# Conductor at 2 of abelian surfaces with real multiplication

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 $$E$$.

• $$\mathfrak{f}_2(A)$$ 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.

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.
• 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.