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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.)

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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 $K_d$).

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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?

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    $\begingroup$ 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. $\endgroup$ – François Brunault Jun 30 at 19:20
  • $\begingroup$ (However, the converse is true if $K$ is a quadratic field.) $\endgroup$ – François Brunault Jun 30 at 19:22
  • $\begingroup$ @FrançoisBrunault I'm sorry, that seems to be a contradiction? $\endgroup$ – David Loeffler Jun 30 at 20:23
  • $\begingroup$ 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. $\endgroup$ – David Loeffler Jun 30 at 20:25
  • $\begingroup$ @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... $\endgroup$ – François Brunault Jun 30 at 20:47

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