User mathman - MathOverflow

why Riemann Hypothesis over curves is easy but 'normal' hypothesis for the Riemann Zeta function $ \zeta (s) $ is hard ,

2012-05-17T16:39:44Z

despite both zetas $ \zeta (s,X) $ and $ \zeta (s)$ have the same functional equation, the same Euler prodcut and the same Riemann-Weil formula

why one of them is 'easy' and can be solved but the other is so hard?? the zeta one $ \zeta (s)$

How ..did Connes get it (trace formula)

2012-04-26T21:54:34Z

i have been reading or at least trying to understand how Connes get the density (approximate) of states

$ N(E)= \frac{E}{2\pi}log \frac{E}{2\pi}- \frac{E}{2\pi}+ \frac{7}{8}+ \frac{1}{\pi}arg \zeta(1/2+iE)$

from the Hamiltonian operator $ H=xp$

the 'smooth' part i know how it is evaluated , simply by computing $ \frac{1}{2\pi}\int dx \int dp H(E-xp) $ and using a certina Maslov index

However , how did he manage to get the oscillating part of the zeros ??? i mean $ \frac{1}{\pi}Arg \zeta (1/2+iE) $

i have been reading the approach http://www.alainconnes.org/docs/bookwebfinal.pdf

first-order linear differential equation with boundary conditions

2012-04-05T15:41:55Z

let be the differential equation

$ -ixDf(x)-if(x)/2= E_{n}f(x) $

with the boundary conditions $ f(x)=f(p^{k}x) $ for 'p' prime and $k=...,-2,-1,0,1,2,...$

is this possible to solve this eigenvalue problem ?? thanks

Trace over the zeros with real part 1/2 Only

2012-04-20T14:31:13Z

If RH is not true, we have that Weil's explicit formula still holds:

$$ \sum_{\gamma} h(\gamma) = h(i/2)+h(-i/2)-2 \sum_{n=1}^{\infty} \frac{ \Lambda(n)}{ \sqrt n}g(logn)+\frac{1}{2\pi} \int_{-\infty}^{\infty}h(r)\Psi(1/4+ir/2)dr.$$

My question is if there are formulae that take into account ONLY the Riemann zeros with real part 1/2 and not the others.. thanks.

Answer by Mathman for Understanding zeta function regularization

2012-04-20T19:35:13Z

in general ONLY the quotient of 2 functional determinant is defined inside the Zeta regularization

SEEK Elizalde zeta regularization theory :) this is the BEST book about the subject i learned ZR from it

in general Zeta regualrization is valid to define QUOTIENT of functional determinants so the infinite constant goes out $ \frac{ det(I-Az)}{det(I-Bz)}$

Answer by Mathman for Is there some explication for the transformation of the eigenvalues in Selberg trace formula

2012-04-20T14:35:34Z

the 'Selberg zeros' appear as momenta of the Laplacian $ p= \sqrt (E-1/4) $ NOT over the Eigenvalues.. a smilar thing happens with Poisson sum formula

$ \sum_{n\in Z}\delta (x-n)=\sum_{n\in Z}exp(2i\pi nx)$

but the Energies are $ E= n^{2}\pi^{2}$ this is the analogy between Selberg and Poisson

selberg trace from classical physics

2012-04-16T18:00:48Z

considering the Hamitlonian for the Selberg Operator $ y^{2} ( \partial _{x}^{2}+ \partial _{y}^{2}) $ given in the Hamiltonian form

$ H=g_{ab}p^{a}p^{b} $ with $ ds^{2} = \frac{dx^{2}+dy^{2}}{y^{2}}$ being the metric

can we obtain from this H the functional determinant expression for the selberg zeta ??

can we obtain from the integral $ d(E)= \iint \delta(E-H(x,y,p) $ the 'smooth' part of the eigenvalue countign fucniton for the Laplacian i mean the factor

$ \int_{0}^{E}dptanh(\pi p) p $ i mean if semiclassical physics can be applied to the problem of the Selberg zeta function.

dilation operator green function

2012-04-07T11:26:34Z

how can i solve $ -ixDG(x,s)-iG(x,s)/2= \delta ( \frac{x}{s}-1) $

i do not know , since it is a first odrder differntial operator, the formal solution i've found would be

$ G(x,s)= \sum_{n} \frac{u_{n}(x)u_{m}(s)}{\lambda _{n}} $

where $ -ixDu_{n}(x)-iu_{n}(x)/2= u_{n} \lambda _{n} $

here G(x,s) is the analogue of the Green function for our operator :)

Answer by Mathman for Looking for the solution of first order non-linear differential equation ($y ′+y^{2}=f(x)$) without knowing a particular solution.

2012-04-07T13:20:01Z

a Riccatti equation can be turned into $ u''+f(x)u=0 $ by the transformation $ y= \frac{u'}{u} $ using the WKB ansatz we have the asymptotic solution to 'u' as

$ u(x)\sim C(f(x)^{-1/4}exp( -\int dx \sqrt -f(x)) $

What is the Connes Trace ??

2012-04-04T19:21:18Z

in http://www.math.osu.edu/lectures/connes/zeta.pdf Connes gives the Trace

$ Tr(R_{\Lambda}U(h))=2h(1)log\Lambda + \sum_{Q_{p}}\int \frac{h(u^{-1})}{|1-u|}d^{*}u+o(1) $

the second integral is over the p-adic and the Real place

for the first integral connes gives $ log\Lambda = \int d^{*}u $

what does it mean ? what is teh connes operator exatly $R_{\Lambda}U(h) $

invariance under dilations

2012-03-31T09:46:07Z

we have that the function (for suitable f)

$ F(x)= \sum_{-\infty}^{\infty}f(x+n) $ is INVARIANT under any integer traslation

$ y=x+n$ for integer 'n'

however my question is can we find a lattice which is invariant under DILATIONS i mean under the transformation $ y=qx$ for integer (positive) or rational 'q' ??

so i am looking a formula like $ F(x)= \sum f(qx) $ so F(x) is invariant under transformation of the form $ y=qx$ thanks.

Answer by Mathman for Riemann Zeta Function connection to Quantum Mechanics.

2012-03-06T18:48:44Z

the Xi function appears as the functional determinant $ \frac{ \xi(s)}{\xi(0)}= \frac{det(H+s(s-1)+1/4)}{det(H+1/4)}$ of a certain Hamiltonian with the potential $ V^{-1}(x)= A \sqrt D n(x) $

with $ n(x) \pi = Arg\xi (1/2+ i \sqrt x) $

here 'n' plays the role of Eigenvalue staircase $ n(x)= \sum_{n=0}^{\infty}H(x-E_{n})$ and 'H(x)' is the Heaviside function.

Comment by Mathman

2012-05-11T18:27:30Z

however is correct, see wu-sprung potential :) for more info

Comment by Mathman

2012-04-21T13:56:19Z

$ det(A.B)=detA.detB $ in case its commutator is zero (i think) $ [A,B]=0 $

Comment by Mathman

2012-04-20T17:51:13Z

OK OK thanks.. i have seen it :) , however i am referring to the fact that there are only traces over ALL the zeros, i was wondering if there are traces which involve only the ZEROS on the line 1/2 to compare this trace with weil's one.

Comment by Mathman

2012-04-06T10:12:55Z

umm then perhaps my condition is too restrictive , how about only invariance under $ f(x)= fp^{k}x) $ here 'p' means all the primes $ p =2 ,3,5,7,..... $ and k=1,2,3,4,5,.. $ is still $ f=0$ ?? thanks :)

Comment by Mathman

2012-04-05T21:39:16Z

aja.. however if we assume that $ f$ is a DISTRIBUTION instead of a funciton is there a possibility to get a different result to $ f=0 $ ?? thanks for your answers

Comment by Mathman

2012-04-05T21:00:35Z

wouldn't the function $ F(x)= \sum_{q}f(qx) $ with a sum taken over all the positive rational would satisfy the boundary conditions ?? with $ F(0)=0 $ and $ \int_{0}^{\infty} F(x)dx =0 $

Comment by Mathman

2012-04-05T16:15:27Z

here D is the derivative operator with respect to 'x' and the boundary conditions apply to ALL the primes $ p=2,3,5,7,.... $

Comment by Mathman

2012-03-07T14:37:39Z

sorry.. A is a constant and $ D= \frac{d}{dx} $ derivative operator , here $ \sqrt D $ stands for the fractional derivative of N(x)