Half integral weight modular forms that reduce to a nonzero constant modulo a given prime

Question

Let $B_k = \frac{N_k}{D_k}$ be the reduced numerator and denominator of the $k$-th Bernoulli number. For a given prime $p>2$, the (unconventionally normalized) Eisenstein series $E_{p-1}(z) = N_{p-1} − 2(p-1)D_{p-1}\sum_{n=1}^{\infty} \sigma_{p-2}(n)q^n$ has the property that its Fourier coefficients are all integers, and the constant coefficient is the only coefficient not divisible by $p$.

Given a prime $p$, I am interested in the existence of a modular form $f_p$ with analogous properties but whose weight is half integral. In other words, $f_p$ should have integral Fourier coefficients, the constant coefficient should be prime to $p$, and all other coefficients should be divisible by $p$.

The exact weight and level are unconstrained, but preferably should be kept small enough to compute with for moderately small values of $p$.

The only example I know of is the theta function $\theta(z) = 1 + 2q + 2q^4 +...$ for $p=2$.

Answer 1

If such a form does exist, then its level must be a multiple of $p$.

If $f = \sum a_{n} q^{n}$ is a half-integer weight modular form with integer coefficients with $a_{i} \equiv 0 \pmod{p}$ for all $i > 0$ and $\gcd(a_{0},p) = 1$, then by multiplying $f$ by an integer relatively prime to $p$, one can assume $a_{0} \equiv 1 \pmod{p}$. If $k+1/2$ is the weight of the modular form, then $f^{2}$ would be an integer weight modular form for $\Gamma_{1}(N)$ (for some $N$) with the property that $f^{2} \equiv 1 \pmod{p}$. The form $f^{2}$ would have weight $2k+1$.

There's a well-developed theory of integer-weight modular forms modulo $p$ for $\Gamma_{1}(N)$ with $p \nmid N$ and the best reference is the paper ``A tameness criterion for Galois representations associated to modular forms (mod p)'' by Benedict Gross (Duke Math Journal, 1990, pages 445-517). In this paper it is proven that if $f$ and $g$ are modular forms for $\Gamma_{1}(N)$ with $f \equiv g \pmod{p}$, then the weights of $f$ and $g$ must be congruent modulo $p-1$. This creates a contradiction in the situation above because we cannot have $2k+1 \equiv 0 \pmod{p-1}$, since $p-1$ is even and $2k+1$ is odd.

Unfortunately, I do not currently have access to the paper of Gross (the link above is behind a paywall), and there may be some caveats in the theorem I've stated above (EDIT: The OP has confirmed that one needs the hypothesis that $p \nmid N$.). In case it's useful, here's another MO post asking a question about modular forms modulo $p$ in level $> 1$. The $N = 4$ case of the theorem I quoted above was independently proven by Tupan. (See the paper "Congruences for $\Gamma_{1}(4)$ modular forms of half-integral weight" in the Ramanujan Journal in 2006.) In particular, level $4$ Cohen-Eisenstein series and Rankin-Cohen brackets of level $4$ forms with $E_{p-1}$ cannot produce examples of half-integer weight forms $\equiv 1 \pmod{p}$. (Thanks to the OP for looking up the paper and pointing out the necessity of the hypothesis that $p \nmid N$.)

If $p \geq 5$ is prime, there is a form of weight $\frac{p-1}{2}$ for $\Gamma_{1}(p)$ that is $\equiv 1 \pmod{p}$, namely $\frac{\eta^{p}(z)}{\eta(pz)} = \prod_{n=1}^{\infty} \frac{(1-q^{n})^{p}}{1-q^{pn}}$. One can search for examples and come tantalizingly close. For example, there's a weight $9/2$ form for $\Gamma_{0}(20)$ whose Fourier expansion beings $$ F(z) = 1 - 20q^{13} - 40q^{14} + 90q^{16} - 40q^{17} - 40q^{18} + \cdots. $$ The coefficient of $q^{n}$ is a multiple of $5$ for $1 \leq n < 100$, but the coefficient of $q^{100}$ is $15292$. (If $a_{n} \equiv 0 \pmod{5}$ for $1 \leq n \leq 108$, Sturm's theorem would force it to always be true.)