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Fix an odd N>0. Let M consist of all odd elements of Z/2[[x]] that are the mod 2 reductions of elements of Z[[x]] arising as the Fourier expansions of modular forms for (Gamma_0)(N); it's easy to see that M is closed under addition. For i in {1,3,5,7} let M_i be the subspace of M consisting of those g in which all the exponents that appear are congruent to i mod 8.

QUESTION: Do M_1, M_3, M_5 and M_7 span M?

Remark 1: When N=1 the answer is yes. For in this case M is spanned by the odd powers of x+x^9+x^25+x^49+..., and each such power lies in some M_i.

Remark 2: When N is prime, I think this will follow from an affirmative answer to another question I've posted on MO--Level p characteristic 2 modular forms and thetas (#121506). But I haven't tried to write things out, and this approach seems awfully complicated.

EDIT: The question above and the further question at the end of this edit arose from an attempt to experimentally understand the action of the Hecke algebra generated by the T_p with p prime and (2N,p)=1 on M (and on a certain subspace C=M(plus)+M(minus) of M when N is prime.)

So suppose N is prime. Let M(plus) consist of those g in M for which for each exponent j that appears in g the Legendre symbol (j/N) is either 1 or 0. Define M(minus) similarly with (j/N) being -1 or 0. Let F=x+x^9+x^25+x^49+... and G=F(x^N). Then M is a module over Z/2[G^2], and I think I can show that the rank is N+1. The subspaces M(plus) and M(minus) are submodules, and C=M(plus)+M(minus) is stable under the Hecke algebra. When N=3,5 or 7 I can prove:

(*)---- M(plus),M(minus) and C have ranks (N+1)/2, (N+1)/2 and N. Moreover C admits a basis of the form A_j, j=1,3,5,...2N-1 where:

(a) A_j=x^j+higher order terms

(b) A_j lies in M_j

(c) When j=N, A_j=G. For the other j, A_j is in M(plus) when (j/N)=1, and in M(minus) when (j/N)=-1.

(For example when N=5, A_1=F, A_3=F*(F+G)^2, A_5=G, A_7=F^2*G, A_9=F^4*G)

FURTHER QUESTION: Does (*) hold for all odd prime N?

Remark: When N=11 I haven't proved (*). But I can show that M=M_1+M_3+M_5+M_7. And I can produce candidates for A_1, ..., A_21---it appears experimentally that the module generated by these candidates is stable under the Hecke algebra.

EDIT(8/12)___When N=11 I can now answer my FURTHER QUESTION, showing that the answer is yes(just as it is when N=3,5 or 7). To view the A_j in these 4 cases go to my question 138495--"Are these two subspaces of Z/2[[x]] the same?" (But I write Cj for A_j there). In that question I indicate that when N=3,5,7, or 11 then the Cj span a certain space M_0 consisting of "odd trace zero level N mod 2 modular forms" and that M_0 is contained in C. I'll now sketch a proof, using modular forms, that M_0=C when N=3,5, or 11. And I'll also show that for general N something very close to the first sentence of (*) holds. I'll use the notation of 138495 throughout, writing C(plus), C(minus), and M(odd) rather than M(plus), M(minus) and M.

Lemma 1___If h lies in C(plus) and in C(minus) then h is a sum of odd powers of G.

Proof___Evidently h=g(x^N) for some g in Z/2[[x]]. Now there is a formal Hecke operator U_N from Z/2[[x]] to Z/2[[x]]. Using modular forms one sees that U_N stabilizes M(odd) so that g=(U_N)(h) lies in M(odd). So both h and g are in Z/2(F,G). And h=g(x^N) is not only in Z/2(F,G) but in Z/2(G,H) where H=G(x^N).

___Now let K be an algebraic closure of Z/2. Since 1/F, 1/G, and 1/H are the mod 2 reductions of the expansions of j(z), j(Nz) and j((N^2)z), Igusa's theory of mod 2 modular forms shows that K(F,G,H) is a Galois extension of K(G) with Galois group PSL_2 (Z/N), and that K(F,G) and K(G,H) are the fixed fields of Borel subgroups. Since these subgroups generate the Galois group, our h lies in K(G). Since h is in M(odd) it is integral over K[G], and lies in K[G]. The rest is easy.

Lemma 2___Suppose M_0 is contained in C. Then either C(plus) or C(minus) is contained in M_0.

Proof___ Let U and U' be the intersections of M_0 with C(plus) and with C(minus). Then U and U' span M_0 and each contain G. So the sum of their Z/2[G^2] ranks is at least 1+rank(M_0)= 1+N. If the lemma fails, the sum of the ranks of C(plus) and C(minus) is at least 3+N. But by Lemma 1 the rank of their intersection is 1. So the rank of C is at least 2+N. But the rank of M(odd) is 1+N, giving a contradiction.

Theorem___Suppose that (2/N)=-1, and that M_0 is contained in C (e.g when N=3,5, or 11). Then M_0=C.

Proof___Otherwise C contains an element h of C(plus) (or of C(minus)) with non-zero trace. Then (h^2)G has non-zero trace and, since (2/N)= -1, lies in C(minus) (or C(plus)). This contradicts Lemma 2.

Lemma 3___C(plus) and C(minus) each have rank at least (N+1)/2 over Z/2[G^2], while C has rank N or N+1.

Proof___Fix an odd r not equal to N with r between 0 and 2N. Fix d so that 2dN is -1 mod r, and write 1+2dN as rs. Then f=(G^2d)F lies in C(plus) and is x^(rs)+ ... Write s as a product of not necessarily distinct primes p_j, compose the T_(p_j) and apply the resulting operator to f. Then we get a B_r of the form x^r+ ... The B_r with (r/N)=1 (resp. -1) lie in C(plus) (resp. C(minus)) and are linearly independent over Z/2[G^2], giving the result.

___Now by Lemma 1 the sum of the ranks of C(plus) and C(minus) is 1+rank C. So either C(plus), C(minus) and C have ranks (N+1)/2, (N+1)/2 and N as in (*) or C has rank N+1, and C(plus) and C(minus) have ranks (N+1)/2 and(N+3)/2 (or vice versa). Surely the two last possibilities don't occur, but I don't see how to rule them out.

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What about the modular form of an elliptic curve with some arbitrary mod 2 galois representation? Where should we look for forms that split it into $M_1$, $M_3$, $M_5$, and $M_7$ pieces? –  Will Sawin May 24 '13 at 20:00
Here's what happens when N=11. There's a weight 2 cusp form corresponding to an elliptic curve; let t in Z/2[[x]] be the reduction(of its expansion). Let f1 be x+x^9+x^25+... and f11 be f1(x^11). Then t+t^3=f1+f11, and there is a weight 4 Eisenstein series whose reduction r satisfies r+r^2=t+t^3. Let m5=t^4(f1), m7=t^4(f11), (t^8)m1=(r^8)(f1) and (t^8)m3=(1+r^8)f11.(One checks that m1=(r^2)(t+t^5+t^7)+t+t^5, and that m3=(r^2)(t+t^5+t^7)+t^7, so that the m_i are all in M.) Since m5+m7=t^4(t+t^3) the m_i sum to t. I admit this doesn't give much insight! –  paul Monsky May 25 '13 at 1:19

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