# Could Riemann Hypothesis be completely false?

1. Nov 3, 2005

### eljose

Heilbronn proved that the Epstein Zeta function did not satisfy RH...but the Zeta function $$\zeta(s)$$ can be put in a form of an Epstein function but a factor k..let be the functional equation for Epstein functions:

$$\pi^{-s}\Gamma(s)Z_{Q^{-1}}(s)=|Q|^{1/2}\pi^{s-n/2}\Gamma(n/2-s)Z_{Q}(n/2-s)$$ (1)

then if we set the Quadratic form with associated Matrix Q so $$Q=Q^{-1}=R$$ with DetQ=1 if we set n=2 (bidimensional) and make the change of variable s=p/2 the functional equation (1) is the functional equation for the Riemann Zeta function or what is the same:

$$\frac{Z_{R}(p/2)}{Z_{R}(1-p/2)}=\frac{\zeta(p)}{\zeta(1-p)}$$

or what is the same the riemann Zeta function is (but a factor) an epstein zeta function with Q^2=I being Q the matrix of a bidimensional quadratic form...but if Heilbronn proved that RH is falso for any Epstein function then RH is false for Riemann zeta function...

2. Nov 3, 2005

### matt grime

Without wishing to give you too much credence, don't you think that the author might have spotted this simple "deduction"? I like the way you refer to THE epstein zeta function, followed by AN epstein zeta function. Which is it?Is there more than one or not? (the phrase "but a factor k" is completely nonsensical; what do you mean to say?)

3. Nov 3, 2005

### eljose

I Know that Heilbronn should have addressed this fact..i do not know why he did not..the question is that if No Epstein Zeta function satisfy RH then the Riemann Zeta function also doesn,t satisfy the RH.

The question is perhaps there is a contradiction and that for the cases Q=Q^{-1} the Epstein Zeta function is supposed to satisfy RH.

when i say is equal but a factor k i am referring to this f and g are equal but a factor k in the sense that f(x)=kg(x).

4. Nov 5, 2005

### HallsofIvy

Staff Emeritus
Interesting!

Just a few days ago you told us that those "snobbish" math journals refused to publish your proof of the Riemann Hypothesis. Now you are suggesting that it is not true?

Perhaps you could get a "non-snobbish" journal to publish both your proof of the Riemann hypothesis and your proof that it is not true!

5. Nov 6, 2005

### mathwonk

well it isn't "completely false" since a decent percentage (more than 30%), in a statistical sense, of the zeroes have been shown to lie on the critical line (by Brian Conrey and others).

6. Nov 7, 2005

### shmoe

Please provide the reference where Heilbronn says that no Epstein Zeta functions satisfy an analogue of RH. I haven't read his work but my understanding was that it doesn't aplly to all Epstein Zeta functions, specifically those that reduce to a Dedekind Zeta function, and in this case we expect the zeros to be on the critical line. (The dedekind zeta function of the rationals is the usual riemann zeta function)

RH could most certainly be false. As mathwonk mentioned, it's definitely not "completely false", Conrey has shown 40% of the zeros are on the critical line (Selberg was the first to get a positive proportion of the critical line). An even larger percentage has been shown to be "close" to the critical line.

7. Nov 7, 2005

### matt grime

And we know the correct analogue of the Riemann hypothesis is true for certain generalizations to some fields, don't we? I notice eljose didn't see the difference between using 'an' and 'the' that I pointed out, so you might be onto a loser there, shmoe, in asking him to clarify anything at all.

8. Nov 19, 2005

### Robokapp

No offense, but how do you know how many zeroes the Zeta has? i don't claim to know what it is about, but isn't a function offering infinite coordonates?

9. Nov 20, 2005

### matt grime

Statistically we know a large percentage (being improved all the time) do lie on the line, just as mathwonk said. We don't even need to know how many zeroes there are to show that.

10. Nov 20, 2005

### fourier jr

why do you guys care about what percentage of the zeros lie on the line? isn't the riemann hypothesis a "for all...." type of thing?

11. Nov 20, 2005

### matt grime

Why care about anything in maths then? Why bother to prove anything? Why even try to prove RH, it's not like it'll help with anything... I thought you were above that kind of thinking.

12. Nov 20, 2005

### fourier jr

even if someone proves that 100% of the zeros of the riemann zeta function it wouldn't mean anything. it's not really even an interesting result. andrew granville proved in 1985 that fermat's last theorem is true for 100% of integers n>2 & it meant nothing, because he didn't prove it for ALL exponents. look here on his page:
http://www.dms.umontreal.ca/~andrew/PDF/FLT100.pdf
i don't know that fussing over what %age of zeros of the riemann zeta function is a total waste of time, but that percentage sure seems misleading. apparently someone called helmut meier proved that even with a proposition that is 100% true, one could still find infinitely many counterexamples to it. i haven't seen it myself but granville says it's great. so even if 100% of the zeros of the riemann zeta function are on the critical line there could still be infinitely many not on the critical line, if we believe meier. makes me wonder why people would pay any attention to what percentage of the zeros are on the critical line.

13. Nov 20, 2005

### fourier jr

more like a distraction or red herring than anything else...

14. Nov 20, 2005

### matt grime

Well, why wouldn't they pay attention? What if RH is false? The percentage result is still a result. Indeed, if one can show that there a nonzero percentage of the nontrivial zeroes are not on the line then RH is false (did you think of that?). No one is saying that the percentage result has anything to do with proving or disproving RH, but it is still a nice result. We all know that 100% doesn't mean somethign must happen for atomic distributions but you are completely missing the point of what mathematics is it seems.
how can a result that adds to our knowledge of the otherwise mysterious be a bad thing?
We cannot yet prove broue's abelian defect group conjecture, yet we can do it for all symmetric groups. does that mean that the result proving it for symmetric groups is pointless because it doesn't prove (or disprove) broue's conjecture?

Last edited: Nov 20, 2005
15. Nov 20, 2005

### fourier jr

proving a conjecture for certain groups partly solves the problem, but finding out what percentage of the zeros of the riemann zeta function doesn't. if you prove that 100% of the zeros are on the critical line you can't get from there to showing that ALL of them are on the line. it doesn't get someone any closer to solving the real problem. i think a far more interesting result is simply the fact that even if you prove that 100% of them are on the line you could still have infinitely many not on the line.

16. Nov 20, 2005

### matt grime

So, proving the result is true for for some proportion of groups is more interesting than proving something true for 100% of zeroes? Interesting, if contradictory position to hold.

Let's put it this way. Conrey, and others, have shown that a strictly positive proportion of the zeroes are indeed on the critical line. That is quite a result. Quite a big one. Certainly it is a lot weaker than proving RH but no one is claiming it as anything more than evidence to back up the belief that RH is true.

It is a big result because it is a hard result. Just as it is a big result that RH is true in certain number fields, but that doesn't prove RH.

Last edited: Nov 20, 2005
17. Nov 20, 2005

### shmoe

It may. If someone manages to prove that 100% are on the critical line (in this asymptotic sense), how do you know that the method of proof won't lead to the whole shebang?

If RH turns out to be false, it will still be interesting to know just how close it came to being true (conversely, just how false it is). % on the line, % "close" to the line, zero free regions, and the like are all different versions of partial results that do give information about the zeros.

No one is dancing in the streets at Conrey's 40% (he may have though...), but it's still a step in the right direction. To dismiss it or other partial results would be very shortsighted.

By the way, the possibility that there are infinitely many off the critical line even if 100% are on it is hardly what I'd call interesting. This is a possibility by default given the nature of asymptotic density (I'd be interested to see where you read about Maier saying this if he's actually proven something non-trivial). Consider that 100% of integers are composite in this sense, yet there are infinitely many primes.

18. Nov 20, 2005

### fourier jr

andrew granville, canada research chair in number theory @ u of montreal mentioned meier's result. it was in reference to selberg's theorem that said the probability was 100% that in short intervals the number of primes is exactly as one would predict.

19. Nov 21, 2005

### matt grime

we know who andrew granville is,we don't need his cv.

20. Nov 21, 2005

### shmoe

Ahh, this would be Helmut Maier's '85 paper "Primes in Short Intervals". You're just using it as an example that 100% true doesn't mean always true, and this has really nothing to do with why Andrew would say this paper is interesting. Maier's paper essentially gives a lower bound on how small we can take our intervals if we want a chance at results "everywhere" and not "almost everywhere", ie. Selbergs result can't be strengthened to "everywhere" unless we make some more assumptions on the length of the intervals.

A much simpler example for your purposes is my statement about 100% of naturals being composite.

Last edited: Nov 21, 2005