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Let $X$ be a projective algebraic curve over some number field $K$, and let $\varphi:X\hookrightarrow \mathbb{P}^n_K$ be an embedding of it (defined over $K$) into some projective space.

Now let $X'$ be a twist of $X$; namely, a curve over $K$ that is isomorphic to $X$ over $\bar{\mathbb{Q}}$. Assume that $X'$ contains a $K$-rational point. Then does $\varphi$ induce an embedding (defined over $K$) of $X'$ into $\mathbb{P}^n_K$?

Here is a basic example I'm thinking of. Let $X$ be a smooth elliptic curve over $\mathbb{Q}$, given on an affine patch by $y^2=x^3-1$. The equation $y^2=x^3-1$ gives an embedding of $X$ into $\mathbb{P}^2_{\mathbb{Q}}$. One way to generate twists of $X$ is to look at the curves $X_a$, for $a\in\mathbb{Q}$, given on an affine patch by $ay^2=x^3-1$. The equation $ay^2=x^3-1$ gives an embedding of $X_a$ into $\mathbb{P}^2_{\mathbb{Q}}$, which, in some sense, is induced by the given embedding of $X$ into $\mathbb{P}^2_{\mathbb{Q}}$. The question is: can this be formulated in a natural way?

Here is a naive attempt. The embedding $\varphi:X\hookrightarrow \mathbb{P}^n_K$ is given by a very ample line bundle $\omega\in H^1(X,O_X^{\times})$, and a basis of global sections of $\omega$. Any twist $X'$ of $X$ is given by an element $\gamma\in H^1(K,\underline{Aut}(X))$, where $\underline{Aut}(X)$ is the sheaf of automorphisms of $X$. A potential reformulation of my question is whether there is a natural way to "twist" $\omega$ by $\gamma$; and furthermore, if $\omega$ is very ample, and if $X'$ has a $K$-rational point, then does this imply that its twist by $\gamma$ is very ample?

If my attempt seems misguided to you, then focus on the example, which inspired my attempt to begin with...

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    $\begingroup$ Take $X$ to be $\mathbb{P}^1$ and $\phi$ the identity. Any non-trivial twist of $X$ is isomorphic to a plane conic, which does not admit an embedding into $\mathbb{P}^1$. $\endgroup$
    – Martin Bright
    Commented Jul 12, 2017 at 17:51
  • $\begingroup$ Mmmmm. That is a correct statement, and is a pretty trivial counter-example. Let me amend the question appropriately. $\endgroup$
    – Quinlan Aktaş
    Commented Jul 12, 2017 at 17:53
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    $\begingroup$ In your case, the key thing is to observe there is a natural way to lift the twists to cohomology classes in $H^1(K, Aut(X, \omega))$, which you can see is the same as the group of automorphisms of $X$ fixing the point at $\infty$. For other automorphisms of $X$, those related to the Tate-Shafarevich group, the embedding would not be preserved. $\endgroup$
    – Will Sawin
    Commented Jul 12, 2017 at 17:57
  • $\begingroup$ Oh, interesting. So let me get this straight. You're saying that $H^1(K,\underline{Aut}(X,\omega))\rightarrow H^1(K,\underline{Aut}(X))$ is injective (obviously), and that $\underline{Aut}(X,\omega)(L)$, for any $L/K$, can be interpreted as those automorphisms for which $\varphi_L^{-1}(\mathbb{P}^n_L\smallsetminus \mathbb{A}^n_L)$ is preserved? Furthermore, $H^1(K,\underline{Aut}(X,\omega))$ is exactly those twists for which the corresponding twists have an induced twist of $\omega$? Do I have the story straight? $\endgroup$
    – Quinlan Aktaş
    Commented Jul 12, 2017 at 18:22

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