Let $A/\mathbb{Q}$ be an abelian surface such that $\text{End}_{\mathbb{Q}}(A)\otimes \mathbb{Q}$ is a real quadratic field $E$. I am interested in bounding the conductor of $A$ at $2$. Let $\mathfrak{f}_2(A)$ be the exponent of the conductor of $A$ at $2$. A general bound [Theorem 6.2, BK94] shows that $\mathfrak{f}_2(A)\leq 20$; I would like to do better than this using the quadratic field $E$.

What I already know:

  • $\mathfrak{f}_2(A)$ is even.
  • $\mathfrak{f}_2(A)\leq 16$ if $E\not\simeq \mathbb{Q}(\sqrt{2})$. This follows from [Theorem 5.5, BK94] (and a little bit of work)
  • There exists abelian surfaces with $E = \mathbb{Q}(\sqrt{2})$ and $\mathfrak{f}_2(A) = 20$.

I would be very interested if someone knows whether the bound $\mathfrak{f}_2(A)\leq 20$ can be improved at all if $E = \mathbb{Q}(\sqrt{2})$ and if we additionally assume that $\text{End}(A_{\bar{\mathbb{Q}}})$ is an order in a nonsplit indefinite quaternion algebra. Any help is appreciated!

Brumer, Armand; Kramer, Kenneth, The conductor of an abelian variety, Compos. Math. 92, No. 2, 227-248 (1994). ZBL0818.14016.


1 Answer 1


(Edit: answer below was given before the QM assumption was added)

There are newforms of weight 2 and level 1024 (and trivial nebentypus), e.g., 1024.2.a.a. The associated abelian surface has conductor the square of the level, so $2^{20}$ (and endomorphism ring $\mathbb Z[\sqrt{2}]$). So the bound 20 for the conductor exponent is sharp, even for abelian surfaces with RM.

  • $\begingroup$ Thanks! I realize I forgot to add an assumption, namely that $A_{\bar{\mathbb{Q}}$ has quaternionic multiplication. I'm sorry for the bad form but I will edit the question and see if someone still has something to say about this. $\endgroup$
    – Jef
    Jan 29 at 20:45

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