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this may be a very naive question, but has the degree conjecture (namely "the degree of any function of the Selberg class is a non negative integer") been proven for automorphic L-functions? Thank you in advance.

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up vote 3 down vote accepted

Going off wikipedia, it is true that automorphic $L$-functions for $GL_n$ over a number field have non-negative integral degree, where by degree I mean the number $2\sum_{i=1}^k \omega_i$, where the $\omega_i$ are the coefficients of $s$ appearing in the gamma factor, which is more-or-less $$L_\infty(s,F)=Q^s\prod_{i=1}^k\Gamma(\omega_i s+\mu_i)$$ We know that for $GL_n(\mathbb R)$ and $GL_n(\mathbb C)$ the $\omega_i$ are either $1/2$ or $1$ (see, e.g., Knapp's "Local Langlands Correspondence: The Archimedean case", in Motives, vol 2), so twice the sum will always be an integer (of course, only a few of these $L$-functions are known to be in Selberg's class).

For general $G$, it depends on whether someone has written done the $L$-factors for general real reductive groups. I don't know if this has been done, or if it is technically known but difficult to write out, or if it is not known.

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Dear BR, Since Langlands proved the local Langlands correspondence for real groups, doesn't this mean that (at least in principle) the $L$-factors at infinity for an arbitrary group are worked out, and in fact will reduce to $L$-factors for standard reps. of $GL_n$? Of course, one can imagine that in particular situations, there are alternative definitions of the $L$-function than via local Langlands, and I don't know if these are always verified to coincide with local Langlands at the infinite primes. Regards, Matthew –  Emerton Nov 22 '11 at 5:04
Matthew, your comment was almost exactly what I was thinking. Since I couldn't find a source that said anything definite, I decided to be cautious. Regarding alternative definitions, I can't help but think that it is a relatively recent result that the gamma factors for $L$-functions on $GL_n\times GL_m$ are "correct" (see Cogdell and PS's Shalikafest article, math.osu.edu/~cogdell/rorsc-www.pdf), and, even then, I believe there is still room for improvement (i.e. choosing elements in the representation space such that the zeta integral is exactly the $L$-function). –  B R Nov 22 '11 at 6:32
Dear BR, Thanks a lot for your additional remarks, and for the Cogdell--PS reference. Best wishes, Matthew –  Emerton Nov 22 '11 at 7:22
Matthew, you are welcome. And thank you for bringing up these important points, which I had elided. –  B R Nov 22 '11 at 7:39
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To the best of my knowledge, we are very far from proving such a conjecture. If $\mathcal S_d$ is the subclass of the Selberg class $\mathcal S$ consisting of the functions of degree $d$ then it is known that

1) $\mathcal S_0=\{1\}$ (Conrey-Ghosh, 1993);

2) $\mathcal S_d=\emptyset$ for $0\lt d\lt1$ (Richert, 1957 and others);

3) $\mathcal S_1$ consists of the Riemann zeta function $\zeta(s)$ and the shifted Dirichlet $L$-functions $L(s+i\tau,\chi)$ with $\tau\in\mathbb R$ and $\chi$ a primitive character (Kaczorowski-Perelli, 1999);

4) $\mathcal S_d=\emptyset$ for $1\lt d\lt2$ (Kaczorowski-Perelli, 2002 and 2011).

Apart from these results, I think that nothing has been established in general. A nice survey of the results obtained so far can be found in the introduction to

J. Kaczorowski, A. Perelli, "On the structure of the Selberg class, VII: $1\lt d\lt2$", Ann. of Math. (2) 173 (2011), 1397-1441.

Note that, in fact, Kaczorowski and Perelli prove their results for functions in the so-called extended Selberg class $\mathcal S^\sharp$, whose elements are not required to satisfy the Ramanujan hypothesis and the Euler product property.

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Thank you for your answer. So nothing has been proven for L-functions that are known to be both automorphic and elements of the Selberg class? –  Sylvain JULIEN Nov 20 '11 at 0:35
Not being an expert in the field, I'm not sure about that. In any case, my feeling is that almost nothing is known even in the automorphic case. –  Stefano V. Nov 20 '11 at 1:45
Can anyone fix my LaTeX? For some reason, it's not working properly... –  Stefano V. Nov 20 '11 at 1:46
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