Timeline for Ring of closed manifolds modulo fiber bundles
Current License: CC BY-SA 2.5
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 7, 2014 at 11:01 | vote | accept | Andreas Thom | ||
| Nov 2, 2014 at 23:58 | answer | added | Qiaochu Yuan | timeline score: 6 | |
| Aug 26, 2010 at 23:20 | comment | added | Paul | The odd dimensional spheres are all zero, since S^1 acts freely (and similarly lens spaces, most compact Lie groups, etc). I suspect the only non-homeomorphic equivalent simply connected 4-manifolds are $CP^2\# -CP^2$ and $S^2 \times S^2$. This should follow from the fact that the ideal is graded, so that equivalent 4 manifolds should arise from fibrations with 1 or 2 diml fibers or base, and the calculations mentioned for 1&2-manifolds. | |
| Aug 26, 2010 at 14:56 | comment | added | Andreas Thom | @Andre: I think that without being able to ignore mapping tori (using the argument above), the ring $R$ would be surely very large (without being able to prove this). | |
| Aug 26, 2010 at 13:19 | comment | added | André Henriques | If one restricts attention to fiber bundles with connected fibers, does the question then become more interesting? | |
| Aug 26, 2010 at 12:13 | answer | added | Igor Belegradek | timeline score: 3 | |
| Aug 26, 2010 at 6:30 | comment | added | Sean Tilson | That was not so much the issue. I was forgetting that we want $\Chi$ to not distinguish between bundles and products... how embarrassing. Would it be uncool to delete my comments? is it obvious I don't spend a lot of time thinking about euler charateristics? | |
| Aug 26, 2010 at 6:04 | comment | added | Andreas Thom | @Paul: $-[M]$ is the just the formal additive inverse of $[M]$. ̯@Sean: I think that the notion of a ring generated by variables subject to relations is standard. Of course it can happen that variables get identified. More formally, it is the quotient of the (a priori non-commutative) polynomial ring with variables indexed by homeomorphism classes modulo the relations $[M] + [N] - [M \cup N]$, $[E] - [B][F]$ and $[pt]=1$. | |
| Aug 26, 2010 at 5:34 | comment | added | Sean Tilson | Thanks Ryan, that must be what Andy is trying to say. It just felt such a "degenerate" product would not be too interesting. | |
| Aug 26, 2010 at 5:17 | comment | added | Ryan Budney | @Sean, read damiano and Agol's answers, they're addressing your concern. The ring is defined via two equivalence relations, first you take homeomorphism types of manifolds, then you form the free commutative ring on the homeormophism types of the manifolds, then you mod out by the ideal generated by all $[M]+[N]-[N \sqcup M]$ and $[E]-[M][N]$. | |
| Aug 26, 2010 at 5:02 | comment | added | Sean Tilson | I guess I mean that usually the product structure is supposed to help refine the information. In this situation, why keep it around at all? Also, $S^1$ is its own double cover so now we have two bundles with the same base space and the same fiber, the trivial bundle and the double cover. This gives us that $S^1+S^1=S^1 \in R$. So ... 2=1? I think that things likes this will crop up. | |
| Aug 26, 2010 at 4:57 | comment | added | Sean Tilson | I thought each equivalence class was a different homeomorphism type. Am I misreading that? Also if we do that then we are really only looking at pairs of manifolds, and the order does not matter and addition is component wise. Is this what was intended? | |
| Aug 26, 2010 at 4:13 | comment | added | Andy Putman | @Sean : Can't all of them be the answer? All that happens is that in Andreas's ring, all the total spaces you are talking about get identified. | |
| Aug 26, 2010 at 4:05 | comment | added | Sean Tilson | It looks like multiplication is not well defined in your ring. There are non-homeomorphic bundles with the same fiber and the same base space, take any non-trivial bundle and then the trivial bundle with the same fiber and the same base. Which one gets to be the product of the fiber and the base? | |
| Aug 26, 2010 at 3:52 | answer | added | Ian Agol | timeline score: 8 | |
| Aug 26, 2010 at 0:45 | comment | added | Ryan Budney | I think he's likely taking the free commutative ring with integer coefficients, with generators the isomorphism types of manifolds, and modding out by the ideal generated by the equivalence relation $[M]+[N]\sim[M\cup N]$ and $[M]\cdot[N]\sim[E]$. So in general −[M] is just a formal thing, it's generally not the isomorphism type of a manifold, for example, $−[1]=−[pt]$ isn't an isomorphism type of any manifold. | |
| Aug 25, 2010 at 23:54 | answer | added | damiano | timeline score: 7 | |
| Aug 25, 2010 at 23:14 | comment | added | Paul | what is -[M] ? | |
| Aug 25, 2010 at 22:20 | comment | added | David Steinberg | If you are interested in the algebraic category, I suggest Bridgeland's Introduction to Motivic Hall Algebras (arxiv.org/abs/1002.4372) which develops similar rings for varieties, schemes, and stacks. | |
| Aug 25, 2010 at 21:50 | history | edited | Andreas Thom |
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| Aug 25, 2010 at 21:40 | history | asked | Andreas Thom | CC BY-SA 2.5 |