Timeline for Why are cup-i products and Steenrod Squares often (always?) unary?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 16, 2013 at 8:09 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
fixed minor issue with math display
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| Aug 12, 2012 at 19:42 | comment | added | Sean Tilson | While the $\cup_i$'s are useful, they may see less attention due to the fact that they do not satisfy a nice list of properties, but the steenrod operations do. | |
| Aug 11, 2012 at 14:42 | answer | added | Peter May | timeline score: 9 | |
| Aug 10, 2012 at 21:39 | vote | accept | Joseph Victor | ||
| Aug 10, 2012 at 9:46 | comment | added | Pelle Salomonsson | The cup_i-product of two closed chains may be nonclosed, if I remember correctly. | |
| Aug 10, 2012 at 7:32 | answer | added | Andrew Ranicki | timeline score: 13 | |
| Aug 10, 2012 at 0:16 | comment | added | David White | I think it does get used as a binary product. For instance, $u \cup_i v$ measures how far $u \cup_{i-1} v$ is from being commutative. So all the $\cup_i$ together are telling you information about the classical $\cup$ product (which is $\cup_0$) in the same way that all the levels of $A_\infty$ together give you a homotopy associative product. I think the main reason to move from $\cup_i$ to $Sq^i$ is that for the application Mosher and Tangora want (division algebras) they care about operations on cohomology, i.e. from $H^*$ to $H^*$. | |
| Aug 9, 2012 at 23:39 | history | asked | Joseph Victor | CC BY-SA 3.0 |