2.7. EISENSTEIN SERIES AND THEIR L-FUNCTIONS 21

2.7.3. The only singularity of Eχ,c(ν; −) lying on Re(ν) 0 is a possible simple

pole at ν = 1. Indeed,

Lemma 2.13. The residue at ν = 1 of Eχ,c(ν; g) is

eχ,c,−1(g) =

δ(χ2

= 1, c = oF )

2 vol(F

×\A1)

vol(ZAGF \GA)

χ−1(det

g), g ∈ GA.

We have

2 vol(F

×\A1)

vol(ZAGF \GA)

=

DF

−1/2

RF

ζF (2)

.

with RF the residue of ζF (s) at s = 1.

Proof. (cf. [20, Proposition 6.13]) Let us examine the function m(ν) =

DF

−1/2

Aχ,c(ν) L(ν,

χ2)/L(ν

+

1,χ2),

which controls the singularity of the constant

term Eχ,c(ν; ◦ −) at ν = 1 by Lemma 2.10. The L-function L(ν, χ2) has a possible

simple pole at ν = 1 which occurs if and only if

χ2

= 1. When

χ2

= 1, the factor

Aχ,c(ν) has a zero at ν = 1 unless S(c) = ∅. Thus, the possible simple pole of

Eχ,c(ν; −) at ν = 1 occurs only when

χ2

= 1 and c = oF .

The Maass-Selberg relation applied to our Eisenstein series takes the form (cf.

[12, Formula (5.13)]) :

∧T

Eχ,c(σ)

2

vol(F

×\A1)−1

=

T

σ

σ

fχ,c) (σ 2

−

T

−σ

σ

M(σ)fχ,c) (σ 2

(2.28)

+ 2

δ(χ2

= 1)

{fχ,c)|M(σ)fχ,c) (σ (σ

log T −

fχ,c)|M (σ (σ)fχ,c)},σ(

where

∧T

is the truncation operator and M(σ) : I(χ||A

σ/2

) →

I(χ−1||A

−σ/2

) is the

global intertwining operator and M (σ) its derivative at a point σ ∈ R − {1}.

On the left-hand side, the norm is the

L2-norm

of

L2(ZAGF

\GA), while, on the

right-hand side, the norm (or the hermitian pairing) is considered for elements of

I(χ||A

σ/2

) ⊕

I(χ−1||A

−σ/2

) by the formula (2.23). Suppose

χ2

= 1, c = oF and write

m(ν) =

R

ν−1

+ O(ν − 1). Then, by

M(ν)fχ,c) (ν

= m(ν) f

(−ν)

χ−1,c

, the formula (2.28)

allows us to compute

lim

σ→1

(σ −

1)2 ∧T

Eχ,c(σ)

2

vol(F

×\A1)−1

= 2R + O(T

−1)

on the one hand. On the other hand, from Lemma 2.10, Eχ,c(ν; g) =

fχ,c)(g) (ν

+

m(ν) f

(−ν)

χ−1,c

(g), which yields eχ,c,−1(g) = R f

(−1)

χ−1,c

(g) = R χ−1(det g). By this, the

same limit is computed as R2 vol(ZAGF \GA). Thus,

R = 2 vol(F

×\A1)/vol(ZAGF

\GA).

This completes the first assertion. Since

R = Resν=1m(ν)

=

δ(χ2

= 1) DF

−1/2

A1,c(1)RF /ζF (2)

=

δ(χ2

= 1, c = oF ) DF

−1/2

RF /ζ(2),

we also have the second statement.