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 |
6
votes
Deequivariantisation of indecomposable sheaves
The equivariant derived category in this case is equivalent to modules for the homology of the circle, ie exterior algebra on a generator in degree -1. … The (dg-enhanced) equivariant derived category of a point is equivalent to modules for the dg algebra $C_*(G)$ of chains on $G$ under convolution (the ``topological group algebra"). …
2
votes
Equivariant homology of $\Omega X$\/-space (references needed)?
Until the relevant experts arrive, I can mention that one relevant key phrase is "Waldhausen A-theory" (or K-theory of spaces, cf. Waldhausen's paper with this title), which is the K-theory of $\Omega …
2
votes
Accepted
For a G-variety, what could one say about the motif of the corresponding simplicial variety
The simplicial variety you write is a presentation of the stack $X/G$ as a simplicial sheaf, and the motive you obtain is just be the motive of $X/G$, no? Of course you might use this as the definitio …
8
votes
Is this a definition of equivariant derived category?
The idea that the equivariant version of the derived category consists simply of complexes with equivariance structure is correct, once you work in a refined enough setting such as differential graded, … But at this enhanced level, the equivariant derived category is indeed just the derived invariants of the group acting on the derived category of X. …
14
votes
Accepted
Are these notions of strongly equivariant D-modules equivalent?
More precisely, there are two ways to think of D-modules on a stack $X/G$ (aka $G$-equivariant D-modules on $X$). … the second step, fixing the action of the Lie algebra (or equivalently the formal group) tells you which sheaves on $X_{dR}/G$ actually come from $X_{dR}/G_{dR}$, ie are strongly equivariant.. …