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Given a quadratic form $F$ in $n$ variables, there is an associated theta function $\theta_F(z) = \sum_{m \in \mathbb{Z}} q^{F(m)}$, which is a modular form of weight $n/2$. Letting $F(m) = m^2$ gives a theta function $\theta_F(z) = 1 + 2 \sum_{n = 1}^\infty q^{n^2} \equiv 1 \pmod{2}$ with weight $1/2$.

Does there exist a theta function $\theta_F(z) \equiv 1 \pmod{4}$ with half-integer weight?

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    $\begingroup$ Maybe the following approach might be helpful.(I'm guessing that the answer to your question is no.) If there is such a theta, raising it to an appropriate odd power, multiplying what you get by 1+2(x+x^4+x^9+...), subtracting off an Eisenstein series and dividing by 2 would give a modular form of integral weight whose mod 2 reduction, g, is x+x^4+x^9+.... Then a theorem of Serre would imply that for any k almost all the coefficients of g^k are 0. I wonder what the evidence is for or against this claim about g. $\endgroup$
    – paul Monsky
    Commented Mar 19, 2013 at 23:03
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    $\begingroup$ @Peter, half the front page is old questions that you have bumped by editing the tags. Please, do this kind of thing 3 or 4 per day, not 30 in half an hour. $\endgroup$
    – Gerry Myerson
    Commented Mar 11, 2015 at 22:37

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This is nothing like a complete answer (EDIT-now it seems to be--see below), but it may suggest a fruitful attack, based on the comment (see below for a version incorporated into this answer) that I made earlier. Your question is related to my question 124243 on this site about a certain ring $M$ consisting of the mod $2$ reductions of elements of $\mathbb{Z}[[x]]$ that are the Fourier expansions of modular forms for $\gamma_0 (N)$. Suppose first that $N$ is an odd prime $p$.

Theorem(?) The element $g=x+x^4+x^9+x^{16}+\dots$ of $\mathbb{Z}/2[[x]]$ does not lie in $M$.

Proof(?) In the notation of my question, and the other question of mine that it links to, let $h=g+g^2+r$. Then $$ h(1+h)=(g+g^4)+(r+r^2)=f1+(f1+fp)=fp . $$ (Again, see below). Now I conjecture in my question that in the field of fractions of $M$, $fp$ has a zero of order $p$, a zero of order $1$ and $p+1$ poles counted with multiplicity. If this is true then both $h$ and $h+1$ have $\frac{p+1}{2}$ poles, counted with multiplicity. Furthermore one of them must have a zero of order $p$, giving a contradiction.

Now I'm reasonably sure that this conjecture is correct, is known to the experts, and follows from results proved by Igusa in the 1950's. Unfortunately no expert has yet responded to my question. I also think that similar techniques will show that $g$ is not the mod $2$ reduction of any modular form for $\gamma_0 (N)$ when $N$ is odd. But the theory, if I understand correctly, is much harder when the level is divisible by the characteristic, and if a proof is to be given based on the idea of my comment this case has to be considered.

Edited comment--Suppose that $F$ is a positive definite quadratic form over $\mathbb{Z}$ in $s$ variables where $s$ is odd. Here's an idea for showing that the power series $\theta_F$ in $\mathbb{Z}[[x]]$ attached to $F$ (I'll defy convention and write $x$ instead of $q$) cannot be congruent to $1$ mod $4$. Choose $S$ so that $s\times S=8m-1$ and let $G$ be a direct sum of $S$ copies of $F$ and one of the one variable form $z^2$. Then if $\theta_F$ is $1$ mod $4$, $\theta_G$ is $1+2x+2x^4+2x^9+2x^{16}+\dots$ mod $4$. Now (see Schoeneberg's paper from the 1930's for example) there is an $N$ such that $\theta_G$ is the "expansion at infinity" of a weight $4m$ modular form, $\phi$, for $\gamma_0 (N)$. Then $(1/2)(\phi-(E_4)^m)$ is a weight $4m$ modular form for $\gamma_0 (N)$ whose expansion at infinity lies in $\mathbb{Z}[[x]]$, and has mod $2$ reduction equal to $g=x+x^4+x^9+x^{16}+\dots$. There are perhaps reasons to believe this can't happen--at any rate it appears to be an approachable problem of a mainstream variety.

One more remark. In my proof? given above I use the fact that the mod $2$ reduction $f_1$ of the expansion at infinity of the cusp form delta is $x+x^9+x^{25}+\dots$. This lovely if well-known fact comes from the infinite product formula for delta and Jacobi's triple product identity.

EDIT---MARCH 24 I now believe I have an argument showing that $g$ is not the mod $2$ reduction of a modular form for any $\gamma_0 (N)$, completing the answer to your question. I'll dispense with any ideas of Igusa, and instead use Hecke operators and classical results of Gauss on ternary quadratic forms. The idea is this. For each odd prime $q$ there is a "Hecke operator" $T_q$ on $\mathbb{Z}/2[[x]]$. If $h$ is the reduction of a modular form for $\gamma_0 (N)$ then the $T_q(h)$ span a finite dimensional subspace of $\mathbb{Z}/2[[x]]$. Now let $h=g^{11}$. I'll use this last result to show that $h$ cannot be the reduction of a modular form.

To this end, fix $K$ and primes $p_1,...,p_K$ each of which is $5$ mod $8$. Then choose primes $q_1,...,q_K$ each of which is $7$ mod $8$ so that the Legendre symbol $(q_i/p_j)$ is $-1$ if $i=j$ and is $1$ otherwise. Now the coefficient of $x^{p_j}$ in $T_{q_i}(h)$ is the coefficient of $x^{p_j}(q_i)$ in $h$. By the last paragraph of the comments following my answer to MO question 28462-why are there usually an even number of representations as a sum of $11$ squares--(an answer based on Disquisitiones Arithmeticae)--, this coefficient is $1$ if $i=j$ and $0$ otherwise. So the $T_{q_i} (h)$ are linearly independent. Since $K$ may be chosen arbitrarily large, the criterion of the last paragraph gives the result.

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  • $\begingroup$ $g$ satisfies the equation $g^4+ g= \Delta$, where $\Delta$ is the reduction mod $2$ of the $\Delta$ function and generator of the forms of level $1$ mod $2$. This is a Galois extension of Galois group $\mathbb Z/2 \times \mathbb Z/2$. Maybe one could look for this extension inside coverings of the modular curve? $\endgroup$
    – Will Sawin
    Commented Mar 22, 2013 at 3:00
  • $\begingroup$ @Will-My proof? is more or less a version of this idea, as I think my conjectures when the level is odd should follow from results of Igusa on the modular curve. But I have no feeling for what happens when the level is even. (I took Katz and Mazur out of the library last year, but it went unread--it's not for amateurs). $\endgroup$
    – paul Monsky
    Commented Mar 22, 2013 at 3:49
  • $\begingroup$ Here's the argument that if u is the mod 2 reduction of an element of Z[[x]] that is the expansion at infinity of some modular form phi of weight w for gamma_0 (N) then the space spanned by the image of u under the formal Hecke operators "T_q" ,q prime, has finite dimension. For fixed N and w the Z-module of such elements of Z[[x]] has finite rank and is stable under the Hecke T_q where (N,q)=1. So the image of this module under mod 2 reduction contains u and has finite Z/2 dimension. The T_q with (2N,q) reduce to "T_q". These "T_q" stabilize the image. And only finitely many q divide N. $\endgroup$
    – paul Monsky
    Commented Mar 25, 2013 at 11:48
  • $\begingroup$ I'm convinced that this works for the element $a(x) = 1 + x + x^4 + x^9 + \dots \in Z/2Z[[x]]$, but I think removing the 1 to get $g(x)$ might mess it up. In the MO question about sums of 11 squares, everything is about $a(x)$. In particular, I'm not convinced that you can go between $x_0^2 + x_1^2 + \cdots + x_{10}^2$ and $x_0^2 + 2 x_1^2 + 8 x_3^2$ when you restrict to $x_i > 0$ instead of $x_i \geq 0$. The principal / lemma that Greg Kuperberg brought up in his answer if only for unipotent formal power series. $\endgroup$
    – stl
    Commented Mar 27, 2013 at 5:04
  • $\begingroup$ Ah, I think I understand it now. We can just add an appropriate power of $1 + 2 \sum q^{n^2} \equiv 1 \pmod{2}$ to $g(x)$ to get $a(x)$. $\endgroup$
    – stl
    Commented Mar 27, 2013 at 14:54

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