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Jun 15, 2020 at 7:27 history edited CommunityBot
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Sep 29, 2011 at 1:13 answer added Piotr Achinger timeline score: 5
May 31, 2011 at 7:13 vote accept Piotr Achinger
May 31, 2011 at 2:51 answer added Matt timeline score: 9
May 30, 2011 at 11:23 comment added Piotr Achinger Great! Why don't you post this as an answer?
May 29, 2011 at 19:10 comment added Matt I believe that even if Frobenius doesn't lift you still get that a Frobenius split variety (with nice assumptions) will lift to $W_2(k)$. This is because even though the obstruction to lifting Frobenius $\eta\in Ext^1(\Omega^1, B^1)$ is non-zero, the obstruction to merely lifting $X$ is the image of $\eta$ under the connecting homomorphism $Ext^1(\Omega^1, B^1)\to Ext^2(\Omega^1, \mathcal{O}_X)$. By the splitting assumption this map is $0$ and hence there is no obstruction to lifting $X$. See V. Srinivas Decomposition of the de Rham Complex.
Mar 21, 2011 at 17:50 history edited Piotr Achinger CC BY-SA 2.5
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Mar 21, 2011 at 15:05 history asked Piotr Achinger CC BY-SA 2.5