delta(z) + delta (13z) is a weight 12 modular form of level Gamma_0 (13). Let A in Z/2[[q]] be the mod 2 reduction of the Fourier expansion of this form. (The exponents appearing in A are the odd squares and their products by 13).

If n is odd and positive let b_n be A^n and c_n be b_n/(1+A)^(1+n). For each odd prime p one has a formal Hecke operator T_p: Z/2[[q]] --> Z/2[[q]]. Is it true that T_3 takes b_n to a sum of c_k?


1.___The T_p, p not equal to 13, stabilize the space spanned by all the b_n and c_n. One can prove this by identifying this space with the space of odd mod 2 modular forms of level Gamma_0 (13) fixed by the mod 2 Fricke involution.

2.___I've verified that when n < 55, T_3 takes b_n to a sum of c_k (and in fact each k is less than 4n/3). For any particular k this is an easy calculation using the Sturm bound. But is it true in general? There is no reason I can see why it should be true, but I find the empirical evidence convincing. Can anyone help?


A sudden inspiration struck!

  1. Let B be A(q^3). Then the following identity, * , holds: (AB+A+B)^4 = AB. To see this, note that A is the mod 2 reduction of the expansion of the weight 12 cusp form delta(z) + delta(13z) for Gamma_0(13). So B is the reduction of a weight 12 cusp form for Gamma_0 (39). Then both sides of * come from modular forms for Gamma_0 (39) of weight 96. Since the Sturm bound in this weight and level is (96/12)*39=312, it suffices to show that the expansions appearing in * agree through q^312; this is quickly verified.

  2. For n non-negative, let d_n be T_3 (b_n). One calculates that d_1 = d_5 = 0, d_3 = c_1 + c_3, and d_7 = c_5 + c_7. The standard argument used by Nicolas and Serre uses * above to show first that d_(n+4) is (Ad_(n+1) + (A^4)d_n)/(A^4 + 1), and then that d_(n+8) is ((A^2)(d_(n+2)) + (A^8)d_n)/(A^8+1). Since multiplication by A^2/(A^2 + 1) stabilizes the space spanned by the c_n, so does multiplication by 1/(A^2 + 1), A^2/(A^8 + 1), and A^8/(A^8 + 1). An induction then shows that all the d_n, n odd, lie in this space.


  1. I had earlier looked for a relation between A and B and by a general method found a symmetric one of total degree 176 of the form (1+B^64)A^112 + a sum of monomials, each of degree < 112 in A, is equal to 0. This led to an impenetrable recursion. I imagined that my relation might be reducible, but couldn't factor it on the computer, and it took me a while to find * .

  2. Suppose instead B is A(q^p) with p a prime other than 2 or 13. Is there a nice relation between A and B? I looked briefly at p=7. The general method I used in in 3. then gives a similar symmetric relation of total degree 352, with 64 and 112 replaced by 128 and 224. But I'm sure that there's something simpler, giving a useful recursion for the T_7 (b_n). I wonder what the general result is.


  1. The following experimental results aren't directly relevant to the original question, but they do address 4. above, and indicate that there are identities much like * when B is A(q^p) rather than simply A(q^3).

Let fp in Z/2[x,y] be the unique irreducible polynomial with fp(A(q),A(q^p)) = 0, so that f3 is x^4*y^4+x^4+y^4+x*y. I've calculated fp for p=3,5,7,11,17,19 and discovered the following in each case:

a. fp(x,y) = fp(y,x)

b. The degree of fp in x is p+1

c. The coefficient of x^(p+1) in fp is (y+1)^k for some k. (In the six cases, k is 4,4,4,8,12,8 respectively).

d. fp(1,y) is (y+1)^k for some k. (In the six cases k is 1,3,5,6,9,14 respectively)

Things do get messier as p grows--f17 and f19 are sums of 46 and 48 monomials. It's reminiscent of the modular equation for the j-invariant, and an approach to showing that a,b,c,d hold in general might be to compare the mod 2 reduction of the uniformizer for the genus 0 curve attached to Gamma_0 (13) with A.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.