# L functions of Langlands Quotients of essentially-square-integrable representations

I'm reading Kudla's Article on the Local Langlands Conjecture for $p$-adic general linear groups, and specifically I'm trying to understand how the ideas of Bernstein-Zelevinski yield show that you only need to prove the LLC for supercuspidal representations (on the automorphic side) and irreducible representatons (on the Galois side).

The part I can't find a reference on is: getting the $L$-functions to match. In particular, if $\tau_1,\,\ldots\,\,\tau_r$ are essentially-square-integrable representations, then they correspond to indecomposable representations $\rho_1',\,\ldots,\,\rho_r'$ on the Galois side. Then the correspondence gives $$Q(\tau_1,\ldots,\, \tau_r) \mapsto \rho_1' \oplus\ldots\oplus \rho_r'$$ (here $Q$ is the Langlands quotient, i.e. the unique irreducible quotient of the representation parabolically induced from $\tau_1\otimes \ldots\otimes \tau_r$). and so for match of $L$-functions we need $$L(Q(\tau_1,\ldots,\, \tau_r),\,s) = \prod_i L(\rho_i',\,s) = \prod_i L(\tau_i,\,s).$$

The fact that $$L(Q(\tau_1,\ldots,\, \tau_r),\,s) = \prod_i L(\tau_i,\,s)$$ Is stated in Kudla's article, but is not proved, and a reference is not provided. I was wondering if anyone could point me in the right direction, or explain why this is true?

(I guess the same question goes for $\epsilon$ factors but these are much more mysterious to me).

EDIT: unnecessary question removed, then edited for clarity.

Thanks!

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Either I still don't understand your question, or I have already answered it. The fact stated in Kudla's article, but not proved, is the content of Theorem 15.7.1 in Goldfeld-Hundley: Automorphic Representations and $L$-Functions for the General Linear Group, Volume 2. –  GH from MO Jul 25 '13 at 21:09
@GHfromMO - I believe we are still missing each other. Just so we are clear, the fact I'm citing is (3.1.3) in Kudla's article in "Motives," Proc. Symp. Pure Math. This is not the same as 15.7.1 from Goldfeld-Hundley. If the $\tau_i$ come from unlinked segments, then the Langlands quotient is precisely the induced representation and we are done. In general, however, the induced representation need not be irreducible and so the Langlands quotient is a nontrivial quotient of the induced representation. –  John Binder Jul 26 '13 at 16:54
John: See my updated response, hope it helps. –  GH from MO Jul 26 '13 at 18:04

The identity $L(Q(\tau_1,\ldots,\, \tau_r),\,s) = \prod_i L(\tau_i,\,s)$ is part of Theorem 3.4 in Jacquet: Principal $L$-functions of the linear group, Proc. Symp. Pure Math. 33 (1979), Part 2, 63-86.