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?


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.

  • $\begingroup$ So I take this as an indication that there are no closed formulas for $\lambda_0$, thx. $\endgroup$ – Marc Palm Feb 8 '12 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 Feb 13 '12 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 Jun 28 '14 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 Nov 1 '16 at 23:33

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

  • 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 Feb 10 '12 at 10:31
  • $\begingroup$ I hadn't thought of that, excellent point! $\endgroup$ – B R Feb 10 '12 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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