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Consider one of the simplest non-abelian examples of modularity. Let $$\eta(6z)\eta(18z) = q\prod_{n=1}^\infty (1 - q^{6n})(1 - q^{18n}) = q - q^7 - q^{13} -q^{19} + q^{25} + 2q^{31} - q^{37} + 2q^{43} - q^{61} - q^{67} - q^{73} - q^{79} + q^{91} - q^{97} - q^{103} \dots = \sum_{n=1}^\infty a_n q^n$$

Then $a_p + 1 =$ the number of solutions of $\{x^3 - 2 = 0\}$ in $\mathbb{Z}/p\mathbb{Z}$, for prime $p > 3$.

1) What might be the best method to prove this?

2) Is it possible to prove it without knowledge of modular forms? i.e. without using the dimension / basis of the relevant space of moduli forms of some weight and level and nebentypus. For example, can it be explained using Weyl–Kac character formula etc.?

3) Unfortunately LMFDB does not yet have a list of weight $1$ modular forms and their corresponding Artin representations. Where can we find more of such weight $1$ examples?

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  • $\begingroup$ Small note: the same problem only with $x^3-x-1$ and $\eta(z)\eta(23z)$ is much more common in the literature, including on mathoverflow, so consider looking for that. Second, I think the level is 108 - the discriminant of the field. Third, the proofs using modular forms are pretty nice: use Hecke's results to prove that the polynomial "is modular", prove the eta product is modular, prove they are equal. $\endgroup$
    – Dror Speiser
    Commented May 12, 2016 at 12:42
  • $\begingroup$ @DrorSpeiser Thanks. I was wondering if there can be a more direct proof, without using the dimension / basis of the relevant space of moduli forms of some weight and level and nebentypus. $\endgroup$
    – Serendipity
    Commented May 12, 2016 at 12:45
  • $\begingroup$ That's your second question, I was more focused on the first ;) $\endgroup$
    – Dror Speiser
    Commented May 12, 2016 at 12:47
  • $\begingroup$ As for more examples, LMFDB has a list of Artin representations, in which those with group equal to $S_3$ and trace of complex conjugation equal to 0 are all modular of weight 1 by the same method of proof, and satisfy a similar solution count property. In fact, all Artin reps. of dimension 2 with trace of complex conjugation equal to 0 are modular, and listed in the LMFDB. This follows from the Langlands-Tunnel theorem for "small" galois extensions, and from some more recent theorems that I don't know to whom to attribute to. (most of these by far will not have an eta product expansion) $\endgroup$
    – Dror Speiser
    Commented May 12, 2016 at 13:04
  • $\begingroup$ For weight 1 modular forms you can look at people.maths.ox.ac.uk/lauder/weight1 $\endgroup$
    – znt
    Commented May 13, 2016 at 21:04

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Here's an answer to 2. You can tell me if it's the "best method".

Euler's Pentagonal number theorem gives that $$ \eta(24z) = \sum_{n \in \mathbb{Z}} (-1)^{n} q^{(6n+1)^{2}}. $$ This yields the formula $$ \eta(6z) \eta(18z) = \sum_{m, n \in \mathbb{Z}} (-1)^{m+n} q^{\frac{(6n+1)^{2} + 3(6m+1)^{2}}{4}}. $$ If $m=x$ and $n = x+2y$, then $m+n$ is even and we get $$ \sum_{x, y \in \mathbb{Z}} q^{(6x+3y+1)^2 + 27y^{2}}. $$ On the other hand if $m=x$ and $n=x+2y+1$, then $m+n$ is odd and we get $$ -\sum_{x,y \in \mathbb{Z}} q^{4(3x+y+1)^{2} + 2(3x+y+1)(2y+1) + 7(2y+1)^{2}}.$$

The upshot (skipping a few small details) is that $$ \eta(6z) \eta(18z) = \frac{1}{2} \left[ \sum_{x,y \in \mathbb{Z}} q^{x^{2} + 27y^{2}} - q^{4x^{2} + 2xy + 7y^{2}} \right]. $$

Now, Gauss proved that if $p \equiv 1 \pmod{3}$ is prime, then there is some $z$ so that $z^{3} \equiv 2 \pmod{p}$ if and only if $p = x^{2} + 27y^{2}$ for some $x, y \in \mathbb{Z}$. It follows from this that the $p$th coefficient of $\eta(6z) \eta(18z)$ is $2$ if $p \equiv 1 \pmod{3}$ and $2$ is a cube modulo $p$.

If $p \equiv 1 \pmod{3}$, but $2$ is not a cube modulo $p$, then $p$ is represented by $4x^{2} + 2xy + 7y^{2}$ in two ways, and this means that the $p$th coefficient of $\eta(6z) \eta(18z)$ is $-1$.

If $p \not\equiv 1 \pmod{3}$, the $p$th coefficient of $\eta(6z) \eta(18z)$ is clearly $0$.

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    $\begingroup$ Thanks! It seems that all such "classic" solutions I have seen involve theta series. Will it work for all $S_3$ representations (or maybe even more)? $\endgroup$
    – Serendipity
    Commented May 12, 2016 at 13:38
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    $\begingroup$ I believe this method will work for all dihedral representations, and no others. The rough idea is that the space of weight 1 cusp forms spanned by differences of theta series should be decomposable into linear combinations of forms obtained from ray class characters of imaginary quadratic fields - these correspond by class field theory to dihedral extensions of $\mathbb{Q}$. $\endgroup$
    – Jeremy Rouse
    Commented May 12, 2016 at 13:47
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    $\begingroup$ More precisely, dihedral extensions where complex conjugation acts by an element of determinant $-1$ -- otherwise you have to deal with ray class characters of real quadratic fields and you get Maas forms instead of modular forms. $\endgroup$
    – David E Speyer
    Commented May 12, 2016 at 13:48
  • $\begingroup$ Yes, that's right - only odd dihedral representations. $\endgroup$
    – Jeremy Rouse
    Commented May 12, 2016 at 14:10

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