Timeline for Given an embedding of $X$ into $\mathbb{P}^n_K$, do you get an induced embedding of any twist of it into $\mathbb{P}^n_K$?
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| Jul 12, 2017 at 18:22 | comment | added | Quinlan Aktaş | 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? | |
| Jul 12, 2017 at 17:57 | comment | added | Will Sawin | 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. | |
| Jul 12, 2017 at 17:54 | history | edited | Quinlan Aktaş | CC BY-SA 3.0 |
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| Jul 12, 2017 at 17:53 | comment | added | Quinlan Aktaş | Mmmmm. That is a correct statement, and is a pretty trivial counter-example. Let me amend the question appropriately. | |
| Jul 12, 2017 at 17:51 | comment | added | Martin Bright | 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 $\mathbb{P}^1$. | |
| Jul 12, 2017 at 17:32 | history | asked | Quinlan Aktaş | CC BY-SA 3.0 |