19
$\begingroup$

Let $\zeta_F$ denote the Dedekind zeta function of a number field $F$.

We have $\zeta_F(s) = \frac{\lambda_{-1}}{s-1} + \lambda_0 + \dots$ for $s-1$ small.

Class number formula: We have $\lambda_{-1} = vol( F^\times \backslash \mathbb{A}^1)$, where $\mathbb{A}^1$ denotes the group of ideles with norm $1$.

What is known or conjectured about $\lambda_0$?

Tate's thesis can be copied word by word for function fields with the same class number formula, so:

What is known for the zeta function of a function field?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
23
$\begingroup$

This is called the (generalised) Euler constant of the number field $K$, denoted $\gamma_K$, as for $K = \mathbb{Q}$ we have $\gamma_K = \gamma_0$, the Euler--Mascheroni constant. There are many estimates known for $\gamma_K$. For example, Theorem 7 of this paper has an upper bound for $|\gamma_K|$. Some other useful references are Theorem B of this paper as well as this paper.

As for a function field, I am not so sure, but I'm guessing things should be similar but slightly easier (as there are only finitely many zeroes for $\zeta_{C/\mathbb{F}_q}(s)$).

EDIT: This paper deals with bounds for $\gamma_K$ for function fields.

Improve this answer
$\endgroup$
4
  • $\begingroup$ So I take this as an indication that there are no closed formulas for $\lambda_0$, thx. $\endgroup$
    – Marc Palm
    Commented Feb 8, 2012 at 11:35
  • $\begingroup$ That's a neat paper, thanks for finding it. Since it confused me, I'll note that Ihara's $\gamma_K$ is $\lambda_0/\lambda_{-1}$ in the OP's formulation. $\endgroup$
    – B R
    Commented Feb 13, 2012 at 4:53
  • 1
    $\begingroup$ As B R notes, Peter's $\gamma_K = \lambda_0$ and Ihara's $\gamma_K = \lambda_0/\lambda_{-1}$ (in last link). For a number field $F$, from $\zeta_F(s) = \lambda_{-1}/(s-1) + \lambda_0 + O(s-1)$ we get $\zeta_F'(s)/\zeta_F(s) = -1/(s-1) + \lambda_0/\lambda_{-1} + O(s-1)$. The constant terms in these Laurent expansions differ by the factor $\lambda_{-1}$, which is probably not 1 unless $F = \mathbf Q$. For $\mathbf Q$ we have $\zeta(s) = 1/(s-1) + \gamma + O(s-1)$ and $\zeta'(s)/\zeta(s) = -1/(s-1) + \gamma + O(s-1)$, which can be misleading for generalizations to other number fields. $\endgroup$
    – KConrad
    Commented Jun 28, 2014 at 1:31
  • 1
    $\begingroup$ Here is a location that functions for the paper of Murty and Van Order. math.uni-bielefeld.de/vanorder/papers.html However the factor in the upper bound for $| \gamma_K |$ is not $2$, but $\Phi_0^n$ for some really gigantic $\Phi_0$ -- probably the size of the $n$-th power of the regulator, very roughly. So ... large. It is a very neat paper indeed, making concrete a series of constants that oftern occur in the context. $\endgroup$
    – Preda
    Commented Nov 1, 2016 at 23:33
20
$\begingroup$

We can actually do a good bit in the function field case, because the zeta function is of the form $$\zeta_F(s)={P(q^{-s})\over (1-q^{-s})(1-q^{1-s})}$$ where $P$ is a polynomial of degree equal to twice the genus of the underlying curve. When the genus of the curve is zero (e.g., $F=\mathbb F_q(t)$), $P(x)=1$. In this case, we can calculate the Laurent expansion of $\zeta_F(s)$ to be (using WolframAlpha to avoid thinking) $${q\over (s-1)(q-1)\log(q)}+{(q-3)q\over 2(q-1)^2}+O(s-1)$$ For the general case, we can multiply the above by the Laurent expansion for $P(q^{-s})$. For a genus-$g$ curve, the corresponding polynomial is $P(q^{-s})=1+a_1q^{-s}+\ldots+a_{2g}q^{-2gs}$. The Laurent expansion of $P(q^{-s})$ is $$\big(1+a_1 q^{-1}+\ldots+a_{2g}q^{-2g}\big)-(s-1)\log(q)\big(a_1 q^{-1}+2a_2q^{-2}\ldots+2g\cdot a_{2g}q^{-2g}\big)+O\big((s-1)^2\big)$$ Multiplying through, we get that the zero-th term in the Laurent expansion of $\zeta_F(s)$, where $F$ is the function field of a genus-$g$ curve, is $${(q-3)q\over 2(q-1)^2}\cdot P(q^{-1})+{q\over (q-1)\log(q)}\cdot {d\over ds}P(q^{-s})\bigg|_{s=1}$$

Improve this answer
$\endgroup$
2
  • 3
    $\begingroup$ And of course the functional equation implies that $P(q^{-1}) = q^{-g} P(1) = q^{-g} h_F$, where $g$ is the genus of the underlying curve and $h_F$ is the class number of $F$. $\endgroup$
    – Peter Humphries
    Commented Feb 10, 2012 at 10:31
  • $\begingroup$ I hadn't thought of that, excellent point! $\endgroup$
    – B R
    Commented Feb 10, 2012 at 15:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .