Timeline for Steenrod squares in terms of chain maps
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 25, 2022 at 13:40 | comment | added | Denis Nardin | @Student You can try to look at this. It's very conceptual, but it's far from elementary. There is no way of systematically compute cohomology operations, as far as I know, even if the answer is known for ordinary cohomology. | |
| Feb 25, 2022 at 4:26 | comment | added | Student | I asked this question because in Lurie's note the construction doesn't seem natural to me. Is there a better explanation of why Steenrod operations exist (in that way)? Or rather, what's a good way to understand the space $[K(\mathbb{F}_q; n), K(\mathbb{F}_q; n+i)]$ systematically? | |
| Feb 24, 2022 at 21:41 | comment | added | Denis Nardin | @Z.M Actually given that the Steenrod squares are quadratic maps, lack of linearity is very unsurprising. The fact that they are $\mathbb{S}$-linear is surprising, but this comes from the fact that they are an avatar of the Frobenius (another quadratic map which turns out to be linear) | |
| Feb 24, 2022 at 18:29 | comment | added | Z. M | Steenrod operation could be defined for "homology" classes of $E_2$-$\mathbb F_2$-algebras (or even for complexes $V\in D(\mathbb F_2$ equipped with a symmetric multiplication $\operatorname{Sym}^2(V)\to V$ (as in Lurie's notes). It looks like the whole thing lives in $D(\mathbb F_2)$ and it seems strange at first glance of nonlinearity. | |
| Feb 24, 2022 at 18:06 | history | answered | Tim Campion | CC BY-SA 4.0 |