Timeline for Why does one consider the dual of the Steenrod algebra?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2016 at 10:16 | answer | added | მამუკა ჯიბლაძე | timeline score: 3 | |
| Jun 14, 2011 at 5:36 | answer | added | John Palmieri | timeline score: 12 | |
| Jun 13, 2011 at 20:12 | answer | added | Hal Sadofsky | timeline score: 27 | |
| Jun 13, 2011 at 19:49 | comment | added | Pierre | Let me add that, if $A$ is the Steenrod algebra and $A^*$ it dual, then there is a map $\lambda : H^*(X)\to H^*(X)\otimes A^*$ (in fact you need to take a completion somewhere, doesn't really matter). You can recover the action of $A$ on $H^*(X)$ easily from this. The great thing about $\lambda$ though, is that it is a map of rings. There are situations when this makes horrendous computations trivial. | |
| Jun 13, 2011 at 18:28 | comment | added | Dylan Wilson | Alternatively, the coproduct on the Steenrod algebra is much easier to work with than the product (in other words: the Adem relations are more complicated than the Cartan formula). On the other hand, working with coproducts are awkward... so we take the dual and then we have a nice product, and the coproduct is ugly but that's alright. | |
| Jun 13, 2011 at 18:18 | comment | added | Todd Trimble | Please read "how to ask". | |
| Jun 13, 2011 at 18:06 | comment | added | Eric Peterson | The dual of the Steenrod algebra is commutative while the original is not (though this means that the diagonal on the dual is fairly complicated). We have many more tools for dealing with commutative things than with noncommutative things. This is sort of a cheap answer, and certainly there are more reasons than just this, but imho it alone is enough. | |
| Jun 13, 2011 at 18:00 | history | asked | user12832 | CC BY-SA 3.0 |