Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
rigid analytic varieties, affinoid varieties, strictly convergent power series over non-archimedean fields
1
vote
Meaning of dagger cohomology $H^{1 \dagger}(G^\dagger)$ in "Frobenius and Monodromy Operator...
I think $X^{\dagger}$ must mean the dagger space associated to the weak formal scheme you get by taking the weak completion of the model $\mathcal{X}$ along the special fibre $\mathcal{X}_{k}$. Then $ …
6
votes
Accepted
Rigid versus log-rigid cohomology for semistable varieties
$\require{AMScd}$I'll expand a little on my comment to give an answer to David's follow up question:
Firstly, the general relationship is described in Chiarellotto's Duke 1999 paper "Rigid cohomology …