Iwaniec and Kowalski, in their famous book Analytic Number Theory states a strong form of the effective Chebotarev density theorem page 143, and prove it assuming both GRH for Artin's $L$-function and Artin's conjecture (that the Artin $L$-function have no pole except maybe at $s=1$). But they also add, with no argument nor reference, that the same theorem is true assuming only GRH, not the Artin's conjecture. I'd like to know how this can be proved.

Let me explicit my question for a reader that would not want to open Iwaniec and Kowalski's book. Let $L/\mathbb Q$ be a finite extension Galois group $G$, and let $\rho$ be an irreducible complex representation of $G$, that one can as well suppose non-trivial (as Artin's conjecture is known for the trivial representation). For $x>0$ a real, let $\psi (\rho,x) = \sum_{p^n < x} tr\ \rho(Frob_p^n) \log p$, where the sum is on all prime power of the form $p^n$, where the prime $p$ is unramified for $\rho$. Then the crucial point in proving Iwaniec-Kowalski's effective chebotarev is the following estimate $$(1) \ \ \ \ \ \psi(\rho,x) = O(x^{1/2} \log x \log x^d q(\rho)),$$ where $q(\rho)$ is the Artin's conductor of $\rho$ and the implied constant is absolute. Estimate (1) is proved under GRH and Artin in Iwaniec-Kowalski (Theorem 5.15)

How to prove (1) under GRH alone, as Iwaniec and Kowalski suggest is possible ?

I have already given some thoughts to the question, but I am not able to solve it. Surely one of the ideas involved should be the following: even assuming only GRH for the Artin's $L$-function, not Artins conjecture, we know that $L(\rho,s)$ has no pole except maybe on the critical line, for by the theorem of Brauer, $L(\rho,s)$ is a quotient of products of Hecke $L$-functions, and those functions have no poles on the critical strip except parhaps at $s=1$ by a deep result of Hecke, and no zeros either, under GRH, except maybe on the critical line. Moreover, one also knows using the same argument of Brauer that $L(\rho,s)$ is meromorphic function of order $1$, that is quotient of two entire functions of order $1$, and therefore that $L(\rho,s)$ has a nice Weierstrass product formula (like (5.23)). Then the natural way to go is to try to follow the proof given by Iwaniec and Kowalski under GRH and Artin, using the above remarks to avoid using Artin. I see several issues with that method, the most important being the following: a crucial step in the method is the estimate of the number of zeros $N(T)$ (with their positive multiplicty) of $L(\rho,s)$ on the critical segment between $s = 1/2 -iT $ and $s=1/2 + iT$ - see Theorem 5.8. This estimate is obtained by integrating $L'/L$ on a suitable rectangle intersecting the critical line on that segment. Yet in the presence of poles this method will not count the number $N(T)$ of zeros, but the difference $N(T)-P(T)$ where $P(T)$ is the number of poles (with their multiplicty) on the same segment. So even a good estimate for $L'/L$ hence of $N(T)-P(T)$ will not prevent $N(T)$ and $P(T)$ to be arbitrary large. Then, when one computes $\psi(\rho,x)$ by some explicit formula (such as (5.53)), both zeros and poles contribute and if these are too many, that is if $N(T)+P(T)$ is too large, the precise estimate (1) will be completely ruined.

Edit after Sausage and Frank's comments and Denis's answer : I will answer Frank's question and this will allow me to explain why Sausage and Denis's suggestion are not sufficient (or so I think -- perhaps I am missing something). The estimate (1) of Lagarias-Odlyzko leads easily by linear combination and then applying Cauchy-Schwartz and the fact the sum of the square of the dimension of irreducible representation of $G$ is $|G|$ to the following form of Chebotarev, for $C$ a subset of $G$ invariant by conjugation, and $M$ the product of primes ramified in $L$ : $$(2) \ \ \ \psi(C,x) = \frac{|C|}{|G|}Li(x) + O \left( \sqrt{x} \sqrt{|C|} \log x (\log x + \log M + \log |G|)\right).$$ Here $\psi(C,s) = \sum_{p^n < x, Frob_p^n \in C} \log p$.

Formula (2) is stated in a slightly weaker form page 144 of Iwaniec-Kowalski, weaker just because they use a substandard majoration for discriminant. Formula (2) was also stated and proved earlier, by Murty, Murty, and Saradha under GRH and Artin: see Modular Forms and the Chebotarev Density Theorem, American Journal of Mathematics, Vol. 110, No. 2 (Apr., 1988), pp. 253-281. Curiously, this paper is not mentioned in Iwaniec-Kowalski.

Now let us compare (2) with the form of effective Chebotarev proved by Lagarias and Odlyzko, and slightly improved soon after by Serre (because Lagarias and Oslyzko use the same substandard lower bound on the discriminant than later Iwaniec and Kowalski, as Serre use better bounds).
$$ (3) \ \ \ \psi(C,x) = \frac{|C|}{|G|}Li(x) + O \left( \sqrt{x} |C| \log x (\log x + \log M + \log |G|)\right)$$

It is clear how (2) is better than (3): we gain a factor $\sqrt{|C|}$. How comes Lagarias and Odlyzko and Serre didn't see it? Well, essentially this is because they use, for counting the zero of $L(\rho,s)$, the zero on $\zeta_L(s)$. The problem is that while $L(\rho,s)$ is an $L$-function of degree $\dim \rho$ (and the various $L(\rho,s)$ therefore have in quadratic average a degree equal to $\sqrt{|G|}$), the function $\zeta_L$ is an $L$-function of degree $|G|$. So to get (1) or (2), one needs to get an optimal bound for each $L(\rho,s)$ individually, as a function of degree $\dim \rho$, and the argument suggested by Sausage can not be sufficient, nor can the one given by Denis, since it is already used by Lagarias and Odlyzko.

So the question remains :

How to prove (1), or (2), without assuming Artin ? Or is it possible that there is a mistake in Iwaniec-Kowalski and proving (2) without Artin is not possible ?

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    $\begingroup$ Deal Joël, assume that $\rho$ factors through $\mathrm{Gal}(F/\mathbf{Q})$. Did you know that any zero or pole of $L(s,\rho)$ must be a zero of $\zeta_F(s)$? I think this will give the estimates that you need. I believe that Foote and Murty even have a nice elementary argument to show that, assuming $\rho$ is irreducible, that $\zeta_F(s) L(s,\rho)$ and $\zeta_F(s) L(s,\rho)^{-1}$ are both holomorphic. $\endgroup$ May 23 '13 at 3:41
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    $\begingroup$ You don't compare (1) to Lagarias and Odlyzko's result. Would you please remind me of how they fall short of what you are looking for? (I suppose the dependence on $\rho$ is worse?) $\endgroup$ May 23 '13 at 4:45
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    $\begingroup$ I see. Interesting question! (As I recall you asked me something closely related by e-mail once.) I don't have any deep insight to share, but it seems plausible that IK were just referring to being able to get square root cancellation in x, i.e. to Lagarias and Odlyzko's result. $\endgroup$ May 23 '13 at 21:41
  • $\begingroup$ Dear Frank: yes, I have been playing with that circle of ideas since a while. IK state without ambiguity that the strong formula (the one which implies an error term in square root of the size of the conjugacy set) holds with just GRH, without Artin. They may have meant otherwise but this is what they write. I have started a bounty to give this question more visibility, hoping that someone would help us resolve this situation: a strong result appearing in the (main stream) literature but for which no proof is to be found. $\endgroup$
    – Joël
    May 25 '13 at 15:22

Since no one answered my question, I have asked the author of the book. Emmanuel Kowalski told me that this remark they make (namely that the form (1) or (2) they give of Chebotarev can be proved using GRH alone, without Artin) is mistaken. In the state of our knowledge, Artin is necessary to get such a precise form.


The trick to avoid Artin's conjecture is due to Deuring; it basically reduces the problem to the case of abelian (Hecke) $L$-functions: one writes the characteristic function (say $\epsilon_c$) of a fixed conjugacy class $c$ as


where $X$ is the cyclic subgroup generated by some element $x\in C$ and $\psi$ runs over the characters of $X$.

(I've only seen this explained in old notes of Serre for a course at Harvard which can presumably still be found in Harvard's library.)


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