Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What does the principal L-functions on GL(n), $n \geq 3, n \in \mathbb{Z}$, look like? Where can I find materials about principal L-functions on GL(n)?

share|improve this question
I suggest you read the book of Godement-Jacquet or Jacquet's survey article in PSPM 33. –  user4245 Nov 3 '10 at 10:59
You could also look in Goldfeld's book. –  David Hansen Nov 3 '10 at 14:17
add comment

1 Answer

up vote 14 down vote accepted

There are several ways to attack standard L-functions (I prefer "standard" over "principal" because it is associated to the standard representation of GL(n) on an n-dimensional space).

I'm going to assume that by "look like" you mean the formula for the local factors as a function of the local data. At primes where the local representation is spherical (which happens for all but finitely many primes), the local factor is $$1\over(1-\alpha_1q_v^{-s})\ldots(1-\alpha_nq_v^{-s})$$ where the $\alpha_i$ are the Satake parameters for the local representation, which are essentially of the form $q_v^{-s_i}$, where the $s_i$ are the exponents of the character of the unramified principal series into which the (spherical) local representation injects (by the Borel-Matsumoto theorem), up to some normalization issues. I'm not going to touch non-spherical or archimedean primes.

On to the different methods (in what follows, all cusp forms are assumed to generate irreducible representations):

1) Godement-Jacquet: In the spirit of Tate's thesis, take a cusp form f on G=GL(n) (and f' in the dual representation) and a Schwartz-Bruhat function $\Phi$ on M(n) and integrate $$\int_{Z_{\bf A}G_k\backslash G_{\bf A}} \langle g\cdot f,f'\rangle \Phi(g)|det(g)|^s dg$$ where the brackets are the inner product pairing. This converges for Re(s) sufficiently large.

The uniqueness of the inner product makes this factor over primes into local integrals of similar shape. You then prove that the local integrals give the correct factor for supercuspidal representations, and then you prove an induction formula to show that the integral gives the correct factor for all representations parabolically induced from supercuspidal. (It may be possible to calculate the factors directly for unramified principal series, but I don't remember.) And then you have to deal with the archimedean primes. This proves that the integral produces the L-function for judicious choice of f, f', $\Phi$.

To prove analytic continuation and functional equation, you invoke a version of Poisson summation for M(n) and proceed as in Tate's thesis. (You end up with lots of weird terms that you show are zero by invoking cuspidality of f.)

2) "Doubling method" Rankin-Selberg integrals: Take a cusp form f on G=PGL(n), a cusp form f' in the dual representation, and a specific Eisenstein series on PGL(n$^2$) and integrate $$\int_{{C_{\bf A}G_k\times G_k\backslash G_{\bf A}\times G_{\bf A}}}E((g_1,g_2))f(g_1)f'(g_2) dg_1dg_2$$ where C is the center of PGL(n$^2$). The embedding of $G\times G$ in PGL(n$^2$) is defined by the natural bi-regular action of $G\times G$ on M(n). The Eisenstein series gives the integral a meromorphic continuation and functional equation. Unwind it and out pops the Godement-Jacquet integral.

When dealing with Eisenstein series, you usually have to do some extra work to get that there are only finitely many poles. However, if I recall correctly (and I'm not sure that I do), the Eisenstein series here is the mirabolic one, which you can prove has nice properties (essentially using the Godement-Jacquet arguments).

Added: To extend to GL(n), you have to add in an extra term to deal with the fact that the center of $G\times G$ is not compact after you mod out by the center of GL(n$^2$), so the naive integral would diverge for trivial reasons. (And you can't just mod out by more in the integral because the Eisenstein series is not constant on the center of $G\times G$, even though it has trivial central character)

3) Cogdell's Eulerian integrals: Take f on GL(n) and f' on GL(n'), with n>n'. Integrate $$\int_{Z_{\bf A}GL_{n'}(k)\backslash GL_{n'}({\bf A})}Pf(g)f'(g)|det(g)|^{s-1/2}dg$$ where P is a projection operation on f. P is brilliantly designed to let this integral unwind completely into what is essentially a Mellin transform of Whittaker functions (which then factors over primes by the uniqueness of Whittaker models). More work shows that this is the L-function of the product.

Note that the decay of the cusp forms implies that the integral is entire. A little observation gives the functional equation.

4) Langlands-Shahidi method: (edited) The constant term of an Eisenstein series on GL(n+n') attached to cusp forms on GL(n) and GL(n') contains the L-function for the product (in your case take n'=1). In fact, it is $${L_S(s,f\otimes f')\over L_S(1+s,f\otimes f')}\cdot A_S(s,f\otimes f')$$ where $A_S$ represents bad-prime factors, which generally you have to work to control (when it is even possible). The meromorphic continuation of the Eisenstein series is equivalent to that of the constant term, so you get that this quotient of L-functions is meromorphic.

When n'=1, you have the mirabolic Eisenstein series, so you know that the only poles in the constant term come from the zeroes of $L(1+s,f\otimes f')$, giving you, up to the $A_S$, the analytic continuation of the L-function you care about.

I never studied the method, so I hope everything I said was correct.


1) Godement-Jacquet (Zeta functions of simple algebras), Jacquet has two survey articles (in PSPM 26 and 33)

2) Gelbart, Piatetski-Shapiro, and Rallis (Explicit construction of automorphic L-functions)

3) Cogdell's website has his expository articles on his integral representation

4) Shahidi has a new book out on the method, and he has expository articles here and there.

share|improve this answer
Thanks for this informative discussion. –  Emerton Nov 5 '10 at 4:43
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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