(June 22, 2012)
Analytic continuation, functional equation: examples
Paul Garrett garrett@math.umn.edu
1.
2.
3.
4.
http://www.math.umn.edu/egarrett/
L(s, χ) for even Dirichlet characters
L(s, χ) for odd Dirichlet characters
Dedekind zeta function ζo (s) for Gaussian integers o
Grossencharacter L-functions L-functions for Gaussian integers
We try to imitate the argument used by Riemann for proving the analytic continuation of ζ(s) and its
functional equation
π −s/2 Γ( 2s ) ζ(s) = π −(1−s)/2 Γ( 1−s
2 ) ζ(1 − s)
from the integral representation
π
−s/2
Γ( 2s ) ζ(s)
∞
Z
y −s/2
=
0
in terms of the basic theta series [1]
θ(iy) =
X
2
e−πn
θ(iy) − 1 dy
2
y
y
n∈Z
whose functional equation
i
1
θ(iy) = √ · θ
y
y
is proven by Poisson summation.
The discussion of L-functions L(s, χ) for Dirichlet characters (over Q) bifurcates into two families, depending
upon the parity of χ, that is, whether χ(−1) = +1 or χ(−1) = −1.
Analogous discussion of the zeta function of the Gaussian integers extends this discussion in a new direction.
Treatment of the grossencharacter L-functions [2] for the Gaussian integers have no counterpart among
Dirichlet L-functions for Z.
L-functions of ideal class group characters deserve parallel treatment, but would require more background,
concerning rings of algebraic integers that fail to be principal ideal domains, and other background. This
background is important, but is not the immediate point, so we defer this part of the discussion.
These examples are simple, but the book-keeping quickly becomes fragile. This should motivate receptiveness
to the more abstract, but very clear, Iwasawa-Tate viewpoint, which we consider soon.
1. Dirichlet L-functions L(s, χ) for even Dirichlet characters
Let χ be a non-trivial Dirichlet character mod N > 1, with L-function
L(s, χ) =
X χ(n)
ns
n≥1
[1] The reason that the argument is given as iy rather than y is that iy can be replaced by any z in the complex
upper half-plane H. This aspect is important later.
[2] The grossencharacters and their L-functions are also called Hecke characters and L-functions, as Hecke first
studied them, about 1920.
1
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
By imitation of the corresponding discussion for ζ(s), we expect to define a theta series θχ with a functional
equation provable via Poisson summation, to exhibit an integral representation of L(s, χ) in terms of θχ , and
to use this integral representation to prove the analytic continuation and see the functional equation in the
symmetry of the rewritten integral representation.
We will find that the most obvious imitative approach succeeds only for even χ. Further, to have a
symmetrical form of the functional equation, we must be able to determine the absolute value of the Gauss
sum
X
(with ψ(b) = e2πib/N )
hχ, ψi =
χ(b) ψ(b)
b mod N
This requires that χ be primitive mod N , that is, its conductor must be exactly N , not a proper divisor.
That is, the non-zero parts of χ must not be well-defined modulo N 0 for any proper divisor of N .
[1.1] The theta series, parity of χ The obvious imitation of the theta series for ζ suggests defining
θχ (iy) =
X
2
χ(n) e−πn
y
n∈Z
Non-trivial χ is extended by 0 to integers having a common factor with N , so χ(0) = 0. By design,

0
(for χ odd)

X χ(n) + χ(−n)
2
θχ (iy)
=
e−πn y =
P
2
2
−πn2 y
n≥1
(for χ even)
n≥1 χ(n) e
That is, this version of θχ cannot be correct for odd χ, since it vanishes identically.
[1.2] The integral representation The integral representation is easy: for Re (s) > 1,
Z
0
∞
Z ∞
X
2
s
θχ (iy) dy
dy
=
χ(n)
y
y 2 e−πn y
2
y
y
0
s
2
n≥1
X χ(n) Z ∞ s
dy
= π −s/2 Γ( 2s ) L(s, χ)
=
y 2 e−y
s/2 ns
y
π
0
n≥1
by replacing y by y/πn2 . This is the desired integral representation.
[1.2.1] Remark: The measure dy/y is dilation-invariant, and also invariant under any change of variables
y → y/C for C > 0. Since these are the changes of variables the computation will require, it is vastly better
to keep the y in the denominator of the measure, rather than absorb it into the y s/2 .
[1.3] Functional equation of θχ
θ(iy) =
P
n
−πn2 y
e
We anticipate that θχ has a functional equation similar to that of
, proven by application of the Poisson summation formula
X
X
f (n) =
fb(n)
(for nice functions f )
n∈Z
n∈Z
The minor but meaningful obstacle is that in this classical guise θχ is not simply a sum of a nice function
over a lattice, so Poisson summation is not immediately applicable. However, since χ(n) depends only upon
n mod N , we can break the sum expressing θχ into a finite sum of sums-over-lattices, and then apply Poisson
summation.
Specifically,
θχ (iy) =
X
2
χ(n) e−πn
y
=
n
X
b mod N
2
χ(b)
X
`∈Z
2
e−π(`N +b)
y
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
Thus, Poisson summation is to be applied to the function
2
f (x) = e−π(xN +b)
y
2
Since this f differs from the Gaussian e−πx merely by dilations and translation, and the Fourier transform
behaves cogently with respect to such, something reasonable will come out, by changing variables:
Z
2
2
e−π(xN +b) y b(ξ) =
e−π(xN +b) y e−2πξx dx
RZ
2
1
=
e−π(x+b) y e−2πξx/N dx
(replacing x by x/N )
N R
Z
2
e2πiξb/N
e−πx y e−2πξx/N dx
=
(replacing x by x − b)
N
R
Z
√
2
e2πiξb/N
√
e−πx e−2πξx/N y dx (replacing x by / y)
=
√
N y R
e2πiξb/N −πξ2 /N 2 y
√ e
N y
=
(taking Fourier transform)
Thus, by Poisson summation,
θχ (iy) =
X
χ(b)
b mod N
=
X
2
e−π(`N +b)
y
=
X
χ(b)
b mod N
`∈Z
1 X −π`2 /N 2 y
e
√
N y
`∈Z
X
X e2πi`b/N
2
2
√ e−π` /N y
N y
`∈Z
χ(b)e2πi`b/N
b mod N
The obvious thing would be to replace b by b`−1 mod N , but the sum over ` does not obviously exclude `
with gcd(`, N ) > 1. This is the first point where the primitivity of χ mod N enters, to make the inner sum
vanish for gcd(`, N ) > 1:
[1.3.1] Claim:
X
χ(b) e2πi`b/N = 0
(for primitive χ mod N , for ` not invertible mod N )
b mod N
Proof: Let gcd(`, N ) = N/m > 1. Then, replacing b by b · (1 + xm for any x mod N/m gives
X
b mod N
X
χ(b) e2πi`b/N = =
X
χ(b(1 + xm)) e2πi`β(1+xm)/N
b mod N/m β mod m
= χ(1 + xm)
X
X
χ(b) e2πi`β/N
b mod N/m β mod m
This holds for all x. The primitivity of χ is exactly that x → χ(1 + xm) cannot be trivial. Thus, the sum is
0, as claimed.
///
Returning to the functional equation for θχ , now we can make the change of variables: replace b by b`−1
mod N :
X
2
2
1 X −1
θχ (iy) =
χ (`) e−π` /N y
χ(b)e2πib/N
√
N y
b mod N
`∈Z
3
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
The inner sum has become a Gauss sum hχ, ψi not depending on `,
X
hχ, ψi =
(with ψ(b) = e−2πib/N )
χ(b) ψ(b)
b mod N
so we have the functional equation:
θχ (iy) =
hχ, ψi
i √ · θχ−1
N y
N 2y
(for even χ primitive mod N )
[1.3.2] Remark: The most striking thing is that the functional equation relates θχ to θχ−1 , rather than to
θχ itself. This is typical of a larger reality.
[1.3.3] Remark: Note that the flip in the argument y is y → 1/N 2 y, not the earlier y → 1/y. The fixedpoint of this map is not 1, but 1/N . This influences the break-up of the integral in proof of the analytic
continuation and functional equation, next.
[1.4] Analytic continuation and functional equation Starting from the integral representation,
π −s/2 Γ( 2s ) L(s, χ) =
Z
∞
θχ (iy) dy
=
2
y
s
y2
0
∞
Z
θχ (iy) dy
+
2
y
s
y2
1/N
Z
1/N
s
y2
0
θχ (iy) dy
2
y
As for ζ(s), the integral from 1/N to ∞ is nicely convergent for all s ∈ C, and gives an entire function.
The goal is to use the functional equation to convert the integral from 0 to 1/N to the other sort, and then
symmetrize the resulting expression as much as possible. That is, replace y by 1/N 2 y in the problemmatical
integral from 0 to 1/N , obtaining
1
2
Z
∞
s
(1/N 2 y) 2 θχ
1/N
i dy
=
N 2y y
1
2
N −s
Z
∞
s
y − 2 θχ
1/N
i dy
N 2y y
Replacing χ by χ−1 in the functional equation above gives
θχ−1 (iy) =
Rearranging,
θχ
hχ−1 , ψi
i · θχ
√
N y
N 2y
√
N y
i θ −1 (iy)
=
N 2y
hχ−1 , ψi χ
Substituting this gives
N 1−s
hχ−1 , ψi
Z
∞
y
1−s
2
1/N
θχ−1 (iy) dy
2
y
That is,
π −s/2 Γ( 2s ) L(s, χ) =
Z
∞
s
y2
1/N
N 1−s
θχ (iy) dy
+
2
y
hχ−1 , ψi
Z
∞
y
1/N
1−s
2
θχ−1 (iy) dy
2
y
This gives the analytic continuation, but is not quite symmetrized. Not only is there the inescapable feature
that sending s to 1−s requires that χ becomes χ−1 , but, also, the Gauss sum and the power of the conductor
N appear.
Recall that for χ primitive,
|hχ, ψi| =
4
√
N
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
Since we do not pretend to know the argument of the Gauss sum, only its size, let
√
N
hχ−1 , ψi
ε(χ) =
Then |ε(χ)| = 1 and
π
−s/2
Γ( 2s ) L(s, χ)
∞
Z
=
dy
1−s θχ−1 (iy)
1
θχ (iy)
+ ε(χ)N 2 −s y 2
2
2
y
s
y2
1/N
Ignoring ε(χ) for a moment, we can symmetrize the powers of the conductor N , by multiplying through by
N s/2 , giving
N
s/2
π
−s/2
Γ( 2s ) L(s, χ)
Z
∞
=
s
(N y) 2
1/N
dy
1−s θχ−1 (iy)
θχ (iy)
+ ε(χ)(N y) 2
2
2
y
The seeming asymmetry in ε(χ) is an illusion: since χ is even,
X
hχ−1 , ψi = hχ, ψi =
χ(b)ψ(−b) = χ(−1)
b
X
χ(b)ψ(b) = χ(−1)hχ, ψi = hχ, ψi
(for χ even)
b
That is,
N
N
= 1
=
hχ−1 , ψi · hχ, ψi
hχ, ψi · hχ, ψi
ε(χ) · ε(χ−1 ) =
Thus, there is the symmetry
ε(χ) · ε(χ−1 ) = 1
(for χ even)
Thus, we have the functional equation
s
s
N 2 π − 2 Γ( 2s ) L(s, χ) =
Z
∞
= ε(χ)
Z
∞
1/N
s
ε(χ−1 (N y) 2
1/N
= ε(χ) · N
1−s
2
s
(N y) 2
π−
1−s
2
dy
1−s θχ−1 (iy)
θχ (iy)
+ ε(χ)(N y) 2
2
2
y
dy
1−s θχ−1 (iy)
θχ (iy)
+ (N y) 2
2
2
y
−1
Γ( 1−s
)
2 ) L(1 − s, χ
(for χ even)
In summary,
s
s
N 2 π − 2 Γ( 2s ) L(s, χ) = ε(χ) · N
1−s
2
π−
1−s
2
−1
Γ( 1−s
)
2 ) L(1 − s, χ
(for χ even)
where
√
ε(χ) =
N
hχ−1 , ψi
and
|ε(χ)| = 1,
ε(χ) · ε(χ−1 ) = 1
(χ even)
[1.4.1] Remark: The appearance of the conductor N is inescapable, as is the appearance of ε(χ).
5
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
2. Dirichlet L-functions L(s, χ) for odd Dirichlet characters
Now accommodations are made for odd χ, primitive mod N .
[2.1] The theta series, parity of χ Again, the obvious imitation of the theta series for ζ suggests defining
θχ (iy) =
X
2
χ(n) e−πn
y
(bad: vanishes identically for odd χ!)
n∈Z
2
but for odd χ this fails, being identically 0 due to cancellation of the ±n summands. The Gaussian e−πx
2
must be modified to be odd: the function xe−πx will do. Thus, let [3]
θeχ (iy) =
X
2
√
χ(n) n y e−πn y
(for odd χ)
n∈Z
[2.2] The integral representation
The integral representation is easy, assuming that we keep track of
e for Re (s) > 1,
the little shifts created by the alterations in θ:
Z ∞
Z ∞
X
e
2
s
s θχ (iy) dy
dy
√
=
χ(n)
y 2 n y e−πn y
y2
2
y
y
0
0
n≥1
=
X
n≥1
χ(n) n
ns+1
Z
π (s+1)/2
∞
y
s+1
2
e−y
0
s+1
dy
= π − 2 Γ( s+1
2 ) L(s, χ)
y
by replacing y by y/πn2 . This is the desired integral representation.
[2.2.1] Remark: The shifts in the exponent of π and argument of Γ are inescapable and correct, since the
functional equation will turn out to be s → 1 − s with these shifts in the gamma factor.
[2.3] Functional equation of θeχ
θ(iy) =
P
n
2
e−πn
We anticipate that θeχ has a functional equation similar to that of
y
, proven by Poisson summation.
As with even χ, the minor obstacle is that θeχ is not simply a sum of a nice function over a lattice, so Poisson
summation is not immediately applicable. However, since χ(n) depends only upon n mod N , we can break
the sum expressing θeχ into a finite sum of sums-over-lattices, and then apply Poisson summation.
Specifically,
θeχ (iy) =
X
2
√
χ(n) n y e−πn y =
n
X
χ(b)
b mod N
X
2
√
(`N + b) y e−π(`N +b) y
`∈Z
Thus, Poisson summation is to be applied to
2
√
√
f ((xN + b) y) = (xN + b) y e−π(xN +b) y
2
(with f (x) = xe−πx )
[3] Given the existing literature, it is important to understand that this is not the standard classical-style
normalization of the corresponding theta function, whence the modifying tilde. Specifically, the usual classical
√
theta series would omit the y. Of course, such variations can be compensated-for in companion normalizations.
The choice here is motivated by somewhat longer-range considerations.
6
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
This differs from f (x) merely by dilations and translation, and the Fourier transform behaves cogently with
respect to such, so something reasonable will come out. By changing variables:
Z
√
√ f ((xN + b) y) b(ξ) =
f ((xN + b) y) e−2πξx dx
RZ
1
√
=
f ((x + b) y) e−2πξx/N dx (replacing x by x/N )
N R
Z
e2πiξb/N
√
f (x y) e−2πξx/N dx (replacing x by x − b)
=
N
R
Z
√
e2πiξb/N
√
=
f (x) e−2πξx/N y dx (replacing x by / y)
√
N y R
=
e2πiξb/N b
√
√ f (ξ/N y)
N y
(taking Fourier transform)
2
The function f (x) = xe−πx is an eigenfunction [4] of Fourier transform:
2
2
xe−πx b(ξ) = −i · ξe−πξ
Thus, by Poisson summation,
X
X e2πi`b/N
X
X
` √
χ(b)
θeχ (iy) =
χ(b)
f ((`N + b) y) = −i
√ f
√
N y
N y
b mod N
b mod N
`∈Z
`∈Z
−i X
` X
=
f
χ(b)e2πi`b/N
√
√
N y
N y
b mod N
`∈Z
As already proven in the treatment of even χ, for primitive χ the inner sum vanishes for gcd(`, N ) > 1.
Returning to the functional equation for θeχ , now we can make the change of variables: replace b by b`−1
mod N :
−i X −1
` X
θeχ (iy) =
χ(b)e2πib/N
χ (`) f
√
√
N y
N y
b mod N
`∈Z
The inner sum has become a Gauss sum hχ, ψi not depending on `,
hχ, ψi =
X
(with ψ(b) = e−2πib/N )
χ(b) ψ(b)
b mod N
so we have
−ihχ, ψi X −1
` −ihχ, ψi e
i θeχ (iy) =
·
χ (`) f
=
· θχ−1
√
√
√
N y
N y
N y
N 2y
(odd χ primitive mod N )
`∈Z
giving the functional equation
−ihχ, ψi e
i θeχ (iy) =
· θχ−1
√
N y
N 2y
(odd χ primitive mod N )
[4] This Fourier transform is readily computed: view the integral defining the Fourier transform as a contour integral
along the real axis. Moving the contour upward by iξ immediately computes the Fourier transform, much as was
done with the Gaussian itself.
7
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
[2.3.1] Remark: The functional equation relates θeχ to θeχ−1 , rather than to θeχ itself.
[2.3.2] Remark: The flip in the argument y is y → 1/N 2 y, not simply y → 1/y. The fixed-point of this
map is 1/N , influencing the break-up of the integral in proof of the analytic continuation and functional
equation, next.
[2.4] Analytic continuation and functional equation Starting from the integral representation,
π
− s+1
2
Γ( s+1
2 ) L(s, χ)
Z
∞
=
0
Z ∞
Z 1/N e
e
s θχ (iy) dy
s θχ (iy) dy
θeχ (iy) dy
2
y
=
y
+
y2
2
y
2
y
2
y
1/N
0
s
2
As for ζ(s), the integral from 1/N to ∞ is nicely convergent for all s ∈ C, and gives an entire function.
The goal is to use the functional equation to convert the integral from 0 to 1/N to the other sort, and then
symmetrize the resulting expression as much as possible. That is, replace y by 1/N 2 y in the problemmatical
integral from 0 to 1/N , obtaining
1
2
Z
∞
s
(1/N 2 y) 2 θeχ
1/N
i dy
=
N 2y y
1
2
N −s
Z
∞
s
y − 2 θeχ
1/N
i dy
N 2y y
Replacing χ by χ−1 in the functional equation above gives
i −ihχ−1 , ψi e
· θχ
θeχ−1 (iy) =
√
N y
N 2y
Rearranging,
√
−iN y e
i =
θ −1 (iy)
N 2y
hχ−1 , ψi χ
θeχ
Substituting this gives
−iN 1−s
hχ−1 , ψi
Z
∞
y
1−s
2
1/N
θeχ−1 (iy) dy
2
y
That is,
π
− s+1
2
Γ( s+1
2 ) L(s, χ)
=
∞
Z
θeχ (iy) dy
−iN 1−s ∞ 1−s θeχ−1 (iy) dy
y 2
y
+
2
y
hχ−1 , ψi 1/N
2
y
1/N
Z
s
2
This gives the analytic continuation, but is not symmetrized.
Recall that for χ primitive,
|hχ, ψi| =
√
N
We do not pretend to know the argument of the Gauss sum, only its size: let
√
−i N
ε(χ) =
hχ−1 , ψi
Then |ε(χ)| = 1 and
π−
s+1
2
Γ( s+1
2 ) L(s, χ) =
Z
∞
1/N
s
y2
dy
e
1−s θχ−1 (iy)
1
θeχ (iy)
+ ε(χ)N 2 −s y 2
2
2
y
s
Ignoring ε(χ) for a moment, symmetrize the powers of the conductor N , by multiplying through by N 2 ,
giving
Z ∞ dy
e
e
s+1
1−s θχ−1 (iy)
s
s θχ (iy)
2
2
N 2 π − 2 Γ( s+1
)
L(s,
χ)
=
(N
y)
+
ε(χ)(N
y)
2
2
2
y
1/N
8
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
Regarding ε(χ): since χ is odd,
hχ−1 , ψi = hχ, ψi =
X
χ(b)ψ(−b) = χ(−1)
b
X
χ(b)ψ(b) = χ(−1)hχ, ψi = −hχ, ψi
(for odd χ)
b
Thus,
ε(χ) · ε(χ
−1
√
√
−i N · −i N
−N
= 1
) =
=
−1
hχ , ψi · hχ, ψi
−hχ, ψi · hχ, ψi
giving the symmetry
ε(χ) · ε(χ−1 ) = 1
(for χ odd)
Thus, we have
s
N 2 π−
s+1
2
Γ( s+1
2 ) L(s, χ) =
Z
∞
= ε(χ)
∞
Z
s
(N y) 2
1/N
dy
e
1−s θχ−1 (iy)
θeχ (iy)
+ (N y) 2
2
2
y
s
ε(χ−1 (N y) 2
1/N
= ε(χ) · N
1−s
2
(1−s)+1
2
π−
dy
e
1−s θχ−1 (iy)
θeχ (iy)
+ ε(χ)(N y) 2
2
2
y
) L(1 − s, χ−1 )
Γ( (1−s)+1
2
(χ odd)
giving the functional equation
s
N 2 π−
s+1
2
Γ( s+1
2 ) L(s, χ) = ε(χ) · N
1−s
2
π−
2−s
2
−1
Γ( 2−s
)
2 ) L(1 − s, χ
(odd χ primitive mod N )
where
√
−i N
ε(χ) =
hχ−1 , ψi
|ε(χ)| = 1
and
(for χ odd, primitive modulo N )
3. Dedekind zeta function ζo(s) for Gaussian integers o
[3.1] Integral representation For o = Z[i], with N α = αα = |α|2 , the zeta function
ζo (s) =
X
06=α∈o mod
o×
1
1
1X
=
s
(N α)
4 m,n (m2 + n2 )s
(summed over m, n not both 0)
is readily expressed as an integral transform of the theta function
X
θ(iy) =
e−π(m
2
+n2 )y
=
m,n∈Z
X
2
e−πn
y 2
n∈Z
by a computation similar to those above:
Z
0
∞
ys
1X
θ(iy) − 1 dy
=
4
y
4 m,n
Z
∞
2
y s e−π(m
+n2 )y
0
dy
1X
1
=
y
4 m,n π(m2 + n2 ) s
= π −s Γ(s) ζo (s)
9
Z
0
∞
y s e−y
dy
y
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
2
2
The theta function is made from a Gaussian f (u, v) = e−pi(u +v ) in
two variables. The fact that this function is a product of two single-variable functions is useful on other
occasions, but not essential at present.
[3.2] Functional equation of θ
The Poisson summation formula
X
X
ϕ(x) =
x∈Z2
will be applied to
ϕ(x)
b
x∈Z2
√
√
ϕ(u, v) = f ( y · u, y · v)
The Fourier transform of ϕ is readily computed
ϕ(u,
b v) =
1
u v f √ ,√
y
y
y
Thus,
2
X
θ(iy) =
e−π(m
+n2 )y
X 1
2
2
1 i
e−π(m +n )/y = θ
y
y y
2
=
m,n∈Z
x∈Z
[3.3] Analytic continuation and functional equation
As in all previous cases, the integral in the
integral representation breaks into two pieces, one already entire, and the other convertible to the same form
via the functional equation of θ, with elementary leftovers describing the poles. That is, first,
π −s Γ(s) ζo (s) =
∞
Z
ys
0
θ(iy) − 1 dy
=
4
y
Z
1
∞
ys
θ(iy) − 1 dy
+
4
y
Z
0
1
ys
θ(iy) − 1 dy
4
y
and the integral over [1, ∞) is entire. Use the functional equation and replacement of y by 1/y to obtain
Z
0
Z
=
1
θ(iy) − 1 dy
y
=
4
y
Z
s
∞
y −s
1
y
0
1
θ( yi )
s y
Z ∞
− 1 dy
yθ(iy) − 1 dy
=
y −s
4
y
4
y
1
Z ∞
Z
y(θ(iy) − 1) + y − 1 dy
θ(iy) − 1 dy 1 ∞ 1−s
dy
=
y 1−s
+
y
− y −s
4
y
4
y
4 1
y
1
Z ∞
1
1
θ(iy) − 1 dy 1
+
−
=
y 1−s
4
y
4 s−1 s
1
Altogether,
π
1
−s
Z
Γ(s) ζo (s) =
1
∞
y s + y 1−s
θ(iy) − 1 dy 1 1
1
+
−
4
y
4 s−1 s
This gives the meromorphic continuation and functional equation
π −s Γ(s) ζo (s) = π −(1−s) Γ(1 − s) ζo (1 − s)
The residues of the completed zeta function π −s Γ(s) ζo (s) at s = 0, 1 are 1/4. Thus, the residues of ζo (s)
itself are 1/4 at s = 0, and π/4 at s = 1.
4. Grossencharacter L-functions L(s, χ) for Gaussian integers
Another kind of L-function appears for every number field larger than Q, called Hecke characters or
größencharakteren, anglicized to grossencharacters.
10
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
Rather than give a general definition, we consider examples for Gaussian integers o = Z[i]. These characters
and L-functions are essential to applications of harmonic analysis to number theory. For example, study
of the distribution of Gaussian primes in angular sectors uses Hecke characters and L-functions in a style
parallel to the use of Dirichlet characters and L-functions in the study of the distribution of primes modulo
N . In one case, we have harmonic analysis on the circle, in the other case, harmonic analysis on (Z/N )× .
[4.1] Hecke L-functions
The simplest Hecke characters on Z[i] are the multiplicative group
homomorphisms
χ : α −→ (α/α)`
(` ∈ Z, group homomorphism C× −→ S 1 )
For χ to be well-defined on ideals it must give the same value on every generator of ideals, which requires
that χ be trivial on units: χ(η) = 1 for every η ∈ o× . In this example, triviality on units is the condition
i `
= (−1)`
−i
1 = χ(i) =
so ` ∈ 2Z. Let N α = αα = |α|2 as in the discussion of the zeta-function of the Gaussian integers. For χ
trivial on units, there are the corresponding Hecke L-functions
X
L(s, χ) =
06=α∈o mod o×
=
X
06=α∈o mod o×
(α/α)`
=
N αs
Y
$ prime
χ(α)
=
N αs
Y
$ prime
1
($/$)`
1−
N $s
1
χ($)
1−
N $s
(with χ(α) = (α/α)` , for ` ∈ 2Z)
[4.2] Harmonic theta series To obtain any L-function via an integral representation from a theta series,
the theta series must mimic the formation of the L-function. For Hecke L-functions with χ(α) = (α/α)` , we
might contemplate something like
X
θχ (iy) =
m,n∈Z
m + in ` −π(m2 +n2 )y
e
m − in
(???)
This fails: although the function
(u, v) −→
u + iv ` −π(u2 +v2 )
e
u − iv
has the right shape to produce the L-function by an integral, this function is not a Schwartz function, so
Poisson summation does not easily apply. After some experimentation, one might try something like
θχ (iy) =
X
2
(m + in)` e−π(m
+n2 )y
(??? (for ` ≥ 0))
m,n∈Z
As with the odd Dirichlet characters, to apply Poisson summation, we might want the theta function to be
of the form
X
√
θ(iy) =
ϕ(v · y)
v∈Z2
If so, then we’d take, instead,
θχ (iy) =
X
(m + in)` y `/2 e−π(m
m,n∈Z
11
2
+n2 )y
(??? (for ` ≥ 0))
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
That is, there would be an additional factor of y `/2 . Both conventions, with and without the factor of y e/2 ,
2
2
make sense. For immediate purposes, we include this factor. For ` ≥ 0, the function (u + iv)` e−π(u +v ) is
Schwartz, but an adaptation is needed for ` < 0. Further, noting that the exponents of m + in and m − in
should differ by `, we anticipate that probably a good choice is [5]
θχ (iy) =






X
(m + in)2` y ` e−π(m
2
+n2 )y
(for ` ≥ 0)
m,n∈Z
X

2
2


(m − in)−2` y ` e−π(m +n )y


(for ` ≤ 0)
m,n∈Z
That is, the exponent in the polynomial part is 2`, not `, even though ` is already even. The polynomials
(u, v) → (u + iv)` for ` ≥ 0 and (u, v) → (u − iv)−` for ` ≤ 0 are holomorphic on R2 ≈ C. Thus, they are
harmonic. [6] For this reason, these theta series are called harmonic theta series. The functional equations
will be proven via Poisson summation, below.
[4.3] Integral representation
It is reasonable to see what happens when we write a simple integral
involving the theta functions. Take ` > 0 for simplicity. Then the lattice point (0, 0) does not contribute to
the sum, and no subtraction is necessary, unlike the case ` = 0 arising for the zeta function of o itself. Since
|o× | = 4, divide through by 4, as in the case of the zeta function:
Z
∞
0
1X
θχ (iy) dy
=
(m + in)2`
y
4
y
4 m,n
s
=
= π −(s+`) Γ(s + `)
Z
∞
y s y ` e−π(m
2
+n2 )y
0
1X
1
1
(m + in)2` · s
4 m,n
π (m2 + n2 )s+`
∞
Z
y s+` e−y
0
dy
y
dy
y
1
π −(s+`) Γ(s + `) X [(m + in)/(m − in)]`
1X
(m + in)2` ·
=
2
2
s+`
4 m,n
(m + n )
4
(m2 + n2 )s
m,n
= π −(s+`) Γ(s + `) L(s, χ)
(for χ(α) = (α/α)` and 0 < ` ∈ 2Z)
Note that omission of the factor y ` in θχ (iy) would yield L(s − `, χ) rather than L(s, χ). To have confidence
that the latter expression is a good normalization for the L-function, we need to see the theta-function’s
functional equation and the consequent functional equation for L(s, χ).
[4.4] Poisson summation, Hecke’s identity, functional equation of θχ
ϕ(u, v) = (u + iv)2` e−π(u
2
+v 2 )
Letting
(for 0 < ` ∈ 2Z)
[5] Regarding the question of including an extra factor of y ` , or not: when the extra factor is omitted, then θ (z) is
χ
a holomorphic function in z ∈ H. This is one argument in favor of dropping that factor.
[6] A twice-differentiable function f on Rn is harmonic if it is annihilated by the Laplacian ∆ = ∂ 2 + . . . + ∂ 2 . In
∂x2n
∂x21
two variables, the Laplacian factors
∂2
∂2
+
∂x2
∂y 2
=
∂
∂ ∂
∂ +i
−i
∂x
∂y
∂x
∂y
The second linear differential operator is the Cauchy-Riemann operator. Annihilation by the Cauchy-Riemann
operator is equivalent to holomorphy.
12
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
√
the theta function is exactly the Schwartz function ϕ, dilated by y, summed over a lattice:
X
√ θχ (iy) =
ϕ (m, n) · y
m,n∈Z
Poisson summation gives
X
θχ (iy) =
ϕ (m, n) ·
m,n∈Z
1 X
1 √ y =
ϕ
b (m, n) · √
y
y
m,n∈Z
To compute the Fourier transform ϕ,
b use coordinates w = u + iv and w = u − iv. The Fourier transform is
Z
Z
−2πi(uξ+vη)
ϕ(u,
b v) =
e
ϕ(ξ, η) dξ dη =
e−πi(wz+wz) z 2` e−πzz dξ dη
(with z = ξ + iη)
R2
R2
Now use
∂
=
∂w
1
2
∂
∂ −i
∂u
∂v
∂
=
∂w
1
2
∂
∂ +i
∂u
∂v
and compute
Z
e
−πi(wz+wz)
∂ 2` Z
1
dξ dη =
e−πi(wz+wz) e−πzz dξ dη
(−πi)2` ∂w
R2
2` −πzz
z e
R2
=
∂ 2`
1
1
e−πww =
(−πw)2` e−πww = i−2` w2` e−πww
2`
(−πi)
∂w
(−πi)2`
since the Gaussian e−πzz is its own Fourier transform. That is, ϕ
b = i−2` ϕ. That is,
2
2 2
2
Fourier transform (u + iv)2` e−π(u +v ) = (−1)−` · (u + iv)2` e−π(u +v )
By Poisson summation,
X
θχ (iy) =
(−1)` X
(−1)`
1
1 √ y =
·
=
· θχ
ϕ (m, n) · √
y
y
y
y
ϕ (m, n) ·
m,n∈Z
m,n∈Z
[4.4.1] Remark: If the factor of y` were omitted the functional equation of the corresponding theta function
would be
X
(m + in)2` e−π(m
2
+n2 )y
m,n∈Z
=
2
2
(−1)` X
(m + in)2` e−π(m +n )/y
1+2`
y
m,n∈Z
This can be obtained from the result above by dividing through by y ` and rearranging slightly: these two
forms of the functional equation are not truly different.
[4.5] Analytic continuation and functional equation
As in all other examples, break the integral of
the integral representation into two pieces, one of which is entire, the other converted to the same sort of
entire integral by using the functional equation of θχ . Treat just the case 0 < ` ∈ 2Z.
π
−(s+`)
Z
Γ(s + `) L(s, χ) =
0
∞
θχ (iy) dy
y
=
4
y
s
Z
1
∞
θχ (iy) dy
y
+
4
y
s
Z
0
1
ys
θχ (iy) dy
4
y
The integral on [1, ∞ is entire in s. The integral on [0, 1] is transformed via the functional equation of θχ and
replacing y by 1/y. Since ` > 0, the theta function has no constant term to subtract, so the book-keeping is
simpler:
Z 1
Z 1
Z ∞
1
i
θχ (iy) dy
θχ (iy) dy
y θχ ( y ) dy
ys
= (−1)`
ys
= (−1)`
y 1−s
4
y
4
y
4
y
0
0
1
13
Paul Garrett: Analytic continuation, functional equation: examples (June 22, 2012)
That is,
π −(s+`) Γ(s + `) L(s, χ) =
Z
∞
y s + (−1)` y 1−s
1
θχ (iy) dy
4
y
The right-hand side is entire, and the functional equation gives a symmetry under s ↔ 1 − s:
π −(s+`) Γ(s + `) L(s, χ) = (−1)` · π −(1−s+`) Γ(1 − s + `) L(1 − s, χ)
(for 0 < ` ∈ 2Z)
The computation for χ(α) = (α/α)` with ` < 0 is similar. The two computations can be put into a common
form
π −(s+|`|) Γ(s + |`|) L(s, χ) = (−1)` · π −(1−s+|`|) Γ(1 − s + |`|) L(1 − s, χ)
(for ` ∈ 2Z)
[4.6] Where’s the contragredient? For Dirichlet L-functions, the functional equation relates L(s, χ)
and L(1 − s, χ), with the complex conjugate character χ rather than χ. That was the simplest example of
the appearance of a contragredient. Where is the contragredient in the functional equation for the Hecke
character L-functions above? The answer is that, although the Hecke character χ(α) = (α/α)` and its
conjugate χ(α) = (α/α)−` are obviously distinct, their L-functions are equal. This can be seen from the
defining sum, by replacing α by α:
L(s, χ) =
X
06=α∈o mod o×
(α/α)`
=
|N α|s
X
06=α∈o mod o×
(α/α)`
=
|N α|s
X
06=α∈o mod o×
(α/α)−`
= L(s, χ)
|N α|s
Thus, here, the functional equation does relate L(s, χ) and L(1 − s, χ), but, also, L(s, χ) = L(s, χ).
For L-functions combining congruence conditions with Hecke-character features as above, the L-function of
the contragredient would rarely match the original.
14