symmetric_square

1806 days ago by jimlb

Note the code that is implemented computes $L(j,\textrm{Sym}^2,f)$ directly, but not the algebraic part.  This means we need to calculate $\langle f, f \rangle$ first using the formula $\langle f, f \rangle = \frac{(k-1)! L(k,\textrm{Sym}^2 f)}{2^{2k-1}\pi^{k+1}}$. 
R=RealField(750);R 
       
def Cusp_Forms(N,k): S=CuspForms(N,k) return S 
       
S=Cusp_Forms(1,16);S 
       
f=S.basis()[0];f 
       
L=f.symsquare_lseries(prec=750);L 
       
def Petersson_Product(L,k): PProduct=factorial(k-1)*L(k)/(2^(2*k-1)*pi^(k+1)) return PProduct 
       

Now that we have $\langle f, f \rangle$ we can calculate the algebraic values $L_{\textrm{alg}}(r+k-1,\textrm{Sym}^2 f) = \frac{L(r+k-1,\textrm{Sym}^2 f)}{\langle f, f \rangle \pi^{2r+k-1}}$. 
def symsquare_lseries_alg(L,r,f): A=L(r+f.weight()-1)/(Petersson_Product(L, f.weight()) * pi^(2*r+f.weight()-1)) return A 
       

We start by checking that we recover the Loeffler-Zerbes example. 
R(Petersson_Product(L,16)); 
        
verbose -1 (370: dokchitser.py, __call__) Warning: Loss of 2 decimal
digits due to cancellation

verbose -1 (370: dokchitser.py, __call__) Warning: Loss of 2 decimal digits due to cancellation

The above value for $\langle f, f \rangle$ agrees with what Loeffler-Zerbes obtain for their example. 
psi = DirichletGroup(7).0^2;psi 
       

We want to calculate $L_{\textrm{alg}}(j,\textrm{Sym}^2 f \otimes \psi) = \frac{(j-1)! (j-16)! G(\psi^{-1})^2}{2^{2j+1} \pi^{2j-15} \langle f, f \rangle} L(j, \textrm{Sym}^2 f \otimes \psi).$
G=DirichletGroup(7); for j in [0..5]: G[j]; 
       
chi=G[4]; 
       
for j in [0..5]: chi(j)*psi(j) 
       
psi(3)*chi(3); 
       

This shows that $\chi$ is the inverse of $\psi$, which is necessary for the algebraic value calculation. 
def symsquare_lseries_twisted_alg(L1,L2,j,chi,f): A=factorial(j-1)*factorial(j-16)*(chi.gauss_sum_numerical(prec=200))^2*L1(j)/(Petersson_Product(L2, f.weight()) * pi^(2*j-15)*2^(2*j+1)) return A 
       
L2=L;L2 
       
L1=f.symsquare_lseries(psi,prec=750);L1 
       
C=ComplexField(750);C 
       
C(symsquare_lseries_twisted_alg(L1,L2,22,chi,f)) 
       

Note the above value agrees with Loeffler-Zerbes. 
A=Rational(R(C(symsquare_lseries_twisted_alg(L1,L2,22,chi,f)).real())) 
       
B=Rational(R(C(symsquare_lseries_twisted_alg(L1,L2,22,chi,f)).imag()/sqrt(3))) 
       
lcm(A.denominator(),B.denominator()); 
       

I am not sure how they get their value in ${\bf Q}(\sqrt{-3})$.  I tried taking the real part of our value and asking SAGE to convert it to a fraction and then divide the imaginary part by $\sqrt{3}$ and do the same thing, but that didn't work. 
continued_fraction(7.83232263839494) 
       

We will now try this with $\psi = \chi_{K}$ for $K = {\bf Q}(\sqrt{-3})$.  This is a quadratic character of conductor $3$, so it is constructed a follows. Note that $\psi$ is its own inverse so it is less work than the above calculation.
G=DirichletGroup(3); for j in [0..1]: G[j]; 
       
psi=G[1];psi 
       
L1=f.symsquare_lseries(psi);L1 
       
Rational(CC(symsquare_lseries_twisted_alg(L1,L2,21,chi,f)).real())