# Q-curves and twisting

An elliptic curve $$E$$ over $$\overline{\mathbb{Q}}$$ is called a $$\mathbb{Q}$$-curve if it is isogenous (over $$\overline{\mathbb{Q}}$$) to all its Galois conjugates -- see Are Q-curves now known to be modular? for example.

If I take a finite Galois extension $$K / \mathbb{Q}$$ and an elliptic curve $$E / K$$ whose base-extension to $$\overline{\mathbb{Q}}$$ is a $$\mathbb{Q}$$-curve, then all the Galois conjugates $$E^{\sigma}$$ are also defined over $$K$$, but the isogenies between them might not be. Supposing $$E$$ to be non-CM for simplicity, then what you get instead is a $$K$$-isogeny from each conjugate $$E^{\sigma}$$ to some possibly non-trivial quadratic twist of $$E$$. Let me say $$E$$ is a strong $$\mathbb{Q}$$-curve over $$K$$ if it's non-CM and it's actually $$K$$-isogenous to all its Galois conjugates. (Clearly any $$\mathbb{Q}$$-curve over $$K$$ becomes a strong $$\mathbb{Q}$$-curve over some finite extension $$L / K$$, but I want to keep $$K$$ fixed here.)

It's easy to produce examples of $$\mathbb{Q}$$-curves which aren't strong $$\mathbb{Q}$$-curves, by taking a strong $$\mathbb{Q}$$-curve and applying a quadratic twist by an element of $$K^\times / K^{\times 2}$$ that's not stable under $$Gal(K / \mathbb{Q})$$. However, I can't find any examples which aren't of this form.

Are there $$\mathbb{Q}$$-curves which are not twists of strong $$\mathbb{Q}$$-curves?

(I'm chiefly interested in the case when $$K$$ is a real quadratic field here.)

I think there are examples of $$\mathbb{Q}$$-curves defined over a quadratic field $$K$$ which are not strong $$\mathbb{Q}$$-curves over $$K$$ in Jordi Quer's paper "$$\mathbb{Q}$$-Curves and Abelian Varieties of $$\mathrm{GL}_2$$-Type". He uses the term $$\mathbb{Q}$$-curves completely defined over $$K$$ instead of strong $$\mathbb{Q}$$-curves (I will stick to Quer's terminology). The key result is Corollary 3.3, which then can be applied for instance to the family of $$\mathbb{Q}$$-curves with an isogeny of degree $$3$$ that he writes in Section 6: $$C^{(a)}\colon Y^2 = x^3 -3\sqrt{a}(4+5\sqrt{a})X+2\sqrt{a}(2+14\sqrt{a}+11a).$$ Suppose that $$C^{(a)}$$ is not CM (the curve is CM only for 9 values of $$a$$). In the terminology of the article, the sets {a} and {3} are dual bases with respect to the degree map, and $$K_d = \mathbb{Q}(\sqrt{a})$$. The curve $$C^{(a)}$$ is defined over $$K_d$$ and by Corollary 3.3 if the quaternion algebra $$(a,3)_\mathbb{Q}$$ is different in the Brauer group from $$(-1,a)_{\mathbb{Q}}^x$$ for all $$x\in\{0,1\}$$ then there is no curve $$\overline{\mathbb{Q}}$$-isogenous to $$C^{(a)}$$ completely defined over $$K_d$$ (when he writes isogenous in Corollary 3.3 he means over $$\overline{\mathbb{Q}}$$, not just over $$K_d$$).

It looks like your notion of strong $$\mathbb{Q}$$-curve over $$K$$ is what Peter Bruin and Andrea Ferraguti refer to as a $$\mathbb{Q}$$-curve being completely defined over $$K$$. Such curves have $$L$$-function factoring as a product of $$L$$-series of newforms for $$\Gamma_1(N)$$. This then seems to coincide with the definition of strongly modular given by Xevi Guitart and Jordi Quer. This latter set of authors provide an explicit example of an elliptic $$\mathbb{Q}$$-curve (which they call a building block after Elisabeth Pyle's thesis) over $$K = \mathbb{Q}(\sqrt{-3})$$ which is not strongly modular, and state that no curve isogenous to it over $$\overline{\mathbb{Q}}$$ and defined over $$K$$ can be strongly modular:

$$Y^2 = X^3 + 4aX^2 + 2(a^2 + b\sqrt{-3})X,$$

for $$a,b \in \mathbb{Q}$$. I haven't checked the details, but this might give you what you're after?

• It is known that strongly modular implies completely defined over $K$ (I'm excluding CM to be safe), but the converse is not always true. I think that the example by Guitart and Quer is an instance of a building block which is not strongly modular, but building blocks are completely defined over $K$ if I understood correctly. Jun 30 '20 at 19:20
• (However, the converse is true if $K$ is a quadratic field.) Jun 30 '20 at 19:22
• @FrançoisBrunault I'm sorry, that seems to be a contradiction? Jun 30 '20 at 20:23
• You are saying that (a) strongly modular $\Leftrightarrow$ completely def / K when K is quadratic; (b) the Guitart--Quer example is a building block that is not strongly modular, and (c) building blocks are completely def / K. Since the Guitart-Quer example is over a quadratic field, (a), (b), (c) can't all be true at once. Jun 30 '20 at 20:25
• @DavidLoeffler Right, sorry these things always get me confused. I think that for this particular curve, the field of complete definition is $\mathbb{Q}(\sqrt{-2}, \sqrt{-3})$. This is explained in Section 3 of arxiv.org/abs/math/0611663 One would have to write down the isogeny, I haven't done that... Jun 30 '20 at 20:47