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I wonder why the Artin conjecture is so important. The only reason I could figure out is that one could use the holomorphy of Artin L-series and Weil's converse theorem to show modularity of two-dimensional Galois representations.

Are there other reasons?

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Holomorphy, or meromorphy? –  Anweshi Jan 11 '10 at 18:47
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Very likely s/he means holomorphy. Meromorphy was proved by Brauer. –  Pete L. Clark Jan 11 '10 at 18:53
    
What is the conjecture precisely? –  Ryan Budney Jan 11 '10 at 19:03
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There is a statement available on Wikipedia, although maybe it should be included in the OP: en.wikipedia.org/wiki/… –  Qiaochu Yuan Jan 11 '10 at 19:05
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2 Answers 2

up vote 17 down vote accepted

From a modern view-point, its importance stems from the relationship with modularity/automorphy. Namely, a stronger conjecture, due to Langlands, should be true: if $\rho:G_K \to GL_n({\mathbb C})$ is a continuous irreducible representation, for some number field $K$, then there is a cuspidal automorphic representation $\pi$ of $GL_n({\mathbb A}_n)$ such that $\rho$ and $\pi$ have the same $L$-function; in short, $\rho$ is determined by $\pi$.

This is a non-abelian class field theory. In the case $K = {\mathbb Q}$ and $n = 2$, it says that $2$-dimensional irred. cont. reps. of $G_{\mathbb Q}$ are classified by either weight one holomorphic cuspidal eigenforms (if the rep. is odd) or Maass cuspidal eigenforms with $\lambda = 1/4$ (if the rep. is even). The odd case is now a fully proved theorem (of Deligne--Serre to go from the modular forms to the Galois reps., and of Khare, Wintenberger, and Kisin to go the other way), while the even case is still open. (If the Maass form is dihedral, one can construct a corresponding dihedral Galois rep. --- this is due to Maass himself --- but otherwise that direction is open; if the image of $\rho$ is solvable, then one can construct the corresponding Maass form --- this is due to Langlands and Tunnell.)

As you note, in the two-dimensional case over $\mathbb Q$, Weil's converse theorem shows that Langlands' conjecture is equivalent to the Artin conjecture. (In fact, proving this over any number field $K$ was one of the goals of the famous book of Jacquet--Langlands). In general, the two are also very closely linked, so that no modern number theorist thinks about one without the other. In general, one working principle of modern number theory is that the only way to establish holomorphy of $L$-functions arising from Galois representations is by simultaneously proving automorphy.

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"of Khare, Wintenberger, and Kisin to go the other way". And Emerton. Cf. Berger's recent Bourbaki talk (1017) arxiv.org/abs/1002.4111. –  Chandan Singh Dalawat Mar 4 '10 at 4:04
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Dear Chandan, In fact, going the other way in this instance is a case of 2-dimensional Fontaine--Mazur with Hodge--Tate weights (0,0), and so is not covered by my results on Fontaine--Mazur (which, in addition to other technical caveats, apply only in the distinct Hodge--Tate weights situation). Rather, it follows directly from Serre's conjecture (proved by Khare--Wintenberger--Kisin; see Inventiones 178), with regard to the proof of which I was merely an astonished onlooker! –  Emerton Mar 4 '10 at 4:13
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I am waking up an old and already well-answered question, to offer another point of view,

The Artin's conjecture appears very naturally in the context of Chebotarev's density theorem. In fact, we can see Cheobtarev's contribution as a clever trick to circumvent the Artin conjecture by reducing the proof to cases where it is known (by works of Dirichlet and Hecke). But the proof of Chebotarev will be much simpler and more natural if we had the Artin's conjecture, which moreover would give better results as far as the error term is concerned. This is, I think, a good justification of the importance of Artin's conjecture.

To explain the role of Artin's conjecture, let us also assume for simplicity GRH. Then, for $G=Gal(K/\mathbb Q)$ a finite Galois group, and $\rho$ an irreducible Artin representation of $G$, the $L$-function $L(\rho,s)$ has no zero on Re $s>1/2$ (by GRH) and no pole either (by Artin), except for a simple pole at $s=1$ if $\rho$ is trivial. Thus the logarithmic derivative, $L'/L(\rho,s)$ has no pole on Re $s>1/2$ (except perhaps...): this illustrates clearly the symmetric and complementary role played by Artin and Riemann's conjectures; both poles and zeros of $L$ contribute to simple poles of $L'/L$, and Artin eliminates some of them, Riemann the others. Now standard techniques of analytic number theory allows us, by integrating $L'/L$ on a vertical line $2+i \mathbb R$ and moving it near the critical line, to $1/2+\epsilon + i \mathbb R$, to get an estimate of the quantity: $$\pi(\rho,x) = \sum_{p^n < x} \log(p) tr \rho(frob_p)^n$$ where the sum is on prime power less than $x$. This estimate is $O(x^{1/2+\epsilon})$ if $\rho$ is non trivial, and $x + O(x^{1/2+\epsilon})$ if $\rho$ is trivial, because of the pole at $s=1$.

Now let $C$ be subset of $G$ stable by conjugaison, $1_C$ its characteristic function. Since $1_C$ is a central function, it is a linear combinaison of character of irreducible representation of $G$, say $$1_C = \sum_\rho a_\rho tr \rho.$$ Hence $\pi(C,x) := \sum_{p^n< x} \log p 1_C(frob_p) = \sum_\rho a_\rho \pi_\rho(x)$. Since $a_1$ is easily computed as $|C|/|G|$, we get $\pi(C,x)=|C|x/|G| + O(x^{1/2+\epsilon})$, which is up to standard manipulation Chebotarev's density theorem.

Hence we can say that Artin's conjecture play to Chebotarev's density theorem a role analog to the role played by the standard conjectures for the Weil's conjecture proved by Deligne. In both cases, a clever and beautiful trick was used (by Chebotarev and Deligne, respectively) to prove a theorem (Chebotarev's theorem, aka Frobenius' conjecture, and Deligne's theorem, aka the last Weil's conjecture) without proving the conjectural statement that makes the theorem limpid (Artin's conjecture, resp. standard conjectures). This is great, but doesn't make the conjectures any less interesting.

Important addendum A friend of mine made me notice something that kind of weakens significantly the point I was trying to make above. Indeed, for the argument outlined above, one doesn't need GRH + Artin's conjecture: GRH is enough. Or, more precisely, the part of Artin's conjecture that is needed is already known under GRH. Indeed, it is clear that in the argument the absence of poles of $L(\rho,s)$ is only used in the region Re $s>1/2$. But by Brauer's theorem, we know that $L(\rho,s)$ is meromorphic on $\mathbb C$, and by elementary reasoning that $\prod_\rho L(\rho,s) = \zeta_K(s)$, where $\rho$ runs amongst Artin's representations of Gal$(K/\mathbb Q)$. Moreover, by Hecke $\zeta_K$ has no poles (except a simple one at $s=1$). Hence if all the $L(\rho,s)$ have no zero on any given region, then they don't have a polo either on that region -- expected for $L(1,s)$ with its simple pole at $s=1$.

I leave the argument above, because it shows that, the absence of poles for $L(\rho,s)$ is important for questions on the distribution of primes, even if it is a theorem rather than a conjecture. Moreover, this argument has an historical interest, as Artin's conjecture was made in the 1920's, and Brauer's theorem proved in 1946.

I ignore if the question of absence of poles on the critical line for the $L(\rho,s)$ (the only part of Artin's conjecture that is still open) has any direct application on the distribution of primes.

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