture
about the distribution of prime constellations.
The strategy
The basic question is whether there exists some constant C so that p n+1?p n
…when C=7×10 7.
Here is the basic proof strategy, supposedly familiar in analytic number
theory. A subset H={h 1,…,h k} of distinct natural numbers is admissible if
for all primes p the number of distinct residue classes modulo p occupied
by these numbers is less than p. (For instance, taking p=2, we see that the
gaps between the h j must all be even.) If this condition were not satisfied
, then it would not be possible for each element in a collection {n+h 1,…,n
+h k} to be prime. Conversely, the Hardy-Littlewood conjecture contains the
statement that for every admissible H, there are infinitely many n so that
every element of the set {n+h 1,…,n+h k} is prime.
Let θ(n) denote the function that is log(n) when n is prime and 0 otherwise
. Fixing a large integer x, let us write n～x to mean x ≤ n<2x. Suppose we
have a positive real valued function f—to be specified later—and consider
two sums:
S 1=∑ n～xf(n)
S 2=∑ n～x(∑ j=1 kθ(n+h j))f(n)
Then if S 2>(log3x)S 1 for some function f it follows that ∑ j=1 kθ(n+h j)
>log3x for some n～x (for any x sufficiently large) which means that at
least two terms in this sum are non-zero, i.e., that there are two indices i
and j so that n+h i and n+h j are both prime. In this way we can identify
bounded prime gaps.
Some details
The trick is to find an appropriate function f. Previous work of Daniel
Goldston, János Pintz, and Cem Yildirim suggests define f(n)=λ(n) 2 where
λ(n)=∑ d∣P(n),d
where ?>0 and D is a power of x.
Now think of the sum S 2?(log3x)S 1 as a main term plus an error term.
Taking D=x ? with ?<14, the main term is negative, which won’t do. When ?=
14+ω the main term is okay but the question remains how to bound the error
term.
Zhang’s work
Zhang’s idea is related to work of Enrico Bombieri, John Friedlander, and
Henryk Iwaniec. Let ?=14+ω where ω=11168 (which is “small but bigger than
?”). Then define λ(n) using the same formula as before but with an
additional condition on the index d, namely that d divides the product of
the primes less that x ω. In other words, we only sum over square-free d
with small prime factors.
The point is that when d is not too small (say d>x 1/3) then d has lots of
factors. If d=p 1?p b and R
a+1>R. This gives a factorization d=rq with R/x ω
break the sum over d into two sums (over r and over q) which are then
handled using techniques whose names I didn’t recognize.
On the size of the bound
You might be wondering where the number 70 million comes from. This is
related to the k in the admissible set. (My notes say k=3.5×10 6 but maybe
it should be k=3.5×10 7.) The point is that k needs to be large enough so
that the change brought about by the extra condition that d is square free
with small prime factors is negligible. But Zhang believes that his
techniques have not yet been optimized and that smaller bounds will soon be
possible.
Posted at May 14, 2013 8:44 PM UTC |
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