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Nov 8 |
awarded | Nice Question |
Nov 8 |
accepted | Pro-algebraic versus continuous Galois cohomology, and schematic homotopy types |
Nov 8 |
answered | Pro-algebraic versus continuous Galois cohomology, and schematic homotopy types |
Nov 7 |
revised |
Pro-algebraic versus continuous Galois cohomology, and schematic homotopy types
added 263 characters in body |
Nov 7 |
comment |
Pro-algebraic versus continuous Galois cohomology, and schematic homotopy types
No, I don't think so. By cohomology, I really mean $Hom_{D}(1,V[i])$, where $D$ is either of the triangulated categories, so $t$-structures don't come into it at all. I'll edit the question to make this clearer. |
Nov 6 |
revised |
Pro-algebraic versus continuous Galois cohomology, and schematic homotopy types
added 16 characters in body |
Nov 6 |
comment |
Pro-algebraic versus continuous Galois cohomology, and schematic homotopy types
Sorry, which particular section of the thesis did you have in mind? It seems to be talking about pro-finite homotopy types to me... |
Nov 5 |
asked | Pro-algebraic versus continuous Galois cohomology, and schematic homotopy types |
Oct 11 |
comment |
“Extended” Weil Cohomology Theories
Ah, I didn't know about Bloch-Ogus, that might be pretty much what I'm looking for. |
Oct 10 |
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“Extended” Weil Cohomology Theories
No, their notion of a mixed Weil cohomology theory doesn't say anything about compact supports or Poincaré duality. It follows from their results that given a Weil cohomology theory, all the extra stuff pops out (by realising the motivic version), but it doesn't form part of their definition of a mixed Weil cohomology theory. |
Oct 10 |
asked | “Extended” Weil Cohomology Theories |
Oct 1 |
comment |
on a characterisation of regular D-modules
Are you sure? If we let $w=z^{-1}$, then the connection becomes $\nabla(f)=df+f\frac{dw}{w^2}$, so it doesn't have logarithmic poles around $w=0$. |
Sep 30 |
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on a characterisation of regular D-modules
An easy way to see that there is no $\mathrm{Ext}^1$ on the de Rham side is to note that $M$ is rank one, i.e. invertible. We have the dual connection $M^{\vee}=(\mathcal{O}_X,\nabla^\vee)$ where $\nabla^\vee(f)=df+fdz$, and $M\otimes M^\vee$ is just the constant integrable connection $\mathcal{O}_X$. Hence we can pass between $\mathrm{Ext}_\nabla^1(M,M)$ and $\mathrm{Ext}_\nabla^1(\mathcal{O}_X,\mathcal{O}_X)$. |
Sep 30 |
answered | on a characterisation of regular D-modules |
Sep 16 |
awarded | Enthusiast |
Aug 27 |
comment |
What is the classification of characters in $p$-adic Hodge theory?
Crystalline characters, in your case, are exactly the twists of unramified characters, see for example this MO question mathoverflow.net/questions/61998/crystalline-characters. |
Aug 16 |
comment |
Categories of sheaves and Kan Extensions
Yes, exactly, for $f^*$ at least. |
Aug 16 |
comment |
Categories of sheaves and Kan Extensions
When you unwind the definitions, doesn't the right Kan property just boil down to the fact that you can glue quasi-coherent sheaves over a Zarski (hyper-) cover? So it should hold for any Zariski local category, like $\ell $-adic sheaves or D-modules. Or does strictifying mess things up? The argument doesn't dualise, though, so I don't know about the left Kan property for pushforward. I also don't know enough about $\infty $ categories to be sure, but presumably in this case it will again just boil down to the fact that the $\infty$-categorical derived caregory is Zariski local? |
Jun 27 |
answered | comparison theorem for connections with regular singularities |
Jun 25 |
comment |
local systems, duals, cohomology
I don't think so. If $n=0$, then $U$ is just $\mathbb{P}^1$, so the cohomology is just that of the sphere, and so we have $h^0=1$, $h^1=0$, $h^2=1$. But if $n\geq 1$ then $U$ is homotopic to the wedge product of $n-1$ spheres, and so we will have $h^0=1$, $h^1=n-1$ and $h^2=0$. |