Timeline for How do we handle the symmetry condition in nCob and TQFTs?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2014 at 23:17 | vote | accept | Josh | ||
| Jul 1, 2014 at 17:13 | comment | added | Zhen Lin | While it is possible to arrange for $M \amalg N = N \amalg M$ as objects, it is not in general possible to make the "swap" isomorphism $M \amalg N \to N \amalg M$ equal to the identity morphism. So there's really no point in forcing $M \amalg N = N \amalg M$ anyway. | |
| Jul 1, 2014 at 15:04 | comment | added | Noah Snyder | Any symmetric tensor category can be strictified, so there's certainly some version of the bordism category where $M \sqcup N = N \sqcup M$. Scrictification is not a very natural thing to do though, and Oscar's right that with standard definitions $M \sqcup N \neq N \sqcup M$. | |
| Jul 1, 2014 at 15:00 | comment | added | Josh | Well, by "name" I just mean their index in some indexing set $\mathcal{I}$, but where $\mathcal{I}$ is not ordered. So I suppose we'd index the "left" $M$ by $i$ say, the "right" $M$ by $j$, and just have $M\sqcup M = M \times \{i,j\}$. But I think this is starting to become more set-theoretic (possibly a new question) than this current question calls for, maybe we can start a discussion or a new question. Thanks very much @Oscar! | |
| Jul 1, 2014 at 14:59 | answer | added | Noah Snyder | timeline score: 12 | |
| Jul 1, 2014 at 14:57 | comment | added | Oscar Randal-Williams | Not quite, labelling by the set's "name" is not a good idea: what would $M \sqcup M$ be in that case? | |
| Jul 1, 2014 at 14:54 | comment | added | Josh | I see. So in taking disjoint unions one can choose to index the constituent sets in such a way as to either keep track of, or forget, their ordering (i.e. index by a set's "name" versus its ordinality). It's interesting how I find the first option more intuitive ("they're just sets, ordering in their $\sqcup$ shouldn't matter!") but in this case it's unhelpful to think about it that way. | |
| Jul 1, 2014 at 14:36 | comment | added | Oscar Randal-Williams | Yes, that's exactly what I mean: there are then canonical isomorphisms $M \sqcup N \to N \sqcup M$, which are not the identity maps, and which give the cobordism category a symmetric monoidal structure. | |
| Jul 1, 2014 at 14:34 | comment | added | Josh | Another definition of disjoint union I've heard is that $M\sqcup N = M\times \{0\} \cup N\times \{1\}$, under this definition $M\sqcup N$ indeed is different from $N\sqcup M$, is this what you're alluding to? | |
| Jul 1, 2014 at 14:30 | comment | added | Josh | So one way of defining the disjoint union $M\sqcup N$ is to simply take the set of points of $M$ and $N$ separately, "label" each point (with a subscript say) in order to distinguish which set it came from, and then take the ordinary set union. But this doesn't distinguish $M\sqcup N$ and $N \sqcup M$. | |
| Jul 1, 2014 at 14:23 | comment | added | Oscar Randal-Williams | You should think about what, really, you mean by the collection of symbols $M \sqcup N$. If you want this construction to give you an actual abstract manifold, it should also give you an actual abstract underlying set: which set is it? | |
| Jul 1, 2014 at 14:17 | history | asked | Josh | CC BY-SA 3.0 |