Timeline for Finding decomposition of Steenrod operators?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 4, 2018 at 11:30 | comment | added | user51223 | @JohnPalmieri Thanks. I should have remembered Wood's paper. But, I am now arriving at the conclusion that despite existence of various bases, there is no clear `algorithm' to do this! | |
| Mar 3, 2018 at 16:29 | comment | added | John Palmieri | This is related: math.stackexchange.com/questions/2312884/… | |
| Mar 3, 2018 at 12:20 | comment | added | user51223 | @DrewHeard Thanks for the reference. This is an interesting example where for the same Adem relation where one uses $Sq^1Sq^4Sq^1=Sq^5Sq^1=Sq^3Sq^3$ and $Sq^3=Sq^1Sq^2$. But, there are different factors of powers of $2$ appearing in the expression! | |
| Mar 3, 2018 at 10:47 | comment | added | Drew Heard | Such decompositions are not unique, as (for example) $Sq^6 = Sq^2Sq^4 + Sq^1Sq^2Sq^1Sq^2$ as well. Sage can do these types of change of basis for you. But you might also find what you want in mathweb.scranton.edu/monks/pubs/bases.pdf (Look at Wall's basis and Arnon's A basis) | |
| Mar 3, 2018 at 10:34 | history | asked | user51223 | CC BY-SA 3.0 |