Twists of elliptic curves

Question

I have a few questions regarding twists of elliptic curves.

1. In the context of the Shafarevich group, I see people refer to the group of twists of an elliptic curve $E/\mathbb{Q}$ by $H^1(\mathbb{Q}, E)$. I know that the group of torsors of a curve $C$, which is in bijection with its set of twists, is isomorphic to the Galois cohomology set $H^1(\mathbb{Q}, \text{Aut}_{\overline{\mathbb{Q}}}(C))$, where $\text{Aut}_{\overline{\mathbb{Q}}}(C)$ is the geometric automorphism group of $C$. Since any automorphism of an abelian variety is a composition of a translation by a homomorphism, $E(\mathbb{Q})$ embeds in $\text{Aut}_{\overline{\mathbb{Q}}}(E)$, so it is safe to say that the group $H^1(\mathbb{Q}, E(\overline{\mathbb{Q}}))$ is only a subgroup of the group of twists/torsors of $E$. Is this correct? What is precisely the difference between these two cohomology groups?

2. Are all twists of an elliptic curve elliptic curves themselves? I guess they all have to be smooth projective of genus $1$, but it isn't clear that they have a rational point, which is necessary to define the group law via the Abel-Jacoby map.

3. A common set of twists are "quadratic twists", which I understand are representatives of 1-cocycle classes of $H^1(\mathbb{Q}, \mathbb{Z}/2\mathbb{Z})$, where the latter may be realized as a subgroup of the automorphism group of $E$ generated through multiplication by minus 1. However, the non-trivial $2$-torsion points of $\overline{E}$ are also order $2$ elements of $\text{Aut}(E)$. Does any of them induce a "quadratic twist"? Or is the group $H^1(\mathbb{Q}, E)$ quadratic twist free?

4. Say $P$ is a geometric point of $E$, and denote by $\gamma_P$ the translation by $P$-geometric endomorphism, and let $c_P$ denote the associated Galois cocycle, which is explicitly given by $g\mapsto \gamma_P^{-1}\circ {}^g\gamma_P$ if I understand correctly. How do I construct a rational model for the twist associated to $c_P$?

5. Is every element of $H^1(\mathbb{Q}, E)$ a finite linear combination of the cocycles $c_P$?

6. It is known that every curve over $\mathbb{Q}$ embeds in $\mathbb{P}^3_{\mathbb{Q}}$. An elliptic curve is obtained as the intersection of a pair of quadratic hypersurfaces. All its twists must therefore also embed in $\mathbb{P}^3_{\mathbb{Q}}$, and a geometric isomorphism between a pair of twists must be parameterized by a degree $1$ projective map, which may then lift to an automorphism of $\mathbb{P}^3_{\mathbb{Q}}$, i.e. be given by an element of $\text{PGL}_4(\overline{\mathbb{Q}})$. This shows that the group of all twists must then be identified with the moduli $\text{PGL}_4(\overline{\mathbb{Q}})/\text{PGL}_4(\mathbb{Q})$. Is it known which subset of this moduli space corresponds to elements of $H^1(\mathbb{Q}, E)$?

Answer 1

Some answers to your questions:

As R. van Dobben de Bruyn explains in the comments, "twist" is used to mean multiple different things. Twists as a variety are classified by $H^1$ of the automorphism group of the variety, twists as an elliptic curve are classified by $H^1$ of the group of automorphisms fixing the identity, and torsors are classifeid by $H^1(\mathbb Q,E)$. The three groups in question are related by a short exact sequence.

1. $H^1(\mathbb Q, E)$ admits a map to the set of twists of a variety. Since the automorphism group as a variety is not abelian, the set of twists as a variety do not form a group and so this is not a group homomorphism. The map is not injective - a class in $H^1(\mathbb Q, E)$ and its additive inverse always correspond to isomorphic curves, but are not isomorphic torsors unless the class is $2$-torsion.

2. A nontrivial element of $H^1(\mathbb Q, E)$ never gives an elliptic curve, since a torsor with a rational point is the trivial torsor.

3. Thus no nontrivial element of $H^1(\mathbb Q,E)$ gives a quadratic twist. Of course, the trivial element gives the trivial quadratic twist. (Also for elliptic curves of $j$ invariant $1728$ the quadratic twist by $-1$ is isomorphic to the curve itself and hence corresponds to the trivial element of $H^1(\mathbb Q,E)$.)

4. The cocycle you have written down is just the coboundary of $P$ inside $E(\overline{\mathbb Q})$ which is the $0$th cocycle for the cohomology of $\mathbb Q$ with coefficients in $E$, i.e. the cohomology of $\operatorname{Gal}(\mathbb Q)$ with coefficients in $E( \overline{\mathbb Q})$. So this cocycle always corresponds to the trivial cohomology class and so a model is just $E$.

5. Not always, since $H^1(\mathbb Q,E)$ is not always trivial.

6. The flaw in your argument is when you say "and a geometric isomorphism between a pair of twists must be parameterized by a degree $1$ projective map" - this is not true. An isomorphism could be given by a map of higher degree as well.

Also, for your question in the comments on if one can parameterize all torsors under $E$: It's not a parameterization exactly but one can classify torsors using the map $H^1(\mathbb Q, E) \to H^1(\mathbb R,E)\times \prod_p H^1( \mathbb Q_p, E)$ where the target can be calculated and then the kernel is the Tate-Shafarevich group.