Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

We have $\zeta(s) = \frac{\lambda_{-1}}{s-1} + \lambda_0 + \dots$ 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?

share|improve this question
add comment

2 Answers

up vote 15 down vote accepted

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, page 61 of this paper has an upper bound for $|\gamma_K|$, which basically states that $$|\gamma_K| \leq 2 \mathrm{Res}_{s = 1} \zeta_K(s).$$ Some other useful references are here and here.

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.

share|improve this answer
The last link is broken. Thank you. This is really helpful. –  plusepsilon.de Feb 8 '12 at 8:50
Hopefully the link should be fixed now. –  Peter Humphries Feb 8 '12 at 8:57
So I take this as an indication that there are no closed formulas for $\lambda_0$, thx. –  plusepsilon.de Feb 8 '12 at 11:35
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. –  B R Feb 13 '12 at 4:53
add comment

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 $${(q-3)q\over 2(q-1)^2}\cdot P(q^{-1})+{q\over (q-1)\log(q)}\cdot {d\over ds}P(q^{-s})\bigg|_{s=1}$$

share|improve this answer
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 $F$. –  Peter Humphries Feb 10 '12 at 10:31
I hadn't thought of that, excellent point! –  B R Feb 10 '12 at 15:29
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.