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Mar 26, 2013 at 11:47 history edited ChrisLazda CC BY-SA 3.0
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Mar 26, 2013 at 10:57 answer added D.-C. Cisinski timeline score: 7
Mar 26, 2013 at 10:47 comment added ChrisLazda At least for F-isocrystals, 6 operations has now been worked out by Caro - he has a good theory of overholonomic F-D modules which are stable under all operations and contain the category of overconvergent F-isocrystals. On quasi-projective varieties he has proved stability of holonomicity (with F-strucure). I am specifically curious as to whether one might hope to make 6 operations work within le Stum's framework, because this base change business seems to suggest not (to me anyway).
Mar 26, 2013 at 3:19 comment added Emerton Dear David, Thanks for this clarification. My understanding was that some of these open problems had been solved, at least partially, by some of the results of Kedlaya over the last several years (at least if one considers $\mathcal D^{\dagger}-F$ modules, i.e. $\mathcal D^{\dagger}$-modules with a Frobenius structure), but I haven't kept up with the details, and so don't know the current status. Best wishes, Matt
Mar 26, 2013 at 2:51 comment added David Zureick-Brown @Emerton: le Stum's site is a "big site", analogous to the big Zariski site (so for the big Zariski site the underlying category is just schemes), so base change does hold. I'm not sure what the status of Berthelot's theory of $D^{\dagger}$-modules is now, but there used to be open problems along the lines of "$f!_$ of overholonomic is overholonomic". Googling now it looks like there are some recent papers by Daniel Caro on this.
Mar 26, 2013 at 2:28 comment added Emerton P.S. I should add that $\mathcal D^{\dagger}$-modules were introduced by Berthelot to provide a six operations formalism that incorporated rigid cohomology. I'm not sure how they interact with Le Stum's theory.
Mar 26, 2013 at 2:26 comment added Emerton Dear Chris, Are you sure that your base-change statement is correct? What if $f$ is the inclusion of $X = \mathbb A^1$ in $Y = \mathbb P^1$ and $g$ is the inclusion of $Y' =$ point at infinity. Then $X' = \emptyset$, so the right hand side of the base-change morphism vanishes. Is this really what happens on the left-hand side? Also, Berthelot's theory of $\mathcal D^{\dagger}$-modules is supposed to give the theory of coefficients and six operations. Regards,
Mar 26, 2013 at 2:06 history asked ChrisLazda CC BY-SA 3.0