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Let D in Z/3[[x]] be sum ((a_n)(x^n)) where the sum runs over all n prime to 6 and a_n is the mod 3 reduction of the number of ideals of norm n in the ring of integers of Q(root(-3)). (So D=x+2(x^7)+2(x^13)+2(x^19)+x^25+(higher degree terms).)

There are formal Hecke operators T_p: Z/3[[x]]-->Z/3[[x]] for all primes p other than 3. Clearly the T_p with p=2 mod 3 annihilate D. Experimentally I find that they annihilate D^7, D^61 and D^547 as well.

Question: If k=(1+3^m)/4 with m odd, do the T_p with p=2 mod 3 annihilate D^k?

Remark: D is the reduction of the expansion of the modular form (eta(2z))^12, but it's not clear that this helps.

EDIT: If I've made no mistake, Noam's formula can be tweaked to show that D^k=V-U where U and V are the mod 3 reductions of the theta series attached to the binary quadratic forms 36xx+6xy+kyy and 36xx+30yy+(k+6)yy. This raises the question of which linear combos of theta series of primitive binary quadratic forms lie in the space spanned by the powers of D. I think similar questions have been answered by Nicolas and Serre (ell=2, level=1) and perhaps by Bellaiche and or Serre (ell=3, level=1).

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Yes, and $D \equiv \eta(2z)^{12}$ certainly helps: it means $$ D^{(1+3^m)/4} \equiv \eta(2z)^{3+3^{m+1}} \equiv \eta(6z) \, \eta(2\cdot 3^{m+1}z). $$ Now $\eta = \sum_a \chi(a) q^{a^2/24}$ where $a$ runs over integers congruent to $1 \bmod 6$ and $\chi$ is the even Dirichlet character $\bmod\,12$. Since $m$ is odd, say $m=2n+1$, we have $$ \eta(6z) \, \eta(2\cdot 3^{m+1}z) = \mathop{\sum\sum}_{a,b \equiv 1 \bmod 6} \chi(ab) q^{(a^2 + 3^{2n+1}b^2)/4}. $$ All the exponents are odd, so our form is in $\ker T_2$. If $p \equiv 5 \bmod 6$ then $(a^2 + 3^{2n+1}b^2)/4$ is a multiple of $p$ iff $a,b \equiv 0 \bmod p$, whence $T_p$ multiplies our form by $p^{w-1} + 1$ where $w$ is the weight. Since $w = (1+3^m)/4$ is odd and $p \equiv 2 \bmod 3$, we have $p^{w-1}+1 = 0$ in ${\bf Z}/3{\bf Z}$ and we're done.

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  • $\begingroup$ Thanks, Noam. The argument also occurred to me just after I submitted the question. I wonder if you have any ideas about my deeper question--is the space spanned by the D^j with j prime to 6 stable under the T_n with n prime to 6? It certainly seems true experimentally. $\endgroup$
    – paul Monsky
    Commented Oct 22, 2013 at 6:52
  • $\begingroup$ Just to make the comment of mine that I'm replacing clearer: When p is 2 mod 3 and w is odd, p^(w-1)+1 is 2, not 0 in Z/3Z. So the end of Noam's proof is wrong. A revision is: ... ,whence c_(np)=c(n/p) where c_n is the coefficient of x^n in D^k. Now the coefficient of x^n in T_p(D^k) is c_(np)+p*c(n/p); since p is 2 mod 3, this is 0. (One thing to note is that in characteristics ell=2 or 3, when p isn't ell then for even w, p^(w-1)=p. So one can define formal Hecke operators on Z/ell[[x]] without taking the weight into effect, that will act well on reductions of forms of any even degree. $\endgroup$
    – paul Monsky
    Commented Oct 25, 2013 at 16:44

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