Timeline for Cohomology ring $H^*(BG,\mathbb{Z}_2)$ for $G=\mathbb{Z}_2\ltimes B^2\mathbb{Z}^2$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2021 at 16:07 | comment | added | user43326 | $F$ is a free $Z/2[Z/2]$-module, $T$ is a module with trivial $Z/2$ action. So, if $a_i$'s form a basis of $H^*(X)$, $a_i\otimes a_i$ form a basis of $T$ and $a_i \otimes a_j + a_j \otimes a_i$ form a basis of $F$. | |
| Nov 17, 2021 at 5:59 | comment | added | wonderich | @user43326 thanks! --- what does it mean: the free F and trivial T in your comment? | |
| Nov 9, 2021 at 10:24 | comment | added | user43326 | Anything that treats the homology of infinite loop spaces will do, for example there is May's book, or this paper people.math.harvard.edu/~dwilson/research/classical-power.pdf by Dylan Wilson | |
| Nov 9, 2021 at 10:19 | comment | added | user43326 | $Q=\Omega ^{\infty }\Sigma ^{\infty}$ | |
| Nov 9, 2021 at 3:30 | comment | added | wonderich | (2) What is your $Q$ in the $𝑄𝐾(𝑍,3)$? | |
| Nov 9, 2021 at 3:30 | comment | added | wonderich | (1) Thanks so much! I voted up. Could you provide a great ref on Nishida relations? | |
| Nov 8, 2021 at 9:50 | comment | added | user43326 | This is standard. As a matter of fact you can replace $B^3Z^2$ by $X^2$ (so $X$ will be $K(Z,3)$). Just write $H^*(X^2)=F\oplus T$ where $F$ is free, $T$ is trivial $Z/2$-module, and $E^{\infty}$ looks like $H^0(Z/2,F)\oplus H^*(BZ/2)\otimes T$. I guess normally you can deduce the Steenrod algebra's action just using the Steenrod diagonal, but at worst, the cohomology of the space you are after is detected by $X\times X$ and $X\times BZ/2$. Another way of determining the A-module structure is to consider your space as a stable summand of $QK(Z,3)$ and use the Nishida relations. | |
| Nov 8, 2021 at 6:14 | history | edited | Gregory Arone | CC BY-SA 4.0 |
deleted 10 characters in body
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| Nov 8, 2021 at 5:21 | history | asked | wonderich | CC BY-SA 4.0 |