Timeline for Integral transform on noncommutative spaces
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 6, 2015 at 22:56 | answer | added | Qiaochu Yuan | timeline score: 1 | |
| Apr 6, 2015 at 20:53 | comment | added | AAK | @bananastack, I learned this from Marco Robalo's thesis, but it is probably in some paper of To\"en. | |
| Apr 6, 2015 at 20:50 | comment | added | bananastack | @Adeel ah, sweet, did not know that. reference? (thanks) | |
| Apr 6, 2015 at 20:33 | comment | added | AAK | @bananastack, saturated dg-categories also come from dg-algebras, by the way (from dg-algebras of finite type, even). | |
| Apr 6, 2015 at 19:46 | vote | accept | Romie Banerjee | ||
| Apr 6, 2015 at 18:39 | comment | added | Romie Banerjee | @QiaochuYuan, ofcourse, thanks for pointing out. | |
| Apr 6, 2015 at 18:39 | comment | added | bananastack | Here's a less ambitious way to think about this. All these fancy QC(X) categories are secretly A-Mod for some dg-ring A (A = End(G), for a generator G. all separated schemes admit one by Bondal-vdB if I'm not mistaken). So, if you are willing to restrict to dg-categories equivalent to A-Mod for some dg-ring A, then Functors are given by bi-modules (and you need to stick an op superscript as QY points out). | |
| Apr 6, 2015 at 18:04 | comment | added | Qiaochu Yuan | In the commutative case note that $\text{QC}(X)$ is in a suitable sense self-dual, and there's no reason for this to be true in the noncommutative case; more simply, already something must be wrong with your formula because it's covariant in $A$ on the LHS but contravariant in $A$ on the RHS. Even in the simplest setting where we have an equivalence like that, namely finite-dimensional vector spaces, we need to take $A^{op} \otimes B$ on the LHS. | |
| Apr 6, 2015 at 16:51 | answer | added | AAK | timeline score: 2 | |
| Apr 6, 2015 at 12:44 | answer | added | Zhaoting Wei | timeline score: 1 | |
| Apr 5, 2015 at 22:19 | comment | added | Will Sawin | In the case where $B$ is a point, this requires a natural map from $A$ to functors from $A$ to a point. Basically a map $A\times A\to D^b(Vect)$. Given one of these, you should be able to produce the described map by tensoring this map with $B$. So you want a self-duality statement. | |
| Apr 5, 2015 at 19:51 | history | edited | Romie Banerjee |
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| Apr 5, 2015 at 18:09 | history | edited | Romie Banerjee |
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| Apr 5, 2015 at 17:52 | history | asked | Romie Banerjee | CC BY-SA 3.0 |