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?


Remarks:


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 strange (possible) fact about the Hecke operator T_3 in level 13 and characteristic 2
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.


Remarks:


3. 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 \* .
4. 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.


EDIT(10/16)


5. 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.


