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

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 $$p=2$$.

• Note once you have one such form you get infinitely many by adding appropriate cusp forms. You can also construct examples (at least in weight 2) from certain Eisenstein congruences. For half-integral weight, have you looked at Cohen-Eisenstein series? My guess is that you can either use these directly, or possibly combined with a congruence. May 15, 2020 at 4:31
• @Kimball I had not heard of Cohen-Eisenstein series before, thank you for bringing them to my attention. From the definition it's not clear to me how they might be useful. Would you mind elaborating a bit on your thoughts?
– BHT
May 15, 2020 at 5:06
• By using Rankin-Cohen brackets you can build many other examples of half-integral modular forms of higher weight which are 1 mod 2 by bracketing with the theta function you have. Similarly you can bracket $E_{p-1}$ with a half-integral modular form to get something that is constant modulo p.
– GTA
May 15, 2020 at 13:17
• Cohen-Eisenstein series are a half-integral weight version of the classical even weight Eisenstein series. See Cohen's 1975 Math Ann paper, or search "Cohen-Eisenstein" to find some more recent references. May 15, 2020 at 14:09
• @GTA Naively it seems to me as though the only way to get a form this way which is constant mod $p$ would be to bracket $E_{p-1}$ with a half integral weight form that is already constant mod $p$, which isn't helpful. I only need a single form for each prime.
– BHT
May 15, 2020 at 20:31

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.)

• I'm not sure if you were misled by the title (which isn't clear), but the OP seems to be looking for forms with a_n = 0 mod p for all n > 0 but a_0 not 0 mod p. May 15, 2020 at 17:33
• Thank you very much for the answer! In the Gross paper the result you state is proven in proposition $4.9$, by showing that the kernel of reduction mod $p$ is the principal ideal $(1-A)$, where $A$ is the Hasse invariant. The author does indeed assume $p$ does not divide $N$ generically throughout section $4$, but determining whether or not this assumption is actually needed for proposition $4.9$ is well beyond my capabilities.
– BHT
May 15, 2020 at 21:24
• Could you explain how you found that example form? I tried to write some Sage code to search for these forms by generating a basis and then reducing mod p and solving a matrix equation, but when I ran it on weight $9/2$, $\Gamma_1(80)$ (Sage requires $16|N$ for some reason), it didn't find the form you mentioned, so I must be doing something wrong.
– BHT
May 18, 2020 at 2:28
• I used Magma with can compute the spaces $M_{k/2}(\Gamma_{0}(N),\chi)$ and $S_{k/2}(\Gamma_{0}(N),\chi)$. It appears that Sage actually computes bases for spaces of cusp forms. There's a trick you can use to verify the presence of this form - the function $g(z) = \eta^{2}(4z) \eta^{2}(20z) \in S_{2}(\Gamma_{0}(80))$. You can then look in the space of weight $13/2$ forms for $\Gamma_{1}(80)$ (with trivial character) for a form that is congruent to $g$ (up to $q^{102}$) modulo $5$. May 18, 2020 at 14:12