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Suppose we are given a field $k$ and a finite Galois extension $L$ with Galois group $G$. We consider the projection $\pi:X\otimes_k L\rightarrow X$ for a smooth projective variety $X$. The object $\pi^*\pi_*\mathcal{F}$ for a coherent sheaf $\mathcal{F}$ should be $\bigoplus_{\lambda}\mathcal{F}^{\lambda}$ where $\mathcal{F}^{\lambda}$ is the Galois conjugate of $\mathcal{F}$ and $\lambda\in G$. I now have two questions:

  1. Suppose $\mathrm{Ext}^i(\mathcal{F},\mathcal{F})=0$ for $i>0$. Does this imply $\mathrm{Ext}^i(\mathcal{F},\mathcal{F}^{\lambda})=0$ for $i>0$?

  2. What happens in the derived world i.e. for a given object $\mathcal{K}$ in $D^b(X\otimes_k L)$ what can one say about $L\pi^*R\pi_*\mathcal{K}$ ?

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    $\begingroup$ Crossposted from math.SE (math.stackexchange.com/q/607761/264) but seems appropriate here. $\endgroup$ Commented Dec 15, 2013 at 15:01
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    $\begingroup$ I'm not really sure. I'm still learning this stuff myself, but for some reason, I feel like your (1) should be true. Somehow, the Galois conjugates of $\mathcal{F}$ should be like twists of $\mathcal{F}$, and all of them should provide effective descent data, with the twisted $G$-action, if $\mathcal{F}$ does. $\endgroup$ Commented Dec 15, 2013 at 17:20

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For 2): the same decomposition holds in the derived category as well: note that $\pi $ is flat and affine, so $L\pi ^*$ and $R\pi _*$ of a complex are just obtained by applying $\pi ^*$ and $\pi _*$ to each component.

I don't see any reason for 1) to be true but it is not so easy to think of a counter-example...

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    $\begingroup$ so in the derived case we have something like $\pi^*\pi_*\mathcal{K}\simeq \bigoplus_{\lambda}\mathcal{K}^{\lambda}$? So what is the Galois conjugate of $\mathcal{K}$? Is this just $\lambda_*\mathcal{K}$? $\endgroup$
    – Aleksa
    Commented Dec 15, 2013 at 18:08
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    $\begingroup$ Yes, and yes. No difference with the case of coherent sheaves. $\endgroup$
    – abx
    Commented Dec 15, 2013 at 18:21

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