Page 6, Definition of Slash operator does not satisfy
Φ|ργγ′=Φ|ργ|ργ′,(i.e., does not define a right action). This correction was brought to our attention by Fan Zhou. To correct it, the slash operator should be defined as
(Φ|ργ)(z)=Jρ(γ,z)⋅Φ(~γ.z)
instead of
(Φ|ργ)(z)=Jρ(γ,z)−1⋅Φ(~γ.z).
(Delete the −1.)
Page 91, two references to Theorem 14.8.5 should actually refer to Theorem 14.8.11 (there is no Theorem 14.8.5).
Page 142, statement of Theorem 15.8.6 when r=1, we get Lp(s,| |d−12p⋅π), not Lp(s,| |1−d2p⋅π). (It may also be worth noting that if r=1, then d=n.)
Page 142-43, proof of Theorem 15.8.6 in the case r=1 Beginning from equation (15.8.7), explicit formulae are written which are only valid if π (which is now a character) is unramified. When r=1 and π is a ramified character, both (15.8.7) and (15.8.8) become Zp(s,Φ,β)=Zp(1−s,ˆΦ,∨β). Roughly the same argument used in the unramified case shows that in this case Lp(s,π′)=1. But Lp(s,| |n−12p⋅π) is also equal to 1, so the statement of the theorem is still correct.
Page 143, line 10 The expression (1−π(p)p−s−n−12) is equal to Lp(s,| |n−12p⋅π), or Lp(s,| |d−12p⋅π), but not Lp(s,| |1−d2p⋅π). Also we remind the reader that this holds only if π is unramified.
Page 124, Proposition 15.3.2Before one can apply Theorem 13.7.3 to an irreducible cuspidal automorphic representation of GL(nAQ), one must know that it is an admissible (g,K)×GL(n,Afinite)-module. The proof is the same as in the GL(2) case, requiring a suitable analogue of theorem 3.12.1. A good reference is Borel, A.; Jacquet, H. Automorphic forms and automorphic representations. With a supplement "On the notion of an automorphic representation'' by R. P. Langlands. Proc. Sympos. Pure Math., XXXIII, Automorphic forms, representations and L-functions (Proc. Sympos. Pure Math., Oregon State Univ., Corvallis, Ore., 1977), Part 1, pp. 189–207, Amer. Math. Soc., Providence, R.I., 1979.
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