Timeline for When can we tell if PROPs, Algebraic Theories, etc. are faithfully detected in a given category?
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
Sep 9, 2011 at 19:17 | comment | added | Chris Schommer-Pries | All of this is sort of tangential to my main question. | |
Sep 9, 2011 at 19:17 | comment | added | Chris Schommer-Pries | Good point. The sum of two identical Euler theories does distinguish them as one gives value 2 and the other value 4. But I see your point. A better counter example would be the disjoint union of two tori and the disjoint union of the sphere and the genus 2 surface. To distinguish these we need a sum of Euler theories with different constants. I think a sum of an Euler theory with constant x not a root of unity and a trivial Euler theory with constant y=1 does distinguish everything. Have I messed up again? | |
Sep 9, 2011 at 14:31 | comment | added | Oscar Randal-Williams | It seems to me that this theory does not distinguish between one torus (as a 2-morphism $\emptyset \to \emptyset$) and the composition of two such morphisms. | |
Sep 9, 2011 at 14:01 | comment | added | Chris Schommer-Pries | @Oscar: The tqft I had in mind is just an "Euler theory". It assigns the trivial one dimensional vector space to every 1-manifold. The value assigned to a general surface is just x raised to the Euler characteristic. The Euler characteristic is additive under gluing, so this gives a tqft. One might object that this isn't faithful because it sends different 1-morphisms to the same vector space! This is probably a fair objection. It can be eliminated by taking the direct sum of the Euler theory with itself. Then different 1-manifolds are sent to different dimensional vector spaces. | |
Sep 9, 2011 at 13:54 | history | edited | Chris Schommer-Pries | CC BY-SA 3.0 |
erroneous claims removed, exposition changes slightly.
|
Sep 8, 2011 at 18:29 | comment | added | Noah Snyder | It seems to me that over finite fields you have no hope for the strong detection property because non-semisimple Frobenius algebras will kill all large genus bordisms, while semisimple ones will be periodic and hence have collisions. | |
Sep 8, 2011 at 18:19 | comment | added | Noah Snyder | I'm also a bit confused about why you need x to be irrational... Either you're allowing linear combinations of bordisms in which case you'd need it to be transcendental, or you're not and it should be enough just for it not to be a root of unity. (Or more likely I'm just totally confused.) | |
Sep 8, 2011 at 18:16 | comment | added | Noah Snyder | A similar question comes up in Conjecture 9.1 from Kuperberg's paper arxiv.org/abs/math/9201301. See also my blog post (and the comment thread): sbseminar.wordpress.com/2007/10/11/… | |
Sep 8, 2011 at 18:06 | comment | added | Oscar Randal-Williams | I wonder if you could explain your example in a bit more detail. (In particular, what is the value on $S^1$ and what does one associate to a general surface?) | |
Sep 8, 2011 at 16:53 | history | edited | Chris Schommer-Pries | CC BY-SA 3.0 |
corrected typo.
|
Sep 8, 2011 at 16:23 | history | asked | Chris Schommer-Pries | CC BY-SA 3.0 |