Timeline for From the perspective of bordism categories, where does the ring structure on Thom spectra come from?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2020 at 21:17 | comment | added | Noah Snyder | If I take the cartesian product of bordisms I don't think I end up with a bordism with corners in the usual sense, because the usual notion of bordism with corners should be constant when restricted to the boundary of the bordisms. Is there some quick fix to this? | |
| Nov 7, 2014 at 3:11 | vote | accept | Qiaochu Yuan | ||
| Nov 7, 2014 at 3:02 | answer | added | Qiaochu Yuan | timeline score: 17 | |
| Oct 6, 2014 at 5:26 | comment | added | Qiaochu Yuan | @Thomas: yes, but my comment above is not about the cartesian product of manifolds. It is about the cartesian product in $\text{Bord}$. | |
| Oct 5, 2014 at 7:33 | comment | added | Thomas Kragh | @Qiaochu: yes, but I am not saying anything about bifunctors. I am asking: is the weird structure you want not induced by the usual Cartesian products of manifolds: $M\times M'$ is a $k+k'$ morphism when $M$ is a $k$ and $M'$ a $k'$? and as I said above, a first hint that this might be meaningful is that it distributes as $M\times (M' \circ N')=(M\times M') \circ (M\times N')$. | |
| Oct 4, 2014 at 8:33 | comment | added | Qiaochu Yuan | @Thomas: as I said above, for any category $C$, a bifunctor (e.g. the cartesian product) $C \times C \to C$ takes as input, say, a pair of $1$-morphisms and returns as output another $1$-morphism. It does not increase the morphism number. | |
| Oct 3, 2014 at 6:05 | comment | added | Thomas Kragh | @Qiaochu (or is it Yuan): But why is the 1-morphism $I$ (from point to point) cartesian product itself not a 2-morphism from $I$ to $I$? | |
| Oct 1, 2014 at 7:38 | comment | added | Qiaochu Yuan | @Thomas: the cartesian product of smooth manifolds in the usual sense is their cartesian product as objects in the category of smooth manifolds and smooth maps. $\text{Bord}$ is a very different category; in particular, manifolds appear as morphisms, not objects, in it. | |
| Sep 30, 2014 at 23:10 | comment | added | Thomas Kragh | I realize that my previous comment was way off (removed - hope that is OK). I think I was maybe just getting my feet wet when it came to understanding the $E_\infty$ structure you where referring to, but not sure. Now I am stuck at the point where I don't see why the Cartesian product of manifolds does not give the structure Hiro is describing? Indeed, it takes a $k$ and $k'$ morphism and spits out a $k+k'$. The usual composition even distributes over this as if it is a second weirdly graded monoidal structure making it all look bimonoidal. | |
| S Sep 27, 2014 at 4:32 | history | bounty ended | CommunityBot | ||
| S Sep 27, 2014 at 4:32 | history | notice removed | CommunityBot | ||
| S Sep 19, 2014 at 3:14 | history | bounty started | Qiaochu Yuan | ||
| S Sep 19, 2014 at 3:14 | history | notice added | Qiaochu Yuan | Draw attention | |
| Jul 10, 2014 at 15:09 | comment | added | Ilias A. | Your observation seems to be correct about non existence of the cartesian product! | |
| Jul 10, 2014 at 8:57 | comment | added | Qiaochu Yuan | @Fedotov: what do you mean by that? Note that functors $\text{Bord} \times \text{Bord} \to \text{Bord}$ take as input, say, a $j$-morphism and a $j$-morphism and return another $j$-morphism, whereas the product structure we want, as Hiro said, takes a $j$-morphism and a $k$-morphism and returns a $j+k$-morphism. In particular the cartesian product can't be responsible for this structure (I don't think $\text{Bord}$ even has cartesian products), but more generally no bifunctor can. | |
| Jul 9, 2014 at 20:18 | comment | added | Hiro Lee Tanaka | You can take a $k$-morphism and a $k'$-morphism to obtain a $k+k'$-morphism. In some sense this structure is already present for the $(\infty,n)$-category of cobordisms up to dimension $n$, but I don't know if there's a name for this structure. It seems like another notion for a symmetric monoidal structure on $(\infty,n)$-categories. | |
| Jul 9, 2014 at 20:15 | comment | added | Hiro Lee Tanaka | Some technical comments: First, the version of the cobordism hypothesis you seek is one for $(\infty,\infty)$-categories. I don't think this has been developed: The usual formulation cuts off at the $(\infty,n)$-level, declaring that all higher morphisms are isotopies of cobordisms. You seem to want all higher cobordisms, because you in particular want to capture the Thom spectrum associated to framed cobordism (which see all higher framed cobordisms between $\emptyset$ and $\emptyset$, for instance). Then the ring spectrum structure you want seems to correspond to a strange property... | |
| Jul 9, 2014 at 19:51 | comment | added | Ilias A. | How about the cartesian product in Bord^fr. | |
| Jul 9, 2014 at 17:27 | history | asked | Qiaochu Yuan | CC BY-SA 3.0 |