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I am interested in computing the algebraic parts of $L(f, n, \chi)$ for a primitive form $f \in S_k(\Gamma_0(N))$ with wieght $k > 2$ twisted by a primitive Dirichlet character $\chi$ of conductor $m$. It is known that there are some non-zero complex numbers $\Omega^{\pm}$ depending on the sign of $\chi$ such that for any positive integer $n$ with $1 \le n \le k-1$, \begin{equation*} \Lambda(f, n, \chi) := \frac{c(m, n, \chi)}{\Omega^{\pm}}L(f, n, \chi) \in \mathbb{Z}[\chi], \end{equation*} for some constants $c(m,n,\chi) \in \mathbb{C}$ depending on $m, n \text{ and } \chi$ and $\mathbb{Z}[\chi]$ is the ring of integers generated by the values of $\chi$.

For $k = 2$, it is well known to compute $\Omega^{\pm}$. For $k > 2$, For a given such $f$ and $\chi$, I can compute the values of $L(f, n, \chi)$ with the Dokchitser's calculator as well as $c(m,n,\chi)$ but can not find a way to compute $\Omega^{\pm}$.

Question: Can anyone let me know how to compute $\Omega^{\pm}$ directly? A reference for this computation is also welcome.

Update #1: Following Chapter 1. on "On $p$-adic analogues of the conjecture of Birch and Swinnerton-Dyer" by Mazur, Tate and Teitelbaum, we have \begin{equation*} c(m, n, \chi) = \frac{(n-1)!m^n}{(-2\pi i)^{n-1}\tau(\chi)}, \end{equation*} where $\tau(\chi)$ is the Gauss sum of $\chi$. For the computations of $L(f, n, \chi)$ one way is to use the functional equation and evaluate the series of integrals involving the Fourier coefficients of $f$ within some precision (This is exactly how Dokchitser built in his L-value calculator). In Mazur, Tate and Teitelbaum, they presented some numerical values of $\Lambda(f, n, \chi)$ on page 11 for quadratic twists of the primitive form $f \in S_6(\Gamma_0(3))$. From there, I can obtain $\Omega^{+}$ with the values of $L(f, n, \chi)$ computed by Dokchitser's $L$-function calculator.

I hope this update can explain it more explicitly as reuns wants.

Update #2: As David pointed out, the form $f$ should be assumed that its Fourier coefficients are in $\mathbb{Q}$.

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  • $\begingroup$ No, I am able to compute $L(f, n, \chi)$ for those $n$'s. What I want to compute is $\Lambda(f, n, \chi)$ by computing $\Omega^{\pm}$. $\endgroup$ – user115356 Oct 4 '17 at 15:40
  • $\begingroup$ Can you be more explicit how you compute it and what you get, maybe take an example ? $\endgroup$ – reuns Oct 4 '17 at 15:55
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I think your question is actually not well-posed, because it is not clear if such $\Omega_{\pm}$ actually exist, and if they do exist they are very far from being uniquely determined.

Firstly, it is not true that the quantity you call $\Lambda(f, n, \chi)$ is actually in $\mathbf{Z}[\chi]$. Shimura's theorem on periods of modular forms states that you can choose the $\Omega_{\pm}$ such that the ratio is in $\mathbf{Q}(f, \chi)$ -- the field extension generated by the values of $\chi$ and the coefficients of $f$, which is in general much larger than $\mathbf{Q}(\chi)$; coefficient fields of modular forms are often pretty big. In Shimura's construction, the $\Omega_{\pm}$ are only unique up to scaling by elements of $\mathbf{Q}(f)^\times$.

However, you want more than just rationality: you want integrality. For each prime $\mathfrak{p}$ of the coefficient field, Vatsal showed how to pick a particular canonical $\mathfrak{p}$-integral normalisation of $\Omega_{\pm}$ such that the $\Lambda(f, n, \chi)$ were integral at $\mathfrak{p}$ (more precisely, at the primes of $\mathbf{Q}(f, \chi)$ above $\mathfrak{p}$). However, there's a nasty caveat here: because $\mathbf{Q}(f)$ might have non-trivial class group, there is no guarantee that you can choose a period that is $\mathfrak{p}$-canonical for every prime $\mathfrak{p}$ of $\mathbf{Q}(f)$ simultaneously!

So the objects you are looking for don't necessarily exist, and when they do exist, they are only well-defined up to multiplication by units in a potentially quite large number field. So in what sense is it meaningful to "compute" them?


Having spent a while pedantically criticising your question, let me now try to answer it. For simplicitly, suppose that $f$ has coefficients in $\mathbf{Q}$; then all my criticisms above go away, and $\Omega_{\pm}$ are well-defined up to signs.

In this case, I believe your question can be answered with modular symbols. For each sign $\pm$, take a basis over $\mathbf{Z}$ of the space of (homological) modular symbols of the appropriate weight, level and sign. Over $\mathbf{Q}$, there is a unique 1-dimensional quotient of modular symbols where the Hecke operators act as they do on $f$; the image of the integral mod symbols in this quotient must therefore be a copy of $\mathbf{Z}$, and you can choose one of your basis elements whose image generates this group.

If I remember correctly, the two numbers $\Omega_{\pm}$ are given by integrating $f$ against these symbols -- concretely, this will be a finite linear combination of integrals $\int_{\alpha}^\beta f(z) z^r dz$, where $\alpha, \beta$ are cusps and $0 \le r \le k-2$.

(The issue when $\mathbf{Q}(f)$ is larger is that the $f$-eigenspace quotient of modular symbols is now 1-dimensional over $\mathbf{Q}(f)$, and the image of the integral modular symbols is a rank 1 torsion-free module over the ring of integers of $\mathbf{Q}(f)$, but it needn't be a free module.)

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  • $\begingroup$ David, thank you for your answer in detail! Yes, the form $f$ in my question should have been mentioned with its Fourier coefficients in $\mathbb{Q}$. I will update the question as criticised. I have checked it with Sage and saw the modular symbol corresponding to $f$ with respect to the Hecke operators. I am not sure if you are familiar with the space of modular symbols on Sage. There is a method called "integral_period_mapping" for the 1-dimensional space of modular symbols for $f$. With this mapping, can I get the algebraic parts of $L(f, n, \chi)$ maybe upto a multiple of rational number? $\endgroup$ – user115356 Oct 5 '17 at 3:44
  • $\begingroup$ This is now a Sage question, not a mathematical one, and is probably better asked elsewhere. Anyway, if you read the docstring for integral_period_mapping you'll see that it's only implemented for weight 2 at present; but in the weight 2 case, it does indeed precisely calculate the integrally normalised algebraic parts. $\endgroup$ – David Loeffler Oct 5 '17 at 5:35

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