Timeline for When is a TQFT the dimensional reduction of a higher dimensional TQFT?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2010 at 21:00 | comment | added | Chris Schommer-Pries | @Theo and Daniel, In fact for certain easy G and level $\tau$ it is not hard to show that $K_G^\tau$ cannot be realized as a full 2-1-0 TQFT (with values in one of the usual suspect 2-cats like Abelian Cats or Algebras, Bimod, Bimod maps). | |
| Oct 14, 2010 at 20:02 | answer | added | Greg Kuperberg | timeline score: 15 | |
| Jul 26, 2010 at 16:17 | comment | added | Daniel Pomerleano | @Theo: I'm not sure anyone knows how to define twisted K-theory over Z anyways as a 2-1-0 theory...(the issue obviously being what to assign to a point) at least I don't :) | |
| Jun 12, 2010 at 16:14 | comment | added | Theo Johnson-Freyd |
Part of the idea that "defined over $\mathbb Z$ means wants to be a dimensional reduction" is as follows. It should be generally true about any TQFT $F$ that $F(M\times S^1) = \dim F(M)$, where $\dim$ is defined appropriately for your target $n$-category. But in most situations, dimensions are integer things.
|
|
| Jun 12, 2010 at 16:12 | comment | added | Theo Johnson-Freyd |
Nice question, although I expect the answer is "this is in many cases still open". I prefer the notation in which $K^\tau_G$ is the 2-1-0 TQFT defined over $\mathbb Z$; then I would say that $CS(-\times S^1) = K^\tau_G(-) \otimes \mathbb C$. So the $\mathbb Z$-version of K-theory wants to be a dimensional reduction, but currently fails without tensoring --- CS theory doesn't see the torsion in K-theory.
|
|
| Jun 9, 2010 at 12:26 | history | asked | skupers | CC BY-SA 2.5 |