User david farmer - MathOverflow

Answer by David Farmer for Hejhal's algorithm and computational methods for non-classical Maass wave forms

2013-04-18T17:46:50Z

I don't think it is practical to directly find higher rank Maass forms along the lines of what Hejhal did, because they are functions of several variables and their Fourier expansions involve multiple sums. Not to mention the need to implement the appropriate special functions that appear. Even if you have the Fourier expansion on the computer (which has been done for SL(3,Z) (by Boris Mezhericher) but not Sp(4,Z)), using the relations to make equations in the coefficients and/or the analog of integrating along horocycles is probably computationally prohibitive.

central/critical/special values of L-functions terminology

2013-04-17T14:06:02Z

I have a question about the terminology for special values of L-functions. Is the following a correct description of standard usage:

Suppose L(s) is an L-function which satisfies a functional equation relating $s$ to $w+1-s$, where $w$ is the (motivic) weight. ADDED LATER: I am assuming the L-function is motivic, otherwise (please correct me if I am wrong) there is nothing special about the value at any integer.

1) If $m$ is an integer then $L(m)$ is a special value of the L-function.

2) If $m$ is an integer and neither $m$ nor $w+1-m$ is a pole of a $\Gamma$-factor of the L-function, then $m$ is a critical point and $L(m)$ is a critical value of the L-function.

3) $L(\frac{w+1}{2})$ is the central value of the L-function.

4) If $\frac{w+1}{2}$ is not an integer, then the central value is not a special value.

I am pretty sure 2) is correct, unless Deligne's notion of critical point is not the only one. I am also pretty sure 3) is correct, since the central point of the functional equation is pretty unambiguous. It is 1) and 4) that I am hoping the experts can clarify.

Answer by David Farmer for Is there an explicit expression for the imaginary part of some non-trivial zero of zeta,

2013-03-18T18:16:27Z

Write $\rho = \frac12 + i \gamma$ for a nontrivial zero of a primitive L-function. ("Primitive" means that it can't be written as the product of other L-functions.)

It is generally believed that:

a) If $\gamma\not=0$ then $\gamma$ is transcendental.

b) If $\gamma\not=0$ then $\gamma$ is algebraically independent of every well-known constant and every other zero of every primitive L-function (except when the L-function has real coefficients, in which case $\frac12 - i \gamma$ is also a zero).

As far as I know, nobody has any clue how to prove these conjectures.

Clarification added later: what definition of L-function are we using?

Greg Martin's comment (below) refers to $L(s+ i y)$ where $L(s)$ is an L-function and $y$ is real. While it is true that for some definitions of "L-function" the set of L-functions is closed under that operation, that is not what I intended.

For the L-functions in my answer above, the Euler product axiom can be written as:

There is a Dirichlet character $\chi$, the "central character" of the L-function, such that \begin{equation} L(s)= \prod_{p \, {\rm prime}} F_p(p^{-s})^{-1}, \end{equation} where $F_p$ is a polynomial of the form \begin{equation} F_p(z)=1-a_p z + \cdots + (-1)^d\chi(p) z^d . \end{equation}

Here $d$ is the degree of the L-function. Note that I have normalized the L-function so that the functional equation relates $s$ to $1-s$.

All known L-functions satisfy that axiom, and this formulation tells you how to select the distinguished member of the family $L(s+i y)$.

Small index subgroups of SL(3,Z)

2013-02-25T14:58:19Z

I would like to know the smallest index subgroups of SL(3,Z).

The smallest I could find has even entries $a_{3,1}$ and $a_{3,2}$, along the bottom row. I could not figure out whether there are subgroups of index 2 or 3.

A search found lots of information about SL(2,Z) but not SL(3,Z).

Answer by David Farmer for Are there known non-real zeros of derivatives of Riemann zeta with 0 < Re(s) < 1/2?

2012-03-21T13:37:17Z

Mathematica claims that the 5th derivative of the Riemann zeta function has a zero at approximately

0.2876 + 4.6944 i.

I don't think it should be too hard to resolve the case of the next several derivatives. The techniques in the paper referenced by you or by Micah Milinovich should let you find an explicit upper bound on the imaginary part of the nonreal zeros of $\zeta^{(k)}(s)$ in the left half-plane. Then it is just a numerical calculation to find all the zeros below that bound.

Answer by David Farmer for Distinct simple zeros of Dirichlet L-functions

2012-03-02T17:53:17Z

You probably guessed this from the lack of responses, but the answer is that there is no hope of making progress on this question using current methods.

One measure of the complexity of an L-function is its degree, where the Riemann zeta function and Dirichlet L-functions have degree 1, the L-function of a holomorphic cusp form has degree 2, the standard L-function of a GL(n) automorphic form has degree n, etc. The precise definition of degree is the number of $\Gamma$-factors in the functional equation, where a $\Gamma(s+A)$ counts as two $\Gamma$-factors.

These days we have very few tools for dealing with degree 3 and higher. Your product of Dirichlet L-functions is like one degree $r$ L-function, and so you are stuck once $r$ is bigger than 2. In particular, the fact it is a product doesn't seem to help much.

It also doesn't help that you want only one simple zero, unless you have a particular explicit case in mind, in which case you can show it by direct calculation.

Comment by David Farmer

2013-04-19T10:05:19Z

I think the paper on his home page is the same as the one he posted to the ArXiv: sites.google.com/site/mboris

Comment by David Farmer

2013-04-17T16:41:15Z

I wrote the question in the "arithmetic" normalization, with $s$ going to $w+1-s$ in the functional equation. I know that "most" L-functions are not motivic, but my understanding is that the concept of special/critical values only makes sense for motivic L-functions. You wouldn't say, for example, that every integer is a critical point for the L-function of a Maass form. There are times when you want to normalize the L-function so that $s$ goes to $1-s$, but this is not one of those times.

Comment by David Farmer

2013-03-21T19:26:49Z

Greg, Not even close: we can't even prove that for degree 2. It is not even known that the members of the Selberg class, of degree 2 or more, have Euler products whose local factors are the reciprocal of a polynomial. (I saw something along those lines -- I forget the details -- from Kaczorowski and Perelli, but what I wrote is definitely true for "degree more than 2.")

Comment by David Farmer

2013-03-18T20:28:12Z

Paul, It is conjectured that the factorization of L-functions into primitive L-functions is unique. This follows from the Selberg Orthonormality Conjecture. Unique factorization implies that you can't "split up" the cancellation of zeros of one L-function by part of the zeros of several different L-functions. So (assuming appropriate conjectures) the Artin L-function can only be entire if there is a cancellation of actual L-functions in the numerator and denominator.

Comment by David Farmer

2013-03-18T19:06:59Z

If an Artin L-function was genuinely a ratio, and not a product, of other L-functions (in other words, if it had infinitely many poles) then it does not deserve to be called an L-function! For this discussion I think it makes sense to assume all reasonable conjectures.

Comment by David Farmer

2013-02-28T18:59:32Z

Great answer. I wish I could "accept" two answers.

Comment by David Farmer

2013-02-25T20:55:42Z

A step I am missing is why the subgroup of $SL(3,Z)$ has to be the pullback of a subgroup of $SL(3,n)$. Is maximality being used? If $G$ is the subgroup, I get that there exists $H<G$ so that $H$ is the matrices congruent to the identity mod $n$, for some integer $n$. Probably I am missing something easy for the next step.