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Apr 13, 2017 at 12:58 history edited CommunityBot
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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