Timeline for The Steenrod Algebra of the Dihedral Group $D_{2n}$, $n=0 \pmod{4}$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Dec 24, 2016 at 19:27 | comment | added | Riccardo | @CharlesRezk I think I managed to prove everything. Thanks a lot. Do you mind making your first comment the answer so I can close this question? | |
| Dec 24, 2016 at 19:26 | history | edited | Riccardo | CC BY-SA 3.0 |
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| Dec 24, 2016 at 16:25 | comment | added | Charles Rezk | Yes yup yes yup | |
| Dec 24, 2016 at 16:20 | comment | added | Riccardo | @CharlesRezk Using the iso $H^1 \cong \hom(H_1,\mathbb{Z}_2)$, we can define $x,y$ as the dual of the respective generators of the abelianization of $D_{2n}$. If I'm following your reasoning, everything boils down to prove what is the image of the first SW class of the representation $\rho \colon D_{2n} \to O(2)$ which defines $w$? | |
| Dec 24, 2016 at 15:34 | comment | added | Charles Rezk | So $w$ is the pullback of the second Stiefel-Whitney class $w_2\in H^2(BO(2),Z_2)$? Then the "Wu formula" computes Steenrod squares in $H^*(BO(2))$. I believe in this case $Sq^1(w_2)=w_1w_2+w_3=w_1w_2$ (since $w_3=0$ in $O(2)$). You didn't characterize $x$ and $y$ enough so that I can say what $w_1$ pulls back to. | |
| Dec 24, 2016 at 14:52 | history | asked | Riccardo | CC BY-SA 3.0 |