Let $L/K$ be an abelian extension of number fields with Galois group $G$ and let $\chi : G \to \{\pm 1\}$ denote a real linear character of $G$. Denote $L(\chi,s)$ the Artin L-function associated to $\chi$, $n=\mathrm{ord}_{s=0} L(\chi,s)$ and $$ L^\ast(\chi,0)=\lim_{s\to 0} L(\chi,s)s^{-n} $$ its special value at $s=0$. It is known this L-functions is also the L-function associated to a Hecke character. My question is: what are the conjectures and results regarding the sign of $L^\ast(\chi,0)$ ? My current work has gotten me to conjecture that it is $-1$ for the trivial character and $1$ otherwise, and I wonder if this is something known or conjectured.
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4$\begingroup$ Why should it be a real number? $\endgroup$– David LoefflerCommented Jan 20, 2022 at 14:55
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1$\begingroup$ Is your conjecture 1 for trivial character or -1 for trivial character? $\endgroup$– David FarmerCommented Jan 20, 2022 at 15:20
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4$\begingroup$ Are you asking for the factor of absolute value 1 in its polar decomposition? Unless $\chi$ is real-valued, that factor is unlikely to be $\pm 1$. $\endgroup$– KConradCommented Jan 20, 2022 at 15:23
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1$\begingroup$ @DavidLoeffler I was looking at Artin L-functions of rational representations initially, and wanted to reduce to Hecke L-functions by Artin induction. But I overlooked that the special value is then not necessarily a real number indeed. $\endgroup$– Adrien MORINCommented Jan 21, 2022 at 11:26
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1$\begingroup$ In general, these signs (or arguments) will be controlled by the root numbers of the $L$-function, which are analogue to Gauss sums and can be quite difficult to determine (one may think of the problem of determining the sign of the classical quadratic Gauss sum). $\endgroup$– François BrunaultCommented Jan 21, 2022 at 20:42
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2 Answers
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If $\chi$ is real-valued, then the question makes sense. Using the functional equation, it reduces to computing the sign of the non-zero real number $L(\chi, 1)$ if $\chi$ is non-trivial, or the residue of $L(\chi, s) = \zeta_K(s)$ at $s = 1$ if $\chi$ is trivial. In either case, $L(\chi, s)$ tends to $+1$ for $s$ large and real, and it cannot vanish on $Re(s) > 1$, so $L(\chi, 1+\epsilon) > 0$ for all positive $\epsilon$. This shows that $L(\chi, 1) > 0$ for $\chi \ne 1$, and that $Res_{s=1} \zeta_K(s) < 0$.
answered Jan 21, 2022 at 12:04
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Let me provide a slightly different proof to that of David Loeffler, using only that $\zeta_K^\ast(0)<0$ for any number field $K$. A real valued linear character factors through $\mathbb{Z}/2\mathbb{Z}$, corresponding to a quadratic extension $L/K$, so we can suppose that $G=\mathbb{Z}/2\mathbb{Z}$ ; denoting $\chi$ its non trivial character we have the relation $\mathrm{ind}~1=1+\chi$ thus $$L^\ast(\chi,0)=\frac{\zeta^\ast_L(0)}{\zeta^\ast_K(0)}>0$$