Timeline for Steenrod operations from the delooping viewpoint
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 19, 2020 at 18:40 | comment | added | Student | I see.. I should not take $\Omega$ and assume it loses no information. I should leave it as $B^* \to B^{*+i}$. | |
| May 19, 2020 at 18:33 | comment | added | Connor Malin | @Student F is a discrete set of points. $\Omega^i F$ is the space of pointed maps from the i-sphere into $F$. | |
| May 19, 2020 at 18:30 | comment | added | Student | I didn't know $\Omega^iF$ vanishes for $i>0$.. instead I thought $\phi_i$ and $\psi_i$ contain the same amount of information.. am I wildly wrong?! | |
| May 19, 2020 at 18:22 | comment | added | Connor Malin | @Student While what you've written is correct $\Omega^i F =0$ if $i>0$. The point is that this cohomology operation will always be 0 on 0th cohomology, but it can deloop to something nontrivial. So any additive way of delooping it will give the trivial operation. | |
| May 19, 2020 at 18:17 | comment | added | Student | Ah my apology! I have edited accordingly. | |
| May 19, 2020 at 18:16 | comment | added | Connor Malin | If you are interested in more no go results, the only stable cohomology operations for rational cohomology is multiplication by a rational number. | |
| May 19, 2020 at 18:13 | comment | added | Connor Malin | @Student Well in your post your map should be $F \rightarrow B^i F$ (you have the Yoneda lemma the wrong way around). You then need to find a delooping of this (nullhomotopic) map, and then a delooping of that map, etc. The issue with taking your delooping to be $B(F \rightarrow B^i F)$ is that this is always trivial. I would be surprised if you can get a good geometric description of any of the Steenrod operations. | |
| May 19, 2020 at 18:01 | comment | added | Student | Thanks for your answer! However, do you find the delooping viewpoint makes sense? If there's a good definition of $Sq^i: Id \to \Omega^i$, we don't need it to square at the right degree. | |
| May 19, 2020 at 17:57 | history | edited | Connor Malin | CC BY-SA 4.0 |
added 104 characters in body
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| May 19, 2020 at 17:52 | history | answered | Connor Malin | CC BY-SA 4.0 |