# Does weight $2$ cuspidal Bianchi modular form have infinitely many zero Fourier coefficient at prime ideals?

Let $$K$$ be an imaginary quadratic field. Let $$f \in S_2(\mathfrak{n})$$ be a weight $$2$$ cuspidal cof level $$\Gamma_0(\mathfrak{n})$$ over $$K$$ (for definitions one can see http://www.lmfdb.org/knowledge/show/mf.bianchi.bianchimodularforms ), consider its $$L$$-function

$$L(f,s)=\sum_{\mathcal{P} \not =0 \in Spec(O_K)} c(\mathcal{P}) (N\mathcal{P})^{-s}$$

here $$c(\mathcal{P})$$ are the Fourier coefficients at the cusp $$\infty$$ and we assume they are all rational numbers.

The question is, do we know that $$c(\mathcal{P})=0$$ for infinitely many prime ideals $$\mathcal{P}$$? How about $$c(\mathcal{P})=0 \mod N(\mathcal{P})$$?

Motivation: if we are considering similar problems over $$\mathbb Q$$, then we know $$a_p=0$$ for infinitely many $$p$$ because the corresponding elliptic curve over $$\mathbb Q$$ must have supersingular reduction at infinitely many primes.

• Do these (conjecturally) correspond to elliptic curves over imaginary quadratic field? I don't think there is a single non-CM elliptic curve over a field with no real places where we know there are infinitely many supersingular primes. If you pick your favorite no-CM elliptic curve over an imaginary quadratic field, and check it is modular, then I think we have no clue how to check it has infinitely many vanishing Fourier coefficients. – Will Sawin Sep 11 at 19:30
• @Will Sawin I hope there is an automorphic approach to this problem, because modularity is so difficult.. – sawdada Sep 11 at 23:32
• I'm trying to argue that there is no automorphic approach. Checking modularity for a given elliptic curve is easy nowadays (I think). But there is no progress on this problem in either the geometric or automorphic settings. – Will Sawin Sep 12 at 1:40
• There is indeed no automorphic approach, AFIAK, which is why the corresponding statement is completely unknown for elliptic modular forms of weight $> 2$. – David Loeffler Sep 12 at 9:43