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 $\langle f, f \rangle = \frac{(k-1)! L(k,\textrm{Sym}^2 f)}{2^{2k-1}\pi^{k+1}}$.
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Now that we have $\langle f, f \rangle$ we can calculate the algebraic values $L_{\textrm{alg}}(r+k-1,\textrm{Sym}^2 f) = \frac{L(r+k-1,\textrm{Sym}^2 f)}{\langle f, f \rangle \pi^{2r+k-1}}$.
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We start by checking that we recover the Loeffler-Zerbes example.
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.
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We want to calculate $L_{\textrm{alg}}(j,\textrm{Sym}^2 f \otimes \psi) = \frac{(j-1)! (j-16)! G(\psi^{-1})^2}{2^{2j+1} \pi^{2j-15} \langle f, f \rangle} L(j, \textrm{Sym}^2 f \otimes \psi).$
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This shows that $\chi$ is the inverse of $\psi$, which is necessary for the algebraic value calculation.
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Note the above value agrees with Loeffler-Zerbes.
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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.
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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.
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