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2

$begingroup$

I'm trying to work out the details of Theorem 4.7.1 in Miyake's book Modular Forms, which shows how to construct modular forms from suitable pairs of Dirichlet $L$-functions. I seem to have made some minor computational error that I'm having trouble tracking down.

Let $chi$ be an odd, primitive Dirichlet character modulo $N$. For $n geq 1$ let
$$a_n = sum_d mid nchi(d).$$
Then we have an $L$-function
$$L(s) = sum_n=1^inftyfraca_nn^s$$
which satisfies
$$L(s) = zeta(s)L(s,chi),$$
where $zeta(s)$ is the Riemann zeta function and $L(s,chi)$ is the Dirichlet $L$-function of $chi$. Set
beginequation*
beginaligned
Lambda_N(s) &= (2pi/sqrtN)^-sGamma(s)L(s) \
Lambda(s) &= (pi)^-s/2Gamma(s/2)zeta(s) \
Lambda(s,chi) &= (pi/N)^-s/2Gamma((s+1)/2)L(s,chi)
endaligned
endequation*
After some computation we find that
$$Lambda(s)Lambda(s,chi) = 2sqrtpiLambda_N(s).$$

I need to compute the residues at the simple poles at $s = 0,1$. I've taken care of $s = 0$, but there seems to be a minor problem for the residue at $s = 1$. Now I know from the book that this residue is supposed to equal
$$-fraciG(chi)B_1,chiL(0,chi)2sqrtN,$$
where $G(chi) = sum_a=1^Nchi(a)e^2pi ia/N$ is the Gauss sum of $chi$.
I have worked this out using the functional equation for $L(s,chi)$, but the problem is that it relates the the values of $L(s,chi)$ to those of the function $L(1 - s,overlinechi)$ of the conjugate character, so I wind up with $L(0,overlinechi)$ instead. All of the other pieces of the desired result are there, so I must have made a mistake unless $L(0,chi) = L(0,overlinechi)$. But I know how to compute these values with Bernoulli numbers, and from this one sees this cannot be the case unless $L(0,chi)$ happens to be real.

complex-analysis number-theory

share|cite|improve this question

asked Mar 14 at 7:56

Ethan AlwaiseEthan Alwaise

6,456717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure $Lambda(s)=Lambda(1-s)$ has two simple poles at $0,1$ of residue $1$ and $Lambda(s,chi)=Lambda(1-s,overlinechi)$ is entire so you are right that $Lambda(s)Lambda(s,chi)$ has two simple poles at $0,1$ of residue $L(0,chi)$ and $L(0,overlinechi)$. Then you can look at $f(z,chi) =C+ sum_n=1^infty e^2ipi n z sum_d chi(d) $ the inverse Mellin transform of $N^-s/2 Lambda_N(s)$ to find $f(-1/(Nz),chi) = G(chi) N^1/2 z^-1 f(z,overlinechi)$
  $endgroup$
  – reuns
  Mar 14 at 17:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also from $ chi(n) = fracG(chi)Nsum_k=1^N overlinechi(k) e^2ipi kn/N$ and $frac1e^2ipi z-1=frac12+sum_l frac1z+l$ you obtain $f(z,chi)=-G(chi)sum_msum_l fracoverline)Nmz+l$ an Eisenstein of weight $1$ for $Gamma_0(N)$ and character $overlinechi$
  $endgroup$
  – reuns
  Mar 14 at 17:02
  
  

  
  
  
  

add a comment | 

2

$begingroup$

I'm trying to work out the details of Theorem 4.7.1 in Miyake's book Modular Forms, which shows how to construct modular forms from suitable pairs of Dirichlet $L$-functions. I seem to have made some minor computational error that I'm having trouble tracking down.

Let $chi$ be an odd, primitive Dirichlet character modulo $N$. For $n geq 1$ let
$$a_n = sum_d mid nchi(d).$$
Then we have an $L$-function
$$L(s) = sum_n=1^inftyfraca_nn^s$$
which satisfies
$$L(s) = zeta(s)L(s,chi),$$
where $zeta(s)$ is the Riemann zeta function and $L(s,chi)$ is the Dirichlet $L$-function of $chi$. Set
beginequation*
beginaligned
Lambda_N(s) &= (2pi/sqrtN)^-sGamma(s)L(s) \
Lambda(s) &= (pi)^-s/2Gamma(s/2)zeta(s) \
Lambda(s,chi) &= (pi/N)^-s/2Gamma((s+1)/2)L(s,chi)
endaligned
endequation*
After some computation we find that
$$Lambda(s)Lambda(s,chi) = 2sqrtpiLambda_N(s).$$

I need to compute the residues at the simple poles at $s = 0,1$. I've taken care of $s = 0$, but there seems to be a minor problem for the residue at $s = 1$. Now I know from the book that this residue is supposed to equal
$$-fraciG(chi)B_1,chiL(0,chi)2sqrtN,$$
where $G(chi) = sum_a=1^Nchi(a)e^2pi ia/N$ is the Gauss sum of $chi$.
I have worked this out using the functional equation for $L(s,chi)$, but the problem is that it relates the the values of $L(s,chi)$ to those of the function $L(1 - s,overlinechi)$ of the conjugate character, so I wind up with $L(0,overlinechi)$ instead. All of the other pieces of the desired result are there, so I must have made a mistake unless $L(0,chi) = L(0,overlinechi)$. But I know how to compute these values with Bernoulli numbers, and from this one sees this cannot be the case unless $L(0,chi)$ happens to be real.

complex-analysis number-theory

share|cite|improve this question

asked Mar 14 at 7:56

Ethan AlwaiseEthan Alwaise

6,456717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure $Lambda(s)=Lambda(1-s)$ has two simple poles at $0,1$ of residue $1$ and $Lambda(s,chi)=Lambda(1-s,overlinechi)$ is entire so you are right that $Lambda(s)Lambda(s,chi)$ has two simple poles at $0,1$ of residue $L(0,chi)$ and $L(0,overlinechi)$. Then you can look at $f(z,chi) =C+ sum_n=1^infty e^2ipi n z sum_d chi(d) $ the inverse Mellin transform of $N^-s/2 Lambda_N(s)$ to find $f(-1/(Nz),chi) = G(chi) N^1/2 z^-1 f(z,overlinechi)$
  $endgroup$
  – reuns
  Mar 14 at 17:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also from $ chi(n) = fracG(chi)Nsum_k=1^N overlinechi(k) e^2ipi kn/N$ and $frac1e^2ipi z-1=frac12+sum_l frac1z+l$ you obtain $f(z,chi)=-G(chi)sum_msum_l fracoverline)Nmz+l$ an Eisenstein of weight $1$ for $Gamma_0(N)$ and character $overlinechi$
  $endgroup$
  – reuns
  Mar 14 at 17:02
  
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to work out the details of Theorem 4.7.1 in Miyake's book Modular Forms, which shows how to construct modular forms from suitable pairs of Dirichlet $L$-functions. I seem to have made some minor computational error that I'm having trouble tracking down.

Let $chi$ be an odd, primitive Dirichlet character modulo $N$. For $n geq 1$ let
$$a_n = sum_d mid nchi(d).$$
Then we have an $L$-function
$$L(s) = sum_n=1^inftyfraca_nn^s$$
which satisfies
$$L(s) = zeta(s)L(s,chi),$$
where $zeta(s)$ is the Riemann zeta function and $L(s,chi)$ is the Dirichlet $L$-function of $chi$. Set
beginequation*
beginaligned
Lambda_N(s) &= (2pi/sqrtN)^-sGamma(s)L(s) \
Lambda(s) &= (pi)^-s/2Gamma(s/2)zeta(s) \
Lambda(s,chi) &= (pi/N)^-s/2Gamma((s+1)/2)L(s,chi)
endaligned
endequation*
After some computation we find that
$$Lambda(s)Lambda(s,chi) = 2sqrtpiLambda_N(s).$$

I need to compute the residues at the simple poles at $s = 0,1$. I've taken care of $s = 0$, but there seems to be a minor problem for the residue at $s = 1$. Now I know from the book that this residue is supposed to equal
$$-fraciG(chi)B_1,chiL(0,chi)2sqrtN,$$
where $G(chi) = sum_a=1^Nchi(a)e^2pi ia/N$ is the Gauss sum of $chi$.
I have worked this out using the functional equation for $L(s,chi)$, but the problem is that it relates the the values of $L(s,chi)$ to those of the function $L(1 - s,overlinechi)$ of the conjugate character, so I wind up with $L(0,overlinechi)$ instead. All of the other pieces of the desired result are there, so I must have made a mistake unless $L(0,chi) = L(0,overlinechi)$. But I know how to compute these values with Bernoulli numbers, and from this one sees this cannot be the case unless $L(0,chi)$ happens to be real.

complex-analysis number-theory

share|cite|improve this question

asked Mar 14 at 7:56

Ethan AlwaiseEthan Alwaise

6,456717

$endgroup$

share|cite|improve this question

asked Mar 14 at 7:56

Ethan AlwaiseEthan Alwaise

6,456717

share|cite|improve this question

asked Mar 14 at 7:56

Ethan AlwaiseEthan Alwaise

6,456717

asked Mar 14 at 7:56

Ethan AlwaiseEthan Alwaise

6,456717

asked Mar 14 at 7:56

Ethan AlwaiseEthan Alwaise

6,456717