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I am really embarrassed to ask this question since I am supposed to be an expert on modular forms, but here goes anyway since I have been stuck on this for several days.

Let $q$ be a prime divisor of $N$ such that $q^2\nmid N$, and let $\chi$ be some Dirichlet character modulo $N$. If $F$ is a normalized eigenform (newform of course) in $S_k(\Gamma_0(N),\chi)$ with Fourier coefficients $a(n)$, it is known from Atkin--Lehner--Li that if $W_q$ denotes an Atkin--Lehner involution then $F$ is an eigenform for the action of $W_q$ with eigenvalue $-a(q)/q^{k/2-1}$, and also that $a(q)^2=\chi(q)q^{k/2-1}$.

I am implementing this in Pari/GP and found a contradiction. Most probably my implementation is wrong, but please follow the following simple example (I know that David Loeffler had a ticket about a similar problem in Sage, but that was solved).

Consider the space $S_3(\Gamma_0(12),\chi_{-4})$, where as usual $\chi_{-4}$ is the odd character mod 4. It has a single eigenfunction defined over $\mathbb Q(\sqrt{-3})$, together with its conjugate. I can choose $W_3=[3,-1;12,-3]$, and $\tau=i=\sqrt{-1}$, so that $W_3(\tau)=(3i-1)/(12i-3)$. I then compute $1000$ Fourier coefficients of $F$, and the value $F(i)$ and $F|_3W_3(i)$. Instead of finding $-a(3)/3^{1/2}$ as ratio, I find $+a(3)/3^{1/2}$ (equal to $i$ and $-i$, but in the wrong order or equivalently with the wrong sign).

I did many other experiments, both in odd and even weights, and with nonreal characters, and my implementation tells me that the eigenvalue should be $-a(q)\overline{\chi(q)}/q^{k/2-1}$.

Now I went through Winnie Li's proof which I essentially reproduced in my book with F. Stromberg, and it seems correct. So I am stuck. Can someone at least check the above example, or tell me what I am doing wrong ? Note that I am not mixing the two possible embeddings of my form: both give me the wrong sign.

ADDED March 27: I am very surprised not to have any (positive or negative) answer. Since the error is systematic, let me give another example in even weight with numerical values. Consider $S_2(\Gamma_0(35),\chi_5)$. It is of dimension $2$ generated by an eigenform and its conjugate. One specific embedding of the eigenform is (sorry for the GP notation) $$F(\tau)=q + 2*I*q^2 - I*q^3 - 2*q^4 + (-2 - I)*q^5 + 2*q^6 - I*q^7 + 2*q^9 + O(q^{10}).$$ I take a random point, say $\tau=I/2+1/Pi$. Simply by summing sufficiently many terms I find $F(\tau)=-0.015178...+0.03676...*I$.

I choose as Atkin--Lehner matrix for the prime $q=7$ the matrix $W_7=[7,-3;35,-14]$ which has the required properties. I compute $F|_2W_7(\tau)$ (here I use 2000 terms since the imaginary part is small) and I find that it is equal to $0.03676+0.015178...*I$, so the eigenvalue is $-I$. However, Atkin--Lehner-Li tells me it should be equal to $-a(7)$, which as one sees from above is equal to $+I$, contradiction. Can someone please check this in Sage, Magma, or for that matter Pari/GP ?

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    $\begingroup$ Here's a shot in the dark. Scanning over the paper of Atkin and Li quickly, I see that $f|W_N = \lambda_N(f) \overline{f}$, where $W_N$ is the full Atkin-Lehner/Fricke involution, and $\lambda_N(f)$ is the pseudo-eigenvalue of $W_N$. In general, what they claim is that $f|W_Q = \lambda_Q(f) g$, for some other form $g$. It's not quite clear which form $g$ is, I suppose in your level $12$ case there are two possibilities. Just calculating one point $i$ and taking the ratio of $f(i)$ and $f|W_3(i)$ is not enough information to pin down if $f|W_3$ is proportional to $f$ or its conjugate, right? $\endgroup$ – Matt Young Mar 24 '18 at 1:21
  • $\begingroup$ You say that $a(q)^{2} = \chi(q) q^{k/2-1}$. This isn't always true. In particular, $|a(q)| = q^{k/2-1}$ if $q$ doesn't divide the conductor of $\chi$ (which means the local representation at $q$ is the Steinberg or a twist thereof), while in the case that $q$ does divide the conductor of $\chi$, one should have $|a(q)| = q^{(k-1)/2}$ (and the local representation at $q$ is a ramified principal series). $\endgroup$ – Jeremy Rouse Mar 24 '18 at 2:45
  • $\begingroup$ @Jeremy: sure, but I assume that $q^2\nmid N$ and $\chi$ is defined modulo $N/q$, so we are not in that case. I added that well-known property of $a(q)^2$ for completeness, but this is not my contradiction. $\endgroup$ – Henri Cohen Mar 24 '18 at 6:40
  • $\begingroup$ To add to my earlier comment, the two forms $f$ and $\bar{f}$ take complex-conjugate values on the imaginary axis. If you evaluate at different points than $i$ do you get consistent values of the pseudo-eigenvalue? $\endgroup$ – Matt Young Mar 24 '18 at 13:36
  • $\begingroup$ @Matt: yes, I get consistent values. Note that in my situation the pseudo-eigenvalues are really eigenvalues since $\chi$ is defined modulo $N/q$. In particular, we really have $F|W_Q=\lambda_Q(F)F$ with the same $F$. I tried many other examples, including when the Galois orbit of $F$ has more than $2$ elements and/or $\chi$ is of order greater than $2$. $\endgroup$ – Henri Cohen Mar 24 '18 at 14:23
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Let's examine the case $N=12$, weight $3$, with central character $\chi_{-4}$. According to the LMFDB, there are two newforms $f$ in this space. I will take it as numerically-verified that for both of these forms, $f|W_3$ is proportional to $f$.

Theorem 2.1 of Atkin-Li (Twists of newforms and pseudo-eigenvalues of W-operators. Invent. Math. 48 (1978), no. 3, 221–243) then states $f|W_3 = \lambda f,$ where $$\lambda = -\frac{|a_f(3)|}{a_f(3)}.$$ Here the minus sign comes from their convention that the Gauss sum of a trivial character is $-1$, under the conditions at hand. Here I multiplied by $|a_f(3)|$ because the pseudo-eigenvalue has absolute value $1$. Indeed, the earlier Proposition 1.1 from that paper shows $\lambda^2 = -1$, so the only question is the sign of $\lambda$. Now the above formula for $\lambda$ becomes $\lambda = - \frac{\overline{a_f(3)}}{|a_f(3)|} = \frac{a_f(3)}{|a_f(3)|}$, since $\lambda = \pm i$. Isn't this the numerical value that is claimed in the original question?

added later: I consulted Theorem 3 of Li's 1975 Math. Annalen paper (Newforms and functional equations). I will continue the discussion of this example. Li defines a version of the Atkin-Lehner operator, denoted $V_3^M$ (I will drop the superscript in what follows). However, I think that her choice of $V_3$ is inconsistent with $W_3$. According to Li's definition, we may choose $V_3 = \begin{pmatrix} -3 & -1 \\ 12 & 3 \end{pmatrix}$. She requires the lower-right entry to be $3$. With $W_3$ given as in the original question, we have $$V_3 W_3^{-1} = \begin{pmatrix} 7 & -2 \\ -24 & 7 \end{pmatrix}.$$ Since $\chi_{-4}(7) = -1$, we have $$f|V_3 = - f |W_3.$$

It should be mentioned that the definition of the Atkin-Lehner operator in the Atkin-Li paper has slightly different congruence conditions. In particular, I checked that the choice of $W_3$ in the original question is consistent with the definition of the Atkin-Lehner definition.

Unless I made some mistakes, I believe the above discussion clears up the problem, at least in this level $12$ example.

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  • $\begingroup$ I am sorry to insist, but I am not referring to pseudoeigenvalues but to true eigenvalues ($f|W_3/f$), so to Theorem 3 (iii) in the original paper of Li 1975 Inventiones, which I believe is rewritten in more recent language in her book (which I don't have with me) $\endgroup$ – Henri Cohen Mar 27 '18 at 22:35
  • $\begingroup$ Pseudoeigenvalue reduces to eigenvalue in this case, then. I don't have access to Li's 1975 paper at the moment. But do you agree that your calculation is in accord with the result of Atkin-Li? $\endgroup$ – Matt Young Mar 27 '18 at 22:40
  • $\begingroup$ I added some discussion in light of the notation from Li's 1975 paper. $\endgroup$ – Matt Young Mar 28 '18 at 2:35
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    $\begingroup$ To Matt and everybody: sorry for wasting bandwidth and time: Matt is right, it is all a question of normalization of the $W_Q$ matrix: the eigenvalue is indeed $-a(q)/q^{k/2-1}$, but ONLY if you choose a matrix $[qx,y;Nz,q]$ with a $q$ in the lower right corner. Thus in my initial example, I must choose $[-3,-1;12,3]$ and NOT $[3,-1;12,-3]$. This probably clears up every problem. Thanks again! $\endgroup$ – Henri Cohen Mar 28 '18 at 9:24

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