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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?

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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.

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The last link is broken. Thank you. This is really helpful. –  plusepsilon.de Feb 8 '12 at 8:50
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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
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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 $${(q-3)q\over 2(q-1)^2}\cdot P(q^{-1})+{q\over (q-1)\log(q)}\cdot {d\over ds}P(q^{-s})\bigg|_{s=1}$$

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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
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