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I need the dimensions of the Atkin-Lehner eigenspace for the paper I'm writing.

As is well known, the cuspidal space $S_{k}(\Gamma_{0}(N))$ can be decomposed by Atkin Lehner involution. For example, when $N=21=3\times7$, we have the decomposition $$ S_{k}(\Gamma_{0}(N))=S^{(++)} \oplus S^{(+-)} \oplus S^{(-+)} \oplus S^{(--)} $$ where $S^{(++)}$, $S^{(+-)}$, $S^{(-+)}$, and $S^{(--)}$ are the subspaces of $S_{k}(\Gamma_{0}(N))$ for which pairs of eigenvalues for $W_3$, $W_7$ are (+1,+1), (+1,-1), (-1,+1), (-1,-1) respectively.

In addition, dimensions of each summand spaces are 0,0,1,0. This is from Table 5 of Antwerp IV.

For k=2, the dimension of each Atkin-Lehner eigenspace has already been given by Table 5 of Antwerp IV and David Kohel.

And for any weight k, I found the following documentation. However, using this, it is difficult for me to find the dimensions of the eigenspaces.

So, is there any idea or data?

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2 Answers 2

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You can compute these dimensions using modular symbols (an auxiliary space which has the same Hecke action as modular forms, but is easier to compute). Here's a Sage example for weight 4 cusp forms of level Gamma0(17):

sage: S=ModularSymbols(Gamma0(17), weight=4, sign=1).cuspidal_submodule()
sage: S.atkin_lehner_operator().charpoly().factor()
(x + 17) * (x - 17)^3

Sage normalises the Atkin–Lehner operator so that its square is multiplication by $(-N)^{k-2}$, since that normalisation works better for odd weights (it avoids introducing square roots). If you want to normalise so that the operator becomes an involution, as is usual for $\Gamma_0$ levels, you need to scale by $N^{(k - 2) / 2}$. So, in my example, the scaling factor is 17, and the output means that the +1 eigenspace has dimension 3, and the -1 eigenspace has dimension 1.

You can compute the local AL operators for each prime $p \mid N$ similarly, and get the simultaneous eigenspaces by taking intersections of kernels – here's a code snippet which does this:

sage: def AL_eigenspace_dims(N, k):
....:     facs = [p^r for (p, r) in N.factor()]
....:     S = ModularSymbols(Gamma0(N), sign=1, weight=k).cuspidal_submodule()
....:     Wmats = [S.atkin_lehner_operator(f).matrix() for f in facs]
....:     for s in cartesian_product([ [-1, 1] for f in facs ]):
....:         Vs = [(Wmats[i] - s[i]*facs[i]**((k-2)/2)).kernel() for i in range(len(facs))]
....:         V = reduce(lambda u,v: u.intersection(v), Vs)
....:         print(s, V.dimension())
....: 
sage: AL_eigenspace_dims(21, 2)
(-1, -1) 0
(-1, 1) 1
(1, -1) 0
(1, 1) 0
sage: AL_eigenspace_dims(21, 12)
(-1, -1) 7
(-1, 1) 6
(1, -1) 7
(1, 1) 8

This modular symbol method will be slow if k or N is large, because it involves linear algebra with matrices having about $kN$ rows and columns. (For $k N$ around 500, it takes roughly ten seconds on my machine). In contrast, the algorithm in Lloyd Kilford's draft code from your link uses the Riemann–Roch theorem, and the running time is dominated by factorising N, so it should be practical for N up to about 100 digits; but his implementation only covers the case of weight 2. In principle it should be possible to extend the Riemann–Roch method to higher weights, but I don't know if anyone has implemented this in Sage.

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  • $\begingroup$ For the first code, in case Gamma0(21), wt=12, I got (x + 4084101)^13 * (x - 4084101)^15, as I expected. (we know that $\dim S_{k}(\Gamma_{0}(21))=28$ and $\dim S_{k}(\Gamma_{0}^{+}(21))=13$.) However, for the second code, I got [1, 1] 0, [1, -1] 0, [-1, 1] 0, [-1, -1] 0. I expect [1, 1] 8, since $\dim S_{k}(\Gamma_{0}^{*}(21))=8$. Here $\Gamma_{0}^{+}(21)=<\Gamma_{0}(21) , W_{21}>$ and $\Gamma_{0}^{*}(21)$ is the group generated by $\Gamma_{0}(21)$ and all A-L involution. What's the problem? Is it because I didn't understand the modular symbol deeply and just followed the code you gave me..? $\endgroup$
    – kslhg
    Commented Sep 21, 2022 at 3:03
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    $\begingroup$ I'm guessing you didn't compensate for the normalisation which makes $W_Q^2$ a nontrivial power of Q when the weight is $>2$. $\endgroup$
    – David Loeffler
    Commented Sep 21, 2022 at 5:49
  • $\begingroup$ Oh thank you! I fixed it. $\endgroup$
    – kslhg
    Commented Sep 21, 2022 at 6:14
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    $\begingroup$ I've edited the answer to include a function which allows you to input any combination of levels and weights. $\endgroup$
    – David Loeffler
    Commented Sep 21, 2022 at 7:21
  • $\begingroup$ Wow! It’s great! Thanks a lot!! $\endgroup$
    – kslhg
    Commented Sep 21, 2022 at 8:05
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Exact formulas for dimensions of Atkin-Lehner eigenspaces follow from trace formulas of Yamauchi and Skoruppa-Zagier. Skoruppa-Zagier corrected some clerical errors in Yamauchi's paper. See:

Nils-Peter Skoruppa and Don Zagier, Jacobi forms and a certain space of modular forms, Invent. Math. 94 (1988), no. 1, 113–146.

In the case of squarefree level, I worked things out explicitly in this paper:

Kimball Martin, Refined dimensions of cusp forms, and equidistribution and bias of signs, J. Number Theory 188 (2018), 1–17.

The main focus is dimensions of new parts of Atkin-Lehner eigenspaces, but dimensions which include oldspaces can be computed similarly. Either you can take the newspace dimensions, and add in the old form contribution, or just use the trace formulas on the full spaces $S_k(N)$ to get a formula mimicking what I did for the newspaces.

Code to compute dimensions (and a link to my paper) in the squarefree level case is available here:

https://math.ou.edu/~kmartin/data/

Using these formulas should be faster than the direct calculations in David Loeffler's answer. You can also code up the non-squarefree level case using Skoruppa-Zagier's trace formula without too much trouble.

Also, in case you're not familiar with it, the LMFDB modular forms page has tabulated a lot of data for newforms, including Atkin-Lehner signs. You can use this check dimensions in many cases.

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    $\begingroup$ That's nice, I didn't know about this work of yours! $\endgroup$
    – David Loeffler
    Commented Sep 20, 2022 at 13:06
  • $\begingroup$ Thank you for your answer. I already knew about your work, but it was a study about new subspace, and what I wanted was the case including old subspace, so it wasn't the answer I wanted. It seems that exact dimension formulae including old one have not been made yet. I need dimensions for other reasons, but anyways, I recently made some conjecture to get dimensions, and I tried indirect calculations to verify them, but all failed. However, looking at your answer, I think something can be obtained by using the trace formula. I'll do some more study. Thank you again for your kind answer. $\endgroup$
    – kslhg
    Commented Sep 20, 2022 at 23:42
  • $\begingroup$ @kslhg Maybe it's helpful to see how the computation goes first for $N=p$, then $N=pq$ for distinct primes. For $N=p$, the dimension will be the Atkin-Lehner dimension of the newspace, plus the dimension in full level. For $N=pq$, say for the $(++)$ space, you want the new $(++)$ dimension, plus the $+$-space dimensions for newforms of levels $N_1 = p$ and $N_1 = q$, plus the level 1 dimension. $\endgroup$
    – Kimball
    Commented Sep 21, 2022 at 2:39
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    $\begingroup$ @kslhg I edited my answer to indicate two ways you can treat the case of full cusp spaces, following what I did for newspaces. $\endgroup$
    – Kimball
    Commented Sep 21, 2022 at 2:44
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    $\begingroup$ @kslhg What you need to know is that if $M$ is coprime to $p$, then each newform in $S_k(M)$ gives one oldform in $S_k(pM)$ with $W_p$ eigenvalue $+1$ and similarly for $-1$. This should be written down somewhere in the original Atkin-Lehner paper, where I believe they explicitly construct the lifts with these eigenvalues. In your example, this means you get $dim S_8^{--}(22) = dim S_8^{new,--}(22) + dim S_8^{-}(11) = 4$, as the other 2 oldspaces contribute nothing. This matches what David Loeffler's code gives me. $\endgroup$
    – Kimball
    Commented Sep 21, 2022 at 5:15

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